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thesis entitled

Teaching Genetics-An Activity Based Approach

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Michael Matthew Sampson

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TEACHING GENETICS-AN ACTIVITY BASED APPROACH
By

Michael Matthew Sampson

A THESIS
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Master of Science

Division of Mathematic and Science Education

2002

ABSTRACT
TEACHING GENETICS-AN ACTIVITY BASED APPROACH
By
Michael Matthew Sampson

When teaching a unit on genetics there are a few “traps" into which one can
fall. The first is the reliance on repetitious paper and pencil worksheets. To
compound this problem, students come into class with misconceptions and a poor
understanding of basic genetics concepts. Finally, students lack an understanding
of human inheritance and the application of genetics. The purpose of this thesis is
to document the improvement of student performance of a redesigned genetics
unit. The content being taught in this redesigned genetics unit included; genes
and chromosomes, human inheritance patterns and applied genetics. The effect
of replacing paper and pencil worksheets with laboratory activities, online activities
and problem solving activities is studied. The effectiveness of these techniques
was evaluated by assessing student performance on pre and posttests, activities
and student weekly reflections. The data show that student performance was
positively impacted by these changes. In addition, student attitudes toward the
processes implemented were documented. The results show that student learning

was enhanced by these changes.

This thesis is dedicated to my family, thank you for your love and support.

ACKNOWLEDGMENTS

The author wishes to express a great deal of appreciation to Drs. Merie
Heidemann and Ken Nadler for their help in the preparation this thesis. A special
note of thanks to Wayne County RESA for the financial support to complete this
research. Thanks also to the students of Redford Union High School for the
valuable data they have provided. In addition, thanks to KTP, the head of the
Redford Union science department. You have been the source of many great
ideas and made me a better teacher. Finally, a special thanks to Matt Withers, you
were a great help in designing some of these activities during our research

summer.

TABLE OF CONTENTS

Number Page
LIST OF TABLES ................................................................................ vii
LIST OF FIGURES .............................................................................. viii
INTRODUCTION ................................................................................. 1
CHAPTER ONE- Implementation ............................................................. 7
CHAPTER TWO- Data Collection Methods ............................................... 17
CHAPTER THREE- Results and Evaluation ............................................... 20
CHAPTER FOUR- Discussion ............................................................... 31
APPENDICES .................................................................................... 34
Appendix A- Pre- and Post-tests ........................................................ 35
Genes and Chromosomes Pretest ................................................ 36
Genes and Chromosomes Posttest ................................................ 38
Human Heredity Pretest ............................................................... 43
Human Heredity Posttest ............................................................. 44
Applied Genetics Pretest ............................................................. 49
Applied Genetics Posttest ............................................................ 55
Appendix 3- Weekly Reflections ........................................................ 60
Weekly Reflection Form ............................................................... 61
Appendix 0- Guided Note Handouts ................................................... 62
Genes and Chromosomes Outline ................................................. 63
Royal Pedigree .......................................................................... 66
Sex Determination Outline ............................................................ 67
Mutations Outline ....................................................................... 70
Human Heredity Outline ............................................................. 72
Human Genetic Disorders Outline ................................................. 75
Breeding Techniques Outline ....................................................... 78
Gel Electrophoresis Outline ......................................................... 80
Genetic Engineering Outline ......................................................... 82
Applied Genetics Outline ............................................................. 84
Appendix 0- Activities ........................................................................ 86
Sex-Linked Trait Review Sheet ...................................................... 87
Mutations Activity ....................................................................... 89
Colorblindness Activity ................................................................. 93
Karyotyping Activity ..................................................................... 95

V

Appendix D (can't)

Human Genetics Activity Form 1 ................................................... 99
Human Genetics Activity Form 2 ................................................... 102
Genetic Disorders Paper ............................................................. 105
Apple Lab ................................................................................. 107
Apple Paper ............................................................................. 1 10
DNA Extraction .......................................................................... 1 1 1
Recombinant DNA Activity 1 ........................................................ 114
Recombinant DNA Activity 2 ........................................................ 118
Food Dye Electrophoresis ............................................................ 123
Gel Electrophoresis .................................................................... 126
Transformation of E. coli .............................................................. 131
REFERENCES .................................................................................. 141

vi

LIST OF TABLES

Number Page
Table 1. Genes and Chromosomes Unit Outline ....................................... 8
Table 2. Human Genetics Outline ......................................................... 10
Table 3. Applied Genetics Unit Outline .................................................. 13
Table 4. Pre-test and Post-Test Scoring Rubric ....................................... 18
Table 5. Student Misconceptions— Genes and Chromosomes Pre-test .......... 21
Table 6. Summary of Weekly Reflections for Genes and Chromosomes ....... 23
Table 7. Student Misconceptions- Human Heredity Pre-test ....................... 25
Table 8. Summary of Weekly Reflections for Human Heredity .................... 26
Table 9. Student Misconceptions- Applied Genetics Pre-test ...................... 28
Table 10. Summary of Weekly Reflections for Applied Genetics ................. 30

vii

LIST OF FIGURES

Number Page
Figure 1. Genes and Chromosome Pre-test Data .................................... 20
Figure 2. Genes and Chromosome Post-test Data .................................. 22
Figure 3. Genes and Chromosome Weekly Reflections Data ..................... 23
Figure 4. Human Heredity Weekly Pre-test Data ........................................... 24
Figure 5. Human Heredity Weekly Post-test Data ................................... 26
Figure 6. Human Heredity Weekly Reflections Data ................................ 27
Figure 7. Applied Genetics Weekly Pre-test Data ..................................... 28
Figure 8. Applied Genetics Weekly Post-test Data .................................... 29
Figure 9. Applied Genetics Weekly Reflections Data ................................. 30

viii

INTRODUCTION

The teaching of genetics in high school can be problematic. Many teachers
use traditional methods that are archaic and, without a doubt, in need of
improvement. These traditional teaching methods, such as an objectivist
methodology, lead to students that are passive learners who look to the teacher or
the textbook as the only “correct answer.” In addition, many students begin to
learn this topic without the prior knowledge required to succeed. When provided
with new information, many students are unable to apply information to solving
genetics problems. This lack of prior knowledge, combined with student
misconceptions about genetics, make learning difficult (Banet and Ayuso, 2000).
The purpose of this study is to show that utilizing a methodology that is consists of
both a social constructivist and objectivist approach impacted student learning of
genetics.
Background

There are numerous studies about the importance of linking pn'or knowl-
edge to learning new information. The more links learners can make with their
prior knowledge base the easier it is for them to intemalize the concept. In order to
make these links between new and prior knowledge, science must be a collabora-
tive activity that utilizes problem-solving activities that change during knowledge
construction (Finkel, 1996). Finally, Banet and Ayuso (2000) also state that when
developing strategies to teach genetics, one must tie into the students' prior

knowledge so they can construct their own Ieaming and problem solving skills.

Students enter any Ieaming environment with preconceived notions and
misconceptions. According to Hogan (1999), student Ieaming occurs in many
social settings including their community and their school. Misconceptions can be
attributed to these social contexts and their personal experiences. Driver, Asoko,
Leach, Mortimer and Scott (1994) state that “commonsense explanations”
(misconceptions) are constructed, communicated and validated in social settings.
This would lead one to believe that in order to change misconceptions, a social
context is required. Gil-Perez and Carrascosa (1990) examined misconceptions in
a physics classroom and called for a change in both teaching and Ieaming
practices to improve "meaningful Ieaming.” These changes were the basis for this
research. Furthermore, they state that when making these changes students need
to know their thinking resembles that of a scientist (ibid, 1990). Banet and Ayuso
(2000) discuss the work of many researchers that looked at the misconceptions
students have about genetics when entering a high school biology class. These
common misconceptions include the relationship between genes, chromosomes
and alleles, the relationship between probability and expression of traits, and the
difference between heterozygous and homozygous). Their study showed that
misconceptions existed even after students were taught the basics of biology.

There is a great deal of research in the areas of how to teach science to
students. There are two traditional methods of instruction used by many
educators: a constructivist approach or an objectivist approach. The constructivist
approach leans heavily on students ”doing" science in cooperative group settings.

According to Driver, Asoko, Leach, Mortimer and Scott (1994), science is a social

process and social interaction promotes Ieaming via the discussion of different
points of view. These group activities allow social interaction between peers which
enable learners to discuss, correct misconceptions and exhibit "meaningful
learning” (Chin and Brown, 2000). Conversely, the objectivist approach is more
passive. In an objectivist approach, the learners look to the book and instructor as
the "fountains of knowledge.” This approach utilizes an instructor driven
approach. Osbourne (1996) points out that both of these methodologies have their
shortcomings, "...just as traditional, objective teachers have often failed to
recognize the need to stmcture Ieaming so that it is active, the advocates of
constructivist methods of teaching have failed to recognize that there is a role for
telling, showing and demonstrating.” Passive learners memorize and regurgitate
facts, resulting in learners that cannot assimilate newly acquired information and
apply it in a problem solving capacity. On the other hand, students who are taught
utilizing only a constructivist methodology can become too dependent on group
feedback. This may lead into deepening their misconceptions if student groupings
are not carefully monitored. Both of these methodologies have their merits.
Hence, educators need to utilize the best of both methods to better reach their
learners and reap the rewards of improved student Ieaming.

There are two Ieaming styles that students utilize, rote Ieaming and
meaningful Ieaming. According to Cavallo and Schafer (ibid), students who learn
by rote are driven by memorizing facts and definitions, cannot apply this
information to new problems, and they do not demonstrate the ability to interrelate

concepts. These authors also state that students who exhibit meaningful Ieaming

have the ability to personalize concepts, link these concepts to their prior
knowledge, and apply these newly learned concepts to solve problems. Thus
there are three requirements for meaningful Ieaming to occur". educators must
make the content meaningful to the students, students must have relevant prior
knowledge, and students must actively attempt to make relationships and
inferences to new concepts.

Another problem area in teaching genetics is the inability of students to
relate biological concepts to genetics. This is another area where a great deal of
educational research is being performed. Cavallo and Schafer (1994) state that
students learn individual biological concepts associated with genetics but lack the
ability to relate them to the “big picture.”

Rationale

The research cited in the preceding paragraphs led to the development of
this question. How can student Ieaming of genetics be improved? The summer of
2001 was spent researching an answer to this question. The result was a
redesigned high school genetics unit. The first change made involved the
elimination of repetitive paper and pencil worksheets. These worksheets were
replaced with group activities that required the actions of active Ieamers. As
Driver, et al (1994) stated, science is a social process and social interaction
promotes Ieaming via the discussion of different points of view. Many of these ac-
tivities required the learner to interact with peers and discuss the reinforcement of

the concepts being taught.

The second change involved the method by which information was
presented to the class. The PowerPoint lecture notes used in the past were been
updated with more current and relevant information, and new eye-catching
graphics. In addition, students were provided with an outline that they used as a
tool to record pertinent information. The outline format was used to free time so
the class could discuss what was being learned.

Cavallo and Shafer (1996) and Banet & Ayuso (2000) state that to increase
meaningful Ieaming educators must make the content personally relevant. Making
human heredity a common thread throughout the unit met this goal. Pre-tests
(Appendix A) were used as a guide to assess on which topics students needed
remediation. Bell and Cowie (2001) state that a pre-test can serve as a formative
assessment tool if it is used to evaluate and make changes that improve on
teaching practices. Another assessment tool utilized to modify teaching style were
the Weekly Reflections (Appendix B). These forms provided feedback on how the
students felt about what they learned and the processes used. This feedback
provided data that was uSed to make changes in teaching methods. According to
Etkina (2000), providing a vehicle for student feedback on the process gives
insight to if there is meaningful Ieaming taking place. The effectiveness of these
changes was evaluated by focusing on these three questions:

. With what information do the students enter the class?

. What information have the students learned?
. How do the students feel about the Ieaming process?

The data gathered during this research will show that these changes discussed in
this introduction have made a positive impact on student Ieaming and understand
in the area of genetics.
Study Group

In this study, the test group consisted of sixty-six suburban students in a
public high school. The students were enrolled in three General Biology classes.
The classes consisted of sophomores and juniors. There were thirty-nine females
and twenty-seven males in the group. Approximately eighty-six percent of the
students were Caucasian. Twelve percent were African-American and two percent
other minority groups. About eight percent of the group has a special education
designation. Most of the students are entering their second year of high school
science. They have taken the Freshmen Level Science Survey courses. This
course introduces the students to three areas of science: Earth, Life and Physical

Science.

CHAPTER 1
Implementation

This document will focus on changes made in three main areas of a
genetics unit: genes and chromosomes, human genetics, and applied genetics.
The first change made involved eliminating repetitive worksheets had replacing
them with activities that required the actions of an active learner. The new ac-
tivities required the Ieamer to take on a cognitive approach because they required
the application and reinforcement of the concepts taught. The second change
involved the method by which information was presented to the class. The
PowerPoint lecture notes used in the past were updated with more current and
relevant information, and new eye-catching graphics. In addition, students were
provided with an outline that they used as a tool to record pertinent information.
The final change involved the addition of more laboratory activities. The addition
of these labs activities allowed the students to play an active roll in how the
science of genetics is applied to everyday life. All of the applied genetics labs
involved the manipulation of DNA. These changes will be addressed in more
detail in the following paragraphs.
Genes and Chromosomes

The genes and chromosome unit lasted two weeks (I' able 1). The unit
began with a pre-test (Appendix A). A discussion of the answers to the pretest
questions ensued. The pre-test served two functions; to determine the prior
knowledge of each student upon entering the class and to reveal those areas that

needed to be covered in more depth due to the lack of prior knowledge. With the

level of prior knowledge determined, and areas of weakness revealed, classroom

discussions were designed.

 

Table 1. Genes and Chromosomes Unit Outline

 

Day One Genes and Chromosomes Pre-test*

Review Pre-test*

 

Day Two Introduce Genes and Chromosomes with guided note handout‘

 

Day Three Quiz on Genes and Chromosomes
Introduce Sex-determination and Sex-Linked Traits with guided note handout“

Sex-Linked Traits worksheet

 

Day Four Check Sex-Linked Traits Worksheet
Quiz on Sex-Linked Traits

Introduce mutations and discuss the effects of mutations with guided note

 

 

 

 

 

handout’
Day Five Pre-lab discussion of Mutations Activity
Day Six Mutations Activity
Day Seven Review Mutations Activity results
Quiz Mutations
Day Eight Jeopardy Review Game
Day Nine Post-test

 

 

 

* denotes newly developed material or activity

 

All new information was presented via a PowerPoint slideshow. The information

presented in the slideshows included the relationship between genes and
chromosomes, sex determination, sex-linked inheritance and mutations. The
students were provided with a guided outline (Appendix B) that required that they
fill in important points during the each discussion. The first topic discussed in class

was the relationship between genes and chromosomes. On day three, the

 

students’ mastery of the content was assessed by a short quiz. The quiz
contained some of the questions from the pre-test. In an effort to provide the
students with instant feedback, and correct any confusion or misconceptions, the
quiz was corrected in class. On day three, after the quiz, the class was introduced
to the topic of sex-linked inheritance and sex chromosomes. This topic was
followed by a cooperative group assignment that required the students to review
the basic information related to sex-linked inheritance. On day four, the class
reviewed the relationship between chromosomes and sex-linked inheritance. In
addition to this review, the class took a quiz to determine their level of
comprehension for this topic. On day five, the topic of how mutations affected
individuals was discussed. A cooperative group activity involving mutations was
performed on day six. The mutations activity, Mutations-Get the Point, allowed the
students to work in groups and use manipulatives to produce sentence models of
different types of mutations. This activity simulated frame shift, insertion and
deletion mutations. Day seven served two functions; to discuss the data collected
during the mutation activity and to quiz the class on their knowledge of mutations.
Day eight was the review day. Students played a review game whose format was
similar to the game show Jeopardy”. The review format was enjoyed by the
class. In addition, this game provided students with opportunities to gain a few
extra credit points towards their test. On day nine, the class took the post-test.
This post-test consisted of an assortment of multiple choice, true/false, and open-

ended questions taken from the pre-test.

Human Heredity

The unit focusing on human inheritance patterns unit lasted about three-

weeks (Table 2). Similarly to the previous unit, this unit began with a pre-test

(Appendix A) and a discussion related to the answers to the pretest questions.

