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University students’ conceptual understanding and application of meiosis Lai, Patrick K. T. 1996

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UNIVERSITY STUDENTS' C O N C E P T U A L UNDERSTANDING AND APPLICATION OF MEIOSIS BY PATRICK K.T. LAI B.Sc. (Biology), The University of Hong Kong, 1978 Certificate in Education, The University of Hong Kong, 1983 Advanced Diploma in Education, The University of Hong Kong, 1990 Master of Education, The University of Hong Kong, 1991 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in THE FACULTY OF GRADUATE STUDIES DEPARTMENT OF CURRICULUM STUDIES IN EDUCATION We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA JANUARY, 1996 © PATRICK K.T. LAI, 1996 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of The University of British Columbia Vancouver, Canada DE-6 (2/88) A b s t r a c t Studies i n the general l i t e r a t u r e of genetics learning have i d e n t i f i e d the students' understanding of meiosis for d i p l o i d organisms at both the high school and college l e v e l s . I t was generally believed by these researchers that a canonical, conceptual understanding of meiosis i s s u f f i c i e n t for students to advance i n genetics. No p r i o r research has documented the students' application of knowledge of meiosis to a r e a l i s t i c genetics problem that involves haploid organisms. This study has extended the genetics learning l i t e r a t u r e by i d e n t i f y i n g some of the problems which students encounter when asked to explain t h e i r conceptual understanding and when solving a problem from t h e i r undergraduate genetics programme. In p a r t i c u l a r , the study focused on the topic area of meiosis. Three areas of students' understanding were scrutinized: t h e i r conceptualizations of meiosis for a d i p l o i d c e l l ; t h e i r approaches to addressing a genetics problem which involved a haploid organism; and, the possible relationships between the approaches and the meiosis conceptualizations. The research approach taken was embedded i n a research perspective known as "phenomenography" (Marton, 1981). Phenomenography i s the study of the q u a l i t a t i v e l y d i f f e r e n t Ill ways i n which people conceptualize various aspects of phenomena. The data source f or the study included a set of interviews conducted with ten undergraduate genetics students using two d i f f e r e n t task .contexts. These included i) a meiosis task and i i ) an applied genetics question that involved the concept of meiosis, using a haploid organism. The outcomes of the study yielded an i d e n t i f i c a t i o n and description of the students' conceptualizations of meiosis and a description of t h e i r approaches to addressing the genetics problem. The findings indicated that a good understanding of meiosis i s a necessary but not s u f f i c i e n t condition for the students to solve the novel problem. A close examination of the successful approaches revealed that a sound understanding of the biology of the haploid organism and of the processes of non-disjunction and crossing-over were necessary for the students to generate reasonable hypotheses about the data given i n t h i s problem. Implications for further research and for the teaching of genetics are discussed. i v TABLE OF CONTENTS ABSTRACT. ' i i LIST OF TABLES v i i LIST OF FIGURES . v i i i ACKNOWLEDGEMENTS. x i i CHAPTER 1: INTRODUCTION TO THE STUDY 1 1.1 I n t r o d u c t i o n t o the Problem 1 1.2 Background of the Study 4 1.2.1 I n t r o d u c t i o n 4 1.2.2 R a t i o n a l e f o r Problem-Solving Research 5 1.2.3 C l a s s i c a l S t u d i e s of Problem-Solving Research...6 1.2.4 Current Work i n Problem-Solving Research 9 1.3 The Study 9 1.3.1 S p e c i f i c Research Questions 10 1.3.2 Context of the Study 11 1.3.3 D e l i m i t a t i o n of the Problem 13 CHAPTER 2: LITERATURE REVIEW AND THEORETICAL FRAMEWORK.... 14 2.1 I n t r o d u c t i o n . . 14 2.2 Problem-Solving Research i n Ge n e t i c s 15 2.2.1 The P i a g e t i a n P e r s p e c t i v e 15 2.2.2 The Expert-Novice P e r s p e c t i v e 17 2.2.3 Other S t u d i e s of Students' Conceptions of M e i o s i s 23 2.3 Contemporary Research on Teaching and L e a r n i n g 25 2.4 R e l a t i o n s h i p of the L i t e r a t u r e to Present Study 2 8 2.5 Choice of T h e o r e t i c a l P e r s p e c t i v e 29 2.6 Two Forms of Phenomenographic Research 30 2.7 C a t e g o r i e s of D e s c r i p t i o n 32 CHAPTER 3: DESIGN OF STUDY AND METHODOLOGY 3.1 I n t r o d u c t i o n 3 7 3.2 The Design of the Study . 37 3.2.1 F a c t o r s I n f l u e n c i n g Design of the Inter v i e w P r o t o c o l 38 3.2.2 The Sub j e c t s 42 3.2.3 The S t r u c t u r a l Format of the Inter v i e w P r o t o c o l 43 V 3.3 The Procedures of Analysis 55 3.3.1 Categories of Description Characterizing Conceptualizations and Approaches (Chapters 4 and 5) 56 3.3.2 Data Analysis: The Relationship between the Meiosis Conceptualizations and the Approaches (Chapter 6) 58 3.3.3 Review of A n a l y t i c a l Outcomes 58 CHAPTER 4: CONCEPTUALIZATIONS OF MEIOSIS. , .59 4.1 Data Analysis for Research Question 1: Introduction . 59 4.2 Conceptualizations of Meiosis..... 60 4.2.1 The 'Canonical' Conceptualization of Meiosis...61 4.2.2 The 'Event-Modified' Conceptualization of Meiosis 79 4.2.3 'Whole-Process Modified' Conceptualization of Meiosis 94 CHAPTER 5: PROBLEM-SOLVING APPROACHES OF UNIVERSITY GENETICS STUDENTS 100 5.1 Introduction... 100 5.2 The 'Problem-Solving' Approach 100 5.2.1 The 'Direct Success' Approach 101 5.2.2 The 'Hypothesis-Formulation and Reframing' Approach 107 5.2.3 The 'Hypothesis-Formulation and Rejection' Approach 114 5.2.4 The ' P a r t i a l Recognition and Forced F i t ' Approach 122 CHAPTER 6: RELATIONSHIP BETWEEN PROBLEM SOLVING APPROACHES AND CONCEPTUALIZATIONS OF MEIOSIS. 129 6.1 Data Analysis for Research Question 3: Introduction 12 9 6.2 Procedures of Analysis 13 0 6.2.1 Identifying the Meiosis Models applied i n Ascobolus Problem. 131 6.2.2 Mapping of Meiosis Models used with the Meiosis Task and the Ascobolus Problem....'....139 vi CHAPTER 7: CONCLUSIONS, IMPLICATIONS AND RECOMMENDATIONS . . . . 145 7 .1 I n t r o d u c t i o n 145 7.2 S p e c i f i c Conclusions from the Study 145 7.2.1 Conclusions f o r Chapter 4 146 7.2.2 Conclusions f o r Chapter 5 147 7.2.3 Conclusions f o r Chapter 6 148 7.3 D i s c u s s i o n of Results 149 7.3.1 C o n t r i b u t i o n to the L i t e r a t u r e on Learning Genetics 149 7.3.2 Proposed I n s t r u c t i o n a l Sequence 157 7.4 L i m i t a t i o n s of the Study 162 7.5 Recommendations f o r Further Research 163 REFERENCES 169 APPENDIX A: Glossary 177 APPENDIX B (Part One): L i f e Cycle of Neurospora ( s i m i l a r to Ascobolus) 179 APPENDIX B (Part Two): S o l u t i o n to the Ascobolus Problem.180 APPENDIX C: Results of the 1994-1995 Class Survey us i n g the Meiosis Task... 182 APPENDIX D: Samples of Complete Interview T r a n s c r i p t s and Flow Charts of Students' Problem-Solving Approaches 183 APPENDIX E: Proposed I n s t r u c t i o n a l Sequence..... 210 vii L I S T O F T A B L E S 1 Mapping of Meiosis Models to Problem-Solving Approaches 140 2 A broad lay-out of topics of f i v e genetics textbooks 158 C-l Frequency and Percentage of each type of meiosis model i d e n t i f i e d from the 1994-1995 class survey and from the ten students interviewed 182 vii i L I S T O F F I G U R E S 1 The meiosis task 45 2 The 'Canonical' model of the process of meiosis as i t appears i n a d i p l o i d c e l l 51 3a A l i n e r octads i n an ascus of Ascobolus 53 3b The haploid l i f e cycle of Ascobolus 54 4 Chris's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis 64 5 Kirk's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis... 67 6 Sharilyn's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis 70 7 May's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis 74 8 L a i l a ' s i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis 78 9 Mike's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis 83 10 Sam's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis 86 11 Doug's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis 89 12 Kamyar's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis 93 13 Jason's i l l u s t r a t i o n showing a 'Whole-Process Modified' conceptualization of meiosis .....96 14 Flow chart summarizing the 'Direct Success' Approach 101 15 Kirk's i l l u s t r a t i o n of non-disjunction for case one ... ; f 103 16 Kirk's i l l u s t r a t i o n of crossover for case two ..104 17 Kirk's example of the 'Direct Success' Approach 106 18 Flow chart summarizing 'Hypothesis-Formulation and Reframing' Approach 108 19 Mike's i l l u s t r a t i o n of non-disjunction for ...110 case one 20 Mike's i l l u s t r a t i o n of crossing over for case two I l l 21 Mike's example of the 'Hypothesis-Formulation and Reframing' Approach .113 22 Flow chart summarizing 'Hypothesis-Formulation and Rejection' Approach.. 115 ix 23 Sam's i l l u s t r a t i o n of crossover model f or case two 118 24 Sam's i l l u s t r a t i o n of crossover f or case one 118 25 Sam's i l l u s t r a t i o n of no crossing over f or case one •'. 119 26 Sam's example of the 'Hypothesis-Formulation and Rejection' Approach f or case one 121 27 Flow chart summarizing the ' P a r t i a l Recognition and Forced F i t ' Approach 122 28 Doug's meiois model (I) for case one 123 29 Doug's i l l u s t r a t i o n of crossover f or case two 124 30 Doug's one-stage meiosis model for case one 125 31 Doug's example of the ' P a r t i a l Recognition and Forced F i t ' Approach 12 8 32 Kirk's meiosis model applied to case one of the Ascobolus problem (Chapter 5, Figure 15 relabelled) 132 33 Kirk's meiosis model applied to case 2 of the Ascojbolus problem (Chapter 5, Figure 16 relabelled) 132 34 Sharilyn's meiosis model used with case one of the Ascobolus problem (Appendix C, Figure C-3 relabelled) 133 35 Sharilyn's_meiosis model used with case two of the Ascobolus problem (Appendix C, Figure C-2 relabelled) 133 36 May's meiosis model used with case one of the Ascobolus problem (Appendix C, Figure C-8 relabelled) 134 37 L a i l a ' s meiosis model used with case one of the Ascobolus problem (Appendix C, Figure C-5 relabelled) 134 38 L a i l a ' s meiosis model used with case two of the Ascobolus problem (Appendix C, Figure C-G relabelled) 134 39 Mike's meiosis model used with case one of the Ascobolus problem (Chapter 5, Figure 19 relabelled) 135 40 Sam's meiosis model used with case one of the Ascobolus problem (Chapter 5, Figure 25 relabelled) 136 41 Doug's meiosis model used with case-one .pf the Ascobolus problem (Chapter 5, Figure 30 relabelled) 13 6 X 42 Kamyar's meiosis model used with case one of the Ascobolus problem (Appendix C, Figure C-12 relabelled) 137 43a Jason's meiosis model used with case one of the Ascobolus problem (Appendix C, Figure C-20 relabelled) .•• 13 8 43b Jason's model from his notes (Appendix C, Figure C-19 relabelled).... 139 44 The Refined meiosis task 166 B-l L i f e cycle of Neurospora 179 C-l Sharilyn's i l l u s t r a t i o n of crossing over for case two of the Ascobolus problem 184 C-2 Sharilyn's i l l u s t r a t i o n of crossing over for case two of the Ascobolus problem 185 C-3 Sharilyn's i l l u s t r a t i o n of non-disjunction for case one of the Ascobolus problem 187 C-4 Sharilyn's example of the 'Direct Success' Approach .' 188 C-5 La i l a ' s i l l u s t r a t i o n of non-disjunction of case one for the Ascobolus problem 190 C-6 La i l a ' s i l l u s t r a t i o n of crossing-over for case two of the Ascobolus problem 191 C-7 La i l a ' s example of the 'Direct Success' Approach.... 192 C-8 May's i l l u s t r a t i o n of non-disjunction for case one of the Ascobolus problem 194 C-9 May's i l l u s t r a t i o n of crossing over for case two of the Ascobolus problem 195 C-10 May's example of the 'Hypothesis-Formulation and Reframing' Approach .196 C - l l Kamyar's i l l u s t r a t i o n of crossover for case two of the Ascobolus problem 198 C-12 Kamyar's i l l u s t r a t i o n of non-disjunction for case one of the Ascobolus problem 199 C-13 Kamyar's example of the 'Hypothesis-Formulation and Reframing' Approach. 200 C-14 Chris's i l l u s t r a t i o n of crossover for case two of the Ascobolus problem 202 C-15 Chris's i l l u s t r a t i o n of inversion for case one of the Ascobolus problem .203 C-16 Chris's f i r s t i l l u s t r a t i o n of deletion for case one of the Ascobolus problem 203 C-17 Chris's second i l l u s t r a t i o n of deletion for case one of the Ascobolus problem 204 C-18 Chris's example of the 'Hypothesis-Formulation and Rejection' Approach.. 205 xi C-19 Jason's i l l u s t r a t i o n of meiosis for the haploids from his notes. . . 207 C-20 Jason's i l l u s t r a t i o n of his s i m p l i f i e d meiosis model for case one of the Ascobolus problem 207 C-21 Jason's i l l u s t r a t i o n of crossover for case two of the Ascobolus problem..... 208 C-22 Jason's example of the ' P a r t i a l Recognition and Forced F i t ' Approach 209 Xll ACKNOWLEDGEMENTS I would l i k e to express my sincere thanks and gratitude to Gaalen Erickson, Anthony G r i f f i t h s and J o l i e Mayer-Smith, my thesis committee, for a l l t h e i r guidance and encouragement. I would also l i k e to thank my mother, for her f a i t h i n me, and her support for my oversea venture, that made i t possible for me to complete t h i s study. 1 CHAPTER 1 INTRODUCTION TO THE STUDY 1.1 Introduction to the Problem At the heart of education i s the three-way transaction . between the student, the teacher, and the material being learned. The ov e r a l l aim i s to produce a s e l f - s u f f i c i e n t learner and a problem-solver. University t r a i n i n g i s not designed to produce lawyers, doctors, engineers, or architects who can work only within what i s already known: with a case straight from the textbook, with a blueprint for an established design. In other words, u n i v e r s i t y educated students are expected to master d i s c i p l i n a r y knowledge so that they are able to use the content learned constructively: to solve problems that depart from textbook algorithms, to weigh c o n f l i c t i n g commentaries and ar r i v e at t h e i r own resolution. A review of the l i t e r a t u r e , however, show.s that the above aims are not achieved i n most instances. A large number of students i n u n i v e r s i t i e s appear to be learning an imitation of the d i s c i p l i n e s they are studying, a counterfeit amalgam of terminology, algorithms, unrelated facts, "right answers"1 , and manipulative s k i l l s that 1 Throughout the thesis, "" is used to denote opinions extracted from either the literature or the students of this study. w is used to denote the terms that are being emphasized by the author. 2 enable them to survive the process of assessment (Ramsden, 1992, p.37). Studies i n higher education have shown that students' d i f f i c u l t i e s i n learning are i l l u s t r a t e d by t h e i r i n a b i l i t y to apply rules or concepts to novel problems. Gunstone and White (1981) described several experimental demonstrations with Physics I students, who had just completed a highly competitive matriculation examination to enter one of Australia's most prestigious u n i v e r s i t i e s . They were asked to predict and then explain what would happen when presented with a common physical event. In one example, the majority of students predicted that i f two b a l l s (one heavy and one lig h t ) held i n front of them were released simultaneously, the heavy one would reach the ground f i r s t . Their reasons for t h i s prediction included comments such as: because "heavy things have a bigger force", or "gravity i s stronger nearer the earth". These Physics students could give excellent d e f i n i t i o n s for the term "gravity", but f a i l e d to solve problems requiring the concept of gravity. The d i f f i c u l t i e s might be accounted for by the students' i n a b i l i t y to use t h e i r knowledge of gravity. Studies by B a l l a with medical students (Balla 1990a; 1990b) re f l e c t e d the same kind of student d i f f i c u l t y i n applying knowledge. He observed that medical students often used basic science knowledge i n c o r r e c t l y or not at. a l l i n formulating and rev i s i n g c l i n i c a l diagnoses. When these students became p r a c t i c i n g c l i n i c i a n s , some of them rarely 3 knew how to apply learned t h e o r e t i c a l knowledge to novel c l i n i c a l s i t u a t i o n s . One prominent explanation to account for t h i s type of students' behaviour stems from the t r a d i t i o n a l dominant perspective on teaching and learning i n t e r t i a r y science (Tobin, 1993). In t h i s perspective, knowledge i s viewed as a commodity to be transmitted to students, whose sole r e s p o n s i b i l i t y i s to learn i t . Teaching i s therefore viewed as transmitting knowledge to students and learning as though i t were a process of receiving and storing knowledge. This perspective on teaching and learning emphasizes the passive nature pf the mind (Gilbert & Watts, 1983) where students accumulate knowledge provided by the teacher. Thus, knowledge i s detached from the learners, who w i l l never know how to 'apply' i t to novel problem situations. A review of some entry-level college science c u r r i c u l a by Wartell (1984) further i l l u s t r a t e s the t r a d i t i o n a l perspective on teaching and learning. The c u r r i c u l a mostly focused on " d e f i n i t i o n t r a i n i n g " and "technique t r a i n i n g " rather than a "mode of thought" or "conceptual framework" approach. In general, students' d i f f i c u l t i e s discussed above may be accounted for by the learning h i s t o r i e s of the students who were continuously exposed to,the t r a d i t i o n a l perspective on teaching and learning, where applica t i o n of knowledge to novel problem situations was not emphasized. 4 1.2 Background of the Study 1.2.1 Introduction The previous section provides an overview of the t r a d i t i o n a l perspective on learning and approach to teaching. T r a d i t i o n a l l y , teaching has been thought of as the transmission of knowledge. The role of the teacher i s to t e l l , to be i n control of the pace and content of lessons and to be the purveyor of the truth and knowledge. As a d i r e c t consequence of t h i s transmission model of pedagogy, the role many students assume i s that of the passive recipient of knowledge, dependent upon the teacher and non r e f l e c t i v e . The outcome of such an approach i s rote learning of laws, formulae and problem solving algorithms i n order to pass examinations. Seldom, i f ever, do students attempt to use t h e i r knowledge to solve problems other than those i n t h e i r textbooks. Contemporary perspectives on learning and approaches to teaching recognize the learner as an 'active' agent. These perspectives are based on the idea that learning involves the 'active' construction of knowledge by i n d i v i d u a l s as they engage thoughtfully with information (Cobb, 1994; Driver et a l . , 1994). Unlike i n the t r a d i t i o n a l approach to teaching, knowledge i s not a commodity that can be simply transmitted from teachers to students. Knowledge, i n t h i s view, i s rather a product of the student's processing and making connections between new information they receive and 5 t h e i r p r i o r knowledge. This learning process requires students to add to or enhance t h e i r previous store of ideas (construction) or to replace p r i o r views and ideas with more plausible, useful ones (reconstruction). In order to enhance the 'construction' and 'reconstruction' aspects of learners, i t i s important to enable learners to be s e l f - r e f l e c t i v e : to think about what they do and do not understand; how new information f i t s with what they already know; what they are doing and why they are doing i t . In t h i s view of learning, the r o l e of teaching i s to provide experiences that w i l l a s s i s t the learner i n making useful (in t h i s research, s c i e n t i f i c a l l y correct) constructions. 1.2.2 Rationale for Problem-Solvincf Research I t i s generally accepted that problem solving i s one of a set of meaningful a c t i v i t i e s that teachers should encourage. Most teachers view problems as representing opportunities for students to learn how to process information and demonstrate t h e i r understanding. In other words, problems should present situations i n which the student does not have a ready answer for the question set by the teacher. The student, very often, knows only some of what i s required to generate a solution and so he/she must work out the rest of what i s required. This a c t i v i t y requires that cognitive work be carried out; work which includes the l i n k i n g of novel problem situations with p r i o r understanding and the possible reorganization of these ideas 6 to s u i t the problem s i t u a t i o n . The 'construction' and 'reconstruction' aspects occur most read i l y i f the student i s "metacognitive" (Novak & Gowin, 1984) -- able to think about his/her thinking. Most 'passive' learners, however, view learning as involving remembering, not independent thinking, hence the reasons for answers i n problems are of l i t t l e value or i n t e r e s t . They tend to focus on 'how to do problems and get the right answer' rather than on 'understanding and applying the underlying concepts' for any given problem. Thus there i s a mismatch of expectations between teachers and students. Sharing the teachers' view with students i s often not enough to change t h i s . Helping students examine t h e i r own approach may a s s i s t them i n examining t h e i r view of problem solving. Analyzing students' problem-solving a c t i v i t y may provide f r u i t f u l information for teachers (and for learners, i f teachers use t h i s information to inform t h e i r i n s t r u c t i o n ) . 1.2.3 C l a s s i c a l Studies of Problem-Solving Research That many students f i n d i t d i f f i c u l t to apply knowledge obtained from a lecture s e t t i n g to solve problems i s f r u s t r a t i n g to both teachers and students. Students claim that they understand the lectures and enjoy reading the texts, but f i n d i t hard to solve problems using the concepts outlined i n the text. In other words, the apparently l o g i c a l sequence of the events: 7 'learn p r i n c i p l e s • do problems • pass examination' does not work for many students. C l a s s i c a l studies i n problem solving o f f e r some preliminary insights into the above phenomenon. These studies provide an overview of the differences between experts and novices when they solve problems. They emphasize ways to c l a s s i f y problems and general problem solving strategies. One of the e a r l i e s t and important studies dealt with ways i n which experts and novices c l a s s i f i e d problems. Experts tend to c l a s s i f y problems according to solution modes or 'schemas' whereas novices are more l i k e l y to use s u p e r f i c i a l features to c l a s s i f y problems. For instance, Chase and Simon (1973) showed that the apparent difference between novice and expert chess players was not t h e i r a b i l i t y to think and to consider a l l possible moves i n advance, but rather that the experts could remember and mentally access a large number of r e a l i s t i c game configurations ( i . e . schemas), whereas for novices each game was a novel experience. Chi, Feltovich, and Glaser (1981) discovered s i m i l a r findings i n physics. Experts, for example, grouped a series of problems as being s i m i l a r because they could be solved using the conservation of energy concept. Novices, on the other hand, grouped these problems because they a l l dealt with i n c l i n e d planes. Other researchers believed that the above differences affected the strategies used by novices and experts to solve 8 problems. Larkin et 1 a l . (1980) found that the most common "weak" h e u r i s t i c used by novices was means-ends analysis or "working backward". Novices usually lacked the domain-s p e c i f i c knowledge that enabled them to generate new "steps" from what was given i n the problem. They had to use the "weak" h e u r i s t i c by f i r s t comparing the "givens" with the "goals" of the problem. They would then work backward from the goals. The knowledgeable expert t y p i c a l l y takes a "working forward" or "knowledge development" approach and generates moves from the givens to the goal of the problem. For example, the expert problem solver started from the givens of the problem, i d e n t i f i e d what was asked and then applied successive equations to solve the problem u n t i l the desired values were found (Larkin et' a l . , 1980, Sweller J., 1991). These experts were believed by researchers to possess domain-specific knowledge about what new information could be developed from the givens. Although the c l a s s i c a l studies described above attempted to provide some explanations to account f o r the d i f f i c u l t i e s experienced by students i n applying t h e i r knowledge to problem-solving situations, they have t h e i r l i m i t a t i o n s . These studies only i d e n t i f i e d and l a b e l l e d discrete components of problem-solving s k i l l s such as ways to c l a s s i f y problems. Categorization and q u a n t i f i c a t i o n of results were the major tasks achieved. Seldom, i f ever, did these researchers attempt to analyze the entire problem-9 solving process of students when they were working on a p a r t i c u l a r problem. 1.2.4 Current Work i n Problem-Solving Research Contemporary research i n education helped researchers to analyze the problem-solving a c t i v i t i e s of students i n d e t a i l . Instead of i d e n t i f y i n g and l a b e l l i n g the in d i v i d u a l problem-solving components as i n the c l a s s i c a l studies, these studies look at the entire problem solving process. For example, studies i n medical diagnostic problem solving (Ramsden, Whelan & Cooper, 1989) examined i n d e t a i l how medical students verbalized t h e i r thoughts as they were analyzing cases. Records of the sequence of verbalized thoughts revealed differences between the medical students i n making the diagnosis and i n using domain-specific knowledge. Chapter 2 w i l l provide a more detailed discussion of the relevant l i t e r a t u r e and t h i s research approach. 1.3 The Study -This study adopts the contemporary research approach and serves to i d e n t i f y , analyze and address some s p e c i f i c content-related problems uni v e r s i t y students experience when attempting to learn genetics and solve genetics problems. In a National Science Foundation sponsored conference on genetics teaching (Teaching Genetics: Recommendations and Research), the following goals for the conference were explicated: 10 (1) to develop a consensus of recommendations/alternatives for teaching genetics at the precollege and college l e v e l which w i l l enhance student understanding of genetics concepts and problem-solving s k i l l s , - and (2) to develop a research agenda for improving classroom teaching practices; and (3) to develop a p r i o r i t i z e d l i s t i n g of recommendations for research about e f f e c t i v e genetics teaching practices. (Smith, Simmons & Kinnear, 1993) 1.3.1 Specific Research Questions S p e c i f i c a l l y , t h i s study sought answers to the following questions:' 1. What are the q u a l i t a t i v e l y d i f f e r e n t ways i n which undergraduate students conceptualize meiosis? 2. What are the q u a l i t a t i v e l y d i f f e r e n t ways i n which undergraduate students approach genetics problem solving? 3. Is there any rela t i o n s h i p between the students' approaches to problem solving and t h e i r knowledge of meiosis? The three research questions for t h i s study provide information that i s d i r e c t l y applicable to the f i r s t goal indicated by Smith, Simmons and Kinnear (1993). The f i r s t question served to i d e n t i f y students' understanding of the process of meiosis. The second research question offered insights into d i f f i c u l t i e s that prevented the students from applying t h e i r understanding of meiosis to a s p e c i f i c problem which featured Ascobolus, a haploid organism 11 f a m i l i a r to many g e n e t i c i s t s . 2 The problem-solving approaches used by the students were i d e n t i f i e d and dissected into d i f f e r e n t stages. Each stage served to reveal the students' understanding as they were t r y i n g to untangle the complexity of the genetic representations i n the Ascobolus problem -- for example, the f l i p p i n g from one meaning of a p a r t i c u l a r genetics symbol to another by students. The f i n a l research question linked the findings of the previous two research questions. Looking into the relationship between the problem-solving approach and students' a b i l i t y to 'apply' t h e i r meiosis models 3 generated from the meiosis task provided information concerning t h e i r success i n solving the Ascobolus problem. The analysis i s done i n three chapters coinciding with the three research questions l i s t e d above. 1.3.2 Context of the Study A t h i r d year un i v e r s i t y course i n introductory genetics which based i t s assessment purely on problem solving provided the context of t h i s study. In a t y p i c a l genetics problem an experiment i s described, result s are given, and the students are asked to interpret the data using the 2 Two interview tasks used in this study: the meiosis task and the Ascobolus problem. The meiosis task explored students' conceptions of meiosis. The Ascobolus problem explores students' application of their knowledge of meiosis in a problem context. 3 Meiosis models represent the mental models used by students in this study. These models describe the sequence of events in meiosis. 12 genetics concepts learned i n the course. Problems were used to test student understanding of the genetics concepts rather than an evaluation method based on w r i t i n g short answers or essays. Many students had d i f f i c u l t i e s solving the problems and demonstrating t h e i r understanding of the concepts i n the problems. The subject nature of genetics creates d i f f i c u l t i e s for most students. The a c q u i s i t i o n of genetics concepts i s a complicated task for most beginners. One serious hindrance i s the ''multiple representations'' of genetic symbols (Mayer-Smith & G r i f f i t h s , 1995). Probably the single most confusing image i n teaching and learning genetics i s the practice of representing chromosomes (generally autosomes) as t h e i r mitotic metaphase X - l i k e appearance ( G r i f f i t h s & Mayer-Smith, 1994). . I n i t i a l interviews with high and low-scoring students enrolled i n t h i s course suggested that the major difference between them was that the high scorers were able to assess t h e i r a n a l y t i c a l d i f f i c u l t i e s with an i n t e r n a l check system, and then respond to problem-solving by a metacognitive or self-questioning mode. They also had l i t t l e d i f f i c u l t y i n distinguishing and unravelling the large amount of meaning imbued with each genetics representation. The above ideas suggested that both an understanding of genetics concepts and an i n t e r n a l check system are essential components of successful problem-solving. In order to support t h i s claim, the present study was designed to' 13 examine the relationship between student understanding of genetics concepts and t h e i r success i n genetics problem solving. 1.3.3 Delimitation of the Problem Genetics i s a complex d i s c i p l i n e that'involves a large number of concepts and problems. Numerous studies conducted with high school and university students have documented students' understanding of major concepts during t h e i r study of genetics: mitosis and meiosis; inheritance; sex linkage; translocation; p r o b a b i l i t y ; and human genetics terms (Smith, Simmons & Kinnear, 1993). The present study focuses on meiosis, a topic which, according to many studies, students have d i f f i c u l t y i n understanding. Other factors that serve to delimit the claims of the study include the use of a small number of students, the selection of one genetics problem and the use of only interview data. This study should of f e r researchers insights about students' d i f f i c u l t i e s not only i n describing the process of meiosis but also i n manipulating genetics data for the purpose of solving complex genetics problems. 14 CHAPTER 2 LITERATURE REVIEW AND THEORETICAL FRAMEWORK 2.1 Introduction In the previous chapter, i t was claimed that students experienced d i f f i c u l t i e s i n applying t h e i r understanding of meiosis to novel 'applied' problems. There may be several reasons to account for t h i s i n a b i l i t y . For example, the students may either (a) have not yet acquired s u f f i c i e n t understanding of the concept, (b) have s u f f i c i e n t understanding of the concept but do not see i t as relevant to the p a r t i c u l a r problem, or (c) have s u f f i c i e n t understanding of the concept but are unable to apply i t to the problem. To f u l l y diagnose students' d i f f i c u l t i e s i n problem-solving, two areas need to be investigated. They are: i ) the students' background knowledge of a p a r t i c u l a r concept and i i ) t h e i r a b i l i t y to apply that concept to an 'applied' problem. No study i n the f i e l d of genetics has considered these two areas together. An understanding of these two areas should help teachers i d e n t i f y students' d i f f i c u l t i e s i n genetics problem solving. This chapter i s divided into two sections. F i r s t , i t provides an overview of problem-solving research i n genetics, a discussion of the l i m i t a t i o n s of t h i s research, and possible extensions of the work. Second, i t introduces the th e o r e t i c a l perspective used to frame the study: the Gothenburg "research s p e c i a l i z a t i o n " (Marton, 1981) ca l l e d 15 phenomenography which takes an interpretative e x p e r i e n t i a l perspective to frame an understanding of knowledge from students' perspectives. 2.2 Problem-Solving Research i n Genetics Although documentation of c l a s s i c a l research on problem-solving appeared i n the early 1970s, very l i t t l e research on genetics problem-solving was reported u n t i l the 1980s. The research t r a d i t i o n s on genetics problem solving are divided into (1) Piagetian and (2) "expert-novice" perspectives. 2.2.1 The Piagetian Perspective Early research based on the Piagetian perspective i n the b i o l o g i c a l sciences has been extensive. The postulation of four mental stages with corresponding age l e v e l s such as sensori-motor, preoperations, concrete operations, and formal operations, was commonplace i n Piaget's work. Research on problem-solving within t h i s perspective viewed problem-solving success as closely related to these mental stages (Gipson & Abraham, 1985; Gipson, Abraham, & Renner, 1989; Walker, Hendrix, & Mertens, 1980; Walker, Mertens, & Hendrix, 1979). In p a r t i c u l a r , these researchers postulated a general corre l a t i o n between formal operational thought and academic success i n genetics (Stewart & Hafner, 1994). Researchers believed that p r o b a b i l i s t i c , combinatorial and proportional reasoning s k i l l s i n problem-solving represented 16 the products of formal operational thought. Walker, Mertens, Hendrix (1979) and Walker, Hendrix, Mertens (1980) even conducted c o r r e l a t i o n studies and found that genetics i n s t r u c t i o n emphasizing the development of formal operation thought could increase students' performance. There are l i m i t a t i o n s to research framed within the Piagetian perspective. Other researchers challenged both the methods and the results of t h i s kind of research (Smith, 1986; Smith & Good, 1983; Driver & Erickson, 1983). At the methods l e v e l , research concentrated on discovering cause-and-effect relationships i n teaching and learning. Researchers tended to believe that i n s t r u c t i o n designed to promote certain types of Piagetian mental stages should lead to problem-solving success. This kind of research tended to disregard the learners' cognitive a c t i v i t y , and to look at the learning process as a "black box", examining only the inputs (teaching methods) and outputs (students' outcomes) (M i l l a r , 1989, p.587). Students' understanding of concepts and approaches to problem solving were never the subjects under exploration. Other researchers challenged the results of research within the Piagetian perspective. Smith (1986) and Smith and Good (1983) compared the problem-solving performance of 11 novices (undergraduate science and nonscience majors who had just completed t h e i r f i r s t genetics instruction) and 9 experts (genetics graduate students and genetics faculty members) on a set of complex genetics problems which 17 r e q u i r e d formal o p e r a t i o n a l reasoning. R e s u l t s demonstrated that formal o p e r a t i o n a l thought alone was i n s u f f i c i e n t t o account f o r a l l the problem-solving s k i l l s r e q u i r e d f o r ge n e t i c s . P i a g e t i a n researchers were a l s o c r i t i c i z e d f o r t r y i n g to i d e n t i f y g e n e r a l i z e d and content-independent forms of thought without c o n s i d e r i n g the r o l e that domain-specific knowledge p l a y s i n l e a r n i n g ( D r i v e r & E r i c k s o n , 1983) . This domain-specific knowledge has been the emphasis of conception research i n science education ( f o r example, D r i v e r & Easley, 1978; E r i c k s o n , 1979; H a c k l i n g & Treagust, 1982; Kargbo, Hobbs, & E r i c k s o n , 1980). 