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This is to certify that the

thesis entitled

Exploring Genetics and Developmental
Biology Using Multidemensional Manipulatives and
Biotechnology Laboratories

presented by

Danida Dawn Saffron

has been accepted towards fulfillment
of the requirements for

Master of Science legreein' Interdepartmental

Biological Science

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7-23-01
Date

 

0-7639 MS U is an Affirmative Action/Equal Opportunity Institution

PLACE IN RETURN BOX to remove this checkout from your record.
To AVOID FINES return on or before date due.
MAY BE RECALLED with earlier due date if requested.

 

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6/01 c:/ClRC/DateDue.p65vp. 15

EXPLORING GENETICS AND DEVELOPMENTAL BIOLOGY USING
MULTIDEIMENSIONAL MANIPULATIVES AND BIOTECHNOLOGY
LABORATORIES
By

Danida Dawn Saffron

AN ABSTRACT OF A THESIS
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
MASTER OF SCIENCE
Division of Science and Mathematics Education

2001

Professor Merle K. Heidemann

 

ABSTRACT
EXPLORING GENETICS AND DEVELOPMENTAL BIOLOGY USING
MULTIDIMENSIONAL MANIPULATIVES AND BIOTECHNOLOGY
LABORATORIES
By

Danida Dawn Saffron

This study examines the effectiveness of biotechnology laboratories and
manipulative models in increasing student understanding of genetics. The major goals
for this unit included improving students’ personal experiences by incorporating models
to translate abstract biological concepts in the text into concrete learning and conducting
biotechnology laboratories at research facilities, including the use of their own DNA.
Data sources for evaluating the effectiveness of the unit included responses to
questionnaires, written essays, pre and posttests performance in laboratory techniques,
and field observations. The practice of using these student centered modeling activities
and laboratory experiments produced significant gains relative to the use of critical
thinking skills, laboratory techniques, and general attitude in learning genetic related

topics.

 

ACKNOWLEDGMENTS

I dedicate this thesis to all my Advance Placement biology students, past and fiiture. I
thank my former students for providing me with insight into how to be a better teacher,
and I hope that these laboratories and activities will inspire future students with an
appreciation of this exciting science. I wish to express my deepest appreciation to my
husband, Chris, for his constant encouragement and patience. To Dr. Merle Heidemann,
Ken Nadler, Roger Herr, and Dan O’Mally I wish to express my gratitude for their
generous time and advice, without which, this thesis would not be possible. Finally, I’d

like to thank David Thome for his comprehensive and thoughtful editing.

iii

 

TABLE OF CONTENTS

LIST OF TABLES

LIST OF FIGURES

INTRODUCTION
STATEMENT OF PROBLEM AND RATIONALE
LITERATURE REVIEW OF PEDAGOGY
DEMOGRAPHICS OF CLASSROOM
HISTORICAL PERSPECTIVES OF TOPICS TAUGHT
LITERATURE REVIEW OF SCIENTIFIC BACKGROUND

INIPLEMENTATION OF UNIT

EVALUATION

CONCLUSION

APPENDICES

APPENDIX A: ACTIVITIES AND EXPERIMENTS

I.
II.
III.
IV.

V.

VI.
VII.
VIII.
X.

Activity 1: BREAKFAST DNA

Experiment 1: Y CHROMOSOME

Activity 2: MANIPULATING DNA REPLICATION
Activity 3: BACTERIAL TRANSFORMATION PRE-
LABORATORY

Experiment 2: BACTERIAL TRANSFORMATION
LABORATORY

Activity 4: GEL ELECTROPHORESIS PRE-LABORATORY
Experiment 3: GEL ELECTROPHORESIS LABORATORY
Experiment 4: ALL YOU HAVE ALU LABORATORY
Experiment 5: ZEBRAFISH EMBRYO LABORATORY

APPENDIX B: ASSESSMENT TOOLS AND RUBRICS

I.

II.

III.
IV.
V.

VI.

STUDENT BIOTECHNOLOGY AND DEVELOPMENTAL
BIOLOGY SURVEY

DNA STRUCTURE AND REPLICATION PRE AND POTS-
TEST

BACTERIAL TRANSFORMATION PRE AND POST-TEST
GEL ELECTROPHORESIS PRE AND POST-TEST
POLYMERASE CHAIN REACTION PRE AND POST-TEST
DEVELOPMENTAL BIOLOGY PRE AND POST-TEST

iv

vi

vii

Nde—d

20

35

48

51
58
60
66

72
82

90
111

. 121

131

132

137
139
142
145

APPENDIC C: STUDENT TEST AND SURVEY RESULTS

1. Table 1 INDIVIDUAL STUDENTS’ LABORATORY
SCORES

II. Table 2 INDIVIDUAL STUDENTS’ PRE AND POST-TEST

III. Table 3 T-TEST RESULTS FOR ALL PRE AND POST-
TESTS

IV. Table 4 AVERAGE PRE AND POST-TEST
PERCENTAGES

V. Table 5 DNA STRUCTURE AND REPLICATION PRE AND
POST-TEST

VI. Table 6 BACTERIAL TRANSFORMATION PRE AND
POST-TEST

VII. Table 7 GEL ELECTROPHORESIS PRE AND POST-TEST

VIII. Table 8 POLYMERASE CHAIN REACTION PRE AND

POST-TEST

IX. Table 9 DEVELOPMENTAL BIOLOGY PRE AND POST-
TEST

X. Table 10 STUDENTS’ GENETICS AND DEVELOPMENTAL
BIOLOGY SURVEY

BIBLIOGRAPHY

147

148
149

150

151

152

153
154

155

156

157

Table 1: Sequence of topics in unit

LIST OF TABLES

vi

20

LIST OF FIGURES

Figure 1: DNA Replication model

Figure 2: The Polymerase Chain Reaction

Figure 3: DNA Structure & Replication Test Results
Figure 4: Bacta'ial Transformation Pre & Post-test Results
Figure 5: Electrophoresis Pre & Post-test Results

Figure 6: Student work form post-test question 8

Figure 7: Second student response to post-test question 8
Figure 8: PCR Multiple Choice Pre and Post-test Results

Figure 9: Developmental Biology Pre and Post-Test Results

vii

13

17

36

38

39

41

42

43

45

INTRODUCTION

STATEMENT OF PROBLEM AND RATIONALE

Advanced Placement Biology is a relatively new course at Grand Blanc High
School. During the 1999-2000 school year I piloted a semester of the Advanced
Placement College Board curriculum and found the lecture driven unit on genetics to be
very difficult for the students, especially the biotechnology section. Student’s grapple
with the concepts of molecular genetics because the details are abstract and most students
have very little background knowledge of the subject. In addition, I was constrained by
time the first year and did not teach developmental biology, which is part of the
Advanced Placement curriculum. Therefore, my goal of this thesis was to research some
effective teaching techniques and develop molecular genetics and developmental biology
laboratories and activities. The tools and methods used to meet the objective included
pretests; class discussions; lab drawings, labels, interpretation and assessment questions;
an essay; and post-tests. In general, I chose to study the teaching of these topics because
they are socially popular, conceptually difficult, vastly growing, and consistently asked
on the AP exam.
Students can find aspects of genetics in the newspaper or magazines almost daily.
Not only engaging because of importance in modern culture, genetics has the power to
inspire. What once seemed only to be science fiction is coming within grasp as we move
closer to understanding how the human genome is sequenced, genetic diseases are
caused, animals are cloned, and organisms are genetically modified. New technology,

clever choices of model organisms, and collaboration have made molecular genetics a

necessary tool for researching almost any field related to biology, including
developmental biology.

These really are the best and most challenging times for teaching and learning
about the “new” biology. The challenge emerges as a paradox because the geyser of
information that makes molecular genetics so exhilarating also threatens to drown
students under the volume of details. In addition, the details are abstract, which makes
teaching them even more difficult. It seems like the biology texts keep getting bigger,
but the academic year remains relatively constant in comparison. “So much biology, so
little time!” (Schofield, 2000)

Moreover, the study of genetics and animal development is important for students
taking the Advanced Placement exam. These two topics comprise over ten percent of the
AP Program’s course outline. Considering the fact that questions on the Advanced
Placement Biology exam are generated from the Advanced Placement College Board

course outline, it is more than advantageous to teach them well!

LITERATURE REVIEW OF PEDAGOGY

I realized during my first year teaching AP biology that my students found the
abstract details of the molecular genetics and developmental biology section very
difficult. During the summer of 2000, through a course of study with the Division of
Math and Science Education within the College of Natural Science at Michigan State
University, I researched and implemented several teaching methods and to encourage
students to experience science by active participation. In other words, to give students the
opportunity to “do” quantitative and qualitative science, ensuring that they get the “big

picture” and major insights into the science of life.

Piaget:

The first instructional theory incorporated into the teaching unit was by Piaget.
Piaget and others have demonstrated that the sequence of topics and activities and their
degree of familiarity to the students must be taken into consideration in constructing
curriculum (Thompson, 1998). Providing activities or situations that challenge and
engage learners and require adaptation to new materials facilitates cognitive
development. This adaptation includes both assimilation and accommodation.
Assimilation involves the interpretation of events in terms of existing cognitive structure,

Whereas accommodation refers to changing the cognitive structure to make sense of the

environment (Kearsley, 200] ).

Learning Cycle:

In the Learning Cycle, students become familiar with new topics and phenomenon
through some concrete experience or lab activity (Thompson, 1998). These “hands-on”
explorations produce the familiarity needed for students to fisrther construct or
understand important abstractions. Research has documented the problems traditional
college freshmen have understanding abstract concepts (Gottfried et al. 1993). Moreover,
studies have revealed that some students are still concrete thinkers and the majority of
college sophomores still function in transitional stages that have not attained the abstract
thinking level (Malacinski & Zell, 1996). When students have the opportunity to actively
construct their own understanding through experimentation and discussion of
information, rather than rote memorization of information, they can construct meaningful

understanding. The process is more interesting to them and produces increased retention

and comprehension.

Authentic Laboratories:

Given this explosion of biotechnology information, effective teaching methods

are needed for advanced students. Students who conduct biotechnology laboratories have

opportunities to acquire the techniques that scientists use when they study molecular
biology and to improve their knowledge genetics and microbiology. Authentic
laboratories are defined as exercises that utilize materials, equipment, and technology that
are used in research, industrial and medical laboratories (Gillen and Chiappetta, 1996).
Authentic laboratories are a powerful strategy for instruction because students are
captivated by this technology and are eager to learn real world applications. Students

often ask, “What good will this information do me? Why do I have to learn this?” They

prefer authentic laboratories to dry laboratories, which might include simulations, cut-
and-paste paper activities, and computer exercises because they usually see the
connection between school science and skills they may use in a technical career (Gillen

and Chiappetta, 1996).

Processing Framework:

Lastly, The pedagogical literature research on strategies and methodologies for
teaching science shows clear evidence that students in process-approach programs learn
more than do students in traditional textbook-based programs (Ostlund and Thompson,
1998). Craik and Lockhart presented an interesting processing framework in 1972.
According to their theory, stimulus information is processed at multiple levels
simultaneously depending upon its characteristics (Kearsley, 2001). Teaching techniques
that involve strong visual images or many associations with existing knowledge will be
processed at a deeper level. Each new activity incorporates several enticing visual models
or living organisms and the sophisticated laboratories require greater processing skills.
Hopefiilly, as students internalize the experiences and process the information on genetics
and development, they will remember it. In this time of ever-expanding biology, teachers
have no choice but to make their courses less encyclopedic in content and more

CXperiential in process: “less is more” (Schofield, 2000).

DEMOGRAPHICS OF CLASSROOM

The unit discussed in this thesis for teaching genetics and developmental biology

to Advanced Placemem students was implemented at Grand Blanc High School, a
member of the Genesee County Intermediate School District in Michigan. AP biology is
intended to be equivalent to an introductory biology course found at the freshman
university level. It is a two-semester course and reflects the outline provided by the
College Board emphasizing molecules and cells, heredity, molecular genetics,
evolutionary biology, diversity of organisms, structure and function of plants, and
ecology. There was only one class of Advanced Placement biology and this was the
second year in a row it has been taught at Grand Blanc by Ron Calo, a biology teacher
and myself. We divided the course contents into two semesters.

Grand Blanc is in southeast Genesee County just a few miles south of Flint,
Michigan. The city and township have a combined population of about 35,000 people.
The individuals enrolled in this class were all senior students, with grade point averages
above 3.5, who received an A in biology and chemistry, and were recommended by their
biology teacher. Out of the 20 students in the class, there were 8 males and 12 females.
The students in the class also had diverse ethnic backgrounds. Most of the students were
motivated to enroll into AP biology because they are expecting to pursue careers in

nursing, dental, medicine, engineering, and environmental science. Moreover, many of
the students were also enrolled in several other Advanced Placement classes at the high

801100] and a few were even taking college classes at the University of Michigan in Flint

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HISTORICAL PERSPECTIVES OF TOPICS TAUGHT

One of the best books I have read as a teacher is The Science Class You “fish
You Had, by David and Arnold Brody. It clearly documents the seven greatest scientific
discoveries in history and the people who made them. Another great book I used for
references below was the Random House We ster’s Diction of Scientists. I have
chosen these two books and the American Biology Teacher journal article
“Biotechnology Outlines for Classroom Use” by Mary Jane Paolella as primary resources
for the chronology, discoveries, and quotes that led to our present understanding of
molecular biology, biotechnology and development.

Like so much else, the systematic study of living things began with the Greeks.
Aristotle (384-323 BC) wrote several biological works, which laid the foundations for
comparative anatomy, taxonomy, and embryology. A few millennia later, around 1672,
Marcelle Malphigi undertook the first microscopic studies in embryology by describing
the development of a chicken egg. Five years later Anton van Leeuwenhoek greatly
improved the microscope and used it to describe spermatozoa as well as many
microorganisms. In 1830, Johannes Muller discovered proteins. Three years latter
Anselme Payen and IF Persoz first isolated an enzyme. In his 1858 book Cellular
Pathology, Rudolf Virchow explained that “cells are the link in the great
chain. . .forrnations that form tissues, organs, systems and the individual”. In 1869,
Friedrich Mieschler discovered DNA, although he did not know it was the genetic
material. August Weismann wrote that the sperm and egg cells of animals must contain

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1886.

Later in the 1900’s, a Russian-hem chemist, Phoebus Levene identified ribose as
the sugar in RNA, and he identified certain components of DNA also. Starting in 1928,
Frederick Griffith discovered that a certain unknown substance fi'om the cells of one
strain of dead pneumococci was able to enter a different and living strain and cause the
live strain to pass the dead strain’s hereditary characteristics to the live strain’s offspring.
Avery, MacLeod, and McCarty first demonstrated the transforming role of DNA in
genetic inheritance in 1943. In 1950 Erwin Chargaff determined the proportionate
amounts of the deoxynucleotide bases in each molecule of DNA: guanine equals cytosine
and adenine equals thymine. Genetic engineering consists of a collection of methods
used to manipulate and transfer genes. The first of these techniques dates back to 1952
when Joshua Lederberg found that bacteria exchange genetic material contained in a
body he called a plasmid. In 1953 Francis Crick and James Watson suggested that
deoxyribonucleic acid was the hereditary material based on its structure. That same year
William Hayes established that plasmids were rings of DNA flee from the main DNA in
the chromosome of the bacteria. George Beadle, Edward Tatum and Joshua Leaderberg
shared a Nobel price in 1958 for the formation of “one gene one enzyme hypothesis”,
which means each gene dictates the production of one enzyme. Beadle also showed how
a bacteriophage could bring about the transfer of DNA by transduction. Working at the
Matthew Meselson and Franklin Stahl carried out their successfiil experiment with E. coli
‘0 prove the semiconservative nature of DNA replication in 1957. In the

semicontiervative model the two strands of the parental molecule separate, and each

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functions as a template for synthesis of a new complementary strand. Watson and Crick
first suggested the concept. In 1959, Arthur Kornberg won the Nobel Prize for his
discovery of the enzyme DNA polymerase, which enabled molecules of the genetic
material DNA to be synthesized.

In the next decade, 1962, Maurice Wilkens shared the Nobel Prize with Francis
Crick and James Watson for his work on the molecular structure of nucleic acids,
particularly DNA using X-ray diffraction. In 1965, researchers realized that genes
conveying antibiotic resistance in bacteria are often carried on plasmids and two years
later DNA ligase, which joins DNA segments together, was isolated. In 1969, Reiji
Okazaki, suggested that one strand of DNA is synthesized as small pieces that are later
joined together. Werner Arber, who studied viruses which infect bacteria (called
bacteriophages), took the next step. He found that bacteria resist phages by splitting the
phage DNA using enzymes. By 1968 Arber had discovered the enzymes produced by
bacteria that split DNA at specific locations. In addition, he found that different genes
— that have been split at the same location by one of the restriction enzymes, as they are
called, would recombine under certain conditions when placed together in the absence of
the enzyme. The resulting product is called recombinant DNA.

Two major events related to molecular genetics occurred in 1970: Hamilton Smith
discovered restriction endonucleases in bacteria that degrade phage DNA without
harming its own cellular DNA; and Daniel Nathans produced the first restriction map by
using restriction enzymes to cut DNA In 1972, Paul Berg used the restriction enzyme
ECORI to produce the first recombinant molecule. Berg spliced and combined into a

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virus. The following year, Stanley H Cohen and Herbert W Brown transplanted a
functioning gene between organisms and produced a hybrid with dual antibiotic
resistance using restriction enzymes and ligases. They cut a chunk out of a plasmid
found in the bacterium Eschericia coli and inserted it in the gap a gene created from a
different bacterium.

The polymerase chain reaction, which allows scientists to amplify a specific DNA
sequence, was invented by Dr. Kary Mullis in 1983. In addition, the sequence of lambda
phage DNA was published that same year. In 1985, Alec Jeffreys designed genetic
fingerprinting. The Human Genome Organization was established in 1988 in
Washington, DC, with the aim of mapping the complete sequence of human DNA. 1988
Telomere (chromosome end) sequence having implications for aging and cancer research
is identified.

In 1995, Craig Venter [The Institute for Genomic Research (TIGR)] and Hamilton
Smith (Johns Hopkins University) announced the complete sequencing of two bacterial
genomes. The 1.9-Mb Haemophilus irrfluenzae genome was finished and all gaps closed
in less than a year; firnding fi'om the DOE microbial genome project, administered by Jay
Grimes, allowed the 580-kb Mycoplasma genitalium genome to be completed in 3
months by Claire Fraser's team at TIGR. In September 1997, a team of scientists led by
Frederick Blattner (University of Wisconsin, Madison) reported completing the sequence
of the 4.6-Mb Escherichia coli K-12 genome. Obtaining the complete DNA sequence of
the E. coli genome has been a goal of the Human Genome Project, both to help
develop sequencing and gene-finding technology and to facilitate studies on gene

function and organization. More than 4200 E. coli genes have been identified, although

10

the functions of over one-third of them remain unknown. Finally, on Febuary 12, 2001 a
working draft of the human genome sequence was published in major science journals.

