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thesis entitled

USING PROBLEM-BASED LEARNING AND HANDS-ON
ACTIVITIES TO TEACH MEIOSIS AND HEREDITY IN A
HIGH SCHOOL CLASSROOM

presented by
TRACIE DIANNE KRAWCZY K

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USING PROBLEM-BASED LEARNING AND HANDS ON ACTIVITES T0
TEACH MEIOSIS AND HEREDITY IN A HIGH SCHOOL BIOLOGY
CLASSROOM
By

Tracie Dianne Krawczyk

A THESIS

Submitted to
Michigan State University
In partial fulfillment of the requirements
For the degree of

MASTER OF SCIENCE

Biological Science - Interdepartmental

2007

ABSTRACT
USING PROBLEM BASED LEARNING AND HANDS ON ACTIVITIES TO
TEACH MEIOSIS AND HEREDITY IN A HIGH SCHOOL BIOLOGY
CLASSROOM
By
Tracie Dianne Krawczyk

While the students learn how to solve heredity problems quickly, they Often do so in a
way that is disconnected from meiosis, the process that creates variation. This unit was
designed to address the issue of connection between meiosis and Punnett squares. I
developed a thematic heredity unit that included problem-based learning activities that
allowed students to work in small groups as well as activities that utilized models to teach
difficult concepts. Students were assessed using pre- and post-tests. Students were also
given pre- and post-surveys to measure their learning preferences before and afier the
unit as well as to allow them the opportunity to self evaluate their own learning. By the

conclusion of the unit, students were able to solve difficult genetics problems with little

guidance and were introduced to the connection between meiosis and heredity.

DEDICATION

Without the support of a special group of people, this paper would not be possible.

Thank you to my wonderful husband, Jason, and my sons, Jacob and Aidan. Any
success I enjoy I owe to the three of you for your continuing love and encouragement.

I promise we’ll be able to go on vacations now!

I would also like to thank my parents and in-laws: Daniel and Diane Laird, and Sam
and the late Susan Krawczyk. Whether it was helping out with babysitting or being an
open car, you were all there. I can’t thank you enough. I also wish to thank my brother,
Dan, for welcoming me into his home on weekends and in the summer so that I could
continue in the program.

Finally, I would like to thank the dedicated people in the Division of Math and
Science Education: Dr. Ken Nadler, Dr. Chuck Elzinga, Margaret Iding, and especially
Dr. Merle Heidemann. I feel blessed to have had the opportunity to work under the

guidance of such amazing people.

iii

TABLE OF CONTENTS

Page Number
LIST OF TABLES ............................................................................ v
LIST OF GRAPHS ........................................................................... vi
INTRODUCTION .............................................................................. 1
IMPLEMENTATION ........................................................................ 10
RESULTS ...................................................................................... 19
DISCUSSION ................................................................................. 43
APPENDIX A ................................................................................. 51
1. Parental Consent Form ................................................................ 52
APPENDD( B .................................................................................. 54
1. Unit pre/post test ....................................................................... 55
2. Unit pre-survey ........................................................................ 61
3. Unit post survey ........................................................................ 62
4. Scoring rubric and answer key for pre/post-test ................................... 64
APPENDIX C ................................................................................. 66
1. PBL 1: Lucky Stripes ................................................................. 67
2. Bert and Ernie Go To the Cat Breeder ............................................. 69
3. PBL 2: Graycie ......................................................................... 73
4. Punnett Square fill-in-the-blank notes .............................................. 75
5. Sockosomes ............................................................................. 77
6. PBL 3: Charlie the Cat ............................................................... 78
7. Cat Coat Color Genetics handout .................................................... 80
8. Meiosis and Fertilization .............................................................. 86
9. The Reebop Family: An Activity 1n Meiosis and Variation ..................... 91
10. Human Chromosomes ................................................................. 94
11. PBL 4: Spot the Cat ................................................................... 98
12. PBL 5: Garfield: A Tail of Two Kitties ............................................ 99
WORKS REFEREN CED ..................................................................... 100

iv

LIST OF TABLES

Page Number
Table l — Scope and Sequence of Heredity Unit ......................................... 11
Table 2 — Pre-test Rubric Sample ........................................................... 12

Table 3 — Statistical Analysis of Open-ended Questions from Pre and Post-test. . . .21

Table 4 — Results of objective questions, meiosis cluster .............................. 28
Table 5 — Results of objective questions, heredity vocabulary cluster ................ 30
Table 6 - Results of objective questions, Punnett square cluster ...................... 31

LIST OF FIGURES

Page Number
Figure l — Comparison of pre and post-test scores ............................................ 20
Figure 2 — Students correctly answering questions in meiosis cluster ...................... 28
Figure 3 - Correct student responses to questions, heredity vocabulary cluster ........... 30
Figure 4 -- Correct student responses to questions, Punnett square cluster ................. 31

Figure 5 - “I am able to correctly predict the outcome of a breeding experiment.” ...... 32

Figure 6 — “ I can explain how parents pass on their traits to their Offspring.” ............ 33

Figure 7 — “I would rather use traditional lecture and book work activities to learn
difficult concepts than models, problem-based learning, and other group

activities.” .................................................................................. 33

Figure 8 — “Participating in ‘Bert and Ernie go the Cat Breeder’ helped me to
understand the terms phenotype, genotype, dominant, recessive, homozygous,

and heterozygous.” ........................................................................ 34
Figure 9 — “Using models helps me to explain material covered in class better.” ......... 35
Figure 10 - “The Reebop activity (marshmallow creatures) was enjoyable.” ............. 36

Figure 11 — “After completing the Reebop activity, I was able to understand how
Law of Independent Assortment can lead to differences in the population.” ...... 36

Figure 12 — “I was able to explain the laws of segregation and independent assortment
after completing the group sockosome activity.” ...................................... 37

Figure 13 — “After participating in the meiosis game (hats), I was able to explain how the
law of independent assortment can lead to differences between individuals in the

population.” ................................................................................. 37
Figure 14 — “Problem based learning was challenging.” ...................................... 38
Figure 15 — “Problem based learning was enjoyable.” ........................................ 38

Figure 16 - “Lessons designed as a problem to solve are more interesting than
traditional worksheet problems.” ........................................................ 39

vi

Figure 17 - “After working through the problem-based learning activity about Charlie
the cat, I was able to predict genotypes and work through Punnett squares to make
predictions about offspring.” ............................................................ 39

Figure 18 — “After completing the PBL activity about Spot the Cat, 1 was able to
predict genotypes and phenotypes of co-dominant traits in breeding
problems” ................................................................................... 40

Figure 19 — “Alter completing the PBL activity about orange cats, I was able to
predict genotypes and phenotypes of sex-linked traits in offspring in
breeding problems” ........................................................................ 40

Figure 20 — “Alter completing a think-pair-share activity, I could explain the concept
being taught.” ............................................................................... 41

Figure 21 — “After completing an activity, I am able to take what I learned and
complete analysis questions with little or no assistance” ............................. 42

Figure 22 — When answering questions for the test, thinking about the activities I
completed during the unit helps me remember the answer.” ........................ 42

vii

INTRODUCTION

Genetics, while fascinating, is a complex and abstract field of biology. Most aspects
of this discipline are very challenging to ninth graders learning it for the first time.
Meiosis and heredity rank in the top five of important concepts that are most difficult
biological concepts for students to master (Finley, et al., 1982). Understanding the
nuances of genetics requires both an application of p1ior knowledge and dedication. The
focus of the study reported here was to teach how traits are passed on through generations
by an integration of meiosis and heredity using hands-on activities and problem based
learning opportunities. The students at the focus were ninth graders taking general
biology.

While common knowledge to most now, the scientific basis of heredity was many
years in development. As early as 4000 BC, people were using applied genetics to
improve the quality of crops and domesticated animals. Ancient Hebrew laws indicate
that this knowledge was applied to humans as well: baby boys born to women who bled
excessively were exempted from circumcision. Boys of fathers who bled excessively
were not; indicating that people of that time recognized what is now referred to as X-
linked traits (Winchester, 1977).

An understanding of the mechanics of heredity came much later. Aristotle wrote that
children could look more similar to a grandparent than a parent, an idea that disagreed
with the Hippocratic view that traits were inherited fi'om all parts of the body. Yet, he
did not understand how this could happen (W hitehouse, 1965). William Harvey later
proposed that, like birds, mammals must produce eggs as well in the body of the female.

Anton van Leeuwenhoek described the presence of “animalcules” in human semen,

which were suggested by some to be parasites found in that particular sample.
Leeuwenhoek disproved this idea by demonstrating that the small, wriggling organisms
were found in the semen of every man in the study and several other types of animals as
well (Winchester, 1977).

For many years after, there was much debate about the role that each parent plays in
the development of an embryo. Jan Swammerdam hypothesized that each sperm
contained a preformed “person” and that the woman’s role in the development of the
embryo was to provide a protection and nourishment in the womb. Pierre Louis Moreau
de Maupertuis instead supported the role of epigenesis and, in a throwback to earlier
ideas, suggested a blending of semen fi'om each parent. Each parent’s semen would have
characteristics for each body part and, thus, the resulting offspring would have two
“particles” for each part fi'om each parent. Some particles could dominate the
corresponding one. Therefore, the offspring may Show a characteristic found only in one
parent (Winchester, 1977). While certainly ahead of his time, it took the work of Gregor
Mendel, an Austrian monk who studied pea plants at his monastery garden, for these
ideas to become more accurate and concrete.

Mendel examined seven characteristics that affected seeds, pods, flowers, and stems
of the plant, noting that characteristics came in two forms. Over the course of 8 years, he
used some 28,000 pea plants in his breeding experiments and carefiilly recorded the
results of his crosses, studying one pair of traits at a time. Mendel found that when he
crossed plants with smooth seeds with plants with wrinkled seeds, all of the offspring had
seeds that were smooth. It did not matter which plant provided the pollen, the results

were always the same. He called the parent plants the Pl generation and the offspring the

F 1 (first filial) generation. When F1 plants were self-fertilized, the offspring (F2)
produced smooth and wrinkled seeds in a ratio of 3:1 of smooth to wrinkled. When he
repeated this test for other traits, he found the same results: the trait which “disappeared”
in the F1 generation “reappear ” in about 25% of the F2 generation. He referred to
these reappearing traits as “recessive.” Traits that appeared in each generation were
referred to as “dominant” (Cummings, 2003).

Mendel realized that while the P1 and F1 smooth seed plants must have a different
genetic makeup, observable traits appeared to be the same. Mendel described the
physical appearance of the plant as its “phenotype,” while the genetic makeup of the
plant was its “genotype.” Different genotypes could produce the same phenotype
because each plant contained two genes for the same trait, one from each of its parents.
In order for the recessive condition to reappear in the F2 generation, the F1 plants must
have the genetic information for it, but not express it. It is only when two copies of the
recessive trait are found together in the F2 generation that the phenotype is seen in the
population (Cummings, 2003).

Mendel then sought to understand the transmission of these copies, now called alleles.
Mendel reasoned that during gamete formation one allele of each gene is segregated into
separate gametes. This became Mendel’s First Law (Cummings, 2003). This explains
why the F2 generation had a 3:1 ratio of smooth to wrinkled seeds. When the F1
generation self-pollinated, either a dominant or a recessive segregated into each gamete.
If the offspring received two dominant alleles (now referred to as homozygous), the
resulting phenotype would be smooth. Ifthe Offspring received one of each (now

referred to as heterozygous), the resulting phenotype would be smooth, also. However, if

the offspring received two recessive alleles (homozygous), the resulting phenotype would
be wrinkled.’

Once he understood the mechanics of dominant and recessive, Mendel followed the
crosses of two traits simultaneously. He found when he repeated his experiments,
following the heredity of two traits, that the F2 generation demonstrated a 9:3 :3:1 ratio
for combinations of both traits. Using this information, Mendel developed his second
law, the Law of Independent Assortment. Mendel proposed that the alleles for each gene
separate into the gametes independently of all of the other alleles of the other genes
(Cummings, 2003).

Molecular biology allows a more complete explanation of how these laws work.
Meiosis is the division of nuclear material into gametes. While meiosis has many of the
same characteristics of mitosis (somatic cell division) there are two sets of division
instead of one in order to reduce the number of chromosomes by half. Each cell begins
with a firll compliment of chromosomes (2n). While mitosis produces genetically
identical daughter cells (2n), meiosis produces genetically different daughter cells that
typically have different combinations of alleles (11). During the first stage of meiosis I,
chromosomes coil and become identifiable within the cell. Every chromosome then pair
with its homologous chromosome, a structure that has the same genes on it but not
necessarily the same alleles. Each chromosome has duplicated to form two identical
sister chromatids attached together by one centromere. All four sister chromatids then
line up across the equator of the cell. During the first meiotic division, one of each
homologous pair separates into two different cells. The division of these homologous

pairs is random, leading to different combinations of chromosomes and, thus, alleles,

explaining Mendel’s Law of Independent Assortment. During the second meiotic
division, sister chromatids separate forming a total of four gamete cells, each of which
has one allele for each gene. This division explains Mendel’s Law of Segregation
(Lewis and John, 1964). When the gametes fuse together, the offspring will have the hill
complement of chromosomes in a combination that will most likely not be like either
parent.

An additional source of genetic variability is produced during meiosis I after
homologous chromosomes line up along the equator of the cell. Sometimes pieces of
sister chromatids can be switched between the homologous chromosomes. This allows
for new combinations of alleles to be possible.

Predicting the outcome of genetic crosses has much to do with predicting the
probability that a gamete will have a particular set of alleles and match up with a gamete
that has another particular set of alleles. This is usually represented as Punnett squares.
Understanding meiosis as it relates to heredity is important in understanding the
fundamentals of each and their importance. However, this can be difficult for students.
Often these concepts are presented in a disconnected way; meiosis is taught along with
mitosis and heredity is taught later on. This may lead to results like those found by
Stewart, et al. (1990): When students were asked to explain how they arrived at the
solution for genetics problems, their answers were only infrequently connected with
Mendel’s Laws or an accurate understanding of the process of meiosis. Many could not
explain why their Punnett squares were set up as they were. Indeed, many students do
not reach a deeper understanding of the concept and rely purely on algorithm to solve the

problem. (Stewart, 1982) This is not a strategy that teachers want their students to use.

Experience in my own classroom has led me to this conclusion as well. Many students
found simple Mendelian genetics and monohybrid (l trait) cross problems “easy”
because they were only expected to use an algorithm to solve the problems in the past.
However, there was a real lack of connection to deeper understanding of complex
biological concepts. As a result, many of my students have had difficulty in the past
solving dihybrid (two trait) cross problems. Stewart (1982) found this to be true as well.
Because students relied on algebraic functions to solve dihybrid crosses, they were often
confused when asked how these solutions related to gametes. Students could not explain
why incorrect combinations of alleles were incorrect. They could only identify correct
solutions. He also found that students had trouble relating associated concepts together,
such as gene-chromosome and allele-trait.