2.2.2 The Expert-Novice P e r s p e c t i v e Instead of r e l y i n g on P i a g e t i a n r i g i d a g e - s p e c i f i c models, other researchers w i t h i n the "expert-novice" p e r s p e c t i v e began to design s t u d i e s to examine two other important areas i n genetics l e a r n i n g : (1) understanding of s u b c e l l u l a r processes (domain-specific knowledge) and (2) s t r a t e g i e s i n s o l v i n g "pedigree" problems. Many s t u d i e s reported students', understanding of s u b c e l l u l a r processes. There were roughly two stages of development of t h i s k i n d of research, w i t h p r o g r e s s i v e refinement of research methods t o e l i c i t more f r u i t f u l f i n d i n g s . Research i n the i n i t i a l stage used textbook problems that could be solved by formulae and algorithms (Stewart, 18 1983; Stewart & Dale, 1981, 1989; Stewart, Hafner & Dale, 1990). Since these problems required only the simple 'plugging i n ' of formulae, they did not necessarily r e f l e c t students' understanding of subcellular processes such as meiosis. There was a progressive awareness of the need of research to look into students' understanding of s p e c i f i c subcellular processes such as meiosis. C o l l i n s , Stewart and Slack (1987) noticed that Ph.D. geneticists used the concept of meiosis to solve genetics problems. S i m i l a r l y , Stewart (1988) also predicted that an understanding of the mechanisms of meiosis was essential for the solving of r e a l i s t i c genetics problems. Following from these findings and predictions, researchers began to c o l l e c t data on students' a b i l i t y to j u s t i f y t h e i r solutions i n terms of the mechanism of meiosis at both the high school and college l e v e l . At the high school l e v e l , Stewart and Dale (1989) did a representative study to i d e n t i f y students' understanding of the mechanism of meiosis. They used either monohybrid or dihybrid problems. Following the students' solution to the problems, the interviewer asked an open-ended question: "Do chromosomes have anything to do with your problem solution?" Of the 50 students interviewed, 41 obtained correct answers to both monohybrid and dihybrid problems. T h i r t y - f i v e of those 41 were able to construct the physical relationships between chromosomes and genes, and used them to explain the 19 solutions to dihybrid cross problems. This study i d e n t i f i e d students' understanding of the s p a t i a l r e l a t i o n s h i p between genes and chromosomes as they undergo meiosis. I t did not, however, examine i n d e t a i l the sequence of events of meiosis. There are also studies that analyzed i s o l a t e d student misunderstandings about meiosis, and recognized that students often did not make fundamental connections between Mendelian genetics and the process of meiosis (Collins & Stewart, 1989; Hafner, 1991; Hilderbrand, 1985; Slack & Stewart, 1990). K i n d f i e l d (1994) extended the above studies by the i n c l u s i o n of individuals ranging i n expertise from introductory college students to p r a c t i c i n g g e n e t i c i s t s , and undertook a more detailed analysis of t h e i r understanding of meiosis, leading to a systematic characterization of in d i v i d u a l s ' models of meiosis. These models described the sequence of events and the p a r t i c i p a t i n g chromosomal structures i n meiosis. In Kindfield's study, she used a genetics problem which consisted of a written representation of a p a r t i c u l a r eukaryotic c e l l , three d i f f e r e n t sets of gametes that could result from t h i s c e l l going through meiosis, and a request to describe the s p e c i f i c meiotic events that would lead to the formation of each gamete set. Follow-up probing questions were t a i l o r e d to the individual's behaviour while working on the problem. Results of t h i s study revealed a 20 number of meiosis misunderstandings that involved chromosome misunderstandings, event-specific misunderstandings (knowledge about i n d i v i d u a l components of the process) and whole-process misunderstandings,(knowledge about the process of meiosis as a whole). The above studies reported experts' and novices' understanding of meiosis, but did not investigate students' a b i l i t y to apply t h e i r knowledge of meiosis i n a complex problem-solving task. In other words, the 'application' component of the meiosis concept had never been addressed. A useful extension of Kindfield's study could include an investigation of both the understanding of meiosis and i t s application. For example, students could be asked to describe the s p e c i f i c meiotic events using a task s i m i l a r to the one used i n Kindfield's study. They would then be given new problems that could be solved r e a d i l y using the same meiosis events, but which, nevertheless, are more complicated than the f i r s t problem. Sweller (1991) suggested that students usually f i n d the second 'applied' problem either d i f f i c u l t or insoluble because they lack certai n types of problem-solving s k i l l s . Other researchers within the expert-novice perspective, instead of studying the understanding of subcellular processes, began to examine problem-solving s k i l l s of individuals. Smith and Good (1984) compared the successful and unsuccessful problem-solving performances of experts and novices. Their study reported that experts perceived a 21 problem as a task requiring analysis and reasoning and that they tended to use a "forward-working" approach, whereas novices used a "means-ends" strategy (by working backward). This re s u l t coincided with the findings of other c l a s s i c a l studies i n problem-solving (Larkin, et. a l . , 1980). Smith and Good (1984) also found that successful problem-solvers often made frequent checks on the correctness of t h e i r work. Based on the work of Smith and Good (1984), began to study the problem-solving s k i l l s of experts and novices by using "pedigree" problems. Pedigrees are diagrammatic displays of patterns of inheritance of a p a r t i c u l a r t r a i t within generations of a family. Genetics pedigrees represent a d i s t i n c t category of genetics problems because the data are presented to the solver i n the form of a diagram (the pedigree or family tree) and the problem-solving process involves generation and testing of hypotheses (Hackling & Lawrence, 1988) . Smith (1988) studied the problem-classification schemes used by various subjects. He had genetics faculty, genetics counsellors, and university genetics students sort a set of 28 problems into d i f f e r e n t classes. Genetics counsellors categorized problems more s i m i l a r l y to students than they did to genetics instructors. Counsellors and students categorized on ' s u p e r f i c i a l ' features whereas genetics instructors usually c l a s s i f i e d problems on solution modes. These findings coincided with results reported by an e a r l i e r c l a s s i c a l study i n physics (Chi, Feltovich & Glaser, 1981). 22 Hackling and Lawrence (1988) conducted a study to compare the problem-solving performance of genetics professors and students to show that the experts demonstrated a better a b i l i t y to recognize cues i n the task than the novices. Embracing the notion that pedigree diagrams, l i k e chess board positions, consisted of a v i s u a l display of cues, he speculated that successful solvers should be able to recognize cues i n the task and to generate systematic hypotheses to explain the inheritance patterns i n the pedigree. For example, a simple statement such as "affected individuals are found i n every generation" was a representative example of a s p e c i f i c cue i n the "pedigree" task. Their study showed that experts used more s p e c i f i c cues than novices to solve problems. This r e s u l t agreed with those found by Chase and Simon (1973) that expert chess-players recognized chess board configurations more successfully than novices. Another finding of Hackling and Lawrence (1988) and Hackling (1990) concerned experts' and novices' a b i l i t y to generate and to test hypotheses. For example, one way to test a hypothesis was by assigning genotypes to in d i v i d u a l s i n the pedigree. If i t was possible to assign a genotype to a l l individuals i n the pedigree and i f these genotypes were l o g i c a l l y consistent with each other, then the hypothesis was tenable (Hackling & Lawrence, 1988). Experts tended to make greater use of hypothesis-testing than the novices. 23 One l i m i t a t i o n of the above studies was that they generally applied 'quantitative' methods i n t h e i r analysis of problem-solving approaches. For example, Hackling and Lawrence (1988) found that experts recognized, i n terms of the mean number, more cues than novices. This focus on quantifying aspects of the problem solving provides a l i m i t e d view of the problem-solving process. To obtain a more complete picture, i t i s important to consider 'qu a l i t a t i v e ' aspects of the problem-solving process during which students interact with a s p e c i f i c task. This 'qu a l i t a t i v e ' method includes a detailed analysis of the in d i v i d u a l steps of the problem-solving process. The r e s u l t i n g information may help instructors i d e n t i f y d i f f i c u l t i e s students experience when they are addressing a problem. Smith and Good (1984) recommended a careful step-by- step analysis of the problem-solving process as an extension of t h e i r work. Section 2.3 provides a more detailed description of the studies which have adopted t h i s research approach. 2.2.3 Other Studies of Students' Conceptions of Meiosis There are other studies of students' conceptions of meiosis (other than those employing the expert-novice perspective). These studies have examined students' alt e r n a t i v e conceptions of ploidy i n meiosis. Brown (1990), Fisher et. a l . (1986), Stewart and Dale (1989), Stewart, 24 Hafner and Dale (1990), and Thomas (1988) (cited i n Brown, 1990) have reported that both high school and college students sometimes believe that two-DNA-molecule chromosomes (a chromosome made up of two DNA molecules) can contain i n the absence of crossing-over, two d i f f e r e n t a l l e l e s of the same gene. These authors noted at least one form of a rather common student b e l i e f concerning chromosome structure: namely, that chromosome structure i s somehow a function of chromosome number (also reported by Hildebrand, 1989) . These studies provide foundations for other researchers to examine students' understanding of the o r i g i n of chromosomes. Ki n d f i e l d (1991) provided a more detailed analysis of students' b e l i e f about the o r i g i n of the "two DNA-molecule" i n d i p l o i d c e l l s . In her students' model of meiosis, unreplicated "two DNA-molecule" chromosomes i n d i p l o i d c e l l s formed from the j o i n i n g of two "single-DNA-molecule" chromosomes contributed by each haploid gamete at f e r t i l i z a t i o n (rather than by a s e l f - r e p l i c a t i o n process). Thus, each zygote can receive d i f f e r e n t a l l e l e s from each . contributing gamete. This o r i g i n of the "two DNA-molecule" chromosomes allows one to explain why some students perceived that ploidy can change as a result of r e p l i c a t i o n . For example, these students believe that the d i p l o i d (2n) parent c e l l , which o r i g i n a l l y contains two-DNA-molecule chromosomes replicates to form "four-DNA-molecule" chromosomes and i s thus l a b e l l e d 4n. It i s believed by K i n d f i e l d that understanding the o r i g i n of chromosomes i n 25 meiosis i s essential for students to develop an understanding of the process of meiosis. I t follows that conception research i n meiosis should involve a study of students' understanding of both the chromosome structure (in r e l a t i o n to ploidy) as well as the process i t s e l f . 2.3 Contemporary Research on Teaching and Learning Research on teaching and students' learning has shifted i t s focus from what might be termed a "technical" orientation: one which emphasizes cause-and-effect relationships i n teaching and learning ( l i k e studies within the Piagetian perspective) to one which recognizes the important role of the p r i o r understandings of learners. The chief aim of these studies i s to report how students think about s p e c i f i c problems and phenomena from t h e i r own viewpoint (see e.g. Marton & Saljo, 1984). The t h e o r e t i c a l stance of one strand of contemporary research derives from studies of learning that have been c a l l e d "phenomenographic" because t h e i r focus i s on i d e n t i f y i n g i n d e t a i l the ways i n which learners understand phenomena or aspects of the world around them. Phenomenographic studies i d e n t i f y categories of description which characterize how people conceptualize phenomena, and process information as they are working on a p a r t i c u l a r problem. These categories represent students' explanations as they are addressing the problem. Findings 26 from these types of studies serve as a r i c h data set for arranging teaching and learning experiences. During the past decade, several studies of student learning i n higher education i d e n t i f i e d students' approaches to complicated problem solving tasks i n di f f e r e n t subject d i s c i p l i n e s (Marton & Saljo, 1976; Svensson, 1977; Ramsden, Whelan & Cooper, 1989). The two e a r l i e r c l a s s i c a l studies of Marton and Saljo, and Svensson were related to the reading of academic texts. Marton and Saljo (1976) asked students to read a text and to answer a series of questions. They reported that some students viewed the learning task as one that required them to "understand" and to "extract" meanings from the a r t i c l e . "Understanding" was e s s e n t i a l l y the cha r a c t e r i s t i c of t h i s group of students who demonstrated what was termed the "deep" approach. Conversely, the other group of students using what was c a l l e d the "surface" approach only intended to reproduce the materials being studied with no intention of understanding the a r t i c l e . A l t e r n a t i v e l y , some of them f a i l e d to recognize the words and phrases i n the a r t i c l e and used t h e i r own pre-determined way to interpret the information. They misunderstood the a r t i c l e and t h e i r way i n approaching the task was defined as "surface". . Svensson (1977) also asked students to read an academic text. He shi f t e d the focus and t r i e d to i d e n t i f y the "integration" component of the "deep" approach. "Integration", according to Svensson, was the a b i l i t y to use 27 a l l the information i n the task. Students using the "surface" approach did not appear to pay attention to the entire task. In other words, they focused on parts of the text rather than the entire a r t i c l e . They memorized information of in d i v i d u a l parts and indicated a lack of understanding of the message conveyed by the a r t i c l e as a whole. In a recent study i n the area of diagnostic problem-solving, Ramsden, Whelan and Cooper (1989) described the metacognitive behaviour of "checking and monitoring responses" that characterized the "deep" approach. Their study was designed to probe fourth-year medical students' approaches i n addressing a data base which contained s i g n i f i c a n t facts about a patient. In other words, the patient's case was presented i n the form of a problem to the student who was interviewed. Non-directive questions were designed to c o l l e c t students' interpretation of the cases, i n p a r t i c u l a r , a diagnosis or set of diagnoses and the students' reasons for t h e i r interpretation. The above study also reported examples of students adopting "deep" and "surface" approaches. Students using the "deep" approach read, understood, and used a l l the information i n the problem. They always "checked and monitored responses" before and af t e r making t h e i r hypotheses. Students using the "surface" approach made several separate hypotheses when they read parts of the information. "Checking and monitoring" hypotheses was rarely 28 considered by t h i s group of students. As a re s u l t , they were unable to discover a fin a l - best analysis that s a t i s f a c t o r i l y accounted for a l l the symptoms i n the task. In summary, a "deep" approach might be described i n terms of a search for "understanding" by employing strategies which attempt to "integrate" features of the task and by "checking and monitoring t h e i r own responses". Students using the "deep" approach tend to extract meaning from the task by considering and understanding a l l the information. More importantly, they constantly "check or monitor" t h e i r interpretations. "Checking or monitoring" one's responses i s considered to be the p i v o t a l c h a r a c t e r i s t i c of metacognitive behaviours. Students, using the "surface" approach, however, address the task without the intention of understanding i t . Thus they ignore most or part of the information of the task and make t h e i r own interpretations. They never check or monitor t h e i r interpretations. In t h i s way, they usually give up the task or use ' f o r c e - f i t t i n g ' strategies to f i t t h e i r interpretations to the data of the task. 2.4 Relationship of the Literature to Present Study The above l i t e r a t u r e review summarizes studies done i n the area of genetics problem-solving and more general problem-solving approaches. In general, these studies tend to concentrate on i d e n t i f y i n g either students' conceptual understanding or the problem solving approaches they used. 29 The present study attempts to look at both of these aspects and i n so doing makes two important contributions. The f i r s t contribution extends the genetics problem-solving research by examining students' understanding of the entire problem-solving process instead of i d e n t i f y i n g or l a b e l l i n g the i n d i v i d u a l s k i l l s within the process. The second contribution extends the phenomenographic studies of student approaches by exploring i n d e t a i l the "checking and monitoring" component of the "deep" approach. The following sections describe the theoretical perspective used to frame the present study. 2.5 Choice of Theoretical Perspective Phenomenography i s considered to be the appropriate perspective to use i n t h i s study for two reasons. F i r s t , phenomenography i s concerned with "relations between the experiencing individuals and the experienced phenomena" (Lybeck, et. a l . , 1988, p.5). In t h i s study, students' understanding of meiosis and t h e i r problem-solving approaches were explored when they were responding to two interview tasks. The f i r s t task explored t h e i r conceptions of meiosis and the other one explored t h e i r application of t h e i r knowledge i n a problem context. Second, phenomenography offers an interpretative framework which a s s i s t s i n describing a person's perceptual world; that i s , i t provides an interpretative experiential perspective. The aim of t h i s study i s not to i d e n t i f y "incorrect" meiosis 30 conceptualizations or problem-solving approaches but rather to describe how genetics students conceptualize the phenomenon of meiosis and address an 'applied' genetics problem. Results of such a study should help instructors to become f a m i l i a r with students' alternative ways of approaching a p a r t i c u l a r problem or d i f f e r e n t ways of understanding a concept. Being aware of these al t e r n a t i v e approaches and conceptual understandings should help a teacher plan and implement more ef f e c t i v e teaching strategies. Such strategies, which focus on students' actual l e v e l of understanding, d i f f e r i n important ways from the t r a d i t i o n a l 'transmissive' mode of teaching, as discussed i n Chapter 1. 2.6 Two Forms of Phenomenographic Research Phenomenography was f i r s t described by researchers at Gothenburg University i n Sweden and has been used extensively i n Sweden and elsewhere to frame a wide range of studies. As mentioned above, research i n the phenomenographic perspective claims that knowledge i s "constituted" by the relationship between the experiencing ind i v i d u a l and the experienced phenomenon. During recent years, there have been two l i n e s of development i n phenomenography. Although phenomenography i s a research approach which focuses on a person's understanding of phenomena, i n a broader sense i t also includes an investigation of the q u a l i t a t i v e v 31 approaches/processes l e a d i n g to the c o n s t r u c t i o n of knowledge. Various s t u d i e s probed people's understanding of p a r t i c u l a r phenomena. These phenomena in c l u d e d concepts i n academic subjects and phenomena experienced by people i n everyday l i f e . Studies reported i n E n g l i s h i n c l u d e research i n t o students' c o n c e p t u a l i z a t i o n s of: matter (Renstrom, 1987, 1988); the "mole concept" i n chemistry (Lybeck, e t . a l . , 1988); f a c t o r s a f f e c t i n g a c c e l e r a t i o n and v e l o c i t y (Johansson, Marton, & Svensson, 1985); r e l a t i v e speed (Walsh, e t . a l . , 1993); and knowledge ob j e c t s ( E n t w i s t l e & Marton, 1994). The second form of phehomenographic study i n v e s t i g a t e d people's approaches during t h e i r engagement w i t h p a r t i c u l a r academic questions (Marton & S a l j o , 1976; Svensson, 1977; Ramsden, Whelan & Cooper, 1989). Results of these s t u d i e s generated the "deep" and "surface" dichotomy used to describe the v a r i o u s approaches to problem s o l v i n g . In order to f u l l y explore students' a b i l i t i e s to approach a genetics problem, both forms of phenomenographic st u d i e s were incorporated i n the present study. The f i r s t p a r t of t h i s study i d e n t i f i e s the students' background knowledge of the phenomenon known as meiosis. The second part of the study i d e n t i f i e s approaches/processes used by the students i n addressing the 'applied' g e netics problem. 32 2.7 Categories of Description Phenomenographic studies i d e n t i f y categories of description which characterize how people conceptualize phenomena and process information... For example, the discovery of q u a l i t a t i v e l y d i f f e r e n t conceptions of force and motion constituted the main findings of Johansson, Marton and Svensson's (1985) investigation of technology students' understanding of bodies i n constant and decelerated motion. The present study i d e n t i f i e s categories of description which characterize students' understanding of meiosis and problem-solving approaches to the applied question. The invention of these categories of description stems from an analysis of q u a l i t a t i v e differences found i n empirical data (Johansson, Marton, & Svensson, 1985), and i s based upon the students' thematic explanations. A careful analysis of the "content" of student explanations leads to the generation of characterizations to depict the "thinker's understanding of that which i s thought about" (Johansson, Marton & Svensson, 1985, p.247). An underlying assumption i s that "each phenomenon or concept can be understood i n a l i m i t e d number of q u a l i t a t i v e l y d i f f e r e n t ways" (Marton, 1986, p.30). These in t e r p r e t a t i v e categories are c a l l e d "categories of description" (Dahlgren, 1984; Johansson, Marton & Svensson, 1985). I t must be mentioned that these f i n i t e categories of description are used to sample conceptualizations, not people^ Unlike "concept maps" (Novak 33 & Gowin, 1984) which are considered to be representative of i n d i v i d u a l patterns of reasoning i n terms of propositional linkages, categories of description are rather a kind of c o l l e c t i v e i d e n t i f i c a t i o n of important explanatory-a t t r i b u t e s . I t i s important to note that the construction of categories of description i s based on a premise of the phenomenographic perspective that emphasizes human-world relationships. In other words, what i s focused upon by the researcher i s a function of both the p a r t i c u l a r phenomenon or problem at hand, and that of the p a r t i c u l a r subjects i n the study. Thus, conceptualizations or problem-solving approaches, i n t h i s study, are not s p e c i f i c a l l y inherent to either i ndividuals or to instances of the world, rather they are inherent to both i n an intertwined r e l a t i o n s h i p . The experience of a p a r t i c u l a r phenomenon has a d u a l i s t i c aspect to i t , for instance, we do not just hear, we must hear something. In other words: There are two aspects of the experience of an object, the act of experiencing, and that which i s experienced, i . e . the meaning of the object towards which one's attention i s directed. These two aspects j o i n t l y form the whole, the experience of the object i n question. (Marton & Saljo, 1984, p.54) From t h i s perspective, a conceptualization or problem-solving approach i s characterized as a sense-making construction described as "constitutionalism" (Marton & 34 Neuman, 1988; Renstrom, 1987, 1988) that distinguishes the phenomenographic perspective from constructivism: While the emphasis i n constructivism i s on acts ... constitutionalism has the unity of the act and the acted upon as i t s point of departure. (Marton & Neuman, 1988, pp. 7-8) Whatever primary c h a r a c t e r i s t i c s are used to determine the generation of categories of description, they s t i l l "refer to whole q u a l i t i e s of human-world r e l a t i o n s " (Johansson, Marton & Svensson, 1985, p.249) mentioned above. The retention of the "human-world" relationship i n the construction of categories of description i s based on two premises. F i r s t , when generating the categories, an attempt needs to be made not to impose too much of a personal frame of reference d i r e c t l y onto the students' explanations. Nevertheless the researchers must simultaneously l i n k t h e i r personal understanding (in t h i s case, of genetics) to the students' understanding, i n order to understand and relate to the content of students' explanations. By so doing, they hope to preserve the o r i g i n a l i t y of student explanations i n r e l a t i o n to that p a r t i c u l a r phenomenon or problem. By incorporating the above premise i n the analysis procedures, the important "human-world" relationship emphasized by phenomenographic perspective should be preserved. Constructing categories of description can be depicted as a kind of i t e r a t i v e "hermeneutic c i r c l e " procedure which uses the "parts" to evolve the "whole". Dahlgren (1984) 35 describes the data analysis procedure as a process which p r i n c i p a l l y involves: The reduction of unimportant d i s s i m i l a r i t i e s e.g. terminology or other s u p e r f i c i a l c h a r a c t e r i s t i c s , and the integration of important core c h a r a c t e r i s t i c s which make up the content and structure of a given category. These core characteristics are generalizable to other problem settings, (p.24) In the present study, the framing of categories of description which characterize the various problem-solving approaches of students was based on the integration of important core c h a r a c t e r i s t i c s such as ways of reading problem information, generating hypotheses, t e s t i n g and reframing hypotheses. F i n a l l y , the findings of the categories of description i n o r i g i n a l studies i s a form of discovery. Once the categories have been found, i t must be possible f o r other researchers to use them and to revise them (Marton, 1988) . In t h i s study, the categories concerning problem-solving approaches were constructed based on previous studies i n medical diagnostic problem-solving and text comprehension (Ramsden, Whelan & Cooper, 1989; Marton & Saljo, 1976; Svensson, 1977). The c h a r a c t e r i s t i c s of the "surface" and the "deep" approaches i d e n t i f i e d by the researchers should be applicable to the four approaches to problem solving i n the present study. However, we should bear i n mind that the meaning attached to these c h a r a c t e r i s t i c s w i l l vary i n 36 d i f f e r e n t academic subjects (Ramsden, 1992), and there are always gradations between the "surface" and "deep" approaches. 37 CHAPTER 3 DESIGN OF STUDY AND METHODOLOGY 3.1 I n t r o d u c t i o n This study was framed as a phenomenographic study; an i n t e r p r e t a t i v e , second-order study as discussed i n Chapter 2. The aim of the study was to ob t a i n i n s i g h t s i n t o genetics students' c o n c e p t u a l i z a t i o n s of meiosis and the r e l a t i o n s h i p of these c o n c e p t u a l i z a t i o n s t o t h e i r approaches i n genetics problem-solving. The study i n v o l v e d i n t e r v i e w i n g ten undergraduate genetics students from a t h i r d - y e a r genetics course whose assessment was based s o l e l y on problem-solving and a p p l i c a t i o n of concepts. This chapter i s d i v i d e d i n t o two p a r t s . The f i r s t p a r t describes the design features of the study, i n c l u d i n g the two i n t e r v i e w tasks (the meiosis task and the Ascobolus problem) and the i n t e r v i e w questions. The second p a r t discusses the procedures of a n a l y s i s that l e d to the generation of the r e s u l t s i n the three a n a l y s i s chapters. The chapter a l s o i n c l u d e s a d i s c u s s i o n of the v a l i d i t y of t h i s study. 3.2 The Design of the Study The design of the study was framed i n terms of the f o l l o w i n g two components: the design of a s u i t a b l e i n t e r v i e w p r o t o c o l based on an extensive p i l o t study and the a n a l y s i s of the i n t e r v i e w data from a phenomenographic p e r s p e c t i v e . 3 8 3.2.1 Factors Influencing Design of the Interview Protocol The interview protocol was designed to probe the students' understanding of meiosis and t h e i r a p p l i c a t i o n of t h i s understanding to a novel problem setting. Two contexts: the meiosis task and the Ascojbolus problem were generated. The design of the interview protocols emerged from the results of a p i l o t study which preceded the main study. The p i l o t study consisted of both a written component and an interview with i n d i v i d u a l students. The written component consisted of an analysis of the students' written responses to the two genetics tasks. The interview component consisted of in-depth c l i n i c a l interviews using a structured protocol. The Ascobolus problem was set by the genetics course professor for the 1993 f i n a l examination. A close scrutiny of the students' responses to th i s question r e f l e c t e d that they had s i g n i f i c a n t d i f f i c u l t i e s i n dealing with the question. These d i f f i c u l t i e s s i g n i f i c a n t l y concerned the biology of the haploid organisms featured i n the problem, the genetic symbols, the concepts of .non-disjunction and crossing over and the data themselves. Seven undergraduate biology students who had completed the genetics course volunteered to become involved i n the p i l o t study. During the interviews, the students were asked to explain t h e i r understanding of meiosis and the Ascobolus problem. The audiotaped interviews of the p i l o t study interviews were thoroughly analyzed between interviews. This 39 procedure allowed the re v i s i o n of the interview protocol between subsequent interviews. The r e v i s i o n of the protocol took place i n three areas. The f i r s t area related to the design of the Ascobolus problem which o r i g i n a l l y consisted of three cases. The f i r s t two cases related to two forms of non-disjunction (during Meiosis I and Meiosis I I ) 4 and the t h i r d case related to crossing over. The students usually mixed up the concepts of non disjunction and crossing over. An incl u s i o n of three cases resulted i n a very long interview. Students became involved i n the two "non-disjunction" cases and often had l i t t l e or no time for the t h i r d case. A.decision was made after the p i l o t study to delete the second case on meiosis II (Mil) non-disjunction and to concentrate on the other two cases i n the main study. The second area related to the probing of the students' understanding of meiosis ; During the i n i t i a l stage of the p i l o t study, the students were asked to describe meiosis i n general. The broad general descriptions i n the students' responses led the author to create a context to confine and define the scope of the study. The context consisted of a meiosis task written by the genetics course professor, which required the students to work through the whole process of meiosis of a d i p l o i d c e l l consisting of only a single p a i r 4 Meiosis I and Meiosis II non-disjunction refer to the failure of homologs during the f i r s t and second division of meiosis to separate to opposite poles. 40 of homologous chromosomes. This task succeeded i n generating a more l i m i t e d range of responses from the students. The completion of the p i l o t study led to the development of the revised meiosis task. This question was more complicated than the o r i g i n a l one used i n the p i l o t study for i t involved a d i p l o i d c e l l with two pairs instead of a single p a i r of homologous chromosomes. The question was given i n the form of a questionnaire to the genetics students (1994) at the st a r t of the course. An examination of 286 questionnaires by the author and the genetics professor revealed varying students' responses which included variations both i n the indi v i d u a l events and the p a r t i c i p a t i n g structures of meiosis. For example, the responses concerning the r e p l i c a t i o n or p a i r i n g events before the f i r s t d i v i s i o n of meiosis included the following: • r e p l i c a t i o n without p a i r i n g ; • no r e p l i c a t i o n or p a i r i n g ; • fusion of homologous pairs instead of p a i r i n g ; • r e p l i c a t i o n producing eight i n d i v i d u a l chromosomes; and • two re p l i c a t i o n s The t h i r d area related to the setting up of probing questions. To obtain a r i c h data base, the author attempted to create an environment where an i n t e l l e c t u a l conversation with the students about genetics might occur. I t was decided that the opening question of the interview should be open-ended so that the students could express t h e i r thoughts 41 f r e e l y i n a discussion format. The design of the interview protocol, then, was to a s s i s t the e l i c i t a t i o n and c o l l e c t i o n of students' explanations about t h e i r understanding of meiosis and about t h e i r approaches. Jones (1985) pointed out: An interview i s a complicated, s h i f t i n g , s o c i a l process occurring between two indi v i d u a l human beings, which can never be exactly replicated, (pp.47-48) Jones' assertion above formed the design c r i t e r i a , which were to introduce i d e n t i c a l contexts but not to ask a set l i s t of i d e n t i c a l questions for ind i v i d u a l students. The author asked probing questions to provide the students with an opportunity to repudiate or af f i r m t h e i r o r i g i n a l explanations. It was appreciated that the use of probing questions would help c l a r i f y some of the points made by the students e s p e c i a l l y when they were unclear or undecided about t h e i r o r i g i n a l decision. The p i l o t i n g experience showed the probing questions to be a non-stressful, e f f e c t i v e method of va l i d a t i o n , e s p e c i a l l y when a student's i n i t i a l a r t i c u l a t i o n of his/her understanding had been incomplete or d i f f i c u l t to interpret. Overall, the interview, protocols were designed with the following c r i t e r i a i n mind. The interviews i n the main study were designed to be: 42 • in t e r e s t i n g enough to engage the interviewees i n an honest and open in-depth exploration and r e f l e c t i o n of th e i r reasoning; • basic enough and low stress so that the student being .....interviewed did not perceive i t to be a genetics test; • able to provide the richest set of data for the purposes of the study; and • of a reasonable duration from both the interviewee's and interviewer's perspective; about 45 minutes. 