 

Table 2. Human Genetics Outline

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Day One Human Genetics Pre-test*
Review Pre-test
Day Two Introduce Heredity in Humans with guided note handout"
Day Three Review Sex-linked traits and Introduce Sex-Linked traits in humans with
guided note handout”
Sex-Linked Traits Review worksheet.
Day Four Check Sex-Linked Traits Worksheet
Introduce Human Inheritance with guided note handout"
Day Five Discuss Colorblindness (online colorblindness test)
Quiz on Human Inheritance
Introduce Human Genetic Disorders with guided note handout"
Day Six Online Colorblindness Activity
Day Seven Online Karyotyping Activity
Day Eight Human Mutation Activity’
Day Nine Human Mutation Activity day two
Day Ten Introduce Human Genetic Disorders Project & begin research
Day Eleven Research Human Genetic Disorders Project
Day Twelve Research Human Genetic Disorders Project
Day Thirteen Research Human Genetic Disorders Project
Day Fourteen Turn in Genetic Disorder Project

 

Jeopardy Review Game

 

10

 

 

Table 2 (cont’d)

 

Day Fifteen Post-test

 

 

 

‘ denotes newly developed material or activity

 

All information presented in this section was done via PowerPoint slideshows.
These slideshows covered the topics of Blood types, Rh factors, polygenic traits,
sex-linked traits and genetic disorders. In order to make the content more relevant
to the students, class discussions were used to introduce the topic for each day.
These discussion questions were derived from an article in The American Biology
Teacher (Haddow, Eunpu, Singer, Brant, & Ledwith, 1988). Students used a
handout (Appendix C) while taking notes that required that the students fill in
important points during the discussion. The first topic discussed was the
relationship between genes and chromosomes and how they related to human
inheritance patterns. On day three, the class was guided in a discussion about
sex-linked inheritance and how it related to humans. A pedigree chart from a
University of Virginia web site, (http://www.people.virginia.edul~ rjh9u/roylhema
.htrnl) was used to identify the hereditary patterns in the European royal families
(Appendix C). Day four involved a class discussion on polygenic traits and, more
specifically, human blood types. On day five, the class discussed how some
inherited disorders, such as color blindness and Down's syndrome, occur. The
students participated in an online colorblindness test (http://members.aol.com/
protanope/colorblindtest.html) to determine if any of the students may be
colorblind. This activity was the impetus for a discussion on diagnosing genetic
disorders. The next two days were spent in the computer lab doing two online

activities. The first of these activities, Colorblindness Problem Set (Appendix D),

11

 

required that the students construct a colorblindness pedigree and answer
questions based on patient information provided on the Biology Project web site
(www.biology.arizona.edulhuman_bio/problem_sets/color_blindness/color_blind

ness.html). The next activity, Karyotyping Online (Appendix D) also on the Biology
Project web site, (http://www.biology.arizona.edu/human_bio/activities Ikaryotyping
Ikaryotypinghtml) required the students to diagnose the genetic disorders of three
patients after having manipulated and observed the patients’ karyotypes. These
web-based activities are a method of utilizing technology in the classroom. Using
technology in the classroom is one component of the school districts” school
improvement plan. On day eight, the class began another cooperative Ieaming
activity that linked mutations to genetic disorders. In the Human Mutations Activity
(Appendix D), the students relied on their prior knowledge involving transcription,
translation, and mutations. The students were given a list of codons that code for
one of the proteins that make up a portion of the hemoglobin protein called beta-
globin. Using prior knowledge from previous biology units, each student needed to
transcribe and translate his/her beta-globin codons, then identify which type of
beta-globin he/she had: the mutated strand that causes sickle cell anemia or the
normal, wild type, strand. This activity was a good culminating activity because it
illustrated the link between DNA, RNA, mutations, and genetic disorders. On day
ten, the students chose their topic for their genetic disorder paper (Appendix D).
Over the next three days, the students spent time in the computer lab and library to

research the genetic disorder they chose and wrote their papers. On day fourteen,

12

the students played the JeopardyW-like review game. On the last day of this unit,

the class took the post-test.

Applied Genetics

The applied genetics unit (Table 3) lasted four-weeks. This unit served as

the unifying theme for the entire genetics unit. As with the previous units, this unit

began with a pre-test (Appendix A). In an effort to provide instant feedback about

student strengths and weaknesses, the pre-test was graded in class.

 

Table 3. Applied Genetics Unit Outline

 

 

 

 

 

 

 

 

 

 

 

 

 

Day One Applied Genetics Pre—test*

Day Two Introduce breeding techniques with guided note handout’
Discuss selective breeding of corn, apples and dogs.

Day Three Apple Lab’

Day Four Research types of apples and how they were bred

Day Five Quiz Breeding Techniques and Mutations
Review basics of DNA
Pre-lab discussion of DNA Extraction Activity

Day Six DNA Extraction Lab“

Day Seven Review DNA extraction results
Pre-lab Working with Restriction Enzymes One and Two‘
Discuss discovery and use of restriction enzymes with guided note
handout"

Day Eight Working With Restriction Enzymes Activity One"

Day Nine Working with Restriction Enzymes Activity Two‘

Day Ten Pre-lab Food Coloring Gel Electrophoresis Practice Activity“
Discuss process of gel electrophoresis with guided note handout'

Day Eleven Food Coloring Gel Electrophoresis Practice Activity’

 

 

13

 

 

Table 3 (cont’d)

 

 

 

 

 

 

 

 

 

 

 

 

Day Twelve Review Food Coloring Practice results
Pre-lab Restriction Analysis and Electrophoresis Activity“
Day Thirteen Restriction Analysis and Electrophoresis Activity“
Day Fourteen Review Restriction Analysis and Electrophoresis results
Introduce genetic engineering with guided note handout“
Day Fifteen Video-Genetically Modified Organisms“
Day Sixteen Video-Genetically Modified Organisms“
Day Seventeen Quiz on Electrophoresis
Discuss benefits and danger of altering the genetic code
Discuss the application genetics with guided note handout“
Pre-lab discussion of Transformation Activity“
Day Eighteen Transformation Activity“ Day One
Day Nineteen Transformation Activity“ Day Two
Day Twenty Review results of Transformation Activity
Day Twenty Quiz on Genetic Engineering
Introduce DNA fingerprinting and Human Genome Project
Day Twenty-one Jeopardy Review Game
Day Twenty-two Post-test

 

 

 

“ denotes newly developed material or activity

 

 

All new information presented in this section was done via PowerPoint slideshows.
These slideshows addressed breeding techniques, restriction enzymes,
electrophoresis, genetic engineering and uses of DNA technology. Students used
a guided note handout (Appendix C) that required that they fill in important points
during the discussion. The first topic discussed in this unit included the various

breeding techniques that man uses in the development of desirable characteristics

14

for plants and animals. This information was the springboard for the first activity of
this section, the Apple Lab (Appendix D). In this activity, students observed
fourteen varieties of apples and recorded data about taste, texture and color.
Students also visited web sites, provided by the instructor (Appendix D), to gather
information about the origin of apple varieties, their harvest time, and other facts
about apples in general. The students used this information to correlate breeding
techniques to apple farming. On day five, students reviewed the structure and
function of DNA in preparation for the DNA Extraction Lab (Appendix D). In the
DNA Extraction Lab, students removed DNA from one of three samples of fruit or
vegetable matter. This activity showed that DNA is found in all tissues. On day
seven, students took guided notes and discussed the function and discovery of
restriction enzymes. This information was the basis for Restriction Enzyme Activity
One and Two (Appendix D). In Restriction Enzyme Activity One (httpzllwww.
sc2000.net /~czaremba/labs/recombo.html), students had to locate the restriction
site on a plasmid, “cut” it and insert a gene. In Restriction Enzyme Activity Two
(http://www.chOOO.net/~czaremba/labs/recombinationdna.html), students used
their knowledge to locate a restriction enzyme that both would “cut out” a desired
gene and cut a plasmid so the gene could be inserted. The DNA Extraction
Activity and the Restriction Enzyme activities were used to introduce the technique
of gel electrophoresis, the focus of the next two activities. In the Food Coloring
Electrophoresis activity (Appendix D), students ran a gel to separate the pigments
that make up the different colors of food coloring and to see how and why things

move through the gel. This activity tied in nicely with the use of restriction

15

enzymes and how they are used to "cut" DNA into smaller fragments. In the
activity, Gel Electrophoresis (Appendix D), students ran a gel that contained three
different samples of lambda DNA. The topics of restriction enzymes and
electrophoresis were used as a segue towards a discussion of genetic
engineering. A video on genetically modified organisms was used to show
students the pros and cons of genetically modifying an organism. In the
culminating activity, Transformation of E. coli (Appendix D), students inserted a
glow-in-the-dark gene into E. coli. The Gel Electrophoresis activities and the
Transformation of E. coli activity were developed from kits purchased from
Carolina Biological. This unit was summarized using the Jeopardym-like review

game. A post-test followed.

16

CHAPTER 2
Data Collection Methods

Data Gathering-Pro and Post Tests
Each section of this newly constructed genetics unit began with a pre-test
(Appendix A). The pre-test for each section consisted of open-ended questions.
The open-ended format of the pretest allowed the students to freely respond to the
questions. During the course of each section, the students were quizzed to
evaluate their comprehension of the subject matter. In turn, at the end of each unit
the students were given a post-test (Appendix A). The post-test contained four of
the open-ended questions found on the pre-test. The remainder of the post-test
consisted of multiple choice and true/false questions.
Data Gathering-Weekly Reflections
Another way to test the effectiveness of a unit, data was collected to see how the
students feel about the Ieaming process. According to Etkina (2000), student
reflections should be carefully analyzed to clear up areas of confusion for the
Ieamers. When students are engaged and happy with the process then their
performance should reflect this. Students completed a weekly reflection sheet
(Appendix B) to show how they felt about the Ieaming process. Two sources of
data were collected from this form; anecdotal and numerical. The anecdotal data
are presented in Tables 6, 8 and 10.
Data Analysis-Pre and Post Test

Four open-ended questions from the pre-test and post-test assessment

tools for each unit were selected for analysis. The questions were chosen

17

because they covered the core concepts of each section of the genetics unit.
Each question was worded neariy identically on both the pre- and post-test.
Student responses were placed into one of three categories: no understanding of
the topic, some under-standing of the topic, and complete understanding of the

topic based on the criteria in Table 4.

 

Table 4. Pro-test and Post-Test Scoring Rubric

 

 

 

 

 

Student Score Category

9 — 10 points Complete Understanding
4 - 8 points Some Understanding

3 or > points No Understanding

 

 

If a student left the question blank or scored less than three points out of ten, then
they were placed in the no comprehension of the topic group. This group of
students received no credit for their response on the assessment tool. Students
who scored between four and eight points out of ten had a basic level of
comprehension for the specific concept, but also had some misconceptions. They
were placed in the some understanding of the topic group. This group of students
received at least half credit on the assessment tool. Finally, students who achieved
a score of nine or ten points out often were placed in the complete understanding

of the topic group. The data shows the percentage of students in each category.

18

 

Data Analysis-Weekly Reflections

The weekly reflection form contained a question that asked the student for
three pieces of information: the concept being learned, the process used to learn
the concept, and how they feel about what they are Ieaming. The main point of
interest on this form was how the student felt about what they learned. This data
was placed in the following groups based on student responses: disliked, slightly

enjoyed, or enjoyed.

19

CHAPTER 3
Results and Evaluation
The main areas of content on which this thesis focused included: genes and
chromosomes, human heredity, and applied genetics. The results of each area

will be discussed in the following text.

Genes and Chromosomes
The pre-test data collected for this section showed that the majority of
students came into the class with very little prior knowledge in the topic area of

genes and chromosomes (Figure 1).

 

What type of mutations can be inherited and
why?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Who is more likely to suffer from a sex-linked
trait and why?

What is gene linkage? _
I

Describe the relationship between genes and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

chromosomes.
I I I I I I I I I I
0% 10% 20% 30% 40% 50% 60% 70% 80% 90%
Percent of Students
El no understanding I some understanding I complete understanding

 

Figure 1. Genes and Chromosomes Pro-test Data

 

 

This lack of prior knowledge is surprising because the students in the study group
should have learned some of these concepts in their freshman level science

course. Sixty-five percent of the students lacked a basic understanding of

20

mutations and how they are inherited. Sixty-five percent of the sample group had

no understanding of sex-linked inheritance. Seventy-seven percent of the sample

group had no understanding of gene linkage. Fifty-three percent had no under-

standing of the relationship between genes and chromosomes. A sample of some

of the common misconceptions on the Genes and Chromosomes Pre-test

(Appendix A) are listed in Table 5.

 

Table 5. Student Misconceptions- Genes and Chromosomes Pre-test (N=66)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Misconception Number of Students
Chromosomes are sex cells. 2
Chromosomes make up genes. 4
Chromosomes determine the genes of an organism. 2
Chromosomes are traits that genes carry. 3
Genes and chromosomes cany DNA and RNA. 1
Genes and chromosomes give the appearance to the organism. 4
Linking one gene to the next, wnnecting genes or chromosomes. 9
The linking of genes to form a strand of DNA. 2
Females are more likely to inherit a sex-linked trait. 8
Any mutation can be inherited 6
All mutations are bad 5
DNA mutations are inherited. 7

 

 

The post-test data for this section showed that the largest increase in

student understanding came in two topic areas (Figure 2). The first area that

showed the greatest improvement was the topic of sex-linked inheritance. Eighty-

six percent of the students sampled achieved some or complete understanding of

the topic compared to 35% on the pre-test.

21

 

 

What type of mutations can be inherited and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Why? J I l
I I I I T I I I I I
Who is more likely to suffer from a sex-linked
trait and why?
What is gene linkage? L
L l I I
Describe the relationship between genes and
chromosomes.
0% 10% 20% 30% 40% 50% 60% 70% 80% 90%
Percent of Students
Cl no understanding I some understanding I complete understanding

Figure 2. Genes and Chromosomes Post-test Data

 

 

 

The other content area with a large increase in student performance was that of
the relationship between genes and chromosomes. Eighty-six percent of the
students had some or complete understanding, while on the pre-test that group
was only 47%. There was a significant improvement in the area of gene linkage.
Twenty-three percent of the pretest scores were in the some understanding
category for the relationship between genes and chromosomes topic. On the post-
test, fifty-four percent of students scored in the complete or some understanding
groups. Understanding of the topics of mutations and how they are inherited
showed the least amount of improvement. On the post-test, 45% of the students
scored either complete of some understanding, while on the pre-test 35% were in
the some understanding category.

The weekly reflection data for this section showed that about seventy-five

percent of the subjects at least partially enjoyed this portion of the genetics unit

22

(Figure 3). A summary of the students' responses on the weekly reflections is

contained in Table 6.

 

Table 6. Summary of Weekly Reflections for Genes and Chromosomes

 

"I really don't get what we are Ieaming."

“It was kind of interesting. Hopefully I'll continue to understand this stuff."

“I feel I learned a lot. This chapter is very interesting.”

“I like what we are Ieaming but it somewhat confusing. It is also interesting."
“I'm feeling good about this stuff. everything is clicking.”

“Genetics is only fun if you like Ieaming about it."

 

 

 

 

14%

21%

 

I disliked El somewhat enjoyed enjoyed

Figure 3. Genes and Chromosome Weekly Reflections Data

 

 

 

Human Heredity
The pretest data for this section (Figure 4) indicated that the majority of the
students came into the class with good base of prior knowledge about human

heredity.

23

 

Draw a pedigree chart with the following
information. ..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Identify each of these genetic disorders...as
dominant or recessive, autosomal or sex linked.

 

 

 

 

 

 

 

 

 

 

What is a polygenic trait?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Describe the process of anmiocnetisis?

 

 

 

 

 

 

 

 

IIIIIIIIIII'TII
0% 10% 20% 30% 40% 50% 60% 70% 80% 90%
Percent of Students

El no understanding I some understanding I complete understanding

Figure 4. Human Heredlty Pro-test Data

 

 

 

This strong base of prior knowledge is probably due to the fact that the students
are receiving good instruction in their freshmen level class on this topic. Fifty-four
percent of the students sampled had at least some understanding of how to
construct a pedigree chart. Ninety percent of the students could identify a genetic
disorder as dominant or recessive and autosomal or sex-linked. Sixty-one percent
knew what a polygenic trait was and could identify at least one. The largest area
where the students demonstrated a lack of prior knowledge was in describing
amniocentesis. Seventy-three percent of the students had no idea what this
procedure was. Some of the common misconceptions that the students

demonstrated on the pretest are listed in Table 7.

24

 

Table 7. Student Misconceptions- Human Heredity Pro-test (N=66)

 

 

 

 

 

 

 

 

 

Misconception Number of Students
Polygenic traits are controlled by other genes. 3
Polygenic traits control only physical appearance. 5
Blood tests are most common method of diagnosing genetic disorders. 5
Hemophilia is a dominant genetic disorder. 4
Hemophilia is an autosomal, genetic disorder. 4
Colorblindness is a dominant genetic disorder 3
Sickle-cell anemia is a dominant genetic disorder. , 5
Sickle-cell anemia is a sex-linked genetic disorder. 2

 

 

 

 

The post-test data for this section showed that the largest increase in
student understanding came in three topic areas (Figure 5). The area that showed
the greatest improvement was constructing a pedigree chart. One hundred
percent of the students sampled achieved some or complete understanding of the
topic. On the pre-test, forty-six percent had no idea how to construct a pedigree.
The next content area that saw that largest improvement was amniocentesis. On
the post-test, 68% of the students had a complete understanding of the process
and how it was used. The last area where there was a significant improvement
was identifying polygenic traits. On the pre-test, sixty-one percent of the students
had at least some understanding of the concept. The post-test data showed that
the percentage in those categories increased to 80%. The final content area,
identification of genetic disorders, showed very little improvement. While the
number of students who demonstrated a complete understanding of the concept
increased to sixteen percent, the number of students with no understanding

showed only a slight increase to twelve percent.

25

 

Draw a pedigree chart with the following
information...

 

 

 

 

Identify each of these genetic disorders...as
dominant or recesshe. autosomal or sex linked.

 

 

 

What isapolygenic trait? I I I l I

 

 

 

 

 

 

 

 

 

 

 

Describe the process of anmiocnetisis? I I l l I I I I

 

0% 10% 20% 30% 40% 50% 60% 70% 80% 90%

Percent of Students

:1 no understanding I some understanding I complete understanding

 

 

Figure 5. Human Heredlty Post-test Data

 

The weekly reflection data for this section shows that all but three percent of
the subject group at least partially enjoyed the Human Heredity component of the

genetics unit (Figure 6). Samples of the students’ reflections are listed in Table 8.

once YOU

 

26

 

3%

 

74%

I disliked El somewhat enjoyed El enjoyed

 

Flgure 6. Human Heredity Weekly Reflections Data

 

 

Applied Genetics

The pretest data for this section (Figure 7) showed that the majority of
students come into the class with a poor base of prior knowledge in the area of
applied genetics. Forty-two percent of the subjects did not answer the selected
pretest questions. This was not a surprise because these topics are not covered in
our Freshman Level Science Survey course. Seventy-seven percent of the
students sampled had no understanding of the importance of restriction enzymes
in the field of genetics. About sixty percent of the students had no idea about the
advantages and disadvantages of breeding techniques used to produce
organisms. About sixty percent of the students had no idea that farmers apply
genetic theories to produce food that they consume everyday. Sixty-eight percent
had no prior knowledge of genetic engineering. Some of the common misconcep-

tions that the students demonstrated on the pre-test are listed in Table 9.