Information included initial analysis of the descriptions of the sequence generated by the
publicly sponsored Human Genome Project and reported by the private company, Celera

Genomics.

11

LITERATURE REVIEW OF SCIENTIFIC BACKGROUND

The following background information is needed by students to firlly understand
and appreciate the genetics and development unit. I present this information in the form
of PowerPoint lecture notes, overhead transparencies, and prelaboratories, which are
computer simulations. This information is intended to enhance the information in student
textbooks. The laboratories and activities designed in this unit can easily be inserted into
the Advanced Placement College Board curriculum, which includes information on DNA

structure, replication, assimilation, separation, amplification and vertebrate development

(Table 1).

DNA Structure:

A single strand of DNA consists of a backbone made of phosphate groups
alternating with the pentose sugar deoxyribose. Linked to the I’carbon of each sugar is
one of four nitrogenous bases. The purine bases, adenine and guanine, have two ring
structures whereas the pyrimidine bases, thymine and cytosine, have only one ring
structures. The 5’ phosphate of a deoxyribose is attached to the 3’OH of the adjacent
deoxyribose to form a phosphodiester linkage (Watson and Crick, 1953). Moreover the

5’ and 3’ ends indicate polarity.

DNA Replication:
Before replication, the parent molecule has two complementary strands of DNA.

Each base is paired by hydrogen bonding with its specific partner, adenine (A) with

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thymine (T) and guanine (G) with cytosine (C). The first step in replication is separation
of the two DNA strands. Each original strand now serves as a template that determines
the order of nucleotides along new complementary strands. Nucleotides bond at specific
sites along the template surface according to the base-pairing rules. The nucleotides are
connected to form the sugar-phosphate backbones of the new strands. Each DNA
molecule now consists of one old strand and one new strand. This is called the semi-
conservative model of DNA replication because the two strands of the parental molecule
separate, and each firnctions as a template for synthesis of a new complementary strand.
The result is two DNA molecules identical to the parent molecule. (Campbell 2000).
Figure 1: DNA Replication model

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“replication fork” (Campbell, 2000). It is formed at the junction of the single strands and
the double-stranded region. New strands are synthesized in a 5’9 3’direction near the
vicinity of the fork (Campbell, 2000). The leading strand is synthesized continuously in
the direction toward the replication fork, whereas the lagging strand is synthesized
discontinuously in a direction away from the replication fork. Both the leading and
lagging strands require an RNA primer provided by the primase because DNA
polymerase elongation can only proceed by addition to a free 3’OH end of an existing
polynucleotide strand and cannot initiate a new strand (Solomon, 2000). The new
complementary strands are synthesized by polymerase 111. On the lagging strand,
multiple initiations must occur because of the 5’9 3’ synthesis requirements. As a
consequence, short segments of DNA are synthesized, which are called Okasaki
fragments. To join these fragments the RNA primer is degraded to free nucleotides and
another DNA polymerase fills in the gaps with deooxynucleotides. DNA ligase then
catalyzes formation of the phosphoester bond between the 5’phosphate and 3 ’hydroxyl to

yield a continuous strand of complementary DNA (Atherly, 1999).

DNA Assimilation/ Bacterial Transformation:

Bacterial plasmids are a separate, smaller (than the main bacterial DNA), circular
DNA molecule that can be isolated and introduced into other cells by transformation.
Transformation involves incorporation of genetic material into a cell, thereby changing
its phenotype. The bacterial plasmid pAMP is used as a vector to transfer antibiotic
resistance genes into E. coli (Advanced Placement Biology Development Committee,

1997). Restriction endonucleases are used to insert the gene into the plasmid. Restriction

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enzymes cleave the DNA at specific sequences of nucleotides, producing cuts that are
usually jagged, with one strand of the DNA molecule extending beyond the second strand
(Solomon, 1999). Ifboth the plasmid and the foreign DNA, with the gene of interest, are
treated with the same restriction enzyme, the jagged ends (sticky ends) of the foreign
DNA will match the sticky ends of the plasmid DNA. DNA ligase can then bond the
foreign DNA fragment to the plasmid DNA.

The transformation procedure begins by treating E. coli with Ca”, which makes
them competent to take up DNA (Rapoza, 1999). Once the E.coli is competent, then a
heat-shock is given to facilitate the absorption of DNA (Advanced Placement Biology
Development Committee, 1997). Lastly, transformation is tested by treating the bacteria
with ampicillin because only the E.coli cells that absorbed the pAMP will survive
(Helms, 1994). This is because the plasmid, pAMP, carries genes for resistance to the
antibiotic ampicillin. As a control, a tube of bacteria without pAMP shows growth
cannot occur in the presence of ampicillin and that the untransformed E.coli are not

already ampicillin resistant.

Agarose Gel Electrophresis:

Gel electrophoresis separates macromolecules such as DNA, RNA, or protein
based on the rate at which they migrate through a gel under the influence of an electric
field (Atherly, 1999). First DNA is digested with different restriction endonucleases in a
separate tube to create DNA restriction fragments. These charged fragments are placed
in wells near one end of a thin slab of polymeric gel. The gel is supported by glass plates

and bathed in an aqueous solution (buffer). Electrodes are attached to both ends of the

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gel, and an electric current is applied. The DNA fragments migrate through the gel matrix
toward the electrode of opposite charge at a rate determined mostly by the molecule’s
charge and size (Campbell, 2000). For nucleic acids, the rate of migration- how far a
molecule travels while the current is on- is inversely proportional to molecular size.
Nucleic acids carry negative charges on their phosphate groups proportional to their
length. Smaller DNA fragments move through the gel more rapidly than large fragments
because the matrixes in the gel impede longer fi'agments more than it does shorter ones.
The gel is stained with ethidium bromide to create fluorescent bands when exposed to
ultraviolet light. The size of the unknown fragments can be determined by comparing

them with the migration patterns of fragments with known molecular weight.

Polymerase Chain Reaction (PCR):

The polymerase chain reaction is a method for amplifying a specific segment of
DNA. The starting material for PCR is a solution of double-stranded DNA containing
the nucleotide sequence that is targeted for copying, a heat resistant type of DNA '
POIYmerase (which catalyzes the reaction), a supply of all four nucleotides (for assembly
0f new DNA), and primers (Mullis, 1994).

The steps for PCR are as follows: 1. The double-stranded DNA to be amplified is
briefly heated to separate the two strands. 2. The DNA is then cooled to allow the
sI'Ilthetic primers to hydrogen bond to each strand. The primers define the ends of the
sequences to be amplified. 3. Two complementary strands are synthesized with DNA
”lymerase by adding nucleotides to the 3’ ends of the primers, using the longer DNA

Strands as templates (Atherly, 1999). The solution is then heated again, starting another

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over a million copies of DNA and after 30 cycles over a billion copies. The final number
of DNA copies can be calculated using 2n, where n is the number of cycles (Atherly,

1999).

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Developmental Biology
Fertilization initiates physical and molecular changes in the egg cell. As a sperm
migrates to the follicle cells to the zona pellucida it discharges hydrolytic molecules to
induce an acrosomal reaction, which allows the sperm to penetrate the zona pelucida
(Campbell, 2000). As soon as one sperm enters the egg, two reactions occur to prevent
any additional sperm from entering. In the fast block to polyspermy the egg plasma
membrane becomes depolarized preventing its fiision with additional sperm, whereas an
“fifertilized egg’s cytoplasm is negatively charged relative to the outside. Release of

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reaction causes a hardened fertilization membrane to form, which is called a slow block
to polyspermy-

Early embryonic development charts the course from a single fertilized egg to an
organism made of many differentiated cells organized into specialized tissues and organs.
It includes cleavage of the fertilized egg to form a blastula, gastrulation, and
organogenesis. Cleavage is a succession of rapid cell divisions during which the embryo
becomes partitioned into many small cells, called blastomeres, without a growth phase.
Cleavage parcels different regions of the cytoplasm that contain different cytoplasmic
components into cells and sets the stage for later development. Cleavage leads to the
formation of a solid ball of cells, the morula, and then usually a hollow ball of cells, the
blastula. The isolecithal eggs of most invertebrates and simple chordates have evenly
distributed yolk. They undergo holoblastic cleavage, which involves division of the
‘ entire egg. In the moderately telolecithal eggs of amphibians, a concentration of yolk at
the vegetal pole slows cleavage so that only a few large cells form there compared to a
large number of smaller cells at the animal pole. The highly telolecithal eggs of reptiles
and birds, with a large concentration of yolk at one end, undergo meroblastic cleavage,
which is restricted to the blastodisc (Solomon, 2000).

In gastrulation, three germ layers- the outer ectoderm, the middle mesoderrn, and
inner endoderm that lines the embryonic digestive track form. Each layer gives rise to
SPOCific structures. In the sea urchin, cells from the blastula wall invaginate and
eventually meet the opposite wall. The new cavity formed, the precursser of the digestive
tube, is the archenteron, which has an opening to the exterior, the blastopore. In

amphibians, the yolk cells obstruct invagination at the vegetal pole. Therefore, cells from

18

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the animal pole move down over the yolk cells and invaginate to form the dorsal lip of
the blastopore. In the bird, invagination occurs at the primitive streak, and no
archenteron forms.

In terrestrial vertebrates, the primary germ layers give rise to four extraembryonic
membranes. The chorion surrounds the embryo and the other membranes. The amnion
encloses the embryo in a fluid-filled amniotic cavity, which keeps it moist and acts as a
shock absorber. The yolk-sac membrane, enclosing a small fluid-filled cavity, is the site
of early formation of blood cells. The allantois forms blood vessels that connect the
embryo with the placenta through the umbilical cord. These membranes in humans are
homologous to those of reptiles and birds.

In organogenesis, the organs of the animal body develop from the three
embryonic germ layers. One of the earliest events of organogenesis is the formation of
the nervous system by induction of the dorsal mesoderm along the roof of the archenteron
(Campbell, 2000). In addition, ectoderm, which is above the notochord, forms a neural
tube from which will develop the central nervous system. Mesoderrn along the sides of
the notochord condenses into somites, which give rise to the vertebrae and skeletal

muscles. The brain and spinal cord develop from the neural tube.

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IMPLEMENTATION OF UNIT

One main goal of the genetics unit was to teach molecular genetics and

developmental biology through a variety of procedures used in major research

laboratories. Pre-laboratory discussions, problem-solving activities, simulations,

demonstration of proper techniques, and post-laboratory discussions are essential

instructional elements to ensure that students have successful experiences in genetics.

My aim was to enhance students’ excitement in genetics by alternating the information

and activities of biology, so students connect concepts discussed in lecture and

discovered in the laboratory. A series of manipulative models, current biotechnology

laboratories and developmental model organisms were researched during the summer of

2000 at MSU. Taken together, students learn about microbiology, molecular biology,

genetic engineering, and embryology.

Table 1: Sequence of topics in unit

 

Topics Covered In Lecture

Corresponding Activities

 

DNA Structure

Breakfast DNA
Y Chromosome Demonstration

 

DNA Replication

Manipulating DNA Replication

 

Bacterial Transformation

Bacterial Transformation Prelaboratory
Bacterial Transformation Laboratory

 

Gel Electrophoresis

Gel Electrophoresis Prelaboratory
Gel Electrophoresis Laboratory

 

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20

 

DNA Structure:

1 restructured my lecture—dominated unit covering CH 16: The Molecular Basis of
Inheritance in the students’ Campbell textbook by adding food and craft materials. The
unit begins with a description of the structure of DNA, which some students find difficult
because of the chemistry involved. To help illustrate the structure of DNA, 1 modified an
activity out of the American Biology Teacher journal (Byrd, 2000). Students are given
cereal and pretzels to build unique models of DNA, while learning all the details and
background history in “Breakfast DNA” (Appendix M.)

First students identify their bag of goodies and are given their corresponding
representation: colored Fruit Loops are nitrogen bases, Golden Grahams are deoxyribose,

and pretzels are phosphates. Students are shown how to construct a simple nucleotide by
connecting certain types of foods together. From there, the details of DNA structure are
presented with the aid of an overhead transparency. Students then proceed to make a
polynucleotide chain in the 5’ to 3’ direction, which is reinforced with the activity for
DNA replication. At this point we discuss the double stranded nature of DNA and base
pairing. Students are asked to create their own lS-nucleotide sequence. They are told
that they can now use just the Fruit Loops (nitrogen bases to represent the
deoxynucleotide). Once they record the sequence on their paper they are asked to have
their neighbor bond the corresponding nucleotide using base pair rules. After they check
their neighbor’s double strand they are asked to illustrate the DNA replication process of
their own DNA sequence.
In addition to student questions and personal observations, the written assessment

of this activity was their illustration of DNA replication. By working collaboratively

21

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very few of my students even needed my help on their drawings. Actually, my best
assessment came fiom student’s comments during the activity. One students recognized
there were broken or discolored Fruit Loops and said, “ they could be mutations”. The
same student also noticed the orange Fruit Loops were not being used and asked if they
could eat them. They were asked if the orange Fruit Loops could represent any other
bases in different nucleic acids. Once they remembered RNA has uracil in place of
thymine, they were allowed to eat the orange ones. The most helpfirl comment though,
came fiom one of my female students who recormnended that I use Honeycomb cereal
for the deoxyribose next time because it had six sides for the 5 carbons and 1 oxygen,
which I’m going to do next year. I was also pleased when one of my students in this
class who returns two periods later as a lab assistant brought her left over baggy of snacks
and showed my other lab assistant, not in AP biology, how to make DNA and replicate it
using her food.

My rationale for this activity and many of the following ones is the extensive
research to create a more “hands—on” interactive environment when teaching genetics in
the classroom ( Byrd, 2000). Simply stated, students remember more information when
they are actively participating in the lesson. I’ve had wonderful experience in the past
using food as a tool to teach the structure of cells and the chemical bond formation during
photosynthesis. In this activity, background information can be accomplished without
loss of student attention because they have something to touch, taste, see, and if they
want, smell. Food always gets students attention because they know they will probably
get to eat it at some point. Moreover, this activity contributes to the scientific principles

taught because students use the different colors and shapes to visualize the patterns that

22

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are recognized at the molecular level by the DNA binding proteins, which are covered
later (Byrd, 2000).

To make the structure of DNA even more personal, I asked for some students to
volunteer to prove their manhood (only the presence of their Y chromosome). I
researched and collaborated with Dan O’Mally of the MSU Cytogenetics Laboratory and
came across a wonderfiil fluorescent stain called Quinacrine mustard. This laboratory
was done only as a quick demonstration due to a time constraint placed by the other labs
that they were performing (Appendix AII). Although the procedure is very simple, the
technology needed to view the chromosome is very expensive. Flourescent microscopes
cost tens of thousands of dollars and require extensive training to focus. Fortunately,
students were able to view the pre-focussed chromosomes fi'om the computer monitor
and microscope. In the firture I hope to have more students participate in this laboratory,

especially some girls to use as a control.

DNA Replication:

This year I introduced DNA replication with a ziplock baggy firll of colorful pipe
cleaners, stickers, and an assortment of beads in an activity called “Manipulating DNA
Replication” (Appendix AIII). This way, students could build the model as I explained
the process of DNA replication. In summary, two pipe cleaners are used to represent
parent DNA molecule. A red pipe cleaner for one of the original template strands is
twisted around a green one for the other template strand. Each of the four ends are
labeled with the antiparallel 5’PO4- and 3’OH. Students thread a purple pony bead to

represent DNA helicase and correspondingly unwind the two strands. I use word

23

association phrases like “Helicase hacks”the hydrogen bonds between the double helix to
make two strands” to help students remember the function of the enzymes involved.
When I teach them about lagging strand synthesis I use the phase “I_’olymerase pastes”
nucleotides alongside the template. Students learn about the requirements of polymerase
III ( pink pony head) by adding a small RNA primer (white pipe cleaner) at the free 3’
OH group and an additional shiny red pipe cleaner (leading strand) is wound the same
template strand continuously towards the replication fork (helicase bead). Lagging strand
synthesis is modeled by discontinuously priming and adding smaller green pipe cleaners
(Okazaki fragments) in the opposite direction of the replication fork, which follows the
overall rule of 5’-) 3’ synthesis. Lastly, a lime bead (symbolizing DNA ligase) “links”
the fragments together.

I extended the DNA replication model by showing students a picture of leading
and lagging strand synthesis occurring simultaneously. Afterwards, I had them loop their
pipe cleaners and trace the direction of synthesis using their 5’ and 3’ labeled ends. To
assess their short-term understanding and recall I had them discus in small groups a few
written questions from a study guide. As reinforcement the following day, I incorporated
some animation from the Biology Place website into their class discussion. This paid
subscription to the biology.com web site is a powerful new tool that facilitates the
learning process. Students can view animations to facilitate learning about difficult
scientific processes such as DNA structure, replication, etc (Bodzin, No date). I was also
able to assess their understanding of the process by watching students use the DNA
replication simulations. Most students were able to correctly build an identical DNA

molecule fiom all the “materials” provided on their screen.

24

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DNA Assimilation / Bacterial Transformation:

The basic lesson plan of this section required about 4 days of instruction. First, I
started with a lecture and discussion on human use for DNA transformation technology;
the Griffith experiment, which showed that rough-nonvirulent Streptococcus pneumoniae
became virulent (and smooth) when mixed with heat-killed virulent bacteria; and the
difference between natural and artificial transformation. Next, students went to the
computer lab to begin their pre-laboratory assignment called “Bacterial Transformation
Prelaboratory” using the Biology Place website (Appendix AIV). The website
http://wwwbiologycom has all 12 of the required Advanced Placement Laboratories on
their “lab-bench”. It gives students a good summary of key concepts, lab technique hints,
and experimental design. Next, students proceeded into the authentic lab “ Experiment 1:
Bacterial Transformation”, which relates to the core of pharmaceutical research: to be
one step ahead of the ever-evolving pathogens that infect us daily. (Appendix AV).

This microbiology laboratory invites students to use antibiotic-resistant plasmids
to transform Escherichia coli. Equipped with sterile pipettes, students transfer ampicillin
sensitive E. coli cells to cold CaClz solution. Then pAMP is added only to the
experimental cells. Both control and experimental bacteria are heat shocked at which
time some of the competent cells take up the pAMP and are transformed. Students then
spread the treated bacterial cells on LB agar plates, some with ampicillin and some
without. They wrap their plates, incubate them overnight, and wash their hands. The
next day they record the number of bacterial colonies on both control and experimental
plates. Students compare and contrast the number of colonies on certain pairs of plates to

determine that the cells are viable, the antibiotic is active, only transformed cells grow in

25

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the presence of ampicillin, and transformation is a very rare event. Moreover, they
determine the transformation efficiency of the experiment and describe what factors
might influence it.