3.2.2 The Subjects Ten students, enrolled i n a third-year university genetics course, participated i n the f i n a l interviews for the study. Because of the r e s t r i c t i o n s imposed by the ethics review procedures for U.B.C, the students were asked to volunteer by signing up for interviews during t h e i r t u t o r i a l sessions. Thus the sample obtained was a sample of convenience rather than a random sample and they tended to be students who had performed well i n the course. The interviews were done after the students had completed the course.on genetics. Topics i n the course included the chromosomal theory of inheritance, Mendel's genetics, tetrad analysis and mutation. These topics were introduced separately i n the lectures, and' thus an integration of these concepts by the students was necessary before they could apply them to the Ascobolus problem. 43 3.2.3 The Structural Format of the Interview Protocol The protocol was produced i n 'interview probing phases' using the meiosis task and the Ascobolus problem as the interview tasks for the phases. Examples of possible types of questions were then included to outline the•general tone and d i r e c t i o n of probing that were to take place i n each context. The f i n a l version of the protocol took on the following form: Phase 1: The Introductory Phase of the Interview. In the i n i t i a l phase the interviewer and student chatted informally and the student was asked warm-up questions about t h e i r personal background and views of learning. This phase was i n i t i a t e d with questions such as: • Please t e l l me about your educational background. What i s your major or specialty? Further Probes: • Why did you select t h i s subject as your major? • What subjects do you f i n d easy or d i f f i c u l t i n your un i v e r s i t y study? These questions helped the interviewer learn more about the students and to discover i f they had experienced any problems i n learning genetics. At the same time, the interviewer made i t clear that the aim of t h i s research was to f i n d out factors a f f e c t i n g the students' understanding 44 and a p p l i c a t i o n of the concept of meiosis. I t was c l a r i f i e d that the interview was not meant to be a test of t h e i r genetics knowledge, and that the interviewer wanted to 'see' how the d i f f e r e n t contexts led them to think about meiosis. Generally the i n i t i a l phase was spent developing a relaxed and comfortable atmosphere. This was considered to be a p a r t i c u l a r l y important phase because human beings respond d i f f e r e n t l y i n d i f f e r e n t situations, and to d i f f e r e n t audiences. In giving accounts to others, they are concerned not only with making their, actions comprehensible, but also the presentation of a credible and legitimate ' s e l f (Jones, 1985) . 45 Phase 2 : Probing into the nature of the conceptualizations of meiosis by using the meiosis task. F i r s t Interview Task: The Meiosis Task In t h i s phase, the meiosis task was provided: Starting with the c e l l shown below containing 4 chromosomes (that i s a set of two p a i r s ) . DRAW A SET OF DIAGRAMS and EXPLAIN to a f i r s t year student who i s confused, what happens when meiosis occurs. Label your diagrams, the chromosomes, and the genes, and discuss your diagrams and labeled parts during your meiosis explanation, so that the "confused" student can keep track of what i s taking place. Your FINAL DIAGRAM should c l e a r l y show the end r e s u l t of the events as c l e a r l y as possible. Figure 1 : The meiosis task The meiosis task served as the f i r s t interview task. It provided the following information: a d i p l o i d c e l l with two pairs of homologous chromosomes; and each p a i r of homologous chromosomes carr i e s a d i f f e r e n t kind of gene, designated as A,a and B,b respectively. Students were expected to describe 46 the products of meiosis when asked to consider c e l l d i v i s i o n i n t h i s generalized parent c e l l . This interview task had two purposes. F i r s t , i t enabled the researcher to explore the students' conceptualization of meiosis. Unlike the second interview task that provided s p e c i f i c information about a p a r t i c u l a r organism, the meiosis task provided only a simple symbolic representation of a generalized d i p l o i d c e l l that contained two pairs of homologous chromosomes. Given t h i s information, students were asked to describe the events or steps leading to the products of meiosis. An examination and analysis of the students' responses provided a picture of the conceptualizations of meiosis held by the students. Second, the meiosis task served to complement the Ascobolus problem (the second interview task). Together, they provided appropriate contexts for investigating the a b i l i t y of students to 'apply' t h e i r knowledge of meiosis i n one context to another. Both tasks required that students construct and draw upon mental models of meiosis. The meiosis task provided a simple question that d i r e c t l y probed students' s p e c i f i c knowledge of meiosis. The Ascobolus problem was a more complex task which required students to analyze information before they could 'apply' t h e i r knowledge of meiosis. The students were introduced to the meiosis task. They were asked to study the task c a r e f u l l y for a while and then to provide answers to the task. Before ac t u a l l y going into 47 the process of meiosis, the students were asked a question to test t h e i r understanding of the d i f f e r e n t kinds of genetic symbols i n the c e l l diagram (see Figure A). The i n i t i a l question was: • You are given t h i s question which you had done before at the st a r t of the genetics course. Read t h i s question. As you are looking at the c e l l diagram, t e l l me the kinds of things that come into your mind. What do these chromosomal representations i n the c e l l diagram mean to you? After l i s t e n i n g to the students' perceptions of the chromosomal representations of the c e l l diagram, the interviewer then proceeded to ask the students to describe the process of meiosis. The task was introduced i n the following way: • You w i l l be given ten minutes or so to work on t h i s question. Please say your thoughts aloud as you are working through the question. Draw diagrams to i l l u s t r a t e your thoughts. As the students were describing the process of meiosis, the interviewer asked probing questions which required the students to affirm, c l a r i f y or repudiate t h e i r explanations. Some examples of the probing questions were: • How do you know that the chromosomes r e p l i c a t e and then p a i r up? • Why do you change t h i s diagram? 48 • Why do you think the end products are two c e l l s each with two chromosomes instead of four? After the students had described the whole process of meiosis by drawing diagrams and v e r b a l i z i n g t h e i r thoughts, the f i n a l phase involved probing the students' understanding of the concept of ploidy (the set of chromosomes). The f i n a l question of t h i s phase was: • Do you have any basic framework i n mind as you are constructing t h i s meiosis model? From the students' responses, some additional probing questions were asked. An example included: • What do you mean by one set of chromosomes? A Brief Meiosis Review In order to a s s i s t readers i n understanding the analysis of students' conceptualizations of meiosis i n the next chapter, a b r i e f description of the 'canonical' meiosis model5 i s provided. Meiosis i s the name given to the nuclear d i v i s i o n s i n special c e l l s that are destined to produce gametes (in animals) and sexual spores (in plants and fungi). Such a c e l l i s c a l l e d a meiocyte. Meiosis involves two c e l l 5 The phrase 'Canonical' meiosis model i s used here to denote the meiosis model which i s generally accepted by the s c i e n t i f i c community. It i s used again l a t e r to lab e l those students' meiosis models which demonstrate a sound understanding of meiosis. 49 d i v i s i o n s of each meiocyte and two associated meiotic d i v i s i o n s of each nucleus. Hence, each meiocyte generally produces four c e l l s , which we s h a l l c a l l products of meiosis. In humans and other animals, the products of meiosis are the gametes - spermatozoa and eggs. In plants and fungi, the products of meiosis are the sexual spores -these are c a l l e d ascospores i n the case of the organism Ascobolus. We need to introduce some terminology. Only c e l l s with two chromosome sets undergo meiosis. Cel l s with two chromosome sets are c a l l e d diploid and are represented symbolically as 2 n , where n i s the number of chromosomes i n a set. In d i p l o i d c e l l s , each chromosome has a matching chromosome i n the other set. Together these chromosomes constitute a p a i r and are c a l l e d a homologous p a i r ; the two members of a p a i r are c a l l e d homologs. The generalized parent c e l l i n the meiosis task i s d i p l o i d and contains two chromosomes i n i t s set. In contrast, c e l l s with only one set (n) are c a l l e d haploid. Haploid c e l l s have to fuse together to form a transient d i p l o i d before they can undergo meiosis. Meiosis involves two c e l l d i v i s i o n steps distinguished as meiosis I (MI) and meiosis I I (Mil). Each meiotic d i v i s i o n i s formally described as consisting of four phases: prophase, metaphase, anaphase, and telophase. The correspondence between the phases of meiosis and the events of importance i s as follows: prophase 1^- nuclear membrane 'dissolves', chromosomes have replicated, and each 50 chromosome consists of two chromatids held together at t h e i r centromere; metaphase I - each p a i r of homologous chromosomes (homologs) moves to a position along the equatorial plane of the c e l l (equatorial alignment I ) ; anaphase I - homologs separate and move d i r e c t i o n a l l y towards one of the two poles of the c e l l (chromosome segregation); telophase I - l a s t stage of the f i r s t d i v i s i o n step when two daughter nuclei r e s u l t ; metaphase II - the "replicated" chromosomes arrange themselves on the equatorial plane (equatorial alignment I I ) ; anaphase II-centromeres s p l i t and s i s t e r chromatids are pulled to opposite poles (chromatid segregation); and telophase II -the nuclei re-form around the chromosomes at the poles and four products of meiosis are formed. Although the phase names (prophase, metaphase, and so on) have h i s t o r i c a l and cy t o l o g i c a l importance, they are not p a r t i c u l a r l y 'descriptive' of the events they represent. For t h i s reason, I used the more x d e s c r i p t i v e ' labels for the events that appear i n parentheses rather than the conventional phase names are used i n t h i s thesis. These 'descriptive' labels are r e p l i c a t i o n , homologous pairing, (equatorial) alignment I, chromosome segregation, (equatorial) alignment I I , and chromatid segregation. Figure 2 provides a complete i l l u s t r a t i o n of the canonical model of meiosis. 51 2a Parent c e l l Chromosome replicates but * centromeres do not Homologs separate and homologs pair move to poles 2b Replication 2c Homologous Pairing each pair of homologs take up a position at the equatorial plane 2d Equatorial Alignment I 2e Chromosome Segregation (Meiosis I is completed) Chromosomes arrange themselves on the equatorial plane Chromatids separate at centromere and move*to poles & 7 2f Equatorial Alignment II 2g Chromatid segregation 2h. Four Product Cells (Meiosis II is completed) Figure 2: The 'Canonical' model of the process of meiosis as i t appears i n a d i p l o i d c e l l 5 2 Phase 3: Ask the students to solve an 'applied' question that incorporates the concept of meiosis, and involves using the fungus Ascobolus. Second Interview Task: In the fungus Ascobolus (similar to Neurospora), ascospores are normally black, the mutation (f) producing fawn ascospores i s i n a gene just to the r i g h t of the centromere on chromosome s i x , whereas the mutation (jb) producing beige ascospores i s i n a gene just to the l e f t of the same centromere. In a cross of fawn by beige parents (+f x b+) most octads show four fawn and four beige ascospores, but two rare exceptional octads were found as shown below. In the sketch, black i s wild type phenotype, a v e r t i c a l l i n e i s fawn, a horizontal l i n e beige, and an empty c i r c l e represents an aborted (dead) ascospore. (i) Provide reasonable explanations for these two exceptional octad cases. ( i i ) Diagram the meiosis that gave r i s e to both octad cases. 9 0 • 0 9 9 9 9 0 © Double mutants (bfO 0 © 0 0 0 0 Octad Case 1 Octad Case 2 53 Ascobolus, the organism featured i n the second interview task, i s a dung-fungus which grows on horse feces. It i s a haploid organism and thus contains only one chromosome set. As mentioned previously, meiosis normally requires the p a i r i n g of two homologous chromosome sets. Ascobolus, being haploid, has to create a temporary d i p l o i d stage before i t can undergo meiosis. Meiosis then occurs i n Ascobolus to produce haploid (n) sexual spores i n groups of four. Each group i s c a l l e d a tetrad. Each of the four products of meiosis then undergoes an additional mitotic d i v i s i o n (mitosis) 6 , y i e l d i n g a group of eight c e l l s (ascospores) c a l l e d an octad. In Ascobolus, the octad forms i s i n a l i n e a r arrangement; and the eight c e l l s are contained i n an ascus (see Figure 3a , 3b and the l i f e cycle diagram i n part one of Appendix B). Figure 3 a : A l i n e a r octad i n an ascus of Ascobolus 6 During a mitotic division, a c e l l divides once, making a duplicate copy of i t s e l f . There is no reduction in chromosome number. 54 Jn "Adult" n "Adult"L Mitosis Mitosis Representative n c e l l s Representative n c e l l s Many-mitoses 2n Transient d i p l o i d meiocyte Meiosis n n n n Sexual spores Figure 3b: The haploid l i f e cycle of Ascobolus Many mitoses This Ascobolus problem was complex. The question, l i k e the meiosis task i n phase 2 of the interview, also involved the concept of meiosis. However, t h i s question 'embedded' the meiotic concepts i n the problem context. This question required an understanding of the l i f e cycle of haploid, appreciation of haploidy i t s e l f and the language i n the data set (for example, aborted, ascospores) as well as a knowledge of meiosis. An answer to the Ascobolus problem i s provided i n part two of Appendix B. 55 In t h i s phase of the interview, the students were asked to study the question c a r e f u l l y . The opening question was as follows: • Let's go to t h i s more r e a l i s t i c genetics question. Try to t e l l me your thoughts as you are going through t h i s question. As the students were going through the question, a number of probing questions based on t h e i r responses were asked. Examples of these probing questions were: • You mentioned crossing over. How does crossing over occur? • Do these genetic symbols have any meaning to you? • Why did you give up your o r i g i n a l hypothesis? F i n a l l y , the students were asked to apply t h e i r meiotic knowledge to f i t into the data of the octad cases. 3.3 The Procedures of Analysis The data analysis was an extensive procedure that began by transcribing a l l of the interviews. Since the data consisted of a discussion of genetic symbols and concepts, flow charts appeared to be an ef f e c t i v e method for i l l u s t r a t i n g students' approaches and assisted i n the generation of the categories of description. 56 3.3.1 Categories of Description Characterizing Conceptualizations and Approaches (Chapters 4 and 5) Data were analyzed by the constant comparative method (Lincoln and Guba, 1985). There were three l e v e l s of data transformation (Novak & Gowin, 1984): production of verbatim tra n s c r i p t s , i n i t i a l framing of potential categories of description on the basis of the content of p a r t i c u l a r excerpts and comparison with s i m i l a r l y categorized excerpts, and the development of a framework depicting each category of description. The f i r s t l e v e l of data transformation was the verbatim t r a n s c r i p t i o n of the audio-taped interviews immediately following each interview. Transcriptions were printed and read through with a p a r t i c u l a r research question i n mind. The purpose of doing t h i s was to i d e n t i f y , f or example, di f f e r e n t ways i n which the students approached the Ascobolus problem. During t h i s reading, descriptive words or phrases were attached to short sections of the t r a n s c r i p t . These descriptive words or phrases were l a b e l l e d as elements of potential approaches or conceptualizations. For example, some students had di f f e r e n t interpretations of the genetic symbols i n the Ascobolus problem from those of the instructor. This c h a r a c t e r i s t i c was i d e n t i f i e d as a possible c h a r a c t e r i s t i c of a problem-solving approach (labelled ' p a r t i a l recognition and forced f i t ' ) as i t pertained to the second research question (on genetics problem-solving approaches). Subsequently, sections of the tran s c r i p t s 57 (hereafter r e f e r r e d to as excerpts) which might f i t i n t o t h i s p o t e n t i a l category were l a b e l l e d . These excerpts were then p r i n t e d and organized by hand i n t o p o t e n t i a l sub-c a t e g o r i e s . For example, w i t h i n the ' p a r t i a l r e c o g n i t i o n and for c e d f i t ' approach, sub-categories i n v o l v e d d e s c r i p t i o n s of v a r i o u s stages of the problem-solving process: students' c o n s i d e r a t i o n of the w r i t t e n and symbolic r e p r e s e n t a t i o n s , students' f o r m u l a t i o n of hypothesis, students' f o r c e - f i t t i n g o peration. The mechanical s o r t i n g allowed f o r numerous excerpts from d i f f e r e n t i n t e r v i e w t r a n s c r i p t s to be considered together. The content of one excerpt was compared and contrasted w i t h the content of another s i m i l a r l y l a b e l l e d excerpt and i n r e l a t i o n to the c o n c e p t u a l i z a t i o n under c o n s i d e r a t i o n . My conduct i n the a n a l y t i c a l process was informed by both the research questions and by i n s i g h t s d e r i v e d from the l i t e r a t u r e review. Excerpts were s e l e c t e d p u r p o s e f u l l y and they had a d i r e c t r e l a t i o n s h i p to the . research questions. F i n a l l y , the l a b e l l e d excerpts of i n d i v i d u a l t r a n s c r i p t s were organized i n t o a flow chart. These flow charts represented the students' ' t h i n k i n g processes' and they appeared to be an e f f e c t i v e means f o r generating the categories of d e s c r i p t i o n w i t h each i n d i v i d u a l approach repr e s e n t i n g a category. These cat e g o r i e s of d e s c r i p t i o n were drawn from the data and were, t h e r e f o r e , hot predetermined. 58 3.3.2 Data Analysis: The Relationship between the Meiosis Conceptualizations and the Approaches (Chapter 6) The analysis of the f i n a l research question was based on the findings of the two 'results' chapters. In the f i r s t stage of the analysis, the meiosis models used by each of the students on the tasks described i n Chapters 4 and 5 were i d e n t i f i e d and cross referenced. The mapping exercise allowed one to see whether each student had used these meiosis models consistently. The next stage was to match these meiosis models onto t h e i r approaches. A table was set up to c l e a r l y show the relationship" between the approaches and'the meiosis models. A close scrutiny of these relationships led the author to make conjectures on factors that are related to problem-solving success. 3.3.3 Review of A n a l y t i c a l Outcomes The research method also included a process of checking the author's interpretations of the interviews with the student p a r t i c i p a n t s , to enhance the study's i n t e r n a l v a l i d i t y (Merriam, 1988). Participants were provided with drafts of the analyses to comment on. This member-checking system was performed for the students of both the p i l o t and the main study. While r e p l i e s were received from only seven of these students, these were s a t i s f i e d with the interpretations and portrayals of the interview discussions. 59 CHAPTER 4 CONCEPTUALIZATIONS OF MEIOSIS 4.1 Data Analysis for Research Question 1: Introduction This chapter presents outcomes of analysis of the data for the f i r s t research question: What a r e the q u a l i t a t i v e l y d i f f e r e n t ways i n wh ich undergradua te s t u d e n t s c o n c e p t u a l i z e m e i o s i s ? The present study aimed to see how un i v e r s i t y genetics students 'applied' or 'used' t h e i r concept of meiosis i n a more complex setting. Before doing t h i s , however, i t was necessary to determine t h e i r understanding of the underlying concept of meiosis. The analysis involved constructing categories of descriptions (hereafter these are c a l l e d 'conceptualizations') to characterize undergraduate genetics students' understanding of meiosis. In t h i s connection, a meiosis task and some questions were generated and used i n interviews with ten students. The meiosis task; the interview questions; and the procedures of analysis were described i n Chapter 3, and are not repeated here. The present chapter consists of two sections. The f i r s t section deals with three d i f f e r e n t conceptualizations of meiosis that were constructed from the students' responses. The second section summarizes the data. In addition, a table 60 which summarizes the results of the class survey (using the meiosis task) i n September 1994 i s included i n Appendix C. In t h i s chapter, examples from a l l the students were included to i l l u s t r a t e the di f f e r e n t students' understanding. The purpose for doing t h i s was to i l l u s t r a t e the s l i g h t variations between the meiosis models generated by these students within the same conceptualization. Analysis of the data i s organized around i n d i v i d u a l students i n the three analysis chapters. The f i r s t conceptualization of meiosis, for example, i s i l l u s t r a t e d by excerpts of interviews from students who subscribed to the 'canonical' conceptualization. The data come from interviews with Chris, Kirk, Sharilyn, May and L a i l a ( a l l pseudonyms), whose responses are presented i n consecutive fashion. This presentation also helps to introduce the students to the reader. In p a r t i c u l a r , t h i s introduction to the students themselves enhances the reader's understanding of the significance of a p a r t i c u l a r way of thinking about meiosis and of approaching the Ascobolus problem. 4.2 Conceptualizations of Meiosis In analyzing the data, three q u a l i t a t i v e l y d i f f e r e n t conceptualizations of the process of meiosis were i d e n t i f i e d from the group of students p a r t i c i p a t i n g i n the study. 61 4.2.1. The 'Canonical' Conceptualization of Meiosis The meiosis models within t h i s 'canonical' conceptualization are s i m i l a r to the s c i e n t i f i c a l l y accepted meiosis model described i n Chapter 3. They include two d i v i s i o n steps, r e s u l t i n g i n the formation of four c e l l s from a parent c e l l . Some examples showed s l i g h t v ariations by merging certain events of meiosis or by having 'ploidy confusion'. Ploidy i s the number of set of chromosomes. However, i t s confusion sometimes has to do with n (the number of chromosomes i n a set). Chris' example characterizes t h i s conceptualization. 'Canonical' Conceptualization: Chris's Example PATRICK: CHRIS: 10 PATRICK: CHRIS: 20 Let's look at the meiosis question. As you are going through the question, say aloud a l l your thoughts and explain to me a l l i l l u s t r a t i o n s and symbols (in the question). These are the chromosomes, big strand of DNA that are arranged -- A,B,a,b -- those are the a l l e l e s of the gene. The big A and the small a are the same gene but they are d i f f e r e n t forms of gene to y i e l d d i f f e r e n t products (Figure 4a).. If you are going to describe the c e l l i n terms of haploid and d i p l o i d , the n and the 2n thing, how would you apply the terms here (to the c e l l i n question)? This(the c e l l i n the diagram) i s the 2n because there are two of these ones (A,a) and two of that one (B,b). Well, the A (type of chromosome) and the B (type of chromosome). These two are A's, the big A and the small a; these two are the B's, the big B and the small b. F i r s t thing i s that they replicate, so you get two of the big A's and the two l i t t l e a's; two big B's and two l i t t l e b's 62 25 PATRICK: 3 0 CHRIS: 35 40 55 60 65 70 (Figure 4b). And then they l i n e up along an e q u a t o r i a l plane (Figure 4c) . Why do you have to r e p l i c a t e the chromosomes? Because i t (the parent c e l l ) i s going to s p l i t i n t o two -- two n e s s e n t i a l l y i n the parent c e l l . I f the f i n a l products (the two c e l l s ) of the f i r s t meiosis are 2n, you have to r e p l i c a t e chromosomes here. You have to r e p l i c a t e due to t h i s reason. (He s t a r t e d to draw Figure 4 to show the l i n i n g up of chromosomes). Because these two r e p l i c a t e d p a i r s of chromosomes (aa, bb) are going up here (Figure 4c), and i t could be,-- e i t h e r one of these p a i r s (AA, aa) can go i n e i t h e r d i r e c t i o n s (up or down). You can have the b i g A (AA p a i r ) go t h i s way (up) and the small a (aa pair) go that way (down). But you can have the other way round. Um -- t h i s i s only the one way (to represent the two product c e l l s a f t e r meiosis I, Figure 4d) . Can you say a l i t t l e b i t more? Well, i f no doubling up ( r e p l i c a t i o n of chromosomes), then each of these (two c e l l s a f t e r meiosis I) w i l l only have a copy of these chromosomes (ca r r y i n g genes A,B; A,b; a,B or a,b respect i v e l y ) . A n d then when you s p l i t again ( i n meiosis I I ) , you won't get a whole set of chromosome i n any one c e l l (any one of the four daughter c e l l s ) (Figure 4e) . PATRICK: What do you mean by whole set of chromosome? CHRIS: Well, you only have an A or B i n -- l i k e you w i l l get a c e l l w i t h (a chromosome holding) a b i g A (gene) and a c e l l with a b i g B and a c e l l w i t h a small a and a c e l l with a small b. (Figure 4e) . PATRICK: Why do you think t h i s arrangement i s wrong? CHRIS: Because each of the gamete ( c e l l ) w i l l be l a c k i n g a chromosome, a b i g chunk of DNA i s missing. PATRICK: Could you have the two b i g As go w i t h the two b i g Bs here? 45 ' PATRICK: 50 CHRIS: CHRIS: You mean BB l i n i n g up wi t h AA? Um - - when the 63 75 80 85 PATRICK: CHRIS: 90 • 95 PATRICK: 100 CHRIS: PATRICK: CHRIS: 105 spindle f i b r e s came to grab the centromere here from the poles, because you have (AA pa i r i n g up with BB), AA, BB and you also have bb, aa (bb pairing up with aa), now you can get t h i s -- t h i s f i b r e and then another one (spindle f i b r e at the other pole), you get these product (cells) with AA, aa and BB, bb; then you end up with e s s e n t i a l l y the same problem (as above): you w i l l be lacking the A chromosome or the B chromosome i n the f i n a l product c e l l s . Okay, what happen after meiosis I? They (AA/aa/BB/bb) get s p l i t up here at the second d i v i s i o n , but you s t i l l won't get a big A and the small a i n the f i n a l product c e l l . And these ones (aa,bb) w i l l s p l i t up. If say -- these were big A and small b coming together, and these were big B and small a (Figure 4d) , then these ( f i n a l product cells) w i l l end up i n d i f f e r e n t patterns. When you are constructing the meiosis model, do you have a basic rule or framework i n your mind? It's r e a l l y basic. A l l the products are -- there are four products (daughter c e l l s ) (Figure 4e) . And what about the products? The c e l l should have each type of gene (A/a and B/b). I am not counting the number of li n e s -- the number of chromosomes, but rather the type of gene in each chromosome (Figure 4e) . 64 (a) Parent C e l l (Lines 6-10) (b) R e p l i c a t i o n (Lines 22-24) (c) Homologous P a i r i n g alignment I, and chromosome segregation (Lines 25-26,37-43) cx ^ 7 (e) Daughter c e l l s (Lines 98-99, 103-6) Chromatid segregation (Lines 44-46,89-92) F i g u r e 4: C h r i s ' s i l l u s t r a t i o n showing a ' Canonical'"" c o n c e p t u a l i z a t i o n of m e i o s i s C h r i s ' s example c l e a r l y showed the e s s e n t i a l c h a r a c t e r i s t i c s of the ' c a n o n i c a l ' c o n c e p t u a l i z a t i o n of meiosis (except he combined three events of homologous p a i r i n g , alignment I and chromosome s e g r e g a t i o n ; see F i g u r e 4c) . In h i s example, chromosomes r e p l i c a t e and then the homologous chromosomes p a i r b e f o r e the f i r s t d i v i s i o n . T h i s i s f o l l o w e d by the l i n i n g of the homologous p a i r s a l o n g the equator of the c e l l ( e q u a t o r i a l alignment, I) . 65 During the f i r s t d i v i s i o n , homologs of d i f f e r e n t chromosome pairs move independently of each other to opposite poles of the c e l l (the A homologue could go either with the B or b homologue, r e s u l t i n g i n two combinations; refer to l i n e s 37-46). Then the c e l l divides, producing two c e l l s with one of each homologue. F i n a l l y , during the second d i v i s i o n , single homologs l i n e up along the equator of the c e l l and the centromere s p l i t s and each s i s t e r chromatid moves to an opposite pole of the c e l l . The s l i g h t v a r i a t i o n included the merging of the three events of homologous pairing, alignment I 7 and chromosome segregation (see Figure 4c). 'Canonical' Conceptualization: Kirk's Example 10 15 PATRICK: KIRK: PATRICK: KIRK: Can you t e l l me something about the c e l l diagram in the question? F i r s t of a l l , i t (the c e l l i n the question) shows four chromosomes. Each gene has a dominant and a recessive: A and B are d i f f e r e n t genes (Figure 5a) Mendel's Law, assorted independently of each other and stuff l i k e that (he just described his logic b r i e f l y ) . T e l l me the hints i n the question. This (the question) says the two pairs of chromosomes. F i r s t thing, they (the chromosomes) replicate (Figure 5b) and I keep whatever format (two sets of chromosomes and gene symbols) the question gives me. And I put i n the genes here. Then I l i n e them (the chromosomes) up (Figure 5c). I put the genes in (the.chromosomes) and separate 7 I and II - denote events occurring at the f i r s t and second division steps of meiosis respectively. ,;f 66 20 25 them (Figure 5d). I t r y to put the diagrams i n a flow chart so that I can keep track of everything. I can say that the b i g B (chromosome wit h the B gene) can go onto t h i s side (the other side so that there i s independent assortment, ending up i n two kinds of gene combinations i n meiosis I ) . I t ' s j u s t independent. Then they (chromosomes) go to another separation (Figure 5e) and form the four c e l l s (Figure 5 f ) . 3 0 PATRICK: Do you have any ba s i c framework or r u l e s i n mind during your c o n s t r u c t i o n of t h i s model? Can you take some diagrams of your model and t e l l me your l o g i c ? 3 5 KIRK: Why can't both the A (A and a) chromosomes go to one c e l l and both the B (B and b) go-to the other c e l l . That sort of t h i n g . 40 PATRICK: Do you have answers to such a question? KIRK: I th i n k there i s an answer. In order to have proper f u n c t i o n of the gametes, you got to have a copy of each gene. This i s the r u l e . 67 Parent C e l l (Lines 5-6) Replication (Lines 14-15) A « ^ (c) Homologous Pairing & alignment I (Line 18) Alignment I I & chromatid segregation (Lines 26-27) (f)Daught er c e l l s . A 8 ^ Chromosome segregation (Lines 19-25) (Lines 26-27) .Figure 5: Kirk's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis Kirk's example showed the essential c h a r a c t e r i s t i c s of the 'canonical' meiosis conceptualization. For example, the daughter c e l l s each contained a copy of each gene (A or a and B or b) to make up a set (see Figure 5 f ) . The s l i g h t 68 v a r i a t i o n included the merging of the two events of alignment I and homologous p a i r i n g (Figure 5c) ,- alignment II and chromatid segregation (Figure 5e). Canonical' Conceptualization: Sharilvn's Example: 1 PATRICK: 5 SHARILYN: 10 15 PATRICK: SHARILYN: 20 25 L e t ' s look at t h i s meiosis q u e s t i o n . Can you t e l l me the meanings of a l l i l l u s t r a t i o n s , symbols and words i n the question? So, here i s the r e p r e s e n t a t i o n of the whole c e l l t h a t has the chromosome set of two. t h i s Aa r e p r e s e n t s a p a i r of chromosomes i n the c e l l and t h a t Bb here r e p r e s e n t s the second s e t of chromosomes (Figure 6a). These l e t t e r s r e p r e s e n t genes on the chromosomes. These dots are the centromeres - c e n t r a l p o i n t s of the chromosomes. Can you draw the whole s e r i e s of diagrams f o r meiosis? T h i s i s prophase I, and i s the f i r s t stage of m e i o s i s , a f t e r the DNA d u p l i c a t e s ( F i g u r e 6b). At end of prophase, these are homologous p a i r s w i t h the same s o r t of DNA but w i t h d i f f e r e n t v a r i a t i o n (AA,aa). They p a i r (Figure 6c)- d i f f e r e n t word -- um -- what's the word, an example c o u l d be eye c o l o u r . Eye c o l o u r i s a k i n d of phenotype, perhaps b i g A c o u l d be blue eyes and the l i t t l e b c o u l d be green eyes, t h a t c o u l d be d i f f e r e n t phenotypes. In the f i n a l c e l l s , they have e i t h e r these, the b i g B or the l i t t l e b. SHARILYN: T h i s i s metaphase I (see F i g u r e 6c)and these are homologous p a i r s (AA,aa,BB,bb). They l i n e up i n 30 the middle. The chromosomes a l l l i n e up and p a i r up at the same time. And then t h i s i s anaphase. I (Figure 6d) -- s p i n d l e s come from these s m a l l o r g a n e l l e (rod-shaped c e n t r i o l e s ) and they a t t a c h to the centromeres -- randomly attached, so they 35 (AA/BB) c o u l d go e i t h e r t h i s way (up) or t h a t way (down) and the other one (aa/bb) go the 69 o p p o s i t e (way). And they ( s p i n d l e s ) p u l l these chromosomes (AA/BB/aa/bb) apart, but they are s t i l l d u p l i c a t e d at t h i s p o i n t (Figure 6e). 4 0 And then i n t h i s stage, you have two c e l l s . PATRICK: Can you continue to d e s c r i b e the sequence of events? 4 5 SHARILYN: Okay, so the c e l l -- I make them two separate c e l l s here (Figure 6 f ) . So, t h i s i s metaphase I I . These chromosomes w i l l l i n e up at the c e n t r a l a x i s again (Figure 6g). Okay, but there i s no cr o s s o v e r whatsoever because they are not p a i r e d 50 anymore. Thi s i s anaphase II (Figure 6h), the chromatids are p u l l e d apart and they are not going to be d u p l i c a t e d anymore. So, these s p i n d l e s a t t a c h randomly again. Same t h i n g a g a i n f o r the lower c e l l . So, these (AA,BB i n one c e l l and aa, 55 bb i n another c e l l ) are p u l l e d apart randomly l i k e t h i s . And then telophase I I , so separate (Figure 6 i ) . C y t o k i n e s i s i s j u s t the s e p a r a t i o n of cytoplasm and then the nucleus forms. 60 PATRICK: Do you have any b a s i c framework i n your mind as you are c o n s t r u c t i n g t h i s model? 