27

 

 

 

What is a restriction enzyme? m
I

Compare the advantages and disadvantages of
breeding techniques... I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Discuss the value of applied genetics... in
breeding plants and animals. I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

What is genetic engineering?

 

 

 

 

 

LI I I I I I I I I I I I
0% 10% 20% 30% 40% 50% 60% 70% 80%
Percent of Students

 

El no understanding I some understanding I complete understanding

Figure 7. Applied Genetlcs Pre-test Data

 

 

 

 

Table 9. Student Misconceptions- Applied Genetics Pro-test (N=66)

 

 

 

 

 

 

 

 

 

Misconception Number of Students
Breeding techniques are not as successful as genetic engineering. 3
“Designer kids” could result from breeding techniques 2
Mutations are caused by inbreeding. 3
Genetic engineering has only a negative impact. 3
Restriction enzymes stop stuff from happening. 6
Restriction enzymes separate genes into bases. 2
Restriction enzymes correct errors in DNA. 2
Restriction enzymes change DNA. 1

 

 

NOTE: Thirty-one students did not attempt to answer the sample questions.

 

 

 

The post-test data for this section indicated that student understanding
increased in all four topic areas (Figure 8). About ninety-five percent of the

subjects attempted to answer the four open-ended questions.

28

 

 

What is a restriction enzyme?

 

Compare the advantages and disadvantages of
breeding techniques...

 

 

Discuss the value of applied genetics... in
breeding plants and animals.

 

 

 

 

 

 

 

 

 

 

What is genetic engineering?

 

 

 

 

 

 

I I T T T T 1

0% 10% 20% 30% 40% 50% 60% 70% 80%

Percent of Students

El no understanding I some understanding I complete understanding

 

 

Flgure 8. Applled Genetics Post-test Data

 

Ninety percent of the students demonstrated at least some understanding of
genetic engineering. Another area of improvement in student understanding was
that of the importance and function of restriction enzymes. Eighty-five percent of
the students achieved at least a level of some understanding; sixty-one percent
achieved a level of complete understanding for this topic. Eighty-six percent of the
students demonstrated at least some understanding of the advantages and
disadvantages of breeding techniques. Of this group, sixty-six percent achieved a
level of complete understanding. The final concept in this section was the value of
applied genetics. Seventy percent of the students demonstrated a level of some
understanding. This is about a thirty percent increase over the pretest.

The weekly reflection data for this section shows that a majority of the

subject group, about ninety-one percent at least partially enjoyed the Applied

29

Genetics component of the genetics unit (Figure 9). Samples of the students'

responses on the weekly reflections are in organized in Table 10.

 

9%

 

I disliked El somewhat enjoyed I enjoyed

Figure 9. Applied Genetics Weekly Reflections Data

 

 

 

 

Table 10. Summary of Weekly Reflections for Applied Genetics

 

"I thought it was very boring.”

“I thought it was interesting Ieaming about apples.”

“The Apple Lab was awesome. I'd like to eat during class more.”
“It (gel electrophoresis) was an interesting subject to learn about."
“It was fun getting to do all of the labs. They were very interesting.“

“I like what we are Ieaming (DNA fingerprinting)"

 

 

 

30

CHAPTER 4

Discussion
In this study, the increase in pre- and post-test scores showed that there is
a direct correlation between the changes in teaching style, an increase in the
number of group and lab activities, and student Ieaming. Overall, these changes

have a positive impact on and student Ieaming.

The results of the first section in the genetics unit, Genes and Chromo-
somes, show the smallest student improvement between pre-test and post-test
scores. This section also contains the fewest “hands-on” activities of the entire
genetics unit. The bright spot of this section was the Jeopardy” review game.
The class enjoyed the review game format. Students requested this type of review
format in other biology units. In addition, this game provided students with

opportunities to gain a few extra credit points towards their test.

The Human Heredity section of this unit shows a dramatic increase in
student performance on post-test. In this section, there were more group activities.
Compared to the other sections of the genetics unit there is less dependence on
lecture and more emphasis on students actively participating. Two activities can
be directly related to the increase in student performance. The first activity is the
Online Colorblindness Activity. The student must answer questions about a family
and construct a pedigree based on the information gathered. One Of the areas of
greatest student improvement was constructing a pedigree. The second activity

related to the increase in student performance was the Online Karyotyping Activity.

31

In this activity, the student constructed a karyotype and diagnosed a genetic
disorder. This activity was related to amniocentesis and the procedure’s
importance in the diagnosis of genetic disorders. The Human Mutation Activity was
a good culminating activity because it illustrated the link between DNA, RNA,
mutations, and genetic disorders. Student achievement was most notable in the

area of human heredity.

The Applied Genetics section of this unit showed the greatest contrast
when comparing the pre-test and post-test data. The increase in student
understanding of each concept can be directly related to the activities of this
section. In the Apple Lab, students researched different varieties of apples. This
research directly relates to two of the concepts of this section. The first concept
was breeding techniques and the second concept was the value of applied
genetics. The increase in student understanding of the concepts of restriction
enzymes and genetic engineering could be attributed to the three activities that the
class performed. In one of the activities, students had to identify the proper
restriction enzyme needed to cut a DNA strand. In the second activity, the
students had to identify the correct restriction enzyme that would both remove a
gene from a plasmid and cut the DNA so that the gene could be inserted. In the
final activity, students had to insert a glow-in-the-dark gene into the bacterium E.
coli. These activities served to increase student understanding because they
linked together the importance of restriction enzymes to genetic engineering. This
increase in student understanding is shown in the post-test scores. Student

performancein these two areas shows a dramatic increase (Figure 8). This

32

research shows that as students became more active in “their" Ieaming process

their understanding of key genetics concepts increased.

A few areas can be improved upon in the future for this genetics unit. The
Genes and Chromosomes section needs more group activities and lab activities
that focus on, perhaps, real worid situations. Such a focus can ellicit a greater
interest and, in response to the increased interest level, greater student Ieaming.
In addition, perhaps a more problem-based Ieaming approach could be utilized.
Such an approach could play a role in developing strategies that cross the
curriculum while meeting standards and objectives and, ultimately, result in
improved student Ieaming. Because so few activities were performed, most of the
negative feedback in the Weekly Reflection data occurred during this portion of the
unit. Student anecdotal data showed that some students “didn’t get it” or that the

topic was “confusing” (T able 6).

The Human Heredity section was very successful. Most of all the data on the
Weekly Reflections were positive. One area that could be improved on involves
decreasing the amount of notes and replacing the note taking with more guided
group discussions. Finally, the Applied Genetics component could use some
modification. The reliance on lecturing was the major downfall during this section.

The student feedback from the Weekly Reflections reinforces this point.

33

APPENDICES
Appendix A- Pre-tests and Post-tests
Appendix B- Weekly Reflections
Appendix 0- Note Handouts

Appendix 0- Activities

APPENDIX A

Pro-tests and Post-tests

35

Name Period 1 2 3 4 5 Date:

Genes and Chromosomes Pro-test

1 . Describe the relationship between genes and chromosomes.

2. What is gene linkage?

3. Explain what occurs during crossing over.

4. How does crossing over affect the gene pool of an organism?

5. What is the genotype of a human male? A human female?

6. What is a gamete?

7. What types of gametes can a male produce?

8. If my wife and I had 3 sons in a row what is the probability that the fourth child would be a
male?

9. Which parent determines the sex of a child?

10. What is a sex-linked trait?

36

1 1 . Give two examples of sex-linked traits in humans.

12. Who is more likely to suffer from a sex-linked trait and Why?

13. What is a mutation?

14. What types of mutations can be Inherited and why?

15. What happens to plant that suffers from nondisjunction? An animal?

37

Period 1 2 3 4 5 Date:

 

Genes and Chromosomes Posttest

MULTIPLE CHOICE- On the blank provided, write the answer the following questions. 2 points

each
1)__

Z)

3)

4)

5)

6)

8)

The rod-shaped structures located in the nucleus of every cell of an organism
are the
a. chromosomes. c. cell membranes.

b. traits. (1. genes.

Sex cells are produced by the process of

a. meiosis. c. fertilization.

b. mitosis. d. reproduction.

A change in genes or chromosomes that causes a new trait to be inherited is
called a(an)

a. alteration. c. variation.

b. mutation. d. hybrid.

The scientist who discovered that the X and Y chromosomes determine the sex
of an organism was

a. Mendel. c. Sutton.

b. Morgan. d. De Vries.

Mutations

a. may be caused by radiation. c. may be passed on to the next
generation.

b. are usually harmful. d. all of these

According to the chromosome theory, which are carried on chromosomes?
a. sex cells c. X chromosomes
b. genes d. Y chromosomes

Maleness and femaleness are determined by the X and Y chromosomes. A
female has
a. 2 X chromosomes. c. an X and a Y chromosome.

b. 2 Y chromosomes. d. 2 X chromosomes and 1 Y.
The number of chromosomes in a sex cell is

a. double the number of chromosomes 0. triple the number of chromosomes

found in body cells. found in body cells.
b. one half the number of d. one third the number of
chromosomes found in body cells. chromosomes found in body cells.

38

9)

10)

11)

12)

13)

14)

15)

16)

17)—

The scientist who discovered mutations was

a. De Vries. c. Morgan.

b. Sutton. d. Watson.

Substances that cause mutations are called

a. proteins. c. chromosomes.

b. mutagens. d. mutants.
Chromosomes are made of long strands of
a. DNA. c. uracil.
b. RNA. d. mutagens.
The scientist who discovered chromosomes was
a. Sutton. c. Morgan.
D. De Vries. d. Flemming.
The main function of chromosomes is to control the production of
a. mutations. c. genes.
b. proteins. d. mutagens.
The female sex cell is called the
a. egg cell. c. mother cell.
b. sperm cell. d. none of these
The scientist who showed that genes are carried on chromosomes was
a. De Vries. c. Morgan.
b. Sutton. d. Flemming.
The male sex cell is called the
a. egg cell. c. daughter cell.
b. sperm cell. d. none of these

In tomatoes, cut leaf is dominant over potato leaf and purple stem is dominant
over green stem. In the table below, the result of a mating producing an F1 is
given:

Purple Cut Purple Potato Green Cut Stag—:3,
1 790 620 623 722—
The most probable parental genotypes are,
a. PpCc x PpCc c. PpCc x Ppcc
b. ppcc x ppcc d. PPCc x ppCc

39

13) There is evidence that a certain color in cats is sex-linked. Yellow is recessive
to black. A heterozygous condition results In tortoise shell or calico color. A
calico cat has a litter of
8 kittens: 1 yellow male, 2 black males, 2 yellow females, and 3 calico females.
What was the male parent's probable color?

a. Yellow c. Black

b. Calico d. Yellow and black

Questions 19-22 are based on the following information:

In Drosophila, there is a dominant gene for gray body color and another gene dominant for
normal wings. The recessive alleles of these two genes result in black body color and vestigial
wings, respectively. Flies homozygous for gray color and normal wings were crossed with files
with black bodies and vestigial wings. The F1 progeny were then test-crossed with the following
F2 results:

Gray Body, normal wings 236
Black Body, vestigal wings 253
Gray Body, vestigal wings 50
Black Body, normal wings 61

19) The genes for these characteristics are

a). not linked c. linked without crossing over

b. linked with 18.6% crossing over d. linked with 81.4% crossing over
20) Which statement is TRUE about the individuals in the F1 generation?

a. There are no linked alleles c. The linked alleles are GG and W

b. The linked alleles are Gv and 9V d. The linked alleles are GV and gv
Which of the following is true about the 61 black, normal flies in the F2

21) generation?
a. There are no linked alleles c. The linked alleles are 9V and gv
b. The linked alleles are the same as d. The linked alleles cannot be
the F1. determined.

22) If some of the black bodied, normal winged flies were crossed with black
bodied, vestigial winged individuals, the offspring would probably be:
a. like those obtained in the F2 c. unpredictable
generation
b. 50% black vestigial, 50% black d. 50% gray vestigial, 50% black
normal normal

23) Nondisjunction, whereby a pair of homologous chromosomes does not
separate in the first meiotic anaphase, is responsible for which of the following
disorders?
a. baldness c. Downs Syndrome (trisomy-21)
b. color blindness d. Hemophilia

24) Assuming a 1/2 probability of reproducing a male offspring, the chance for five
successive male births in a family is:
a. 1/32 c. 1/8
b. 1/16 d. 1/64

40

TRUE OR FALSE-Determine whether each statement ls true or false.

25)

26)

27) _
28)
29)

30)

31)

32)

33)

If the body cells of an organism each contain 18 chromosomes, the sex cells of
that organism will each contain 36 chromosomes.

Genes are located on chromosomes.
Proteins determine the traits of an organism.
The process of mitosis produces sex cells.

The X and Y chromosomes are sex chromosomes.

Somatic mutations are genetic mistakes that can affect the way in which traits
are inherited.

All mutations are harmful.
Radiation is an example of a mutagen.

The production of proteins is called replication.

APPLICATION-Answer each question in complete sentences. 40 points possible

34) Describe the relationship between genes and chromosomes. 4 points

35) What is gene linkage? 4 points

36) What is a gamete and why are they important? 4 points

37) What is the genotype of a male? Female? 2 points

38) If my wife and I had 4 sons in a row what is the probability that the fifrth child would be a

39) Which parent determines the sex of a child and why? 4 points

40) What is a sex-linked trait, and give two examples in humans? 4 points

41

41) Who is more likely to suffer from a sex-linked trait and why? 3 points

42) What is the difference between a chromosomal and a gene mutation? 4 points

43) If an individual is born with only one sex chromosome, and it is an X, what sex is it and why?
4 points

44) What types of mutations can be inherited and why? 4 points

42

Name Period 1 2 3 4 5 Date:

Human Heredity Pretest

1. What are the four blood types in humans?

2. What Is a polygenic trait? Give an example.

3 . Identify each of the following genetic disorders as dominant or recessive, autosomal
or sex linked.

A. Hemophilia
B. Color-blindness

C. Sickle-cell anemia

4. Draw a pedigree chart showing the following information:
Grandpa ls color-blind, grandma has normal vision. Grandma and grandpa have
three children, the oldest a son, the next two are daughters. The third daughter ls
married and has three girls, the youngest one is color-blind.

I = color-blind male

= normal male

 

 

 

 

. = color-blind female

Q = carrier female
0 = normal female

5. What Is the most common method used to diagnose genetic disorders? Describe
how this procedure Is performed.

43

Name

Period 1 2 3 4 5 Date:

 

Human Heredity Posttest

MULTIPLE CHOICE-Write the letter of the answer that best completes each statement.

1)

2)

3)

4)

5)

The number of chromosomes in a person's body cells is
a. 23. c. 64.
b. 46. d. 92.

The number of chromosomes in a human sex cell is

a. 23. c. 64.
b. 46. d. 92.
An allele is
a. an exact duplicate of DNA. c. the actual makeup of a gene.
b. a special chart showing gene
combinations. d. one member of a gene set.

Skin color is determined by
a. a pair of alleles. c. multiple alleles.

b. a sex-linked trait. d. the Y chromosome.
If a child inherits a gene for type A blood from one parent and a gene for type

0 blood from the other parent, which blood type will the child have?

6)

8)
cells is

9)

a. A c. AB
b. B d. 0
Which of the following is determined by multiple alleles?

a. sickle cell anemia c. Down syndrome
b. blood group d. hemophilia
The chromosome that carries the gene for colorblindness is the
a. X chromosome. c. twenty-first chromosome.
D. Y chromosome. d. forty-second chromosome.

The blood disorder that results in the manufacture of oddly shaped red blood

a. baldness. c. hemophilia.

b. Down syndrome. d. sickle cell anemia.
Which of the following is an example of a sex-linked trait?

a. male-pattem baldness c. colorblindness

b. Down syndrome d. sickle cell anemia

10) If a female who carries an allele for hemophilia marries a male who does not
have hemophilia, what percentage of their children may inherit the disease?

a. 0 percent c. 50 percent
b. 25 percent (1. 100 percent
1 1) What combination of alleles must a person inherit to have group 0 blood?
a. lAi c. IBI
b. I“II3 d. ii
1 2) The failure of chromosome pairs to separate during meiosis is called
a. nondivision. c. nondisjunction.
b. nondissociation. d. nondetachment.
13) The disorder in which the blood does not clot normally is
a. baldness. c. hemophilia.
b. Down syndrome. d. sickle cell anemia.
14) In amniocentesis,
a. the baby is tested for poisons. c. the baby is operated upon.
b. cells in the fluid surrounding the d. the weight of the baby is
baby are tested for genetic disorders. determined.
15) In Down syndrome, there is an extra chromosome on the
a. fifteenth pair. c. thirteenth pair.
b. twenty-first pair. (1. ninth pair.
16) A person with sickle cell anemia
a. inherits 1 sickle cell gene from his c. inherits 1 sickle cell gene from his
or her mother. or her father.
b. inherits 1 sickle cell gene from each
parent. (I. none of these
17) Sickle cell anemia results from an error in the
a. hemoglobin molecule. c. plasma.
b. white blood cells. (I. blood vessels.
18) Which of the following is not an inherited disease?
a. muscular dystrophy c. malaria
b. Huntington disease d. cystic fibrosis
19) Sex-linked traits are carried on the
a. X chromosome. 0. twenty-first pair of chromosomes.
D. Y chromosome. d. forty-second pair of chromosomes.