Overall the lab was successful with only one group not getting growth on a plate.
Students also did very well on the laboratory analysis questions. The class average was
17.8 out of 19 points (Appendix CI). Explanation of discrepancy from ideal data was
discussed and recorded as probably due to incorrect temperatures in the beakers of water.
In the fixture I will use water baths instead of hot plates and beakers. Some students had a
few points deducted from their lab for not including units in their transformation
efficiency. A few points were also deducted for omitting an explanation of the effect of
factors that might influence transformation efficiency. One group didn’t convey that they
understood one of the results of comparing LB/ Amp- and LB- plates were to show that
the antibiotic was active. However, they did get credit for saying, “normal E.coli can’t
grow on ampicillin”

Although sterile technique and safety issues concerning live pathogens was
discussed before the lab, one student returned to class after school very concerned that
she had ingested some antibiotic resistant E.coli. She said that she accidentally stuck the
same pen she used to count the colonies in her mouth out of habit. She worriedly asked if
she was going to be all right, especially after another student in the class had been out for
a week with food poisoning. After being a little panicked I asked her if she actually
touched her pen to the bacteria. Luckily, she followed the procedure and only contacted

the glass petri dish with her pen, so I didn’t need to call an ambulance!

26

 

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Agarose Gel Electrophoresis:

A short dry laboratory called “DNA Scissors: Introduction to Restriction
Enzymes” was used to begin the next experiment. In this paper/scissors activity, students
are introduced to restriction enzymes and simulate the activity of restriction enzymes
with scissors. They are also introduced to restriction maps and asked to make
predictions. After a brief discussion of “sticky ends”, students are given time to begin the
interactive pre-laboratory for the gel electrophoresis lab using Biologycom (Appendix
AVI). In addition, students are involved with demonstrations of biotechnology
techniques, such as loading gels with a micropipette. The demonstration is made visible
to the class by gathering everyone to the front of the room and modeling the technique
with a few students. Then each group of students are given a sample gel and colored dye
kit to practice before using the expensive biotechnology kits.

The “authentic” lab is done as the first part of an all day field trip to Kedzie Hall
on the campus of Michigan State University with special accommodations made by Dr.
Merle Heidemann. Electrophoresis is the laboratory technique used to separate DNA
fragments based on size. Its basis lies in the fact that DNA’s negative charge when
suspended in a conducting medium will migrate in an electric field. During Experiment
3: Gel Electrophoresis Laboratory, students prepare a sample of undigested DNA for use
as a control (Appendix AVII). In addition they prepare a sample of phage lambda DNA
digested with EcoRI endonuclease. Students prepare a third sample of DNA digested
with a standard restriction enzyme (HindIII endonuclease). The fragment sizes produced
by Hind III are known and provide a standard against which fiagment sizes produced by

Ele endonuclease can be compared. Lab groups load each of their three DNA samples

27

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into individual wells. A tracking dye is added to the sample to show the leading edge of
DNA migration. Students separate the sample fragments using gel electrophoresis. The
electrophoresis apparatus is turned off when the tracking dye nears the end of the gel to
prevent the running off of the actual DNA. The gels are immersed in a dye that allows
the fiagments to be observed. In this case Carolina Blue is used and easily destained with
water. Using semi-log graph paper, students graph a standard curve using the observed
migration distances and known fiagment sizes for the HindIII standard endonuclease.
Since migration distance is inversely proportional to fragment size, plotting migration
distance against the log of the number of base pairs produces a straight line. Using the
best-fit line, students can then interpolate the size of each 13le fiagment produced.
Overall, students seemed to understand the main objectives of the lab with an
average score of 16.2 out of 19 (Appendix CI). In addition, most students’ gels were
readable. There were a few misconceptions found in their lab analysis. A few students
drew the digested DNA fi'agments in different wells after separation by electrophoresis,
instead of bands in the same lane. The same students described the affect of voltage in
electrophoresis as “ if greater, the length of fragments would appear to be more, if less,
the length of fragments would appear to be less”. For the amount of DNA used they said
“ more bands would show up and possibly confuse the data/ readings if more DNA were
used”. When the labs were returned we discussed how increased voltage will increase
the rate of migration and increased amounts of DNA will increase the intensities of the
fragments after staining. For the most part, the class understood the principles of
electrophoresis and restriction analysis to answer the questions on the laboratory and

correct their peers’ misunderstandings.

28

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The rationale for conducting this experiment was to solidifiy students’
understanding of these topics by integrating abstract concepts into a concrete experience.
Using newly acquired genetic knowledge in a real world context helps students
understand the genetic concepts and provides motivation for their learning (Munn, 1999).
Many of the techniques that are fundamental to genome research and biotechnology, such
as restriction fragment analysis and DNA sequencing can be canied out in a high school
classroom. The complexity of the lab does not fit into the high school class periods and
the benefits of taking a group to a university are enormous. Ironically, when students
were asked to comment on the laboratory one wrote, “ we’re missing school”.
Fortunately, another student wrote “ I learned more about research at MSU and got a

chance to use and look at a lot of expensive research tools”.

Polymerase Chain Reaction (PCR):

As with the Gel Electrophoresis Lab, the third authentic laboratory called “All
You Have ALU” was performed at Michigan State University (Appendix AVIII). This
lab could not have been accomplished with out the generous use of equipment and
materials by John Gerlach and collaboration with Roger Herr of the Medical Technology
Department. During my research time at MSU, I spent several days learning the basic
theory and adapting them for high school student use. Roger Herr also gave me an
opportunity to practice the lab procedure on my peers using the thermo-cycler, lkb DNA
Ladder standard marker, gel electrophoresis apparatus, ethidium bromide, and digital
florcscent camera. His help has made this lab one of my best learning and teaching

CXperiences.

29

 

 

 

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To introduce students to the laboratory I had them visit the Cold Spring Harbor
DNA learning center at http://vector.camorg/Shochhole.html. They were
asked to read the introduction on polymerase chain reaction and view the animated
amplification process of PCR. 1 reviewed the information on polymerase chain reaction
with a few PowerPoint slides and had them color code and label the steps of PCR.
Students were given the laboratory exercises to read on the hour-long bus ride from
Grand Blanc to East Lansing. During that time I migrated from group to group on the
bus to discuss the labs and ask if there were any questions. I also delegated specific tasks
to different lab groups.

In summary, the All You Have Alu lab uses the polymerase chain reaction to
amplify a short nucleotide sequence fiom human chromosome 16. The object is to create
a personal DNA fingerprint that shows the presence or absence of the Alu transposable
DNA sequence on each chromosome at the PV92 locus. The source of DNA for this
procedure is a sample of squamous cells obtained from student cheek cells. The students
collect their cells using saline mouthwash, separate them by centrifugation, and
resuspend the cells in Chelex. Students boil the samples to lyse the cells and liberate the
chromosomal DNA The Chelex binds metal ions, released by the cells that inhibit the
PCR reaction. Each student combines their DNA sample in clear supernatant with a
buffered solution of heat-stable Taq polymerase, oligonucleotide primers, the four
deoxynucleotides, and the cofactor magnesium chloride.

I demonstrated the correct use of the three different micropipettes and organized

separate groups to use the microcentrifuge because only 8-10 samples could be run at a

time. In the future, I would borrow more rnicrocentrifuges to reduce the 40-minute wait

30

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time between the first and last group. Fortunately, those students who were done first
could check on the gel electrophoresis lab and clean up. We did have a quick show of the
effects of an unbalanced centrifuge, but no harm was done. Considering the number of
steps in the procedure, rotation of the steps between groups, variety and of new lab
equipment and materials, and keeping an eye on their other experiment, I think this group
did a wonderful job.

Roger Herr generously demonstrated the next part of the lab procedure. He placed
each student’s labeled PCR mixture in the DNA thermal-cycler and programmed the
machine to run 30 cycles. Each cycle consists of 30 seconds at 94°C to separate the
DNA strands, 30 seconds at 58 °C to bind the primers and 30 seconds at 72°C to extend
the deoxynucleotides. Moreover, he loaded the amplified products into the agarose gel
because of the toxicity of Ethidium bromide stain.

The only assessment in this lab was the effort and technique used during the
experiment and the written group analysis of the photographed results. 1 divided the class
into two groups by the coded numbers on their samples. Each group was to determine
the resulting DNA fingerprints on their photo and share it with the class. The resulting
bands of this lab were crystal clear compared to the Carolina Blue stained gels in the Gel
Electrophoresis lab. 15 of the 18 samples could be easily analyzed. Unfortunately, three
student samples were impossible to decipher. One student pippetted her sample
incorrectly and couldn’t retrieve enough of it to show on the gel. The other two students’
samples were lost in the boiling water baths because they did not close their safety locks.
We heard a couple sample tubes pop loudly, but those two tipped over in the water before

we could save them. Of the 15 readable samples the class determined 11 students had a

31

 

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clear band which appeared to be 550 base pairs, which is homozygous negative for the
ALU allele; 3 samples had a band appear at 850 base pairs, which is homozygous
positive for both ALU alleles; and 1 student had a band at both the 550 bp and 850 bp,
which is heterozygous for the ALU alleles.

One of my rationales for doing this laboratory was because “significant learning takes
place when the subject matter is relevant to the personal interests of the student”
(Kearsley, 2001). In this lab students were analyzing their own DNA from their own
cheek cells. Moreover, the rapid accumulation of genetic information generated by the
Human Genome Project and related research has greatly increased student attention of
genetic issues. By collaborating with real scientists in real research labs, my students
have the opportunity to share the excitement of the field of research and hopefully be
inspired to pursue some scientific research themselves or at least make educated
decisions related to genetic public policy. In addition,

“the involvement of scientists benefits the science community directly by
helping to produce a more scientifically literate population, which aflects
how the public perceives and accepts scientific endeavors. It also helps
the university because the political and financial support, the students, the
firture jobs of many of the students-and ultimately, even our research
problems-come from this world not from within the university ” (Munn,
1999) .
Last but not least, our school district career technology coordinator, Gary Towers, agreed

to support the field trip next year with technology education funds.
Developmental Biology

As a model organism for vertebrate development, I chose the zebrafish (Danio

rerio) for students to observe during the “Zebrafish Laboratory” (Appendix AIX). These

32

tropical freshwater fish are easy to maintain, prolific breeders, and have completely
transparent embryos. Because the embryo is transparent, students can follow the
development during cleavage, blastulation, gastrulation, neuralation, and can clearly see
the beating heart. Students helped collect embryos from the bottom of the tank by using
a rubber siphon. Then they carefully pipetted out the eggs fi'om all the debris into a
beaker full of tank water. Individual lab groups transferred several of the eggs into their
own petri dish and then to a dissecting microscope. Students were to observe each egg
and try to determine the stage of development. To take a closer look at the embryos
students were shown how to make a coverslip bridge and view their embryos under a
light microscope. In addition, they needed to sketch their embryos and record the times
of their observations. Later, students needed to describe in their laboratory handout the
features of each type of embryo they observed and predict the age by comparing them
with their textbook and other references. Groups of students were very generous in
sharing their zebrafish at different stages of development with other groups. In this way,
most of the stages were identified. The later stages of development were by far the most
interesting to the students. They were just fascinated by watching the little embryos flop
their tails and bodies around in the egg. Some students watched the zebrafish embryo
hearts beating for about ten minutes. Even the principal of our school was amazed as he

peered into the microscope at a living and developing embryo.
Assessment Tools

The assessment tools and methods used for this unit included a pretest; class

discussions; the lab drawings, labels, interpretation, and assessment questions; an essay,

33

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and posttest. All assessment tools can be found in Appendix B. The pretest and post-tests
were the same except for the DNA structure and replication tests. All tests were multiple
choice, except the gel electrophoresis and PCR tests, which had some short answer
questions. The AP Biology essay question asked students to name and describe the
origin, function, and mechanism of operation of the four extraembryonic membranes of a
bird and to briefly describe one variation between these membranes in a bird and those in
a human in either development or function. Students’ essays were graded by the AP

College Board 10-point rubric for this topic and were given an extra point for

illustrations.

In addition to my assessments of the students, I asked students to evaluate the unit
by completing a survey (Appendix BI). The survey asked them to rate the laboratory
experiences, classroom activities, and the field trip. For example, students were asked to
rate the classroom simulations of laboratory techniques in terms of helping them
understand biotechnology concepts taught in this course. Evaluation was based on a

qualitative scale ranging from excellent to unsatisfactory.

34

.
~Qav.

“I‘ll.

El”
‘ t<uv

a; “.e
H St.

EVALUATION

Although student understanding of this unit was facilitated by essays, class
discussions, laboratory exercises, and manipulative models, the evaluation process was
facilitated primarily by the pre and post-tests. The six main topics evaluated include DNA
structure, DNA replication, bacterial transformation, gel electrophoresis, polymerase
chain reaction, and developmental biology. Each of the students was given a varying
number of multiple choice questions before and alter the unit. In addition, both the gel
electrophoresis and PCR portion of the evaluation process included several short answer
pre and posttest responses. The test questions were targeted to weak areas discovered
during the first year’s AP Biology students or topics that had not been included due to a
lack of time. These pretest and post-test questions can be found in Appendix B. The
questions asked were chosen based on topics that had not been included during the prior

year or had previously shown low student comprehension.

The null hypothesis is that there will be no difference between the pre and post-
test scores of the students. The individual student test results are shown in Appendix CI].
The null hypothesis was rejected at a significance level below 0.001 using the t-test for
all pre and post-test data. Stated another way, there is a 99% probability that the

observed test gains are due to something besides chance. The t-test probabilities and
inverses of the t-distributions are located in Appendix CIII. In addition, the class’s
average pretest percentage for the unit was 16.9%. After the lessons were taught with the

new teaching activities, the average students score on the same questions rose to 77.2%.

35

These statistically significant results are included in Appendix cm. In general, this
year’s student’s scores are better in comparison to the first year’s scores.

The first subject specific set of evaluation data is fiom the DNA structure pre and
post-tests results from questions 1-4 on Table 5 (Appendix CV). The objectives analyzed
on these tests included students being able to identify the three components of a
nucleotide, distinguish between the nitrogen bases, describe base pair rules explain what
kind of chemical bonds connect the nucleotides of each strand, and explain what type of
bond holds the two strands together. The pre and post-tests were identical with a few
exceptions. One difference in the post-test included questions where students had to label
the correct nitrogen base to its complementary base pair. These questions only had the
purine and pyrimidine shapes with the number of hydrogen bonds (questionslB-l 6). See
Appendix BH for the complete tests. The average number of questions students answered
correctly on the

pretest was 0.8 out of

Figure 3: DNA Structure & Replication Test Results
a possible 4

 

compared to the 3 .5

out of 4 on the

 

posttest (Appendix

 

 

C11). Over half the

class received perfect

Number of 88 W! correct

 

 

scores on the post-

Question #

 

 

test. The greatest

 

improvement was seen on question number 1 (Figure 3). Question I asked students to

36

idtmify a then
nucleondc No
correctly ansm
identify a 5‘ ca
Elm though all

The obi
on Table 5. foc
mMMM:
replication (A?
W651 by incre
l-hpendix C1\
the We of repl
mm? on the
Wtonsen 3i
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fame fr0m mm

Milken durir

unable to dm
311de did no!
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identify a chemical group that, together with a sugar and a nitrogen base, makes up a
nucleotide. No students had the correct answer on the pretest, whereas 18 students
correctly answered phosphate group on their posttest. Almost half the class were able to
identify a 5’ carbon of deoxyribose on their pretest which contributed to the lowest gain,
even though all but one student got it on the posttest.

The objectives targeted on the DNA replication portion of the tests, questions 5-7
on Table 5, focussed on students being able to describe the process of DNA replication, -
explain the role of enzymes, and explain the Meselson-Stal semiconservative model of
replication (Appendix BII). Students showed an improvement in their performance on the
pretest by increasing from an average of 25% to an average of 80% on the post-test
(Appendix CIV). The greatest improvement was observed on students understanding of
the type of replication DNA undergoes. Only 0.5% of the students answered question 6
correctly on the pretest, whereas 95 % of the students recalled that DNA replication was
semi-conservative on the posttest (Figure 3). The data on the pretest showed that half of

the students, who had gotten this question wrong, had the misconception that DNA
replication was conservative. The smallest difference between pre and posttest scores
came from number 5, which asked students to identify (on a figure) what is most likely to
be broken during replication (Figure 3). 12 students on the pretest and 16 on the posttest
were able to determine that the dotted lines between the bases were the bonds broken, but
Students did not recognize them to be hydrogen bonds (question 2 in Appendix BII).
Overall, half the class received perfect scores on the DNA replication portion of their
Posttest that matched the pretest. In addition to the questions asked on the pretest,

several other questions regarding DNA replication were assessed on the posttest. These

37

included students’ ability to distinguish the processes that occur during replication, the
composition of primers needed to initiate DNA replication, and which enzymes are
necessary for DNA replication.

The pre and posttest questions for the Transformation unit were the same for both
tests. These test scores were used to assess student’s understanding of natural genetic
recombination in bacteria, the composition of a bacteriophage, the structure of a bacterial
chromosome, the process of transformation, types of vectors used in recombinant DNA
technology, the structure of a plasmid and characteristics of plasmids. The whole class
improved their performance on the eight pretest questions by an average of five points;
each question was worth one point (Appendix CII). Eight students out of 20 received a
perfect score on the transformation posttest (Appendix CH). The greatest amount of

improvement was on question number 5, which every student answered correctly on the
posttest (Appendix CVI). This question asks students to understand that DNA,
ribosomes, cell membranes, and enzymes are all found in prokaryotic cells, whereas

nuclear envelopes are not (Appendix 8111). The question with the smallest difference

 

 

 

 

 

 

from pre to post test

was question 11:31:34: Bacterial Transformation Pre & Post-test
number 7, which 25

asks students to E b 20 -

u g 15 _ l Protest
understand that g g 10 q l Postest
plasmids can be '3 5 ~

0 a
MasaDNA 12345678
, Question #
”0“” (Figure 4).

 

 

 

38

This difference is partially due to the fact that students scored the highest on this question
during the pretest. Oddly, the three students who got this question wrong all said that a
plasmid is a type of bacteriophage.

The objectives of the Electrophoresis portion of the unit focussed on students
being able to describe how restriction enzymes and gel electrophoresis are used to isolate
DNA fragments, and interpret results from gel electrophoresis semi-log graphs. The first
type of data collected was in the form of students responses to multiple choice questions,
whereas the second type of data collected was from student’s responses to short answer
questions (Appendix BIV). On the multiple choice tests, Students showed an
improvement in their performance on the three question pretest by increasing from an
average of 37.5% to an average of 90% on the posttest (Appendix CIV). The greatest
improvement was observed on students understanding of the fundamentals of nucleic
acid separation of gel electrophoresis. Only 5 of the students answered question 6
correctly on the pretest, whereas 19 of the students remembered that nucleic acids are

separated on the basrs Of differences Figure 5: Electrophoresis Pre & Post-test

in their size (Appendix CVII). This Resu'“

 

concept was emphasized repeatedly

on the pre laboratory simulation,

‘d—I—h—h

ONbO’OON#O’Q8

answer

class discussions, and the laboratory

exercise. The smallest difference

I of Students wI correct
41* LIL. #4 4.1 . 4,414 L ,4

between pre and posttest scores

 

Question 8

 

 

came from question 3 (Figure 5).