65 SHARILYN: I do know t h a t these chromosomes f i r s t have to d u p l i c a t e . I know the sequence and events. Well, the f i r s t stage the chromosomes have to p a i r , so -- I know that and I know t h a t i n me i o s i s -- the chromatids aren't to be separated i n the f i r s t d i v i s i o n . The c e l l s formed are h a p l o i d , t h a t i s n. PATRICK: What i s your understanding of n and 2n? 70 SHARILYN: 2n i s - - n i s the f u l l chromosome s e t . 2n i s l i k e the homologous chromosomes. 2n i s l i k e r i g h t here ( i n the c e l l g i v e n i n the q u e s t i o n ) , t h i s chromosome and t h a t one, they have the same s o r t 75 of gene, but d i f f e r e n t a l l e l e s (A,a). N i s the set of chromosome -- f o r example, the A type or the B type. The b i g A and the sma l l a both belong to the A type f o r example.. N i s not the chromosome number as many people would say. 70 (a) Parent c e l l (Lines 5-11) (f) Prophase I I (Lines 45-46) (g) Metaphase I I Alignment I I (Lines 47-48). End Prophase I r e p l i c a t i o n & homologous p a i r i n g (Lines 16-20) T c T Metaphase I Alignment I (Lines 28-31) [e)Telophase I (Line 40) (d) Anaphase I and chromosome segregation (Lines 31-39) (h) Anaphase II Chromatid segregation (Lines 50-55) (i) Telophase I I Cy t o k i n e s i s (Lines 56-58) Figure 6: Sharilyn's i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis 71 Sharilyn's example showed the most complete meiosis model for the ^canonical' conceptualization. The two events of segregation (namely chromosome segregation and chromatid segregation) are represented by a series of diagrams to show the formation of spindle f i b r e s , moving of chromosomes and s i s t e r chromatids to opposite poles, formation of c e l l membrane and nuclear membranes. Chromosome segregation i s represented by three diagrams i n Figure 6 (d) , (e) and (f) . For chromatid segregation, i t i s represented c l e a r l y by two diagrams i n Figure 6 (h) and ( i ) . Replication and homologous pairing were included i n one step (Figure 6b), showing that this student realized that doubling of chromosomal materials should have occurred before meiosis (premeiotic synthesis of DNA materials). Sharilyn had a confused view of ploidy. She considered the homologous chromosomes A,a and B,b as representing a chromosome set rather than the non-homologous chromosomes A,B and A,b(lines 5-10, 70-75). 'Canonical'' Conceptualization: May's Example: 1 PATRICK: Let's go to the meiosis question. As you are working through the question, say aloud a l l your thoughts and explain to me a l l illustrations,diagrams and terms 5 MAY: Here is a c e l l with four chromosomes. The A ones are a pair (A,a) and the B ones are a pair (B,b). I guess this c e l l (the given c e l l in the question)is 2n (Figure 7a). 2n means you have two sets of chromosomes, A set and B set. It is not 72 the number of chromosomes. The two A ones p a i r up d u r i n g m e i o s i s , so do the two B ones. They end up w i t h n. PATRICK: Do you have names f o r t h i s p a i r of chromosomes (A 15 and a)? I guess tha t they are homologous chromosomes. So, what happens i n meiosis i s th a t as opposed to m i t o s i s where you get the daughter c e l l s • e x a c t l y the same as the parent c e l l , you get a r e d u c t i o n i n the -- what I mean i s each f i n a l c e l l formed c o n t a i n s one member of the s e t of chromosomes found i n the parent c e l l . PATRICK: Can you t a l k about the sequence of events which takes p l a c e i n the process of m e i o s i s , s t a r t i n g from t h i s c e l l ? MAY: 20 MAY: What happens are -- there are two stages of 30 meiosis and the f i r s t one, what happens i s th a t --here i s one (chromosomal) arm, and i t r e p l i c a t e s (Figure 7b), so they (the two arms) are e x a c t l y the same. Each one has the b i g A (gene) or the s m a l l a (gene) on i t l i k e t h a t . Okay, so 35 t h a t ' s what the b i g B and the small.b look l i k e . So, b i g A, s m a l l a, then b i g B -- each one i s now a s i s t e r chromatid, e x a c t l y the same (Figure 7b). That's r e p l i c a t i o n and then the next stage, I can't remember how i t i s c a l l e d but -- okay, they 4 0 (chromosomes) l i n e up at the equator -- homologous ones and they l i n e up and p a i r up at the middle of the c e l l (Figure 7c). PATRICK: Why do you have to draw the AA p a i r i n g up w i t h the 45 aa? MAY: Because they are homologous, so th a t the AA cannot p a i r w i t h the BB, because they (BB) are not homologous t o the AA. The r e p l i c a t e d p a i r s (AA,BB,bb,aa) p a i r up f i r s t b e f o r e they l i n e up at the middle of the c e l l . Right a f t e r t h a t , you get j u s t t h i s centromere moves to o p p o s i t e p o l e s , and i t s movement i s independent, l i k e t h i s b i g AA move up or move down. But whichever d i r e c t i o n i s the 55 f i r s t one (AA), the other one (aa) u s u a l l y move to the other p o l e and same wit h the B's (BB and bb). I j u s t draw one case, where they (AA,bb) go up and these ones (aa,BB) go down (Figure 7c), 50 73 so then you get big AA up here, big BB, small aa down at the bottom ( c e l l ) . They form two c e l l s (Figure 7d). The next d i v i s i o n i s the centromere s p l i t s into half and then -- these one (A,b) go up; these ones (a,B) go down (Figure 7d). It could be the other way. I just remember the second one (division) i s when the centromere s p l i t s . Do you have any basic rules or framework for the construction of the meiosis model? Well, I just remember what happens at what stage, so when I draw i t (the meiosis model), i t ' s kind of easier. I remember f i r s t they (chromosomes) replicat e , and they get exactly the same one i n one exact place. Then you remember the p a i r up. I can remember certain steps, then i t ' s extremely easy to draw. So this parent c e l l goes into four ( f i n a l c e l l s ) . These one ( f i n a l four c e l l s , Figure 7e) w i l l be the same (A,b; A,b; a,B; and a,B). They end up with two p o s s i b i l i t i e s . If you s p l i t (the centromere) the other way, i t ends up with the other p o s s i b i l i t y (A,B; A,B; a,b; a,b). Could you assign n and 2n to these d i f f e r e n t diagrams? Um -- so i t starts with 2n and ends up with n. Do you have any rules i n mind as you are constructing t h i s meiosis model? F i r s t , meiosis i s just r e p l i c a t e and then separate. The homologous ones separate (in meiosis I ) . The second d i v i s i o n i s when the centromeres s p l i t . (aj Parent C e l l (Lines 6-9) (b) Replication (Lines 30-38) fit •« ft & ex. -Alignment II and Homologous Pairing chromatid segregation chromos alignment I and lines 59-65) L ome segregation (lines 39-58) CK 8 (e) ' Daughter Cells (Lines 75-77) ^ 3 u r e _ 7 : May's i l l u s t r a t i o n showing a 'Canonical conceptualization of meiosis May's conceptualization of meiosis also consisted of the essential c h a r a c t e r i s t i c s of the 'canonical' meiosis model except there were a number of s l i g h t v a r i a t i o n s . Similar to Chris's example, May combined the three events of 75 homologous pairing, alignment I and chromatid segregation together (Figure 7c). She also combined the two events of alignment II and chromatid segregation together (Figure 7d), May (l i k e Sharilyn) had a confused view of ploidy. She also considered the homologous chromosomes A,a and B,b as representing a chromosome set(p.72, l i n e 10). vCanonical' Conceptualization: L a i l a ' s Example: 1 PATRICK: Let's go onto the meiosis question which you had done. E x p l a i n to me a l l i l l u s t r a t i o n s . LAI LA: So, what I f i r s t do i s to define everything. These are the chromosomes, r i g h t ? These are the a l l e l e s 5 (A,a,B,b) i n each of the chromosomes. PATRICK: What do you mean by a l l e l e s ? LAILA: You got chromosomes with genes on i t . Each gene 10 made up of two or more a l l e l e s . PATRICK: So, what i s that b i g A and l i t t l e a, b i g B and l i t t l e b? 15 LAILA: They are j u s t a l l e l e s of the same gene. So, b i g B, b i g A are dominant a l l e l e s , small a and small b -recessive a l l e l e s . So they (the chromosomes i n the question) are a l l homologous chromosomes and they have a partner (Figure 8a). What happens i s these 20 (chromosomes) r e p l i c a t e (Figure 8b). PATRICK: So, do you mean the centromeres r e p l i c a t e at the same time? 25 LAILA: Of course they (the centromeres) d i d . So, they r e p l i c a t e . And then, homologous chromosomes p a i r (Figure 8c). You got homologous chromosomes, that means they•have the same kin d of gene. I rather remember the l o g i c rather than the names. So, 3 0 homologous chromosomes p a i r . PATRICK: How do you know that the chromosomes r e p l i c a t e and then p a i r i n g up? 76 3 5 LAILA: 40 They r e p l i c a t e and then they p a i r , i s that r i g h t ? Meiosis, you s t a r t w i t h a 2n and you end up wit h an n. M i t o s i s s t a r t s w i t h an n, ends up .with,an n, or 2n and you end up with a 2n. Meiosis halves the no. of chromosomes and m i t o s i s keeps them (the chromosome number) the same. I know you have four (chromosomes i n the question) there, and I've to . end up wi t h two. 45 50 55 60 65 70 75 80 PATRICK: LAILA: PATRICK: LAILA: What's that two? Homologous chromosomes p a i r . So then you have the centromeres, spindle forms and these (homologous pa i r s ) a l i g n and separate. So, t h i s i s one pole, t h i s i s another. You got f i b r e s going up. This i s a p a i r (AA) and s p i n d l e f i b r e s s t a r t a t t a c h i n g to the centromere. Now, here they s p l i t (Figure 8d). I can't remember the names. So, they s p l i t okay and i t depends on how they s p l i t . There're d i f f e r e n t outputs. Am I doing t h i s r i g h t ? Why do you ask t h i s question? Cause I'm not sure whether I am doing t h i s r i g h t . Because I know I should have a c e r t a i n number of chromosomes. Wait a minute. We s t a r t up w i t h eight(chromosomes a f t e r r e p l i c a t i o n ) ( F i g u r e 8b), And we end up with four ( a f t e r the f i r s t m eiotic d i v i s i o n ) ( F i g u r e 8e). Okay, I got i t . (she d i d not s p l i t the centromere of the r e p l i c a t e d p a i r ) (Figure 8 b,c,d,e r e v i s e d ) . PATRICK: Why do you change t h i s diagram? LAILA: Because I r e a l i z e that i t was wrong. I was going through the process i n my mind and I know the output. PATRICK: Can you t e l l me the output? LAILA: Two c e l l s w i t h two chromosomes (Figure 8e revised) PATRICK: Why i s that (Figure 8e) wrong here? LAILA: I f I put i t (Figure 8e) that way, I end up with two c e l l s with four chromosomes, i t should end up with two c e l l s with two chromosomes. From that, you have t h i s one (AA) goes up, t h i s (aa) goes down, t h i s (bb) goes and t h i s (BB) goes down or the 77 85 90 PATRICK: reverse. Here,(In M i l ) , t h i s one (AA) goes up, t h i s one(aa) going down, t h i s one (BB) going up and t h i s (bb)going down. You can have the other way round, the aa, BB i n the lower c e l l and the AA, bb going to the upper c e l l -- over there. Why do you think the end products are two c e l l s each with two chromosomes instead of four? LAILA: Because that's meiosis. Yea, so you s t a r t with four 95 chromosomes (in the o r i g i n a l ) , you end up with two chromosomes i n four c e l l s . 100 105 110 PATRICK: So, i f you have the centromere s p l i t , how does t h i s v i o l a t e the rules of meiosis? LAILA: Because i f you end up with one, two, three, four (chromosomes A,a,B,b) i n the two c e l l s (after MI i n Figure 8e) and each (chromosome) r e p l i c a t e s (AA,aa,BB,bb), r i g h t ! Then s p l i t up i n meiosis I I , you end up with four c e l l s with four chromosomes (Figure 8 f ) . You b a s i c a l l y kept the number (of chromosomes). But i f you have two c e l l s each with four chromatids and that's wrong. Meiosis II can't occur because there i s no s p l i t t i n g of chromosomes anywhere t h i s case. PATRICK: When you are constructing t h i s model, do you have basic framework i n mind? 115 LAILA: 120 I got l i t t l e checkmarks i n mind, n would be two and 2n four (chromosomes). 2n becoming n at the f i n a l product. This i s one of the checks. The other check • i s r e p l i c a t i o n and homologous p a i r i n g before separation. I know that from here (at MI, see Figure 8e), I have to end up with two c e l l s each with two chromosomes each. 125 130 PATRICK: How are these two chromosomes (A, a i n the o r i g i n a l ) d i f f e r e n t from that one (AA, aa at MI)? LAILA: Cause t h i s one (aa, see Figure 8e) i s (made of) two chromatids, t h i s (the parent c e l l , see Figure •8a) doesn't. That one r e p l i c a t e d and t h i s one didn't ( r e p l i c a t e ) . In cl a s s , Dr. G r i f f i t h s always made the d i s t i n c t i o n between chromosomes and chromatids, so when a chromosome r e p l i c a t e s , i t i s a chromosome with two chromatids. So, t h i s i s the basic. 78 (a) Parent C e l l (Lines 3-19). (b)- R e p l i c a t i o n (b' ) Revised (Lines 19-20) . (Lines 63-65] (c')Revised (Lines 63-65! (d) Alignment I and chromosome segregation (Lines 46-52) ~A ST CK W A £ !f) Daughter C e l l s (Lines 104-106) c] Homologous P a i r i n g (Lines 26-27) (d') Revised (Lines 63-65) (e) Chromatid segregation (Lines 62-63,80-89) Figure 8: L a i l a ' s i l l u s t r a t i o n showing a 'Canonical' conceptualization of meiosis The revised and the unrevised parts found i n L a i l a ' s i l l u s t r a t i o n made i t distinguishable from the previous four examples. The revised part showed the framework of the 'canonical' meiosis conceptualization as the other four examples. 79 L a i l a i n i t i a l l y r eplicated the centromeres when she replicated the chromosomal materials i n Figure 8b. As a r e s u l t , single unreplicated chromatids (A,a,B,b) go to opposite poles of the c e l l i n the f i r s t d i v i s i o n of meiosis(see Figure 6d). Her understanding of the structure of a chromosome and a chromatid resulted i n the revisions i n her i l l u s t r a t i o n . She r e a l i z e d that when a chromosome re p l i c a t e s , i t i s s t i l l a chromosome which consists of two chromatids•(lines 127-129). The amount of chromosomal material doubles, but the number of chromosomes remains unchanged. In order to generate two product c e l l s each with two chromosomes at the f i r s t d i v i s i o n (lines 117-118), she r e a l i z e d that she could not r e p l i c a t e the centromere before d i v i s i o n . 4.2.2 The 'Event-Modified' Conceptualization of Meiosis This conceptualization f i t s the fundamental framework of the 'canonical' model of meiosis i n that two d i v i s i o n s lead to the formation of four c e l l s . Deviation from the 'canonical' model appeared as modifications of i n d i v i d u a l events of the 'canonical' model. These modifications included either a change i n the sequencing of the event (for example, p a i r i n g could take place before r e p l i c a t i o n ) , i n the orientation of p a r t i c i p a t i n g structures, or i n both the p a r t i c i p a t i n g structures and the sequencing of the event. The i n d i v i d u a l examples were as follows: 80 (1) Sequencing of the events The event of homologous pa i r i n g occurred before r e p l i c a t i o n of chromosomes (see examples of Mike and Sam) ; (2) Orientation of participating; structures Chromosomes aligned at right angles to, instead of p a r a l l e l to, the axis of the equator before d i v i s i o n (see example of Doug); and (3) The sequencing of events and the participating: structures In t h i s case, both the sequencing of events and the p a r t i c i p a t i n g structures are modified. This student reversed the events immediately before the f i r s t and second d i v i s i o n s . The "homologous pa i r i n g " event took place before the second d i v i s i o n instead of before the f i r s t d i v i s i o n , thus leading to a change i n the p a r t i c i p a t i n g structures within an event. Individual s i s t e r chromatids became the p a r t i c i p a t i n g structures i n the f i r s t d i v i s i o n whereas homologs became the p a r t i c i p a t i n g structures i n the second d i v i s i o n . Thus s i s t e r chromatids separated i n the f i r s t d i v i s i o n and unreplicated homologues separated i n the second d i v i s i o n (see example of Kamyar). 81 The above c o n c e p t u a l i z a t i o n s are now i l l u s t r a t e d and expanded based upon i n t e r v i e w e x t r a c t s and students' flow-chart diagrams. 'Event-Modified' C o n c e p t u a l i z a t i o n : Mike's Example: PATRICK: 10 15 20 MIKE: PATRICK: MIKE : PATRICK: MIKE : 25 PATRICK: Let's go to this meiosis question. Treat me as i f I were a layman of genetics and explain to me every i l l u s t r a t i o n , diagram, symbol as you go along t h i s question. What we have here are two di f f e r e n t homologues. What do you mean by homologues? Homologues are a set of two -- th i s here A i s a chromosome. It has a matching homologue, which i s i d e n t i c a l and i t ' s the l i t t l e a on t h i s side. How can you say that they are identical? Um -- Okay. This b i g A would mean a dominant a l l e l e . A l l e l e i s a part of a gene or part of a chromosome with gene on i t . They (A and a homologues) are i d e n t i c a l because they.have the same genetic material except the l i t t l e a i s recessive compared to the dominant big A, they just got di f f e r e n t a l l e l e s but they are the same chromosomes (Figure 9a). Which symbols or i l l u s t r a t i o n s i n the diagram make you think they (the A,a homologues) are the same kind of chromosomes? MIKE: I know that they are of similar shape, and with 3 0 the centromeres i n simil a r position. PATRICK: Okay, then how about the process of c e l l division? MIKE: What w i l l happen i n meiosis i s that i t w i l l go 35 through -- consider meiosis I. Basically, the homologues -- they are paired up f i r s t , the big A and the l i t t l e a here and then you haye the big B and the small b here (Figure 9b) . And then*'' 8 2 40 45 they d u p l i c a t e , so b a s i c a l l y each one now, t h i s i s what they look l i k e (Figure 9 c). I t h i n k they should be connected here l i k e t h i s . T h i s i s one centromere (connecting the s i s t e r chromatids AA). They are not i n d i v i d u a l chromosomes, they are d u p l i c a t e d form. I c a l l t h i s a chromatid. They separate l i k e t h i s (Figure 9d), one i s AA, i t can be e i t h e r w i t h BB or i t can go w i t h the bb. In the second c e l l , you have aa, bb or i t c o u l d be BB. 50 55 60 PATRICK: And then what happen next? MIKE: And then, here they (the chromosomes) go through the same stages but there i s no d u p l i c a t i o n . They are l y i n g i n the e q u a t o r i a l boundary (of the c e l l ) -- i n the middle. Um -- and then I say t h a t these (AA,BB; aa,bb) separated again (Figure 9e), being p u l l e d by s p i n d l e . a n d these ones (A,B) go up and these one (a,b) go down. You end up w i t h f o u r h a p l o i d c e l l s (Figure 9 f ) . So they are the f i n a l products of m e i o s i s . PATRICK: Do you have any b a s i c framework i n mind as you are c o n s t r u c t i n g t h i s meiosis model? 65 MIKE : I b a s i c a l l y remember d u p l i c a t i o n and then s e p a r a t i o n . F i r s t , they (the chromosomes) have to d u p l i c a t e and then they separate t w i c e . F i n a l l y , the f o u r c e l l s have to c o n t a i n one s e t of chromosomes. 70 PATRICK: What do you mean by one set of chromosomes? 75 MIKE: B a s i c a l l y , each c e l l c o n t a i n s one of the A and B type chromosomes from th a t homologous p a i r s (A,a,-B,b) The number (of chromosomes) i s not my concern here. 83 (a) Parent c e l l (Lines 6-23) (b) Homologous pa i r i n g (Lines 34r38) (c) Replication and chromosome segregation (Lines 38-44) A A - b (f) Daughter c e l l s (Lines 57-59) (d) Products of chromosome segregation (Lines 44-46) (e) Alignment, chromatid segregation (Lines 51-55) Figure 9: Mike's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis 84 Mike's example c l e a r l y showed a change i n the sequencing of the events. Homologous pa i r i n g took place before r e p l i c a t i o n (see Figures 9b and c). 'Event-Modified' Conceptualization: Sam's Example: PATRICK: SAM: 10 15 PATRICK: SAM: PATRICK: SAM: 20 25 PATRICK: SAM: 30 35 Let's look at the meiosis question. As you are going through this question, say aloud your thoughts and explain to me a l l i l l u s t r a t i o n s , diagrams that you go through. Okay, so say th i s i s the c e l l (given i n the question) -- there are chromosomes i n the c e l l which are homologous, l i k e they are partners. These two (A,a) are homologous and these two (B,b) are homologous (Figure 10a). What do you mean by homologous? Um -- I think i t ' s l i k e the same DNA sequence. How does the symbol A remind you that these a) are homologous? (A and Often the size of the chromosome makes me think that they are homologous. So, th i s i s the A chromosome and i t i s also the a. They are of the same size. Can you describe the process of meiosis from this o r i g i n a l c e l l i n the question? In meiosis, chromosomes f i r s t pair, okay, l i k e this one (Figure 10b), small b can be on top and big B can be at the. bottom -- the other way round, independent assortment. They duplicate and they are attached at the centromere. These are the duplicated pairs (AA,aa,BB,bb; Figure 10c). They (the duplicated pairs) separate. These two (AA,BB and aa,bb) go to one' c e l l and then these (AA,BB i n one c e l l and aa,bb i n another c e l l ) separated (Figure 10 d). It can be the other way round, so you get another combination i n the two c e l l s . The separation i s independent. ) 85 40 45 50 55 60 65 70 PATRICK: SAM: PATRICK: SAM: PATRICK: SAM: PATRICK: SAM: PATRICK: SAM: What are you thinking right now? I am tryin g to get four haploid c e l l from a d i p l o i d 2n c e l l . So, what i s the logic you are looking for i n changing the 2n to the n? 2n i s two sets of chromosomes. So, you have two sets of chromosomes at the beginning and when i t (the o r i g i n a l c e l l ) gets to an n i n the products, there must be a set of chromosomes i n the (final) c e l l s . At meiosis II, the centromere doubles and you have a set of two chromosomes i n each of the offspring c e l l (Figure lOe). Do you have any basic framework i n mind as you are constructing t h i s meiosis model? My basic framework i s to get four haploid c e l l s from a d i p l o i d c e l l . Can you define the term haploid? Um -- the organism with one set of chromosomes i s the haploid. What do you mean by one set of chromosomes? Well, l i k e i n human c e l l s , there are fo r t y six chromosomes. Twenty three -- there are two set's of twenty three chromosomes, that i s a 2n c e l l . But for example, l i k e the gametes -- they only have twenty three chromosomes, they have one set. (a) Parent C e l l (Lines 6-10) (b) Homologous pairing (Lines 27-30) A (d) Products of chromosome segregation (Lines 33-38) fir —"—s (c) Replication, alignment & chromosome segregation (Lines 30-32) A A A b A a (e) Products of chromatid segregation (Lines 51-53) Figure 10: Sam's i l l u s t r a t i o n showing an \Event-Modified' conceptualization of meiosis 87 Examples of Mike and Sam both i l l u s t r a t e a s h i f t i n the sequencing of the events. B a s i c a l l y , homologous p a i r i n g occurred before r e p l i c a t i o n . Other than t h i s difference, there was no change i n the in d i v i d u a l p a r t i c i p a t i n g structures. Homologous chromosomes separated i n the f i r s t d i v i s i o n and s i s t e r chromatids separated only i n the second d i v i s i o n of meiosis. Centromeres never doubled u n t i l the second d i v i s i o n . Sam also combined certain events together i n his i l l u s t r a t i o n . In his example, r e p l i c a t i o n , alignment I and chromosome segregation were combined together (Figure 10c). Other than t h i s , the events of alignment I I and chromatid segregation were not shown, they were combined together with the four meiotic products i n Figure lOe. 'Event-Modified' Conceptualization: Doug's Example: 1 PATRICK: Let's look at the meiosis question. Explain to me a l l the i l l u s t r a t i o n s , diagrams that you draw. 5 DOUG: Um, you know any l i v i n g things have basic structures c a l l e d c e l l . The c e l l has a nucleus. In the nucleus, there are genetic materials c a l l e d chromosomes. So, right here, t h i s i s c a l l e d the chromosome. 10 PATRICK: What i s the meaning of the symbols A,a,B,b here? Can they be two A's or two B's? 15 D O U G : The big A represents dominant gene and small a represents recessive. Dominant means you have two big A's you get the dominant phenotype. One big A and one small a, the phenotype of big A shows up. Only when you get two a's together, you get the recessive phenotype (Figure 11a). 88 2 0 PATRICK: Can you show me the steps of meiosis? DOUG: 25 30 PATRICK: F i r s t of a l l , this w i l l duplicate (the chromosome);(Figure l i b ) . Horizontal l i n e means position of the gene. Then they l i n e up here(at the equator) lin e up i n a v e r t i c a l l i n e , with each homologues pairing up, they w i l l separate, one strand (AA and BB) goes to thi s side and the other strand (aa and bb) goes to the other side (Figure 11c) ; (He seemed puzzled) . What i s your puzzle here? DOUG: 35 40 Oh, I remember when the whole thing separates, the centromere here isn't separate for the f i r s t part of meiosis. The whole thing that separates (AA, aa)(Figure l i d ) not the individual strand (A,A, A,a). For the second part, i t w i l l l i n e up again on this l i n e , two over here and two over there. And then right here, two strands (b,b) w i l l separate and one goes to each side (Figure l i e ) . If t h i s i s aa, this i s bb, the next one a and b; a and b. This one you got A,B; A,B (Figure l l f ) . 45 PATRICK: Is that the only one combination you have? DOUG: Nope, i f you have big B on thi s side, then you got small a and big B and that side, big A and small b (Figure l i d ) . 50 PATRICK: When you are constructing this meiosis model, do you have any basic framework i n mind? DOUG: 55 60 Well,. I think for this kind of question, usually chromosomes w i l l l i n e up f i r s t and then separate to two d i f f e r e n t ' c e l l s . So, b a s i c a l l y everything l i n e up f i r s t and separate to two di f f e r e n t c e l l s . And one more'thing i s for the f i r s t part of meiosis, everything lines up but.the individual members of the pairs do not separate, for the second part, the chromatids separate. PATRICK: What do you, mean by n and 2n and put them into the model? 65 DOUG: I know the last one i s 4ri, each one i s one n. If each one i s one n, this one (the o r i g i n a l c e l l ) 2n and thi s one 4n,(the duplicated chromosome 89 70 diagram), there are four parts, so i t is.4n. I know information about n, i n human chromosomes, i s i t twenty three pairs? If i t i s d i p l o i d , you got for t y six. I am d i p l o i d except the gametes. I am re f e r r i n g to number of chromosomes. (a) Parent c e l l (Lines 4-18) (b)Replication (Lines 21-22) U Y n (e) Alignment II (d) End of chromosome & chromatid segregation segregation (Lines 39-40) (Lines 33-36) 1 t • (c)Pairing, alignment . I & chromosome segregation (Lines 24-28) (f) Daughter c e l l s (Lines 41-42! Figure 11: Doug's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis Doug modified the orientation of the p a r t i c i p a t i n g structures i n his i l l u s t r a t i o n . Instead of being p a r a l l e l to the central axis, the p a r t i c i p a t i n g chromosomes at both alignment I and II became perpendicular to the central axis 90 of the c e l l (Figure 11c and e). Further, the four homologues pa i r up i n a v e r t i c a l l i n e a r fashion rather than i n groups of two (Figure 11c). 'Ploidy confusion' accompanied Doug's model. He verbally described a change i n ploidy of the generalized c e l l a f t e r r e p l i c a t i o n . The c e l l (in Figure l i b ) , instead of being l a b e l l e d as 2n, was now_described as a 4n c e l l (lines 66-67) To Doug, n refers to the number of the A (or a) 'species' chromosomes (or chromatid) present i n a c e l l ' rather than the number of chromosomes i n a set. 'Event-Modified' Conceptualization: Kamyar's Example: 1 PATRICK: Let's go to this meiosis question. Explain to me 10 PATRICK: . Why do you have these chromosomes? Can you explain symbols A,a,B,b to me? a l l your i l l u s t r a t i o n s , steps, diagrams. 5 KAMYAR: This i s the d i p l o i d c e l l or haploid? I don't know. We have a c e l l and these are the d i f f e r e n t chromosomes. When you want to go through meiosis... 15 KAMYAR: The big A these are a l l e l e s on certain regions of the chromosomes, which means that these may be certain DNA sequences. That i s what the chromosomes are made up of, right? 20 KAMYAR: When we go to meiosis, we end up with a c e l l which has half the information of the (original) c e l l . You start with d i p l o i d c e l l and want to get haploid c e l l , and i t ' s usually a process we go through... we are looking at ... producing gametes and the gametes need to be haploid so then from a d i p l o i d ( c e l l ) , you make haploid c e l l s . 91 25 PATRICK: Why do you consider t h i s ( c e l l i n the question) as a d i p l o i d c e l l ? KAMYAR: It has these two a l l e l e s , this big A and that small a and the big B and the small b 30 so two copies of- the same a l l e l e . Two copies of the same homologue? The big A and the small a. Oh, two copies of the same gene (Figure 12a). PATRICK: Okay, can you now talk about the steps of meiosis? 35 KAMYAR: F i r s t , there's a duplication of each of these strands (of chromosome). What we get here i s the big A with the centromere. Here we get another one (AA linked by one common centromere) and then we 4 0 have the B, and we get another one (so BB linked by one common centromere)(He duplicated a and b at the same time).(Figure 12b) . PATRICK: Do you consider this ( c e l l with chromosome 45 duplicated) as haploid or diploid? KAMYAR: No, thi s c e l l (Figure 12b) i s a 4n. Because th i s (original) c e l l (Figure 12a) i s d i p l o i d and now we double everything. Then this i s the 4n. Well, n i s number of chromosomes. 2n , you have four 50 chromosomes, n, you only have half the chromosomes. You double that o r i g i n a l set of chromosome (from the 2n c e l l ) , you get 4n, eight chromosomes. What happens i s a l l these (duplicated .chromosomes) l i n e up (at the equator of the c e l l ) . 55 So then t h i s i s the big A, thi s i s the small a, this i s the B thi s i s the small b (Figure 52c). Okay. Now after that (the l i n i n g up), what happens i s spindle fibres attach to the centre of centromeres and p u l l on these strands. We get the 60 big A maybe going to the top and the other (A) go to the bottom. And then what happens aft e r that (here he.got two c e l l s each with four single chromosomes, A,a,B,b) (Figure 12d). The centromeres are r e p l i c a t i n g (during the 65 separation). Then i t ' s the way I presented here. I guess when the chromosomes duplicate, the centromeres duplicate as well. 70 PATRICK: KAMYAR: What happen after .the f i r s t division? Then after t h i s d i f f e r e n t c e l l s . [ f i r s t separation), you have two Perhaps we take the top. c e l l 92 75 (Figure 12e), we have a c e l l that has these genes (A,a,B,b), ri g h t . With th i s c e l l , our o r i g i n a l plan was to go down to n, what we do is that these (A,a,B,b) line up. The A lin e s up with a and the B lines up with the l i t t l e b (Figure 12f). They are pairing up here. 80 PATRICK: KAMYAR: 85 95 Okay, then what happens after the pai r i n g up? Once again, we get the spindle f i b r e s attach to centromeres. With this arrangement (A,a; B,b i n Figure 12f) you have diff e r e n t types of attachment because big A and big B are on top, they are going to end up being on top of the c e l l ; l i t t l e a and l i t t l e b are going to be at the bottom. If th i s i s the other way, l i t t l e a and l i t t l e b on top and big A and big B at the bottom. What happens here you could have the big A, big B on top; l i t t l e a and l i t t l e b at the bottom. Or i t be A, b on top and then l i t t l e a and big B at the bottom. You could have l i t t l e a l i t t l e b on top; big A and big B at the bottom. You could have the opposite, big B and l i t t l e a on top; big A and l i t t l e b at the bottom (Figure 12g). PATRICK: 10 0 KAMYAR: 105 PATRICK: How many c e l l s do you end up with? You end up with... after that, you end up with two c e l l s . What we did here i s we end up with two c e l l s , but we only take one of them (after MI).(Figure 12e), but this one forms two and that one... you end up.with four c e l l s . Do you have any basic framework i n mind as you are constructing this model? KAMYAR: What happens f i r s t i s the chromosomes have to 110 double. You always keep i n mind that you start with a 2n (cell) and end up with n. F i r s t the chromosome number doubles, and then two c e l l s are formed. After the two c e l l s are formed, you end up every c e l l um... every c e l l has d i p l o i d , the same 115 (number) of chromosomes again as the parent and those c e l l s w i l l divide into two and then you end up with (four) haploid c e l l s , n i s the number (of chromosomes) and i t i s two. 93 2n [a] Parent C e l l (Lines 28-32) ( E l Replication (Lines 36-42! A (5 ^ — 0 "A A U Cv. • 1,1 — - — A 19 (e) Top product c e l l s of chromatid segregation (Lines 73-75) A 0 i i r ~b (d) Products of (c) Alignment I chromatid (Lines•54-57) segregation (Lines 58-68) [f) Homologous Pairing (Lines 76-79) Ok. s Jo CL l A b (g) Ce l l s from chromosome segregation (Lines 84-105) • Figure 12: Kamyar's i l l u s t r a t i o n showing an 'Event-Modified' conceptualization of meiosis Kamyar modified both the p a r t i c i p a t i n g structures and the sequencing of events of meiosis. P a i r i n g occurred i n the second d i v i s i o n (Figure 12f) instead of the f i r s t d i v i s i o n , 94 meaning that i n d i v i d u a l rather than paired homologs aligned i n the f i r s t d i v i s i o n (Figure 12c). Kamyar's modification also included the p a r t i c i p a t i n g structures within an event. Thus, i t was the chromatids that separated i n the f i r s t d i v i s i o n (instead of the homologs, see Figures 12d,e) and the unreplicated homologs separated i n the second d i v i s i o n (Figure 12g), giving r i s e to the four f i n a l products of meiosis. 'Ploidy confusion' also accompanied Kamyar's model. He (l i k e Doug) had a confused idea of what n means. To Kamyar, n seems to be the number of the A (or a) 'snecies' chromosomes, or chromatids present i n a c e l l . 4.2.3 'Whole-Process Modified' Conceptualization of Meiosis This conceptualization a l t e r s the_basic framework of the entire process of meiosis. I t considers meiosis as a one-stage rather than a two-stage process of successive nuclear d i v i s i o n . The products of meiosis are two i n number (instead of four), and being more l i k e that expected i n mitosis, contain the same amount of chromosomal material as the parent c e l l . 'Whole-Process Modified' Conceptualization: Jason's Example 1 PATRICK: Let's look at the meiosis question here. What do these representations of chromosomes on the diagram mean to you? 95 10 15 20 25 30 35 40 45 PATRICK: JASON: PATRICK: 50 JASON: There are four chromosomes and the centromeres are c i r c l e s , the gene l o c i are some distance from the centromere. Some l o c i are at a further distance from the centromere; the a locus i s further away from the centromere (Figure 13a). Any more ideas? There are two genes A and B. A i s probably dominant over the l i t t l e a and B probably dominant over the l i t t l e b. As you are working through, say aloud your thoughts. JASON: You have the A chromosome here; 'a' chromosome here. You leave the centromere at one end. It doesn't matter i n this case for the centromere has nothing to do with map distance. And you have big B , l i t t l e b (Figure 13b). From here (Figure 13b) on, I usually draw l i k e t h i s , to double up l i k e that (Figure 13c). And then you have two, as l i k e t h i s two bs (Figure 13d). From here on, •you have A matches with a, B matches with b (Figure 13e). You have two c e l l s . PATRICK: Could there be some other combinations of the two daughter c e l l s ? JASON: Yep. I am not sure. You have, crossing over, you have big A come together with l i t t l e a. That are the results of the crossover (Figure 13e). PATRICK: Do you have any framework i n mind as you are constructing the meiosis model? JASON: The n chromosomes in the parent c e l l (Figure 13b) duplicate into the 2n (Figure 13c) and become the n again by d i v i s i o n (Figure 13e). That's the general rule. PATRICK: What do you mean by the n and 2n here? JASON: They represent the number of chromosomes in the c e l l , n here represents four chromosomes because the parent c e l l has four chromosomes, r i g h t . 