45

20) A chart that shows the relationship among individuals in a family

Man)
a. karyotype. c. allele.
b. amniocentesis. d. pedigree.
21) An example of a sex-influenced trait is
a. colorblindness. c. male-pattem baldness.
b. hemophilia. d. cystic fibrosis.
22) Which of the following shows the size, number, and shape of all the
chromosomes in an organism?
a. a karyotype c. an allele
b. a phenotype d. a pedigree
23) The hormone that controls the development of male
characteristics is
a. estrogen. c. hemoglobin.
b. testosterone. d. growth hormone.
24) A pair of alleles that are equally dominant are
a. nondisjunctive. c. sex linked.
b. multiple alleles. d. codominant.
25) Human blood groups are determined by
a. 1 allele. 0. 3 alleles.
b. 2 alleles. d. 4 alleles.

TRUE OR FALSE-Determine whether each statement is true or false.

26) There are more colorblind females than there are colorblind males.

27) Nondisjunction causes the condition called Down syndrome.

28) Sex-linked traits are passed from parent to child on the Y chromosome.
29) Male-pattem baldness is a sex-linked trait.

30) The trait for skin color in humans is determined by multiple alleles.

COMPLETION-Fill in the word or number that best completes each statement.

31) Hemophilia and colorblindness are both traits.
32) Because 6 pairs of genes are responsible for human skin color, scientists say this trait is

determined by

 

46

33) involves taking some fluid out of the sac that surrounds an unborn baby.

 

34) is a condition in which all the body cells have an extra twenty-first
chromosome.
35) is an inherited disease that causes blood to clot slowly or not at all.

 

36) Each member of a set of genes is called a(an)
37) A(An) shows the size, number, and shape of chromosomes In an
organism.

38) The relationships among the individuals in a family are shown by a(an)

 

 

 

39) A person who is cannot see the difference between certain colors.

40) The presence of a group of 3 chromosomes Is called a(n)
CRITICAL THINKING AND APPLICATION- Discuss each of the following In a brief paragraph.

 

41) A man with type A blood is said to be the father of a child with type 0 blood. The mother
has type B blood. Could the man be the father of the child? Explain your answer.

42) Describe the process of amniocentesis. Why might this procedure be performed?

43) What Is a polygenic trait? Give 3 examples.

44) Identity each of the following genetic disorders as dominant or recessive, autosomal or
sex linked.

Hemophilia
Tay-Sachs

Huntington Disease

47

45)

 

 

 

 

 

 

 

 

 

Draw a pedigree chart with the following Infon'natlon. Include all of the genotypes
that you can. Grandpa is color-blind, grandma has normal vision but Is a carrier.
Grandma and grandpa have three children, the oldest two are daughters and the
youngest Is a son. The son Is married and has two girls and a boy. The oldest
daughter ls colorblind and so is the boy.

= color-blind male

= normal male

= color-blind female

= carrier female

= normal female

Draw a pedigree chart with the following information. Include all of the genotypes that you
can.

Grandpa has type O+ blood and grandma has type B-. They have three children, the
oldest two are boys, the youngest a girl. All the boys have type B- blood, the girl has 0.
The gin marries and has two children, a girl and a boy. Her boy has blood type 0- and
his girl has blood type 3+.

= male

= female

48

Name Period 1 2 3 4 5 Date:

Applied Genetics Pretest

Multiple Choice-Place the letter of the best answer in the blank beside the question number. (1
point each)

1. Substances that cause changes in DNA are
a. restriction
enzyme b. mutagens c. chimeras d. insertions
2. A plant that has polyploidy has .
a. large cells D. four times the c. more than 2n the d. all of the above
haploid number of chromosome
chromosomes number
3. The cross that results in the Rhode Island Red hen contains genes from five
different breeds of fowl. This is an example of .
b. genetic
a. sequencing engineering c. mutagenesis d. hybridization
4. The term that is least closely related to the others is .
b. selective (1. DNA
a. recombinant DNA breeding 0. gene therapy fingerprinting
5. Which of the following techniques listed below relies on the fact that the
chances of two individuals being exactly alike are slim?
0. selective d. DNA
a. cloning b. inbreeding breeding fingerprinting
6. DNA sequences can be cut at specific sites by
d. restriction
a. plasmids b. mutagens c. recombinant DNA enzymes
7. Crossing individuals with similar characteristics so that those characteristics will
appear in their offspring is called
a. inbreeding b. mutagenesis c. hybridization d. polyploidy
8. A series of bands that result in a gel after electrophoresis is the working basis of
0. DNA
a. DNA chimeras b. recombinant DNA fingerprinting d. DNA sequencing
9.
Decisions regarding the human genome are the responsibility of .
a. the federal b. society as a c. the state (I. only scientists.
govemment whole government

10. The mule exhibits desirable traits from its mother, he horse, and it's father, the
donkey. The expression of these desirable traits is known as

a. controlled
characteristics b. a great job c. vital traits d. hybrid vigor
11.
The process In which human genes are repaired is called
a. genetic c. DNA
engineering b. gene therapy fingerprinting d. recombinant DNA
12. A technique used to physically change specific parts of the genome is
a. genetic .
engineering b. hybridization c. mutagenesis d. inbreeding

49

13.
The term least closely related to the others is

a. restriction

enzymes b. clones c. “sticky ends” (1. DNA fragments

14. A product of genetic engineering that helps people who have diabetes
mellitus is .

a. penicillin. b insulin. c interferon. d. HGH

15. The process by which genes, or parts of DNA, from one organism are
transferred to another organism is called

a. genetic d. selective

engineering. b. inbreeding. c. hybridization. breeding.

16. The crossing of two genetically different but related species of organisms is
called

a. genetic

engineering. b. biotechnology. c. inbreeding. d. hybridization.

17. Through genetic engineering, scientists have developed a way to protect

plants from the disease caused by the
d. tobacco mosaic
a. AIDS virus. b. influenza virus. c. hepatitis B virus. virus.

18. An organism that has the traits of both parents is called a
a. purebred. b. cross breed. c. hybrid. d. mutation.
19. Which organisms are used by scientists in genetic engineering?

a. bacteria b. protozoans c. viruses (1. molds
20. In genetic engineering, the new pieces of combined DNA are called

d. recombinant
a. hybrids. b. plasmids. c. mutations. DNA.
Directions- One word makes each of these statements false. Cross out that one word and write a
word on the line provided that will make the statement true. You may not change more than one
word in the statement.

(2 points each)
21. Somatic mutations occur in gametes.
An inversion mutation occurs when a piece of chromosome breaks off and is
22. lost.
23. A deletion mutation is a change in a single nitrogen base in DNA
24. A frame-shift mutation is an example of a chromosome mutation.
In a translocation mutation, a pair of chromosomes fail to separate during
25. cell division.

50

Use the diagram below to answer questions 26-29.

 

/

  

7'6
.6*
c9.

 

 

 

 

6‘
c
Q.

plasmid
2 —
m 2:: -I...
bacterium ‘
26. What process does this diagram show? (2 points)
27. Describe the process that Is taking place at each of the numbered points in the

diagram. (4 points)

28. Briefly discuss why this process is so important. (2 points)

51

Critical Thinking-Answer following questions using complete sentences. If you fail to follow
these directions you will receive no credit for your answers.

29. Discuss the value of applied genetics in breeding plants and animals. (4 points)

30. Compare the advantages and disadvantages of breeding techniques
(hybridization, inbreeding, etc) and genetic engineering. (10 points)

52

31.

32.

33.

35.

What are restriction enzymes and how are they used in genetic engineering? (4
points)

Mongrels, that is, mixed-breed dogs, are generally healthier than purebred dogs that
receive identical care. Explain why and propose a way to counter act the problem in
purebreds. (4 points)

Briefly outline the steps used in DNA fingerprinting. (6 points)

What is a transgenic organism and how is it different from a hybrid? (6 points)

Describe 4 ways DNA fingerprinting is used. (4 points)

53

36.

A plant breeder has thomless rose bushes with pink flowers, thorny rose bushes with
sweet smelling yellow flowers and thorny rose bushes with purple flowers. Describe

how the plant breeder might develop a purebred variety of thomless sweet smelling
purple roses. (6 points)

Name Period Date

Applied Genetics Post-Test 80 points possible
Multiple Choice-Place the letter of the best answer in the blank beside the question number.
(1 point each)

1. Substances that cause changes in DNA are .

a. restriction b. insertions c. chimeras d. mutagens
enzyme

2. A plant that has polyploidy has

a. four times the b. large cells 0. more than 2n the d. all of the above
haploid number of chromosome
chromosomes number

3. The cross that results in the Rhode Island Red hen contains genes from five
different breeds of fowl. This is an example of

a. genetic

engineering b. hybridization c. mutagenesis d. sequencing

4. The term that is least closely related to the others is .

a. recombinant b. selective c. gene therapy d. DNA
DNA breeding fingerprinting

5. Which of the following techniques listed below relies on the fact that the
chances of two individuals being exactly alike are slim?

a. DNA b. inbreeding c. selective d. cloning
fingerprinting breeding

6. DNA sequences can be cut at specific sites by .

a. restriction b. mutagens c. recombinant d. plasmids
enzymes DNA

7. Crossing individuals with similar characteristics so that those characteristics will
appear in their offspring is called

a. inbreeding b. mutagenesis c. hybridization d. polyploidy

8. A series of bands that result in a gel after electrophoresis is the working basis
of .

a. DNA b. recombinant c. DNA chimeras d. DNA sequencing

fingerprinting DNA

9.

Decisions regarding the human genome are the responsibility of .
a. the federal b. society as a c. the state d. only scientis .
government whole government
10 The mule exhibits desirable traits from it’s mother, the horse, and it’s father, the
donkey. The expression of these desirable traits is known as .

a. controlled b. hybrid vigor c. vital traits d. a great job
characteristics
11

The process in which human genes are repaired is called
a. genetic b. gene therapy c. DNA d. recombinant
engineering fingerprinting DNA

55

12
A technique used to change specific parts of the genome is .
a. genetic b. hybridization c. mutagenesis d. all of the above
engineering
13 The term least closely related to the others is

a. clones b. restriction c. “sticky ends” d. DNA fragments
enzymes
14 A product of genetic engineering that helps people is

a. human growth binsulin. cinterferon. d. all of the above.

hormone.

15 The process by which genes, or parts of DNA, from one organism are
transferred to another organism is called

a. selective b. inbreeding. c. hybridization. d. genetic
breeding. engineering.
16 The crossing of two genetically different but related species of organisms is
called
a. hybridization. b. biotechnology. c. inbreeding. d. genetic
engineering.

17 Through genetic engineering, scientists have developed a way to protect plants
from the disease caused by the

a. tobacco mosaic b. influenza virus. c. hepatitis B virus. d. AIDS virus.

virus.

18 An organism that has the traits of both parents is called a

a. hybrid. b. cross breed. c. purebred. d. mutation.

19 Which organisms are used by scientists in genetic engineering?

a. viruses b. protozoans c. bacteria d. molds

20 In genetic engineering, the new pieces of combined DNA are called

a. hybrids. b. recombinant c. mutations. d. plasmids.

DNA.

One word makes each of these statements false. Cross out that one word and write a word on the
line provided that will make the statement true. You may not change more than one word in the
statement. (2 points each)

21 Germ mutations occur in body cells.

A translocation mutation occurs when a piece of chromosome breaks off and is

22 lost.

23 A non-disjunction mutation is a change in a single nitrogen base in DNA.

24 A point mutation is an example of a chromosome mutation.

In an inversion mutation, a pair of chromosomes fail to separate during cell
25 division.

56

Use the diagram below to answer questions 26-28.

3 ,3
.9

m};

l

 

 

\T

plasmid

 

 

 

 

©
c
Q.

 

bacterium ‘
26. What process does this diagram show? (2 points)
27. Describe the process that is taking place at each of the numbered points in the

diagram. (4 points)

28. Briefly discuss why this process is so important (2 points)

57

Critical Thinking-Answer following questions using complete sentences. If you fail to follow
these directions you will receive no credit for your answers.

29. Discuss the value of applied genetics in breeding plants and animals. (4 points)

30. Compare the advantages and disadvantages of breeding techniques (hybridization,
inbreeding, etc) and genetic engineering. (10 points)

31. What are restriction enzymes and why are they important in genetic engineering? (4
points)

32. The flightless turkey eaten on Thanksgiving was developed from several varieties that were
able to fly. Discuss how turkey breeders developed the turkey we eat today. (4 points)

58

33. Briefly outline the steps used in DNA fingerprinting. (6 points)

34. Describe 2 methods of inserting recombinant DNA into an organism.

(6 points)

59

APPENDIX B

Weekly Reflections

60

Name:

 

Weekly Reflections

Week Ending Date:

 

1. Concept: State the concept being Ieamed.

2. Process: Describe the process you (the class) used to learn the concept

3. Attitude: Describe your attitude toward the concept, the process, etc. Tell how you feel
about what you are learning.

61

APPENDIX C

Note Handouts

62

Genes and Chromosomes Guided Note Handout

Background lnfonnation

Between 1884 and 1888 details of mitosl§_a_nd meiosis were reported, the cell nucleus was
identified as the location of the genetic material, and ”qualities" were even proposed to be
transmitted on chromosomes to daughter cells at mitosis.

In 1903 Walter Sutton and Theodore Boveri formally proposed that chromosomes

 

 

contain the genes. The C_hromo§ome Theory of Inheritance is one of the foundations of genetics
and explains the physical reality of Mendel's principles of inheritance.

The Chromosome Theory of Inheritance States...

Chromosomes are the cellular components that contain games

Each 193 occupies a specific place on a chromosome

Genes may exist in several alleles and each chromosome contains one allele for each gene

 

Types of Chromosomes

Autosomes

resemble each other in size and placement of the centromere

Humans have 22 pairs of autosomes

Sex Chromosomes

differ in their size, depending on the M the organism

Humans have 1 pair of sex chromosomes

in humans sex chromosomes are identified as x or 1 (X chromosomes are much smaller than the
X)

Gene Linkage

Genes on the same chromosome that do not undergo independent assortment. These genes are
inherited together

Who discovered that genes were linked? m

How did MEL" discover that genes were linked?

By studying fruit flies Drosophila melanggaster

63

Why did Morgan use fruit flies?

When doing any genetics study you need to choose an organism that has a short generation time
(4 weeks) and observable contrasting traits

Drosophila melanogaster, met both of these requirements

What did Morgan do? Step 1

Morgan crossed purebred flies with gray bodies and normal wings GGWW with purebred flies with

 

black bodies and small wings ggww

In the space below draw a punnet square to show this cross.

GW

 

gw Gng

 

 

 

What results would you expect from this cross?

100% gray normal

What did Morgan do? Step 2

Next Morgan crossed one of the F1 flies with a black, small winged fly ggww

in the space below draw a Punnet square to show this cross.

 

._.~——.-_--m._—.~.-.-.o-.—.. - —-.-. -.‘

GW 9W 9W gw

 

: gw Gng gng gng ggww

 

 

 

 

 

 

What results would you expect from this cross?

25% gray normal 25% black normal
25% black small 25% gray small
Actual Results from Step 2

41.5 % gray normal 8.5% black normal
41.5 % black small 8.5% gray small

A majority of the gray flies had normal wings and a majority of the black flies had small wings.

64

Morgan’s Conclusion
That the gene for wing size and body color were inherited together.
if 83% of the offspring have the same gene combinations as the parents. where did the other
combinations come from?
The remaining 17% of the offspring had new combinations of genes.
These offspring are called recombinants
What is a recombinant?
- Offspring that have a new combination of genes not seen In either parental
genotypes
How does recombination occur?

Recombination occurs when crossing over has broken the linkage groups apart during meiosis.

65

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66

Sex determination and Sex-linked Inheritance Guided Note Handout

What are sex chromosomes?

Sex chromosomes are two chromosomes that determine the sex of offspring. They may differ in
g and M depending on the species of the organism they are from. In m; and
Drosophila, males have a smaller sex chromosome.

Are sex chromosomes of man homologous?

Human sex chromosomes are different in size and s_ham. The smaller sex chromosome is called
the X chromosome, and a larger one, is called the X chromosome.

Males have the genotype fl and females are a

How is sex determined in humans and fruit flies?

During rye—log each parent forms gametes. These gametes contain _h_al_f the number of
chromosomes as a normal body cell.

How many chromosomes does a human gamete contain? Q

How many chromosomes does a fruit fly gamete contain? 4

So how is sex determined In humans and fruit flies?

Each gamete contains an X or a X chromosome.

Gametes produced by a male have an 1 or a X chromosome.

Gametes produced by a female have only an X chromosome.

Th9 99.999? 3‘19??? °' .59? 99“"“3933”

 

 

 

 

 

 

I

I x Y
x xx xv
x xx xv
Male = XY Female = XX

 

What is the probability of having a male child?

What is the probability of having 3 male children in a row?

 

67

What is the probability of a fourth child being a female?

 

Which parent determines the sex of a child?

Since females can only produce gametes with an X chromosome it is the male of the species that
determines the sex of the offspring.

What are sex linked traits?

Traits with an gjfle located only on the X chromosome.

The X chromosome lacks a corresponding allele.

What does this mean?

Sex-linked recessive traits (such as white eyes in fruit flies, hemophilia, and colorblindness in
humans) occur more commonly in m_alg because they have only one x chromosome. Therefore
there is no chance of them having a second copy of the domina_ntlnorma_l allele.

The Genotypes of Colorblindness

 

Normal Male = XBY Colorblind Male = X°Y Normal Female = XBXB

 

 

Colorblind Female = X°X° Carrier Female = XBX15

 

 

 

Cross a normal male and a carrier female.

 

 

xb x3x" xbv

 

 

 

 

Cross 3 colorblind male and normal female.

 

xB x'3xb XBY

 

 

 

 

 

68

 

Cross a colorblind male and carrier female. _ _ _. .__.