Instruction using meiosis reasoning problems has been shown to increase student
understanding and more accurately assess conceptual models they develop (Kindfield,
1994). Problem-based learning models, particularly ones that utilize the Piagetian model
of development, allow students to master concepts in a step wise fashion that encourages
ownership of learning as well as progress to increasingly more difficult problems.
(Walker, et al., 1980) Students use inductive methods to approach and solve these
problems, leading to a richer understanding of the material, not simply rote
memorization. Each new problem requires the student to revisit previously developed
concepts, relate the new problem to the concept, and then work through contradictions
between the new set of data and the previous conceptual model. This is accomplished
through deductive reasoning. Watson’s group found that students who learned genetics

using this method scored 6-10 points higher than a control group on a summative

assessment (Watson, 1991). The control group learned the same information using
traditional lecture and laboratory methods. Watson concluded, “If the objective of
instruction is to produce students capable of critically analyzing complex data, then the
instructional strategy should focus on the form of systematic sequential thought
appropriate to the analytical procedures applied in the discipline.” Such a method of
instruction is problem-based learning.

Problem-based learning involves presenting open-ended problems to small groups of
students. Teachers act as facilitators for these groups as they develop knowledge in a
particular content area (Goodenough & Cashion, 2006). A key feature of this method is
the collaboration, or cooperative learning, that takes place within the group (Evenson &
Hmelo, 2000). This method of instruction was first described by Howard Barrows (as
cited in Nendaz & Tekian, 1999), as early as 1986, and applied to medical school
curriculum. Barrows described four main objectives to problem-based learning: “(a)
structuring knowledge for better recall and application in clinical contexts; (b) developing
an effective clinical reasoning process; (c) developing self-directed learning; and ((1)
increasing motivation for learning” (Nendez & Tekian, 1999). Since then, it has been
applied to other disciplines. This instructional method takes on a cooperative learning
structure, defined as a “structured form of group work where students pursue common
goals while being assessed individually” (Prince, 2004).

By taking on this problem-based learning approach and utilizing cooperative learning
environments, student enjoyment as well as success can be raised. Research on problem
based learning as a learning strategy has found that the 12 proven benefits include not

only higher achievement, but greater use of higher level reasoning strategies, positive

attitudes toward the subject matter, positive attitudes toward teachers, and greater on-task
behavior (Lord 1994). While this method of instruction has been used for many years in
other subject matters like medicine, it is equally effective in the sciences (Watson, 1991).
Students of teachers involved in Lord’s study reported that the students enjoyed learning
from each other and looked forward to attending class. Research also supports the
community learning concept increases the depth of knowledge and understanding as
students discuss answers and learn from each other (Sharan and Sharan, 1990).

Other active learning strategies were also employed including the use of models and
“think-pair-share activities.” “Think—pair-share” is a method of cooperative learning that
allows students to process content information from a formal learning setting such as a
lecture or class discussion. Using this method, students first think and develop a written
response independently to a question posed to the class. Next, students “pair” up to
discuss their answers together. Finally, the groups share their answers with other groups
during a class discussion (D. Walker, 1998).

In this study, problem-based learning and active learning strategies, including those
mentioned above, were employed to teach a unit on genetics and heredity to ninth grade
biology students at Almont High School. Item analysis of pre- and post-test questions
was used to determine the effectiveness of utilizing these learning strategies. Student
self-assessment was also considered using data collected from pre and post-surveys
administered to participants.

Almont Community School district is located in the village of Almont, approximately
32 miles north of Detroit in Lapeer County. Almont has limited racial and ethnic

diversity, with approximately 96% of citizens being Caucasian as of the 2000 census.

Approximately 3% of the population is Hispanic or Latino, with a number of citizens
who are seasonal workers. The population is a mix of upper lower and lower middle
class blue-collar workers with a smaller population of white collar workers. Companies
that are tied to the automotive industry, either as Tier H suppliers or as automotive
assembly plants, employ most workers in Lapeer County. In Almont, the largest
employer is the school district itself. It is a small school district of approximately 1800
students grades K-12. The district has four buildings: a primary (K-2), an elementary (3-
5), a middle school (6-8), and high school (9-12). As recently as 1987, the entire school
district was contained in one building. During the nineties, the area experienced a large
population growth leading to the building of the primary and middle schools. The district
also provides services for children from the surrounding Almont Township and accepts
students from bordering districts as part of 3 “schools of choice” program. In addition,
Almont High School also hosts the Cognitively Impaired and Emotionally Impaired
special education programs for Lapeer County. Almont High School has a student
population of 601 students. The majority of students take a basic college prep curriculum
during the ninth grade. The majority of Almont graduates (60%) will enroll in a college,
university, or trade school. Others enlist in the armed forces or seek employment. The

dropout rate is approximately 4-6%. The daily attendance rate is 96%.

IMPLEMENTATION

I crurently teach Biology to nearly every ninth grader at the high school. Since I
began teaching at Almont eight years ago, I have worked each year to modify, add, and
redesign curriculum for the biology and life science programs. Recently, the eighth grade
life science and ninth grade biology classes have exchanged some of the curricula. Last
year, ninth grade biology included the genetics and heredity unit for the first time. I
chose heredity and meiosis as my focus of research in order to have the opportunity to
concentrate on the unit and sequence curriculum that would be hands-on and relevant for
my students. The students in the study were from two of the five classes of Biology
taught this year that had the highest rates of voluntary participation. Students in these
two classes represent a broad range of past scholarship and, therefore, seemed to best
represent the entire grade. Eleven of 28 students volunteered to participate in the study
from fourth hour and 16 of 21 students volunteered from fifth hour. All students who
volunteered to be part of the study read and signed a consent form (Appendix A) to allow
their data to be analyzed and reported. Each student also had a parent or guardian grant
permission for use of the data as well.

The unit’s major theme was to explain how variation might occur between two
closely related individuals using knowledge of genetics and meiosis. In order to
accomplish this goal, the unit was broken up into sections. Each section focused on a
particular topic with the goal of building knowledge to solve open-ended problems.
PowerPoint presentations were utilized for the directed instruction portions of each
section, and mainly included introductions and summative activities. Students were

largely responsible for the acquisition of knowledge through problem based scenarios and

10

short labs or group activities. In order to achieve the goal of learning both heredity and
meiosis, each section had a particular goal and correlating activities associated with it

(Table 1).

Table One: Scope and Sequence of Heredity Unit

 

 

 

 

 

 

 

Section/Topic Objectives Activities
1. Introduction Review of connections Cat hair lab;
between traits and genes Think-Pair-Share problem
2. Terminology Students will explain the Cat hair lab 2;
following terms: dominant, Bert and Ernie Go the Cat
recessive, homozygous, Breeder
heterozygous, genotype,
phenotype; Students will
recognize homologous
chromosomes
3 &4. Punnett Students will be able to Think-Pair-Share Problem;
Squares predict the outcome of a Group Punnett Square
monohybrid cross Problems;
Charlie the Cat PBL
5. Meiosis Students will describe the Sockosomes;
major events of Meiosis I Reebop building;
and H; Students will explain Humans as Chromosomes;
the laws of segregation and Human Karyotyping On-
independent assortment and Line
how they lead to variation in
gametes
6 & 7. Students will determine the Spot the Cat PBL;
Inheritance of probabilities of traits which A Tail of Two Kitties PBL
co-dominant are sex-linked or co-
and sex linked dominant
traits

 

 

 

 

The heredity and meiosis unit was developed at Michigan State University during the
summer research experience. It was planned to last six weeks of course time. All
activities, except the on-line karyotyping activity, were new to the biology course and

were compiled or developed over the course of the summer at Michigan State University.

Movement within the district science curriculum warranted the addition of this unit in the
biology course, and the unit was written to fulfill this requirement.

A pretest on heredity and meiosis (Appendix B) was administered that included both
open—ended and multiple choice questions to gauge students’ prior knowledge. Due to
the fact that open-ended questions elicit a variety of responses, answers to most of these
questions were scored on a scale of 0-3 points. A few questions were scored on a scale
of 0-2. In either case, zero points were awarded for no response or for responses that
were completely wrong and the maximum points represented a superior answer. Other
answers were rated based on correctness. Each open-ended question had its own scoring
rubric (Appendix B). A sample rubric is included here to provide an example of how
points were awarded on various questions. Questions 2-5 ask students to explain the
outcomes of a cross between two human parents. The father has brown eyes (B) and
straight hair (c). The mother has blue eyes (b) and curly hair (C). Scoring for question 2
is described below (Table 2).

Table 2: Pretest rubric sarmrle

 

 

 

 

 

 

 

 

 

Question 2: Explain why you have eyes like your father and hair like your
mother.
Points Responses
awarded

3 Describe dominant and recessive alleles to explain how
inheritance happened

2 Describe dominant or recessive alleles to explain how
inheritance happened.

1 Parents passed on different genes (for different traits)

OR mentioned “strength” of gene
0 Completely incorrect/no response

 

Multiple-choice questions were scored one point for correct answers and zero points

for all other answers. The pretest had 23 multiple-choice questions (Appendix B). Some

12

of these questions assessed prior knowledge of vocabulary while others assessed student
knowledge of meiosis and Punnett square outcomes. A complete answer key can be
found in Appendix B.

A unit pre-survey (Appendix B) on student preferences for different instructional
methods was also given to determine student preconceptions and perceptions of lecture,
lab, modeling, and problem based learning activities. This information was used to
determine attitudinal changes and preferences after completing the unit and post survey.
Students were given a choice of numbered responses to use for each question ranging
from l=strongly disagree to 5=strongly agree.

The unit itself was broken up into seven sections (Table 1). While the first few
sections were more guided, the activities became more student-centered as the unit
progressed.

Section One:

This section, entitled “Lucky Stripes” (Appendix C), introduced students to the cat-
breeding theme upon which each of the problem based learning activities would be based.
Students were given background information and photos of a striped cat, a black cat, and
their litter of two striped kittens and one solid kitten. Students were then charged with
the task of figuring out why the kittens of these two cats looked like either parent instead
of a mixture. They frrst took part in a “think-pair—share” activity, allowing them time to
think first individually, and then share their ideas with a partner, and finally participate in
a class discussion of the question. In order to get a more complete idea of the differences

between the cats, students examined, using hand lenses, the striped hairs of one cat and

13

the solid hairs of another and made sketches and written observations on the handout for
the lesson.

After discovering that the individual hairs were colored differently (striped cats have
striped hair, not patches of alternating color), Students related the information to what
they had already learned about basic protein synthesis from a previous unit. They were
then given a reading assignment fi'om their text book (Biology: The Dynamics of Life.
Glencoe/McGraw Hill) about Gregor Mendel for homework. This reading assignment
was to give them some background information in formulating a hypothesis about the
day’s problem and to give them a foundation for the next unit on vocabulary.

Section Two:

The second section, “Graycie,” focused on vocabulary acquisition, an often difficult part
of learning genetics due to the large volume of unfamiliar words that sound very similar.
Students examined the hairs of “Graycie,” a grey striped cat, and “Fizzy,” a black striped
cat, with a hand lens. Drawing fiom the previous day’s lesson on stripes and no stripes,
students were asked to conclude why one cat was lighter than the other. In order to learn
why, they paired off to read through “Bert and Ernie Go to the Cattery.” This script,
written by classmate Kelly Joos and me introduces the concepts of dominant and
recessive, homozygous and heterozygous, and genotype and phenotype. Students used
the Bert and Ernie dialogue to answer questions about the key terms and then join in a
class discussion about them. Using what they learned from the script, students then
applied the new terms to explain why Fizzy and Graycie different. They also applied this

knowledge to the section one problem to build an answer about why Fizzy and Baby’s

14

kittens look as they do. A summative assignment (Appendix C) was given for students to
complete individually for homework in order to review the terms learned.

Section Three:

The third section, “Learning Punnett Squares,” introduced the concept of predicting the
outcomes of a monohybrid cross. The instruction began by showing a Punnett square as
chromosomes so that students could relate them to chromosomes and gamete division.
Referring back to the two previous cat problems, a “think-pair-share” activity was
conducted to develop Punnett squares for heredity problems studied in sections one and
two. Students were given practice problems as homework. After students returned very
uncertain about completing the problems on their own, two days of class were spent
practicing problems: one day going over the homework problems and a second day
working on new problems during class where they had access to assistance.

Section Four:

The first student-centered problem-based activity took place during this section. The
problem, titled “Charlie the Cat,” introduced a champion black cat whose lineage was
being challenged by a competitor who did not believe that two all white cats could
produce a black cat. Students were given supplementary materials (Cat Coat Color
Genetics — Appendix C) to work through the problem and produce a Punnett square to
either prove or disprove the competitor’s claim. Using their understanding of simple
dominant and recessive, groups of two worked through the problem.

Section Five:
Section five introduced the concept of meiosis and the resulting genetic variation. The

goal was to develop a deeper understanding of heredity and why Punnett squares allow

15

one to make predictions about the outcome of breeding two organisms. Students used
models, “sockosomes” (Appendix C), for a visual and tactile way learn the major events
and outcomes of meiosis I and H. Four students played the role of “giant sockosomes”
while the remaining students worked in pairs with their models to follow along with the
PowerPoint lesson.

The second activity in the section was small group centered. Four students worked
together based on the handout “Meiosis and Fertilization” (Appendix C). Using the
sockosomes, the small groups discussed the first and second division of meiosis. A group
assessment was administered to determine whether or not group members completely
understood what they had read.

The third activity, “The Reebop F arnily” (Appendix C) brought together the
relationship between gamete formation and Punnett squares. Students were given
envelopes with chromosomes for each Reebop parent, marshmallow creatures with
several phenotypes for body features such as eyes and antennae. Students randomly
choose chromosomes from each parent pair to form a Reebop offspring genome. After
completing the genotype and phenotype of the offspring on paper, various materials were
used to construct the offspring and they were placed into the Reebop “nursery” for
comparison. Because each set of parents had the same genotype, the offspring could be
compared as possible offspring of the same cross.

To reinforce variation of gametes, students participated in the final activity of the
section, “Human Chromosomes.” Each student played the role of a chromosome with
two gene loci. The human chromosomes lined up with their homologous pairs and

completed meiosis I and II by moving to different sections of the room. The genotypes

16

of the gametes that resulted were recorded on the board for each round to compare. Once
this had been practiced a few times, the concept of crossing over was introduced by
showing the students how to switch the labels for one of the genes between two non-
sister chromatids. After Meiosis I and II, the resulting gametes were recorded on the
board to compare with the first rounds. After practicing a few more times, three
additional rounds were completed and a random gamete from each round was chosen to
fertilize a gamete I selected. Students derived the genotype and phenotype for each of
these Offspring and drew a picture of each. Finally, they completed challenge questions
to help them summarize what they had learned from the activity.

Sections 6 & 7:

Sections six and seven consisted of problem based activities that were student led.
Section six, “Spot the Cat” (Appendix C), was a problem that introduced the concept of
co-dominance. Using the resource packet and the concepts learned in the unit, students
determined how it was possible that a spotted cat could have a solid sibling. In the
seventh section, students completed a problem related to sex linked traits. Students
determined why a tortoise shell cat (a “tortie”) had a litter with such a variety of
phenotypes. After these problems, students were given an individual problem to solve
based upon one of the two completed as a team.

At the conclusion of the unit, the students were administered a two part post-test
(Appendix B). The first part was open ended and assessed knowledge of vocabulary,
Punnett squares, and outcomes of Meiosis I and II. This part was a graded assessment for
the students. However, scores from this assessment were not included in data collection

for the study. The second part was an assessment comprised of the same questions as the

17

pre-test, and was scored using the same rubric (Appendix B). This portion was not
assessed for a grade and was used for data collection for this study.