 

This Question asked students to use the semi-log graph results from a gel electrophoresis

39

procedure to approximate how far a DNA fiagment with 3000 base pairs travels on a gel.
9 students on the pretest and 17 on the posttest were able to interpolate the data and
determine the DNA fiagment migrated 2.6cm (Appendix BIV).

The second half of the Electrophoresis pre and post test included two short answer
questions that were taken from Advanced Placement College Board essays on previously
released tests . Each question was, as objectively as possible, evaluated using the
assessment rubric in Appendix BIV. Some samples of students’ responses are included
in this evaluation. All names of the test subjects have been changed to protect their
privacy, as was stated on the parental and student consent forms approved by the
University Committee on Research Involving Human Subjects.

Question 7 asks students to “Explain how the principles of gel electrophoresis
allow for the separation of DNA fragments” (Appendix BIV). Overall, only one student
received any points for her responses to this question, but she was also allowed to take it

home because she was not present when the rest of the class took the pretest (Appendix .,
CII). Most students just left the question blank, few students responded with “l don’t
know”, and only one student said, “small monkeys run with it”. As can be inferred by
these results, students have had no prior experience with gel electrophoresis. On the
post-test another student answered the question by writing, “Gel electrophoresis separates
DNA fragments through the following process: DNA has a negative charge and
Consequently will travel towards the positive electrode in an electrophoresis through the
gel. Because of basic laws of particle physics and common sense, it follows that the
shorter fragments will travel more quickly than the larger fragments. If this process is

Stepped before all the segments of DNA reach the end of the gel, one can see the various

40

strands of DNA in their progression to the (+) end and determine the relative lengths of
these strands, thus separating them based on their lengths.” In accordance with the
grading rubric (Appendix BIV), this response received 3 points compared to the 0 points
he earned on the pretest (Appendix CII). Some of the students included the need for there
to be a current to carry the negatively charged DNA, but no students mentioned the need
for a buffer to carry the current across the gel. Moreover, students overlooked the
necessity of a stain to view the DNA bands in the post-test.

Question 8 asks the students to “Describe the results you would expect from
electrophoretic separation of fragments from the following treatments of the DNA
segment above (see Appendix BIV). Assume that the digestion occurred under
appropriate conditions and went to completion. 1. DNA digested with only enzyme X.
11. DNA digested with enzyme X and enzyme Y combined. All students scored a 0 on

the pretest, even the student who took it home.

Figure 6: Student work form post-test question 8

 

 

 

 

 

 

 

 

41

A senior female, answered question 8 part I by writing, “ I would expect there to be 4
different fragment lengths. The shortest 400, then 1300, then 1500, and then 1700.” On
question 8 part 111, she responded by “ I would expect there to be 5 different fragment
lengths. The shortest 400, then 500, then 1200, the 1300, then 1500 (longest)”. The
student also included the following illustrations (Figure 4) for better explanation, as was
requested.

Although most students could describe the DNA digestion and resulting sequence
of bands on the gel, a couple students drew misleading illustrations (Figure 7). Notice
that the DNA fragments are in different lanes, which gives the misconception that the
DNA was loaded into separate wells instead of the same. This response and the same for
part II, earned this student and a few other students 2 points out of the possible 4. Table

2 shows that the overall improvement in the scores was 3.1 points out of 7 (Appendix

CII).

Figure 7: Second student response to post-test question 8

 

E fingmm-rs :® 4OOL,@®O hp

 

 

+

~
daDQU
C.

JS__

 

 

42

The Polymerase Chain Reaction pre and posttests were used to assess student’s
understanding of the following DNA Technology concepts: how restriction fragment
length polymorphism analysis and PCR can be applied; how the creation of sticky ends
by PCR restriction enzymes is usefirl in producing recombinant DNA; and the purpose of
PCR, and the steps and materials involved. Again, the first type of data collected was in
the form of students’ responses to multiple choice questions and the second type of data
collected was fi'om student’s responses to short answer questions. On the multiple choice
test questions, the average score on the pretest was 1.1 of the possible 5 points compared
to the 4.6 out of 5 on the posttest (Appendix CII). There were 12 students who received a
perfect score on the posttest (Appendix C11). The greatest amount of improvement was
on question number 5, which asks students to recognize the requirements of restriction
fragment length polymorphism to determine parentage (Figure 8). 100 percent of the

students recalled that every band present in a child would be expected to be present in at

Figure 8: PCR Multiple Choice Pre and Post-test Results

 

 

 

 

 

 

i i

ii

t

O L.

i; g I Pretest
a 3 l Postest
(D 0

‘0-

o

a:

 

 

Question #

 

l

43

least one of the true parents compared to only 1 of the students on the pretest (Appendix
CVIII). The posttest question 1 and 3 also showed that all students could master this
information visually by interpreting the “family’s DNA material” (Appendix BV) . Even
after doing the “All you have ALU” laboratory at MSU, 30% of the students did not
understand that PCR technique uses heat resistant DNA polymerase . Most of these
students chose the incorrect answer involving DNA ligase. Unfortunately, the questions
on the AP tests can be very specific.

The second half of the PCR pre and posttest included a two-part short answer
question. In the same way, each question was, as objectively as possible, evaluated using
the assessment rubric in Appendix BV. Questions 6 and 7 asks students to describe
“What the PCR technique is and what the major steps are” Not a single student received

any points on the pretest, but these responses were some of my most entertaining. Some
of the student’s responses for what PCR is and why it is valuable were “people chasing
rabbits”, “because without it, how would we catch the rabbits”; “CPR in a different
order” , “to save lives”. As with the other short answer pretest questions, most of the
students just left the answer space blank because they probably never heard of it before.
A male student answered the question on the posttest by writing, “so that we can isolate
parts of DNA and identify them”. He described the steps of PCR as “1). Denature DNA
at high temperatures, 2). Allow plasmid attaching, 3) Therefore DNA can follow its due
Process”. These responses earned him 0.5, 1, 0.5, and 0 pts respectively, for a total of
two points out of four (Appendix CII). Another student received firll credit for saying “
PCR technique can amplify or make the supply of a particular gene or protein more

abundant. It can do this quickly” She went on to describe the steps of PCR with “1) Heat

the strands at 94°C to denature the strands, 2. Cool to 58°C so the primer can be added to
the separate strands, 3) raise the temp to 72°C so the DNA polymerase can lay down the
proper bases.“ Other students also received full points for describing the change in
temperature, (i.e. “raise, lower, raise”), instead of the actual temperatures used during the
experiment. Because the pretest scores were all zeros, I did not graph the difference
between pre and post-test scores, but instead organized the scores earned into Appendix
CD. From this table, it is easy to see that the most common score earned was a 3 out of 4
with 8 students earning that score. The next most common score was a 4 out of 4. The
total test group improved their performance on this question by 2.9 points, which was the
average on the posttest (Appendix C11).
The developmental biology pre and post-test were used to assess student’s
mastery of the following developmental concepts: fertilization, stages of early

embryogenesis, and

 

 

the I013 0f Figure 9: Developmental Biology Pre and Post-Test Results
extraembryonic T
20 -
membranes. The E I pretest
o 15 , post test
average number of 2 g i
3 310 t 1
questions students 3 fl ‘
a 5‘ I ‘ i
anSWered correctly 't 0 q l ‘
Onthepretestwas 1 2 3 4 5 6 7 8 91011 12
meatlon it
0.9 out of a possible

 

 

 

12 compared to the 8 out of 12 on the posttest (Appendix CII). There was only one

Student who received a perfect score on the post-test. The greatest amount of

45

improvement was on question number 6, which asks students to determine what process
establishes the primary germ layers (Figure 9). 95 percent of the students recalled that
germ layers are produced during gastrulation compared to the 0 percent on the pretest
(Appendix CIX, question 6). The posttest showed that most students could describe the
correct sequence of stages during embryogenesis and describe the major distinguishing
events during fertilization and cleavage, which were heavily emphasized during the
zebrafish lab. The two areas in which they were still weak were illustrating an example
of embryonic induction (Appendix CIX, question 4) and the location of human
fertilization (Appendix CIX, question 1). Most of the students, especially the males who
misunderstood question I, answered that fertilization occurred in the ovary. My only
rationale for students missing this question is the possibility that they were thrown off
track when they didn’t see fallopian tube as a choice. One of the computer simulations
used to help them understand fertilization involved a female reproductive track with
highlighted labels where the egg was passing through, fertilized (in the fallopian tube),
and eventually implanted. In question 2, which asked students to determine what
somites give rise to in mammalian embryos, 50 percent of the students mistakenly choose

“the notocor ” and only 32% correctly answered “muscles and vertebrate”.

Student Survey Results:

Overall, the unit for teaching biotechnology and embryology was successful with
the vast majority of students, according to survey results (Appendix CX). Specifically,
the Students found the Medical Technology and Genetic Testing Field Trip to MSU to be

very interesting and usefirl to the study of biology, with over 72% of the students rating it

46

as excellent. 100% of the students rated the classroom simulations of laboratory
techniques, such as the “Manipulating DNA Replication” activity, as excellent to very

good in terms of helping them to understand biotechnology concepts taught in this class.

One of my students wrote “I still have it, it’s in my earl”. Imagine what the world

would be like if DNA models hung fiom every student’s rear view mirror?

In addition, ninety five percent of the students rated the question on how
laboratory work related to ideas discussed in class as very helpful. By far the students
rated the All You Have ALU laboratory, which involved using the PCR on their own
DNA, the highest in comparison to other science course laboratories. On the other hand,
over 22% of students rated the Embryology of Zebrafish as only fairly good in
comparison to other science course laboratories. Possibly, a few students were
unsatisfied with their grade and consequently did not value some of the course
components. Although one student was less than enthusiastic about biotechnology and

embryology, the majority of students rated each of the components as excellent or very

good.

47

CONCLUSION

The rapid accumulation of genetic information generated by the Human Genome
Project and related research has greatly increased public attention to genetic issues.
Molecular biology is a current hot spot in research activity, which will continue reaching
into every comer of life science. Molecular biology has already become a potent tool for
exploring development and embryology. Developmental biologists continue to unravel
the regulatory networks that tell cells where they are in an embryo and choreograph their
differentiation into specific cell types during transformation of egg to organism. High
school students need to be educated in DNA science, especially those interested and able
to pursue a biology-related career. Therefore, biology teachers, particularly AP biology
teachers, are challenged to help students become successful in understanding these
difficult concepts.

The purpose of this unit was to develop and implement a teaching model for
engaging students in genetics and developmental biology. The unit contains several
teaching techniques to enhance students’ involvement and appreciation of biology. The
overarching theme encompassed by the buzzwords of educational literature (inquiry-
based, authentic laboratories, process skills, and hands-on), is the value of active
learning. The teaching activities developed in this unit are governed by the firndamental
principle that many students learn best by doing. These new laboratories and activities

give students the opportunity to “do” quantitative and qualitative science, and ensure that

they get the “big picture”.

48

The pre and post-tests focused on the concepts of DNA structure, replication,
transformation, separation, amplification, and developmental biology. The data garnered
from these tests statistically show a significant increase in student performance on the
selected objectives. However, the source of improvement is confounded amongst the
activities, pre-laboratories, laboratories, lectures, and class discussions. This means it is
difficult to conclude that one method of teaching is a larger contributor to student
comprehension. It is known that students need several iterations using different method
of teaching. Further study is needed in the mechanisms behind student learning with
regards to molecular genetics and vertebrate development before the most critical
teaching technique can be elucidated.

The student survey results might provide an avenue to intensify the study of
teaching techniques. Students clearly enjoyed the biotechnology laboratory “All you
have ALU” at Michigan State University. This laboratory-field trip combination was
very personalized using their own DNA; required deep processing skills while engaged in
the lab procedure; and involved real research equipment, facilities and researchers.
Interestingly, their evaluation of the different components of this unit closely correlated
with many of the educational theories researched.

There were a few weaknesses in the first attempt at implementing the new
teaching activities in this unit. The major problem was time; there is never enough time

to teach AP biology. Even though two of the labs were done as all day field trips, we
simply didn’t have enough time to have students perform the Y chromosome laboratory.
Instead, they watched a quick demonstration. I plan on using multiple centrifuges during

the All You Have ALU laboratory in the future to ensure extra time for students to

49

 

perform the Y chromosome lab. It is important to have a female’s DNA as a control next
time.

One of the major strengths of the labs was the enthusiasm they generated. One
student asked if he could do an independent research project using the Zebrafish. I
excitedly agreed, because self —initiated learning is the most lasting and pervasive
(Kearsley, 2001). He set out to test various environmental factors to see how
development can be influenced and inherited. Unfortunately, we couldn’t get any of the
embryos to develop into fertile adults. Near the end of the school year, by chance, I came
across an article detailing the conditions needed to raise the little fish. I’m planning to
have students observe the embryos develop into adult fish next year, perhaps as some
kind of case study project. Another benefit gained by analyzing the data for this unit was
that I was able to really focus on students’ rough spots during their review right before
they took the AP test. Students were very interested and appreciative when I shared some
of my results.

Finally, I would like to echo the words of advice from Louis Pasteur, who would
challenge instructors and their students regarding laboratories to:

“Take interest, I implore you, in sacred dwellings, which one designates

by the expressive term: Laboratories. Demand that they be multiplied,

that they be adorned. These are the temples of the future, temples of well

being and of happiness. There it is that humanity grows greater, stronger,

and better.”

50

APPENDIX Al.

BREAKFAST DNA

Instruct r’ ide:

Time Frame: 1 hour

Target Group: Advanced Placement Biology students

Preparation of materials: Pour each of the materials below into a plastic sandwich bag.
Sources of materials: any grocery store

Problems encountered: Students had difficulty relating the square Golden Graham
cereal to the deoxyribose sugar. Replace the Golden Graham with Honeycomb cereal.

Reference: The American Biology Teacher, Volume 62, Number 7, September 2000

51

BREAKFAST DNA

Instructor’s Guide:

Objective: To use cereal & pretzels as a “hands on” tool to explain DNA structure,
replication, transcription, and translation.

Materials:
15 red FruitLoops
15 purple FruitLoops
15 yellow FruitLoops
15 green FruitLoops
15 blue FruitLoops
cup of Alphabits cereal
cup of Golden Graham cereal
'/2 cup of pretzels

DNA Structure:
1. Nucleotide- basic backbone of the DNA
molecule
- F mitLoops - ba_ses
- Golden Grahams- sugars
- Pretzels — phosphates

 

4' “Take one golden graham and place a blue

fruitloop on the upper right hand comer”.

(3’ “Break the pretzel stick in half and place on of the pieces on the upper
left hand corner of the golden graham”

{It Discuss the structure of a nucleotide and how the graham loosely
represents the sugar, the loop represents the base and the pretzel stick
represents the phosphate bond.

(10 In this case the “blue” fruitloop represents thymine.

 

A. The sugar is the 5-car’oon su ar DNA INC] eotide
deoxm'bose. By convention the [J'
carbons on this sugar are labeled | Base
1’ through 5'. '0— P—0—sz I
ll 5 /°\
B. The phosphate is attached to the 4i? H H :1. .
, . Phosphate I I Deoxunhose
5 carbon of the deoxyribose PIES—2' 1 (sugar)

sugar.

C. The M is attached to the Mg of the deoxyribose sugar.

52

II. Type: of nucleotides- determined by their bases
0 Blue- thmine (deogvthymidineaacleotide)
0 Red- adenine (deoxyadenosine nacieotide)

0 Yellow- cy_tosine (deoxycyaidine nucleotide)
0 Green- guanine (deoxygaanidine npcleotide)

A. They are called nitrogenous bases because acach base contains at least two
nitrogen atpmg.

B. There are two classes of bases

0 Pyrimidines: cyaosin§(C)and thymine“)
0 Purines: adenine (A) and gaanine1G)

{it Precede to make the other three nucleotides
4' Figure 5.27

III. DNA Polynucleotide Chain

(3* Note: you may only slide your nucleotides, you may not pick
them up!

4‘ To make a polynucleotide chain, take your deoxythmidine
(blue) nucleotide and attach the pretzel stick (5’ phosphate) to
the bottom left corner (3’ OH) of the golden graham of the
deoxyguanidine (green ) nucleotide.

A. In a DNA polynucleotide chain, nucleotides are joined by
phosphodiester bong formed between the 5_' carbon of one
sugar and the 3_' carbon of the next sugar.

 

4‘ Make a line of loops (any color order) 15 nucleotides long
B. 15 Nucleotide sequence:

 

IV. Double Stranded DNA (Base pairing rules)

A. Edwin Chargaff determined that the ratio of adenine to thymine and the ratio
of guanine to cytosine were always the same in all organisms tested.
1. The amount of A=T and the amount of C=G

B. Hydrogen bonds
1. Two hydrogen bonds form between thmine and adenine
2. Three hydrogen bonds form between cxtpsine and gaanine
(It Figure 16.6 Base pairing in DNA

C. The bonding of bases is complementary.
1. The sequence of one chain dictates the sequence in the other.

D. Record your double stranded DNA molecule sequence (add H bonds)

53

 

4' How could you base pair your nucleotides that are in front of you without
picking them up
(it Figure 16.5

E. The two complementary strands of DNA are antiparallel
1. ma strand runs 5’-) 3’, whgeas the pther runs 3’-) 5’.

V. DNA Replication
(3‘ After Watson and Crick proposed the double helix model of DNA, three
models for DNA replication were proposed

 

A. Cpagflativa Mpdel-the two parental DNA strands are back together after
replication has occurred.

B. Semiggaservative Model-the two parental DNA strands separate and each of

those strands then serves as a template for the synthesis of a new DNA strand.

C. Dispersiva Modal-the parental double helix is broken into double-stranded
DNA segments act as templates for the synthesis of new double helix
molecules.

D. Mmlson aad Stahl’s experiments with E.coli showed the density of the
molecules in each group matches the labeling pattern expected if DNA is

replicated semigznmatively.

E. Model for DNA replication
1. Before replication, the parent molecule has 2 complementary strands of
DNA. Each base is paired by hydrogen bonding w/ its specific partner.

2. The first step in replication is separation of the two DNA strands.

3. Each parent “old” strand now serves a s a template that determines the
order of nucleotides along “new” complementary strands. Nucleotides

plug into specific sites along the template surface according to the base
pairing rules.

4. The nucleotides are connected to form the sugar-phosphate backbones of
the new strands. Each DNA molecule now consists of one “old” strand
and one “new strand. The two DNA molecules are identical to the one
molecule with which you started.