96 n=4 1 (a) Parent C e l l (Lines 5-9) • A A ^A^ & (b)Parent C e l l Redrawn (Lines 20-24) (d) Crossing over & Chromatid segregation (Lines.26-27) -0- n=4 -A a n=4 A (c) Replication (Lines 25-26) (e) Two Daughter C e l l s (Lines 28-29, 35-36) Figure 13: Jason's i l l u s t r a t i o n showing a "Whole-Process Modified" conceptualization of meiosis In addition to modifications i n s p e c i f i c meiotic events, Jason's example exhibits a fundamental misunderstanding of the process of meiosis as a whole. Instead of undergoing two d i v i s i o n s to form four daughter c e l l s i n meiosis, t h i s example showed only a one-stage d i v i s i o n that resulted i n the formation of two c e l l s . The two c e l l s had the same number and kind of chromosomes as the parent c e l l . This p a r t i c u l a r model p a r t i a l l y resembled a mitotic rather than a meiotic process. Jason was d e f i n i t e l y unclear about the purpose of meiosis. 97 Jason was a l s o u n c l e a r on the concept of p l o i d y . He l a b e l l e d both the parent and the product c e l l s of m e i o s i s as n=4 (see F i g u r e s 13a and e; l i n e s 48-50). Here, n represents the t o t a l number of chromosomes i n s t e a d of the number of chromosomes i n a set i n a c e l l . Jason does not recognize that the parent c e l l , which contains two whole sets of chromosomes, was d i p l o i d and thus should be l a b e l l e d 2n i n s t e a d . In summary, t h i s chapter d e s c r i b e s three q u a l i t a t i v e l y d i f f e r e n t m e i o s i s c o n c e p t u a l i z a t i o n s i d e n t i f i e d from students' i n t e r v i e w s u s i n g a phenomenographic p e r s p e c t i v e . These models i l l u s t r a t e d students' ideas on the sequence of events and the p a r t i c i p a t i n g s t r u c t u r e s i n v o l v e d i n m e i o s i s . I t i s assumed t h a t students who can d e s c r i b e a ' c a n o n i c a l ' c o n c e p t u a l i z a t i o n , they d i s p l a y a sound understanding of the concepts of m e i o s i s which i n c l u d e : the p a r t i c i p a t i n g s t r u c t u r e s and the sequencing of events at d i f f e r e n t stages of m e i o s i s . I f these b a s i c concepts were c l e a r , students were able to generate four product c e l l s from a s i n g l e parent c e l l . Each product c e l l , i n t h i s case, c o n s i s t s of one of the two sets of chromosomes as the parent c e l l . M o d i f i c a t i o n s of e i t h e r the sequencing of events or the p a r t i c i p a t i n g s t r u c t u r e s l e d to the c r e a t i o n of the "event m o d i f i e d " c o n c e p t u a l i z a t i o n . Three d i f f e r e n t types of m o d i f i c a t i o n s f i t w i t h i n t h i s c o n c e p t u a l i z a t i o n . In the most b a s i c m o d i f i c a t i o n (see examples of Mike and Sam), students 98 reversed the sequence of ' r e p l i c a t i o n ' and 'homologous pairing' events. A l l the other steps were essentially-i d e n t i c a l to the 'canonical' model. The next l e v e l of modification revealed.^changes i n the orientation of the p a r t i c i p a t i n g structures (see example of Doug). Instead of ali g n i n g p a r a l l e l to the central axis of the c e l l before c e l l d i v i s i o n , homologous chromosome pairs aligned at a right angle to the axis. I t was the t h i r d l e v e l of modification that revealed a l t e r a t i o n s both i n the p a r t i c i p a t i n g structures and the sequencing of events. In connection to the sequencing of events, one student (Kamyar) described the 'pairing' event immediately a f t e r the f i r s t d i v i s i o n . In connection to the p a r t i c i p a t i n g structures, s i s t e r chromatids separate i n the f i r s t d i v i s i o n and paired, unreplicated homologues separate i n the second d i v i s i o n of meiosis. The 'whole process modified' conceptualization indicated a modification i n not only i n d i v i d u a l events or p a r t i c i p a t i n g structures, but also a modification i n the outcome of meiosis. Instead of a process consisting of two di v i s i o n s , t h i s model (see example of Jason) described meiosis as a one-stage process, with the products of meiosis being more l i k e that expected i n mitosis. Three types of 'ploidy confusion' accompanied some mental models of meiosis. Although ploidy refers s p e c i f i c a l l y to the number of set of chromosomes, i t s meaning i s closely associated to n (the number of chromosome 99 i n a s e t ) . Thus the author, i n t h i s thesis, considered a confused idea of n to be also a contribution to 'ploidy confusion'. Doug and Kamyar viewed n as.the number of A (or a) 'species' chromosomes instead of the number of chromosomes o r chromatids present i n the c e l l . Sharilyn and May had a confused idea of what constitutes a set of chromosomes. They both considered the homologous chromosomes A,a and B,b as representing a chromosome set rather than the non-homologous chromosomes A,B and A,b. Jason never considered chromosomes i n a set. He., lab e l l e d both the (diploid) parent c e l l and the f i n a l product c e l l s of meiosis as n=4 for they contain four chromosomes. In other words, he related n to the t o t a l number of chromosomes rather than the number of chromosomes in a set. 100 CHAPTER 5 PROBLEM-SOLVING APPROACHES OF UNIVERSITY GENETICS STUDENTS 5.1 Introduction This chapter presents outcomes of analysis of the data for the second research question: What a r e the q u a l i t a t i v e l y d i f f e r e n t ways i n which undergraduate s t u d e n t s approach g e n e t i c s p r o b l e m s o l v i n g ? The analysis involved constructing categories of description (hereafter these are c a l l e d 'approaches') to characterize undergraduate genetics students' problem solving approach. Data were obtained from interviews with students who were asked to solve a genetics question. The genetics question and interview questions were described i n Chapter 3. In t h i s chapter, one representative student's example i s used to i l l u s t r a t e each approach. The other students' examples are placed i n Appendix D. 5.2 The Problem-Solving Approach Four q u a l i t a t i v e l y d i f f e r e n t approaches to the Ascobolus problem were i d e n t i f i e d from the group of students p a r t i c i p a t i n g i n the study. These w i l l be referred to as the 'direct success', the 'hypothesis-formulation and reframing', the 'hypothesis-formulation and rejection' and the ' p a r t i a l recognition and forced f i t ' approaches. 101 5.2.1 The 'Direct Success' Approach This approach consists of the following steps which include: student demonstrates a detailed consideration of a l l the written representations and the symbolic representations i n the problem; student formulates reasonable hypotheses; student checks and monitors responses; and, student achieves desired outcomes. The sequence of events explicated by t h i s approach i s summarized i n the flow chart below: Student recognizes and is conscious of all written and symbolic representations Student formulal .es reasonable hypotheses Student checks and monitors responses Student achieves desired outcome Figure 14: Flow chart summarizing the 'Direct Success' Approach 102 The 'Direct Success' Approach: Kirk's Example 1 Patrick: Let's go to t h i s question. Kirk: When I go through the question, I usually write down a l l the information. F i r s t , I t r y to read through i t . Mutation f --5 okay. So i t says mutation to the l e f t and right of the centromere. So I draw this (chromosome s i x with a centromere). Just to the r i g h t / l e f t of the centromere; so I draw these (f and b) close to the centromere. Then i t gives the cross and the parents, so I draw out the cross (he drew 10 +f and b+ with a cross i n between). And then i t says usually i t i s black (the normal spores). So, Neurospora, I believe i t i s haploid, i s i t ? Patrick: Do these symbols have any meaning to you? 15 Kirk: Yep. + i s the wild type and i t i s pretty straight forward. Patrick: How about the single +/ two +'s? 20 Kirk: Here the Ascobolus i s the haploid, so +f (the genotype of the parent) means these are two diff e r e n t genes; so they [the two genes]can have two d i f f e r e n t mutations: the fawn and the beige. The question doesn't say they are two d i f f e r e n t genes; 25 i t can be just two d i f f e r e n t mutations with the centromere i n between. Patrick: Can you apply your meiosis model to f i t into the data of case I? 30 •' Kirk: Well, here you have to use a meiosis model. So the chromosome duplicates, and then i t w i l l l i n e up and random assortment.You got +f and b+ i n the two c e l l s , so that i s the f i r s t d i v i s i o n . Here i s the second d i v i s i o n ; the chromosomes 35 l i n e up and you get two +f's and two b+'s.These are not the usual way I draw i t because the ascospores grow out i n an ascus i n an orderly manner (in chains of eight). Patrick: What i s your normal way of drawing i t ? 40 Kirk: It would be l i k e t h i s (the c e l l s placed i n blocks, one on top of another). It should be +f, +f, b+, b+. Look at the question, black i s the normal. Most octads are four fawn and four beige so these are the patterns (+ f, +f , +f, +f , b+ , b+ , b'+ , b+) 103 45Patrick: Why do you double up the products (+f,+f, b+,b+)? Kirk: Well, sometimes they [the questions]give i t t h i s way (+f, + f, b+,' b+: tetrads) and sometimes they give i t that way (+f,+f,+f,+f,b+,b+,b+,b+: octads). I just respond to the 50 pattern given i n the question.(Looking at the question again, he la b e l l e d the spores). These must be ++, these must be fb, bf, whatever. Why would these be aborted? (for case 1) t would come back to this picture. Why would that happen? It would be an abnormal process of meiosis, so I am going to sort out i n 55 my mind what the p o s s i b i l i t i e s are. After doing so many questions, I sort of think of the pattern here. So I am thinking maybe i t i s due to non-disjunction for i t ends up with aborted spores. For crossover you have double mutant. But for this one (case one), i t should be MI non disjunction. 60 We have a +f and a b+ parent c e l l (Figure 15a), followed by duplication of chromosomes (Figure 15b). This one separates normally for the f i r s t time (to +f,+f, b+, b+) and you get one abortion (Figure 15c). When you get a Mil non-disjunction , you w i l l get three d i f f e r e n t types (of c e l l s ) . So I guess 65 i t should be a MI non-disjunction. So we have a +f +f b+ b+ (in upper c e l l after f i r s t division) (Figure 15c) and then you get +f +b,- +f +b (after second division) and then these two give you the wild type (Figure 15e). And then, mitosis here and these four (spores) become eight (Figure 15f). —\-o—\— - v o t-First Division (a) (b) Mitosis A : -± 4 WT WT (c) Second D i v i s i o n (e) [Figure 15: Kirk's i l l u s t r a t i o n of non-disjunction for case one] 104 70 Kirk: 75 80 Let us go to case 2. So these are the + f, + +, bf, b+ (the four kinds of spores produced i n case 2). Now, double mutant (bf)-- these are parental (+f, b+) and these (bf,++) are due to the crossover (Figure 16b). What has happened are crossovers (Figure 16a). Okay, and then they (the chromosomes +f/bf, ++/b+)(Figure 16c) separate (Mil segregation) to a +f, bf, ++ and b+ (Figure 16d). These four (spores) again undergo mitosis to become eight (He did not draw the eight spores). This one i s easier, just one event (double mutant) i s the clue. Double mutant -- you have to get a crossover. Patrick: How about the events before crossing over? Kirk: The chromosomes double and then they pair. +- r + (a) -4—6—Y--t—o-r-(d) (b) t -4—. fa 1 (c) Figure 16: Kirk's i l l u s t r a t i o n of crossover for case two 105 Kirk demonstrated a detailed consideration of a l l the written and symbolic descriptions i n the Ascobolus problem. He recognized the relevant symbols of the ascospores such as 0 = aborted; 0 = beige; 0 = black, normal; 0 = fawh; and 0 = double mutants. He also i d e n t i f i e d the correct positions of the mutant a l l e l e s f and b on chromosome s i x . More importantly, he understood the relevant genetic make-up of the respective types of ascospores such as +f = fawn; ++ = black, normal; b+ = beige and bf = double mutants. After going through these written representations i n the problem, he considered the symbolic representations which included the pattern of the spores given i n the two octad cases of the problem. He recognized the presence of 50% aborted ascospores i n case one and the presence of viable double mutants (bf) i n case two of the problem. From these observations, he started to formulate the hypotheses f or the two cases. He proposed MI non-disjunction would explain the results for case one, and crossing over would explain the results for case two. Kirk generated a reasonable hypothesis for the two cases. F i n a l l y , he used his 'canonical' meiosis model, as was i l l u s t r a t e d i n the meiosis task, to solve the problem. His success could be attri b u t e d to his detailed consideration and understanding of the di f f e r e n t kinds of representations; his understanding of the l i f e cycle of the haploid organism Ascobolus; his a b i l i t y to relate the data to 106 reasonable hypotheses and his use of a 'canonical' meiosis model to check his hypotheses. Consideration of all word-descriptions Consideration of all symbolic descriptions Formulation of reasonable hypothesis 1. Identify relevant symbols of spores 2. Identify position of f and b genes (lines 4-8) 3. Identify relevant genetic make-up of spores (lines 16, 21-24)1 Case One: Aborted spores (line 52) Case Two: Pouble mutants (lines 72-73) /Non-di sj uncti oh Hlines 57-58) Crossing over (line 73.) Checking & monitoring responses Successfully solved problem 'Canonical" meiosis model and mitosis (lines 60-63, 65-68 & Figure 15) YES "Canonical" meiosis model and mitosis (lines 75-84 & Figure 16) Figure 17: Kirk's example of the 'Direct Success' Approach 107 5.2.2 The •' Hypothesis -Formulation and Reframincf/ Approach The 'hypothesis-formulation and refraining' approach i s characterized by the students' .reframing of t h e i r hypothesis. Students using t h i s approach also understood and considered a l l the written representations i n the problem, which required a sound understanding of the biology of the haploid organism Ascobolus. However, they could not relate the data i n the problem to a reasonable hypothesis i n i t i a l l y (for example, a 'non-disjunction' hypothesis for case one) . Fortunately, they possessed an i n t e r n a l checking and monitoring system. This type of metacognitive behaviour allowed them to go back to the problem again and consider a l l the relevant symbolic representations. They then 'reframed' t h e i r o r i g i n a l hypotheses and checked them by t h e i r meiosis models. They f i n a l l y solved the problem. However, students using the 'hypothesis-formulation and reframing' approach took a 'longer' path than students using the 'direct success' approach for they needed to go back to the question again to reframe t h e i r o r i g i n a l hypotheses (refer to Figure 18). 108 Understanding and consideration of all written representations Partial consideration oi 'the symbolic representations •—• \ ' r " * Formulation of original hypothesis Reconsideration of all other symbolic representations Use of genetic knowledge Reframing the original hypothesis Checking and monitoring responses (II) Partial or complete success in solving problem Figure 18: Flow chart summarizing 'Hypothesis-Formulation and Reframing' Approach The 'Hypothesis-Formulation and Reframing' Approach: Mike's Example 1 Patrick: Let's go to this more r e a l i s t i c genetics problem. As you are working through i t , say aloud a l l your thoughts. Mike: As I am reading the question, I try to relate the picture to 5 the words i n the question. .1 don't know why there are eight spores here -- the problems I deal with are. just four spores, so the way to do this problem i s to do meiosis f i r s t to form the four c e l l s and then mitosis to form the eight c e l l s , right? I start picking out chromosome s i x (he then wrote 109 10 down -- fawn just to right of centromere). I wrote these down, because I don't have to go back to the question again. A cross of fawn by beige parent -- beige just to the l e f t (of centromere). And a cross of fawn by beige, most octads show four fawn and four beige ascospores but three exceptional 15 octads are found. Patrick: What do these symbols (+f b+) mean to you? Mike: This (+) means wild type and this (f) means fawn. You can 20 never t e l l which i s dominant, you see -- they are (+f) two diff e r e n t kinds of genes here. Also the + and beige (b) genes in the b+ parent. This + means b a s i c a l l y I can have an f and l i t t l e + here (f +) meaning wild type of the fawn, not showing the fawn phenotype. Now,for these normal black to appear (in 2 5 case 1) means that there has to be one crossover. I don't understand why we get nothing in these empty c i r c l e s (in case 1). Patrick: What i s the meaning of th i s empty c i r c l e (aborted)? 30 Mike: They mean aborted. They are dead. Patrick: What are you going to do next? 35Mike: I'm just going to -- get the aborted. See, the f i r s t thing that came to my mind i s tetrad analysis and now th i s i s octad analysis. But i t almost seems that you can use non-disjunction. I am not sure yet. 40Patrick: Why do you use non-disjunction? Mike: It seems to f i t . When improper segregation of chromosomes during either MI or M i l . 45Patrick: How w i l l the products be diffe r e n t (in MI and Mil non-disjunction) ? Mike: There w i l l be a difference i n products. In meiosis I (non disjunction), you just have one slot where you have one empty 50 c e l l (after MI). You have two empty sl o t s (after MID (Figure 19) .. In the two normal s l o t s , you have -- +f and b+. So, i t ' s MI for this case (case 1) . 110 1 n ' ir (b) duplicat i o n MI e b -v o The two parents (a) Mil (1 empty slot) (c) (2 empty slots) (d) Figure 19: Mike's i l l u s t r a t i o n of non-disjunction f or case ond Patrick: Can you explain how the wild types are formed i n t h i s case (case 1)? 55 Mike: Because you get a beige and wild type and a wild type and fawn i n each one (b+/+f) so that they complement each other You get a ++,the + r i c h a l l e l e s give you the wild type. GOPatrick: Is t h i s one (the wild type) haploid or diploid? Mike: They are haploid. Because the products of the ascospores, they are haploid. Patrick: How about representations? 65 70 Mike: This one you have fawn, beige, black and double mutant. This one - - I just see a single crossover occurs, meaning either (crossover)here or there ( l e f t or right of centromere between two non-sister chromatids), meaning you get + and fawn, giving you the fawn. This (cross) gives you the ++ which is the wild type, this (cross) would give you beige fawn, the double mutant, and then beige and + give you the beige. Patrick: Can you show me the cross with your previous meiosis model? I l l 75 Mike: 80 85 These are the two parents to start with (+f x b+) with the cross over here (Figure 2 0a), what should happen i s these two parents pair and then double. Neurospora fungus I believe should be haploid i t s e l f , so they become +f/+f times b+/b+ and now a crossover takes place (Figure 20b), so end up with +f and bf to the upper c e l l and ++ and b+ go to (the bottom cel l ) (Figure 20c). We have the beige and fawn (bf to top cel l ) and beige and + go down (b+) . From here (the two c e l l s ) , I can simply draw the four ( f i n a l products as b+)(Figure 20d), and then mitosis to form, eight c e l l s (He did not draw out the eight spores). P A R E N T S + f G (a) 0 +\ / * 0 (b) +• +• (d) -f (c). fc . £ ©—3— e B E : 0 ? 4-b Figure 2 0 : M i k e ' s . i l l u s t r a t i o n of crossing over for case two 112 Mike's example i l l u s t r a t e s the c h a r a c t e r i s t i c s of the 'hypothesis-formulation and reframing' approach. He considered a l l the written representations i n the problem. These written representations included the correct positions of the f and b genes on chromosome s i x ; and the relevant genetic make up of the ascospores. The recognition and understanding of these representations should have enabled him to solve the problem. Unfortunately, he missed some important information i n the symbolic representations of the problem. For example, he considered only the presence of normal black spores i n the result of case one, and neglected the presence of aborted spores. The normal, black (non-parental) spores made him select crossing over as the o r i g i n a l hypothesis to explain the pattern of spores i n the results of case one. Mike overcame t h i s d i f f i c u l t y by using an i n t e r n a l check system. He went back to the problem again to reconsider other symbolic representations (l i n e 31). Recognition of the presence of aborted spores other than the normal, black spores i n case one led him to propose a non-disjunction hypothesis to replace the crossing, over hypothesis. Mike checked his f i n a l hypothesis by using his 'event-modified' meiosis model generated from the meiosis task. This meiosis model placed the event of homologous p a i r i n g before r e p l i c a t i o n . A s h i f t i n the sequence of the two events makes the model a b i t inaccurate. He did solve the problem, but his success was obscured by the fact that he used an inappropriate notion of 113 t h e s t e p s l e a d i n g t o t h e r e s u l t s . M i k e s o l v e d the p r o b l e m o n l y p a r t i a l l y f o r he u s e d t h e ' e v e n t - m o d i f i e d ' m e i o s i s m o d e l . Understanding and consideration of all written representations Partial consideration of symbolic representations Formulation of initial hypothesis Checking and monitoring responses (I) Reframing the original hypothesis Checking and monitoring responses (IT) Successfully solved problem 1. Identify position of f and b genes (lines 9-11) 2. Identify relevant genetic make-up of spores (lines 19-24) Case One: 1. Only considered the presence of normal black spores 2. neglected the presence of aborted empty spores (line 24) e Crossing Over (line 25) Reconsider other symbolic representation: aborted empty spores are also present in case 1 (line 31) 'MI type (line 52) ^non-disjunction (lines 37-38) rEvent-Modified' meiosis model and mitosis (lines 49-51) Partial F i g u r e 2 1 : M i k e ' s example o f t h e ' H y p o t h e s i s - F o r m u l a t i o n and R e f r a m i n g ' A p p r o a c h 114 May also used the 'hypothesis-formulation and reframing' approach to address the Ascobolus problem. However, she solved the problem by using the 'canonical' meiosis model to check her hypothesis (see Appendix D). 5.2.3 The 'Hypothesis-Formulation and Rejection' Approach One major c h a r a c t e r i s t i c of the 'hypothesis-formulation and rejection' approach i s the f a i l u r e of the students to reframe to the o r i g i n a l hypotheses. This f a i l u r e could be accounted for'by an i n a b i l i t y of these students to relate a reasonable hypothesis to the data i n the Ascobolus problem. Apart from t h i s major ch a r a c t e r i s t i c , a l l the other ch a r a c t e r i s t i c s of t h i s approach are s i m i l a r to the 'hypothesis-formulation and reframing' approach. Students using t h i s approach also missed some information provided by the problem, so they either have to reconsider other symbolic representations or use accurate or 'naive' genetic concepts to reject t h e i r o r i g i n a l hypothesis. Although they exhibited the "metacognitive" behaviour of checking and monitoring, they f a i l e d to reframe t h e i r o r i g i n a l hypotheses and therefore 'got stuck'; with the end result being that they gave up on the problem. 115 Understanding and considering some written representations Partial understanding and consideration of the symbolic representations Formulation of hypothesis Checking and monitoring responses Reconsideration of all other symbolic representations Use of accurate/inaccurate genetic knowledge Rejection bf original hypothesis Dead End Figure 2 2: Flow Chart summarizing 'Hypothesis-Formulation and Rejection' Approach The 'Hypothesis-Formulation and Rejection' Approach: Sam's Example 1 Patrick: Let's look at this r e a l i s t i c genetics problem. Say aloud a l l your thoughts as you are going through t h i s problem. Sam: Okay. (He c i r c l e d the sentence ascospores are normally black 5 i n the question). I just have to lay out.the information (in the question) for myself when I read the question. And sometimes I have to draw diagrams to understand the question and to do highlighting. Fawn is the gene just to the right of the centromere b to the l e f t (of the centromere). (He 10 drew a chromosome diagram i n response to the posit i o n of the mutant gene b and f on the chromosome).Mutation's on chromosome 6. (When he read through the question up to the point of in a cross of fawn by beige parents (+f x b+), he 116 15 then drew the cross. So, this i s the fawn (he c i r c l e d the two fawn ascospores i n case 2 and labelled them as fawn). Horizontal line i s beige (so he c i r c l e d the beige ascospores and labelled them as beige) and wild type i s black (so he labelled the black ascospores by ++) and empty 20 c i r c l e s represented aborted ascospores (he then labelled the four empty ascospores i n case 1 as aborted). Three exceptional octads were found. (He repeated the wordings of the question). This i s the meiosis for the haploid, right? 25 Patrick:Why? Sam: Because i t says this fungus i s similar to Neurospora. Patrick:Can you define the term haploid? 30 Sam: Um, the organism with one set of chromosomes i s the haploid. Patrick:What do you mean by one set of chromosomes? 35 Sam: Well, l i k e i n human c e l l s , there are forty s i x chromosomes. There are two sets of twenty three chromosomes, that i s a 2N c e l l . But for example, l i k e the gametes they only have twenty three chromosomes, they have one set (of chromosomes). 40 Patrick:So how many set of chromosomes are there i n these two parents? Sam: There i s only one set. The question says chromosome number si x , so we are dealing with one chromosome. There i s not 4 5 another set of chromosome, there i s only one set because i t i s haploid. In my previous meiosis model, there are two sets of chromosomes, but here (Ascobolus) has only one set. Okay, this +f produces sp e c i f i c type of gametes and th i s b+ produces sp e c i f i c type of gametes. So this can produce only 50 one kind of gamete, +f and this can produce one kind (of gamete), b+. And we are dealing with this p a r t i c u l a r chromosome (number s i x ) . Patrick:What do you understand by the, symbols +,f and b? 55 Sam: This + i s the wild type, of b (in +b) and this i s the mutant of f (in +f). You represent the wild type by + and the • mutant by the small f.. So (in case 1) from the eight (ascospores) produced, four are aborted. Four of them are 60 black, which are the normal, ones, the colour. So th i s would produce l i k e t h i s (+f/ b+) fusing together. For th i s stage, I c a l l t h i s 2n. You have something here l i k e -- obviously i t i s 117 a mitosis here to produce these eight ascospores and meiosis that produces these four ascospores. There was a meiosis 65 before the mitosis. Patrick:Why did you c a l l this (+f/b+) 2n? Sam: Because this i s chromosome s i x (+f) and th i s i s chromosome si x (b+) so they are of the same chromosome. So when the 70 gametes (+f and b+) are combining together, they are also combining in a l l those chromosomes. They are a l l now present i n t h i s c e l l -- every thing i n this 2n c e l l . Since we have two sets of chromosomes here, + and f as one set, b and + as another set, i t i s d i p l o i d . 75 Patrick: How would you explain the results of the eight ascospores i n case 1? Sam: I think I should go through meiosis to produce the N c e l l --80 the haploid. And then when I go to meiosis, there i s also probability of crossing over l i k e -- either i n th i s region or that region of the chromosome (in the region between non-s i s t e r chromatids to the right or l e f t of the centromere). Both (crossing over) of them w i l l again produce the same 85 thing (product c e l l s ) . If we consider no crossing over, i t w i l l produce ... -Patrick:Why didn't you consider crossing over? Sam: There's either two p o s s i b i l i t i e s , crossing over and no 90 crossing over. Let's see crossing over. So, i f have +f b+, there i s a crossing over here... So from meiosis, they f i r s t duplicate (+f +f, b+, b+)(Figure 23a), f i r s t double (to form +f, +f, b+, b+). So now we have s i s t e r chromatids. So, . Now, say this i s one crossing over -- there i s no crossing 95 over between these two ( + f,+f) because they are s i s t e r chromatids (Figure 23b). Crossing over between thi s one produces +f, this one produces ++, th i s one produces bf and this one produces b+. Okay,++ the wild type which i s black, the bf i s the double mutant, and b+ the beige and the +f i s 100 the fawn (Figure 23c), so this i s case 2. 118 1 n - f i 1 - t , *J n b — ( — 9 — b 1 a — -£ +f (fawn) fl —t 1— bf (double (3 ++ (black) " V mutant) b+ (beige) (a) .(b) (c) Figure 23: Sam's i l l u s t r a t i o n of crossover model for case two Patrick:Let's go to case 1. Sam: This i s black, ++, the aborted i s (case 1) ... Um, how can you get that (in case 1). You can have a crossing over 105 here (between the two outer non-sister chromatids, +f and b+). (Figure 24a) . So this (the f i r s t crossover) would produce a ++ and a bf, and this (the second crossover) w i l l again produce a ++ and a bf (Figure 13b).But the bf (produced) i s the double mutant, but -- the double mutant i s viable, but 110 they say (in the question) that the four empty spores here are dead. So.I have to give up (crossing over). But I get to the ++ (by the crossing over)somehow, right? (Figure 24b) 115 So, i f there i s no crossover, ... So, t h i s i s the +f and this i s the b+ (parent);(Figure 25a) -- then pairing. And then replication (of +f and b+; see Figure 25b; with r e p l i c a t i o n of centromeres). They (the replicated chromosomes), they go to two c e l l s (Figure 25c). So, that would be +f, b+ (in one c e l l ) and' +f, b+ (in another c e l l ) . 1 don't get i t (++ product c e l l ) . I have to get to the ++ part --' crossing over? 119 120 125 But here, i f I continue the process of meiosis, at the end, I would get four c e l l s , l i k e t h i s i s +f, this i s b+, this would be +f, this would be b+, fawn, beige, fawn, beige (Figure 25d) and then they would go mitosis to get the eight c e l l s (Figure 25e). But this (pattern) doesn't f i t into the case 1, four aborted, four normal, right? I guess something i s screwed up. - 4 -i -f 4 b 0 -r i -(b) -f -©-i f - f -v. -r 4 (d) -^-<XJ~U ITS Figure 25: Sam's i l l u s t r a t i o n of no crossing over for case one 120 Sam's example serves to i l l u s t r a t e the c h a r a c t e r i s t i c s of the 'hypothesis-formulation and rejection' approach. He i n i t i a l l y attempted to consider some written representations i n the problem such as the relevant symbols of ascospores( for example, 0 = dead, aborted; © = beige; (P = fawn and • = wild type). He also considered the r e l a t i v e positions of the f and b a l l e l e s on chromosome s i x . Unfortunately, without considering the information c a r e f u l l y , Sam reversed the location of the two a l l e l e s . Sam also had d i f f i c u l t y i n interpreting the language of the symbolic representations. He recognized the presence of 50% black and 50% dead spores i n the results of case one without r e a l i z i n g that t h i s pattern of spores should be related to a 'non-disjunction' hypothesis. He i n i t i a l l y proposed a 'crossing-over' and -'no crossing-over' hypotheses to account for the res u l t s i n case one. Sam possessed an in t e r n a l check system. After he had formulated his hypotheses, he checked his hypotheses. The presence of viable double mutants i n the results of case two made him r e a l i z e that aborted, dead spores i n case one were not double mutants. Based on t h i s recognition, Sam rejected his 'crossing-over' hypothesis. He also knew the process of no crossing-over well. This process led to the formation of four fawn and four beige spores which did not f i t into the pattern of spores i n case one. Thus he rejected the 'no crossing-over' hypothesis. Rejection of hypotheses i n case one did not help Sam reframe his o r i g i n a l hypotheses. What seems evident i s that he either had 121 l i m i t e d o r no knowledge o f n o n - d i s j u n c t i o n . Thus he c o n s i d e r e d the o n l y mechanism of c r o s s i n g - o v e r and d i d not c o n s i d e r non-d i s j u n c t i o n as an o p t i o n . He seemed to have no o t h e r mechanism to draw upon o t h e r than c r o s s i n g - o v e r f o r case one. Thus , Sam f a i l e d to reframe h i s o r i g i n a l hypotheses 8 . Understanding and consideration of some written representations Partial understanding of symbolic representations Formulation of hypothesis Checking and monitoring Rejection of hypothesis 1. Identify symbols of spores (lines 15-21) [2. Identify position of f and b genes (lines 10-11) Case One: It 50% black and 50% dead spores (lines 59-60) /Crossing-Over^ Wline 90) /tfo crossing over Valine 113) Viable double mutants in case 2 implies aborted| spores are never double mutants (lines 109-111) four fawn and four beige from meiosis and mitosis (lines 121-124J Reject crossing over Reject no crossing over Dead End NO Successfully solved problem F i g u r e 26: Sam's example of the ' H y p o t h e s i s - F o r m u l a t i o n and R e j e c t i o n ' Approach f o r case one Sam adopted the 'Direct-Success' approach and did succeed for case two. In this chapter, only case one was used to illustrate his 'hypothesis-formulation and rejection' approach. 122 5.2.4 The ' P a r t i a l Recognition and Forced F i t ' Approach One major c h a r a c t e r i s t i c of the ' p a r t i a l recognition and forced f i t ' approach i s the misrepresentation of the data i n the problem by the students. Students using t h i s approach had l i m i t e d knowledge of the biology of the haploid organism Ascobolus and could not relate the data to a reasonable hypothesis. These students misrepresented the meaning of the gene symbols i n the problem. Although these students formulated hypotheses, these hypotheses were never checked or monitored. They used only a ' f o r c e - f i t t i n g ' operation to f i t t h e i r hypotheses into the results of the problem. The sequence of events explicated by th i s approach i s summarized by the following flow chart: Figure 27: Flow Chart summarizing the ' P a r t i a l Recognition and Forced F i t ' Approach Misunderstanding representations Partial/Incomplete consideration o f symbolic representations 'Force-fitting' Operation Dead E n d 123 The ' P a r t i a l Recognition and Forced F i t ' Approach: Doug's Example 1 Patrick: Let's go to this r e a l i s t i c genetics question. As you are going through t h i s question, say aloud a l l your thoughts. Doug: Um, because there are two groups of spores here (case 1), one 5 at the top (a group of four normal, black spores) and one at the bottom (a group of four aborted spores) . So t h i s i s a MI pattern, MI means meiotic d i v i s i o n -- f i r s t stage. (He then labelled.the black, normal spores by a and the aborted spores by b). ' lOPatrick: Why do you label the spores' by small a and small b? Doug: Well, just to define two dif f e r e n t types. If you have two gametes right here, a and b (he wrote down a,a; b,b separately). 15 20 25 Patrick:Why do you write down the a, a and b,b separately? Doug: Because i t looks a b i t correct. Because i t ' s just l i k e the duplication diagram (Figure 28a) i n my o r i g i n a l meiosis model. And then, these two separates (aa);(Figure.28b). This i s small a, small a, small b, small b and then- duplicated again might be meiosis (Figure 28c). I mean mitosis. This i s from the book diagram. It's something s i m i l a r . Well, right, just separate into four d i f f e r e n t a l l e l e s by meiosis. This i s the end of meiosis and then af t e r mitosis, t h i s (a) doubles into (a,a) and the t o t a l i s eight. CX 11 '. <2-4 <3v mitosis • a. Figure 28: Doug's meiosis model (I) for case one 124 Patrick:What makes you draw t h i s diagram? Doug: Because I memorize, i t ' s according to the book diagram. I t 3 0 s t a r t s with something r i g h t here (the parents), then i t goes four and then ei g h t . I don't know what question I wrote t h i s . In order to have four chromosomes, I need to have two p a i r s at the s t a r t . So, I need not r e f e r to my meiosis model. Um, I j u s t couldn't remember. 