Students were also given a post-survey (Appendix B). First, these questions were
the same as those on the pre-Survey and meant for comparison to measure changes in
attitudes and preferences for different instructional techniques used in the unit. Second,
students were asked about specific activities from the unit to gauge student opinion on the
usefulness of each. Finally, students completed a self-assessment of their learning based
upon the goals of the unit (Table 1) using the remainder of the questions in the post-
survey.

As with the pre unit survey, student responses to each statement on the post unit
survey were measured from 1-5 and used the same titles as those of the pre-survey
(Appendix B). However, some of the questions were brand new, while others were
different from the pretest but correlated with a pretest question. These questions were

more specific with regard to an activity on the post-test than on the pretest.

18

RESULTS

Several assessment tools were used to determine the effectiveness of the new heredity
and meiosis unit developed for the biology course at Almont High School. The pre- and
post-tests, and pre- and post-surveys were used to make this determination along with
results of various embedded assessments (summative activities and group work).

The pre-test was administered to determine student prior knowledge and to measure
grth in knowledge as a result of the unit. The test consisted of both open ended and
multiple-choice questions. Generally, answers to open ended questions were assigned
points according to a rubric (Appendix B) on a three-point scale, though a few questions
were graded on a two-point scale. By providing more information, students were
awarded more points. Students with completely wrong answers or no response to the
question were awarded zero points for the test item. Scores for the open-ended questions
(Figure 1) were low, as expected. Average scores for each of the problems were all
below 1 point. Objective questions on the test account for the gaps in the numbering on

the figure.

19

Figure One: Comparison ofPre and Post-Test Scores On Open-ended Questions

 

2.5

 
  

2
3 1.5 , Pre-test
or Post-test
3 1
o
>
<

Question

 

 

 

*denotes question scored out of possible 2 points instead of 3

Several questions had results that were quite low for the simple reason that the
question was left unanswered by many or most students. Other questions had quite high
averages (number eight), which can be attributed to knowledge gained from the prior unit
over protein synthesis. Scores overall improved on each open-ended question on the
post-test taken at the conclusion of the unit (Figure 1).

In general, improvement occurred for each question, and the average score for test
increased. Most of the results were statistically significant based on a paired t-test
analysis. Table 3 contains the results of the statistical analysis of the Open-ended

questions.

20

Table 3: Results of Statistical Analysis of Open-ended Questions from Pre/Post-test
(N=27)

 

 

 

 

 

 

 

 

 

 

 

 

 

Question Pre-test mean Post-test mean t-value Statistically
significant at
p=0.05

l 0.22 1.37 -4.62 Yes

2* 0.89 1.74 -5.75 Yes

3 0.74 1.96 -7.53 Yes

4 0.78 1.93 -6.05 Yes

5 0.67 2.19 -12.3 Yes

6* 0.11 1.15 -7.63 Yes

8* 0.89 1.22 -1.73 No

9 0.11 1.85 -10.0 Yes

11 0.19 0.96 -3.61 Yes

14 7.41E-02 1.00 -4.22 Yes

35 0.19 0.93 -3.31 Yes

 

 

 

 

 

The first question dealt with connecting three key terms together: gene, allele, and
chromosome. The average pre-test score, 0.22 out of 3, resulted from most students not
being able to connect even two of the terms together. In fact, the students who scored
any points on this problem did so by simply describing one of the terms without
connecting it to another. On the post-test (Table 3), this score improved to 1.37 out of a
possible 3. More students were able to correctly connect at least two of the terms
together. Very few were able to connect all three without being too general to be able to
assess whether a deeper understanding had been reached. These concepts were first
introduced in the first section, the Lucky Stripes PBL (Appendix C), and were reinforced
throughout the unit. The improvement made is considered statistically significant at the
95% confidence level with P<0.05 (Table 3).

Problems 2 - 6 (Appendix B) were based around a human heredity situation. Out of a

possible two points, the average pre-test score for question 2 was 0.89 points. Student

21

 

responses earning points ofien included information about inheriting genes from certain
parents, however few of these mentioned inheriting genes from both parents for each
gene. This type of response was anticipated due to MCCF benchmarks for heredity that
expect students in middle school to learn that traits are inherited fi'om parents, and that
parents have the traits themselves. Students who mentioned inheriting genes fiom both
parents included in their responses that the “strength” of the genes makes a difference.
They thought that the genes of one parent might have more strength than that of another
parent. However, they did not mention whether this was due to the actual gene itself or
to the parent the gene belonged to. During class discussions, several students mentioned
that they thought the sex of the parent made a difference; either that the same sex parent
largely contributed to the gene or that certain genes were stronger for certain parents in
general, for instance, eye color largely being controlled by one gender of parent in every
case. This score improved on the post-test. The class average for this question improved
to 1.74 out of 2 (Table 3).

Given the results of the paired t-test, the null hypothesis was rejected, indicating that
their increase in score was due to the instruction received during the unit. Students
mentioned the terms “dominant” and/or “recessive” in their answers, demonstrating their
use of new vocabulary and its applications to new situations.

Questions 3 and 4 (Appendix B) dealt mostly with differences and similarities
between siblings and offspring based upon the laws of independent assortment and
segregation, though the problems do not specifically mention these laws by name.
Students averaged 0.74 points out of three on question three and an average of 0.78 for

question four on the pre-test (Table 3). Most students reported that it depended upon

22

which genes each parent gave, earning one out of three points, but none of the students
mentioned dominant or recessive genes being the reason why for either question. At the
conclusion of the unit, students did mention the terms dominant and/or recessive much
more frequently in both answers, resulting in an increase in the average score for the two
questions (1.96 and 1.93, respectively). In both cases, the null hypothesis was rejected
and the results of the t-test are statistically significant. However, students generally did
not use information about the law of independent assortment in their answers, as I
expected. This result was consistent with the results of the Human Chromosome Activity
analysis. The final question of this assessment asked students whether it is possible for
two siblings with the same parents to look differently (this is the same as question four on
the pre and post test). Students generally did not use information about the law of
independent assortment in their answers, even though the activity as well as the question
preceding it was about this law. Most students in this case also relied more on their
understanding of Punnett squares to answer the question or were vague in their response
(“It depends on what each parent gives”) Perhaps a better-structured question is
necessary in the future to direct student thinking, allowing for a better analysis of the
connection in students’ minds between these ideas.

Question five (Appendix B) asked if it were possible to have children with both the
same eye and hair color as one of the parents. Students averaged 0.67 points out of a
possible three, with most mentioning, again, that it depended upon which of the alleles
the parent of interest gave to the child. On the post-test, the class average for this
question was 2.19 of a possible three points (Table 3). The t—test for this question

indicates that the null hypothesis is rejected and that the results are statistically

23

significant. While most students were able to explain that dominant and recessive traits,
6 out of 27 (22%) mentioned that the child’s traits might also depend upon the other
parent due to recessive traits of the first. This answer earned 3 out of 3 points and
represented a superior answer. Like the last two questions, this question required
students to combine their knowledge of vocabulary with their understanding of Punnett
squares and probability. However, they also needed to apply their knowledge of the law
of independent assortment, at least in a broad sense, to understand that the gametes used
to create one sibling may not be the same for the other.

Question six (Appendix B) asked students to connect the process of meiosis with
Punnett squares. As expected, the score for this question was low on the pre-test, with
0.07 out of 2 points being the average. This was mostly due to lack of familiarity with
the concepts or vocabulary used in the question. Students who did earn points on this
question mentioned the word “heredity”, but did not offer any supporting evidence of
their understanding. One of these two students had transferred fiom another school
district where Punnett squares were covered in middle school and was able to offer that it
had “something to do with those boxes.” But this student could not remember what the
boxes were for or why one set them up as they are. On the post-test, many more students
mentioned the word “heredity” (13 out of 27 students). Nine more students were able to
explain either meiosis or Punnett squares. Only one student of 27 could explain how the
two were connected. Perhaps a rewording of the question would have yielded better
results. Student learning did take place, however. The results of the t-test indicate that

the null hypothesis was rejected and that the results were statistically significant (Table

3).

24

Question eight (Appendix B) asked students to explain what a chromosome was.
Chromosomes had been covered in a previous unit and many students could offer that
chromosomes were pieces of DNA. This answer earned 1 out of a possible 2 points. The
pre-test class average for this question was 0.96 out of 2, demonstrating that the majority
of the students came with some of this background knowledge. On the post-test, the
average increased to 1.15 points, showing that some students also knew that
chromosomes had genes on them. However, the improvement on this question was not
statistically significant (Table 3). The null hypothesis is accepted in this case because the
p-value is greater than 0.05. Perhaps more students were able to only give the
information learned previously than those who were able to expand on their prior
knowledge. Having learned the information once, it became the answer they knew and
additional information was not remembered.

Question nine asked about the purpose of meiosis. This concept was brand new for
these students, and the pre-test score was expected to be low. The average score was
0.19 out of a possible three, with many students mentioning that it had something to do
with chromosomes. This answer could have been influenced by the previous question,
because this score, while low, was still much higher than I expected. The post-test
average was 1.74 out of three (Table 3). The results of the t-test indicate that the
null hypothesis was rejected because the p value was well below the 0.05 at the 95%
confidence level. Answers earning two points on the post-test included “meiosis is the
process by which gametes are produced.” About half of the students assessed (13 out of
27) earned two points. Students who mentioned that meiosis also had to do with the

reduction of chromosome number earned more points, as did those who mentioned the

25

“shuffling” of chromosomes. Five of 27 students earned three points for their answer.
Again, a more directed question may have resulted in more specific answers.
Information for this question was presented during the Reebop, Sockosomes, and
Humans as Chromosomes activities (Appendix C).

Each student looked at a picture of a human karyotype in question 11 (Appendix B) to
determine which chromosomes came from the mother. The goal of this question was to
find out if the student knew about homologous pairs, the source of each chromosome of a
pair, and whether or not the student knew the difference between sex chromosomes. The
average score on the pretest was 0.19 points. Most of the students, 14 out of 27, did not
even answer the question. This score improved to 0.96 out of a possible 3 (Table 3).
Statistically speaking, the t-test resulted in the null hypothesis being rejected. Many
students were able to at least identify the sex chromosomes or circle one chromosome
from most of the pairs. While the statistical analysis indicated that the improvement on
the question was significant (Table 3), anecdotally, the question did not allow for good
results. Even after the unit was taught, many students still had difficulty answering this
question. Although the number of students who did not answer the question dropped to
three, nine out of the remaining 24 students who did answer incorrectly by circling whole
pairs of chromosomes. Other students (10 of 27) either circled only one sex
chromosome, only one homologous chromosome from each autosomal pair, or some
combination of both.

Number fourteen (Appendix B) is related to question eleven, asking students to
explain whether or not two chromosomes were homologous and why. The two

chromosomes both featured hair color loci, yet not for the same hair color. Students who

26

correctly answered this question indicated that they were homologous because they had
the same gene (for hair color) in the same location. On the pre-test, the average score for
this answer was 0.30 out of three points. The post-test average was 1.00 out of a possible
three points (Table 3). The number of students who left the answer blank decreased to 3
out of 27 students, and 15 out of 27 students were at least able to explain that they both
had a hair color gene on the chromosome, though only 7 mentioned that it was in the
same location. Results of the paired t-test do not support the null hypothesis.

Question 35 (Appendix B) asked students to explain why only the boys in a family
were affected by a particular disorder. On the pretest, the average score was 0.19 out of
3. The majority of students, 21 out of 27, left the question completely blank. On the post
test, this score improved to 0.93 out of 3. (Table 3). The null hypothesis is rejected
according to the results of the t-test. Six of 27 students left the question blank again, but
12 of the 27 students mentioned gender as a factor in their answer. Four of the 27 were
able to connect the information from the Problem Based Learning activity on sex-linked
traits (A Tail of Two Kitties PBL) to the problem and explained that the key to the
answer was the X chromosome.

The remaining questions on the assessments were objective in nature. Several were
arranged as clusters, and so will be discussed this way. Questions 12 through 18
(Appendix B) concentrated on the details of meiosis. Students scored one point for each
correct answer, and zero points for wrong or missing answers. The results of these

answers are represented in Figure 2 and in Table 4.

27

Figure 2: Students Correctly Answering Questions in Meiosis Cluster (N=27)

 

 

 

 

 

 

 

I Pre-test
DPost-test

 

 

 

 

Number of students

 

 

 

 

 

 

 

 

 

 

 

 

 

Question

 

 

 

Table 4: Results of Objective Questions, Meiosis Cluster

 

 

 

 

 

 

 

Question Pre-test Mean Post-test Mean t-value Statistically significant?
12 0.33 0.85 -4. 19 Yes
13 0.30 0.52 -1.65 No
15 0.15 0.67 -4. 19 Yes
16 0.22 0.52 -2. 13 Yes
17 0.26 0.37 -0.90 No
18 0.26 0.70 —4.00 Yes

 

 

 

 

 

 

 

Question twelve (Appendix B) dealt with vocabulary, specifically the term “haploid.”
Students had to identify the correct description of a haploid cell. The results of a paired t-
test reject the null hypothesis (Table 4).

Differences between the pre and post-test for number 13 (Appendix B) were not
significant, however. The question asked students to apply the definition of diploid to the
combination of gametes. The results of the t-test show p=0.110, far above the p value of
0.05 needed to meet the 95% confidence interval. This concept would have been covered
during the Sockosomes, Reebop, and to a smaller degree, Humans as Chromosomes

activities.

28

 

Question 15 (Appendix B) asked about the products of meiosis. This was covered
primarily during the Sockosomes and Humans as Chromosomes activities. This objective
was met. Four students answered correctly on the pre-test, and 18 answered correctly on
the post-test. The result of the t-test indicated this difference to be statistically
significant.

Question 16 (Appendix B) assessed student understanding of recombination. Students
had to pick out “mitotic division” as a process that would not result in genetic
recombination. The increase in correct responses between the pre- and post-test were
found to be statistically significant (p=0.04) (Table 4). However, when asked to explain
the law of segregation in question 17, the differences in responses were not statistically
significant to show a gain (p=0.38).

Students did seem to learn the basic steps of meiosis. Question 18 (Appendix B) asked
students to determine the outcome of the first meiotic division. Where six students
answered correctly on the pre-test, 18 answered correctly on the post-test. The paired t-
test for student data on this question shows that this increase is statistically significant
(Table 4).

The next cluster of questions dealt with vocabulary used in heredity and genetics:
genotype, phenotype, dominant, recessive, homozygous, and heterozygous. Results of

assessment from this cluster are found in Figure 3 and Table 5.

29

Figure 3: Correct Student Reponses to Questions from Heredity Vocabulary Cluster
(N=27)

 

0)
O

 

N
0|

 

 

N
O

 

 

 

gr I Pre-test
i n Post-test

 

 

 

 

 

 

 

 

 

 

 

_L
O

 

 

U'i
.,.LA- I

 

Students answering
a

 

 

 

 

 

 

 

 

 

O
l

 

 

19 20 21 22 23 24 25
Question

 

 

 

Table 5: Results of Objective Questions, Heredity Vocabulary Cluster

 

 

 

 

 

 

 

 

Question Pre-test Mean Post-test Mean t-value Statistically significant?
19 0.19 0.63 -4.00 Yes
20 0.15 0.78 -5.20 Yes
21 0.22 0.89 -6.24 Yes
22 0.26 0.93 -6.24 Yes
23 0.26 0.85 -6.15 Yes
24 0.30 0.89 -5.38 Yes
25 0.26 0.82 -5.00 Yes

 

 

 

 

 

 

 

Results from this section indicate that the differences between the pre and post-test for
the heredity vocabulary section were statistically significant, with p=0.000 for each one.
These terms were introduced in the “Bert and Ernie” activity and reinforced through each
PBL activity.