F. Illustrate the replication of your DNA below.

54

BREAKFAST DNA

Objective: To use cereal & pretzels as a “hands on” tool to explain DNA structure,

replication, transcription, and translation.

Materials:
15 red FruitLoops
15 purple FruitLoops
15 yellow FruitLoops
15 green FruitLoops
15 blue FruitLoops
cup of Alphabits cereal
cup of Golden Graham cereal
'/2 cup of pretzels

DNA Structure:
1. - basic backbone of the DNA
molecule \
- FruitLoops -
- Golden Grahams-
. Pretzels —

A. The sugar is the

. By convention

 

 

 

 

the carbons on this sugar D-
are labeled 1’ through 5'. — _
O—fi‘ D—5_CH2 0

B. The is attached 0 4'C’H/ \H\
deoxynbose sugar. H [lj—i-2n.l

C. The is attached to the OH H

of the deoxyribose sugar.
11. Types of nucleotides- determined by their bases

0 Blue— ( )
0 Red- ( )
0 Yellow- ( )
0 Green- ( )

 

A. They are called nitrogenous bases because

  

DNA nucleotide

Base I

‘1 .
I-

Deoxur'i hose
H (sugar)

 

 

 

55

B. There are two classes of bases
0 Pyrimidines: and
o Purines: and

 

 

III. DNA Polynucleotide Chain

A. In a DNA polynucleotide chain, nucleotides are joined by
formed between the _ carbon of
one sugar and the carbon of the next sugar.

 

B. 15 Nucleotide sequence:

 

IV. Double Stranded DNA (Base pairing rules)

A. determined that the ratio of adenine to
thymine and the ratio of guanine to cytosine were always
the same in all organisms tested.

1.

 

 

B. Hydrogen bonds
1. Two hydrogen bonds form between
2. Three hydrogen bonds form between

 

 

C. The bonding of bases is .
l. The sequence of one chain dictates the sequence in the other.

D. Record your double stranded DNA molecule sequence (add H bonds)

 

E. The two complementary strands of DNA are antiparallel
1.

V. DNA Replication

A. -the two parental DNA strands are back together after
replication has occurred.

B. -the two parental DNA strands separate and each
of those strands then serves as a template for the synthesis of a new DNA
strand.

C. -the parental double helix is broken into double-stranded

DNA segments act as templates for the synthesis of new double helix
molecules.

56

D. experiments with E. coli showed the density of the
molecules in each group matches the labeling pattern expected if DNA is
replicated .

E. Model for DNA replication
1. Before replication, the parent molecule has 2 complementary strands of
DNA Each base is paired by hydrogen bonding w/ its specific partner.

2. The first step in replication is separation of the two DNA strands.

3. Each parent “old” strand now serves a s a template that determines the
order of nucleotides along “new” complementary strands. Nucleotides
plug into specific sites along the template surface according to the base
pairing rules.

4. The nucleotides are connected to form the sugar-phosphate backbones of
the new strands. Each DNA molecule now consists of one “old” strand
and one “new strand. The two DNA molecules are identical to the one
molecule with which you started.

F. Illustrate the replication of your DNA below.

57

Pl

APPENDIX All.

Y CHROMOSOME

Inst r’ i e

Time Frame: Part of 1 day field trip

Target Group: Advanced Placement Biology students

Preparation of materials:

1. Reagents:

/ 0.1M citric acid (192.1 g/L)/5 = 3.84 g/ 200ml deO

/ 0.2M Na2HPO4 (l41.96g/L)/5 = 5.68g/ 200ml dH20

I 50.4ml citric acid + 69.6ml Nazi-IP04 =100ml buffer

/ 3g Sucrose in 5ml of unused buffer

/ 0.5% quinacrine dihydrochlon'de (0.5g/100ml dH20) or (0.3g/ 60ml dH20)
tr Note: solution is good 1-2wks if refrigerated and kept dark.
tr Caution: Quinacrines are carcinogenic. Use blotting paper and do not ingest.

s
9t
a
g;

Toothpicks

Alcohol burner

Microscope slides and coverslips
Blotting paper

Gloves

Aluminum Foil

Distilled water in water bottle
Stopwatch

\\\\'\'\\\.

Background Information for the teacher:
EQ Heterochromatin has been divided into two types. Constitutive heterochromatin is

always condensed during interphase. F aculatative heterochromatin is composed of
sequences that may be euchromatic in some developmental or physiological states,
and heterochromatic in others, implying that genes may or may not be active

Problems encountered: lack of time

References:

1.

9'95“.“

Dan O’Mally room 3123 cytogenetics lab in Life Science Bldg MSU. Phone 355-
2731

MSU cytogenetics lab manual : qbands (QBAND)
http://www.ud1.es/usuaris/e4650869/docencia/practiques_online/ cariotips/
http://wwwhosufl.edu/mooreweb/AdvafncedGcneties/lo-z1-99/10-21-99.html
Campbell, Neil, et al. 1999. Biflpgy, Benjamin/Cummings, Melano Park, CA.

58

Y CHROMOSOME

Enigma;

Quinacrine mustard solution is a fluorescent stain used to identify specific chromosomes
and structural rearrangements. It is especially useful for distinguishing the Y
chromosome from the other groups of chromosomes. The Y chromosome has the most
heterochromatin, which is highly condensed DNA that is not transcribed. Quinacrine is a
passive stain, which works its way into the chromosome and binds to specific histones
(HI) responsible for DNA packaging. H1 binds to DNA adjacent to a nucleosome; the
string of nucleosomes coil to form a chromatin fiber that is 30m in diameter. The Y
chromosome fluoresces the brightest because it contains the highest concentration of
these histones.

Matefl :

Toothpicks

Alcohol burner

Microscope slides and coverslips
Blotting paper

Gloves

Aluminum Foil

Distilled water in water bottle
Stopwatch

mm

Obtain a sample of check cells by lightly scraping the inside of your cheek with a

toothpick.

Transfer the cheek cells to a microscope slide

3. Quickly run your slide and cheek cells through a low flame to dry and fix your
sample.

4. Add 3 drops of quinacrine mustard stain to your slide and let it stand for 5-15
minutes.

5. Gently rinse the slide with a distilled water bottle for 30 seconds or until color
disappears from H20.

6. Add 5 drops of buffer solution to specimen and let sit for 1 minute.

7. Lightly tap excess buffer from slide.

8. Add 2-3 drops of sucrose solution and mount the coverslip. Remove excess sucrose
from the top of the coverslip with tissue.

9. View slide with fluorescence microscope with appropriate filters. (Note: 445
wavelength is recommended).

10. Observe and compare a cell from someone of the opposite sex.

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59

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June 19

APPENDIX AIII.

MANIPULATING DNA REPLICATION

Instructor’g Guidg

Time frame: 1 class period
Target group: Advanced Placement Biology students

Background information for the teacher: The abstract concepts in molecular biology
are a challenge to concrete thinkers. The “real” pipe cleaners are used as a sensory tool
to understand the abstract process of DNA replication. Each student is given an
unassembled kit (see materials below) to construct as the teacher explains the process of
DNA replication. Note: Kits for each student are very inexpensive and compact enough
to take home for practice.

Preparation of materials;
Place the following materials in a ziploc bag
1-12” red pipecleaner
1-12” green pipecleaner
1-6” shiny red pipecleaner
2-3” shiney green pipecleaners
3-1” white pipecleaners
2- paper hole reinforcement labels
4-self adhesive file folder labels
l-purple pony bead
l-green pony bead
3-pink pony beads

Sources of materials: Michael’s craft store

Problems encountered: showing how polymerase only reads 5’-) 3 ’

References: “Modeling DNA Replication” American Biology Teacher, Vol 60, No. 6,
June 1998

 

MANIPULATING DNA REPLICATION

Objective: To use pipe cleaners and beads to model the process of DNA replication.

I. DNA Structure
A. Two 12” long pipe cleaners are wound around each other to represent the
DNA double helix.
(3! The red pipe cleaner represents one of the original DNA template strands.
(It The green pipe cleaner represents the other parent DNA strand

B. The two strands of DNA are antiparallel (their sugar phosphate backbones
run in opposite directions.
(11 Label one end of your red pipe cleaner as 5’PO4 ‘
4' Label one end of your green pipe cleaner as 3’OH respectively

II. DNA Separation and Unwinding
A. A branch point in a replication bubble at which DNA synthesis occurs is called
a replication fork.
(It Autoradiographic studies have shown that DNA synthesis almost always
proceeds in both directions from the point where replication is initiated
4‘ Figure 16. 6 show picture of bidirectional replication).

B. Helicase attaches to the replication fork and uses energy derived fi'om
hydrolysis of ATP to drive the unwinding and separation of the DNA molecule.
(It Since 2 ATP molecules are required for unwinding each base pair, donut-
shaped paper hole reinforcements labeled “ATP” are attached to the helicase
bead before it is threaded along the length of the wound DNA strands.

1. The replication fork is produced by threading a purple “pony bead” 2/3 of
the way down the double helix, unwinding the two strands, to simulate the
action of the enzyme helicase.

4: DNA Gyrase (grey bead) relaxes supercoils created by helicase. (As
unwinding continues, compensating DNA turns accumulate downstream of
the replication fork.) DNA gyrase cuts both strands, allowing them to
rotate and unwind, relieving the built-up tension. The ends are then
rescaled by the gyrase.

2. After DNA helicases unwind the DNA, the exposed single stranded
segments are subject to nuclease digestion and breakage. To prevent this,
single-stranded segments are immediately bound by single stranded binding
protein (ssb), which hold the unpaired DNA strands apart to be replicated.)

(1! DNA bound to SSB’s is semirigid w/ no bends, which facilitates DNA
synthesis.

(3! “Helicase hacks” our double helix into two strands

61

<31 The DNA is now ready for both leading and lagging strand synthesis

111. Leading Strand Synthesis

A The new complementary strands are synthesised by polymerase 111 (pink bead)
4t “Polymerase pastes” nucleotides alongside the template

B. DNA polymerase has several lirrritations that contribute to the complexity of
the replication process.

1. Nearly all known DNA polymerases can add a nucleotide only to the
fi'ee 3 ’-OH group of a base-paired polynucleotide so that DNA chains
are extended only in the 5’-) 3’ direction.

2. DNA polymerases can add nucleotides only to the 3’end of an existing
polynucletide strand.

C. Initiation of new chain growth is accomplished by synthesizing a short segment
of RNA with a free 3’OH group. A 1” piece of white pipe cleaner (MA
primer) is wound around the far end of the single DNA strand which has the
free 3’OH (not labeled).

4‘ The primase component of the primosome is responsible for synthesis of
the short RNA fragment. Both the leading and lagging strands require an
RNA primer provided by the primase.

D. On the end of the white RNA primer that faces the replication fork, place a pink
pony bead (DNA polymerase III).

E. A 6” shiny red pipe cleaner (leading strand of DNA) is attached to the DNA
polymerase. The leading strand is wound around that same single DNA strand
starting at the white RNA primer and heads into the replication fork

{It The two parent strands are replicated in different ways.

F. The newly synthesized DNA strand that extends 5’-) 3’ in the direction of the
replication fork movement, the leading strand, is continuously synthesized.
Only one priming event is required to initiate synthesis of the leading strand.

IV. Lagging Strand Synthesis

A. The other new strand, the lagging strand, is also synthesized in the 5’93”
direction. However, this new strand, called an Okazaki fragment, can only be
made discontinuously.

(3! Each Okazaki fragment is only 100-200 nt long and requires multiple priming
sites due to the 5’ 9 3’ synthesis requirement of DNA polymerase.

62

B. A second 1’ white RNA primer, which serves to initiate Okasaki fragment
synthesis, is wrapped around the other single DNA strand at the replication fork.
4' Remember, on the leading strand the white RNA primer was at the opposite
end of the single stranded DNA.

C. Next, a 2” shiny green pipe cleaner (Okasaki fragment) is attached to a second
pink bead (DNA polymerase 111). Both are wound around the DNA template,
heading in the direction out of the fork.

D. Beyond that shiny green segment, another white RNA primer is wound, and
further “out” an adjacent shiny Green Okasaki fragment and pink polymerase II is
added.
at The RNA primer between each Okazaki fragment is removed by DNA

polymerase I. (the RNA segment is degraded to flee nucleotides) Then,
deoxynucleotides fill in the gap and are polymerized by DNA polymerase HI.

E. DNA ligase (lime bead) is symbolically attached to the starting end of the
Odasaki fragment. Ligase catalyzes the formation of a phosphoester bond
between the 5’-phosphate and 3’hydroxyl to yield a continuous strand of
complementary DNA

(It “Ligase links” the Okazaki fragments (lagging DNA)

63

MANIPULATING DNA REPLICATION

Objective: To use pipe cleaners and beads to model the process of DNA replication.

1. DNA Structure

A. Two 12” long pipe cleaners are wound around each other to represent the

J

B. The two strands of DNA are (

11. DNA Separation and Unwinding

A. A branch point in a replication bubble at which DNA synthesis occurs is called
a

_

B. attach'bs to the replication fork and uses energy derived from
hydrolysis of ATP to drive the unwinding and separation of the DNA molecule.

1. The replication fork is produced by threading a purple pony bead
( ) 2/3 of the way down the double helix and unwinding the
two strands.

2. Afier DNA helicases unwind the DNA, the exposed single stranded
segments are subject to nuclease digestion and breakage. To prevent

this, single-stranded segments are immediately bound by
_, which hold the unpaired DNA

 

strands apart to be replicated.)
Q

 

III. Leading Strand Synthesis

A. The new complementary strands are synthesised by
Q

 

 

B. DNA polymerase has several limitations that contribute to the complexity of
the replication process.

1. Nearly all known DNA polymerases can add a nucleotide only to the free
3’-OH group of a base-paired polynucleotide so

 

 

2. DNA polymerases can add nucleotides only to the

 

C. Initiation of new chain growth is accomplished by synthesizing a short segment
of RNA with a free 3’OH group. A 1” piece of white pipe cleaner (
) is wound around the far end of the single DNA strand which has the 3”OH
label.

D. On the end of the white RNA primer that faces the replication fork, place a pink
pony bead ( _).

E. A 6” shiny red pipe cleaner ( J) is attached to the
DNA polymerase. The leading strand is wound around that same single DNA
strand starting at the white RNA primer and heads into the replication fork

F. The newly synthesized DNA strand that extends 5’9 3’ in the direction of the
replication fork movement, called the , is
synthesized.

 

Lagging Strand Synthesis

. The other new strand, _, is also synthesized in the 5’93”
direction. However, this new strand, called an _, can only be
made

 

. A second 1’ white RNA primer, which serves to initiate Okasaki fragment
synthesis, is wrapped around the other single DNA strand at the replication fork.

. Next, a 2” shiny green pipe cleaner ( _) is attached to a
second pink bead ( _). Both are wound around the DNA
template, heading in the direction out of the fork.

 

. Beyond that shiny green segment, another white RNA primer is wound, and
firrther “out” an adjacent shiny geen Okasaki fragment and pink polymerase II is
added.

. (lime bead) is symbolically attached to the starting end of the
Okasaki fragment. Ligase catalyzes the formation of a phosphoester bond

between the 5’-phosphate and 3’hydroxyl to yield a continuous strand of

complementary DNA.

Q

 

 

65

APPENDIX AIV.

BACTERIAL TRANSFORMATION PRE-LABORATORY

Instructor-’1 Guide:

I. Key Concepts 1: Bacterial Transformation

Define transformation.
Q. A phenomenon in which external DNA is assimilated by a cell.

What gene are you introducing?
m a gene for resistance to the antibiotic ampicillin is introduced into a bacterial strain
that is killed by ampicillin.

How do you know it was taken up?

is Ifthe susceptible bacteria incorporate the foreign DNA, they will become ampicillin
resistant.

[1. Bacterial Colonies

What are some of the materials required in this lab?
E Escherichia coli, Luria Broth agar and petridishes

How many individual cells are in each colony (roughly)?
ix Each colony in the petri dish is made up of millions of individual cells.

111. E. coli Bacteria

Name one place E. coli is found?
it the human gut

Describe two places E. coli stores it's genetic material? _
*3 It has genetic material in the bacterial chromosome and some E. coli store genetic
material in plasmids

IV. Plasmids

Define plasmids
*3 A small ring of DNA that carries accessory genes separate from those of a bacterial
chromosome. Also found in some eukaryotes, such as yeast.

What are plasmids used for in genetic engineering?

E In genetic engineering, plasmids are one means used to introduce foreign genes into a
bacterial cell.

Differentiate between an +amp R cell and -amp R cell

it Some plasmids have the ampR gene, which confers resistance to the antibiotic
ampicillin. E. coli cells containing this plasmid termed "+ampR" cells, can survive
and form colonies on LB agar that has been supplemented with ampicillin. In
contrast, cells lacking the ampR plasmid, termed "~ampR" cells, are
sensitive to the antibiotic, which kills them.

V. Competent Cells

Define a competent cell.
9: Competent cells are more likely to incorporate foreign DNA because their cell walls
are altered so that DNA can pass through more easily.

How do you make a competent cell?
Q: Cells are made competent by a process that uses calcium chloride and heat shock.

When is it easier to make a cell competent?
is Cells that are undergoing very rapid growth are made competent more easily.

Explain why each phase has it's particular # of cells?

‘3 The growth rate of a bacterial culture rs not constant. In the early hours (lag phase),
growth rs very slow because the starting number of dividing cells rs small. This rs
followed by a time rapid cell division known as the log phase.

Draw & label the 4 phases of cell division

 

 

Numbers! cells (log scale )—>

 

 

Time (hours)

VI. Design of the Experiment

List 5 techniques of sterile procedure.

is Always wash your hands and work surface before beginning.

\ Always keep the lid of the petri dish on it or over it at all times.

Rs Never touch the end of a tool that touches bacteria.

1s Always keep hair pulled back and use goggles when flame is present.

\ Always wash your hands thoroughly with soap and hot water before leaving the lab.

67

VII. Transformation Procedure

Why do you also prepare a second group of E. coli cells?
Q to verify that E. coli will not grow on agar with ampicillin unless it is transformed,
and that nothing in the procedure itself affects the survival of E. coli.

What is the only difference in your control group?
Q The control group lacks the ampR plasmids

Describe the 6 Steps in this procedure

Q Ampicillin sensitive E. coli cells are transferred to cold CaC12.

Q The ampR plasmids are added only to the experimental cells.

Q Both cells are heat shocked at 42°C.

Q The treated cells are spread on LB plates containing ampicillin

Q All plates are incubated for 24 hours

Q Count E. coli colonies and determine if cells have been transformed.

68

BACTERIAL TRANSFORMATION PRE-LABORATORY

1. Key Concepts 1: Bacterial Transformation

Define transformation.

 

 

What gene are you introducing?

 

 

How do you know it was taken up?

 

 

1]. Bacterial Colonies

What are the materials required in this lab?