35 P a t r i c k : L e t ' s go to case 2. Doug: Well, t h i s i s crossover I think. Because i t ' s M i l p a t t e r n and M i l p a t t e r n means that, there i s a crossover and there are 40 four d i f f e r e n t kinds of spores. Patrick:What s p e c i f i c kind of information i n the diagram allows you to i n f e r that t h i s i s a crossing over? 45 Doug: Um, these two(bf) might be crossed from these (two) h o r i z o n t a l (beige) and these v e r t i c a l (fawn). Patrick:Can you t e l l me your idea of crossing over? 50 Doug: I f you have gene A r i g h t here, t h i s i s . . . dominant, that the b i s recessive, another one r i g h t here, B,B, a, a (Figure 29a). These two are recessive (a,a) These two cross over (Figure 29b), and the products w i l l be b i g A, small b; small a, b i g B (Figure 29c). The chromosomes j u s t break here and 55 they (exchange). The break i s h o r i z o n t a l . Supposedly, i f you have the cross here (he d i d a blow up of the cross X), 2 and 3 are o r i g i n a l l y j o i n e d here, i t breaks r i g h t here and then they would j o i n together ( v e r t i c a l break)(Figure 2 9b). 60 Doug: I f the h o r i z o n t a l l i n e represents dominant of A, v e r t i c a l l i n e represents the dominant of B, and the two cross together and you got... you got b i g A and b i g B. Right here, you got both A and B, then you have v e r t i c a l and h o r i z o n t a l r i g h t here (bf). A CK b _ 1 -a " B-A (c) 6 Figure 29: Doug's i l l u s t r a t i o n of c r o s s over f o r case•two 125 65 Patrick:What i s the meaning of the bf symbol here? Doug: Oh! t h i s i s the r e c e s s i v e , cause the + r e p r e s e n t s the dominant. 70 P a t r i c k : S o , how about the (genotypes) of the parents? Doug: T h i s i s dominant f o r b and r e c e s s i v e f o r f (+f). That i s r e c e s s i v e f o r b and dominant f o r f (b+). I mean... a c t u a l l y , i t i s not dominant, t h i s i s mutation, + means w i l d type and f 75 means mutant. I j u s t memorize the p a t t e r n . The q u e s t i o n asks about how I e x p l a i n the o f f s p r i n g s . The q u e s t i o n asks you why they look l i k e t h a t . ( A f t e r l o o k i n g at the q u e s t i o n f o r a s h o r t time, he changed the gene symbols). 80 Doug: 85 T h i s i s the w i l d type of a... oh, i t ' s b. w i l d type of f and and t h i s i s f ( f o r parent +f). T h i s i s the mutant type of b and the w i l d type f o r f . ( f o r the other parent b+) (He drew the two symbols + or b/f on two d i f f e r e n t chromosomes with the centromere t o one side),(Figure 30a) .These two are w i l d type and these two are mutants (Figure 3 0b). Patrick:Have they (+,f, b,+) d u p l i c a t e d ? Doug: No, they d u p l i c a t e ( a f t e r m i t o t i c d i v i s i o n ) . 90 And they separate l a t e r . T h i s i s w i l d type f o r b, t h i s i s mutant f . T h i s i s mutant b and w i l d type f (Figure 30b). So, they then (+ f o r b) d u p l i c a t e s i n t o two +, + 7for b) two f , f ; two b,b; and two +,+(for f) i n m i t o s i s (Figure 30c). 95 P a t r i c k : S o , do the r e s u l t s r e p r e s e n t case 1 or case 2? Doug: I guess case 1 then. P a t r i c k : Are the r e s u l t s ( f i n a l products) four i n d i v i d u a l s or e i g h t i n d i v i d u a l s here? 126 100 Doug: I t h i n k the f o u r +'s are the w i l d types and then the white ones (empty c i r c l e s ) are the mutants. P a t r i c k : A r e these the parents here (the o r i g i n a l + f, b+) here? Are they r e p r e s e n t i n g n or 2n? 105 Doug: Parents, they are 2n. One (chromosome) i s one n. So, together they are 2n i n one parent. Oh!, t h i s one (chromosome +) i s from f a t h e r and t h i s one (chromosome f) i s from mother. l l O P a t r i c k : A r e they (the o f f s p r i n g s ) n or 2n here? Doug: Right here, they are one n. Patrick:How i s t h i s one n d i f f e r e n t from t h i s one n here? 115 Doug: Oh, because r i g h t here, these two chromosomes of the parents are l i n k e d by a centromere, but here the o f f s p r i n g chromosomes, they are separate ? The parents are d i p l o i d . The example of Doug i l l u s t r a t e s the ' P a r t i a l recognition and forced f i t ' approach. Doug misrepresented the information i n the problem. He represented the +f parent of Ascobolus by u__^ _— instead of *-t-e>—t- . He also ignored the information about the re l a t i v e positions of the b and f a l l e l e s oh chromosome s i x . Instead of placing the f and b a l l e l e s on either side of the centromere, he placed the centromere to one side of the -f . chromosome such as u \ . His MI hypothesis was correct and so was his M i l . He recognized the pattern but had l i m i t e d , understanding of i t s significance. For example, he proposed a "MI pattern" hypothesis to account f o r the formation of two groups of spores i n case one. "MI ;pattern", according to Doug, corresponded to the two products of meiosis i n the f i r s t d i v i s i o n . He might have simply memorized i n lecture that the pattern of spores i n case one was explained by MI and that the pattern i n case two was Mil (see l i n e s 38-40): ' 127 The metacognitive behaviour of checking and monitoring was absent i n Doug's analysis. He generated 'naive' models of meiosis and came up with a new set of 'invented' data. By using the ' f o r c e - f i t t i n g ' operation, he f i t t e d h is data into the re s u l t s of the problem. For example, he generated his 'naive' one-stage meiosis model to explain the results i n case one (see Figure 30). This 'naive' model resulted i n the formation of eight spores: w i l d type was l a b e l l e d as "+" and aborted spores was l a b e l l e d as "b" and " f " as i f they were single mutants. This model not only f a i l e d to relate to his o r i g i n a l 'non-disjunction' hypothesis, but also contained his misrepresentations of genetic symbols for the ascospores. Regarding Doug, his proposed mechanisms were correct, although he could not explain these mechanisms. He d e f i n i t e l y had his facts jumbled and confused. 128 Misunderstanding the written representations Partial/Incomplete consideration of pictorial representations 1. Ascobolus is a diploid organism (line 106) 2. +f parent was represented by (Figure 30) (ignored position of f and b genes) Case One: [Two groups of spores (lines 4-6) Case Two: Four types of spores with double mutants (lines 40, 45) Formulation of hypothesis MI pattern JLines 6-7) /Mil pattern & crossing \over (Line 38) Force-Fitting operation One-stage meiosis model and and mitosis (lines 84-85, 92-93 and Figure 30) Eight spores formed: wild type as "+" and aborted spores as single mutants "b" or "f' (Figure 30c) Crossing-over model (Figure 29, lines 45, 50-64) Crossover product as AB which represented the double mutant. Successfully solved problem Dead End NO Fiqure 31: Doug's example of the ' P a r t i a l Recognit ion and Forced F i t ' Approach 129 CHAPTER 6 RELATIONSHIP BETWEEN PROBLEM SOLVING APPROACHES AND CONCEPTUALIZATIONS OF MEIOSIS 6.1 Data A n a l y s i s f o r Research Question 3: I n t r o d u c t i o n The major focus of t h i s study was to i n v e s t i g a t e the a b i l i t y of students to apply t h e i r knowledge of meiosis (Chapter 4) to the Ascobolus problem (Chapter 5). This chapter presents outcomes of an a n a l y s i s of the data p e r t a i n i n g to the t h i r d research question of t h i s i n v e s t i g a t i o n : Is t h e r e any r e l a t i o n s h i p between the s t u d e n t s ' approaches to p rob lem s o l v i n g and t h e i r knowledge o f m e i o s i s ? In order to address t h i s t h i r d research question, a comparison of the meiosis models used by the students i n the i n i t i a l meiosis task (Chapter 4) and i n the Ascobolus problem was undertaken. The purpose of t h i s a n a l y s i s was to determine whether the students were able to d i s p l a y some consistency i n t h e i r understanding and a p p l i c a t i o n of t h e i r meiosis models and to a s c e r t a i n the extent t o which the b i o l o g i c a l content of problem s e t t i n g had an i n f l u e n c e on how they used t h e i r meiosis model. To s i m p l i f y the r e p o r t i n g of t h i s a n a l y s i s , a l a b e l l i n g system was e s t a b l i s h e d f o r the four types of problem-solving approaches and the three types of meiosis models. 130 The l a b e l l i n g system for the four types of problem-solving approaches used i n addressing the Ascobolus problem was as follows: 1 = 'Direct Success' Approach 2 = 'Hypothesis-formulation and Reframing' Approach 3 = 'Hypothesis-formulation and Rejection' Approach 4 - ' P a r t i a l Recognition and Forced F i t ' Approach The names describing the di f f e r e n t meiosis models generated from the meiosis task were also used to describe the meiosis models students used i n the Ascobolus problem. These names from Chapter 4 which are restated below, have been l a b e l l e d as A,B and C where: A = 'Canonical' Meiosis Model B = 'Event-Modified' Meiosis Model C = 'Whole-Process Modified' Meiosis Model In addressing the Ascobolus problem, some of the students altered s l i g h t l y the meiosis models that they had generated i n the meiosis task. These altered models are indicated by adding an asterisk (B*). 6.2 Procedures of the Analysis The analysis focused on determining whether the students were consistent i n t h e i r use of a meiosis model i n both interview tasks (the meiosis task discussed i n Chapter 4 and the Ascobolus problem presented i n Chapter 5). 131 Chapter 4 outlined the conceptualizations used by students i n responding to the meiosis task; these conceptualizations were referred to as models. The second task examined students' responses to the Ascobolus problem and t h i s analysis resulted i n the construction of four d i f f e r e n t approaches that were used by the students. Furthermore, t h i s analysis examined the type of meiosis model that the students used i n t h e i r approach to the problem. When determining the types of meiosis models used with the Ascobolus problem, both genetic diagrams and interview excerpts were used. For example, i t was sometimes d i f f i c u l t to decide upon the sequence of the 'pairing' and the ' r e p l i c a t i o n ' events i n the model since these did not d i r e c t l y appear i n the students' genetic diagrams. In those cases, the interview transcripts provided additional information that was used for establishing the sequence of events. If the interview transcripts are used to c l a r i f y a cert a i n point, the excerpts used are indicated by l i n e numbers (for example, l i n e s 3-7). 6.2.1 Identifying the Meiosis Models Applied i n Ascobolus Problem Chris: Chris displayed a sound understanding of meiosis (model A) when responding to the meiosis task (see Chapter 4, Figure 4), but applied a d i f f e r e n t model, which belonged 132 to 'whole-process modified' category (model C), i n solving the Ascobolus problem (see Appendix C, Figures C-16 and C-17). His model exhibited his misunderstanding and misinterpretation of the process of deletion. He also did not show the sequence of meiosis i n his model. Kirk: This student displayed a good understanding of meiosis (model A) and used t h i s same model on the Ascobolus problem. The models used by t h i s student can be charted as indicated i n Figure 32 and Figure 33. Replication Crossing Over & Pairing • & (Fig. 15b, Chromosome segregation line 56) (Fig. 15c) Chromatid segregation (Fig. 15d) Figure 32: Kirk's meiosis model applied to Case 1 of the Ascobolus problem (Chapter 5, Figure 15 relabelled) Equatorial Products of Replication Crossing over Alignment I chromosome Pairing {Fig. 16a) (Fig. 16b) segregation (line 84) {Fig. 16c) Products of chromatid segregation {Fig. 16d) Figure 33: Kirk's meiosis model applied to Case 2 of the Ascobolus problem (Chapter 5, Figure 16 relabelled) 133 Sharilyn: Sharilyn displayed a good understanding of meiosis (model A) and also used the same model i n the Ascobolus problem. The models can be charted as follows: R e p l i c a t i o n (Fig. C-3a) and P a i r i n g (Fig. C-3b & line 88) E q u a t o r i a l alignment I ( F i g . C-3c) Chromosome s e g r e g a t i o n [Fig. C-3d) E q u a t o r i a l alignment II (Fig. C-3e) Chromatid s e g r e g a t i o n (Fig. C-3f & g) Figure 34: Sharilyn's meiosis model used with Case 1 of the As-cobolus problem (Appendix C, Figure C-3 relabelled) R e p l i c a t i o n (Fig. C-2a) & P a i r i n g _ (Fig. C-2b) & l i n e 54) E q u a t o r i a l alignment I & c r o s s i n g over (Fig. C-2c) Chromosome s e g r e g a t i o n (Fig. C-2d) Chromatid s e g r e g a t i o n (Fig. C-2f & g) Figure 35: Sharilyn's meiosis model used with Case 2 of the Ascobolus problem (Appendix C, Figure C-2 relabelled) < E q u a t o r i a l alignment II (Fig. C-2f) 134 S i m i l a r l y , May and L a i l a also used meiosis model A for both tasks (see Appendix C, Figures C-8, C-5 and C-6). Replication & Pairing Chromosome Chromatid (Line 52 & Fig. C-8a) ^segregation * segregation (Fig. C-8b) (Fig. C-8c) Figure 36: May's meiosis model used with Case 1 of the Ascobolus problem (Appendix C, Figure C-8 relabelled) Replication (Fig. C-5a) Pairing • {Lines 39-40 & Fig. C-5b) Chromosome segregation (Fig. C-5c) Chromatid segregation (Fig. C-5c,d) Figure 37: L a i l a ' s meiosis model used with Case 1 of the Ascobolus problem (Appendix C, Figure C-5 relabelled) Replication ^ Pairing and Chromatid (Fig. C-6a) chromosome segregation segregation (Fig. C-6b & lines 71) (Fig. C-6c) Figure 38: La i l a ' s meiosis model used with Case 2 of the Ascobolus problem (Appendix C, Figure C-6 relabelled) 135 Mike: Mike used the type B meiosis model on the meiosis task (pairing before r e p l i c a t i o n instead the other way around) and used the same model i n Case 1 of the Ascobolus problem. The model was as shown i n Figure 39. Pairing ^Replication Chromosome segregation (Line 77 & (Fig. 19b) (Fig. 19c) Fig. 19a) Chromatid segregation (Fig. 19d) Figure 39: Mike's meiosis model used with Case 1 of the Ascobolus problem (Chapter 5, Figure 19 relabelled) Sam: Sam used the type B meiosis model. His model used with the Ascobolus problem was altered somewhat from his o r i g i n a l meiosis model described i n Chapter 4. His discussion of the segregation of the chromatids instead of the chromosomes i n the f i r s t d i v i s i o n was a v a r i a t i o n on his o r i g i n a l meiosis model. The model used with the Ascobolus problem was as shown i n Figure 40. 136 Pairing d i n e 114 & Fig. 25a) > Replication (Fig. 25b) Chromatid segregation (Fig. 25c) Chromosome segregation (Fig. 25d) Figure 40: Sam's meiosis model used with Case 1 of the Doug used the type B meiosis model i n i t i a l l y but was unable to use the same model i n the Ascobolus problem. The model used with the Ascobolus problem was i d e n t i f i e d as belonging to the type C category because the model exhibits a fundamental misunderstanding of the process of meiosis as a whole. This i s i l l u s t r a t e d i n Figure 41. Figure 41: Doug's meiosis model used with Case 1 of the Ascobolus problem (Chapter 5, Figure 30 relabelled) Doug's model consists of one instead of two nuclear d i v i s i o n s . Therefore, i t was not d i r e c t l y related to his o r i g i n a l meiosis model (model B) which included two 'nuclear d i v i s i o n s ' steps. Doug used the eight meiotic products generated from t h i s model to force a f i t with the data of the Ascobolus problem. Each meiotic product was taken to represent an ascospore. In Case 2, Doug introduced an Ascobolus problem (Chapter 5, Figure 25 relabelled) Doug: Replication (Fig. 30a) • Chromatid segregation (Fig. 3 0b) 137 incomplete meiosis model with crossing over. The model only-included the events of r e p l i c a t i o n , p a i r i n g up and crossing over (see Figure 29) . There was no indi c a t i o n of chromatid segregation occurring. Therefore", t h i s model was also t o t a l l y unrelated to the o r i g i n a l meiosis model used i n Chapter 4 (model B). Kamyar: Kamyar used the type B meiosis model on the meiosis task. The model used with Ascobolus problem also belonged to the type B category, but deviated from the o r i g i n a l model i n that the two s i s t e r chromatids moved into the same c e l l during segregation I I . The model was shown i n Figure 42. • Chromatid segregation I * ( i n c l u d i n g s p l i t t i n g of centromeres) (Fig. C-12b) **Chromosome segregation I I (The two homologs. go to the same c e l l ) (Figr. C-12c) Figure 42: Kamyar's meiosis model used with Case 1 of the Ascobolus problem (Appendix C, Figure C-12 relabelled) t h i s resembles the meiosis model generated from the meiosis task (Figure 12d) t h i s step deviates from the meiosis model generated from the meiosis task (Figures 12f and g) R e p l i c a t i o n -(Fig. C-12a) 138 Jason: Jason used a type C meiosis model i n i t i a l l y . In Case 1 of the Ascobolus problem, he i n i t i a l l y copied a 'meiosis for the haploids' model from his notes taken during the genetics course (see Appendix C, Figure C-19). The model followed the ske l e t a l framework of the 'canonical' meiosis model (type A) by i t s sequence of events: r e p l i c a t i o n , segregation of rep l i c a t e d homologous pairs i n the f i r s t d i v i s i o n , and segregation of s i s t e r chromatids i n the second d i v i s i o n of meiosis. However, when Jason proceeded to apply the model to the data i n the Ascobolus problem, he used a more s i m p l i f i e d meiosis model (see Appendix C, Figure C-20). This s i m p l i f i e d model, which i n no way related to the model i n his notes (see Figure 43b), i s shown i n Figure 43a. R e p l i c a t i o n — • *Chromatid s e g r e g a t i o n Figure 43a: Jason's meiosis model used with Case 1 of the Ascobolus problem (Appendix C, Figure C-20 relabelled) * T h i s event resembles the s i n g l e s e g r e g a t i o n event i n the o r i g i n a l m e i o s i s model generated from the 'meiosis t a s k (see F i g u r e 13d and e) 139 R e p l i c a t i o n and Chromosome Chromatid p a i r i n g * segregation * segregation (Fig. C-19b) (Fig. C-19c) • (Fig. C-19d) Ficrure 43b: Jason's model from h i s notes (Appendix C, Figure C-19 relabelled) In Case 2, t h i s xone-stage r e p l i c a t i o n ' meiosis model was a l s o used by Jason to e x p l a i n the c r o s s i n g over process (see Figure C-21). This model, while i t belongs to the type C category, d i d not resemble h i s o r i g i n a l model si n c e i t l e f t out the event of segregation. 6.2.2 Mapping of Meiosis Models used w i t h the Me i o s i s Task and the Ascobolus Problem The meiosis models used on the two i n t e r v i e w tasks are summarized i n Table 1-. This t a b l e i l l u s t r a t e s which of the ten students were c o n s i s t e n t i n t h e i r use of meiosis models i n the i n t e r v i e w tasks, t h e i r problem-solving approach and t h e i r success i n problem-solving. 140 TABLE 1: Mapping of M e i o s i s Models To Problem-Solving Approaches Student M e i o s i s Models M e i o s i s Models Ascobolus S u c c e s s f u l l y generated from used w i t h the Problem- s o l v e d m e iosis task Ascobolus S o l v i n g Problem problem Approach C h r i s A C 3 No K i r k A A 1 Yes S h a r i l y n A A 1 Yes May A A 2 Yes L a i l a A A 1 Yes Mike B B 2 P a r t i a l Sam B B* 3 No Doug B C 4 No Kamyar B B* 2 P a r t i a l Jason C C* 4 No Problem-Solving Approach: 1-'Direct Success' 2- ' H y p o t h e s i s - f o r m u l a t i o n & r e f r a m i n g ' 3- ' H y p o t h e s i s - t e s t i n g and r e j e c t i o n ' 4- ' P a r t i a l R e c o g n i t i o n and F orced F i t ' R e s u l t s i n Table 1 show th a t , i n a study of ten students, s i x used e i t h e r the ' d i r e c t success' or the 'hypothesis f o r m u l a t i o n and reframing' approach. Among these s i x students, four (Kirk, S h a r i l y n , May and L a i l a ) generated a s i m i l a r 'canonical' meiosis model to the one used i n the meiosis task -- that i s , they c o n s i s t e n t l y used the same meiosis models i n both i n t e r v i e w t a s k s . The four students who used the 'canonical' model succeeded i n s o l v i n g the Ascobolus problem completely. The remaining two students, Mike and Kamyar, used the 'event-modified' model, which i n d i c a t e d q u i t e a reasonable understanding of meiosis when addressing the Ascobolus problem. They succeeded i h p a r t i a l l y s o l v i n g the problem. A c l o s e s c r u t i n y of the 141 responses of these s i x students to the Ascobolus problem showed that they had a sound understanding of both the biology of the haploid organism Ascobolus, and an a b i l i t y to relate the data set (for example, half wild type and half aborted i n Case 1) to a non-disjunction hypothesis. Results i n Table 1 also showed that, out of the ten students, four (Chris, Sam, Doug and Jason) used ei t h e r the 'hypothesis formulation and rej e c t i o n ' or the ' p a r t i a l recognition and forced f i t ' approach. Out of these four students, only Chris generated a 'canonical' meiosis model for the meiosis task, but he f a i l e d to use the same model with the Ascobolus problem.. The other three students (Sam, Doug and Jason) generated either an 'event-modified' or a 'whole-process modified' meiosis model. The example of Chris confirms that a display of a conceptual understanding of the meiosis task involving diploids does not guarantee that students w i l l be able to apply the same model to a problem involving a haploid organism. Chris used the 'hypothesis-testing and re j e c t i o n ' approach. A major c h a r a c t e r i s t i c of t h i s approach was the i n a b i l i t y of students to reframe t h e i r o r i g i n a l hypotheses. Chris, i n p a r t i c u l a r , f a i l e d to relate the pattern of spores i n Case 1 (half w i l d type and half aborted) to a 'non-disjunction' hypothesis. Instead, he proposed the 'deletion' and 'inversion' hypotheses, which he was not that f a m i l i a r with, to account for the pattern of the spores. His un f a m i l i a r i t y with the two hypotheses l e f t him unable to use 142 his o r i g i n a l meiosis model to test these two hypotheses. As a r e s u l t , he ended up with an incomplete and inadequate explanation (see Figure C-18). Sam, s i m i l a r to Chris, also f a i l e d to rel a t e the data set i n the problem to a reasonable hypothesis. He proposed the 'crossing-over' and vno crossing-over' hypotheses to explain the pattern of spores i n Case 1 of the Asco£>olus problem. Although he used the 'event-modified' meiosis model to reject the 'no-crossing over' hypothesis, he could not reframe his o r i g i n a l hypothesis due to an i n a b i l i t y to relate the pattern of the ascospores (half w i l d type and half aborted) i n Case 1 to a 'non-disjunction' hypothesis. The above examples of Sam and Chris i l l u s t r a t e that certain types of domain knowledge are required before students can generate, a reasonable hypothesis about the possible meiotic mechanism i n t h i s problem. This may be one of the key factors i n understanding the success of the students i n solving the Ascobolus problem. For example, students should have a sound understanding of the concept of non-disjunction, (an abnormal phase of meiosis r e s u l t i n g from the f a i l u r e of homologous chromosomes to separate properly to opposite poles of the c e l l ) . In the Ascobolus problem, the presence of half wil d type and half aborted, ascospores i n Case 1 suggests that non-disjunction occurs during the f i r s t d i v i s i o n . A sound understanding of the concept of crossing over also contributes to the generation of a reasonable 143 hypothesis. Some students related crossing over to the formation of non-parental ascospores i n general. Thus they tended to propose the xcrossing-over' hypothesis to explain the pattern of spores i n Case 1. In other words, they seemed to have no other hypothesis to draw upon other than crossing over. They f a i l e d to r e a l i z e that crossing over involves only the exchange of corresponding chromosome parts, leading to the formation of double mutants seen i n Case 2. Thus, i t appears that a clear understanding of the content knowledge of non-disjunction and crossing-over i s an essential precursor to the formulation of a reasonable hypothesis by the students. In other words, content knowledge plays a keen role i n the formulation of hypotheses i n genetics problem-solving . The remaining two examples of Doug and Jason i l l u s t r a t e that a l i m i t e d knowledge of the biology of the haploid organism Ascobolus prevented them from solving the Ascobolus problem. They both f a i l e d to appreciate the biology as well as the genetic symbols used i n Ascobolus. Thus they used the ' p a r t i a l recognition and forced f i t ' approach. Findings from t h i s chapter suggest that content knowledge plays an important role i n successful problem solving i n genetics. Students must have a fundamental understanding of the biology of the haploid organism, Ascobolus and complete f a m i l i a r i t y with the gene symbols used i n the problem. The generation of a reasonable hypothesis also required a clear understanding of how 144 meiosis occurs i n haploid organisms and the processes of non-disjunction and crossing over. Having possessed these knowledge bases, the students could then be able to apply t h e i r understanding of meiosis to the Ascobolus problem. Only those students who consistently used the 'canonical' meiosis model f i n a l l y succeeded i n solving the Ascobolus problem completely. 145 CHAPTER 7 CONCLUSIONS, IMPLICATIONS AND RECOMMENDATIONS 7.1 Introduction There are three different, sections i n t h i s chapter. The 'conclusions' section summarizes the results of the three analysis Chapters (4,5,and 6). The 'discussion' section refers back to the l i t e r a t u r e and compares the results of th i s study with those reported i n the l i t e r a t u r e . I t also discusses how the results observed here may be used to inform teaching. F i n a l l y , the 'l i m i t a t i o n s ' section discusses the generalizabil.ity of t h i s study and the p o s s i b i l i t y of further research that needs to be done i n thi s area. 7.2 Specific Conclusions from the Study This research was a phenomenographic study of ten t h i r d year undergraduate genetics students. The study covered a number of areas dealing with the students' conceptualizations of meiosis and the approaches they used to address an applied question of genetics that involved the concept of meiosis. A n a l y t i c a l methods based upon a phenomenographic perspective (Marton, 1981, 1988) were used to es t a b l i s h categories of descriptions i n the following related areas: students' conceptualizations of meiosis for d i p l o i d organisms and students' approaches i n addressing a complex genetics task that involved the concept of meiosis. 146 These categories of description provided basis for analysis of factors a f f e c t i n g students' successful a p p l i c a t i o n of the meiosis models i n solving the more complicated Ascobolus problem. The conclusions a r i s i n g from the analysis presented i n these areas, discussed i n Chapters 4 through 6, are summarized below. 7.2.1 Conclusions for Chapter 4 CONCLUSION 1 This study i d e n t i f i e d three q u a l i t a t i v e l y d i f f e r e n t types of students' conceptualizations of meiosis. These conceptualizations were l a b e l l e d as the 'canonical' meiosis model, the 'event-modified' meiosis model, and the 'whole-process modified' meiosis model. CONCLUSION 2 The 'canonical' meiosis model was characterized by a s c i e n t i f i c a l l y accurate description of meiosis including the two stages of c e l l d i v i s i o n ; the appropriate sequencing of events and p a r t i c i p a t i n g structures before, during and af t e r the two d i v i s i o n s . CONCLUSION 3: The 'event-modified' meiosis models also included the two stages of c e l l d i v i s i o n . However, these models showed modifications from the 'canonical' meiosis models i n the sequencing of events,- i n the orienta t i o n of the p a r t i c i p a t i n g structures, and i n both the p a r t i c i p a t i n g structures and sequencing of events. 147 CONCLUSION 4 : The 'whole-process modified' meiosis model was characterized by s i g n i f i c a n t modifications i n the process or outcomes of meiosis. This model described meiosis as a one-stage instead of a two-stage process, with the products being more l i k e those expected i n mitosis. 7.2.2 Conclusion for Chapter 5 CONCLUSION 5: This study i d e n t i f i e d four q u a l i t a t i v e l y d i f f e r e n t types of approaches taken when students addressed the Ascobolus problem. These approaches were l a b e l l e d as the 'direct success' approach, the 'hypothesis-formulation and reframing' approach, the 'hypothesis-formulation and rejection' approach and the ' p a r t i a l recognition and forced f i t approach'. CONCLUSION 6 A l l students using the 'direct success' approach; the 'hypothesis-formulation and reframing' approach; and, the 'hypothesis-formulation and re j e c t i o n ' approach appeared to understand the biology of Ascobolus. Among these students, only those using the 'direct success' approach were able to generate a reasonable hypothesis which f i t the data and use t h e i r knowledge of meiosis to confirm the hypothesis. Students using t h i s approach not only understood the biology of Ascobolus, they also displayed a good understanding of the concepts of non-disjunction and crossing-over that allowed them to generate reasonable hypotheses. Students using the other two approaches were 148 unable to generate an i n i t i a l hypothesis to f i t the data. They used t h e i r knowledge of meiosis to 'reframe or 'drop' t h e i r o r i g i n a l hypotheses. Students adopting the 'hypothesis-formulation and reframing' approach were able to draw upon t h e i r knowledge of genetic p r i n c i p l e s to 'reframe' t h e i r i n i t i a l hypothesis. Students using the 'hypothesis-formulation and rejection' approach, however, seemed to have no other mechanism to draw upon other than crossing over for case one. Although they appeared to know the biology of Ascobolus, t h e i r l i m i t e d knowledge of the concepts of non-disjunction and crossing-over prevented them from generating reasonable hypotheses for Case 1. F i n a l l y , students using the ' p a r t i a l recognition and forced f i t ' approach displayed very l i m i t e d knowledge of the biology of Ascobolus as well as the concepts of non-disjunction and crossing over. These l i m i t a t i o n s prevented them from solving the problem. 7.2.3 Conclusion for Chapter 6 CONCLUSION 7 An understanding of the biology of the haploid organism, Ascobolus, as well as the concepts of non-disjunction and crossing over were necessary precursors for the successful application of a meiosis model to the Ascobolus problem. The complete success i n solving the problem also r e l i e d on the application of a 'canonical' meiosis model; 149 CONCLUSION 8 Most students adopting the 'direct success' approach and the 'hypothesis-formulation and reframing' approach consistently used the same meiosis model i n the two interview tasks. 7.3 Discussion of Results This section consists of two parts. The f i r s t part refers back to the l i t e r a t u r e and compares the results of t h i s study with those reported i n the l i t e r a t u r e . The comparison i s done i n two areas: the meiosis models and students' approaches to complicated problem-solving tasks. The second part involves a discussion of how the resul t s observed i n t h i s study may be used to inform teaching. 7.3.1 Contribution to the Literature on Learning Genetics Studies of students' conceptions of meiosis at both the high school and college levels have i d e n t i f i e d students' confusions about ploidy, as i t pertains to meiosis (Brown, 1990; Fisher et a l . , 1986; Smith, 1991; Stewart and Dale, 1989; Stewart, Hafner and Dale, 1990; Thomas, 1988 (cited i n Brown, 1990). These studies have reported that students relate ploidy to the number of chromosomes rather than the number of sets of chromosomes. The present study also i d e n t i f i e s the same kind of confusion. For example, Doug and Kamyar described the change of ploidy as a result of r e p l i c a t i o n of chromosomes (also reported by K i n d f i e l d , 150 1991). Jason l a b e l l e d both the (diploid) parent c e l l and the f i n a l product c e l l s of meiosis as n. Clearly, Jason believed that both the parent c e l l and the product c e l l s contained the same amount of genetic material and i n his model t h i s was true since his meiotic model was l i k e mitosis. He also related n to the t o t a l number of chromosomes i n the parent c e l l rather than the number of chromosomes i n a set. Other types of ^ploidy confusion' were also reported by th i s study, Sharilyn and May had a confused view of what a set of chromosomes i s . Both considered the homologous chromosomes A,a and B,b as representing a chromosome set rather than the non-homologous chromosomes A,B and A,b. This study also i d e n t i f i e d other kinds of confusion i n the students-' meiosis models. In the vevent-modified' meiosis models, Mike and Sam were unclear on the r e l a t i v e sequencing of r e p l i c a t i o n and p a i r i n g before the f i r s t d i v i s i o n step of meiosis. They placed the event of p a i r i n g before the event of r e p l i c a t i o n (also reported by K i n d f i e l d , 1994). Kamyar segregated the chromatids during the f i r s t d i v i s i o n ; and, segregated the chromosomes during the second d i v i s i o n instead of vice versa (also reported by K i n d f i e l d , 1994). Another form of confusion i n the xevent-modified' meiosis models relates to the orientation of the p a r t i c i p a t i n g chromosomes. Doug described a perpendicular rather than a p a r a l l e l alignment of chromosomes r e l a t i v e to the central axis of the c e l l during the f i r s t and second 151 d i v i s i o n s of meiosis. Doug either did not understand spindle formation or had f a i l e d to take t h i s into account. In the 'whole-process modified' meiosis model, Jason described a form of c e l l d i v i s i o n which was not discussed i n the l i t e r a t u r e . This student described meiosis as involving one rather than two successive nuclear d i v i s i o n s . The products of meiosis consisted of only two c e l l s , each with the same amount of genetic material as the parent c e l l ; rather than four c e l l s , each with half the amount of genetic material as the parent c e l l . Jason was confusing meiosis with mitosis. This study also extends the general l i t e r a t u r e of genetics problem-solving by examining students' a b i l i t y to apply t h e i r meiosis models to a more complex genetics question (the Ascobolus problem). Studies i n the general l i t e r a t u r e of problem-solving i n genetics focus on id e n t i f y i n g the students' understanding of meiosis for d i p l o i d organisms. I t was generally believed by these researchers that a 'canonical' meiosis model should be s u f f i c i e n t for students to advance i n genetics learning. No research to date has documented students' application of th e i r knowledge of meiosis to a question that involves haploid organisms. Findings of t h i s study indicate that a good understanding of meiosis alone i s i n s u f f i c i e n t for successful genetics problem solving when a problem concerns a haploid organism. I t also suggests that students possess a sound understanding of other conceptual knowledge i n 152 genetics. Both a fundamental understanding of the biology of the haploid organism Ascobolus and the concepts of non-disjun c t i o n and crossing over are necessary precursors to the successful application of students' understanding of meiosis to the applied problem examined i n t h i s study. Without such precursors, the students would never be able to use t h e i r knowledge of meiosis, even i f they were able to generate a 'canonical' meiosis model from the meiosis task presented i n Chapter 4 . Results from Chapter 5 show that among students adopting the 'direct success' approach and the 'hypothesis-formulation and reframing' approach, only those who possessed these precursors, were able to apply t h e i r knowledge of meiosis to 'confirm' or to 'reframe' t h e i r hypotheses. On the other hand, students adopting the 'hypothesis-formulation and rejection' approach and the 'p a r t i a l recognition and forced f i t ' approach were unable to apply t h e i r knowledge of meiosis. They had l i m i t e d understanding of non-disjunction and crossing over that made them unable to relate the data to a reasonable hypothesis. They also displayed a l i m i t e d understanding of the biology of the haploid organism, Ascobolus. Studies of students' conceptions of meiosis i n the l i t e r a t u r e only i d e n t i f y ploidy confusions accompanying the meiosis models without r e l a t i n g these confusions to other areas of genetics learning. This study suggests that a clear concept of ploidy might be essential for a sound understanding of the biology of the haploid organism 153 Ascobolus. For example, Doug and Jason had ploidy confusions accompanying t h e i r meiosis models. They also displayed a li m i t e d understanding of the biology of Ascobolus and the genetic representations used to describe i t . This find i n g extends the l i t e r a t u r e on genetics problem-solving by emphasizing the important role played by content knowledge i n successful problem-solving. I t has implications for the design of i n s t r u c t i o n a l sequences for the teaching of genetics that involves the concept of meiosis. One proposed sequence i s discussed i n the next section. The present study also i d e n t i f e d four q u a l i t a t i v e l y d i f f e r e n t problem-solving approaches. A scrutiny of a l l the four approaches described i n Chapter 5 suggests that they might be interpreted i n terms of the "surface" and the "deep" approaches i n the l i t e r a t u r e . Marton and Saljo (1976) ; Svensson (1977) ; and, Ramsden, Whelan and Cooper (1989) i d e n t i f y "understanding" and "integration" as ch a r a c t e r i s t i c s of the "deep" approach. The 'direct success' approach exhibits the ch a r a c t e r i s t i c s of a "deep" approach. The students' responses i n Chapter 5 provide a number of examples which i l l u s t r a t e the "understanding" c h a r a c t e r i s t i c . Understanding that Ascobolus i s a haploid organism allowed Kirk and Mike, for example, to further r e a l i z e that the gene symbols such as "+f" should represent two d i f f e r e n t gene l o c i instead of one gene locus i n d i p l o i d organisms. An understanding of the concepts of non-disjunction and crossing-over allowed 154 students using t h i s approach to formulate reasonable hypotheses to explain the pattern of spores i n the problem. The "integration" c h a r a c t e r i s t i c was i l l u s t r a t e d by students considering various sources of information i n the problem. For example, before making the 'non-disjunction' hypothesis for Case 1, Kirk considered a l l the information i n the problem. These included the data of the pattern of spores i n both cases and the gene symbols. He recognized that the pattern of spores i n case one was half w i l d type and half aborted. He also re a l i z e d that the aborted spores were dead and, therefore, were not formed by crossing over. The integration of t h i s information allowed him to generate a reasonable hypothesis about the given data for the organism, which was essential for the successful appli c a t i o n of his understanding of meiosis to the problem. The 'hypothesis-formulation and reframing' approach exhibits c h a r a c t e r i s t i c s of both the generic "deep" and the "surface" approaches. Students adopting t h i s approach recognized that Ascobolus i s a haploid organism. Thus they r e a l i z e d the meaning of the gene symbols, +f and b+, and the p o s i t i o n of the f and b genes. However, these students i n i t i a l l y did not integrate the information provided. Students tended to read the pattern of the spores i n Case 1 of the question without attending to the data i n Case 2. Nevertheless, they went back to the question and reconsidered a l l other information i n the data. They successfully 'reframed' t h e i r o r i g i n a l hypotheses. 