Heredity problems that were not simple dominant and recessive, such as co-
dominance, were the focus of the last cluster of questions (questions 26, 28, 29, 32, 33,
and 34) on the formal assessment (Appendix B). Results of this section are shown in

Figure 4 and in Table 6.

3O

Figure 4: Correct Student Responses to Punnett Square Cluster (N=27)

 

25
3
g 20
g 15 I Pre-test
E 10 DPost-test
0
'U
3
m

 

Question

 

 

 

Table 6: Results of Objective Questions, Punnett Square Cluster

 

 

 

 

 

 

 

 

Question Pre-test Mean Post—test Mean t-value Statistically significant?
27 0.22 0.70 -4.31 Yes
28 0.15 0.82 -7.21 Yes
29 0.19 0.74 -5.00 Yes
32 0.41 0.82 -3.70 Yes
33 0.15 0.59 -3.61 Yes
34 0.22 0.52 -2.53 Yes

 

 

 

 

 

 

Results of paired t-tests for each of these questions indicate that the results were
statistically significant. These questions required that students be able to complete a
Punnett square of a monohybrid cross, and then calculate genotypic and phenotypic
ratios. Students practiced these during each of the PBL activities as well as in practice
problems assigned during the unit. Through this assessment, they demonstrated their
proficiency at monohybrid crosses. The two questions with which they had a little
trouble dealt with co-dominance and recognizing the phenotype related to co-dominance.
While the notation used in the assessment was slightly different, about half of the
students were still able to recognize and distinguish the proper notation given four

choices.

31

Four questions (7, 26, 36, 31) (Appendix B) on the pre- and post-assessment were not
used in the data analysis due to errors in the post assessment and irrelevance. However,
because of repetition in the assessment, other questions covered the same topics and were
used in this analysis.

A pre- and post-survey (Appendix B) was administered to students to measure their
changes in attitude regarding teaching methods before and afier the unit, as well as to
allow them an opportunity to self-assess their learning.

Students reported that they were better able to predict the outcome of a breeding
experiment by the end of the unit (Figure 5). They also reported that they were better

able to explain how parents pass on their traits to their offspring (Figure 6).

Figure 5: “I am able to correctly predict the outcome of a breeding experiment.”

 

   

g 107
E 8’
z 6 IPre—survey
O
E 4 Post-survey
2

E 0
Z i .2. Q) 0 _E 0 2-

2’ E, E, a "5’. 2’3

O

9 3 3 8 a r: ‘3’

H ._ — c (I, a

W U '0 3

Response

 

 

 

32

Figure 6: “I am able to explain how parents pass on their traits to their ofl‘spn'ng.”

 

   

2
S 10
g 8
9 6
“5 4 EPre-survey
g 2 IPost-survey
E o
5 > o o .5 o _>

as s g s as

e co co 8 w 2 cn

.. .9 .‘L’ "" W

‘0 1: '0 g ‘”

Response

 

 

 

Students also felt that the methods of instruction used were preferable to more
traditional methods of deliver, namely lecture and book work (Figure 7). Results of the

post-survey indicate that this view was even stronger after the unit.

Figure 7: “I would rather use traditional lecture and book work activities to learn
difficult concepts than models, problem-based learning, and other group activities.

 

 

8

E

I

'u

a

O

‘6

3

S

2 g
strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

33

By examining student responses to questions, opinions about specific teaching
methods and their effectiveness were also learned. The activity dealing with vocabulary

acquisition was favorably viewed (Figure 8).

Figure 8: “Participating in ‘Bert and Ernie Go to the Cat Breeder’ helped me to
understand the terms phenotype, genotype, dominant, recessive, heterozygous, and
homozygous

 

Number of Students

 

strongly disagree uncertain agree strongly
disagree agree
Response

 

 

 

Students also reported that the use of models was helpful in learning the difficult
concepts covered in the unit. While students had expressed this view initially in the pre-
survey, this viewpoint was even stronger by the end of the unit, as reported in the post-

survey (Figure 9).

34

Figure 9: “Using models helps me to explain material covered in class better.”

 

Presurvey
I Post-survey

Number of students

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Examples of modeling activities included the Sockosome, Reebop, and Humans as
Chromosomes activities (Appendix C). Students reported on the post-survey (Appendix
B) that the Reebop marshmallow activity was enjoyable (Figure 10), however, they also
reported that they were uncertain about the concepts the activity was based on (Figure
11). The sockosome modeling activity also left students uncertain about the concepts

involved (Figure 12).

35

Figure 10: “The Reebop activity (marshmallow creatures) was enjoyable.”

 

Number of students

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

 

Figure 11: “After completing the Reebop activity, I was able to understand how the law
of independent assortment can lead to differences in the population.”

 

Reebops and law of ind. assort.

—\

Number of students
0 N ¥ 0) a O

 

strongly disagree uncertain agree strongly

Responses

 

 

 

36

Figure 12: I was able to explain the laws of segregation and independent assortment after
completing the sockosomes group activity.

 

Number of students
0 —l N (A, A 01 O) \l on

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Students did feel more confident about the concepts learned during the Humans as

Chromosomes activity and reported so on the post-survey (Figure 13).

Figure 13: “After participating in the meiosis game (hats), I was able to explain how the
law of independent assortment can lead to differences between individuals in the
population.

 

Number of students
0 _. N no A or or \1 on

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

37

Students were also surveyed during the post-survey (Appendix B) about problem-
based learning and the activities during the unit that were based upon this type of
instruction. Students found that problem based learning was both somewhat challenging

(Figure 14) and enjoyable (Figure 15).

Figure 14: “Problem-based learning was challenging.”

 

 

Numberol'students
O —t N w # 0| 0) N on

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Graph 15: “Problem-based learning was enjoyable.”

 

Number of students
0 —t N (a) A 0| 0) V (D (D

 

strongly disagree uncertain agree strongly
disagree agree
Response

 

 

 

38

Comparison of data fi'om the pre and post-survey (Appendix B) shows students did
agree that problem-based learning was a more interesting approach to learning new
concepts than more traditional methods (Figure 16).

Figure 16: Lessons designed as a problem to solve are/were more interesting than
traditional worksheet problems.

 

.1

Numberofstudents
DAN onloasrcncoo

 

Strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Students responded favorably to individual problem-based learning activities when
asked about them on the post-survey (Appendix B). Results of these questions are found
in Figures l7, l8, and 19.

Figure 17: “After working through the problem based learning activity about Charlie the

Cat, I was able to predict genotypes and work through Punnett squares to make
predictions about offspring.

 

Vi“)

V

 

Number of students
0 A N w # U! a: \l m

strongly disagree uncertain agree strongly agree
disagree

 

 

 

39

1%?!

Figure 18: “After completing the problem based learning activity about Spot the cat, I
was able to predict the genotypes and phenotypes of co-dominant traits in offspring in
breeding problems.”

 

_l

Numberofstudents
O —B N w A O! O) \l on (O O

 

strongly disagree uncertain agree strongly
disagree agree
Response

 

 

 

Figure 19: “After completing the problem based activity about orange cats, I was able to
predict the genotypes and phenotypes for sex linked traits in offspring in breeding
problems.

 

Number of students

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

40

Think-pair-share (D. Walker, 1998) was also evaluated using the pre and post-survey.
Because students were not familiar with the name of the method, this question was
worded differently on the pre-test (Appendix B) than it was on the post-test (Appendix
B). Students generally thought less favorably about how “think-pair—share” improved

their understanding of a particular concept after the unit than before it (Figure 20).

Figure 20: After completing a think-pair-share activity, I could explain the concept being
taught.

 

 
  

14
3 12
5
3
0 8 Pre-survey
E 6 IPost-survey
g 4
3
Z 2
0
strongly disagree uncertain agree strongly
disagree agree
Response

 

 

 

The remaining questions on the pre and post-survey (Appendix B) dealt with the
students’ ability to complete assessments using the knowledge they gained from
completing the various activities in the unit. The first of these asked students to consider
the difficulty experienced in completing the assessment questions following the activities
(Figure 21). Students felt that they could use new knowledge to answer questions after
an activity. They also felt that they could answer questions on a formal assessment by

remembering what they had done during an activity as well (Figure 22).

41

Figure 21: “After completing activities from this unit, I was able to take what I learned
and complete the analysis questions with little assistance.”

 

Numberofstudents
O A '0 OJ h 01 O) N on to

 

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Figure 22: When answering questions for a test, thinking about activities I completed
during the unit helps me remember the answer.

 

—|

onsmooo

 

Number of students

strongly disagree uncertain agree strongly
disagree agree

Response

 

 

 

Information from these pre and post-surveys was used along with pre and post-test

data to determine the effectiveness of the unit.

42

DISCUSSION

Data collected and analyzed during this unit generally support the conclusion that it
was successful in accomplishing the major goals: acquisition of necessary vocabulary,
accurately predicting the outcome of a monohybrid cross, describing the major events of
meiosis I and 11 (including the laws of segregation and independent assortment), and
predicting outcomes of breeding experiments where the trait is not a simple
dominant/recessive.

Students were successful in learning necessary vocabulary and completing heredity
problems using Punnett squares. Data found in Table 5 and Figure 3 support this
conclusion. Students struggled with vocabulary afier completing the “Bert and Ernie”
activity, where these terms were first used. They continued to work with the terms
through the problem-based activities and heredity problems. Students were never given a
vocabulary quiz, per se. However, because the key terms were reinforced throughout the
rest of the unit, especially in the problem-based learning activities, students learned the
vocabulary using a whole language process.

On the post-test (Appendix B), I expected more students to be able to describe
dominant and recessive alleles in their answers and for overall scores to increase for the
question. This increase in points awarded would indicate advancement in understanding
of the concept from middle school (a score of 1 on the rubric) (Appendix B) to a high
school level (a score of 2 or 3 on the rubric). Students were able to apply these
vocabulary terms correctly and scored well overall (Table 3). Application of these terms
would indicate students understood the vocabulary and did not simply memorize

definitions.

43

Analysis of vocabulary word acquisition occurring introduced later in the unit
(haploid, diploid) was ambiguous (Table 4, Figures 1 and 2). Question twelve
(Appendix B) dealt with vocabulary, specifically the term “haploid.” Many students got
the question right on the pre-test, while some missed it. Because this word was new, it
would be best explained by guesswork on the students’ part. This term would have been
introduced during the activities about meiosis (Sockosomes, Meiosis and Fertilization,
Reebops, and Humans as Chromosomes) (Appendix C). The term “diploid” was also
introduced during these activities. Since most of the focus of instruction was on gamete
formation, fertilization was not well covered, as indicated by this test item. It may also
indicate that while students could pick out the definition from four choices, like question
12, they may not have really understood the meaning of the term. Although the terms
were learned within the context of meiosis, students may have only remembered them as
definitions. These terms were not learned and reinforced in the same way as the others
from the Bert and Ernie activity. Perhaps a second activity like this one and more
integration of the terms in the meiosis portion of the unit is needed.

Students struggled a bit more with the details of meiosis, in particular the laws of
segregation and independent assortment and their role in variation (Figures 1 and 2, Table
4). Therefore, the unit was less successfirl than I had expected. My overall goal was to
teach genetics within the context of meiosis and allow students to develop a clearer
connection between the two. I wanted to seamlessly move between the two big ideas of
meiosis and heredity in a way that allowed students to learn them as a single idea.
However, this proved to be a bit more difficult than I had anticipated. Students needed

more time to practice Punnett squares, which required more class time and additional

44

assignments for practice. Hoping not to fall into the same trap of having the students
simply learn the algorithm of drawing out the Punnett squares without any connection to
“origin” of the boxes, I attempted to tie our lessons on heredity with meiosis (Reebop and
Humans as Chromosomes). Table 4 indicates some success in accomplishing this
connection. However, when asked about details, students had difficulty in explaining the
relationship between variation and the laws of segregation and independent assortment
(Table 3 Figure 1 )

The connection may not have been clear enough to the students. While students
seemed to understand that the law of segregation resulted in genetic variation, they did
not understand the law itself (Table 4). It is also possible that students knew that mitotic
division results in genetically identical cells from a previous unit on mitosis and were
able to pick this out of the choices available in question 16. In any case, students did not
learn enough about the law of segregation during the Humans as Chromosomes or
Sockosomes activities. In this way, my results were not unlike those of Stewart (1980,
1982)

Students typically enjoyed the active learning strategies employed in the unit (Figure
10). However, when asked about whether or not the activities helped them to understand
how the law of independent assortment can lead to variation, the results where lower
(Figure 11 and 12). The added dimensions of the different activities (moving around the
room, permission to wear hats, and using food) may have proved too distracting for the
students and affected their learning. The problem may lie in the word “enjoyment” here.
While playing with the marshmallows was fun, most students did not draw the necessary

connections at the conclusion of the activity.

45

While the problem-based learning activities were less enjoyable to them (Figure 15)
than the Reebops, students self report better success at achieving the outcomes (Figures
17, 18 and 19). Completing fun activities also requires thinking, and new learned
behavior.

However, problem-based learning had many benefits that were perhaps unrealized by
students. Problem-based learning not only provided an opportunity to teach differently,
but it allowed the opportunity to make the class more student centered. Discussions that
occurred during class were substantive, and at times heated! During several of the PBL
activities, students grouped up together to discuss the problems and work through a
solution. “Now, look...” and “But if you try it this way...” could be heard as I walked
around the room. I was impressed by the degree of ownership the students assumed in
order to solve the problems. This took a little time to happen, however. Many of the
students were used to teacher-centered activities and often wanted me to tell them the
correct answer or if they were completing the assignment correctly. I had to learn to step
back a bit and be more Socratic in my approach to their desire to hear the answer.
Eventually, the students became more confident and began to take on the PBL’s with
very little need for me. After solving the sex-linked trait activity, a particularly difficult
one, many of the students shouted out excitedly, “I figured it out! Can I share my answer
when we go over it?” They were pleased with themselves and I was happy for them. It
was, perhaps, the best part of the unit. Given the involvement level of students during
each of the PBL activities, it is more than likely I will use them again when teaching

genetics.

46

 

The structure of the PBL activities allowed for high levels of student motivation that
did not necessarily occur during “think-pair-share activities.” Students felt less confident
about knowledge discussed after a think-pair-share than after a PBL (Figure 20). PBL
had a sense of purpose to the conversation missing in a “think-pair-share” activity.
“Think-pair-share” also seems to work better for shorter questions covering one
particular point that can be addressed quickly. This may also have lead to students
feeling less sure about the content. PBL involves longer conversations concerning many
details that build upon each other.

One drawback to PBL, also described in the literature, is the fact that generally
students who use it do not do as well on standardized testing than students who use more
traditional methods (Evenson & Hmelo, 2000). Because PBL encourages method
thinking instead of fact recall, students often do not have the chance to show what they
have learned on a standardized test. While students did not take standardized tests during
this unit, many did report that they felt it was less helpful to them in remembering
information for the test (Figure 22). However, assessments following activities were
accomplished easily by remembering what had occurred. When the assessments were
more analytically in nature, students were confident that they could complete them using
what they had learned (Figure 21). Perhaps this was due to the length of time between
the activity and the assessment. Short standardized assessments at the conclusion of each
PBL would be interesting to analyze to see if duration between the activity and
assessment made a difference. It would also indicate whether or not my students had the

same difficulties as those described by Evenson and Hmelo.