 

 

 

 

 

 

Characteristics of Sex-linked traits

Phenoggic expression more common in _ma_les_

Sons cannot inherit the trait from their fa_the_rs, but daughters can

Sons inherit their Y chromosome from their father

What purpose does the Y chromosome serve?

Only a few genes have been identified on the Y chromosome, one them is the testis-determining

factor (TDF) promotes development of the male phenotype

69

Mutations Guided Note Handout

Who coined the term mutation?

Hugo de Vries a tum-of-the-century scientist who rediscovered the work of Mendel recognized that
occassional abrupt, sudden changes occurred in the patterns of inheritance in the primrose plant
these were called mutations

De Vries proposed that new alleles arose by mutations

 

So, what is a mutation?

A change in an organism's ge_neti_c information that occurs during replication

Are mutations a bad thing?

Some mutations can be beneficial to an organism while others can be I_eth_a!

What type of cells are affected by mutations?

All ce_lls_ can be mutated

Germ cell mutations affect reproductive cells, these an be inherited.

Somatic cell mutations affect other body cells, these cannot be inherited.

Types of mutations

Chromosomal mutations affect segments of chromosomes, whole chromosomes or entire sets
of chromosomes

Types of Chromosomal Mutations

Inversion -when part of a chromosome is reversed during replication

Translocatlon-when part of a chromosome breaks off and attaches to another chromosome. The
other chromosome exchanges a segment of itself also.

Duplication-when a segment of a chromosome is remated during replication

Deletion- when a segment of a chromosome is lost (not replicated)

Nondisjunction-a mutation that involves the whole chromosome or set of chromosomes. When
involving one chromosome it usually results in an m copy of a chromosome in one cell and the
lg; of that chromosome from the other. This situation results in a number of human disorders.
When involving a whole set of chromosomes the result is a cell that has a 3_N (or 4N) chromosome

number. This situation is fatal in animals and in plants it produces a hardier, large specimen

70

What is a gene mutation?

A change in the DNA of a cell

Types of Gene Mutations

Point Mutation

A mutation that changes g_r_t_e_ base on the DNA template

The result may be change in an amino acid on a protein chain (remember that some amino acids
have multiple codons)

Example: sickle cell anemia

Frameshift Mutation

A mutation that ln_se_rts or 51% one base on the DNA template. Insertion or deletion will change

all of the amino acids that occur after it thus producing a completely different protein

71

Human Inheritance Guided Notes Handout
Multiple Alleles- tr_a__its that have three or more alleles
Examples
> ABO blood groups
> Rh factor
> Eye color
Codominance-rather than expressing a blending of traits the heteroygotes express both
phenotyms
An example is human ABO blood types.
Neither type A or type g is dominant over the other but they are both dominant over type 9 so they
are codominant.
Blood Group Genotypes/Phenotypes
>i¥mfi=ngg
> i":8 or IBi = 1M
> W3 = Tyg_e AB
> ii = M
Blood type AB people manufacture antibodies to both A and B types
Blood Type A people manufacture only anti-B antibodies
Blood type B people make only anti-A antibodies
Blood type 0 people make no A or B antibodies
Codominance Problem 1

Complete the following cross lAi x lBi

 

 

I“ i
1° f 1"]:I3 I°i |
i | I‘i ii |

 

 

What are the possible genotypes and phenotype of the offspring?

Genotypes i‘l’, I'i, i‘i and ii Phenotypes AB, 3, A and o

72

Codominance Problem 2

Complete the following cross IAIA or IBi

A A

 

A
B

 

What are the possible genotypes and phenotype of the offspring cross?

- Genotypes l“ll3 and I‘i Phenotypes AB and A
Codominance Problem 3

Complete the following cross I’l‘lA or W3

 

 

1“ IA
1‘ | I‘I“ I‘I‘ |
1" | I‘IB I‘I" |

 

 

What are the possible genotypes and phenotype of the offspring?

° Genotypes I‘l‘ 8r l‘l" Phenotypes A and AB

Rh Blood Groups

In addition to the ABO blood types there is another blood antigen factor that can be inherited from
parents.

Individuals with the antigen are Rh msitive

Individuals without it are Rh nggatlve

Rh pgsltlve is dominant to Rh ngative

if one parent is heterozygous positive for the Rh antigen and the other is negative, what are their

chances of having negative offspring? 50 %

 

 

 

Rh+ Rh- i
an- Imman- Rh-Rh-I
Rh- |Rb+Rh-Rn-Rifl

 

73

Epistasls
When one gene interferes with the expression of another
Example
> bildneLsMIdow'mk
Polygenic Traits-a trait that is controlled by the effects of Mlle genes
These traits are not expressed as absolute or Mg forms
These traits are recognized by their expression as a gradation of small differences
Example
> There are 6 genes responsible for skin color.
BBBBBB= very dark pigmentation
bbbbbb = very light pigmentation
All the other genotypes are intermediates of these combinations.
Other Examples of Polygenic Traits in Man
> Height
> Lupus
Weight
Eye Color

intelligence

VVVV

Many forms of behavior
Plelotrophy-when a gene has an effect on more than one tr_ai_t
Example Sickle cell anemia

> Sickle-celled individuals have a natural immunity to malaria

74

Human Genetic Disorders Guided Notes Handout
The Human Genome
There are 44 autosomes and 2 sex chromosomes in the human genome, for a total of 46
Chromosomal Abnormalities (nondisjuction)
Trisomy 21 (Down’s Syndrome)
> mid to severe mental retardation
short, stocky body type
large tongue leading to speech difficulties

a propensity to develop Alzheimer's Disease

VVVV

incidence of Down's Syndrome increases with age of the mother
> 25% of the cases result from an extra chromosome from the m
Kleinfelter‘s syndrome (XXY)
> Male In agmarance
> Undngeveloped sex organs
> M
Turner‘s syndrome (X)
> Female in appearance
> Sex organs underdeveloped
> Sterile
Recessive Genetic Disorders
According to Mendel, in order for a recessive trait to be seen both alleles must be M. Here
are some examples of recessive genetic disorders. Albinism, Phenylketonuria, Tax-Sachs
Disease, Sickle-cell anemia 81 ngtic Fibrosis.
Albinism
The lack of the pigment M in skin, hair, and eyes.
Homozygous recessive (aa) individuals make no melanin, so they have face, hair, and eyes that

are white to yellow.

75

Several mutations may cause albinism:

1) the lack of M used to produce the pigment melanin

2) the inability of an pm to convert tyrosine into Mp

Phenylketonuria (PKU)

Individuals lack the ability to synthesize an enzyme to convert the amino acid phenylalanine
(found in aspartame-Nutrasweet‘“) into tyrosine. Individuals homozygous recessive for this
M have a buildup of phenylalanine in the urine and blood. The breakdown products can be

harmful to developing nervous systems and leads to mental retardation.

 

1 in 15 000 infants suffers from this problem. (a very common disorder)

Most states screen newborns for PKU.

Tay-Sachs Disease-an autosomal recessive disorder resulting in dpgeneration of the nervous
system.

Children rarely survive past five years of age. Sufferers lack the ability to metabolize lipids (fats).
These lipids accumulates in m cells, eventually killing them.

Sickle-cell anemia

Nine-percent of African-Americans are heterozygous, while 0.2% are homozygous recessive. The
recessive allele causes a single amino acid substitution in the peg chain of the protein
hempgiobin.

When oxygen concentration is low, sickling of cells occurs.

Heterozygotes make enough ”good beta-chain hemoglobin" that they do not suffer as long as

remain high, such as at sea-level.

 

Cystic Fibrosis-causes the production of mucus that clogs the airways of the lungs and the ducts
of the mnereas and other glands.

The most common genetic disease in Caucasians.

Dominant Genetic Disorders

According to Mendel, in order for a dominant trail to be seen only one allele must be inherited.
Here are some examples of dominant genetic disorders: Polydactly,

Huntington's disease and dwarfism.

76

Polydactly-the presence of an _e_x_tg digit on the hands
in modern times the extra finger has been cutoff at birth and individuals do not know they cany this
trait. The extra digit is rarely functional and definitely causes problems buying gloves.
Huntington's disease-a disorder that causes the progressive destruction of brain cells. If a parent
has the disease, at least l_ia_il_' of the children will have it. The disease usually does not manifest until
after age of 35.
Sex linked Genetic Disorders
According to Morgan, a sex-linked trait occurrs on that part of the X chromosome that lacks a
corresponding location on the X chromosome. These are more prevalent in ma_les_. Some
examples of sex-linked genetic disorders are colorblindness, hemophilia and Duchenne M.D..
Colorblindness
Color blindness afflicts 8% of males and 0.04 % off human females.
Color vision depends on three genes, each producing chemicals sensitive to different colors of light.
32;! and was detecting genes are on the X-chromosome, while the blue detecting gene is on an
autosome.
Hemophilia-A group of diseases in which blood does not clot normally
England's Queen Victoria was a carrier for this disease. The allele was passed to two of her
daughters and one son. Since royal families in Europe commonly interrnarried, the allele spread.
Duchenne Muscular Dystrophy
Muscular dystrophy is a term encompassing a variety of muscle wasting diseases.
DMD affects cardiac and skeletal muscle, as well as some mental functions.
Occurs in 1 in 3500 newborns
Most sufferers die before their 20th birthday
Diagnosing Genetic Disorders
Amniocentesis-Primary method used today.

> A thin needle is inserted into the amniotic fluid surrounding the fetus.

> Fetal cells are withdrawn with the fluid.

> These fetal cells can be used to construct a karyotype of the fetus.

77

Breeding Techniques Guided Note Handout

Breeding Stragegies

Selective Breeding-selecting a few individuals that have desired traits to serve as grants for the

next generation

 

Advantages

Disadvantages

 

produce new characteristics

- loss of some genetic variability

 

 

increase frequency of desired trait

 

susceptibility to disease

 

Inbreeding-breeding individuals with similar traits so that those traits will appear in their offspring

 

 

Advantages

Disadvantages

 

- maintains a certain set of desired traits

- individuals are related

 

- increased chance of genetic disorders

 

 

 

- susceptibility to disease

 

Hybridization-crossing members of different, but related species usually resulting in a hardier

 

 

variety of offspring
Advantages Disadvantages
- produces a hardier variety - none

 

- increases genetic variety

 

increases desired characteristics

 

 

° decreased chance genetic disorders

 

 

MutagenesIs-the introduction of mutagens to cause beneficiall mutptions

What is a mutagen?

A mutagen is anything that causes a mutation

 

 

 

Some Common Examples:

>

>
>
>

Ultraviolet radiation
X-rays
Tars from tobacco

Asbestos

How can mutations be used as a breeding technique?

Beneficial mutations can be maintained by selective breeding

79

Gel Electrophoresis Guided Note Handout
What does it mean?
Gel-a gelatin-like substance
Electra-refers to electricity
Phoresis-a Greek word that means to “carry across"
What is it?
It’s a separation technology in which molecules are forced across a gel by an electrical current.
Electrodes provide the driving force behind the movement
What Is separation dependent upon?
Qlflgp and M of the specimen determines separation.
When placed in a buffer solution and applied to a gel, charge and mass act in concert. The
npgativey charged particles will migrate toward the positive pole, the msitively charged particles
will migrate towards the negative pole.
Does DNA have a charge?
The sugar3phosphate “backbone” of DNA has a 3 charge. DNA will always migrate towards the
mil—W9 pole-
How does it really work?
The frictioniforce of the sample moving through of the gel acts as a sieve, separating the
molecules by gjz_e_.
Loading the gel
Using a micropimt, the samples are placed into each wpfl.
The gel is then covered with running buffer.
Connect the gel reservoir to a ppwer supply using the electrodes.
Running the gel
The electrodes have different charges. The black electrode has a npgative charge. The red
electrode has a Me charge. The electrical current from one electrode will m the sample

which the other electrode attracts.

80

Reading the Gel

Orient the gel so that the wells are on the t_op. The m fragments move slower than the gmlLet:
fragments. One lane usually contains a DNA ladder.

Reading the Gal

The DNA ladder contains fragments of know length and mass.

 

Bands on the gel are compared to the ladder if length and mass are needed

What can the bands tell us?

Cancer types, characterizing genetic dysfunction, identifying who committed a crime, nucleotide
sequences.

A short summary...

Gel electrophoresis can be used to separate and determine the giz_e of the RFLPs. The exact
number and size of fragments produced by a specific restriction enyme varies from individual to

individual.

81

Genetic Engineering Guided Note Handout
What is Genetic Engineering?
Techniques used to manipulate permanent gen—etip changes in DNA Transplanting genes from
one living source into another where it will be expressed.
What are the required materials?
> Restriction Enzymes-an enzyme that cuts a gig molecule at a spgiflc base sequence.
(there are 75 known restriction enzymes) 1
> DNA-the carrier of the genetic code. Specifically the seguence of bases (a gene) that
codes for the my; you wish to produce (“YFG”)
> Host organism-the organism that you will insert the desired gene into
So how do I do It?
> identify “YFG”
> Out “YFG” with restriction enzyme
A restriction enzyme cuts the DNA seguence for “YFG” at a site before and after
“YFG”. (you will need to use the same restriction enzyme to cut your plasmid)
> Combine YFG with host DNA
Cut the host DNA called a mg with the same restriction eme used to cut
“YFG”
The “stic ends” of "YFG" attaches to the “sticky ends” of the plasmid. This fusion
of DNA from different individuals results in recombinant DNA
> Insert recombinant DNA into host
Mix recombinant DNA with millions of flagella. Suspend them in a dense “_Slfl

solution. Some of the bacteria will take in the recombinant DNA Isolate these

 

bacteria and grow them.
Are there other methods of inserting “Y FG”?
> microinjection

> attachment to wire-like pellets that are shot into cells with a microscopic gun

82

How do you know if the host takes up the plasmid?

> YFG is usually m with another gene. This other gene codes for a protein that can be
expressed physically

Examples-
Bioluminescence
resistance to antibiotics

> Sequence DNA to check for YFG
Radioactive phosphorus is placed on a single strand of DNA
DNA strand is divided into 4 groups each group is treated with different chemicals
DNA pieces are separated by electrophoresis this reveals the pm sequences present in
the DNA strand

> How is this technology used?

Transgenic bacteria are used to produce large quantities of proteins.

 

 

 

 

Examples:

HGH insulin vitamins
xantham gum citric acid flavoring agents
vaccines antibiotics antiparasitics
biofueis toxic waste disposal interferon

 

 

 

 

 

Transgenic plants could increase disease resistance, put an end to the need for
fertilizers and insecticides.

Transgenic animals could produce farm animals that produce milk and meat faster.

83

Applied Genetics Guided Note Handout
What is DNA Fingerprinting?
DNA Fingerprinting, is a technique used to analyze regions of human DNA.

Approximately 99% of human DNA is identical.

 

The remaining 1312 is unique to each individual unless Identica_l twins or clones.
Howls DNA fingerprinting done?
> DNA extraction
> Digestion of DNA with a restriction enzymes
> Gel electrophoresis
> Preparation of a "southern blot
DNA extraction
DNA can be extracted from almost any m.
Sources of DNA found at a crime scene might include blood. semen. tlsfl from a deceased
victim, cells In a hair follicle , and even M on a piece of gum.
DNA extracted from items of evidence is compared to DNA extracted from reference samples from
known individuals, normally from blood.
Digestion of DNA
Extracted DNA is treated with a mtriction engvme.

The enzyme most commonly used for forensic DNA analysis is Haeiii, which cuts DNA at the base

 

sequence £3993.

Gel electrophoresis

Following DNA digestion, the resulting DNA fragments are separated by size.

849%; DNA fragments move more rapidly than larger ones.

The smaller DNA fragments move the fart_he_st from the wells.

Preparation of a ”southem blot"

The separated DNA fragments are denatured in a basic solution.

The single stranded DNA fragments are transferred to the surface of a nylon membrane by blotting.

The process is similar to blotting of wet ink on a dry paper.

84

What Is this used for?
Biological Evidence
> FBI and police labs around the US. have begun to use DNA fingerprints to link
suspects to biological evidence - blood or semen stains, hair, or items of clothing -
found at the scene of a crime.
> Another important use of DNA fingerprints in the court system is to establish paternity
in custody and child support litigation. DNA fingerprints bring a nearly perfect match.
Personal Identification
> Because every organ or tissue of an individual contains the same DNA fingerprint, the
US. armed services have just begun a program to collect DNA fingerprints from all
personnel for use later, in case they are needed to identify casualties or persons
missing in action.
Developing Cures for Inherited Disorders
> Research programs to locate inherited disorders depend on the information contained
in DNA fingerprints. This is the first step in designing a genetic cure for these
disorders.
Diagnosis of Inherited Disorders
> DNA fingerprinting is used to diagnose inherited disorders in both prenatal and newborn
babies. These disorders include py_stic fibrosis. hemophjljg, Huntington's disease, familial

Alzheimer's, sickle cell anemia and many others.

85

APPENDIX D

Activities

86

Name Period 1 2 3 4 5 6 Date

Sex-Linked Traits Review:

Work out the following problems completely.

 

 

 

 

1. What are the two sex chromosomes? and

2. in order for an individual to be a male he must contain the __ and _ chromosomes.
3. A female's sex chromosomes are and

4. Of the male and female , the is responsible for the sex of their offspring.

 

5. Mate a male (XX) with a female (XY). Answer the questions that follow:

 

Gametes

 

 

 

 

 

 

 

6. What is the possibility of producing a boy?

 

7. What is the phenotypic ratio?

 

8. If a family contained 5 children, 3 boys and 2 girls, what is the chance of having another girl?

9. What are sex-linked traits?

 

10. Which sex chromosome is the largest?

 

11. Which sex chromosome contains the most genes?

 

12. Name 2 sex-linked diseases.

13. What is the genotype of a carrier female for Hemophilia?

14. Mate a normal male and a carrier female for Hemophilia. Use the Punnett square below to show

your work.
x"v x x“x"

 

Gametes

 

 

 

 

 

 

 

20. What are the chances of having a normal boy?

 

87

21. What are the chances of having a normal girl?

 

22. What are the chances of having a hemophiliac boy?

 

23. Determine the possible blood types from the mating of a person who is homozygous A blood
type and a person who is heterozygous B blood type.