155 The 'hypothesis-formulation and rejection' approach also i l l u s t r a t e s an incomplete "understanding" of the problem c h a r a c t e r i s t i c . For example, Sam displayed a l i m i t e d understanding of the concept of crossing over and non-disjunction, which prevented him from generating a reasonable 'non-disjunction' hypothesis for case one. He got stuck on the 'crossing-over' mechanism and did not come up with another reasonable one to explain the data. F i n a l l y , the ' p a r t i a l recognition and forced f i t ' approach i n t h i s study exhibits most of the ch a r a c t e r i s t i c s of the "surface" approach. Using t h i s approach, Jason and Doug displayed a weak understanding of both the biology of the haploid organism, Ascobolus, and the concepts of crossing over and non-disjunction. Thus they 'invented' t h e i r own gene symbols for the haploid organism and f a i l e d to generate a reasonable hypothesis for the cases i n the problem. A weak understanding of the content knowledge displayed by the problem caused these students to lose track of the whole problem. The present study also extends the general l i t e r a t u r e on approaches to problem-solving by providing a detailed description of the "checking and monitoring" c h a r a c t e r i s t i c of the "deep" approach. I t i s believed by the author that the "checking and monitoring responses", the metacognitive behaviours are one of the c r i t i c a l c h a r a c t e r i s t i c s of the "deep" approach. Students of t h i s study exercised these metacognitive behaviours either i n confirming, i n reframing, 156 or i n re j e c t i n g t h e i r hypotheses. Students adopting the 'direct success' approach exhibited a l l the c h a r a c t e r i s t i c s of the "deep" approach. In the confirming case, i f the students were able to tunderstand the biology of the haploid organism Ascobolus and to generate a reasonable hypothesis which f i t the data, they were able to use t h e i r knowledge of meiosis to test and to confirm the hypothesis. Those who succeeded i n solving the Ascobolus problem used a 'canonical' meiosis model. In the reframing case, the i n i t i a l hypothesis generated by the students did not seem to f i t the data and students drew upon t h e i r knowledge of meiosis to reframe t h e i r o r i g i n a l hypothesis. In t h i s s i t u a t i o n , those who succeeded i n solving the Ascobolus problem also used a 'canonical' meiosis model. In the rej e c t i n g case, the hypothesis generated by the students did not f i t the data. Although they could 'reject' t h i s o r i g i n a l hypothesis, they were unable to use t h e i r knowledge of meiosis to reframe t h e i r o r i g i n a l hypothesis. This s i t u a t i o n appeared to be related to t h e i r l i m i t e d knowledge of the concepts of non-disjunction and crossing over. These findings demonstrate the existence of gradations within the "deep" approach since they a l l display aspects of the "understanding", "integration" and "checking and monitoring" c h a r a c t e r i s t i c s . Those students who did use the "checking and monitoring" c h a r a c t e r i s t i c of the "deep" 157 approach were not necessarily successful. The c r i t i c a l point for success lay i n the combination of the formation of a reasonable hypothesis and the application of a 'canonical' meiosis model to check and monitor the hypothesis. I t c e r t a i n l y implies that content understanding i s essential for meaningful learning. I t also suggests that teaching generic learning approach may be a waste of time. Consolidating the understanding of content knowledge might be a precursor to successful problem-solving. 7.3.2 Proposed Instructional Sequence A survey of f i v e genetics textbooks by the author suggests that meiosis i s introduced as part of a p a r t i c u l a r sequence involving the presentation of Mendelian genetics or chromosomal theory of inheritance, and, genetic v a r i a t i o n . A close examination of t h i s t r a d i t i o n a l textbook sequence suggests that most authors of genetics texts emphasize the importance of Mendelian genetics as the foundational concept. Out of the f i v e texts, four place t h i s topic immediately a f t e r the introductory chapter. The Mendelian laws explain the segregation of the gene pai r s , which i n turn affects the outcome of meiosis. A follow-up discussion of the chromosomal theory of inheritance, which includes meiosis, i s present i n most of these texts. Of the f i v e textbooks, three also include the concepts of haploidy and dipl o i d y i n t h i s 'meiosis' section. Most of these textbooks include a descriptive summary of the process of meiosis, by 158 describing the various stages (for example, prophase, metaphase, anaphase, telophase). This description also introduces many new genetic terms. A f i n a l section deals with genetic v a r i a t i o n , which includes a discussion of the di f f e r e n t kinds of abnormalities occurring i n the c e l l that lead to various kinds of meiotic products. For example, the abnormality of non-disjunction (seen i n Case 1 of the Ascobolus problem) would be regarded as one of these va r i a t i o n s . Table 2 provides a b r i e f lay-out of the sequence of topics introduced i n these f i v e genetics textbooks. Table 2 A broad lay-out of topics of f i v e genetics textbooks Topics Genetics Textbooks 9 I II III IV V 1 Meiosis Mendel Mendel Mendel Mendel (ploidy) 2 Mendel Meiosis Meiosis Meiosis Meiosis 3 Genetic Variations ---9 T e x t b o o k s m a r k e d w i t h a n a s t e r i s k a r e : I . K l u g , W. & Cummings, M. (1991) C o n c e p t s o f G e n e t i c s , New Y o r k , M a c m i l l a n . I I . R u s s e l , P. ( 1 9 9 0 ) . G e n e t i c s , C a l i f o r n i a , H a r p e r C o l l i n s . I I I . G a r d n e r , E . & S n u s t a d , D. ( 1 9 8 4 ) . T h e P r i n c i p l e s o f G e n e t i c s , New Y o r k , J o h n W i l e y a n d S o n s . I V . R o t h w e l l , N. (1988) . U n d e r s t a n d i n g G e n e t i c s , New Y o r k , OUP. V. B u r n s , G. & B o t t i n o , P. (1989) . T h e S c i e n c e o f G e n e t i c s , New Y o r k , M a c m i l l a n . 159 This study shows that students had d i f f i c u l t i e s with a number of aspects of meiosis, including the r e l a t i v e sequencing of pa i r i n g of chromosomes, the sequence of chromosome and chromatid segregation, and, the outcomes of meiosis. The results of t h i s study also show that ploidy confusion may contribute to an incomplete understanding of the biology of Ascobolus. F i n a l l y , an i n a b i l i t y to relate the data i n the Ascobolus problem to a reasonable hypothesis also affects the students' performance i n problem-solving. Based on these findings, the author proposes one possible i n s t r u c t i o n a l sequence, aimed at arranging learning experiences for students so that they may be better equipped to successfully apply t h e i r knowledge of meiosis to an 'applied' problem that involves a haploid organism. The proposed sequence follows the t r a d i t i o n a l i n s t r u c t i o n a l sequence found i n genetics textbooks, with only s l i g h t modifications suggested by the res u l t s of t h i s study. F i r s t , since ploidy understanding may be related to understanding the biology of Ascobolus (see Doug and Jason), i t i s suggested that the concepts of haploidy and diploidy should be integrated with the teaching of meiosis. While t h i s approach was adopted i n a basic manner i n genetics textbook I (see Table 2 ) , a further refinement i s suggested here. I t i s proposed that the introduction of the concept of ploidy take place i n two steps. This may allow the instructor and students to focus on the issues contributing to ploidy confusion, namely the effect of r e p l i c a t i o n on 160 ploidy (see Doug and Kamyar), and the difference of ploidy and i t s p i c t o r i a l representations i n diploid.and haploid c e l l s (see Jason, May and Sharilyn). As an example of how t h i s might be assessed, two sample questions are proposed (see questions 1 and 2 i n Appendix E). After the introduction of the concept of ploidy, the next problematic area concerns students' confusion regarding the outcome of meiosis. The 'canonical' meiosis models of t h i s study show a sound understanding of meiosis as involving two successive nuclear d i v i s i o n s , with product c e l l s each containing half the chromosomal materials as the parent c e l l . Mendel's laws provide the foundational basis which describe the formation of product c e l l s i n meiosis. Traditional sequences found i n genetics textbooks separate Mendel's laws and meiosis into two chapters. I t i s proposed that i t might be useful to integrate the teaching of Mendel's laws with the teaching of meiosis. This integrative approach may help students relate these important concepts. The other problematic area i n the students' meiosis models concerns the sequence of chromatid and chromosome segregation; and, the sequencing of the 'pairing' event. Kamyar segregated the chromatids rather than the chromosomes i n the f i r s t d i v i s i o n of meiosis and described homolog pai r i n g as an event occurring i n the second d i v i s i o n rather than i n the f i r s t d i v i s i o n of meiosis. This student appeared not only to have a l i m i t e d knowledge of the sequence of events i n meiosis, but also may have had a confused or 161 incomplete understanding of the ,terminology of genetics ( f o r example, chromatid and chromosome). Instead of e x p l a i n i n g meiosis d e s c r i p t i v e l y and c h r o n o l o g i c a l l y by the v a r i o u s stages of the process ( f o r example, prophase, metaphase, anaphase, telophase), the present r e s u l t s suggest students should pay a t t e n t i o n on the terminology, s t r u c t u r e s , and reasons behind the steps of meiosis (see Appendix E, questions 3, 4 and 5 f o r example of questions that may be used as part of such an approach). This proposed sequence i s aimed at h e l p i n g students gain a c l e a r concept of p l o i d y . I t should a l s o a l l o w them to understand the b i o l o g y of the h a p l o i d organism Ascobolus and to describe a 'canonical' meiosis model. F i n a l l y , the proposed i n s t r u c t i o n a l sequence, which f o l l o w s the t r a d i t i o n found i n other genetics textbooks, would i n c l u d e the i n t r o d u c t i o n of the t o p i c of genetic v a r i a t i o n . This t o p i c would lead to the p r e s e n t a t i o n of data such as that seen i n the Ascobolus problem (half w i l d type and h a l f aborted ascospores) and provide o p p o r t u n i t i e s f o r students to p r a c t i s e w i t h generating hypotheses to e x p l a i n s i t u a t i o n s such as non d i s j u n c t i o n (see Appendix E, question 6 ) . Most genetics t e x t s only describe the process of non-d i s j u n c t i o n i n a s t r a i g h t f o r w a r d sense. They do not attempt to i n t e g r a t e the concept of no n - d i s j u n c t i o n w i t h data or novel problem s i t u a t i o n s . This sequence i s proposed as a means of enabling students to apply genetics content knowledge to novel problem s i t u a t i o n s . 162 7.4 Limitations of the Study This phenomenographic study emphasizes the sampling of conceptualizations and approaches rather than the students. Rather than trying>-to generalize across the student population of t h i s genetics course, i t attempts to i d e n t i f y as many conceptualizations and approaches as possible given constraints imposed by sample size. These conceptualizations and approaches are a form of generalization that transcend in d i v i d u a l experiences. Although these may s t i l l be conjectures, they are generalizations. There are, however, l i m i t i n g features of the design of thi s study that might r e s t r i c t the scope of the generalizations. These r e s t r i c t i o n s include the selection of interviewees and the design of the interview tasks. The small number of students interviewed creates l i m i t a t i o n s to the study. The participants of t h i s study signed up for the interview v o l u n t a r i l y , and most of them scored highly i n the genetics course. A close scrutiny of the students' conceptualizations and the approaches indicates that half of them belong to the 'canonical' category and the 'direct success' approach. Although results o f f e r some insights into d i f f e r e n t students' understandings of meiosis and approaches to problem solving, there may not have been enough students who experienced considerable d i f f i c u l t y with the course. The design of the interview tasks also creates some possible l i m i t a t i o n s . Since the meiosis task explores only 163 the content knowledge of meiosis, results generated from t h i s interview task can be generalizable only to other research settings that involve the study of meiosis. Within the scope of meiosis, t h i s task serves to i d e n t i f y only the diagnostic features of the standard process of meiosis without considering the abnormal phase of meiosis.' Furthermore, because the task question included the symbolic representations of the chromosomes, i t was not possible to i d e n t i f y students' understanding of symbolic representations of genes and chromosomes. The design of the Ascobolus problem also l i m i t s the scope of g e n e r a l i z a b i l i t y . The haploid organism, Ascobolus, contains only one single set of chromosomes and produces round ascospores (in octads). This problem was used to i d e n t i f y students' approaches and i t i s not known whether s i g n i f i c a n t l y d i f f e r e n t approaches would be generated i f a di f f e r e n t organism ( i . e . , other than Ascobolus) were used. 7.5 Recommendations for Further Research Based on the l i m i t a t i o n s of t h i s study discussed i n the previous section, recommendations for further research are offered. A broader sampling strategy, and extension and refinement of the interview tasks would contribute to increasing the g e n e r a l i z a b i l i t y of the present study. A broader sampling strategy could be f a c i l i t a t e d by introducing the research at a more convenient time i n the academic year, which might result i n more students 164 volunteering to p a r t i c i p a t e . Instead of doing the research at the end of the course, when the students who performed poorly were quite discouraged, the study should be done e a r l i e r i n the course af t e r they had studied the relevant concepts. Also, i t would have been useful to have had the opportunity to discuss d i r e c t l y the nature of the study to the students, for instance i n a t u t o r i a l s e t t ing. These changes might encourage a more diverse group of students to p a r t i c i p a t e i n the study. Their p a r t i c i p a t i o n would provide a rich e r data set on d i f f e r e n t students' understandings of meiosis and t h e i r problem-solving approaches. Elaboration and refinement of the interview tasks used i n the study would also contribute to increasing the g e n e r a l i z a b i l i t y of the research. With respect to elaboration of the tasks, Chapters 2 and 7 point out that the major research studies done i n genetics problem-solving have focused on meiosis conceptualization. They also point out that no research has investigated how students apply the knowledge of meiosis to novel problems that involve haploid organisms. This study i s unique i n that i t focusses on not only the students' understanding of meiosis, but.also t h e i r application of t h i s understanding to a p a r t i c u l a r problem setting. Further elaboration on t h i s work would b u i l d up a repertoire of findings i n t h i s area and provide us with a better understanding of how un i v e r s i t y genetics students apply t h e i r knowledge of meiosis to problems that involve 165 other kinds of situations such as sex-linkage and l e t h a l genes. It would also be useful to f i n d out how these students conceptualize and apply concepts i n other areas of genetics. In p a r t i c u l a r , t h i s study, could represent a s t a r t i n g point for further studies into how students conceptualize and apply other.related concepts of c e l l d i v i s i o n . The author believes that any study of students' conceptualization i n a p a r t i c u l a r topic area should be accompanied by a study of i t s application. Results of such studies should provide information for instructors regarding how to arrange meaningful learning experiences that w i l l a s s i s t students i n learning to apply concepts f l e x i b l y and broadly. The design of the Ascobolus problem e l i c i t e d a number of students' problem-solving approaches. Researchers would also expect d i f f e r e n t approach i f a d i f f e r e n t organism ( i . e . other than Ascobolus) were used. Thus i t i s also desirable to redesign the problem by using a d i f f e r e n t haploid organism. This redesign would allow the researchers to detect whether the organism used i n the problem would have any effect on l i m i t i n g the scope of the g e n e r a l i z a b i l i t y . With regard to the refinement of the meiosis task, Chapters 3 and 4 provide a detailed description of the meiosis task and interview protocols used f o r e l i c i t i n g the students' understanding of meiosis. This task includes the drawing of the parent c e l l with gene symbols and chromosomes assigned. Students were asked to work from t h i s s t a r t i n g 166 c e l l to the meiotic products. The present study shows that some students have d i f f e r e n t understandings of the written and symbolic representations i n genetics questions. Bearing t h i s i n mind, one could refine the meiosis task using only written representations and observing how the students respond. Findings of t h i s rewritten task may provide useful information on what genetic symbols the students use to represent t h e i r understanding and i n t e r p r e t a t i o n of a task which only contains written information. An example of t h i s refined meiosis task could take the form used most recently by the genetics professor who has worked with the author on t h i s research project, and reproduced i n Figure 44. A plant of genotype Aa Bb produces gametes that are 1/4 A B 1/4 A b 1/4 a B 1/4 a b showing that the two l o c i are on d i f f e r e n t chromosomes. Draw diagrams of the chromosomes i n interphase, chromatid formation, and the major types of chromosome movement during meiosis, to show how t h i s r a t i o i s produced. Figure 44: The Refined meiosis task This refined meiosis task, which consists of a s p e c i f i e d gamete set, also serves to i d e n t i f y students' understanding of crossing-over, an essential distinguishing feature of meiosis. The r e s u l t s of the present study imply that content understanding i s essential for successful problem-solving i n 167 genetics. A sound understanding of meiosis alone was i n s u f f i c i e n t for the students to solve the novel problem s i t u a t i o n . They need to possess a clear understanding of the biology of Ascobolus as well as the knowledge of non-disjunction, for example, which enabled them to generate the hypothesis about the meiotic mechanism necessary to solve the problem. Non-disjunction belongs to an abnormal phase of meiosis which was not included i n the o r i g i n a l meiosis task. Thus, i t might be desirable to include an investigation of the concept of non-disjunction i n the meiosis task. A f i n a l design issue that needs to be considered focuses on extending the present research into a longitudinal study. E a r l i e r i n t h i s chapter, the author proposed one i n s t r u c t i o n a l sequence which he suggested would improve the learning experiences for the students. I t would be worthwhile to study the effectiveness of t h i s proposed sequence on students' learning. The present 'snapshot' study reported students' understanding and application of the meiosis knowledge at a certain point of time. Other research studies of a longitudinal nature should be i n i t i a t e d to examine the effectiveness of di f f e r e n t teaching approaches on genetics learning. In summary, the conclusions and recommendations discussed i n t h i s chapter have ref l e c t e d a need to appreciate the idea that students come to class with a diverse set of ideas. To influence students' knowledge construction and reconstruction, educators need to create 168 r e a l i s t i c and exploratory educational environments which e x p l i c i t l y respect students' p r i o r ideas and provide learning experiences which address these. Only when we take into account students' perspectives can we hope to begin to see a s h i f t from the t r a d i t i o n a l views of teaching and learning as discussed i n Chapter 1 and teach i n such a way that students w i l l have the opportunity to engage seriously with the content knowledge being taught. 169 REFERENCES B a l l a , J . I . (1990a). 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L i b e r a l 177 APPENDIX A GLOSSARY OF EDUCATIONAL AND GENETIC TERMS A L L E L E : one of the dif f e r e n t forms of a gene that can exis t at a single locus e.g., A and a a l l e l e s , b and f a l l e l e s ANAPHASE: an intermediate stage of nuclear d i v i s i o n during which chromosomes are pulled to the poles of the c e l l s , ( i t i s referred to as segregation or separation by the students i n the th e s i s ) . CATEGORY OF DESCRIPTION: an interpretative descriptive category which characterizes a conceptualization of meiosis or a s p e c i f i c problem-solving approach; an in t e r p r e t a t i o n of the students' understanding. CENTROMERE: the constricted region of a nuclear chromosome, to which the spindle f i b e r s attach during d i v i s i o n . CHROMATID: one of the two side-by-side replicas produced by chromosome d i v i s i o n . CHROMOSOME: a l i n e a r end-to-end arrangement of genes and . other DNA, sometimes with associated protein and RNA. CONCEPTUALIZATION: t h i s i s a term used to broadly r e f l e c t how someone sees, v i s u a l i z e s , thinks about, understands or makes sense of experiences and phenomena. I t ' i s a q u a l i t a t i v e description of a person-world r e l a t i o n s h i p . In th i s thesis, i t represents how students described t h e i r understanding of meiosis based upon the d i p l o i d c e l l given i n the meiosis question. CROSSING-OVER: the exchange of corresponding chromosome parts between homologs by breakage and reunion. DELETION: removal of a chromosomal segment from a chromosome set. HOMOLOG: a member of a pai r of homologous chromosomes. HOMOLOGOUS CHROMOSOMES: chromosomes that p a i r with each other at meiosis, with corresponding gene l o c i consisting of 178 the same kind of genes. Each member of such genes i s c a l l e d an a l l e l e . D i s t i n c t phenotypes may be determined by dif f e r e n t a l l e l e s at one locus. HOMOLOGOUS PAIRING: a process that occurs a f t e r r e p l i c a t i o n where a replicated p a i r of homologous chromosome aligns side by side. METAPHASE: an intermediate stage of nuclear d i v i s i o n when chromosomes a l i g n along the equatorial plane of the c e l l (referred to as equatorial alignment i n the t h e s i s ) . PHENOMENOGRAPHY: a research approach to "study the q u a l i t a t i v e l y d i f f e r e n t ways i n which people experience and conceptualize the world around them. The experiential perspective i s one of the basic features, various aspects of r e a l i t y and various phenomena are described i n terms of the d i f f e r i n g ways i n which they appear to people." (Lybeck, Marton, Stromdahl and Tullberg, 1988, p . 5 H - This research approach can also extend to include an investigation of the process leading to the construction or application of knowledge such as the problem-solving approach i n t h i s study. PROPHASE: the early stage of nuclear d i v i s i o n during which chromosomes condense and become v i s i b l e . ~ REPLICATION: DNA synthesis (centromeres do not s p l i t ) Telophase: the la t e stage of nuclear d i v i s i o n when daughter nuclei re-form. WILD TYPE: the genotype or phenotype that i s found i n nature or i n the standard laboratory stock for a given organism 1 0 . References: G r i f f i t h s , A.J.H.' et a l . (1993). An Introduction to Genetic Analysis, New York, Freeman.' . Marton, F. (1981). Phenomenography - Describing conceptions -of the world around us. Instructional Science, 10, 177-200. Lybeck, L., Marton, F., Stromdahl, H. & Tullberg, A. (1988). The phenomenography of the 'mole concept' i n chemistry. In P.Ramsden (Ed.) Improving Learning-. New Perspectives. London: Kogan Page. 179 APPENDIX B (PART ONE) L I F E CYCLE DIAGRAM OF NEUROSPORA (SIMILAR TO ASCOBOLUS) NEUROSPORA M e i o s i s Figure B - l 1 * : L i f e cycle of Neurospora. {Ascobolus i s si m i l a r but develops an open cup-shaped apothecium, not a perithelium) 1 1 Reference: Burnett, J.H. (1975). Mycogenetics. London: John Wiley & Sons. 180 APPENDIX B (PART TWO) SOLUTION TO THE ASCOBOLUS PROBLEM12 Case One Cross + c f x h 0 + Students should i) know that abortion i s due to lack of chromosomal material; i i ) know that blacks must have both +'s; i i i ) i n f e r o r i g i n of the f i r s t d i v i s i o n non-disjunction, and iv)know that haploid Ascobolus becomes a temporary d i p l o i d and r e p l i c a t i o n of chromosomes occurs. Non-disjunction at f i r s t d i v i s i o n , Spindle fibres f a i l to p u l l the chromosomes to opposite poles Upper c e l l has a l l the chromosomal material, while the lower c e l l has no chromosomal material O O o o Product., cells undergo further mitosis to form eight products The solution was drafted by the genetics professor who set the same problem for the 1993 f i n a l examination of the course. Intermediate steps between each d i v i s i o n are not shown. 4 181 Case Two: Students should know i) that there i s no loss of chromosomal material (no abortion); i i ) note the presence of ++ and bf combinations i n the data set and i i i ) should be able to i n f e r that an obvious way to explain the data set i s a crossover. -4- -9 o + \ ^ 1 W O Crossover occurs between non-sister chromatids before f i r s t division Products of f i r s t division o CD +f © +f © bf © bf © b+ © b+ Products of of second division which w i l l further divide by mitosis to form eight products 182 Appendix C RESULTS OF THE 1994-1995 CLASS SURVEY USING THE MEIOSIS TASK Table C-l Frequency and Percentage of each type of meiosis models i d e n t i f i e d 13 from the 1994-1995 class survey and the ten students interviewed Types of Models Findings i n the Class Survey/ Interviews 'Canonical ' Models 7 0 (19.2%) 'Event-Modi f i ed ' Models (1) Pa i r ing before 26 rep l i ca t i on (7.0%) (2) Homologs a l ign at 9 0 degrees 16 to centra l axis (4.4%) (3) Repl icat ion without pa i r ing 72 before f i r s t (19.7%) d i v i s i o n ; and, pa i r ing of homologs in second d i v i s i on 'Whole-Process Modif ied' Model 80 Single nuclear d i v i s i on (21.9%) Splitting of centromere 47 in f i r s t division (12.9%) Fusion of homologs 18 (A with a; B with b) (5.0%) before f i r s t division Pairing without 24 replication before (6.6%) f i r s t division Replication in 12 second d i v i s i o n (3.3%) Types of meiosis models identified in both the survey and the thesis are put in bold print. 183 APPENDIX D SAMPLES OF COMPLETE INTERVIEW TRANSCRIPTS AND FLOW CHARTS OF STUDENTS' PROBLEM SOLVING APPROACHES The 'Direct Success' Approach: Sharilyn's Example 1 Patrick: Let us focus on this question. S h a r i l y n : F i r s t , I always write down the d e t a i l s even though they are in the question. I always write them down again because for 5 some reasons, they help me to see the d e t a i l s -- the symbols -- because lots of times I forget, so I write them down. Mutation f produces fawn, so I write f equals fawn, i t i s the gene to the right of the centromere, so I just draw the chromosome 6 and then ju s t to the right, I put the f there. 10 Mutation b = beige, act u a l l y just to the l e f t of the same centromere, so that would be b (so she drew the b gene just to the l e f t of the centromere on the same chromosome 6). In a cross of fawn by beige parents, most octads show four fawn and four beige. I just mark down what the symbols are, this 15 dark one i s the wild type (she marked i t by wt). V e r t i c a l l i n e i s fawn and horizontal li n e i s beige and empty equals dead (so she mark the fawn ascospore by f and the beige by b). Provide explanation to these exceptional octads (so she i s reading through the question again). 20 Patrick: What do these symbols (i . e . , +, f) mean to you? Sharilyn:The + means.wild type, and the f means the mutant a l l e l e . Okay, this i s the double mutant bf, so what 25 probably i s going on here. I am just t r y i n g to write i t out. This is b,f, they are very close together. One i s +f and one is b+(the parents), there must be a crossover (see Figure C-l) occurring somewhere between the b and f to get the double mutant. My picture 30 (Figure C-la) isn't right because the centromere here isn' t i n between the two genes (+ and f, b and +).(She redrew the diagram). This i s the centromere and these are the homologous pairs and these are the chromatids (see Figure C-lb). The centromere has to. be i n the middle, this 35 i s the +f and this i s the b+, so a. crossover occurs here (Figure C - l c ) . Here you get this result -- +f which i s this one, then you get ++ because of crossing over and then you get the bf the double mutant of the same reason and then you have the b+. I got four types of cell's i n equal number 40. (Figure C-ld) . . . ' •>' >' 184 Redrawn to place -+• -y- centromere i n the -\- 4~ () \ h ' r Z centre > l; 0 '( fusion of parents to U •+ form a " d i p l o i d " state (a) (b) -t- £ 1 n * — \ o * — - ' r ^ - ^ - t - 1 — 0 — * ~ (c) (d) ) X Figure C - l : Sharilyn's i l l u s t r a t i o n of crossing over for case two of the Ascobolus problem Patrick: How can you use your meiosis model to explain t h i s l o g i c here ? Sharilyn:What 1s going on i s that these two (parent) c e l l s are fusing, 4 5 one with +b and one with f+ because they are haploid, they have to fuse i n order to get meiosis. They are haploid because they are Neurospora. Patrick: Why do' you use two l i n e s to represent the chromosomes in the 50 parents? Sharilyn:I am just representing these (parents) as having replicated (Figure C-2a). I honestly don't know whether they (parent chromosomes) r e p l i c a t e before or after they fuse. And then i n 55 t h i s c e l l , now you have the homologous'pairs, and this i s chromosome s i x . This (prophase I) is l i k e the one (cell) I draw here (Figure C-2b). This i s metaphase I where they (the chromosomes) l i n e up, and then a crossover occurs i n metaphase. A crossover occurs here (Figure C-2c). So what you 60 end up with i s +f,++,bf,b+. These (chromosomes) are s t i l l l i n e up but they have crossover and um, and then these spindle attach and p u l l these homologous chromosomes (+f/++, bf/b+) apart (Figure C-2d) where you have two c e l l s with + f/.+ +, bf/b+ (prophase 2) (Figure C-2e) . Then metaphase 2. 65 They (+f/++ i n one c e l l and bf/b+ in another c e l l ) l i n e up on the axis (Figure C-2f), they divide -- the spindle attach they divide l i k e that and end up with four c e l l s after cytokinesis where you get +f, ++. These (spores) are linear I guess, so I should have drawn i t as i n an ascospores. You 185 70 also get bf and b+ and they l i n e up i n an as c i (Figure C-2g).Then they undergo mitosis and become the octad (Figure C-2h). fusion to form prophase I metaphase "I Anaphase I (b) - (c) (d) Figure C-2: Sharilyn's i l l u s t r a t i o n of crossing-over for case two of the Ascobolus problem 186 Patrick: How about the f i r s t case i n the question? 75Sharilyn:Okay, i t ' s harder to do i t . I recognize the pattern. It looks l i k e a non-disjunction or maybe that's i t . The pattern just looks familiar. I know that i n non-disjunction, you get l i k e -- depending on when the non-disjunction occurs.You either get a pattern l i k e t h i s there are four c e l l s , four that 80 are the wild types and the other four are aborted. Patrick: What i s the other type of non-disjunction? Sharilyn:I think -- I have to try. The other one would be um, s i x 85 . (spores) that were normal and two (spores) that were aborted. That's p a r t i c u l a r l y when I could recognize a pattern so I would think of what sort of things would make up the pattern and I remember about non-disjunction. 