47

One of the questions on the pre and post-test (Appendix B) did not reveal much
information about what students did or did not know. Rather, it revealed that some of the
questions were not written clearly or were too abstract for students to understand.
Students found the question itself was difficult to understand and they were not clear
about the expectations for the answer. At issue was question 11. Students were asked to
circle maternal chromosomes on a karyotype. While the statistical analysis indicated that
the improvement on the question was significant (Table 3), anecdotally, the question did
not allow for good results. Even after the unit was taught, many students still had
difficulty answering this question. While the goal of the question was important, the
question itself may have been too abstract. Even on the post-test, many students did not
answer it correctly, or at all, simply because they didn’t understand what to do with the
picture. Several students asked during the test what they should do and some of the
answers students gave revealed that they did not know exactly what the question required
them to do. Fewer pairs of chromosomes in the karyotype and more explicit directions
may be more helpfirl in the firture.

In general, I believe that the unit successfully accomplished its goals and allowed
students a unique opportunity to learn in real world, hands on way. The problem-based
learning was the greatest strength of the unit, and something I will maintain in it.
Students may not have “enjoyed” it in the way that they define it, but they found it
challenging and learned well using it. It created substantive conversation and a student-
centered learning environment that engaged the students’ minds. Students were highly
motivated by the method and found it challenging, but not overwhelming. It also allowed

me an opportunity to coach, and I enjoyed that myself.

48

Other aspects of the unit provided learning experiences for my students and myself.
While my students enjoyed using models of all sizes, they had a difficult time learning
using them. Being busy with their hands may have been enjoyable, but their minds
needed to be busy as well. It may also be that models are an abstract application of an
idea. Perhaps the models themselves may have been the problem. Students may have
enjoyed them (Figure 10), but perhaps did so without a sense of purpose. Being able to
connect the model to the concept may have been the key factor missing. In the firture,
better management and clearer connections with the models may be needed.

Assessment may be the key to managing this aspect better. A group assessment after
each activity may not only create a greater sense of responsibility to each student to keep
an engaged mind, but it may also create a sense of responsibility to other students to get
and give needed help. It may also help to break up some of the activities into pieces
where one part is done and discussed before the next one is done, perhaps at a later time.
In this way, the purpose of the models and the parallels between the models and the
concepts they represent can be more clearly defined. Students could imitate me by
moving the sockosomes on the table, but more information is needed to be sure that they
know exactly why we move them as we do. Similarly, in Humans as Chromosomes,
students could follow each other around the room without really understanding why. By
breaking up the activities into segments, the students would have an opportunity to reflect
and think about the information they had learned. In this way, they would learn the

details better and would be able to apply the information to the segment that followed.

49

I would also like to improve the integration between meiosis and heredity. Students
did quite well on the topic of heredity, but seemed to miss its connection with meiosis, as
evidenced by assessment questions involving it (Table 4).

Having completed this study, I will continue to incorporate many of the activities into
my heredity unit, especially problem-based learning. By refining the modeling activities
to include more assessment, this unit could continue to improve and allow students to

master the difficult concepts involved in heredity and meiosis.

50

APPENDIX — A

51

Parental Consent and Student Assent Form
Collection of Data for Master’s Thesis

Dear Parents/Guardians and Students:

I am currently working toward completion of a Masters degree in Michigan State
University’s Division of Science and Mathematics Education (DSME). I have chosen to
do my thesis work on increasing student comprehension in biology through activity based
learning. Students will be learning about cell reproduction and heredity by participating
in simulations, making models, acting, and using problem based learning labs.

In order to evaluate the effectiveness of this unit, data will be collected fi'om students
through pre and post-tests, lab questions, homework assignments, and surveys. With
your permission, I would like to include your child’s data in my thesis. Your child’s
privacy will be a foremost concern and will be protected to the maximum extent
allowable by law. All data generated shall remain confidential. At no time will your
child’s identity be associated with the data nor will they be identified in any pictures
taken to be used in the thesis presentation.

Participation in this study is voluntary. Your child will receive no penalty in regard to
his/her grade should you deny permission for the use of his/her data. Your student will
still be expected to participate in the classroom activities and complete assignments.
However, your student’s data will not be used in my thesis work. Consent forms will
remain in a sealed envelope until after grades for the unit have been issued. At any time
during the unit, you may request that your student’s information not be included, and
your request will be honored. There are no known risks associated with participation in
this study. There will be no benefit to students in terms of extra credit or points for
participating in this study.

Ifyou are willing to have your student participate in this study, please complete the
attached form and return it to me by October 2, 2006. If you have any questions about
this study, please feel free to contact me by email at krawczykt@almont.k12.mi.us or by
phone at 798-8595. Questions about the thesis project can also be directed to Dr. Merle
Heidemann at DSME, 118 North Kedzie Laboratory, Michigan State University, East
Lansing, MI, 48824, by phone: (517) 432-2152, ext 107, or by email: heidma2@msu.edu

Ifyou have any questions or concerns regarding your rights as a study participant, or are
dissatisfied at any time with any aspect of this study, you may contact - anonymously, if
you wish - Peter Vasilenko, Ph.D., Director of Human Research Protections, (517) 355-
2180, fax (517) 432-4503, e-mail irb@msu.edu, or regular mail 202 Olds Hall, Michigan
State University, East Lansing, MI 48824.

Thank you,
Tracie Krawczyk

Biology Teacher
Almont High School

52

Please fill out the following consent information and return it by October 2, 2006

l voluntarily agree to have participate in this
study. (print student name)

Please check all that apply.
Data:

I give Mrs. Krawczyk permission to use data generated from my child’s
work in this class in this thesis project. All data from my child shall remain
confidential.

I do not wish to have my child’s work used in this thesis project. I
acknowledge that my child’s work will be graded in the same manner regardless
of their participation.

Pictures:

I give Mrs. Krawczyk permission to use pictures of my child during her
work on this thesis project. My child will not be identified in these mediums.

I do not wish to have my child’s picture used at any time during this thesis
project.

 

 

(Parenthuardian signature) (Date)

I voluntarily agree to participate in this thesis project

 

 

(Student signature) (Date)

53

 

APPENDIX - B

54

Meiosis and Heredity Pre- and Post-Test
Write one or two sentences explaining how each of the following terms is related.
1. gene-allele-chromosome

For questions 4-6, read the following and answer the questions.
Your father has brown eyes (B) and straight hair (c). Your mother has blue eyes (b)

and curly hair (C). You have brown eyes and curly hair.
2. Explain why you have eyes like your father and hair like your mother.

3. Is it possible for you to have a sibling with blue eyes and curly hair? Explain

your answer.

4. How is it possible that two siblings could have the same parents but look different
from each other?

5. Ifyou were to have children, is it possible that they will have brown eyes and
curly hair like yourself? Explain your answer.

6. What does meiosis have to do with a Punnett square?

7. Which of the following is not an example of a trait that can be passed on?
a. Missing limb due to amputation (parent’s limb lost to amputation)
b. Eye color
0. Hair color
(1. Ability to roll tongue

8. What is a chromosome?

9. What is the purpose of meiosis?

55

10. Which of the following cells is NOT a gamete?
a. Egg
b. Sperm
c. Pollen
d. Leaf

11. Circle the chromosomes from this human karyotype that came from the person’s
mother.

   

21' ' (I 3 ~21 ' . 3:31:

 

12. Gamete cells are haploid. This means that they have
a. Both chromosomes from each homologous pair
b. Neither of the chromosomes from each homologous pair
c. One of the chromosomes from each homologous pair
d. Some homologous pairs, but not others

13. When two gametes combine, they form a
a. Diploid cell
b. Haploid cell
c. Triploid cell
d. None of these

56

For question 14, refer to the following chromosomes:

 

 

h .
B lon de Brown arr

hair

 

 

 

 

 

 

14. Are these chromosomes considered homologous? Explain your answer.

15. At the end of meiosis, there are
a. 2 haploid cells
b. 2 genetically identical cells
c. 4 haploid cells
d. 4 genetically identical cells

16. Which of the following does NOT result in genetic recombination?
a. Law of independent assortment
b. Crossing over
c. Mitotic division
d. Law of segregation

17. The law of segregation of alleles explains that

a. The way that a set of homologous chromosomes line up during meiosis
does not depend upon the other sets of homologous chromosomes.

b. The way that a set of homologous chromosomes line up during meiosis
does depend upon the other sets of homologous chromosomes.

c. Each chromosome in a set of homologous chromosomes separates from
the other during meiosis

(1. Each chromosome in a set of homologous chromosome does not separate
from the others during meiosis.

18. After the first division of meiosis,
a. Sister chromatids have separated
h. Cells contain a diploid number of chromosomes
c. Homologous chromosomes have separated
d. The gametes are ready to be fertilized.

19. A dominant allele

Is found in the majority of the population

Is only found in plants

Is hidden when the recessive allele is present

Is observed whether one or two copies are present

9.09"!”

57

20. The phenotype of a recessive trait is only seen in an individual when
a. One dominant allele is present
b. One recessive allele is present
c. Two dominant alleles are present
(1. Two recessive alleles are present

 

Suppose you crossed two pea plant, one (the mom) with the genotype Tt and the
other (the dad) with the genotype tt. T is the allele for tall height and t is the
allele for short height.

21. Which of the following are the genotypes of the plants?
a. crossed
b. Tall and short
c. Tt and tt
d. Pea plants

22. Which of the following are the phenotypes of the plants
Crossed

Tall and short

Tt and tt

Pea plants

92° 9"!”

23. The genotype of the mom is described as
a. Heterozygous
b. Homozygous recessive
c. Homozygous dominant
d. None of these

24. What would be the meiotic products for this trait be for the mom?
a. T,T,t,t
b. t,t,t,t
c. T,T,T,t
d. T,T,T,T

25. What would be the meiotic products for this trait for the Dad?
T,T,t,t

t,t,t,t

T,T,T,t

T,T,T,T

999‘?”

58

26. Which of these Punnett squares matches this cross between the parents?
92 = mom (3‘ = dad

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+0.0-

 

 

 

 

 

 

 

 

 

6
t t
9 r tt tt
T It It
cf
r t
9 tT tt
tT tt
6
I t
r Tt Tt
t It It
r t
r TT Tt
t IT 11

 

 

 

 

 

27. When the parent plants are crossed, what is the genotypic ratio of the punnett
square?
a. 1TT12Tt: ltt
b. OTT: 2th2tt
c. 2TT: 0th Zn
d. OTT: 4th Ott

28. When the parent plants are crossed, what is the phenotypic ratio of the punnett
square?

2 tall: 2 short

4 tall: 0 short

0 tall: 4 short

3 tall: 1 short

999‘.”

59

29. Iftwo heterozygous tall plants are crossed together, what is the genotypic ratio
that results?
a. lTT:2Tt:1tt
b. OTT: 2Tt22tt
c. 2TT: 0Tt:2tt
d. OTT: 4th Ott

 

30. Cystic Fibrosis (F) is the most common lethal genetic disorder among Caucasians.
The allele that causes this disorder is recessive. If a woman who is homozygous
for CF has a child with a man who does not have the allele, what is the genotypic
ratio that results?

a. 1 FF: 2 Ff: 1 ff
b. 2 FF: 2 Ff: 0 ff
c. 2 FF: 0 Ff: 2 ff
d. 0 FF: 4 Ff: 0 ff

31. Each of their children has a _% chance of getting CF
a. 100%
b. 50%
c. 25%
d. 0%

 

32. A horse with red hair (S) and a horse with white hair (8’) have a baby horse
(foal) with red and white hairs mixed together. What is the genotype of the foal?

a. SS
b. SS’
c. S’S’
(1. ss

33. Which of the following terms is used to describe this foal?
a. recessive
b. dominant

c. incomplete dominance
(1. sex linked

 

34. A husband and wife have three boys and four girls. All of the girls have normal
vision. Two of the boys are colorblind. Which of the following terms is used to
describe this type of inheritance?

a. Recessive
b. Dominant
c. Incomplete dominance
(1. Sex linked

35. Why are the boys the only children affected?

60

Student Pre Survey of Meiosis and Heredity Unit
For each question, use the following scale:

Strongly Strongly
Disagree Disagree Uncertain Agree Agree
1 2 3 4 5

1. I would rather use traditional lecture and book work activities to learn
difficult concepts than models, problem based learning and other group
activities.

2. Lessons designed as a problem to solve are more interesting than
traditional worksheet problems.

3. When answering questions for the test, thinking about activities I
completed during the unit helps me remember the answer.

4. When covering new material in class, I often help other students who have
questions understand the material or ask other students for help with the
material.

5. After talking to another student about material I have learned in class, I am
able better able to explain the concepts.

6. I remember information from a lecture better when I can talk to other
students about it.

7. Using models helps me to explain material covered in class better.

8. After completing a lab or activity, I am able to take what I learned and
complete analysis questions with little assistance.

9. I complete or review notes at home to review the material covered in
class.

10. I am able to correctly predict the outcome of a breeding experiment.

11. I am able to explain how parents pass on their traits to their offspring.

61

 

Student ID:

 

Student Post-Survey of Meiosis and Heredity Unit

For each question, use the following scale:

Strongly Strongly
Disagree Disagree Uncertain Agree Agree
1 2 3 4 5

8.
9.

. Participating in “Bert and Ernie Go to the Cattery” helped me to

understand the terms phenotype, genotype, dominant, recessive,
heterozygous, and homozygous.

Working through the group punnett square problems allowed me to
recognize what I did and did not understand about punnett squares.

After working through the problem based Ieaming activity about Charlie
the Cat, l was able to predict genotypes and work through punnett
squares to make predictions about offspring.

The Reebop activity (marshmallow creatures) was enjoyable.

After completing the Reebop activity, I was able to understand how the
law of independent assortment can lead to differences in the population.

I was able to explain the laws of segregation and independent assortment
after completing the group sockosome assessment activity.

After participating in the meiosis game (hats), I was able to explain how
the law of independent assortment can lead to differences between
individuals in the population.

Problem based learning was challenging.

Problem based learning was enjoyable

10. The problem based Ieaming activities allowed me to show all that I had

Ieamed about heredity and meiosis.

11. After completing the problem based Ieaming activity about Spot the Cat, 1

was able to predict the genotypes and phenotypes of codominant traits in
offspring in breeding problems

12.After completing the problem based activity about orange cats, I was able

to predict the genotypes and phenotypes for sex linked traits in offspring in
breeding problems.

62

In general
13. I would rather use traditional lecture and book work activities to learn
difficult concepts than models, problem based Ieaming and other group
activities

14. Lessons designed as a problem to solve were more interesting than
traditional worksheet problems.

15. When answering questions for the test, thinking about how I solved the
problem based Ieaming activity (Charlie the Cat and The Cat Breeding
Problem) helped.

16. During Think-Pair-Share activities, I either explained the material to other
students or asked other students for help with the material.

17. After completing a Think-Pair—Share activity, I could explain concept being
taught

18. After completing activities during this unit, I was able to take what I
Ieamed and complete the analysis questions with little assistance.

19.l remember information from a lecture better when it includes Think-Pair-
Share activities.