 

 

How many individual cells are in each colony (roughly)?

 

 

III. E. coli Bacteria

Name one place E. coli is found?

 

 

Describe two places E. coli stores it's genetic material?

 

 

IV. Plasmids

Define plasmids

 

 

What are plasmids used for in genetic engineering?

 

 

69

Differentiate between an +amp R cell and -amp R cell

 

 

V. Competent Cells

Define a competent cell.

 

 

How do you make a competent cell?

 

 

When is it easier to make a cell competent?

 

 

Explain why each phase has it's particular # of cells?

 

 

Draw & label the 4 phases of cell division

 

 

 

 

 

 

 

 

 

 

 

 

VI. Design of the Experiment
List 5 techniques of sterile procedure.

 

 

70

 

 

 

VII. Transformation Procedure

Why do you also prepare a second group of E. coli cells?

 

 

What is the only difference in your control group?

 

 

VIII. Describe the 6 Steps in this procedure

 

 

 

 

 

 

71

APPENDIX AV.

BACTERIAL TRANSFORMATION LABORATORY

lngtmgtop’s Guide:

Time Frame: 2 hours

Target Group: Advanced Placement Biology students

Preparation of materials: see Carolina Biological Teacher Guide

Sources of materials: Carolina Biological Supply Company (catalog # 21-1142)

Problems encountered: Students’ water temperatures were not exact. I plan on using a
water bath instead of beakers of water on hot plates.

Reference: Carolina Biological Colony Transformation Kit
AP Biology Laboratory Manual for Teachers and Students
Lab 18: Molecular Genetics: Recombinant DNA

Answers to Questions in the Student Laboratory:

RESULT

1. Observe the bacterial growth through the bottom of the culture plates. Do not
open the plates. Count the number of individual colonies (use a permanent marker
to mark off each colony as it is counted). If cell growth is too dense to count
individual colonies, record "lawn." Share results on Positive Controls with another
lab group.

Positive Control: (LB+) Q lawn Positive Control: (1.8-) Q lawn

Experimental: (LB/Amp+) Q 5-500 Negative Control: (LB/Amp-) Q 0

INTERPRETATIQN AND CONCLUSIONS

1. Compare and contrast the number of colonies on each of the following pairs of plates.
What does each pair of results tell you about the experiment?

a) LB+ and LB- : Q Cells are viable. In the absence of antibiotics both
cells grow equally well.

b) LB/Amp- and LB/Amp+ : Q Only transformed cells grow in the presence of
ampicillin.

c) LB/Amp+ and LB+ : Q Transformation is a rare event, only 1 in 1million

72

cells is transformed.
d) LB/Amp- and LB- : Q Wild-type cells fail to grow in the presence of
ampicillin. The antibiotic is active.

2. Transformation efficiency is expressed as the number of antibiotic resistant colonies
per pg of pAMP DNA Because transformation is limited to only those cells that are
competent, increasing the amount of plasmid used does not necessarily increase the
probability that a cell will be transformed. A sample of competent cells is usually
saturated with small amounts of plasmid and excess DNA may actually interfere with
the transformation process. The object is to determine the mass of pAMP that was
spread on the experimental plate and was, therefore, responsible for the transforrnants
observed.

a) Determine total mass (in ug) of pAMP used in Step 8.
[Mass = Concentration x Volume].

m 0.00»pr * 10p.L = 0.05pg

b) Calculate the total volume of cell suspension prepared. [Steps 2 and 13]
Q 250uL + 250uL = 500uL

c) Calculate the fraction of the total cell suspension that was spread on the plate.
[Volume Suspension Spread/Total Volume Suspension (part b)== Fraction Spread]
Q 100uL/500uL = 0.2

d) Determine the mass of pAMP in cell suspension. [Total mass of pAMP (part a) X
fraction spread (part b) = Mass pAMP Spread].
Q 0.05ug * 0.2 = 0.01ug

e) Determine number of colonies per ug pAMP. Express answer in scientific
notation. [Colonies Observed/Mass pAMP Spread (part c) = Transformation
Efficiency].

Q Ex: (“100 colonies” / 0.01ug) = 1x 104 transformants/ug

3. What factors might influence transformation efficiency? Explain the effect of each
you mention.

Q Technique errors (such as not taking large enough cell masses, not resuspending
the cells, missing or indistinct heat shock). These could decrease or even
eliminate colony growth on plates

Q Growth stage of cells transferred could effect the transformation efficiency
because healthy, rapidly dividing cells are more likely to uptake DNA.

Q As the cell solution becomes saturated with DNA, additional plasmid DNA could
decrease the transformation efficiency.

73

BACTERIAL TRANSFORMATION LABORATORY

INTRODUCTION
Background Information

The bacterium Escherichia coli (E. coli) is an ideal organism for genetic
manipulation and has been used extensively in recombinant DNA research. It is a
common inhabitant of the human colon and can easily be grown in standard nutrient
mediums.

The single circular chromosome of E. coli contains 5 million DNA base pairs (1
/600th the total amount of DNA in a human cell). In addition, the cell contains small,
circular; extrachromosomal (outside the chromosome) DNA molecules called plasmids.
These fi'agments of DNA, 1,000 to 200,000 base pairs in length, also carry genetic
information. Some plasmids replicate only when the bacterial chromosome replicates and
usually exist only as single copies within the bacterial cell. Others replicate
autonomously and often occur in as many as 10 to 200 copies within a single bacterial
cell. Certain plasmids, called R plasmids, carry genes for resistance to antibiotics such as
ampicillin, kanamycin, or tetracycline.

In nature, genes can be transferred between bacteria in three ways: conjugation,
transduction, or transformation. Conjugation is a mating process during which genetic
material is transferred from one bacterium to another "sexually" different type.
Transduction requires the presence of a virus to act as a vector (carrier) to transfer small
pieces of DNA fi'om one bacterium to another. Bacterial transformation involves transfer
of genetic information into a cell by direct absorption of the DNA released from a donor
cell.

Through the process of bacterial transformation, a bacterium can acquire a new
trait by incorporating and expressing foreign DNA. In the laboratory, the DNA used most
commonly for transformation experiments is bacterial plasmid DNA. These plasmids
often carry a gene for antibiotic resistance. The presence of the antibiotic-resistance gene
makes it possible to select bacteria containing the plasmid of interest; the bacteria that
contain the plasmid will grow on a medium that contains the antibiotic, whereas bacteria
lacking the plasmid will not be resistant to the antibiotic and will die. Transformation
can occur naturally, but the incidence is extremely low and is limited to a relatively few
bacterial strains. During the growth cycle of these strains, there exists a short period of
time when the bacteria are most receptive to uptake of foreign DNA. At this stage the
cells are said to be competent (Competence to absorb DNA usually develops toward the
end of the logarithmic growth phase, just before cells enter the stationary phase in
culture.) The mechanism by which competence is acquired is not completely understood,
but in the laboratory, treating bacterial cells with divalent cations such as Ca2+ and Mg2+
can induce the competent state.

In this exercise, you will simulate the construction of a recombinant plasmid.
Plasmids can transfer genes such as those for antibiotic resistance which are already a
part of the plasmid, or plasmids can act as carriers for introducing foreign DNA fiom
other bacteria, plasmids, or even eukaryotes into bacterial cells. Restriction

74

endonucleases are used to cut and insert pieces of foreign DNA into the plasmid vectors
(Figure 1).

Prokaryotic Cell Eukaryotic Cell

/ ’_\
-" “‘
.,‘\_"4~
"'3‘ mt
. .
.1 ' .
__.... .. .- \‘ '
l.—-r—.'< -. . » .
e r . ~
. ,'
'.. , .~ -
“JV-.1 '4'
I

j plasmid is isolated chromosomai DNA is isolated

 

.m‘

   

from the bacterial cell from the eukaryotic cell

U ‘-

1 chromosomal DNA is cleaved by

 

the same restriction endonuclease
used to cleave the plasmid

7 "Sticky" ends \
> "sticky" ends

\ /toreign DNA, in this case. eukaryotic
chromosomal DNA, is inserted

into the plasmid

plasmid DNA is cleaved by
a restriction endonuclease

c.-~\

the recombinant plasmid, with its
segment of foreign DNA, is taken
into the bacterial cell where the
foreign DNA will be expressed, and
its gene products produced, in the
same manner as the cell's own DNA.

recombinant DNA molecule

 

 

 

 

transformed bacterial cell

Figure 1: A typical recombinant DNA experiment using a restriction endonuclease.

75

Each restriction endonuclease “recognizes: a specific DNA sequence (usually a 4 to 6
base-pair sequence of nucltotides in double stranded DNA and digests phosphodiester
bonds at specific sites in the sequence (recall that phosphodiester bonds link one
nucleotide to the next in a DNA polynucleotide chain). If circular DNA is cut at only one
site, an open circle results. Ifthe restriction endonuclease recognizes two or more sites on
the DNA molecule, two or more fi'agments will result. The length of each DNA fragment
corresponds to the distance between restriction sites (restriction sites flank the fragment
at its ends). Some restriction endonucleases cut cleanly through the DNA helix at the
same position on both strands to produce fragments with blunt ends. Other endonucleases
cut specific nucleotides on each strand to produce fragments with overhangs or "sticky
ends" (Figure 2). Using the same restriction endonuclease to cut DNA from two different
organisms produces complementary sticky ends, which can be realigned in a "template-
complement" manner, thus recombining the DNA from the two sources (Figure 2).

Bacterial Plasmid Eukaryotic Chromosome

 

 

 

 

'Mfiimfiic - ‘\ GL3} '1‘ °
c 1' 'r A A G c 'r 'r n“; a
“'“ir
restriction site
a gene or a gene
fragment is flanked by

sticky ends

’ A.:.r. .- . 'c
G 'I'TA

 

 

 
 

bind at their “sticky ends”

 
 

the restriction fragments ;...

Figure 2 EcoRI endonuclease cleaving a bacterial plasmid and a
fragment of eukaryotic DNA and then recombining them.

76

In bacteria, restriction enzymes provide protection by breaking and destroying the
DNA of invaders, such as that of bacteriophage viruses. However, since the recognition
sites for restriction endonucleases also occur within the bacterial DNA itself, bacteria
have a mechanism for preventing their own restriction enzymes from digesting their own
DNA For each restriction endonuclease produced by a bacterium, there is a
corresponding enzyme that methylates the bacterial DNA at that enzyme's specific
recognition site: addition of a methyl group to the nucleic acid base prevents a close
association fiom forming between the restriction endonuclease and the recognition site.
In this way, bacteria can break down the DNA of invaders while protecting their own
genetic material from destruction.

E

To understand the principles of bacterial transformation

To understand the conditions under which cells can be transformed

To understand the process of competent cell preparation

To understand how a plasmid can be engineered to include a piece of foreign DNA
To understand how plasmid vectors are used to transfer genes

To understand how restriction endonucleases function

To understand how antibiotic resistance is transferred between cells

To use sterile technique

To calculate transformation efficiency

999999999

MATERIA : for h f r r rou
J E. coli culture-starter plate

1- marker

2- 15ml tubes

p—AMP plasmid

4- 1m] sterile transfer pipette

500uL- 0.05-M CaClz ice cold

Bunsen burner

inoculating loop

spreading rod

water-bath 42° C

3ml Luria broth

2 Luria broth agar plates

2- LB/ ampicillin agar plates

1- 37°C incubator

\'\'\'\'\'\\'\\'\'\'\\

77

£3

METHQD/ PROCEDURE

Ste
1 .

10.

ll.

12.
13.

p by step
Mark one sterile 15-mL tube ”+AMP "; this tube will have the plasmid added to it.
Mark another tube ”-AMP ”; this tube will not have the plasmid added.

Use a sterile micropipette (Fig. 1) to add 250 microliters (uL) of ice cold 0.05M
CBC]: to each tube.

Place both tubes on ice.

Use a sterile inoculating loop to transfer one or two large (3mm) colonies of E. coli
cells from a starter plate to +AMP” tube using a sterile inoculating loop.
a) Try to get the same amount of bacteria into each tube.
b) Be careful not to transfer any agar from the plate along with the cell mass.
c) Vigorously tap the loop against the wall of the tube to dislodge the cell mass.
d) Hold the tube up to light to observe cell mass fall off of the loop.

. Immediately mix the suspension by repeatedly drawing in and emptying out a sterile

micropipette with the suspension.

Return "+AMP" Tube to ice. Transfer second mass of cells to "-AMP " Tube, and
suspend as described in Step 3.

Return "-AMP " Tube to ice. Both tubes should now be on ice.

Use a sterile inoculating loop to add one loopful of pAMP solution (0.005 [pg/uL)
directly into the cell suspension in tube "+AMP".
a) Note 10 uL is measured when DNA solution forms a “bubble” across the
loop Opening.
b) Swish the loop to mix the DNA. Then mix by tapping the tube with your
finger.
c) This solution contains the antibiotic resistance plasmid.

Return ”+AMP" Tube to ice, and incubate both tubes on ice for 15 minutes.

While the tubes are on ice, label each agar plate (lLB and 2 LB/Amp) on the bottom
with lab group name and date. .

a; Label one LB/Am plate “LB/Am +” This is the experimental plate.

b Label the other L / p plate “L /Amp -“ Thrs rs the negative control.

c) Label one LB plate either “LB+” or “LB-“ as directed by your instructor. This
is a positive control

A brief pulse of heat facilitates entry of foreign DNA into the E. coli cells. Following
the 15 minute incubation, heat-shock cells in both the "+AMP" and "-AMP" tubes.
a) Carry the beaker containing ice and tubes to the water bath.
b) Remove both tubes directly from the ice and immediately immerse them in the
42° C water bath for 90 seconds.

Immediately return cells to ice for two or more minutes.
Use a sterile micropipette to add 250uL of Luria broth to each tube.
a) Mix by gently tapping with your finger

78

b) Set the tubes in a test tube rack at room temperature.
c) Any transformed cells are now resistant to ampicillinbecause they possess the
gene whose product renders the antibiotic ineffective.

14. Use a sterile micropipet to add 100uL cell suspension fiom “-AMP” Tube onto
“LB/Amp-“ Plate. If your group has been assigned the -AMP Positive Control, add
another 100p L onto “LB-“ Plate.

15. Immediately spread cells over the surface of plate(s) using a sterile-spreading rod.
a) Dip cell spreader in ethanol, and briefly pass it through Bunsen flame to ignite
alcohol before spreading each plate. Allow the alcohol to burn off away from
Bunsen flame and alcohol container.
b) Lift the plate lid only enough to insert spreader and cool spreader by touching
it to the side of the agar away fiom cell suspension.
c) Use the spreader to evenly distribute cell suspension over the surface of the
late, and replace the plate lid.
d) Return the cell spreader to ethanol without flaming.
16. Use the sterile micropipet to add lOOuL cell suspension from +AMP Tube onto
+Amp Plate. Ifyour group has been assigned the +AMP Positive Control, add another
100uL onto +LB Plate.

17. Immediately spread cell suspensions as described in Step 15.

18. Allow the plates to set for several minutes. Then wrap them together with tape. Place
the plates upside down in 37° C incubator, and incubate for 12 to 24 hours.

RESULTS

1. Observe the bacterial growth through the bottom of the culture plates. Do not
open the plates. Count the number of individual colonies (use a permanent marker
to mark off each colony as it is counted). If cell growth is too dense to count
ilngividual colonies, record "lawn.” Share results on Positive Controls with another

a group.

Positive Control: (LB+)

 

Positive Control: (LB-)

 

Experimental: (LB/Amp+)

 

Negative Control: (LB/Amp-)

 

79

[\"lIRl
Question

1 Wen
expe

compe
PTOba't
Salurax
the trar
Spread
Obsem

 

 

INTE RET TI AND N SION
Questions

1.

Were results as expected? Ifnot, explain possible reasons for variations from
expected results.

 

 

Compare and contrast the number of colonies on each of the following pairs of plates.
What does each pair of results tell you about the experiment?
a) LB+ and LB- :

 

b) LB/Amp- and LB/Amp+ :

 

c) LB/Amp+ and LB+ :

 

d) LB/Amp- and LB- :

 

Transformation efficiency is expressed as the number of antibiotic resistant colonies
per ug of pAMP DNA. Because transformation is limited to only those cells that are
competent, increasing the amount of plasmid used does not necessarily increase the
probability that a cell will be transformed. A sample of competent cells is usually
saturated with small amounts of plasmid and excess DNA may actually interfere with
the transformation process. The object is to determine the mass of pAMP that was
spread on the experimental plate and was, therefore, responsible for the transformants
observed.

a) Determine total mass (in pg) of pAMP used in Step 8.
[Mass = Concentration x Volume].

 

b) Calculate the total volume of cell suspension prepared. [Steps 2 and 13]

 

c) Calculate the fiaction of the total cell suspension that was spread on the plate.
[Volume Suspension Spread/Total Volume Suspension (part b)= Fraction Spread]

 

(1) Determine the mass of pAMP in cell suspension. [Total mass of pAMP (part a) X
fraction spread (part b) = Mass pAMP Spread].

 

80

e) Determine number of colonies per ug PAMP. Express answer in scientific
notation. [Colonies Observed/Mass pAMP Spread (part c) = Transformation
Efficiency].

 

 

4. What factors might influence transformation efficiency? Explain the effect of each
you mention.

 

 

 

 

 

 

81

APPENDIX AVI.

GEL ELECTROPHRESIS PRE-LABORATORY

Instructor’s Guigg;

1. Key Concepts
What types of organisms have restriction enzymes?
Q bacteria

Define restriction enzymes.
Q A degradative enzyme that recognizes and cuts up DNA (including that of certain
phages) that is foreign to a bacterium.

Define recognition sequence.
Q A specific sequence of nucleotides at which a restriction enzyme cleaves a
DNA molecule.

II. How do restriction enzymes work?
Why use restriction enzymes?
Q They cut DNA only within very precise recognition sequences.

Recognition sites are palendromic, what does that mean?
Q This means that the recognition sequence on one DNA strand reads in the opposite
direction on the complementary strand.

1]]. Cutting DNA with Restriction Enzymes
What does the prefix ”micro" mean?
Q the prefix "micro-” indicates one-millionth. It is symbolized by p,

IV. Gel Electrophoresis
What is gel electrophoresis?

Q a procedure that separates molecules on the basis of their rate of movement through a
gel under the influence of an electrical field.

In gel electrophoresis, what is the direction of movement affected by?
Q the charge of the molecules

82

1n- geE
Q rh
gt‘

What

When
faiths
! Bl

In gel electrophoresis, what affects the rate of migration?

Q the rate of movement is affected by their size and shape, the density of the
gel, and the strength of the electrical field.

What is the charge of DNA?
Q DNA is a negatively charged molecule

When DNA has been cut by restriction enzymes, what fragments will migrate the

farthest?

Q Because the smallest fragments move the most quickly, they will migrate the farthest
during the time the current is on.