Show you work here:

 

Gametes

 

 

 

 

 

 

24. Mate a colorblind male and a carrier female.

Female I I Male 7

 

 

 

 

Gametes

 

 

 

 

 

 

25. What are the chances that their sons will be colorblind?

26. What are the chances that their daughters will be carriers?

27. if a woman's father had hemophilia, what are the chances that she is normal?

28. Explain your answer.

29. If her mother was a carrier, what are the chances that she is norrnal?, a carrier?

30. Explain why are there more males with sex-linked problems than females?

88

 

 

Name Period 1 2 3 4 5 6 Date

MUTATIONS-Get the point?
Problem:
How can mutations change an organism?
Objectives:

1) To define the types of point mutations
2) To describe how mutations produce changes in the cell

Materials:
scissors lab sheet sheet of “bases"

Procedure:

Part A-Organizing the sguence
1) Cut out the letters on the ”Base cut-out sheet”

2) Using your base cut-outs arrange the bases in the following order:

THE CAT SAW THE DOG
(REMEMBER: THIS SENTENCE REPRESENTS A DNA SEQUENCE)

a) Why are the bases in sets of three?

b) What term is used to describe a set of three bases?

Part B-Simulating a substitution point mutation
1) Using your base cut-outs, arrange the bases as in Part A.
2) Replace the letter D with the letter M

o) How does the new sentence read?

b) Does the sentence have a new meaning? Explain.

c) Does the sentence have a new meaning? Explain.

89

Part C-Simulating a frame shift mutation

1) Using your base cut-outs, arrange the bases as in Part A.
2) Add the letter M after the letter C and regroup the letters in groups of threes.

a) How does the new sentence read?

b) Does the sentence have a new meaning? Explain.

c) Does the sentence have a new meaning? Explain.

Part D-Simulating a deletion point mutation

1) Using your base cut—outs, arrange the bases as in Part A.
2) Remove the letter C from the sentence and regroup the letters in groups of three.

a) How does the new sentence read?

b) Does the sentence have a new meaning? Explain.

c) Does the sentence have a new meaning? Explain.

Analfiis

1 . What is a mutation?

2. How can DNA mutate? Describe how this occurs.

90

3. Why is a change in DNA permanent for the cell?

4. How can a point mutation cause severe changes in the organism?

5. Why should we be concerned about mutations?

91

Name Period 1 2 3 4 5 6 Date

Color Blindness Problem Set

This problem set is based on a question received from a woman named Audrei. Audrei is red-green
color blind and so are other members of her family. She wanted to know if we could help her
understand how she inherited her color blindness.

Audrei's family

There are 7 children in Audrei's family, three girls and four boys. Two of he giris, Audrei and Liz, are
red-green color blind. Caroline has normal color vision. Only two of the boys have been tested. Paul
is color blind and David has normal color perception. Andrew and Jason, who have not been tested,
may or may not have normal color perception. Barbara, the mother of the seven children, has
normal color vision, but Sidney, the father, has the red-green color perception defect. Audrei also
has a half brother Stephan. Audrei and Stephan have the same mother, but a different father.
Stephan is also red green color blind.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Barbara Sidney
CB 3) L_.
Audrei Liz David Paul Caroline Andrew Jason
. = affected female - = affected male
@ = carrier female = normal male

 

 

 

O = normal female

Go to this web site and read the background infonnatlon and begin the activity:
http:/fwww.biology.arizona.edulhuman_blolproblem_setslcolor_blindnesslintro.html
Red-green color blindness

Red-green color blindness is an X-linked, recessive trait. in this problem set we will establish the
pedigree of Audrei's family and see how the color perception defect is passed on from one

generation to the next. Once you have identified the genotype of an individual use the key above to
color in their spot on the pedigree.

93

 

1)

2)

3)

4)

5)

6)

7)

3)

9)

Audrei is the family member who contacted us. She and her father Sydney are colorblind, but
her mother, Barbara, has normal vision. What is Audrei's genotype?

What is her father‘s genotype?

What is the mother's genotype? Remember that Audrei's mother does not have a color
perception defect.

What is the genotype of Audrei's sister Caroline, who has normal vision?

In her message, Audrei also tells us that her sister Liz and brother Paul are also colorblind,
but that her brother David has normal vision. How will Liz be represented on the family

pedigree?

Paul, like Liz, is colorblind. How will he be represented in the family pedigree?

Andrew and Jason have not been tested. What is the possibility that they will have the color
perception defect?

We don't know if any of Audrei's grandparents were red-green colorblind, but we can make
some educated guesses. Is it possible that all of Audrei's grandparents have normal vision?

Would we predict mat Stephen's father is colorblind? Why or Why not?

10) What is the probability that Audrei's sons will be colorblind? Explain you answer.

11) What is the probability that Audrei's daughters will be colorblind? Explain your answer.

Name Period 1 2 3 4 5 6 Date
Karyotyping Activity

introduction

This exercise is a simulation of human karyotyping using digital images of chromosomes from
actual human genetic studies. You will be arranging chromosomes into a completed karyotype, and
interpreting your findings just as if you were working in a genetic analysis program at a hospital or
clinic. Karyotype analyses are performed over 400,000 times per year in the US. and Canada.
Imagine that you were performing mesa analyses for real people, and that your conclusions would
drastically effect their lives.

G Banding

During mitosis, the 23 pairs of human chromosomes condense and are visible with a light
microscope. A karyotype analysis usually involves blocking cells in mitosis and staining the
condensed chromosomes with Giemsa dye. The dye stains regions of chromosomes that are rich in
the base pairs Adenine (A) and 'Ihymine (T) producing a dark band. A common misconception is
that bands represent single genes, but in fact the thinnest bands contain over a million base pairs
and potentially hundreds of genes. For example, the size of one small band is about equal to the
entire genetic information for one bacterium.

 

Chromosome smear
Identifying features of a
chromosome
.
‘~ placement of centromere

/,/“(

e0" 9.“

_ banding pattern

I-IQZITII"

 

 

 

 

u,
M

The analysis involves comparing chromosomes for their length, the placement of centromeres (areas
where the two chromatids are joined), and the location and sizes of G-bands. You will electronically
complete the karyotype for three individuals and look for abnormalities that could explain the

phenotype.
Your assignment

This exercise is designed as an introduction to genetic studies on humans. Karyotyping is one of
many techniques that allow us to look for several thousand possible genetic diseases in humans.

You will evaluate 3 patients' case histories, complete their karyotypes. and diagnose any missing or
extra chromosomes. Then you'll conduct research on the intemet to find web sites that cover some
aspect of human genetics.

Go to the following web site and begin the activity:

http://www.biology.arizona.edulhuman__bio/activities/karyotypinglkaryotypingz.html

95

Patient A History-
Patient A is the nearly-fuII-terrn fetus of a forty year old female. Chromosomes were obtained from
fetal epithelial cells acquired through amniocentesis.

Interpreting the karyotype

Lab technicians compile karyotypes and then use a specific notation to characterize the
karyotype. This notation includes the total number of chromosomes, the sex chromosomes, and
any extra or missing autosomal chromosomes. For example, 47, XY, +18 indicates that the
patient has 47 chromosomes, is a male, and has an extra autosomal chromosome 18. 46, XX is
a female with a normal number of chromosomes. 47, XXY is a patient with an extra sex
chromosome.

1. What notation would you use to characterize Patient A's karyotype?

Making a diagnosis

The next step is to either diagnose or mle out a chromosomal abnormality. In a patient with a
nonnai number of chromosomes, each pair will have only two chromosomes. Having an extra or
missing chromosome usually renders a fetus inviable. in cases where the fetus makes it to term,
there are unique clinical features depending on which chromosome is affected. Listed below are
some syndromes caused by an abnormal number of chromosomes.

2. What diagnosis would you give patient A?

 

Diagnosis Chromosomal Abnormality

 

patient's problems are due to something other than an

N I f r m
0"“3 # 0 Ch omoso es abnormal number of chromosomes.

 

 

 

 

 

 

 

Klinefelter's Syndrome one or more extra sex chromosomes (i.e., XXY)
Down's Syndrome Trisomy 21, extra chromosome 21
Trisomy 13 Syndrome extra chromosome 13

Patient B History-

Patient B is a 28 year old male who is trying to identify a cause for his infertility. Chromosomes were
obtained from nucleated cells in the patient's blood.

Interpreting the karyotype
Lab technicians compile karyotypes and then use a specific notation to characterize the karyotype.

This notation includes the total number of chromosomes, the sex chromosomes, and any extra or
missing autosomal chromosomes. For example, 47, XY, +18 indicates that the patient has 47
chromosomes, is a male, and has an extra autosomal chromosome 18. 46, XX is a female with a
normal number of chromosomes. 47, XXY is a patient with an extra sex chromosome.

1. What notation would you use to characterize Patient B's karyotype?

96

Making a diagnosis

The next step is to either diagnose or rule out a chromosomal abnormality. In a patient with a
normal number of chromosomes, each pair will have only two chromosomes. Having an extra or
missing chromosome usually renders a fetus inviable. In cases where the fetus makes it to term,
there are unique clinical features depending on which chromosome is affected. Listed below are
some syndromes caused by an abnormal number of chromosomes.

2. What diagnosis would you give patient B?

 

 

 

 

 

Diagnosis Chromosomal Abnormality
patient's problems are due to something other than an abnormal
Normal #of chromosomes number of chromosomes.
Klinefelter's Syndrome one or more extra sex chromosomes (i.e., XXY)
Down's Syndrome Trisomy 21, extra chromosome 21
Trisomy 13 Syndrome extra chromosome 13

 

 

 

 

Patient C History-
Patient C died shortly after birth, with a multitude of anomalies, including polydactyly and a cleft lip.
Chromosomes were obtained from a tissue sample.

Interpreting the karyotype
Lab technicians compile karyotypes and then use a specific notation to characterize the karyotype.

This notation includes the total number of chromosomes, the sex chromosomes, and any extra or
missing autosomal chromosomes. For example, 47, XY, +18 indicates that the patient has 47
chromosomes, is a male, and has an extra autosomal chromosome 18. 46, XX is a female with a
normal number of chromosomes. 47, XXY is a patient with an extra sex chromosome.

1. What notation would you use to characterize Patient C's karyotype?
Making a diagnosis

The next step is to either diagnose or rule out a chromosomal abnormality. In a patient with a
normal number of chromosomes, each pair will have only two chromosomes. Having an extra or
missing chromosome usually renders a fetus inviable. In cases where the fetus makes it to term,
there are unique clinical features depending on which chromosome is affected. Listed below are
some syndromes caused by an abnormal number of chromosomes.

2. What diagnosis would you give patient C?

 

 

 

 

 

 

 

Diagnosis Chromosomal Abnormality

Normal # of ch as patient's problems are due to something other than an abnormal
'0' ' iosor " number of chromosomes.

Klinefelter's Syndrome one or more extra sex chromosomes (i.e., XXY)

Down's Syndrome Trisomy 21, extra chromosome 21

Trisomy 13 Syndrome extra chromosome 13

 

 

97

Final part of assignment

Now, the final part of this activity is to search the intemet for web sites that cover some interesting
aspect of human genetics. An excellent way to conduct intemet searches is to use a meta-search
engine such as Dogpile at httmllwwwdogpilgcom.

At the Dogpile site, when you type in key words, Dogpile will search all of the major search engines
on the web and compile the result for you in nice, friendly format. After you conduct your intemet
search, at the bottom of your paper, add the web address of a human genetics web site that you like,
the title of that site, and a 2-3 sentence summary of the site's content.

 

98

Name Period 1 2 3 4 5 6 Date

Human Genetics beta-globin Identification Project

Sickle cell anemia is a genetic disorder in which the red blood cells are sickle shaped in the
presence of a low oxygen concentration in the blood stream instead of their normal flattened disk
shape (see figure below). This is caused by a point mutation, a change in one base in the DNA
coding sequence. In this activity your task is three-fold.

In Part One, you must decode the DNA strand below and write out the resulting m-RNA strand in the
space provided.

In Part Two, you must translate the m-RNA strand and determine the amino acid sequence of the
beta-globin protein.

In Part Three, you must search the web to determine if you have the DNA that codes for the normal
beta-globin or the mutated form.

   

Normal Red Blood Cells Comparison of Normal Red
with a Sickle Cell Blood Cell and a Sickle Cell

Normal Red Blood Cells

DNA Sequence-

TAC CAG GTA AAC TAG GGG CAT CTI' TIT AGG CGA CAG TGG CGC GAG ACC CCC 'I'I'C
CAA TTG CAT CTA CTC CAG CCT CCG GTC CGT AAC CCC GCA GAT AAC CAA CAG ATA
GGG ACC TGT GTT TCC AAA AAG C'IT AGC AAA CCC TTG AAT AGT TGG GGG CTA CCC
CAA TAC CCT ‘I'I'G GGG TTT CAG TTC CGT GTA CCG TI’C TTC CAG GAG CCT CGT AAG
AGT CTA CCC AAT CGC GTA AAC CTC TTG GAA TTT CCC TGC AAA CGT TGA AAC AGG
CTl' AAT GTG ACA CTA 'ITI' GAG GTG CAT CTA GGT CTT Tl’A AAG GCG AAC GAT CCC
'ITA CAT GAC CAA ACA CAG CGT GTA GTA AAA CCC ‘ITC CTC AAG TGG GGT GGA CAA
GTT CGT CGA ATA G‘IT T'IT CAA CAG CGA CCT CAG CGC TTG CGT AAT CGG GTG TIT
ATA GTG ATC

99

Part i. m-RNT sequence

Part II. Amino Acid sequence

100

PART III. IDENTIFICATION OF BET A-GLOBIN PROTEIN

After examining the beta-globin chain of hemoglobin, determine if it is the wildtype (normal) or
mutant strain. Write a paragraph to explain how you determined which beta-globin your DNT
sequence is coding for.

101

Name Period 1 2 3 4 5 6 Date

Human Genetics beta-globin identification Project

Sickle cell anemia is a genetic disorder in which the red blood cells are sickle shaped in the
presence of a low oxygen concentration in the blood stream instead of their normal flattened disk
shape (see figure below). This is caused by a point mutation, a change in one base in the DNA
coding sequence. in this activity your task is three-fold.

In Part One, you must decode the DNA strand below and write out the resulting m-RNA strand in the
space provided.

In Part Two, you must translate the m-RNA strand and determine the amino acid sequence of the
beta-globin protein.

In Part Three, you must search the web to determine if you have the DNA that codes for the normal
beta-globin or the mutated form.

     

Normal Red Blood Cells Comparison of Normal Red
”°"“" R“ B'°°d “"5 with a Sickle Cell Blood Cell and a Sickle eerr
DNA Sequence-

TAC CAG GTA AAC TAG GGG CTT C‘IT ‘ITT AGG CGA CAG TGG CGC GAG ACC CCC 'ITC
CAA TTG CAT CTA CTC CAG CCT CCG GTC CGT AAC CCC GCA GAT AAC CAA CAG ATA
GGG ACC TGT G'IT TCC AAA AAG CTT AGC AAA CCC TTG AAT AGT TGG GGG CTA CCC
CAA TAC CCT ‘ITG GGG 'ITT CAG TTC CGT GTA CCG ‘ITC TTC CAG GAG CCT CGT AAG
AGT CTA CCC AAT CGC GTA AAC CTC TI'G GAA TTT CCC TGC AAA CGT TGA AAC AGG
CTT AAT GTG ACA CTA TIT GAG GTG CAT CTA GGT CTT TTA AAG GCG AAC GAT CCC
‘ITA CAT GAC CAA ACA CAG CGT GTA GTA AAA CCC 'ITC CTC AAG TGG GGT GGA CAA
GTT CGT CGA ATA GTT TIT CAA CAG CGA CCT CAG CGC TTG CGT AAT CGG GTG TIT
ATA GTG ATC

102

Part I. m-RNT sequence

Part II. Amino Acid sequence

103

PART III. IDENTIFICATION OF BETA-GLOBIN PROTEIN

After examining the beta-globin chain of hemoglobin, determine if it is the wildtype (normal) or
mutant strain. Write a paragraph to explain how you determined which betaglobin your DNA
sequence is coding for.

 

104

Name Period 1 2 3 4 5 6 Date

Genetic Disease Paper

DIRECTIONS: You represent a particular genetic disease. Your task is to describe to the reader of
this paper a variety of things. First, identify which disease you represent. Second, describe to the
reader how you developed; are you a result of a gene mutation such as a point mutation, frameshift
addition, frameshift deletion or are you a result of a chromosomal mutation such as deletion,
addition, translocation or nondisjunction. Third, when describing how you developed, identify
(using a fictitious name) the individual you developed in and the results of being afflicted with you.
(Remember, mutations are carried from one generation to the next via the gametes.
Hence,errors in meiosis play a role.) Finally, describe to the reader how you cause pain and
suffering to the affected person. In other words, what symptoms are caused by you. Are the
symptoms you cause treatable and/or curable? if so, what is done to treat and/or cure these
symptoms. Do you, as the genetic disorder, affect males only? Do you affect females only?

MECHANICS: Your paper must have an introduction, a body and a conclusion. You must
document the resources you use at the end of your paper. You must write your paper in such a
way that the average seventh grader would understand what you are writing about. (in other
words, take the time to use a dictionary, look up words that you don’t understand and put
them in your own words.) For every “big” word you use that you don’t understand, five points will
be deducted from the total. As a gentle reminder, cheating of any sort, which includes plagiarism,
will not be tolerated. I will fail you for the entire marking period if you choose to perform in such a
dishonest manner.