20. Using models (Reebops, sockosomes) helped me understand the
connection between heredity and meiosis better.

21. Completing fill-in-the—blank notes at home helped me to review the
material.

22. I am able to correctly predict the outcome of a breeding experiment.
23.1 am able to correctly predict the outcome of a breeding experiment.

24.l am able to explain how parents pass on their traits to their offspring.

63

 

 

 

 

 

 

 

 

 

 

 

 

Pre and Post-Test Grading Rubric for Open-ended Questions
Item 3 points 2 points 1 point 0 points
1 Describe and Describe and Describe 1 All
connect all 3 terms connect 2 terms tenn/Generic incorrect/no
answer response
2 Describe Mention only All
dominant/recessive inheritance from incorrect/no
alleles and explain one parent answer
how phenotypes
appear
3 Describe recessive Describe dominant Mentions All incorrect/
traits and or recessive/relate contribution of No response
recombination to inheritance from only one parent or
distant relative “mixing” of genes
4 Describe Describe recessive Only mention All incorrect/
recessive/dominant traits or contribution of no answer
traits and recombination one parent
recombination
5 Describe Describe recessive Only mention All incorrect/
recessive/dominant traits or contribution of no answer
traits and recombination one parent
recombination
6 Connect meiosis, Mention All incorrect!
sex cells, and “possibilities” but no answer
genetics no connection to
meiosis
8 Piece of DNA Piece of DNA All incorrect]
containing genes no answer
9 Cell division of Division of Cell division All incorrect/
gametes where gametes/ right no answer
chromosome number of
number is halved/ chromosomes/passi
Variation in ng of genes
population
1 1 One autosomal One autosomal Some pairs with All incorrch
chromosome chromosome from one circled no answer
circled from each each pair circled
pair and X OR one sex
chromosome chromosome
circled circled
13 “Yes” same genes “Yes” same genes, “Yes” without All incorrect/
in same loci do not mention loci explanation no answer
35 Trait is sex linked Trait is sex linked Related to sex, All incorrect/
and found on X (but mention Y but not specific no answer
chromosome chromosome)

 

 

 

 

 

 

 

Answer key for Pre and Post-Test Objective Items

7. A
10.
12.
13.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.

w>>wow>>wocunnoo>oc

w
:5
U

65

APPENDIX — C

66

 

Introduction to Mendelian Genetics: Lucky Stripes

Brainstorm: What is different about these cats that some have stripes and
some don’t?
Class ideas:

Laboratory Assignment
You and your lab partner will be studying the hairs of the famous cats to
determine differences in the hairs of the striped and solid mts.

Directions:
Examine the hairs of the cats on the slides and draw a picture of each using first
your eyes and then the microscope:

 

Name of Cat
from which the When viewed with hand lens
hair came

 

Fizzy or Curly
(Hair Type #1)

 

Fizzy or Curly
(Hair Type #2)

 

Baby, Larry, or
Moe

 

 

 

 

Discussion:
1. What is different about the three types of hairs you observed?

2. What could account for these differences? (Why are some cats striped and
other cats are not?)

3. Relate the differences between the heirs to our recent discussion of genes
and proteins

67

Fizzy and Curly both have a trait called agouti. Agouti cats have hairs with
individual stripes on them. Agouti is a dominant condition (A).

1. What would a cat that was heterozygous look like?

2. What would a cat that was homozygous recessive look like?

3. What is the genotype of Baby?

4

. What is the genotype of Fizzy?

68

Bert and Ernie At the Cat Breeder
Written by T. Krawczyk and K. Joos

Ernie is in the apartment when Bert arrives home from grocery shopping.

Bert: Hey, Ernie. I’m back. They had 2 for one coupons for soy milk and
organic asparagus at the grocery store. Wait...what’s that?
Ernie: [peering into cardboard box] Hey, Bert. I found something we
really need.
Bert: [suspiciously] Like when we REALLY needed that ceramic Elvis
sculpture? I
l
Ernie: No, this is even better! And I bet that you can’t figure out what it is. 1

I’ll give you a hint... it starts with the letter C!

 

Bert: Ernie, you know how I hate cookie crumbs all over everything. The
last time that Cookie Monster was here I was Swiffering for weeks.

Ernie: Naw, naw, Bert. It’s not a cookie. This thing says, “meow!”

Bert: [hysterically] A CATll Are you crazy, Ernie! A CAT?! You didn’t let
that cat get to my pigeons did you! [running to window] Oh, poor
Bernice - she must be terrified by now. You can’t keep a cat in an
apartment that already has pet birds. Don’t you know the food
chain Ernie?

Ernie: Oh, Bert! Relax. Graycie is just as sweet as can be, aren’t you?
Look at her cute grey stripes. What a delightful phenotype.

Bert: What is a phenotype? It had better not be another cat!

Ernie: A phenotype is a word that means your physical characteristics.

Bert: Well, Ernie, that cat has to go. I am allergic ...... achoo... .to cat hair.
That’s MY phenotype. Come on, Grover can drive us back to the
cattery.

Ernie and Bert arrive at the cattery with Graycie (still in the box).

 

Bert: Okay, Ernie. Now quick put her back where you found her.

Ernie: We can’t just leave her here, Bert. Let’s find someone here to talk
to first. Oh, look...there’s her parents, Fizzy and Fuzzy.

69

Bert:

Ernie:

Bert:

Ernie:

Bert:

Ernie:

Bert:

Bert:

Ernie:

Bert

Ernie

Bert:

Ernie:

Fizzy and Fuzzy don’t look like Graycie, Ernie. Are you sure that
they are her parents?

Sure, I’m sure. They both have stripes. Graycie has stripes.
What’s the problem?

Graycie has grey stripes and those cats have black stripes. Don't
you think that Graycie would have black stripes, too?

Bert, non dilute pigmentation is dominant over dilute pigmentation
in cats. It has to do with the packing of the melanin in the hairs.

Did you fall and hit your head recently?

My head is naturally football shaped, Bert. No, seriously, there was
a terrific program on public television about this very thing. It was
about Ge—NET-ics. Can you say that?

[rolling eyes] Ge-NET-ics.

[cheerfully] Good, Bert! Melanin is the pigment in skin and hair that
makes them the color that they are. Cats with dilute pigmentation
have less pigment packed into the individual hairs. Some cats
have the type of gene that causes dilute and others have the
regular form of the gene.

So, is there something wrong with this cat, then?

No, having less pigment is just something that makes cats different
from each other. It’s just like the way our heads are shaped
differently.

[under his breath] Well, I’m not so sure that’s a good example,
Ernie

What did you say, Bert?

Nothing. Hey, Ernie, I understand why this cat looks grey and why
these cats look darker. But I don’t understand why a cat with dark
striped parents would have light stripes!

Well, Bert, some forms of a gene are called “dominant” while others
are called “recessive." In this case, dark stripes are a dominant
trait. Graycie’s parents have one of these dominant forms of the
gene.

70

Ernie:

Bert:
Ernie:

Bert:

Ernie:

Bert:

Ernie:

Bert:

Ernie:

Bert:

Ernie:

But Ernie, I STILL don’t understand how Graycie ended up, well,
gray.

Recessive forms of the gene can be present in an individual, but
not observable.

So, Fizzy and Fuzzy have hidden genes?
Sort of. They are there, but you can’t tell.

So let me get this straight... Fuzzy and Fizzy each gave Graycie the
dilute form of the melanin pigment gene which made her look gray.
But they are not grey because they also have the dominant trait at
the same time. Right?

Exactly, Bert.

So having the dominant and the recessive forms makes you look
like the dominant form of the trait.

Right. And having the two different forms is called “heterozygous.”
That’s a genotype - the genetic make up of an organism.

And having only the recessive form of the gene makes the cat look
like the recessive form of the trait. What would happen if the eat
only has the dominant form of the trait?

In this case, the cat would be dark also. Having two of the same
type of the gene is called “homozygous.”

Whoa! This stuff is really interesting, Ernie. I should start watching
more public television. So Graycie’s genotype must be
homozygous for the recessive trait and her parents’ genotype is
heterozygous for the trait.

Good job, Bert. Let’s go home and watch some more educational
programming. I have it recorded on TWO. There’s going to be a
great show on the history of rubber duckies. Hee hee hee.

[sighing] Oh, Ernie. Let’s go home. I need a cold soy milk.

71

Name: Hour:
Cat Coat Color Genetics: Bert and Ernie Review lnforrnation

1. What did Bert find out from Ernie about dilute cats like Graycie? (What is
different about their fur?)

2. What does the term phenotype mean?

3. Give an example from the story of a phenotype.

4. What does the term genotype mean?

5. Give an example from the story of a genotype.

6. What does the term dominant mean?

7. Give an example from the story of a dominant version of a gene.

8. What does the term recessive mean?

9. Give an example from the story of a recessive version of a gene.

10. What does the term heterozygous mean?

11.Explain why Fizzy and Fuzzy are considered to be heterozygous.

12. What does the term homozygous mean?

13. Explain why Graycie is considered to be homozygous.

14. Explain how it is possible for Graycie to be the kitten of Fizzy and Fuzzy.

72

Diamonds and PUURRLS
Problem #2: Graycie
Objectives:
1. Identify genotype and phenotype.
2. Describe the terms dominant and recessive
3. Describe the terms heterozygous and homozygous
4. Explain why an offspring may have different traits than its parents

 

Objective: To determine the difference between grey and black cat hair.
Directions:
1. Obtain a sample of each cat’s hair
2. Examine each with a hand lens. Draw a picture and write a description of
each in the chart below.

 

 

 

Cat Sketch of hair Written description of
hair
Fizzy
Graycie

 

 

 

 

 

Discussion: Discuss with your partner what you found out about the differences
between the two kinds of cat hair.

Statement: Write a two to three sentence statement below to describe the
difference between the two kinds of hair.

 

 

 

 

Class Discussion: In the space below, include any information from the class
discussion not included in your statement above.

73

Name: Hour: __
Problem #2: Graycie

Complete the following table with the characteristics of each cat:

 

 

Cat Fizzy Baby Larry Moe Curly
Fur Agouti Solid
characteristic

 

 

 

 

 

 

 

 

Problem #1: Lucky Strips Vocabulary Review

1. What is the phenotype of Baby?
2. What is the phenotype of Fizzy?

3. What would a cat that is heterozygous for agouti look like?

4. What would a cat that is homozygous recessive for the agouti trait look
like?

5. What is the genotype of Baby?

6. What is the genotype of Fizzy? (HINT: What are the phenotypes of the
kittens: Larry, Moe, and Curly)

7. Explain how Curly ended up with stripes while Larry and Moe ended up
solid black.

74

Punnett Squares Notes
Curly’s chromosomes:

O O\ .0 O

Curly

Dominant versions are represented by a
Recessive versions are represented by a

 

 

What are the other possible combinations for future brothers and sisters of
Curly?

00 0 0

Offspring

An easier way to do this? Punnett Squares

0 Each row/column represents a version of the gene each parent could
donate.

0 The result is a chart with all of the possible combinations for one offspring.

Mom -)
Dad 1 a a
A
a

How many different combinations are there?
What are the different combinations possible?

 

 

 

 

 

 

 

 

 

Identify the phenotypes that match each combination:

 

99°F)?

lf Fizzy and Baby had another kitten, what are the chances that it
would have dark stripes?

 

75

REMEMBER:
o the chart represents the possibilities for each kitten.
o The possbilities start over each time a new kitten is created
0 The next kitten could have striped OR solid fur. It does not have to
be striped just because two siblings are already solid and one is
already striped.

Let’s review the Graycie problem using Punnett Sguares
This will be a Think-Share-Pair activity.

Fizzy and Fuzzy are black agouti cats. They have a kitten named Graycie, who
is a grey agouti cat. Dilute fur is recessive to non-dilute fur. Complete a Punnett
square to show that Graycie could be the kitten of Fizzy and Fuzzy.

First, figure out the phenotypes and genotypes of the parents and Graycie:

 

 

 

 

 

 

 

 

 

 

Graycie Fuzzy Fizzy
phenotype
genotype
Next, complete the Punnett Square:
Fuzzy ->
Fizzy;

 

 

 

 

 

 

 

What are the chances that Fizzy and Fuzzy could have a kitten with dilute fur?
T-P-S: How do we know for sure that Fizzy and Fuzzy are both

heterozygous?

Homework Assignment: Complete the Punnett square story problem set. Be
sure to look back at your notes to help you with vocabulary.
This will be due TOMORROW.

76

Sockosomes

Sockosomes are construction paper “socks” that are used to represent chromosomes and
chromatids to learn about Meiosis I and H.

Each sockosome set is decorated differently with one allele labeled with letters. Other
alleles are represented by the decorations on the sockosome.

Sockosomes each have a small piece of Velcro to attach sister chromatids together.

Students practice lining up and separating the sockosomes on their tables to represent the
divisions that occur during Meiosis I and II.

To increase durability, the sockosomes are covered in contact paper.

Velcro
attaches
them
together

    

Letters represent alleles

Decorations make it easy to find homologous chromosomes and can represent other
genes on the chromosome

77

Problem #4: Charlie The Cat

The Diamonds and PUURRLS Cattery has a new champion cat named
Shiver-Me- Timbers lll (nicknamed Charlie). He is a cat with solid dark fur.
When cats are shown, paperwork must be provided to prove lineage. The
paperwork for Charlie the Cat says that his parents are Cloud (Dad) and
Marshmallow (Mom). Both cats are pure white.

Two days alter the championship, the owner of the second place cat
challenged the pedigree of Chartie the Cat. She contends that Cloud and
Marshmallow can ’t possibly be Charlie ’s parents.

Is the she being a sore loser, or is she pointing out a cheatefl!!!
What do you think?

Goals for this problem:
1. Determine the possible genotypes of the parents and of Charlie
2. Determine whether it is possible for Chanie to be the kitten of Cloud
and Marshmallow.

Questions to get started with:
1. What are the genotypes/phenotypes of the cats involved?

2. Which genotypes cause each of the phenotypes?
To continue, you will need to ask Mrs. K for research materials.
You may consult with other goups during the research portion of the problem.

3. What are the dominant/recessive versions of each of the traits?

 

Cloud Marshmallow Charlie

 

phenotype

 

 

 

 

 

genotype

 

Research information

78

 

 

Punnett Squares

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Results:
What are the possible genotypes of Charlie and his parents?

Is it possible for Charlie to be the kitten of Marshmallow and Cloud?
Explain your answer.

79

Cat Coat Color Genetics
Adapted from: http://sirlou.best.vwh.net/catgenetics.html
Courtesy of Dr. Merle Heidemann, Michigan State University, DSME

 

The Color-Conformation Genes

The color—conformation genes determine the color, pattern, and
expression of the coat. Since these characteristics are among the most
important of the cat's features, at least from a breeding point of
view, more emphasis is given the color conformation genes than the
others. These genes fall into three logical groups: those that control
the color, those that control the pattern, and those that control the
color expression. Each of these groups contains several differing but
interrelated genes.

The Orange-Making Gene

The second of the genes controlling coat color is the orange-
making gene. This gene controls the conversion of the coat color into
orange and the masking of the agouti gene and comes in two alleles:
nonorange and orange.

The non-orange allele, "0", is wild and allows full expression of
the black or brown colors. The orange allele, "0", is mutant and
converts black or brown to orange and masks the effects of the non—
agouti mutation of the agouti gene (all orange cats are tabbies). This
gene is sex-linked -- it is carried on the "X" chromosome beyond the
limit of the "Y" chromosome. Therefore, in males there is no homologous
pairing, and the single orange-making gene stands alone.