V. Design of the Experiment

Where are the 3 samples of DNA taken from
Q a virus, the bacteriophage lambda

What are the 3 samples of DNA being used
Q One sample will be uncut DNA, one will be incubated with the restriction
enzyme HindIII, and one will be incubated with EcoRI.

VI. Preparing the Gel

List the 4 main steps in preparing the gel

Q Mix agarose & buffer

Q boil mixture in microwave

Q cool mixture to 65°C and pour into gel tray
Q let gel solidify at room temperature

What end are the wells placed in the electrophoresis chamber?
Q the wells are placed at the negative electrode end

VII. Loading the Gel

What two mixtures are added to the DNA and why?

Q tracking dye- the molecules move at a faster rate than that of the DNA so you can see
when to shut off the apparatus.

Q sucrose or glycerol to make the mixture dense enough to sink into the well.

83

X.Ri

When
8 W

Siairiir
“hat c
1 ad;

mt

XL An
How dc
”Shim
Q. 1179‘;

DX.

Complez

 

VIII. Filling the Wells

What do you want to avoid doing while filling the wells?
Q do not puncture the bottom of the well with the micropipette!

IX. Electrophoresis
When you turn on the power, what should move toward the positive electrode?
Q the DNA / tracking dye

X. Running the Gel

When do you turn off the power?
Q When the tracking dye has moved near the end of the gel.

Staining and Photographing the DNA
What do you have to do to see the DNA

Q add methylene blue, which will bind to the DNA and then rinse the gel repeatedly
with water, so that the dye washes off the gel.

XI. Analysis of Results 11

How do researchers determine the size of DNA fragments produced with a particular

restriction enzyme?

Q they run the unknown DNA alongside DNA with known fragment sizes. The known
DNA acts as a marker.

Complete the following table

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0:“ Actual Base Pairs (hp) {Measured Distance (mm)
our DNAII HindIII * ‘ ' ‘ '
L. _L J W 23,130 11
297;}? a at: I-{ii :: is ‘ ‘
' ' . 9416 f6
_ ‘ - .or , .
. — .j t 39 . k—
"'."' — .22.: 2322 29
u in.=-,'r'.’.‘34.".s‘fi'i.'-‘r-I‘-.’.‘§¢>‘:IL.L‘-J-' win-E :33 ' I 32
564 l , 46

 

 

 

 

 

 

Read the number of base pairs in the unknown fragment fiom the graph. Record your
answers in the table below.

 

Distance Interp elated
Migrated (mm) Fragment Size
(in base pairs)

 

 

 

 

 

 

Fragment 1 l 20 N I 4796 ‘
Fragment 2 [TN Flo-5-
Fragment 3 I 36 l I 1268 N

 

 

 

 

 

 

 

'— WW.— . ‘

85

GEL ELECTROPHRESIS PRE-LABORATORY

1. Key Concepts
What type of organisms have restriction enzymes?

 

Define restriction enzymes.

 

 

Define recognition sequence.

 

 

II. How do restriction enzymes work?

Why use restriction enzymes?

 

Recognition sites are palendromic, what does that mean?

 

 

III. Cutting DNA with Restriction Enzymes
What does the prefix "micro" mean?

 

IV. Gel Electrophoresis

What is gel electrophoresis?

 

 

In gel electrophoresis, what is the direction of movement affected by?

 

In gel electrophoresis, what affects the rate of migration?

 

 

What is the charge of DNA?

 

When DNA has been cut by restriction enzymes, what fragments will migrate the
farthest?

 

 

V. Design of the Experiment

Where are the 3 samples of DNA taken fi'om

 

What are the 3 samples of DNA being used

 

 

VI. Preparing the Gel

List the 4 main steps in preparing the gel

 

 

 

 

What end are the wells placed in the electrophoresis chamber?

 

VII. Loading the Gel

What two mixtures are added to the DNA and why?

 

 

87

VIII. Filling the Wells

What do you want to avoid doing while filling the wells?

 

1X. Electrophoresis

When you turn on the power, what should move toward the positive electrode?

 

X. Running the Gel

When do you turn off the power?

 

Staining and Photographing the DNA
What do you have to do to see the DNA

 

 

XI. Analysis of Results [1

How do researchers determine the size of DNA fi'agments produced with a particular

restriction enzyme?

 

 

Complete the following table

 

Measured Distance (mm) A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

”1“ Actual Base Pairs (lip)
DNA! DNAII HindIII
up»; ”at ~-:-.-.".-":-. :3
‘ 7 E 94 l 6 a
- L '13
- = .* 6557
.. :... 4361 [:1
_. "’ g3 , . -
"" - .2— 2322
-- 1’53 _ ,
564

 

 

 

 

 

 

 

88

 

Read the number of base pairs in the unknown fragment from the graph. Record your
answers in the table below.

 

Distance 1 Interp elated
Mgrated (mm) Fragment Size
(in base pairs)

 

 

 

 

 

 

 

Fragment 1 r [——
Fragnent 2 r‘ ’ ,..............
Fragment 3 r.- l “l

 

 

 

 

 

89

APPENDIX AVII.

GEL ELECTROPHORESIS OF A. DNA LABORATORY

Instggggr’g Guide:

Time Frame: 1 day field trip + 1 class period

Target Group: Advanced Placement Biology Students

Preparation of materials: The classroom set includes materials for 5 groups of students
to perform the lab experiment.

1. Supplied ready to use:

\'\'\\'\'\\'\'\\\

Instant Lambda DNA (long white tubes)

Instant Restriction Enzyme EcoRl (red tubes)

Instant Restriction Enzyme HindIII (green tubes)
Instant Restriction Enzyme BamHl (blue tubes)
Empty Microtubes (colored yellow)

Electrophoresis Box Gel Comb

Microsyringe Tips & Microsyringes

Bromophenol Blue Loading Dye (screw-topped tubes)
Foam Tray, for Holding Microtubes

Wires with Alligator Clips

0.8% agarose gel, made by dissolving the agarose powder in diluted
electrophoresis buffer: 1.2g of agarose in 150ml 1x TbE solution will give
approximately 15 gels (each gel box holds about 10 mL agarose).

(3! Note: Do not dissolve the agarose in water. The easiest way to
dissolve the agarose is to use a microwave oven. Heat to
boiling to dissolve agarose. Stir fi'equently. Cool to 60 deg C
before pouring if solution is prepared immediately before use.
Caution: loosen all caps on containers when heating in a
microwave.

2. Supplied, but requiring preparation:

I

J

Carbon Fiber Electrode Tissue, cut to approximately 42 x 22 mm

4! Note: The carbon fiber material may release small fibers, which can
cause minor skin irritations. It is a wise precaution to wear protective
gloves.

Electrophoresis (TBE) Buffer Solution, prepared from the 20x concentrate
supplied in the kit; add 25mL of 20x TBE to 475mL distilled or deionized

water
(3! Note: the buffer solution may be saved and reused several times.

90

J Carolina BLU tm concentration DNA stain

3. Materials needed, but not provided:

I
J
I

J

9V Batteries, 3 per gel
Distilled water to rehydrate the lambda DNA

Water bath set at 37°C, so that the DNA may be incubated with the
restriction enzymes

Marking pens, to label the wells

Background information for the teacher
1. Casting agarose gels:

a.

Ensure that the comb teeth are set completely down on their "shoulders"
when placed on the gel box, and covered to the precise depth with the
agarose gel. Too shallow a well will result in some DNA splashing out
due to vigorous unloading of the digital micropipette by students with
unsteady hands. Wells should be able to accommodate 15 to 50 uL
samples. Too deep a well may have a "shouldered connection" to the next
well, allowing mixing with DNA cut by other restriction enzymes - or the
controls - and the resulting bands will be blurred and indistinct.

Pour the gels on unmovable counters, because it is inevitable that some
students will bump tables while the gel is setting.

Removing the comb afier gel solidification must also be done very
gently. Slowly lifting one end at a time reduces the number of ripped gel
beds. (Note: Adding the running buffer before comb removal also helps.)

2. Loading gel beds:

a.

Be sure that students have identified their well sequence as starting from
a particular side and have recorded this information in their lab handout,
as well as what sample they have put into each well. Remind them to
change pipette tips after loading each well.

Using two hands on the pipette will help students hold them steady (one
should be close to the base and the other on the handle with a thumb on
the button). Instruct them to slowly expel the contents of the pipette,
being sure that the tip is not poking through the bottom of the well.

3. Electrophoresis:

a.

b.

Connect the red lead to the positive pole (anode - red) of the battery setup
and the black lead to the negative pole (cathode - black) of the batteries.
Be sure to watch for dye movement as the experiment begins to ensure
that the apparatus is working properly and that the electrodes have been
plugged into the appropriate pole.

When finished, turn the power off first; unplug the power supply from
the outlet, and unplug the leads to the electmphoresis chamber.

91

d. Gingerly move the gel to a disposable plastic tray for staining. Make sure
that students wear gloves.

4. Disposal of culture materials:

a. At the end of the laboratory period, the teacher should collect all bacterial
cultures and other instruments that have come into contact with the
cultures. Disinfect materials and glassware using a 10% bleach solution.

Sources of materials: Carolina Biological Supply Co (catalog # 21-1010)

Table 1. 13le and HindIII m” HM” Non
Fragment srze m Base Parrs

 

 

 

22011 11111111
Acmtlbp 11mm
21.226 23,130
1,421 9,416

5304* 6.557
5,643‘ 4,361

 

 

 

 

 

4,878 2,322
3.530 2.027
’ 564
lb .
‘Mfmsinslebmd Sfiiiirid3§§es‘“c"°“
Figure 1. Ideal Restriction
Digest of it DNA

Problems encountered: Some of the students’ gels produced staircase bands very close

to the wells. One reason for this result might have been the lack of buffer on an uneven
desk.

92

References:

l. Student’s Guide to Exploring Restriction Analysis and Electrophoresis of
DNA

2. Teacher’s Guide to Exploring Restriction Analysis and Electrophoresis of
DNA

3. Advanced Placement BIOLOGY Laboratory Manual for Students Exercises 1-
12 Edition D

4. Advanced Placement BIOLOGY Laboratory Manual for Teachers Exercises
1-12 Edition D

Answers to Questions in the Student Laboratory

Graph Title:
§ Migration Distance of Restriction Fragments

1. For which fragment size(s) was your graph most accurate? For which fragment size(s)
was it least accurate? What does this tell you about the resolving ability of agarose-
gel electrophoresis?

Q» The “best fit” line drawn on the data passes closes to the points representing
medium-length fragments. Therefore, the graph was most accurate for these
fragments. Points/fragments at either extreme of the range were least accurate.
Larger fragments might be resolved on a less concentrated agarose gel, while
smaller fragments would resolve on a more concentrated gel.

2. Discuss how each of the following factors would affect the results of electrophoresis:
a) Voltage used.
Q Increased voltage will increase the rate of migration for all fi'agments (too
high a voltage will melt the gel)

b) Running time
Q. Increased running time will increase the distance that the fragments migrate

c) Amount of DNA used
‘3 Increased DNA amounts will increase the intensities of the fi'agment bands
afier staining.

d. Reversal of polarity
fix DNA fragments will move in opposite direction.

3. Two small restriction fragments of nearly the same base-pair size appear as a single
band, even when the sample is run to the very end of the gel. What could be done to
resolve the fragments? Why would it work?

‘3 The fragments could be separated on a more concentrated agarose gel. At higher
concentration, the pore size of the gel would be more suited to separation of

93

smaller fragments of similar size. The fiagments could be transferred to another
electrophoresis chamber that contains a gel other than agarose, such as
polyacrylamide. The different affinities between the media and the molecules
might allow for a better separation. Another possibility would be to digest the
fragments with different restriction enzymes and then utilize the reconstruct of the
fragments.

. What is a plasmid? How are plasmids used in genetic engineering?

\ A plasmid is a circular piece of double-stranded DNA capable of self replicating
and being expressed in a bacterial cell. When DNA is spliced into the plasmid a
recombinant plasmid is formed. These plasmids may be taken up by bacteria by
transformation.

. What are restriction enzymes? How do they work? What are recognition sites?
ix Restriction enzymes are proteins capable of recognizing a specific DNA sequence
(the restriction site) and catalyzing the hydrolysis (cutting) of this DNA molecule.

. What is the source of restriction enzymes? What is their fimction in nature?

ix Restriction enzymes are generally isolated from prokaryotic organisms. In nature,
they function to protect bacteria fi'om foreign DNA by cutting it into
nonfunctional pieces.

. Describe the firnction of electricity and the agarose gel in electrophoresis.

‘3 In electrophoresis, agarose acts to sieve the DNA. The electricity creates a
positive charge towards which the negatively charged DNA fragments migrate.

. If a restriction enzyme digest resulted in

 

 

 

 

DNA fragments of the following sizes: "“9“”
4,000 base pairs, 2,500 base pairs, 2,000
base pairs, 400 base pairs, sketch the
resulting separation by electrophoresis.
Show starting point, positive and negative fl.
electrodes, and the resulting bands.
2500
3:.
J39...
Poslive

94

9. What are the functions of the loading dye in electrophoresis? How can DNA be
prepared for visualization?

t». The loading dye functions to increase the density of the DNA so that it will
remain in the wells, and to provide a dye which will run at a more rapid rate thatn
the DNA fragments. For visualizaiton, DNA fi'agments can be stained using
methylene blue or other stains.

10. Use the graph you prepared from your lab data to predict how far (in cm) a fragment
of 8,000 bp would migrate.
ix A fragment of 8,000bp would migrate a distance between the fragments of 9,416
and 6,557 on the HindIII table (Answers for the standard curve in graph will

vary).

11. How can a mutation that alters a recognition site be detected by gel electrophoresis?
t. A point mutation causes a change in the base-pair sequence in the recognition
size. The altered sequence cannot be recognized by the active site of the
restriction enzyme so that the enzyme will not cut a particular size. Thus a larger
fiagment will be generated.

95

GEL ELECTROPHORESIS OF A. DNA LABORATORY

INTRQDUQTIQN

Background Information

Bacteriophages are viruses that invade bacteria. Bacteriophages must take over the
molecular machinery of their bacterial hosts to reproduce. The phage lambda (7L) preys
upon Escherichia coli. It can either multiply within its host destroying it (the lytic cycle),
or the A DNA can insert into the bacterial chromosome and remain dormant there for
several generations (the lysogenic cycle). An environmental trigger, such as ultraviolet
light, activates the lytic cycle.

Phage A is a relatively simple organism. It consists of a double-stranded length of DNA
wrapped around a core of protein and encapsulated in a proteinaceous coat. The entire
genetic makeup (genome) of A has been sequenced and is 48,502 base pairs long. Within
this genome are genes that code for the virus's protein coat, bursting (lysis) of the
bacterial cell, integration of 1 DNA into the host's chromosome and so on. The order in
which these genes are activated is important. For example, it would be of little benefit to
the virus if the host bacterial cell were lysed before new virus particles had been
assembled. Consequently, 1 has evolved an elaborate system of gene regulation that has
been studied in great detail.

Relatively little of the A genome is required to package DNA and deliver it into bacterial
cells. About 20,000 base pairs can be deleted from its central region and replaced with
DNA from another organism, without affecting the phage's viability. Several specially
constructed forms of A can be used by molecular biologists to transfer new genes into
bacteria.

Enzymes that cut up DNA are called restriction endonucleases. Bacteria make them to
restrict the proliferation of invading viruses (such as the bacteriophage A). Different
restriction enzymes cut at specific sequences of bases in the DNA. The bacterium's own
DNA is protected by the addition of methyl (-CH3) groups to adenine (A) or cytosine (C)
bases at the sites that normally are ”recognized” by the enzymes. Figure 1 shows
recognition sites for three such enzymes.

Restriction enzymes or restriction endonucleases are essential tools in recombinant
DNA methodology. Several hundred have been isolated fiom a variety of prokaryotic
organisms. Restriction endonucleases are named according to a specific system of
nomenclature. The letters refer to the organism (bacteria) from which the enzyme was
isolated. The first letter of the name stands for the genus name of the organism. The next
two letters represent the second word or the species name. The fourth letter (if there is
one) represents the strain of the organism. Roman numerals indicate whether the
particular enzyme was the first isolated, the second, or so on.

96

Examples:

13le E = genus Escherichia
co = species coli
R = strain RY 13
I =first endonuclease isolated from this bacteria

BamHl B= Bacillus
am = amyloliquefaciens
H= strain
1 =first endonuclease isolated from this bacteria

Hind] 11 H = Haemophilus
In = influenzae
Rd = strain
III= third endonuclease isolated from this bacteria

Restriction endonucleases recognize specific DNA sequences in double-stranded DNA
(usually four to six base pair sequences of nucleotides) and digest the DNA at these sites.
The result is the production of fragments of DNA of various lengths. Some restriction
enzymes cut cleanly through the DNA helix at the same position on both strands to
produce fragments with blunt ends (Figurel). Other endonucleases cleave each strand
off-center at specific nucleotides to produce fragments with "overhangs" or sticky ends

(Figure 2).

 

 

[Cleavage by EcoRI produces sticky ends i l Cleavage by HaeIII produces blunt ends

A}: n4}—

 

 

'11» me me ' ‘
3:12: :23:
CTTAA G CTTAA G CG CG CG CG
11.1-1- géfi 1: g:
m GGCC
GAATTC cccg
TTAAG u
I: GGCC
21111: flew
‘3 Figure2

Figure 1

97

By using the same restriction enzyme to "cut” DNA from two different organisms,
complementary "overhangs" or sticky ends will be produced and can allow the DNA

from two sources to be ”recombined.”
l I h i

Gel electrophoresis can be used to separate DNA fragments of different sizes. First a gel
is cast from agarose, a very pure form of polysaccharide, which is obtained fi'om
seaweed. At one end of the slab of gel

are several small wells, made by the

teeth of a comb that was placed in the

gel before it sets. A buffer solution is
poured over the gel, so that it fills the
wells and makes contact with the
electrodes at each end of the gel. Ions

in the buffer solution conduct

electricity. The buffer also stops the gel
from drying out. 1

An electrical current is applied to the
electrodes, setting up an electrical field
across the gel. When any molecule
enters an electrical field, the mobility or speed at which it will move is influenced by the
charge of the molecule, the strength of the electrical field, the size and shape the
molecule, and the density of the medium (gel) through which it is migrating Phosphate
groups in the backbone of nucleic acids give the DNA fragments a negative electrical
charge, so that the DNA migrates through the gel towards the positive electrode. In
agarose, the migration rate of linear fiagments of DNA is inversely proportional to their
size. Small fragments move quickly through the porous gel, whereas larger fragments
travel more slowly. In this way the pieces of DNA are separated by size.

 

The invisible DNA fragments are mixed with a small volume of loading dye. This dye is
dissolved in a dense sugar solution, so that when it is added to the wells, it sinks to the
bottom, taking the DNA with it. The loading dye/ DNA moves through the gel, so that
the progress of the electrophoresis can be seen.