90Patrick: Go ahead. Can you go on with your meiosis model again? Sharilyn:So, the (parent) c e l l s fuse. Um, you have +f, b+, these fuse (Figure C-3a). I'm not sure whether they replicate here or after they fuse (she then put a ? there). These (+f/+f and 95 b+/b+) are paired and then they go to this (prophase 1)(Figure C-3b) and then they go to metaphase 1 (Figure C-3c) which i s exactly the same as before (case 1). Um, this (stage after metaphase l ) i s supposed to be when they (the chromosomes) are pulled apart (anaphase 1). But i n th i s 100 case, they (the chromosomes) don't,they both go to the same pole.instead of becoming separated. So, you end up with a c e l l here that has +f/+f and b+/b+ and nothing i n that one (Figure C-3d). This (empty c e l l after meiosis I) would just end up with two zeros. These ( + f/ + f and b+/b+) w i l l l i n e up 105 at metaphase 2 (Figure C-3e) and then they are pulled apart (anaphase 2). This .is anaphase, +f, b+, they are pulled apart (Figure C-3f) and you end up with 2 c e l l s (Figure C-3g)., you get +f/b+ (in each one of them) and they undergo mitosis (Figure C-3h). You end up with four wild type because they 110 complement each other (the two +'s i n +f/b+). The two +'s are complements to each other. They are s t i l l haploid. Figure C-3: Sharilyn's i l l u s t r a t i o n of non-disjunction for case one of the Ascobolus problem Sharilyn only showed one of the two cells that got a l l the chromosomal materials. She did not show the two aborted cells after non-disjunction. 188 Patrick: Why are the daughter c e l l s haploid? Sharilyn:They are s t i l l . h a p l o i d because they behave l i k e a haploid. When they divide again, they would fuse and divide i n the 115 same manner as the parents. And also because these (+f and b+) won't pair i n meiosis, so they are n+1 set of chromosome, because they have an extra one, but not the 2n (diploid). Consideration of all written representations Consideration of all symbolic representations Formulation of reasonable hypothesis Checking & monitoring responses Successfully solved problems 1. position of f and b genes (lines 7-12) 2. relevant symbols of genes (lines 14-17) 3. relevant genetic make-up of spores (lines 7-8, 10 & 23) Case One 50% empty, aborted spores (lines 75, 79-80) it MI non-disjunction HneJZfi} Meiosis for haploids & mitosis (lines 93-111 & Figure C-3) Case Two Viable double mutants (line 24) crossing over (lines 28-29) Meiosis for haploids and mitosis (Lines 52-67 & Figure C-2) YES Figure C-4: Sharilyn's example of the 'Direct Success' Approach 189 The 'Direct Success 7 Approach: L a i l a ' s Example lPatrick: Let's look at this r e a l i s t i c genetics question. Say aloud your thoughts as you are going through t h i s question. L a i l a : Okay, (she read the question and drew a diagram showing the 5 cross between the two parents +f x b+. She also drew a chromosome with the b and f mutant genes to the l e f t and right of the centromere).Okay,(she then looked at the diagram) horizontal i s beige (she labelled the ascospore 10 beige) v e r t i c a l i s fawn (again she labelled the ascospore fawn) empty c i r c l e i s aborted and the double mutant i s bf, and wild type i s black. So, you got +f and b+ (she wrote down +f x b+ again to start the cross). 15Patrick: What do you mean by +f and b+? L a i l a : So, f i s fawn, b i s beige, so wild type (she wrote down the word black beside the wild type ascospores), and (she looked at the empty circ l e s ) these are blank, aborted (she 20 wrote down the word aborted beside the empty c i r c l e s ) . Okay, I guess, meiosis i s needed for th i s question. So, okay what happens here we have that (she drew the duplicated version of the +f and b+ each with two separate centromeres to start with). Okay, we are ta l k i n g about Neurospora, so th i s i s n 25 (haploid). So, they (+f and b+) replicate, and they (duplicated chromosome) pair up, right? So, you got wild type (in case one) and there's a non-disjunction (when she looked at the empty ascospores again). 30 Patrick:Why do you think i t ' s a non disjunction? L a i l a : Oh, because I know non-disjunction gives you monosomies who die. Well, i n humans, monosomies are l e t h a l , so they die. Monosomy i s l e t h a l . Monosomy i s one chromosome less, and 35 chromosome doesn't appear here (in the empty c e l l s ) . I t ' s not here. With Neurospora, we don't c a l l them (the organism) monosomy, I guess, because they don't have a pair (of chromosomes). And so, non-disjunction. They (The chromosomes +f and b+) replicate and they pair, right? (She paired up 40 the duplicated chromosomes +f/ + f, b+/b+).. Something i s wrong, I am not doing this right. (She recognized that the # of centromere within each duplicated pair should be one). Yea, you see, I do i t again (she merge the centromere)(Figure C-5a). So,you have t h i s (+f/+f) going to 45 one (cell) and that (b+/b+) going to the other ( c e l l after the f i r s t division)(Figure C-5b). And then, you got after s p l i t t i n g i n M i l , one, two, three, four (cells) 190 50 55 right? (Figure C-5c) But I guess -- and then they (the four c e l l s ) (Figure C-5d) replicate by mitosis and then you got eight c e l l s (Figure C-5e).Now, i n thi s case(one),what happens i s non-disjunction, so both (duplicated pairs +f/+f, b+/b+) went over there (up to one c e l l ) , they (the +f/+f, b+/b+) s p l i t and +f, b+ came together, so they are a l l black. But nothing was here (in the bottom c e l l ) , so these are a l l empty. (a) (b) Figure C-5: L a i l a ' s i l l u s t r a t i o n of non-disjunction of case one f o r the Ascobolus problem 191 L a i l a : 60 65 And then here, you have two bf (double mutants i n Case 2) so you must have recombination. Yep, that's r i g h t . (She then drew the products +f, +f, ++, ++, bf, bf, b+, b+). Okay, we have recombination that gives us ...(she drew the +f/++ bf/b+ after crossing over)(Figure C-6a) and then here, normal disjunction (normal segregation) +f/++ (to one cel l ) and bf/b+(to the other cell)(Figure C-6b) and then they (+f/++, bf/b+)' separate (Figure C-6c)and they ( + f, ++, bf, b+) replicate by mitosis (Figure C-6d) so that you get these two (+f +f) here, these two (++ ++) here, these two (bf bf) there and these two (b+ b+) over there ( a l l the eight products). Patrick: How about the events before crossing.over? 70 L a i l a : Oh! They (The chromosomes) replicate and then pair. 4 e-4 G-b -©--t—e-4 -- 4 -— t -f--1--t -4-4r - i — Mitosis Figure C-6: Lai l a ' s i l l u s t r a t i o n of crossing-over for case two of the Ascobolus problem 192 Consideration of all written representations Consideration of all symbolic representations Formulation of reasonable hypothesis Checking & monitoring responses 1. Identify relevant symbols of spores (lines 9-12, 18-20) 2. Identify position of f and b genes (lines 6-7) Case One Empty spores (line 28) Non-disjunction ne 27) Meiosis for haploid and mitosis (lines 44-50 & Figure C-5) YES Case Two [Double mutants bf (line 56) /l^ ecorn Recombination or \C4^ssingover (line 57) Meiosis for haploid and mitosis (lines 58-71 & Figure C-6) Successfully solved problem Figure C-7: L a i l a ' s example of the 'Direct Success' Approach The 'Hypothesis-Formulation and Reframing' Approach: May's Example 1 Patrick: Let's look at thi s question. Go through the question and say aloud your thoughts as you are reading through th i s question. 5 May: 10 So f i r s t step I write down the symbols that are i n the 5 question. (She wrote down • = normal, O = fawn and 6 = beige) . The ascospores are normally black,, so I write down a black ascospore to represent a normal. A mutation f producing fawn ascospores , so f i s a mutation (so she wrote down f: mutation (fawn)-- producing fawn i s a gene just to the right of the centromere (so she drew a chromosome with the f mutation to the right of the centromere).Um, on 193 chromosome 6, whereas mutation b producing beige i s a gene just to the l e f t of the centromere (so she drew 15 mutation b to the l e f t ) so b here and f here ( l e f t and right of centromere). In a cross of fawn by beige parents, fawn times beige (so she drew the cross),most octads show four fawn and four beige. Empty c i r c l e s (in case 1) are dead. Okay, so the f i r s t case you get four normal and four dead, 20 you know somehow that there was crossing over between the two parents, so you end up with a ++ and a bf. Patrick: Why do you think i t i s a crossing over? 25May: I don't know. I just quickly thought about i t , maybe double mutants (bf) are l e t h a l but then i t might be some sort of abnormal meiotic d i v i s i o n where both (of these duplicated chromosome set +f/+f, b+/b+) go to one side. I don't know • what i s the name of the process. 30 Patrick: Before you are going to explain (case 1) to me, could you t e l l me the meanings of these symbols +f, b+? May: For the fawn (parent, +f), the + means the wild type for 35 the b mutation, so,that i t ' s not beige, but i t has mutation for the f, and so i t has a fawn phenotype. Patrick: Are the parents haploid or diploid? 40May: They are haploid, because there i s only one a l l e l e for each kind of mutation (f and b). Patrick: But how about i f i t i s a diploid? How would you write down the genotype then ? 45' May: Um, you probably get small f small f (ff) for the fawn parent here. You probably get two a l l e l e s . Okay. (She went through.her o r i g i n a l meiosis model).So that i s the +f/+f and the b+/b+ (duplication)(Figure.C-8a). I 50 think i t ' s the case where both (+f/+f, b+/b+) both go to the same pole and then the other pole you get nothing (Figure C-8b). Patrick: Why do you suddenly s h i f t your thoughts from .crossing over? 55 . May: I just notice here that the double mutants (bf) here (in case .2) i s diffe r e n t ( s t i l l viable, not dead as i n case 1). So double mutant i s not l e t h a l . 60Patrick: How about i f i t i s lethal? 194 May: I t s t i l l probably won't be because then you get (from c r o s s i n g over of the parents i n case 1), +f, ++, bf, b+, you get four d i f f e r e n t types of c e l l s . So I thi n k what happens i s instead of normal d i v i s i o n -- sort of separating l i k e t h i s --65 l i k e one up here (+f/+f or b+/b+) and one down (+f/+f or b+/b+), both (+f/+f and b+/b+) go up so that i n one c e l l you end up wit h both (+f/+f,b+/b+), so you get two sets of chromosomes i n one c e l l (+f/+f, b+/b+) and then the other c e l l you get nothing here (Figure C-8b). I think what 7 0 happens i s that you get complementation ( a f t e r the chromatids separate and the +f and b+ go to one c e l l ) ( F i g u r e C-8c), l i k e that the w i l d types ++, so that you get normal. And here these are l e t h a l , that means they have no chromosomes i n s i d e them. And then you get one d u p l i c a t i o n by m i t o s i s , so you get 75 the four products of meiosis -- l i k e that -- end up with eight ( c e l l s ) (Figure C-8d). f 4— 4-b, 4 i f -4- O -V— x4 -irr- o —t-r k4r (b) (d) F i g u r e C - 8 : M a y ' s i l l u s t r a t i o n o f n o n - d i s j u n c t i o n f o r c a s e o n e o f t h e Ascobolus p r o b l e m May: For t h i s one (case 2) you end up with fawn, normal, double (mutant), beige. So, t h i s one looks l i k e c r o s s i n g over, because you get four products. 80 P a t r i c k : Why four products means cros s i n g over? May: Oh, four d i f f e r e n t products, you get the two par e n t a l ones (+f, b+) and then the two non-parental ones (++, b f ) . So 85 t h i s i s the answer f o r that one (case 2), you get fawn, normal, double (mutant) and then-beige. 195 -r-~ ^ — ; r . f b. -t • - \ -i -Figure C-9: May's i l l u s t r a t i o n of crossing over for case two of the Ascobolus Problem Patrick: Can you show me how you get this by diagram? May: I guess the crossing over i s on either side of the centromere 90 (so she drew the crossing over diagram between the two parents)(Figure C-9a), crossing over can be here or there (l e f t side or right side of the centromere). Then these two separate (+f/+f and b+/b+, Figure C-9b). And these (+f/+f and b+/b+) separate again (Figure C-9c), so you end up 95 with +f, which i s fawn, ++ which i s the wild type, bf which is the double mutant, b+, the beige (Figure C-9d). And then the four c e l l s duplicate by mitosis to end up with eight. 196 Understanding and consideration of all written representations Partial consideration of symbolic representations 1. Label the spores (lines 6-8) 2. Identify position of f and b genes (lines 8-16) 3. Identify relevant genetic make-up of spores (lines 34-36) Case One: 1. Only considered the presence of dead aborted spores 2.. neglected the viability of double mutants in case two Formulation of initial hypothesis Checking and monitoring responses (I) Reframing the original hypothesis Checking and monitoring responses (II) Successfully solved problem T Crossing Over"" Qine_2p)_ 1. Reconsider other symbolic representations: double, mutants in case two are viable (lines 56-58) 2 . Using genetic knowledge: crossover forms four types of products (lines 61-63) Jon-disjunction lines 27-28, 64J Meiosis for haploids and mitosis flines 48-52, 63-69, 74 and Figure C-8) YES Figure C-10: May's example of the 'Hypothesis-Formulation and Reframing' Approach. The 'Hypothesis-Formulation and Reframing' Approach: Kamyar's Example 1 Patrick: Let's go through th i s more r e a l i s t i c genetics problem. Say aloud your thoughts as you are going through th i s problem Kamyar: Um, I guess i t ' s important to know that th i s i s the 5 Neurospora. Neurospora i s haploid, so just keep that i n mind. Um, they are normally black and then'mutation.... Those mutations are going to end up with different colours. So, mutation i s i n chromosome si x . So, okay, I usually draw the chromosome with the centromere, so this i s mutation b, this 10 i s mutation f. Mutation b producing beige ascospores where 197 15 20 thi s gene i s just to the l e f t of the centromere. A cross of fawn by beige parents -- the +f and the b+ (He labelled the fawn spores as O and the beige spores as 9). Most octads show four fawn and four beige, but three rare octads.'.'. are shown below. (He then looked at case 2). Okay, these empty c i r c l e s represent the aborted ascospores (he labelled them aborted) . Okay, um -- you are making t h i s cross, you are expecting these genes at a distance according to the question, so you are doing th i s crossover. Patrick: How do you know that the genes are a certain distance apart? Kamyar: Well, i t ' s -- they are saying that the f mutation i s to the right of the centromere and b to the l e f t . I am also seeing 2 5 that these empty c i r c l e s are aborted, so I am thinking maybe crossover to cause these end up with. Um,these are double mutants,because of double mutants (in case 2) so they (empty aborted spores) wouldn't be. ... (puzzling). 3 0Patrick: So, what i s your question right now, after you see the (words) double mutant i n case 2? Kamyar: After I see the words double mutant; I think to myself that the double mutants can be explained by the crossing over. 35 So, what about these aborted ones? They are not viable,so they're probably um, the aborted ones could be... Patrick: 40 Kamyar: 45 Let's do case 2 f i r s t , to solve this? Do you have to use your meiosis model What happens here i s you end up with fusion of the two parents (Figure C - l l a ) . What we get after t h i s (fusion) i s again duplication.I am going to attach my centromere and not to duplicate the centromere.So you have a crossover here spread over here between these two (chromosomal) strands (inner +f and b+)(Figure C - l l b ) . Patrick: Why do crossovers have to take place between these two strands? 5 0Kamyar: 55 That's always the question I ask. I don't know that. Okay, so then I am going to draw the products of the crossover. You have +f, ++,and then you get a bf and then a b+. After you go through t h i s , t h i s i s the f i r s t d i v i s i o n (Figure C - l l c ) , then you go through mitosis,that they double everything (Figure C - l l d ) . 198 Patrick: How about the second stage of meiosis, do they have to go through the second stage of meiosis i n thi s case? Kamyar: What happens i s -.- but then you started out with haploid, 60 parents.You double the number of chromosomes and then'you've gone back to that haploid. Neurospora, start with haploid, and you are going to end up the same haploid state i n the ascospores. To me, i t seems that i t Neurospora i s not going through this b i t (Mil) of meiosis. These four ascospores now 65 go through mitosis,so then you end up with two of each (+f,+f,++,++,bf,bf,b+,b+) which are exactly of what we have here (ascospores shown i n case 2)(Figure C - l l d ) . Parents — \ © H X 2 O r (b) (a) + f + f 4~ + + r - — e — i i — t — + + •4 b \t L i bf -t bf b+ b+ Figure C - l l : Kamyar's i l l u s t r a t i o n of crossover for case two of the Ascobolus problem Patrick: Let' s go to case 1 70 Kamyar: Half of them are aborted. Patrick:What do you mean by abortion? Kamyar: They are not viable -- could be deletion. 199 75 Patrick:What do you mean by deletion? Kamyar: Part of the chromosome i s missing. But i f deletion happens, a l l are dead, i t ' s not just half (that are dead). 80 Patrick:Could you use your own meiosis model to explain case 1? 85 90 95 100 Kamyar: What happens is probably non-disjunction. When I look at my model again, I just remember that this (case 1) can be solved th i s way. Um, i t ' s non-disjunction. Patrick:At which stage of meiosis? Kamyar: I guess what happens i s this i s meiosis I here. And meiosis I, these two go on top (+f,+f) and these two go to the bottom (b+,b+) (Figure C-12b) and then i n meiosis II, t o t a l l y separate, which means they become four d i f f e r e n t things (+f, +f, b+, b+ i n four c e l l s ) . Now, i f a non disjunction occurs, both of them are going to end up on top, and nothing ends up at the bottom. So, with non-disjunction, you have nothing for the bottom c e l l (after meiosis I ) . I draw something l i k e t h i s +f +f b+ and the b+ and t h i s i s aft e r meiosis one (Figure C-12b). I get that (meiosis I) into two d i f f e r e n t c e l l s . I n meiosis II, I get these (+f, +f) going to the top c e l l and these (b+, b+) to the other. And then th i s one (empty c e l l a f t e r meiosis I) gives r i s e to two empty ones (cel l s after meiosis II)(Figure C-12c) They go through mitosis to give r i s e to four empty ones ( c e l l s ) . The four normal c e l l s , they are going through mitosis and produce two that are +f,+f and two with b+,b+ (Figure C-12d). Parents -t "t 4 — « — r b (a) Figure C-12: Kamyar's i l l u s t r a t i o n of non-disjunction for case one of the Ascobolus problem 200 Understanding and consideration of all written representations Partial consideration of symbolic representations Formulation of initial hypothesis Checking and monitoring responses (I) Reframing the original hypothesis Checking and monitoring responses (H) Successfully solved problem 1. Identify position of f and b genes (lines 9-11, 23-24) 2. Identify relevant symbols of spores (lines 12-13, 16-17) Case One: 1. Only considered dead aborted spores (line 25) |2. neglected the viability of double mutants in case two 1. Reconsidered other symbolic representations: double mutants are viable in case 2 (line 34-36) 2. Using genetic knowledge deletion led to formation of all dead spores (lines 77-78) Jon-disjunction (lines 82, 84) Meiosis for haploids and mitosis (lines 88-104) NO Figure C-13:- Kamyar's example of the 'Hypothesis-Formulation and Reframing' Approach 201 The 'Hypothesis-Formulation and Rejection 7 Approach: Chris's Example 1 Patrick: Let's go to this more r e a l i s t i c genetics question. Say aloud a l l your thoughts as you are going through th i s question. Chris: I'm just trying to get an idea of what i t ' s asking. Well, 5 crossover happens somewhere. Well, because you can you are getting a non parental types here. Patrick: Which are the non-parental types? lOChris: Well, these ones are dead (the aborted ascospores), these are the wild types (in case 1). (He labelled the black spores as ••, dead, aborted spores as O, beige spores as G and fawn as O ) . They are the non-parental types. So, these are the parents +f and b+. Um, the only gametes that are made are 15 +f and +b. I mean thi s gamete (+f) i s wild type of the b and this + (in gamete b+) i s the wild type for the f. Now, when these (gametes) gametes fuse to form a diploid.The parents are haploid, but the gametes are fused to form a meiotic d i p l o i d . 20 Patrick: Why do you think that the parents are haploid? Chris: Because we were tol d here that (Ascobolus) i s sim i l a r to Neurospora which i s haploid. 25 Patrick: You mentioned about crossing over. How does crossing over occur? Chris: If you get one crossover here, then these products are going to be mixed. This one (crossing over) i s occurring 30 here, then you w i l l get ++, this i s going to be the wild type -- and these ones-- the double mutants. Double mutants are le t h a l , so why aren't these (in case 2) are dead? And here (in case 1), the same thing happens (crossover), but how come these aborted ones are dead. 35 Patrick: Could you go back to your own meiosis.model to explain case 2? Chris: The parents (he drew +f, b+)(Figure C-14a)-- when these two 40 (parents) cross, and then, when these (+f, b+) duplicate, then you are going to get a crossover right here (to y i e l d bf) (Figure C-14b) which i s dead. Then these two (f+ fb) are going to the top c e l l , these two (++, +b) are going down 202 (Figure C-14c). Further d i v i s i o n form f+, fb from top c e l l 45 and + +, +b from bottom c e l l (Figure C-14d).The products of meiosis (f+, fb, ++ and +b) undergo further mitosis to form the octad (he just drew the f+,f+; fb, fb)(Figure C-14e). (e) Figure C-14: Chris's i l l u s t r a t i o n of crossover for case two of the Ascobolus problem Patrick: Let's go back to case 1. 50Chris: These ones (the aborted ones) are dead. These (aborted ones) won't have a l l of them (the chromosomes), they might have some (chromosomal materials). So, there i s a crossover occurring somewhere .(along the chromosome). 55Patrick: S t i l l thinking of crossing over (for case 1)? Chris: Well, other than that, a chromosome loss. Well, not necessarily a deletion, i t could be an inversion. Like in inversion, um, these (chromosomes) w i l l f l i p around (he drew 203 60 a crude diagram)(Figure C-15), but I don't think t h i s i s probable. Figure C-15: Chris's i l l u s t r a t i o n of inversion for case one of the Ascobolus problem Patrick: Why not? Chris.- Because i f you have an inversion, you get t h i s kind of loop 65 thing,and then i f you are going to get a crossover here (between the loops), a l l the crossovers w i l l die and these ones (wild types)are going to be dead too. So, inversion i s not a ... Well, i t could be deletion. 70Patrick: Can you draw another diagram to explain t h i s ? Chris: If t h i s i s a deletion, then say you are getting t h i s diagram with a deletion here -- f gene and whatever (deleted s i t e shows up as A). Then this one (bA) and t h i s one (bA) would 75 die (after crossing over), but these ones (++,++)are the wild types (Figure C-16). Even though these (+f, +f strands) are deleted, the deleted end are joined and crossover can occur, then here the wild types are formed. So that explains the results in case one, half dead and half wild type, but there 80 should.be other (types) of crossing over. b •* bA (die) -• bA (die) ++ (Wild type) -• ++ (Wild type) Figure C-16: Chris's f i r s t i l l u s t r a t i o n of deletion for case one of the Ascobolus problem ..^  204 Patrick: What's that? Chris.- There should be another crossover between the two inner strands (b+ and +A). You get a second (type of) crossover and 85 you could get -- th i s one i s b+ (from the outer top strand) and then these (bA and +A) would die, and then th i s would be ++. But I can see that th i s (combination) i s pretty u n l i k e l y (Figure C-17). b+. Z Z Z Z bA (die) A + + +A (die) Figure C-17: Chris's second i l l u s t r a t i o n of deletion for case one of the Ascobolus problem Patrick: Why not? 90 Chris: Because we are to l d half of them (ascospores i n case 1) are dead. Well, because two crossovers could occur anywhere, l i k e you would expect.to see that in here but also expect to see thi s there, l i k e crossovers could occur random, so th i s 95 would s t i l l be b+. We are not probably seeing that (b+ i n case 1), right. So, crossover i s not probable. 205 Understanding and consideration of some written representations Partial understanding and consideration of symbolic representations Formulation of hypothesis Identify symbols of spores (lines 11-13) Case One: Checking and monitoring Rejection of original hypothesis Successfully solved problem Non Parental spores (lines 11, 13) OR Dead aborted spores (line 50) + Viable double mutants in case 2 Aborted spores not double| mutants (lines 31-34) Deletion model b+ beige spores] (lines 83-86) I^nversion model dead spores formed (lines 64-67) i Reject crossing over Reject deletion Reject inversion Dead End NO Figure C-18: Chris's example of the 'Hypothesis-Formulation and Rejection' Approach 206 " P a r t i a l Recognition and Forced F i t " Approach: Jason's Example 1 P a t r i c k : Say aloud your thoughts as you are doing t h i s question. Jason: Black i s the w i l d type (he wrote down • = w i l d type, Q = beige and O = aborted). 5 P a t r i c k : You j u s t have to go through case 1 and 3. Jason: Okay. These ones ( •, • i n the question) you see I guess two w i l d type. This one w i l l be beige (© i n the question). This 10 one would be v e r t i c a l l i n e -- fawn (Q) i n the question). Double mutant w i l l be beige and fawn (©). I am not sure about t h i s one a c t u a l l y -- how you get double mutants. I am not sure how to approach t h i s problem. This one i s n ' t quite l i k e most ascus fungus problem I used to work on. 15 P a t r i c k : Maybe the main emphasis f o r you i s not to solve the problem but to t e l l me the thoughts as you are going along the d i f f e r e n t parts of the problem. Jason: I am not sure why you have a centromere here other than the 20 f a c t that you are going to f i g u r e out map distance, r i g h t ! I mean but I see how... P a t r i c k : What i s t-he centromere reminding you then? Can you say a b i t more? 25 Jason: I am expecting map distance I am expecting no crossover i f the gene i s close to the centromere or I am expecting how t h i s (the centromere) i s helping me to f i g u r e out a map (gene). Well, you see, t h i s (statement i n the question) makes 3 0 sense: four fawn and four beige ascospores, because i f you have that four fawn, four beige, they would be l i k e these ones (6000 and (D(D(D©) , one two three four, r i g h t ? So I guess t h i s i s what they say you normally get i n a cross and these are the rare ones (cases 1 and 2). I am not sure to have 35 abortion ( i n case 1) Black i s w i l d type. . P a t r i c k : Maybe you can go through case 1. Jason: Okay. I am not sure why you get the w i l d type. I don't know 40 how the three genes (+,f and b) how you get ascospores working w i t h these three genes. I haven't seen the three gene ascospores. U s u a l l y i t i s the two genes I am working with. So, f o r t h i s one, I mean I would -- okay, + f.cross with 207 b+. I get the wild type (in case 1), obviously I have to get 4 5 the four + and the aborted -- Like t h i s question doesn't say which gene gives the aborted. Empty c i r c l e s represent the aborted. Would that be due to non-disjunction? 50 55 60 Patrick:Can you use your previous meiosis model to f i t into t h i s case? Jason: I wouldn't use that one. I would use meiosis and haploid . Meiosis and haploid -- you start with the haploid (parents). So, i n this case, what happens i s the two haploids w i l l come together to form a d i p l o i d (2n) (Figure C-19a) and (the chromosomes) w i l l replicate (Figure C-19b) and then separate (Figure C-19c). The meiosis for the haploid and there i s the meiosis for the d i p l o i d . So, a vari a t i o n of the (meiosis) model. But the way I think of non-disjunction i s to get these products. You have +,+,+,+ (spores) and four empty cells.Probably you w i l l get t h i s , the two (set of duplicated chromosomes of the two parents) w i l l go up here (Figure C-20) . r K f n n °—(— T - r— h p.st ' /division K h A-ft o — J 0-(a) (b) A- (d) Figure C-19 Jason's i l l u s t r a t i o n of meiosis for the haploids from his notes \ -v -V o o o [Figure C-20: Jason's i l l u s t r a t i o n of his s i m p l i f i e d meiosis model for case one of the 'Ascobolus problem 208 Patrick:How about case 2? 65 Jason: In t h i s case, you have f o u r groups of spores, each i n the number of two - - a n equal s p l i t , f i s fawn, b i s beige, [ f f = fawn; + = ++; f b = double mutant and bb = b e i g e : he as s i g n e d genotype t o the f o u r types of spores i n case 2) 70Patrick:What do you understand by the symbol f here? Jason: I t j u s t r e p r e s e n t s fawn. Patrick:How i s your f d i f f e r e n t from the f i n the qu e s t i o n ? 75 Jason: Umh, they are the same o b v i o u s l y . T h i s f r e p r e s e n t s the fawn here ( i n case 2 ) . I would w r i t e down i n the t e s t . But t h i s one i s the one i n t u i t i o n .... I have to w r i t e down and then I had t o say what g e n e t i c s p r o p e r t y I would, use t o e x p l a i n why 80 t h i s o c c u r s . But I. am sure f o r I had not seen such q u e s t i o n i n the textbook. P a t r i c k : C a n you manipulate t h i s model [meiosis] to f i t i n t o the data? 85 Jason: +f i s c r o s s i n g w i t h the b+.So these are (bb and f f ) probably double r e c e s s i v e . They are bb, these and the=f are f f . Then these are bf, the + are ++. So, i f to get these (four types of products i n case 2) •, you c r o s s them the par e n t s and the 90 products are 2 f f , 2++, 2bf and 2bb (Figure C-20). These are the products and they are the 1:1:1:1 and i t seems that they f o l l o w the simple Mendelian g e n e t i c s r a t i o , r i g h t ! ! You know, to w r i t e down these might get me some marks f o r t e s t s . But I am not sure why i t goes from t h e r e . D u r i n g the-95 c r o s s , so you p r o b a b l y have something l i k e t h i s . I am again not sure how the centromere i s u s e f u l i n the q u e s t i o n , r s o I am not b o t h e r i n g about i t . You pro b a b l y have +f. ( & \ 1—•) and by t h i s you might be able t o e x p l a i n these products by going through meiosis,. but I am not sure (how) and the o n l y 100. way t o e x p l a i n t h i s ( r e s u l t ) i s to have c r o s s o v e r . T h i s u -4 (. o-t™--»-*-> ) double up and t h a t i s how you get the bb (Figure C-21a), and t h i s doubles up ( * ) and t h a t i s how you get the f f ( ' ) (Figure C-21b) . . . Figure C-21: Jason's i l l u s t r a t i o n of crossover f o r case two of the Ascobolus .problem 209 Patrick:How about the + gene here (in the parents) 105 Jason: I am not sure about the + because the + and f; wild type for the locus of the f and th i s (+ type for the locus of the b (in the b+ parent) that ( + + in the product cells) but I am not sure how I get 110 there and that's often screw me up. this i s the i s the wild So you need Misunderstanding the written representations Partial/Incomplete consideration of symbolic representations 1. Ascobolus is a diploid organism 2. labelled spores as bb=beige, ff=fawn, +=wild type & o = aborted (Figures C-20 & C-21); 3. interpreted gene symbol "+" as the wild type locus of f gene in a +f parent (lines 106-107) 4. Ignored the relative position of the b and f genes by placing centromere to one side (lines 19-20, 97) Case One: T Case Two: Aborted empty spores (lines 46-47) Four types of spores with double mutants (lines 90-92) Formulation of hypothesis Force-fitting operation Non-disjunction (line 58) Prepared meiosis model on his notes (Figure C-l9) and 'reinvented' meiosis model (Figure C-20) Crossing over (line 100J Ignore crossing over used 'reinvented' meiosis model (Figure C-21) Successfully solved problem Dead End NO Figure C-22: Jason's example of the ' P a r t i a l Recognition and Forced F i t ' Approach 210 Appendix E PROPOSED INSTRUCTIONAL SEQUENCE The Concept of Ploidy Question 1: Given two c e l l s 1 and 2 below, explain the meaning attached to the labels (n and 2n) n 2n C e l l 1 C e l l 2 Answer to the above question should say that: C e l l 1 i s a haploid c e l l with one copy of two d i f f e r e n t types of chromosomes. The two different chromosomes form a set of chromosomes of the c e l l . C e l l 2 i s a d i p l o i d c e l l with two copies of two d i f f e r e n t types of chromosomes. The two copies denotes the two sets of chromosomes of the c e l l . Question 2: (Include replication) Given four c e l l s ( 3 , 4 , 5 and 6 ) . Label these c e l l s using the above terminologies. Attach meaning to your labels. 211 C e l l 3 C e l l 4 C e l l 5 a4 C e l l 6 Answer to the above question: C e l l 3 i s l a b e l l e d n for i t consists of one set of the chromosome carrying the a l l e l e A. C e l l 4 i s l a b e l l e d n for i t also consists of one set of the chromosome carrying the a l l e l e A. Though the chromosome i s replicated, i t i s s t i l l considered as ONE set. C e l l 5 i s l a b e l l e d 2n for i t consists of two sets of chromosomes carrying the a l l e l e s A and a. C e l l 6 i s l a b e l l e d 2n for i t also consists of two sets of chromosomes carrying the a l l e l e s A and a, though the chromosomes are replicated. In other words, r e p l i c a t i o n has no effect on the assignment of n to a c e l l . Genetics Terminology used i n Meiosis Question 3 Given the following c e l l ( c e l l 7) which c a r r i e s two chromosomes carrying the a l l e l e s A and a respectively. Box I i l l u s t r a t e s the pai r i n g up of the two chromosomes. Box II i l l u s t r a t e s the r e p l i c a t i o n of the two chromosomes. Label 212 the two diagrams by appropriate terms such as chromosomes, s i s t e r chromatids, homologous chromosomes and a l l e l e s A - a * i C e l l 7 Block I Block II Answer to the above question: Sister chromatids allele k: A f - -la W allele Homologous.chromosomes Block I Chromosome Block II B i o l o g i c a l Significances of Meiosis Question 4 What i s the b i o l o g i c a l significance of meiosis i n d i p l o i d organisms? Answer to t h i s question: Meiosis i s to provide genetic v a r i a b i l i t y to the c e l l s by the r e s h u f f l i n g of genes or the exchange of genetic materials. 213 Question 5 Given a parent c e l l with two homologous pairs of chromosomes carrying the a l l e l e s A,B and a,b respectively, show by diagram(s) how the exchange of genetic materials can occur. Parent A „ B A a- b a B a b — & Product Answer to the question: The parent c e l l , with two sets of chromosomes, end up i n four product c e l l s each with one set of chromosomes. In order to achieve t h i s , the parent c e l l must double i t s chromosomal materials before d i v i s i o n . The parent c e l l contains a homologous p a i r of chromosomes. Since i t i s a d i p l o i d , the two homologs come from two parents, therefore, we l a b e l the two homologs as the paternal and the maternal homologs. Exchange of genetic materials should occur between the homologs. However, a f t e r r e p l i c a t i o n , they become s i s t e r chromatids and so exchange occurs between non-s i s t e r chromatids as shown below: The Four Products Parental Recombinants Parental A fit X 6 214 Up to now, students should be able to figure out why c e l l s have to undergo r e p l i c a t i o n and homologous p a i r i n g before meiosis I. Replication i s to provide s u f f i c i e n t amount of chromosomal materials to be d i s t r i b u t e d i n the product c e l l s . P a i r i n g i s to ensure genetic v a r i a t i o n . This a n c i l l a r y question serves to clear the problems facing those students who used the event-modified models and had trouble i n the timing of the "pairing" event i n meiosis. After doing t h i s question, they should be clear why p a i r i n g has to occur and when should i t occur. The next a n c i l l a r y question i s to convince the students why c e l l s have to divide twice to form only four products. Abnormality i n the Meiotic Process Question 6 Given a c e l l at the f i r s t d i v i s i o n of meiosis. By means of diagrams, indicate the state of the intermediate c e l l (at the second division) and the product c e l l s when spindle fibres f a i l to attach to the homologous pairs i n the f i r s t d i v i s i o n . 215 The C e l l at f i r s t d i v i s i o n : 

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