As a result there is no dominance effect in males: they are either
orange or non-orange. If a male possesses the non-orange allele, "0",
all colors (black, dark brown, or light brown) will be expressed. If
he possesses the orange allele, "0", all colors will be converted to
orange.

In females there is homologous pairing, one gene being carried on
each of the two "X" chromosomes. These two genes act together in a
very special manner because of X inactivation. If the female is
homozygous for non-orange, "00", all colors will be expressed. If she
is homozygous for orange, "00", all colors will be converted to orange.
It is when she is heterozygous for orange, "00", that interesting
things begin to happen: through a very elegant process, the black—and-
orange tortoiseshell or brindled female is possible.

Shortly after conception, when a female zygote is only some dozens
of cells in size, a chemical trigger is activated to start the process
of generating a female kitten. This same trigger also causes the zygote
to "rationalize" all the sex-linked characteristics, including the
orange-making genes. In this particular case, suppression of one of
the orange-making genes in each cell takes place in a not-quite-random
pattern (don't worry about this). Each cell will then carry only one
orange—making gene.

Since the zygote was only some dozens of cells in size at the time
of rationalization, only a few of those cells will eventually determine
the color of the coat (the orange-making genes in the other cells will
be ignored). If the zygote were homozygous for non-orange, "00", then
all cells will contain "0", and the coat will be non-orange. Likewise,
if the zygote were homozygous for orange, "00", then all cells will
contain "0", and the coat will be orange. If, however, the zygote were
heterozygous, "00", then some of the cells will contain "0” and the
rest of the cells will contain "0". In this case, those portions of the
coat determined by "0" cells will be orange, while

80

those portions determined by "0" cells will be non-orange. Voila! A
tortoiseshell cat!

A female kitten has two "X" chromosomes, and therefore two
orangemaking genes, one from each parent. Assuming for the sake of
discussion an equal likelihood of inheriting either allele from each
parent -- an assumption that is patently false, but used here for
demonstration only —- then one quarter of all females would be non-
orange, one quarter would be orange, and one-half would be
tortoiseshell. A male kitten, on the other hand, has only one "X"
chromosome, and therefore only one orange-making gene. Keeping the same
false assumption of equal likelihood, then one—half of all males would
be non-orange and one—half would be orange. This means that there would
be twice as many orange males as females if our assumption were
correct.

Since a male has only one orange-making allele, there cannot be a
male tortie. An exception to this rule is the hermaphrodite, which has
an "XXY" genetic structure. Such a cat can be tortie, since it has two
"X” chromosomes, but must invariably be sterile. In fact, despite the
presence of male genitalia, a hermaphrodite is genetically an
underdeveloped female, and may have both ovaries and testes, with
either fully functional.

The Color-Density Gene

The third and last of the genes controlling the coat color is the
color-density gene. This gene controls the uniformity of distribution
of pigment throughout the hair and comes in two alleles: dense, "D",
and dilute, "d". The dense allele, "D", is wild, is dominant, and
causes pigment to be distributed evenly throughout each hair, making
the color deep and pure. A dense coat will be black or orange. The
dilute allele, "d", is mutant, is recessive, and causes pigment to be
agglutinated into microscopic clumps surrounded by translucent
unpigmented areas, allowing white light to shine through and diluting
the color. A dilute coat will be blue (gray) or cream.

Geno I Sex I Color
_____ +__--____+_______-
ooDD I Female l Black
_____ +__-_____+_-_-____
ooDd I Female | Black
_____ +---_____+.._______
oodd I Female I Blue
_____ +-__--_--+_-__-__-
l

OoDD Female I Blk/Red
oYDDI Male I Black
OYDDI Male I Red

_____ +____.-_.._+__---_-_
OoDd I Female I Elk/Red
oYDdI Male | Black
OYDdI Male I Red

..... +_____-__+_--_____
Oodd I Female I Blu/Crm
oYddI Male I Blue

OYddI Male I Cream

81

The dilute colors are rarer (hence generally more prized)
because they are recessive. Note that although tortoiseshell females
are two-color they introduce no new colors.

The Albinism Gene

The first of the color-conformation genes affect coat pattern is
the albinism gene. This gene controls the amount of body color and
comes in three alleles: full color, "C", blueeyed albino, "ca", and
albino, "c". The full color allele, "C" is wild, is dominant, and
produces a full expression of the coat colors. This is sometimes called
the nonalbino allele.

The blue-eyed albino allele, "ca", is mutant, is recessive to the
full color allele and dominant to the albino allele, and produces a
nearly complete albinism with a translucent white coat and very washed-
out pale blue eyes.

The albino allele, "c", is mutant, is recessive to all others, and
produces a complete albinism with a translucent white coat and pink
eyes.

The albanism genes combine in some rather interesting ways:

C | full color full color full color
ca I full color B-E Albino B-E Albino
c | full color B-E Albino Albino

The Agouti Gene

The next gene controlling the pattern of the coat is the agouti
gene. This gene will control ticking and comes in two alleles: agouti,
"A", and non-agouti, "a". The agouti allele, “A", is wild, is
dominant, and produces a banded or ticked (agouti) hair, which in turn
will produce a tabby coat pattern. The non-agouti allele, "a", is
mutant, is recessive, and suppresses ticking, which in turn will
produce a solid-color coat. This gene only operates in conjunction with
the non-orange allele of the orangemaking gene and is masked by the
orange allele of the orange-making gene.

The Tabby Genes

The last of the genes affecting the coat pattern is the tabby
gene. This gene will control the actual coat pattern (striped,
spotted, solid, etc.) and comes in two alleles: mackerel or striped
tabby, "T", and blotched or classic tabby, "tb". The mackerel-tabby
allele, "T", is wild, is co-dominant with the spotted tabby allele and
dominant to the classic-tabby allele, and produces a striped cat, with
vertical non-agouti stripes on an agouti background. This is the most
common of all patterns and is typical grassland camouflage, where
shadows are long and straight.

A spotted tabby is genetically a striped tabby with the stripes
broken up by polygene influence. There is no specific "spotted-tabby"
gene. This spotted coat is a typical forest camouflage, where shadows
are dappled by sunlight shining through the trees. Do not confuse the
spots of our domestic cats with the rosettes of the true spotted cats:
entirely different genes are involved.

The blotched- or classic-tabby allele, "tb", is recessive to the
mackerel-tabby allele and will produce irregular non—agouti blotches or
"cinnamon-roll" sworls on an agouti background. When the "cinnamon-
rolls" are clean and symmetrical, and nicely centered on the sides, a
strikingly beautiful coat is achieved.

82

The Spotting Gene

The next gene controlling color expression is the white-spotting
gene. This gene controls the presence and pattern of white masking the
normal coat pattern, and has four alleles: non-spotted, "s", and
spotted,"S". The non-spotted allele, "5", is wild, is recessive, and
produces a normal coat without white.

The spotted allele, "8", is mutant, is dominant, and produces
white spotting which masks the true coat color in the affected area.
This is a variably-expressed allele with a very wide expression range:
From a black cat with one white hair to a white cat with one black
hair.

Unlike the white gene or the albinism gene, the white—spotting
gene does not affect eye color: if your all white cat has green eyes,
it is most definitely a colored cat with one big white spot all over.

The Dominant-White Gene

The final gene controlling color expression is the dominant-white
gene. This gene determines whether the coat is solid white or not,
and comes in two alleles: non-white, "w", and white, "W". The non-
white allele, "w", is wild, is recessive, and allows full expression of
the coat color and pattern.

The white allele, "W", is mutant, is dominant, and produces a
translucent all-white coat with either orange or pale blue. Blue-eyed
dominant—white cats are often deaf, orange-eyed cats occasionally so.
Interestingly, a white cat may be odd—eyed, having one blue and one
orange eye. Such a cat is often deaf on the blue side.

It is important to remember that, genetically speaking, white is
not a color, but rather the suppression of the pigment that would
normally be present. A heterozygous white cat can an often does produce
colored kittens, sometimes with no white at all.

The Standard Solid Colors

The solids form the basis for all other colors in nomenclature and
genotypes. The subtle differences possible in blues (grays) has made
this one of the most popular colors among breeders, with several breeds
being exclusively blue. Blues, regardless of pattern, are often
referred to as "dilutes."

Since the orange allele of the orange-making gene also masks the
nonagouti allele of the agouti gene, red and cream solids are
genetically identical to red and cream tabbies. Careful selective
breeding has made cause the non-agouti areas (the stripes) to widen and
overlap, effectively canceling the paler agouti background and
obscuring the tabby pattern. A generation or two of random breeding,
however, and the stripes will return.

The patched solids, solid—and—whites or bi—colors, are formed by
adding the white-spotting gene, "8*", to the solids. Cats with less
than 50% white are expected to be 85, while those with more than 50%
white are expected to be SS.

The tortoiseshells or torties are formed by combining both the
dominant and recessive sex-linked orange genes, "00", with the solids.
Because of the sex-linking of the orange genes, the tortie is always
female. A tabby pattern may be visible in the orange areas, with any
tabby pattern being permitted. In some individuals, the agouti and
non-agouti orange areas may offer such contrast as to produce a false
tri-color (black-orange-cream).

The patched tortoiseshells or calicos are formed by combining both
the dominant and recessive sex-linked orange—making genes, "00", to the

83

solids and adding the white—spotting gene, "8*". Like the torties,
the calicos are always female, and like the patches, any whitespotting
pattern is permitted.

Color I Genotype I Usual eye color

_____________________ +____._._________._____..____+___..__..__-_____..__
Black I ooD* C*aa** ssww I cpr org grn

Blue I oodd C*aa** ssww I cpr org grn

Red I OOD* C***T* ssww I cpr org

Cream I OOdd C***T* ssww I cpr org

Black patch I OOD* C*aa** S*ww I cpr org grn
blue patch I oodd C*aa** S*ww I cpr org grn
red patch I OOD* C***T* S*ww I cpr org

cream patch I OOdd C***T* S*ww I cpr org

The Standard Tabby Colors

The tabbies are formed by adding the agouti gene, "A*", to the
solids. This causes the otherwise solid color to show the pattern
dictated by the tabby gene: light and dark stripes (mackerel allele,
"T*") or blotches (blotched allele, "tbtb").

The brown tabby corresponds to the black solid: sufficient
undercoat color shows in the agouti areas to provide a brownish cast.
When in mackerel pattern, this is the "all wild" genotype, and
represents the natural state of the cat.

The red tabby, when in mackerel pattern, presents an alternate
stable coat often found on feral domestic cats, usually as a pale
ginger. The patched tabbies or tabby—and-whites are formed by adding
the white spotting gene, "8*", to the tabbies. Like the patched solids,
any white spotting pattern is permitted. The tabby-tortoiseshells or
torbies are formed by combining both the dominant and recessive sex-
linked orange genes, "00", with the tabbies colors. Like the torties,
the torbies are always female.

Color I Genotype I Usual eye color

_______________________ +______--_._.___________+____________.____
tortie | OOD* C*aaT* ssww I cpr org

blue tortie I Oodd C*aaT* ssww I cpr org grn
_______________________ +_______.________.______+_______.....________
calico I OOD* C*aaT* S*ww I cpr org

blue calico I Oodd C*aaT* S*ww | cpr org grn
_______________________ +_________--__....-_____+________________
brown tabby | ooD* C*A*T* ssww I cpr org yel hzl

blue tabby I oodd C*A*T* ssww I cpr org yel hzl

red tabby I OOD* C***T* ssww | cpr org yel hzl

cream tabby I OOdd C***T* ssww I cpr org yel hzl
_______________________ +____.._.--___.__._____-__+__-___________.__
brown patched tabby | OOD* C*A*T* S*ww I cpr org yel hzl

blue patched tabby I oodd C*A*T* S*ww I cpr org yel hzl

red patched tabby I OOD* C***T* S*ww I cpr org yel hzl

cream patched tabby I OOdd C***T* S*ww | cpr org yel hzl
_______________________ +__________.___________+____._._._-_.____.—__
torbie I OOD* C*A*T* ssww | cpr org yel hzl

blue torbie I Oodd C*A*T* ssww | cpr org yel hzl
_______________________ +____.__.____._.__________+____._______.__.__...
torbico I OOD* C*A*T* S*ww | cpr org yel hzl

blue torbico I Oodd C*A*T* S*ww I cpr org yel hzl

84

The patched tabby-tortoiseshells, or patched torbies or torbicos, are
formed by combining the dominant and recessive orange-making genes,
"00", with the standard tabbies and adding the white spotting gene,
"8*", to the torbie colors. Like the patched solids, any whitespotting
pattern is permitted.

The Whites

White is not a color, but rather a masking of the color genes
resulting in an absence of color. There are four ways a cat can have an
all white coat: be full-spotted white, be dominant white, be blue-eyed
albino, or be albino. Each of these ways is genetically different.

The full-spotted white coat comes from extreme expression of the white
spotting gene in the dominant homozygote, SS, masking all colors and
patterns. This coat may have a few non-white hairs, especially on a
kitten. The eyes will be the proper color for the masked "true" coat
colors, and may be anything except dominant-white blue, albino blue, or
pink.

The dominant white coat comes from expression of the dominant-
white gene, "W*", masking all colors and patterns. The eyes are always
orange, dominant-white blue, or odd (one of each). The blue—eyed
albino comes from expression of the blue-eyed albino allele of the
albino gene, "ca*", masking all colors and patterns. The eyes are
always albino blue. The albino coat comes from expression of the
albino allele of the albino gene, "cc", masking all colors and
patterns. The eyes are always pink.

Color I Genotype I Usual eye color

______________________ +-____.____.____-____________+___.__.________.___
full-spotted white I ****** ****** **SS** I not wbl/abl/pnk
dominant white I ****** ****** ****W* I org wbl odd

blue-eyed albino I ****** ca***** ****** I alb

albino I ****** CC**** ****** I pnk

85

Meiosis and Fertilization
Adapted from 'Mitosls, Meiosis, and Fertilization"
Copyright, 2005, by Dr. R. Scott Poethig, Dr. Ingrid Waldron, and Jennifer Doha/1y
Department of Biology, University of Pennsylvania. Used with Permission.

Meiosis

Mitosis gives rise to almost all the cells in the body. A different type of cell
division called meiosis gives rise to sperm and eggs. During fertilization the
sperm and egg unite to form a single cell called the zygote which contains
chromosomes from both the sperm and egg. The zygote undergoes mitosis to
begin development of the human embryo which eventually becomes a baby.

a. In humans, how many chromosomes should a zygote have, so the baby's
body cells will each have a normal set of chromosomes?

b. If the sperm and egg each contribute all of their chromosomes to the zygote,
how many chromosomes should each sperm and egg have to produce a normal
zygote?

0. Suppose sperm and eggs were produced by mitosis. If a sperm of this type
fertilized an egg of this type, how many chromosomes would the resulting zygote
have?

d. Why would this be a problem?
e. How could this problem be avoided?

Meiosis reduces the number of chromosomes by half, so in humans each sperm
and each egg has only 23 chromosomes, including one chromosome from each
pair of homologous chromosomes. Therefore, fertilization results in a zygote
which has 23 pairs of homologous chromosomes, one in each pair from the
sperm and one from the egg. When the zygote undergoes mitosis to begin to
form an embryo, each cell will have the normal number of 46 chromosomes.