Your paper must be typed. Let me know if you do not have access to a computer (or typewriter).
Do not inform me the day your paper is due that you did not have access to a computer (or
typewriter). I will not accept hand written papers. If you are planning to be absent the day your
paper is due, make certain your paper is turned in beforehand. LATE PAPERS WILL NOT BE
ACCEPTED.

105

 

THIS PAPER WILL BE DUE ON FRIDAY, MARCH 1, 2002.

 

CONTENT
Content is age appropriate 15 points
Proper identification of genetic disease 10 points
Description of disease 15 points
Description of symptoms 15 points
Is the disease treatable or curable? 10 points
MECHANICS
introduction 10 points
Body 10 points
Conclusion 10 points
Grammar (spelling, punctuation, etc.) 10 points
Bibliography 10 points
TOTAL 115 points

106

Name Period 1 2 3 4 5 6 Date

APPLE GENETICS LAB

DIRECTIONS:

Use the websites below to gather information about the following types of apples. Use the blank
spot on the second page to include the information about a variety of apple not included on this list.
Instead of a lab report, you will be required to type a one-page, single-spaced essay about apples,
your favorite apple and all the information about your favorite apple.

 

Taste
Apple Name and genetic Harvest Appearance Texture

history (hybrid, mutation, etc.) Time (swg'etgrstgiur, (firm, mealy,

etc.)

 

Paula Red

 

Cameo

 

Ginger Gold

 

McIntosh

 

Red Delicious

 

 

Golden Delicious

 

 

 

 

 

107

 

 

Taste
Apple Name and genetic Harvest Appearance Texture

. . t. r
history (hybrid, mutation, etc.) Time (swoeremxur (firm, mealy,

etc.)

 

Fuji

 

JonaGold

 

Jonathon

 

Ida Red

 

Rome

 

Empire

 

Gala

 

 

Braebum

 

 

 

 

 

108

 

 

Apple Name and genetic Appearance Taste

history (hybrid, mutation, etc.) Harvest (sweet, sour, Texture
Time 0' ta" (finn. mealy,
etc.)

 

 

 

 

 

 

 

 

 

More about apples than anyone would ever want to knowi

US Apple Association www.usapple.org

Apple Varieties www.greatlakesfruit.comlvarietie.html

NY Apple Association www.nyapplecountry.comlmain.html
Washington Apples www.bestapples.comlnewlvarletieslindex.html
Virginia Apples www.virginiaappies.orglvarietlesl

Apples and More www.urbanextuiuc.edulapplesl

109

 

Name Period 1 2 3 4 5 6 Date

THE APPLICATION OF GENETICS-THE APPLE GENETICS PAPER

Seeing a variety of fruits and vegetables at a grocery store is not a phenomenon. All of these items
selected to be sold at a grocery store because of some desirable genetic traits. Apples, too, are
selected and sold for their desirable characteristics.

Very few people dislike apples. Because there are so many varieties of apples, most individuals
can say that they have a favorite variety. Could this variety of apple be a favorite because of
texture or flavor? The answer to this question is probably “yes.” Let us look a little bit deeper,
though. From where did this apple get its traits?

Your task is to provide a one—page, single-spaced typed paper that describes the apple of your
choosing. The margins are to be one inch. In your paper you will describe the traits that make your
apple pleasing to your palate and any history that relates to your apple, including its heritage. Your
paper must include an introduction, a body, and a conclusion. Your paper must also be
grammatically correct.

One day will be spent in the computer lab as a means of gathering information about your favorite
apple variety. Any sources used must be cited. if you fail to cite your resource(s) you will
automatically lose 20 points. IF YOU PLAGIARISE YOU WILL RECEIVE AN ”E” FOR THE
MARKING PERIOD.

The following checklist should help you with organization. The due date for your paper will be
Monday March 11‘h 2002 upon your arrival to class. Excuses such as “my printer isn't working” will
not be accepted. Plan ahead! LATE PAPERS WILL NOT BE ACCEPTED. You may email the
paper to me for 10 extra credit points. To receive the extra credit your paper must arrive to me by
6:00 pm Sunday evening.

 

General infonnatlon about apples
Introduce your apple

Describe its desirable traits
Heritage of apple (it’s a hybrid or mutation)
Contains introduction

Body

Conclusion

Single Spaced

One Inch margins

Grammatically correct

Resources cited

110

Name Period 1 2 3 4 5 6 Date

DNA Extraction Teacher Copy
BACKGROUND INFORMATION:

 

The purpose of this lab is to extract DNA from fruit and vegetable matter. First of all, as a review.
let’s discuss DNA. DNA, which stands for deoxyribonucleic acid, is a double -stranded, helical
acid molecule capable of replicating and determining the inherited structure of a cell's proteins.
The importance of DNA is that it is the substance of genes, the units of inheritance that transmit
information from parents to offspring. DNA is found in any cells that are considered living.

in order to extract DNA from different samples we will need a buffer solution. A buffer is a
substance that minimizes change in the concentration of H’ and OH'. As a result of a buffer,
biological fluids resist change to their pH when acids or bases are introduced. Buffer works by
accepting hydrogen ions from the solution when they are in excess and donating hydrogen ions
to the solution when they have been depleted. Most buffers are weak acids or weak bases that
combine reversibly with hydrogen ions. A buffer is necessary in this experiment for its ability to
break apart the double helical structure of DNA. A buffer carries with it a negative charge. A
strand of DNA gets its structure as result of the hydrogen bonds between nitrogenous base pairs.
DNA also contains histones, which are electrically charged. The negative charge of the buffer
attracts the histories and breaks apart the bonds. This allows the DNA to be separated and easily
extracted. Baking soda (NaHCOa) was also used in this experiment because when placed in an
aqueous solution, the baking soda is converted to Na’ + HCO3'. The negative charge on HCO;
attracts the electrically charged amino acids and H‘. This helps the buffer solution to separate the
DNA, and at the same time it maintains the pH of the solution. Dishwashing liquid aids in breaking
apart and separating DNA from the proteins.

PROBLEM:

How can DNA be extracted from fruits and/or vegetables?

MATERIALS:

Buffer solution: FruItIVegetabie Samples:

600 mL distilled water 500 mL distilled water

25 9 baking soda 50 g of fruit/vegetable matter

7.5 9 table salt mix on low speed in a blender for 10 seconds and

1M ' '
25 mL Dawn drshwashrng detergent then on high speed for 20 more seconds

Buffer solution Distilled water 3 samples
mortar and pestle 3 clean test tubes 3 rubber stoppers
3 coffee filters 3 wooden sticks ice cold ethanol (95%)

111

PROCEDURE:

1) Obtain 30 ml of a fruit/vegetable sample.

2) Pour 15 ml of the buffer solution into the test tube.
3) Flatten the coffee filter and fold it in half three times.
4) Insert the coffee filter into the test tube.

5) Pour the sample into the filter and allow the filterate to double the volume in the test tube.
After the volume in the test tube triples, remove the filter and throw it out.

6) Cap the test tube and gently invert for 2 minutes. DO NOT SHAKE THE TEST TUBE!

7) Slightly tilt the test tube and gently pour 10 mL of ice cold ethanol on top of the DNA
solution.

8) insert a narrow wood rod through the alcohol layer - just below the boundary of the alcohol
and buffer. Gently twist the wood stick and spool the DNA around the stick

9) Repeat Steps 1-12 for the remaining samples.

10) Clean up your lab station. NO SOLID MATERIALS IN THE SINK. All liquid material may
be washed down the drain with water.

 

 

 

 

 

 

 

 

 

 

DATA TABLE:

SAMPLE COLOR INTERFACE SPOOLABLE
YES OR NO
YES OR NO
YES OR NO

OBSERVATIONS:

112

1

CONCLUSION:

Type a 2-page, single-spaced lab report explaining your lab. Remember to include the following

sections: Problem, Materials, Procedure, Data and Conclusion. Refer to your lab and textbook.

to address the following questions in the conclusion of your lab report.

Summarize your data.

Why was it necessary to smash up the onion before adding the solutions?

What is the purpose of the buffer?

Which sample produced the best DNA? Which sample produced the poorest DNA?
Why?

Why is the liquid soap important?

What is the function of DNA in an organism's body?

How is DNA extracted from plant material?

PIP.” 95".”?

113

fl

 

Name Period 1 2 3 4 5 6 Date

RECOMBINANT DNA ACTIVITY ONE

Introduction:

In order to remove a gene from one cell and insert it into another cell, the gene must be cut from the
original chromosome and implanted into the one in the recipient cell. This is accomplished by using
special chemicals called restriction enzymes. These enzymes recognize a specific sequence of
nucleotides and cutting the DNA at this specific location leaving "sticky ends.“ if the cell receiving
the gene is cut with the same enzyme, it will have the same sticky ends, and will accept the new
gene. Another enzyme called ligase, is used to glue the spliced ends together.

This technique is used to place mammalian genes into bacterial cells, so that the bacteria can
produce the material coded for by the mammalian gene. For example, if the human gene for the
production of insulin is inserted into a bacterial cell, the altered bacterium will produce insulin. As
the bacterial cell divides, the offspring will also have the ability to produce insulin.

Objective:
0 In this laboratory exercise the student will simulate how genes are removed from one cell
and inserted into another.

Materials:
0 Scissors
- Tape
0 Handout
Procedure:
1. Using your scissors, cut the PLASMID DNA into strips, cutting along the DASHED lines
separating them.
2. Using clear tape, arrange the PLASMID DNA in the proper order, forming a long chain of
DNA.
3. Attach both ends of the PLASMID DNA to make a loop.
4. Use the same procedure for the CHROMOSOME segment. DO NOT CONNECT THE
FREE ENDS.
5. For this exercise we will use the restriction enzyme known as EcoRI. EcoRl recognizes the

following sequence of DNA and cuts it accordingly:

5' ...GAATI‘C... 3' CUl'S ...G AND AA'ITC...
3' ......CI'I‘AAG 5' ...CTTAA G...

0 Find the correct hue sequence on the chromosome and cut it accordingly.
- Follow the example below as you cut the base sequence.

5' ...G A A T T C... 3‘
3' ..-C T T A A G... 5'

Find the same base sequence (GAATTC) further along on the chromosome.

Cut following the example.

Now locate the second occurrence of this base sequence (GAATTC) and cut again.

You have a portion of the chromosome containing YFG plus two ”sticky ends."

0. Find the correct base sequence in the circular plasmid of the bacterium. Cut it in the same
way you cut the previous chromosome.

‘99.“?

114

11. You can now open the plasmid. Notice that one cut of the plasmid gives you two sticky
ends!

12. Insert YFG into the cut plasmid.

13. Check to see that your base pairs match.

14. Tape in place.

Use these questions as a guide when writing your conclusion:

What is the purpose of putting a human gene in a bacterial cell?

What happens to the donor DNA when it gets into the bacterial cell?

What does the scissors represent in this process?

What does the tape represent in this model?

What would be the advantage of inserting the human gene into a human cell?

9'99””?

 

 

115

PLASMID DNA

_TTAAGTTAGCCGGAGTCAT_

_ _4
_AATTCAATCGGCCTCAGTA_

'l'l'l'l'l'l'l'l'l'l'l']

ACTGTAATATTCGTAAAGC
WTGACATTATAAGCATTTCGW

'I'I'I'I'I'l'l'l'l'l'l'l

_TCATGTAACTGGGTCCATA_

_2

_AGTACATTGACCCAGGTAT_

'I'I'I'I'I'l'l'I'l'l'l'J

:ATCGTAATATTCGTAAAGC
WATACCATTATAAGCATTTCG

m3

116

CHROMOSOME

WWCTGCGATAACCGTTCCAGT
WWGACGCTATTGGCAAGGTCA

'I'Ill'l'Ill'l'l'l'l'l'l

.AAGGAATTCTGCATTACGA"

_TTCCTTAAGACGTAATGCT_

'Ill'l'l'l'l'l'l'l'l'l']

W.CATCGACTAGGAATTCCCTm
WWGTAGCTGATCCTTAAGGGA

m3

2

117

Name Period 1 2 3 4 5 6 Date

RECOMBINANT DNA ACTIVITY TWO

Introduction:

Biotechnology has come a long way in the past ten years. Biologists are now wrrying out ideas that
they could just dream about. The idea of producing human insulin without humans, or being able to
splice a gene into a defective cell to correct a major enzymatic defect are now coming to light. The
idea that genes were untouchable has now been disproved with the idea of recombinant DNA.

Objectives:
0 Students will recognize the major restriction enzymes used in DNA technology.
0 Students will construct a sample DNA molecule and a plasmid from the materials present.
. Students will apply the Recombination process to the DNA and plasmid producing a model
of a plasmid containing the human gene.

Materials:

. Scissors

0 Clear tape

0 Cell DNA

0 Plasmid DNA
Procedures:

1. Using your scissors, cut the Cell DNA into stn'ps, cutting along the DASHED lines

separating them.

Using clear tape, arrange the DNA in the proper order, forming a long chain of DNA

Use the same procedure for the plasmid DNA Since there is no correct order you may tape

the plasmid pieces in any order you so choose.

Cut out each of the restriction enzymes and place them aside for later use.

Examine the Cell DNA. Locate the area marked with dark lines. This is the Gene for insulin.

You want to cut this area out and transplant it to the plasmid DNA. Do not randomly cut this

out. You must first locate the proper restriction enzyme that will work on both the cell DNA

and the Plasmid.

6. Use the restriction enzymes, one by one, and locate the areas that they will react with on
the cell DNA. Mark these areas with pieces of paper and tape, taging the area.

7. Now do the same to the Plasmid. Make sure you include at least one antibiotic resistant
gene and the origin of replication in your final plasmid. These will be pointed out to you at
the beginning of the lab.

8. Find the areas where the plasmid, including the origin of replication and one antibiotic
resistant gene, and Cell DNA match and cut and tape them together. You will now have a
larger plasmid holding the piece of cell DNA within it.

5"? 9"!”

Use these questions as a guide when writing your conclusion:

What is the function of a restriction enzyme?

Which restriction enzyme(s) did you use to complete your plasmid?

Is it possible that your classmates used different enzymes? Explain.

What was the purpose of making sure the origin of replication and at least one antibiotic
resistant gene was present in the final plasmid?

How might a virus be used to transmit this gene into human cells?

5" 999.”?

118

Cellular DNA

TAGCTTGCCCCGGGATCCTGG.6
_TTCGAACGGGGCCCTAGGACCh

GGAGGAATTCTTAW
CCTCCTTAAGAATW

      

5

3 4
E = YFG YOUR FAVORITE GENE (insulin)

 

WAAGCTTCC
mTTCGAAGG

_CTC.T.AAG.AATTCAGTTCGTCC_2
_GAGATTCTTAAGTCAAGCAGG_2

lilililéllililllllilllllili.

WHACCCGGATCCGTGTCCCGGGC
WTGGGCCTAGGCACAGGGCCCGm

 

m1

119

Plasmid DNA

_ATTCGGCATCCAAGCTTGAGG_
_TAAGCCGTAGGTTCGAACTCC_

“““““““
”””””””””

”””
“““““““

””””””””
“““““““

””””””””
“““““““

“““““““

IIl'I'IiIl.el?i1|l|lt'ti|||l'e—
.............

_AGAAAATETGTGTCCAGTAGG_

—_..
l||I'l'lrtr.t..EEr..II'I'III'I'I'Ih

WWGGGGTCTCAAAGAATTCCAGA

 

: 151:: antibiotic resistance gene

 

5
"331* = antibiotic resistance gene

 

 

 

 

= antibiotic resistance gene

 

35:3; = origin of replication

 

 

 

120

Restriction Enzymes

 

 

-. I r
w u n m
a
.m P m
H H F.

 

WAAGCTT CCGGGAATTT.
G_Gc

MLTCGAAW

CWLTTAAGW

............................h ...............................................

a ll
am HI
Bgl ll

WGGAC_C GGATC_C AGATCTW
MLCTGG C_CTAGG .LCTAGAW

 

 

 

 

 

121

Restriction Enzymes

 

I.. .Ia E
m m m
u S . x
u G
mGAGCT_C .WCCCGG_G H.

 

 

mc_TCGAG. .WG_GGCCC

 

122

Name Period 1 2 3 4 5 6 Date

Food Dye Electrophoresis Teacher Copy

BACKGROUND INFORMATION:

Gel Electrophoresis is a separation technique that can be used to separate particles of different
molecular weight and electric charge. The smaller, less massive the molecular weight of a
substance the faster it will move through the gel. Also, negatively charged substances will move
toward the positive (red) electrode. In this activity you learn how electrophoresis works and you
will be identifying the dye(s) present in an unknown sample of food dye by comparing them to the
dyes present in known samples.

PROBLEM:

How is gel electrophoresis used to identify the dye(s) present in an unknown food coloring
sample?

PRE-LAB PREP:

BUFFER SOLUTION PREPARATION:

10.8 g tris base

5.5 g boric acid

20 ml 0.5 M EDTA at a pH 8.0 (3.722 9 EDTA fill to a volume of 20 ml)
1000 ml distilled water

Combine ingredients in a 2 liter flask.
Store at room temperature in a sealed plastic (or glass) container.

AGROSE PREPARATION:
3 g of agrose powder
100 ml of buffer

In a 500 ml flask mix agrose powder in 100 ml of water.
Heat to boiling in microwave or on hot plate.
Mix until it is free from clumps.

Store at room temperature in a sealed glass container.