Cells that have two copies of each chromosome (i.e. cells that have pairs of
homologous chromosomes) are called diploid cells. Most of the cells in our
bodies are diploid cells. Cells that only have one copy of every chromosome are
called haploid cells.

f. Which types of cells in our bodies are haploid?

Meiosis consists of two cell divisions, meiosis l and meiosis II. This reduces the

chromosome number by half and produces four haploid daughter cells (instead of
two diploid daughter cells as in mitosis).

86

Meiosis l

Meiosis | is different from mitosis because homologous chromosomes line up
next to each other and then separate, as shown below. This produces daughter
cells with half as many chromosomes as the parent cell, i.e. haploid cells. Notice
that each of the daughter cells has a different chromosome from the homologous
pair of chromosomes. This means that the alleles

in each daughter cell are different.

Meiosis ll

Meiosis II is like mitosis. Each chromosome splits in half, so that each daughter
cell inherits one chromatid from each chromosome.

Using your sockosomes, go through each step of meiosis until you are confident
that you understand the difference between Meiosis I and Mitosis and the
difference between Meiosis land Meiosis ||.

87

Qroug Assessment
For this activity, you will be working with the members of the group at your table.

First, practice reviewing the steps of meiosis together. Make sure that everyone
in your group can explain all of the steps.

Your grade on this assessment will be a group grade. Each person will be asked
a different question about the process of meiosis. The response to these
questions will determine the group ’8 grade on this assessment.

Now, use your sockosomes to model meiosis in a cell which has two pairs of
homologous chromosomes. Find another group that has the two different
versions of the gene you do not have, either albinism (A for pigmented skin and a
for albinism) or thumb bending (H for straight thumb and h for the hitchhikers
thumb). Put these four sockosomes in a pile in the middle of a circle which
represents a cell. This pile of sockosomes will represent the two pairs of
homologous chromosomes, each with the DNA copied, so the cell is ready to
undergo meiosis. Model the steps in meiosis.

g. List all of the different combinations of alleles that can be observed in the
daughter cells (sperm or eggs) produced by meiosis in the chart below.
Combinations of Albinism and Thumb Bending Alleles

 

 

 

 

 

 

 

 

 

h. Describe the differences between the mother cell that undergoes meiosis and
the daughter cells produced by meiosis.

88

 

Fertilization

a. The following diagram provides an overview of the information covered thus
far. Review the diagram, and fill in the correct number of chromosomes per
human cell in each blank.

 

Mother Father
Meiosis ,1, Meiosis l
999 sperm __
Ndilization

zygote

Mitosis l
Embryo

Mitosis ,1,

baby

Now we will consider fertilization in more detail, in order to provide a basis for
understanding genetics. Events during meiosis and fertilization determine the
genetic makeup of the zygote, which in turn determines the genetic makeup of
the baby mat develops from the zygote.

You already know that sisters or brothers can have different characteristics, even
when they have the same parents. One major reason for these different
characteristics is that the processes of meiosis and fertilization result in a
different combination of genes in each child.

To begin to understand this genetic variability, you will model meiosis and
fertilization for a very simplified case where there is only one pair of homologous
chromosomes per cell, and the two homologous chromosomes carry different
alleles of the same genes. One person in your group will be the mother with two
sockosomes that are the same color that represent the two maternal homologous
chromosomes with different alleles of the same labeled gene. Another person
will be the father with two sockosomes that are a different color than the mother’s
carrying the same two labeled alleles as the mother’s sockosomes. (The
different colors for the mother‘s and father‘s sockosomes represent the fact that,
although the labeled alleles are the same for the mother’s and father’s
chromosomes, there are many genes on each chromosome and the mother’s
and father’s chromosomes will have different alleles for many of these genes.)

b. In this simple example, how many different types of eggs will be produced by
meiosis?

89

c. How many different types of sperm will be produced by meiosis?

The different types of sperm can fertilize the different types of egg to result in
zygotes with different combinations of chromosomes from the mother and the
father. Fertilization can be demonstrated by having the mother and father each
contribute one chromatid from one of their sockosomes, representing one
chromosome from the egg and one chromosome from the sperm. Try to produce
as many different types of zygotes as you can by pairing each type of sperm with
each type of egg. To demonstrate fertilization, it works best to lay the
chromosomes out on the table, so you can more easily see the multiple different
possible combinations.

d. How many different types of zygotes can be produced by fertilization in this
simple case?

e. What different combinations of the labeled alleles can be observed in the
zygotes?

A pair of human parents could produce a great many more different genetic
combinations than observed in this simplified example. For example, humans
have 23 pairs of homologous chromosomes, so many, many different
combinations of chromosomes can be found in the eggs or sperm produced by
one person, and the different combinations of eggs from one mother and sperm
from one father could produce zygotes with approximately 70 trillion different
combinations of chromosomes! You can see why no two people are genetically
alike, except for identical twins who are derived from the same zygote.

Questions
1. How many chromosomes are there in a human skin cell produced by mitosis?

How many chromosomes are there in a human sperm cell produced by
meiosis?

2. Describe the differences between mitosis and meiosis.

3. What are the similarities between mitosis and meiosis?

9O

 

Name: Hour:
The Reebop Family: An Activity In Meiosis and Genetic Variation

Reebops are an imaginary species that are prolific and require minimal care. In this
activity you are to simulate the process of Reebop meiosis and fertilization. The activity
will result in the construction of Reebop offspring. You are to sort Mom and Dad Reebop
chromosomes using the concepts of meiosis in forming the gametes, decode the
chromosomal code found on the baby. After the Reebops are born, the Reebop siblings
will be assembled together in the nursery to be analyzed. Chromosomal analysis of this
unusual species has revealed that Reebops have 7 pairs of chromosomes, a total of 14
chromosomes.

Objectives:
A. To simulate a model of meiosis
B. To interpret the results of the role of meiosis in sexual reproduction

Procedure:
1. Meet Mom and Dad Reebop. They are on the front desk. Note their general
anatomy.

2. On your desk is an envelope that contains a copy of both Mom and Dads
chromosomes. Remove the chromosomes from the envelope and separate them
by sex. In this simulation, Dad Reebop’s chromosomes are blue and Mom
Reebop’s chromosomes are pink. Of course, this is not true in real organisms.

3. Your partner will get the 7 pairs of chromosomes of one sex and you will get the
7 pairs of the other sex. Pair up each set of chromosomes.

a. Are the parents haploid or diploid? . Explain.

b. How did you know how to pair up the chromosomes? (2 ways)

0. Write the term that describes these paired chromosomes.

 

4. You will simulate the results of meiosis by forming gametes from Mom Reebop’s
and Dad Reebop’s chromosomes. After the homologous chromosomes are lined
up you will flip a coin to determine which of mom’s chromosomes are segregated
in to the gamete. For the first pair of chromosome, choose the chromosome on
the left if you flip heads, and choose the chromosome on the right if you flip tails.
Be sure to segregate mom and dad’s chromosomes separately. Put this
chromosome to the side and return the chromosome not chosen to the envelope.
Repeat for pair two. Repeat until all of the chromosome pairs have been used.

5. The other person should repeat the process in #4 for dad’s chromosomes.

6. Put the two gamete chromosomes together.

a. What process does joining the gametes simulate?

b. What do we call the fertilized egg?
o. Is the fertilized egg haploid or diploid?

7. Congratulationsll A baby Reebop has just been born.

91

Each baby should have pairs of chromosomes, one of each pair from
and the other from

 

 

8. Turn the Baby Reebop’s chromosomes over and use the key to decode its
chromosomes code as you construct this new offspring.

9. Put your finished Baby Reebop in the nursery with its siblings and compare the
babies.
a. Are any two Baby Reebops alike?

b. Why do they differ?

c. What makes identical twins identical?

d. Does each parent contribute the same amount of genetic information to a
child?

9. Why do siblings in a given family look similar yet are all different?

Data and Results
Allele
Ma
Q/
E/e
D/d
M/m
T/t
Ul

Class XX Xx Xx Genotypic Ratio Phenotypic
Data Ratio

 

92

Challenge Questions:.
1. Give an example of a genotype and a phenotype from this activity.

2. Would it be more likely that two siblings would look alike if reproduction
occurred by mitosis or meiosis. Explain.

3. How do the Mom and Dad Reebops compare to one another? Based upon
observation, can you tell if they’re genetically identical? Why or why not?

4. When you choose one chromosome from Mom and one from Dad and put them
in the baby pile, what processes does this action represent in actual
reproduction?

5. Mom and Dad Reebop are heterozygous for every trait. This means that they
are hybrids/purebred (circle one)

93

“Human Chromosomes” -- Teacher Edition

Adapted from Joesph P. Chinnici, Joyce W. Yue, and Kieron M. Torres

“Students as ‘Human Chromosomes’ in Role-Playing Mitosis &
Meiosis.” The American Biology Teacher, 66(1): 3539.

Objectives:

1.

Explain these terms: homologous chromosomes, sister chromatids,
genome, diploid, haploid, zygote, allele

 

2. Explain how homologous chromosomes segregate and non-
homologous chromosomes assort independently during meiosis

3. Predict the role of meiosis in sexual reproduction.

Materials:

1. baseball cap (or other kind of hat)

2. 2 plastic nametags (one with letter inserted, one with 4X4 in. cardboard
piece attached)

3. “Guide to being a human chromosome”

Procedure:

1. Attach your “alleles” to your hat and shirt. Introduce the idea of alleles.

2. Our classroom today will represent a cell. To begin with our cell will be
the largest square. This may seem kind of large, but remember, a cell
must grow large enough before it divides.

3. First, one person with each type of letter (including upper and lower case)
is going to be involved. Come to the middle part of our cell, not lined up
yet. The rest of you will be working in a second.

4. Before cells divide, what must happen? DNA replicates.

5. In order to show replication of DNA, we are going to link arms with our
replicate. Those of you on the outside, find your replicate and link arms.

6. Once they are replicated, what do we call each replicated chromosome?
Sister chromatids

7. The first division in meiosis is the separation of homologous
chromosomes. So homologous pairs need to line up together on the
equatorial plane of the cell (line A). Be careful how you are lined up!

8. Divide!!! Area 1 and area 2 represent separate cells (either side of line A).
Is this the only way you can arrange yourselves?

9. Who can tell me which part of what we did represented segregation?

Law of independent assortment?

10. Are these gametes ready to be fertilized?
11. The second division needs to occur. Sister chromatids line up along the

equatorial plane of cells 1 and 2 (line B)

12. Divide!!! Areas 3-6 represent the four separate cells (separated by lines A

and B). Segregation? Law of independent assortment?

13. Let’s take a look at the alleles found in each “cell.” Fill in the chart on the

board.

14. Let’s do this a few more times. Each time, I am going to give you less and

less help remembering what to do. Remember, you do not have to go the

94

same way each time. We will fill in our gamete information each time on
the board.

15. We will choose one gamete from each round to fertilize a gamete that I
give you. You will use this information to draw 3 “people.”

Classroom Set Up

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3
<— A 6
l 2
B
4 5
# # #of # # #of
students homologous correction students hOITIOIOQOUS correction
Pairs enzymes Pa'rs enzymes

32 8 O 26 6 2
31 7 3 25 6 1
30 7 2 24 6 O
29 7 1 23 5 3
28 7 0 22 5 2
27 6 3 21 5 1
Letter 32 students 28-31 student 24-27 20-23
Pairs

A I H G F

B L l H G

C K J I H

D J K J |

E M L K J

F N M L - '

G O N

H P ................................................................................................................

 

 

 

 

 

 

 

95

Name: Hour:
Students As Human Chromosomes Activity
Results
Round Cell 3 Cell 4 Cell 5 Cell 6

 

 

 

 

 

 

 

 

 

 

 

eyes taster
taster

peak widow's fingers/toe
Hairline 3

thumb thumb (nearsighted) vision

over left over

hair middigital

hair

eaflobes afiached
eaflobes

 

chin in chin

Other parent’s genotype:

 

96

Directions:

1. Draw and label a picture of each of the “children” that result from the cross of
the “chosen” gamete created during each round of our game and our parent.

2. Answer the analysis questions.

 

 

Round One Round Two Round Three

 

 

 

 

 

Challenge Questions:

1. Briefly describe what occurs during the process of meiosis. Use the

homologous chromosomes A a and T t to help you explain.

2. Compare the law of segregation with the law of independent assortment.

3. How is it possible that two siblings could have the same parents but look

different from each other?

97

Name: Hour:
Pannen

 

Spot the Cat?

Bert and Ernie missed their friend Graycie. For just a brief time, she had
touched both of their lives. Even Bert missed her a little from time to time. They
decided to visit the Diamonds and PUURRLS cattery one afternoon to see her.
Once there, they met some new friends, Bub and his sister, Spot.

Bub was lounging with his big white belly showing. Bert and Ernie noticed
that more than half of Bub was made up of white spots. Perched on a counter
nearby, they saw Spot. “What a clever name,” said Ernie. “She’s about half
made up of white spots.” Bert wondered what made a cat have spotted white fur.

“Last time we were here, we found out that having solid white fur was a
dominant trait,” he said. “Do you remember Marshmallow and Cloud?”

“Of course I do,” replied Ernie. “But they didn’t have any spots, Bert. Having
spots must be a different type of gene.”

“Maybe the amount of spotting is controlled by this other gene. You know,
Ernie, we’ve been getting pretty good at genetics,” Bert said proudly. “I’ll bet
that we could figure out the genotypes and phenotypes of Bub and Spot.”

“Won’t we need to know what their parents look like?” asked Ernie.

“Naw, Ern. I’ll bet we can even figure out what their parents might look like
just by using what we know about Bub and Spot. In fact,” he continued, “I’ll bet
we can figure out what the kittens would look like if Bub and Spot were bred
together.”

Can you do the same? (Concentrate on the spotting trait) Be sure to use your
research materials and text book to help you.

98

A “Tail” of Two Kitties

Sometimes, a client of the Diamond and PUURRLS cattery has a specific
cat in mind...a cat that doesn’t even exist. They may see some of the parent
cats at the cattery in online pictures and think, “I would love a kitten that looks
exactly like that!” Such was the case when a very particular client called the
cattery and explained that she would love to take home a cat that looked like
their award wining Kat. Kat’s parents were no longer available for breeding,
so the breeders needed to use the cats that were available at the cattery. A
cat named Whiskers was chosen and bred with a cat named Scout. Two of
the kittens, Calli and Sundae were tortoise cats like Whiskers. The client
bought both of them because they looked so much like Kat. One kitten was a
cute white cat with spot named Sprinkles. Two of the cats, Garfield and
Sherbert, were orange.

“Oh, I love the little orange ones. They are so precious,” said the
client. “I would take one of them, too, but they are both boys. Since Calli and
Sundae are girls, I don’t think it would be a good idea. Do you have any
kittens available that are orange girls?”

“Actually, most orange cats are boys,” said the breeder. “It is very rare
to find a female orange cat.”

“How strange,” said the client. “I thought all cats could be all colors.”

“Well, orange pigmentation is an unusual gene,” said the breeder. “In

fact, tortie cats are always girls.”

“I think I may know why!” said the client.

Can you figure out why?

Goals for this PBL:
1. Research how orange pigmentation is passed on.
2. Determine the genotypes of Whiskers, Calli, Sundae, Garfield, and
Sherbert.
3. Determine why only female cats appear as torties.
4. Determine why most orange cats are males.
5. Explain whether or not a female cat could be orange.

99

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