FOOD COLORING PREPARATION:

Sample 1 Sample 2 (ORANGE) Sample 3 Sample 4 (UNKNOWN)
(GREEN) (PURPLE)

40 ml of buffer 40 ml of buffer 40 ml of buffer 40 ml of buffer

10 ml of glycerine 10 ml of glycerine 10 ml of glycerine 10 ml of glycerine

10 drops of blue 10 drops of red 10 drops of blue 10 drops of blue

10 drops of yellow 10 drops of yellow 10 drops of red 10 drops Of red

10 drops of yellow
Store at room temperature in a sealed plastic (or glass) container.

123

MATERIALS:

 

 

 

 

Buffer solution gel electrophoresis box gel comb
5, 9 volt batteries 2 electrical leads 2 carbon electrodes
4 Food coloring samples 1.5 % agrose solution metric ruler
1-ml syringe 4 micropipette tips 10-ml graduated cylinder
PROCEDURE:
1) Cut two pieces of the carbon fiber tissue 42 mm x 22 mm. These will be used in step 11 as
the electrodes.
2) Obtain 10 ml of melted agrose gel and pour it into the gel box well area (see figure 1).
3) Insert the comb into the gel box. Make sure notches in the comb and the gel box are
properly aligned.
H (+)
n Well Area D
Figure 1 .
4) Allow the gel to cool.
5) Once the gel has hardened, cover the gel with 10-12 ml of running buffer.
6) Gently remove the comb from the gel. Be sure not to rip or tear the gel.
7) Place one of the yellow tips on the syringe.
8) Draw 10 pL of a known sample of food coloring.
9) Place the tip of the micropipette below the surface of the buffer but above the first well and

gently push down the plunger of the syringe and fill the well. Using figure 2, label the sample
in each of the wells.

 

 

 

 

Lone1=5ompl¢1

Lone2=$omplcz
Lone3=$omple3

 

 

 

 

 

(+)

 

Figure 2.

10) Remove the yellow micropipette tip and put a clean one on the syringe.
1 1) Repeat steps 8 -10 for samples 2 and 3 and the unknown.

124

12) Place carbon fiber electrodes in each end of the box (see figure 3).
(-) (+)

U I

_nu fl!

Fiaurc 3.

 

 

 

 

13) Attach the red alligator clip to the carbon fiber electrode farthest from the wells.
14) Attach the black alligator clip to the carbon fiber electrode closest to the wells.
15) Attach the other red alligator clip to the positive (+) terminal of the battery.

16) Attach the other black alligator clip to the negative (-) terminal of the battery.
17) Allow the gel to run for 30-40 minutes.

18) Clean up your lab station. NO SOLID MATERIALS IN THE SINK. All excess liquid material
may be washed down the drain with water.

OBSERVATIONS: Draw a color diagram of your gel.

 

(d

 

 

 

RESULTS

Lane 1 = Blue 8. Yellow
Lane 2 = Red 8. Yellow
Lane 3 = Blue 8. Red

Lane 4 = Blue, Red & Yellow

 

 

 

 

(fl

 

 

CONCLUSION:

Type a 2-page, single-spaced lab report explaining your lab. Remember to include the following
sections: Problem, Materials, Procedure, Data and Conclusion. Refer to your lab and textbook
to address the following questions in the conclusion of your lab report.

Summarize your data.

Based on the background information and your data, which of the food coloring samples
are negatively charged and why?

Based on the background information and your data, which of the food coloring samples
is the most massive and why?

Based on the background information and your data, which of the food coloring is least
massive and why?

How is gel electrophoresis used to identify an unknown sample of food coloring?

5’95”!“3‘

125

Name Period 1 2 3 4 5 6 Date

Gel Electrophoresis of DNA
BACKGROUND lNFORMATlON:

Gel Electrophoresis is a separation technique that can be used to separate DNA fragments of
different molecular weight and electric charge. The smaller, less massive DNA fragments will
move through the gel faster than the larger, massive ones. Also having a negative charge, DNA
will move toward the positive (red) electrode. The steps of electrophoresis are listed below:

1) Pour an agrose gel.

2) Cover the agrose gel with buffer.
3) Mix DNA samples with loading dye.
4) Load the gel with DNA samples.

5) Run the gel.
6) Stain the gel.
7) Read the gel.

Using the steps above, you will learn how electrophoresis works to separate ADNA fragments cut
by the restriction enzymes EcoRI and Hindlll and comparing them to the uncut ADNA.

PROBLEM:
How can the length of ADNA samples be determined by electrophoresis?

PRE-LAB PREP:

BUFFER SOLUTION PREPARATION:

10.8 g tris base

5.5 g boric acid

20 ml 0.5 M EDTA at a pH 8.0 (3.722 9 EDTA fill to a volume of 20 mi)

1000 ml distilled water

Combine ingredients in a 2 liter flask.
Store at room temperature in a sealed plastic (or glass) container.

AGROSE PREPARATION:
0.8 g of agrose powder

100 ml of buffer

in a 500 ml flask mix agrose powder in 100 ml of water.
Heat to boiling In microwave or on hot plate.
Mix until it is free from clumps.

Store at room temperature in a sealed glass container.

126

MATERIALS: (for 8 groups of 3)

200-ml TBE Buffer solution 200-ml 0.8 °/o agrose solution 8 gel boxes
8 gel combs 16 carbon electrodes 40 9-volt batteries
16 electrical leads Uncut ADNA XDNA cut with EcoRI
ADNA cut with Hindlll 8 1-ml syringe 48 yellow micropipet tips
8 10-ml graduated cylinder 200-ml Gel Stain 8 metric rulers
PROCEDURE:

1) Cut two pieces of the carbon fiber tissue 42 mm x 22 mm. These will be used in step 11 as

2)
3)

4)
5)

the electrodes.

Obtain 10 ml of melted 0.8% agrose gel and pour it into the gel box well area.

Insert the comb into the gel box. Make sure notches in the comb and the gel box are
properly aligned (see figure 1).

H M

 

 

 

 

 

 

H m I, I H

Figure 1.

 

Allow the gel to cool.
Once the gel has hardened, cover the gel with 10-12 ml of running buffer.

Loading the Gel-

1)
Z)
3)
4)

Gently remove the comb from the gel. Be sure not to rip or tear the gel.
Place one of the yellow tips on the syringe.
Draw 10 uL of the ADNA sample into the micropipet.

Place the tip of the micropipet below the surface of the buffer but above the first well and
gently push down the plunger of the syringe and fill the well (see figure 2).

 

 

 

 

 

 

 

Figure 2.

127

5)

6)
7)

Using figure 3, label the sample in the well.

 

 

 

 

 

(-)
- _ _
Lane 1 = uncut l—DNA
Lone Z = k—DNA/Ecdu
Lane 3 = A—DNA/ HI'IdIII
(+)

 

 

 

Figure 3.

Remove the yellow micropipette tip and put a clean one on the syringe.
Repeat steps 1—6 for the DNA cut with EcoRland cut Hindlll.

RUNNING THE GEL-

1)

2)

3)

4)
5)
6)

Place carbon fiber electrodes in each end of the box (see figure 4).
(-) (+)

l U l
\III: [I_

Figure 4.

Attach the md alligator clip to the carbon fiber electrode farthest from the wells. One part of
the clip on the outside of the box and the other on the inside in the buffer solution holding the
carbon fiber to the box.

Attach the black alligator clip to the carbon fiber closest to the wells. One part of the clip on
the outside of the box and the other on the inside in the buffer solution holding the carbon
fiber to the box.

Attach the other red alligator clip to the positive (+) terminal of the battery.
Attach the other black alligator clip to the negative (-) terminal of the battery.
Allow the gel to mn for 3 hours.

Clean up your lab station. NO SOLID MATERIALS lN THE SINK. All excess liquid material
may be washed down the drain with water.

 

 

 

 

 

 

128

 

OBSERVATIONS: Draw a color diagram of your gel. Measure the distance that each band
migrated from its well.

(-)

 

 

 

 

RESULTS

 

 

 

 

 

 

CONCLUSION:

Type a 2-page, single-spaced lab report explaining your lab. Remember to include the following
sections: Problem, Materials, Procedure, Data and Conclusion. Refer to your lab and textbook
to address the following questions in the conclusion of your lab report.

1)
2)
3)
4)
5)
6)

7)

Summarize your data.

Explain your results compared to that of the restriction map.

Why do you not have the same number of bands?

Name the 3 restriction enzymes we are using, their purpose and the sequence they each
recognize.

What kind of results would we have if we changed the number of batteries to utilize different
time periods?

This lab was a lot of work for you and me. Also, it is quite expensive. Did you enjoy it. . .was
it worth it? Be brief.

How can the length of ADNA samples be determined by electrophoresis?

129

 

 

 

Restriction Map of Lambda DNA

 

 

 

 

 

 

Uncut Lambda
l 1
48,502 base pairs
Lambda cut with (5le
l l l I l l l
21 ,226 4878 5643 7421 5804 3530
Lambda Cut with Hindlll
2027 125 4361
l l l l l l l I 1
23,130 2322 9416 564 6557 4361

 

130

 

 

Name Period 1 2 3 4 5 6 Data

Transformation of E. Coli

BACKGROUND INFORMATION:

The process of transformation refers to the techniques used to transplant genes from one living
source into another organism. This gene will be expressed in the organism. The procedure of
transformation is difficult in higher plants and animals but can be done simply in bacteria. The
transformation process includes the following steps:

1) Isolation of two kinds of DNA.

2) Treatment of plasmid and foreign DNA with the same restriction enzyme.
3) Mixture of foreign DNA with clipped plasmids.

4) Addition of DNA ligase.

5) Introduction of recombinant plasmid into bacterial cells.

6) Final screening for transformed cells.

Using the steps above, you will insert a gene from one bacterium into another bacteria. The
bacteria, Vibrio fischeri, found in deep-sea animals. These bacteria contain a gene that causes
the mouth area of a species of fish, Monocentn's japom'cus, to glow. You will insert this “glow-in-
the-dark” gene and another gene for resistance to the antibiotic ampicillin into E. coli. You will be
able to detect if the E. coli has taken up the “glow-in -the-dark" gene, if it grows on the petri dish
that contains agar treated with the antibiotic ampicillin.

PROBLEM:

How can DNA from one organism be introduced into another organism?
MATERIALS:

E. coli culture tube plasmid DNA 0.005 ug/ul
6 vials of calcium chloride 6 vials of sterile LB broth
12 15-ml sterile transformation tubes 38 sterile 1-ml pipettes
glass beads 20 sterile transfer loops
wire inoculating loop 26 sterile petri dishes
700 ml sterile LB agar 3.5 ml ampicillin solution
boiling water bath 12 600-ml beakers w/ ice
6 culture tube 6 glass marking pencils
roll of masking tape plastic wrap

plastic wrap Bunsen burner

42°C water bath 10% bleach solution

'E. coli culture tube, plasmid DNA, ampicillin and poured agar plates must be refrigerated for
not more than 8 weeks.
*Plasmid DNA and ampicillin can be frozen for up to 1 year.

All other material can be stored at room temperature.

131

PROCEDURE:
STREAKING A STARTER PLATE-

1) Label and date a starter petri dish on its lid. (Figure 1)

  

Name
Date
Period

 
     
 

E cal/Starter Plate

  

Figure 1.

2) Hold the wire-inoculating loop like a pencil and sterilize the loop and wire in the flame of a
Bunsen burner until red-hot. DO NOT TOUCH THE LOOP, IT IS VERY HOTl

3) With the other hand, pick up the E. coli culture tube. Remove the cap with the last finger of
the hand holding the inoculating loop. Quickly pass the mouth of the open culture tube
through the flame. BE SURE NOT TO SET THE CAP DOWN ON THE TABLETOP.

4) Stab the loop into the side of the agar in the culture tube to cool it off. Drag the loop through
the culture several times. Remove the loop, flame the culture tube, replace the cap and set

the culture tube down.

5) Lift the lid of the starter plate (Figure 2) and streak the plate as described below. DO NOT

SET THE LID ON THE TABLETOP.

 

 

 

 

6) Glide the loop in a zigzag pattern across ‘/4 of the plate (Figure 3A).

7) Close the lid, rotate the plate ‘/4 turn (Figure 3B).
8) Flame the inoculating loop.

132

   

Figrre 3A. Figure 3B.

9) Lilt cover as shown in Figure 2.

10) Cool the loop by gently stabbing it into the surface of the agar.

11) Place the loop through the first streak and glide the loop in a zigzag pattern across V. of the
plate.

12) Flame inoculating loop.

13) Repeat steps 8 — 12 two more times.

14) Flame inoculating loop.

133

DAY TWO MATERIALS:

E. coli starter plate plasmid DNA 0.005 pg/ul

1 vial of calcium chloride 1 vial of sterile LB broth 2 15-ml sterile
transformation tubes 5 glass beads 4 sterile transfer loops
7 sterile transfer pipets 2 LB agar plates

2 LB/amp agar plates 1 600-ml beaker wlcrushed ice

1 test tube rack glass marking pencil

roll of masking tape 42°C water bath

10% bleach solution

PREPARING THE E. COLI SAMPLE-

1) Mark 1 sterile 15—ml tube “+”. Mark the other sterile 15-ml tube “-“.

2) Use a sterile transfer pipet to add 250 ml of ice-cold calcium chloride to each tube (Figure
4).

 

Figure 4.

3) Place each tube on ice.

4) Using a sterile plastic inoculating loop, transfer a mass of E. coli about the diameter of a
pencil eraser from the starter plate to the tube marked “+”. DO NOT PICKUP ANY AGAR
WITH THE CELLS.

5) Immerse the loop in the calcium chloride solution and spin the loop to remove all cells.
(Figure 5)

134

6)

7)
8)

5| 0| 6‘ u

 

Figure 5.

Immediately suspend the cells by repeatedly drawing up the entire volume into a sterile
transfer pipet gently reinjecting the mixture into the tube. Repeat until there are no clumps in
the mixture.

Return the “+" sample to the ice
Repeat steps 4-7 for the “-“ sample.

Adding the Plasmid DNA-

1)

2)
3)
4)

5)

6)

Use a sterile inoculating loop to add 1 loopfull of the plasmid DNA to the sample marked “+"
(Figure 5).

Spin the loop to mix the DNA with the cells.

Return the tube to the ice and incubate for 15 minutes.

While your sample is incubating label you petri dishes. (Figure 6). Make sure to also
include your name, class and hour.

LB/amp- LB/amp+
3 4

Figure 6.

After the 15—minute incubation time, “heat shock" the cells by placing them in a 42°C water
bath for 90 seconds. Gently agitate the tubes while they are in the water bath. After the
“heat shock", return the samples to the ice bath for 1 minute.

Use a sterile transfer pipet to add 250 pl of Lauria broth to each tube (Figure 7). Gently
mix the tubes by flicking with your finger.

135

 

7)

 

 

Figure 7.

Place the tubes in the test tube rack for a 10-minute recovery.

PLATING THE TRANSFORMED BACTERIA-

1)

Z)
3)

4)

“Clam shell” (Figure 2) the lid and use a sterile transfer pipet to add 100 pl of the “-"
transformation tube to the petri dish labeled LB- 1 plate.

“Clam shell” the lid and carefully pour 4-6 glass beads onto the agar surface.

Using both a back-and-forth and up-and-down motion move the beads across the entire
agar surface.

Allow the petri dish to rest for 5 minutes so the suspension can be absorbed by the agar.

136

5) Remove the glass beads by holding the dish vertically over a beaker. “Clam shell” the lid
and gently tap the dish to roll the beads out (Figure 8).

 

 

 

 

 

 

33003;

Figure 8.

6) Repeat steps 1-5 for the “-" transformation tube on the LBIamp 3 plate.

7) “Clam shell” (Figure 2) the lid and use a sterile transfer pipet to add 100 pl of the “+"
transformation tube to the petri dish labeled LB+ 2 plate.

8) “Clam shell” the lid and carefully pour 4-6 glass beads onto the agar surface.

9) Using both a back-and-forth and up-and—down motion move the beads across the entire
agar surface.

10) Allow the petri dish to rest for 5 minutes so the suspension can be absorbed by the agar.

11) Remove the glass beads by holding the dish vertically over a beaker. “Clam shell” the lid
and gently tap the dish to roll the beads out (Figure 8).

12) Repeat steps 7-11 for the “+” transformation tube on the LBIamp 4 plate.

13) Wrap the plates together with masking tape and place them upside down in the incubator
at 37°C for 24 to 48 hours.

14) Place all contaminated materials in the bleach solution on the demonstration table.
15) Wipe down lab table with disinfectant.

137

DATAIOBSERVATIONS:

1) In the space provided below, predict the results of the transformation activity.

 

 

 

 

 

Predicted results: Predicted results:
Reason for prediction: Reason for prediction:
Observed results: Observed results:

 

 

138

 

 

 

 

 

 

Predicted results: Predicted results:
Reason for prediction: Reason for prediction:
LB/de‘ LB/amp+
"' 4
Observed results: Observed results:

 

 

2) Observe the colonies from the bottom of the petri dish. DO NOT OPEN THE PET RI DISHI

3) Record your observed results in the spaces provided on this page. If your observed results
are different from the predicted results explain what you think may have happened.

4) Count the number of individual colonies found on each plate. If the bacterial growth is too
dense to count individual colonies record “Too Dense”.

Plate Number of Colonies

LB +
LB -
LBIAMP +
LBIAMP -

 

 

139

 

 

CONCLUSION:

Type a 2-page, single-spaced lab report explaining your lab. Remember to include the following
sections: Problem, Materials, Procedure, Data and Conclusion. Refer to your lab and textbook
to address the following questions in me conclusion of your lab report.

Summarize your data and observations.

What was the purpose of plating your “+” and “-“ cells on the LB plates?
What was the purpose of plating your “+” and “-“ cells on the LBIamp plates?
What tells you that the transformation you performed has been successful?
How can DNA from one organism be introduced into another organism?

9'95”.”3"

140

 

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