After electrophoresis, the gel is stained to reveal the DNA, either as a smear (many
fragments of a wide range of sizes) or bands (each band is comprised of numerous DNA
fragments of a similar size). Within a smear, specific bands can by highlighted using
probes, which bind to particular sequences of DNA (or RNA). Some probes are
radioactive and for reasons of safety cannot be used in schools.

A. DNA, restricted with the enzyme I-IindIII, is often run on gels alongside other DNA.

The A fragments, which are of known sizes, can then be compared to other DNA
fragments, thereby giving an indication of their sizes.

98

Objectives
V Understand how gel electrophoresis separates DNA molecules present in a mixture

and use electrophoresis to separate DNA fiagments.

G’ Understand how restriction endonucleases function and demonstrate how they are

used
<7 Understand the importance of restriction enzymes to genetic engineering experiments.
0' Determine unknown DNA fragment sizes when given DNA fragments of known size.

METH D/ PR ED
Microsyringe

l.
The microsyringes are designed to be used with

disposable tips. Each tip should be used only once,
then thrown away. The tips are marked at 2 and 10 ':
microliters (uL), allowing small volumes of liquid to (
be dispensed with precision. Best results will be

obtained if you observe the following precautions:

2.

Using the microsyringes

 

Never pull the plunger right out of the
microsyringe (when the plunger is re-inserted the

seal may become damaged).
Before you load the microsyringe, pull the plunger out a little (1-2 mm). This will
give you some extra air with which to expel the last drop of liquid from the tip.

0 When you dispense liquids, hold the microsyringe as near to vertical as
possible, and at eye level so that you can see what you are doing.

Remove the droplet of liquid from the end of the microsyringe tip by touching
the inner wall of the nricrotube.

0 Do not touch the point of the microsyringe tip with your fingers. There are
proteases and DNases in your sweat, which may contaminate and degrade the

reactants.

Rehydrating the dried lambda DNA
a) Add 100 uL of distilled water to the 1 DNA tube (the narrow white tube). Use the

louL graduation on the tip 10 times to dispense this.

b) Cap the tube lightly and allow it to stand for 5 minutes

c) Hold the closed tube firmly at the top, then flick the side repeatedly with a finger
to mix the contents. Do this for one full minute. [)5 by accident, you drop the tube

or find that drops of liquid are scattered inside the tube, tap it firmly on the bench
several times to return the liquid to the bottom.

99

3.

d) Allow the tube to stand for an additional 5 minutes.

e) The A DNA solution should look slightly opaque. Note: It is vital that the DNA is
thoroughly mixed with the water. To ensure that this is done, always draw the

DNA solution up and down in the microsyringe tip a few times.

Notes

0 DNA sticks to glass, For this reason, disposable polypropylene tubes (to
which DNA does not arfliere) are used in this kit. The white tube contains
long of dried 2. DNA. A buffer in the tube helps to maintain the, stability of
the DNA once it is in solution. Blue dye has also been added to help you
judge when the DNA has dissolved. It is very important to rehyrb'ate the DNA
using the correct volume of water so that the final concentration of the bufi'er

is appropriate for the activity of the enzymes.

0 A common cause of failure is inadequate mixing of the DNA with water. A
"blank ” gel usually indicates that the DNA has not been dehydrated at all.

Lanes that are overloaded alongside empty ones are often due to uneven
mixing of the solution; most of the DNA lies at the bottom of the tube, so it

ends up in the last enzyme tube filled

Cutting DNA w/ restriction enzymes

0)

b)

Add a flesh tip to the microsyringe. Put 20uL of
A DNA solution into an enzyme tube of your
choice. Mix the liquid and the dried enzyme by
carefully drawing the liquid up and down in the
tip a few times. The liquid in the engine tubes
should have a distinct blue hue, but there should
be no concentration of dye at the bottom of the
tube.

Repeat this for each enzyme tube and the yellow

“control" tube, using a fresh tip each time to
prevent cross-contamination between the tubes.

Cap each tube with a matching colored lid.

IMPORTANT!
,_ Mix well before
L9". and after
' dispensing

   

DNA

solution . .
‘5"3» .

20 1.1 L

I
EcoR BamHl Hindlll Uncut

1 DNA

(Control)

Put the tubes in the foam rack provided (this may be cut into sections, if desired).

Incubate the tubes at 37°C, in a water bath or incubator, for between 30 and 45

minutes.

Place the tubes in a hot water bath at 65°C for 10 minutes to denature the

enzymes.

100

Notes
. Each tube contains at least 10 units of restriction enzyme. A unit of enzyme
will cut lug of A DNA in one hour at 37°C

The enzyme tubes are color coded:
EcoRl BamHl Hindlll Empty
(Pink) (blue) (green) (yellow)

0 Like the dried DNA, the enzymes have been preserved with an appropriate
buffer and a blue dye. It is essential to add the correct volume of liquid to the

tubes, and ensure that their contents are thoroughly mixed. This is because

EcoRI and BamHI can be less specific if used in excess or with the wrong
concentration of buffer. For instance, in an inappropriate buffer concentration

EcoRI may cut DNA molecules at the sequence AATT instead of the longer,
more specific sequence, GAATTC,

4. Preparing the agarose gel
a) Use a microwave oven, hotplate or a boiling water bath to melt some agarose gel

1?)

d)

(0.8% made up in TBE buffer). Ensure that no lumps or fibers remain in the
molten agarose. Store the molten agarose in a water bath at 55-60°C until it is

needed.

Place the gel box on a level surface, where you can leave it undisturbed for the
next 20-30 minutes. This is necessary because if the gel sets at an angle, the DNA
fi'agments will not run evenly through the gel. Place the comb at one end of the

get box. ,
1 Molten agarose
Pour about 10 mL of molten agarose 55-60°C
into the gel box so that it fills the , ,
central cavity and flows under and f/ i"
between the teeth of the comb. Try
not to spill liquid into the areas at
either end (if some gel does flow
over, just leave it to set-you can

scoop it out later).

Gel
Box

 

Leave the box, undisturbed, until the
get has set hard (agarose gel is opaque when set).

Cut two pieces of the carbon fiber tissue, each about 42 mm x 22 mm. These will
be the electrodes at either end of the gel box. Put the electrodes to one side until

you have loaded the gel.

101

5. Loading the gel . .
8) Pour slightly more than 10 mL of TBE buffer solution into the gel box. The liquid
should just cover the surface of the gel and flood into the areas that will hold the

electrodes.
b) Very gently ease the comb from the gel, allowing the buffer solution to fill the
wells lefi behind. Take care not to tear the wells.

c) Put the gel box where it will remain undisturbed while the gel is being run

It is easier to see what you are doing next if the gel box is placed on a dark surface.
A (tentatively, a strip of black insulating tape can be stuck onto the bottom of the box

beneath the wells.

d) Label on the side of the gel box with. a marker pen which DNA you will put into
the well, include your undigested lambda phage DNA (control).

 

e) Put a clean tip on the microsyringe. Add 2uL of Loading
loading dye to the tube containing the DNA you 3d”
wish to load. Mix the dye and the DNA sample 2 WA 3“” A
very thoroughly by drawing the mixture up and 5:,
down in the microsyringe tip. \b,’ (— i W
fl Pipette the loading dye/DNA mixture into one of 15., 1";
the labeled wells, holding the tip above the well ‘1 i
but under the buffer solution The loading dye is 7:7 \J
M! r We?

denser than the bufler and will move into the
well. Take great care not to puncture the bottom

of the well with the microsyringe tip.

g) Repeat steps d-f with Max well ..
each DNA digest and before 'Oadi'ig l
the “control" tube with .1 If; _.;t_.: . , .

   

no restriction enzyme.
Remember to use a new
tip for each digest, to
avoid cross-
contamination.

\
TBE buffer solution
2—3 mm depth over gel

   

6. Running the gel

0) Fit one electrode at each
end of the box as shown in
the illustration. Join the
electrodes to batteries using

102

 

 

wires with alligator clips. Sufficient batteries (in series) should be used to give a total
voltage of no more than 45 volts. Ensure that the positive terminal of the battery is

connected to the electrode farthest from the wells.

b) Three nine-volt batteries are used for 3.5
hours. Discard batteries afier 2 runs. Serious
or lethal electrical shock could result if the
equipment is connected directly to a power

Use no

supply.
more than

~15 volts

    
 

WARNING!

c) Check that contact is made between the
buffer solution and the electrodes (add a
little more buffer if necessary). When the
current is flowing you should see bubbles on
the electrodes. Allow electrophoresis to
proceed undisturbed, for several hours. In a

warm room it may be necessary to put the
gel box in a plastic bag to reduce evaporation of the bufler.

d) When the blue loading dye has reached the end of the gel disconnect the batteries. If
you leave the batteries connected the DNA will run of the end of the gel!

e) Rinse the alligator clips in tap water and dry them thoroughly to prevent corrosion.

Notes
Ions in the TBE bufl‘er solution used in and above the gel conduct electricity. This

buffer solution is alkaline. Under alkaline conditions the phosphate groups of the
DNA are negatively charged and therefore the DNA moves towards the positive
electrode (anode) when a current is applied. EDTA in the buffer "mops up" or
chelates divalent cations, which helps to prevent damage to the DNA because
such ions are required (as co- factors) by enzymes that degrade DNA.

The loading dye does not affect the DNA, but moves through the gel slightly
ahead of all but the smallest of DNA fragments (here, those smaller than about
300 base pairs).

At higher voltages the DNA fragments will move faster, but the separation of
bands will not be so clear. This is because the movement of fragments of greater
molecular mass is increased differentially at higher voltages, so that they "catch
up " with the smaller fragments. At higher voltages the DNA bands may also

become distorted.

7. Staining the DNA
a) Remove and dispose of the electrodes. Pour off the buffer solution.

b) Wear some plastic gloves to prevent the stain from touching your skin.

103

c) Pour about 10 mL of DNA-staining solution (Carolina BL Um concentrate) onto
the surface of the gel. Leave it for exactly 4 minutes, then return the stain to a

bottle for re-use.

d) Rinse the surface of the gel very carefully with cold water. Repeat this 3 or 4
times to remove any excess stain. Pour the water away. Use distilled or deionized

water, as this will help to preserve the DNA bands.

e) Put the gel box in a plastic bag, to prevent the gel from drying out, then leave the
gel to "develop" overnight.

1) The next day, destain the gel again, this time leaving the water in the box and
changing it about 4 times (once every 2 hours). Destaining will gradually wasb

away any "background" stain.

Notes
Uncut DNA does not move far through the gel, and forms a single, broad band.

A DNA cut with a restriction enzyme will form distinct bands of known sizes
Stained gels may be stored indefinitely in a sealed plastic bag if kept in a

refiigerator.

Picture and diagram: Determining Fragment Size
a) Examine your stained gel.

b) Afler observing the gel, carefully wrap it in plastic wrap and smooth out all
wrinkles.

c) Using a marking pen, trace the outlines of the sample wells and the location of the
bands.

d) Remove the plastic wrap and flatten it out on a white piece of paper on the
laboratory bench Save the gel in a Ziplock" plastic bag. Add several drops of
buffer. Store at 4'C. You can make your measurements directly from the marked

plastic wrap.

LEM
Background Information:
The size of the fragments produced by a specific endonuclease (15le in this laboratory)

can be determined by using standard fragments of known size (fragments produced by

HindIII). When you plot the data on semilog graph paper, the size of the fragments is
expressed as the log of the number of base-pairs they contain. This allows data to be
plotted on a straight line. The migration distance of the unknown fragments, plotted on

the x-axis, will allow their size to be determined on the standard curve.

104

Standard Curve for Hind Ill
1. Measure the migration distance (in cm) for each HindIII band on your gel. Measure
from the bottom of the sample well to the bottom of the band. The migration distance

for the largest standard fragment (approximately 23,120 base-pairs) nearest to the
origin does not need to be measured. Record these measurements in the Table below.

Table 1: Distance Hind II produced fggments migrate in ggawrosegellcm)
Acutal bp Measured Distance (cm)

23,130

 

 

 

 

[I
[ 9,416
[ 6,557
[

 

4,361
I 2,322 (may form a single band)
I 2,027 (may form a single band)
{ 570 (may notbedetected)
[ 125 (may not be detected)

2. Plot the measured migration distance for each band of the standard HindIII digest
against the actual base pair (bp) fiagment sizes given in Table 1 using the semi-log

 

 

 

 

 

 

 

graph paper-
3. Draw the best-fit line to your points. This will serve as a standard curve.

B. Interpolated Calculations for EcoRI
From the standard curve for HindIII, made fi'om known fi'agment sizes, you can calculate

fragment sizes resulting from a digest with EcoRI.

1. Measure the migration distances for each 13le band. Record the data in Table 2.

 

 

 

 

 

 

 

 

 

 

 

Table 2: Distance EcoRIJroduced fragments migrate in agarose gel (cm)
L Band # Measured Distance (cm) Interpolated bp Actual hp
L Band 1
L Band 2
L Band 3
L Band 4
L Band 5
L Band 6

 

2. Determine the sizes of the fi'agments of phage lambda DNA digested with EcoRI.
Locate on the X-axis the distance migrated by the first EcoRI fragment. Using a ruler,
draw a vertical line from this point to its intersection with the best-fit data line. Now
extend a horizontal line fi'om the intersection point to the Y-axis. This point gives the

105

base-pair size of this EcoRI fragment. Repeat this procedure and determine the remaining
EcoR] fragments. Enter your interpolated data in Table 2.

Your teacher will provide you with the actual base pair data

3. Compare your results to these actual sizes. Note: this interpolation technique is not
exact. You should expect as much as 10% to 15% error.

106

Graph Title:

 

  
 

log Base Pair length

Distance Migmted in Cm

107

INTERPRETATION AND CONCLUSIONS
Questions

1. For which fiagment size(s) was your graph most accurate? For which fiagment size(s)
was it least accurate? What does this tell you about the resolving ability of agarose-
gel electrophoresis?

 

 

 

 

2. Discuss how each of the following factors would affect the results of electrophoresis:

(1) Voltage used.

 

 

 

e) Running time

 

 

 

t) Amount of DNA used

 

 

 

e. Reversal of polarity

 

 

 

3. Two small restriction fiagments of nearly the same base-pair size appear as a single
band, even when the sample is run to the very end of the gel. What could be done to
resolve the fi'agments? Why would it work?

 

 

 

 

4. What is a plasmid? How are plasmids used in genetic engineering?

108

 

 

 

 

. What are restriction enzymes? How do they work? What are recognition sites?

 

 

 

 

. What is the source of restriction enzymes? What is their function in nature?

 

 

 

 

. Describe the function of electricity and the agarose gel in electrophoresis.

 

 

 

 

. If a restriction enzyme digest resulted in DNA fragments of the following sizes: 4,000
base pairs, 2,500 base pairs, 2,000 base pairs, 400 base pairs, sketch the resulting
separation by electrophoresis. Show starting point, positive and negative electrodes,
and the resulting bands.

 

 

 

 

109

9. What are the fianctions of the loading dye in electrophoresis? How can DNA be
prepared for visualization?

 

 

 

 

10. Use the graph you prepared from your lab data to predict how far (in cm) a fragment
of 8,000 bp would migrate.

 

 

 

 

11. How can a mutation that alters a recognition site be detected by gel electrophoresis?

 

 

 

 

110

APPENDIX AVIII.

ALL YOU HAVE ALU LABORATORY

Inst r’ i
Target Group: Advanced Placement Biology students

Time Frame: Alu Insertion Polymorphism Kit requires several different activities. Plan
your time as follows:

 

 

 

 

 

 

 

 

 

Dal Time Activity
One or more days 30 min. / Pre-Lab: Mix and aliquot Chelex
before lab 30 min. / Pre-lab: Set-up work stations
40 min. / Pre-Lab: Prepare gel solution and cast Jels
Lab period 1 30 min. / Isolate Squamous Cell DNA
10 min. J Set-up PCR reactions
70 min. / Amplify DNA in thermal cycler.
Lab period 2 15 min. / Load DNA samples into gels
20+ min. Electrophoresis
10 min. / Post-lab: Photograph gels
Lab period 3 40 min / Results and Discussion

 

Background information for the teacher:

/ Upon receipt of the kit, store proteinase K, PV92 Primer/Loading Dye Mix, and
pBR322/ BstN I markers in freezer (approximately -20°C). Other materials may be
stored at room temperature (approximately 25°C).

\/ Arrange to borrow or use thermal cycle from nearby college. Another method would
be to use three separate water baths.

/ prossible arrange an in house field trip and have students do all of the lab in one
day.

/ Prior knowledge of basic methods of gel electrophoresis and staining of DNA is
presumed.

Materials:
/ The materials in the Alu Insertion Polymorphism Kit are suflicient for 25 reactions.
The kit includes:
0 Chelex~ 100 resin, 1.5 g
. Ready-to-Go PCR Beadsm, 25* (*Ready-to-Go PCR Beads incorporate Taq
polymerase, dNTP's, and MgCIz Each bead is supplied in an individual O.5-ml
test tube.)

111

o Ready-to-Go PCR Beadsm, 25“ (‘Ready-to—Go PCR Beads incorporate Taq
polymerase, dNTP's, and MgCI; Each bead is supplied in an individual O.5-ml
test tube.)

PV92 Primer/Loading Dye Mix, 700u1

PBR322/Bst N I markers, .075ug/ml 1301.11

Instructions, student

Instructions, teacher

5 g agarose

4 latex gloves

/ Items of equipment needed but not provided:

Camera for photographing gels (optional)

Centrifuge, clinical, for 15 ml tubes, minimum 500 x g
DNA thermal cycler, programmable

Electrophoresis chambers for agarose gels
Electrophoresis power supplies

Paper cup, 1 per student

1.5 ml polypropylene tubes, 2 per student

15 ml polypropylene tube, 1 per student

Micropipets, l-lOpJ, 1-20ul, l-200ul, one for instructor or several shared by
students

Micropipets, lOO-lOOOul (or l-ml transfer pipets)
Forceps, 1

Laboratory markers, 1 per student

Micropipet tips, for 1-20ul pipets, 3 per student
Micropipet tips, for loo-1000111 pipets, 3 per student
Water bath, boiling, per 12 students

/ Reagents and supplies needed but not provided:
0 Beakers containing ice
Electrophoresis buffer
lul Ethidium Bromide (stock 0.5) / 1m] buffer
Saline solution, 0.9% NaCl in water, 10 ml per student

Prelgb mg
I Make up a 10% Chelex solution: 1.5g Chelex + 15 ml distilled or deionized water.

1 For each student, aliquot 500p] of 10% Chelex suspension into a 1.5 ml
polypropylene tube. (Be sure to shake the stock tube or draw liquid in and out of pipet
tip several times to re-suspend the Chelex beads each time before pipetting a student
aliquot).

/ For each student, prepare and aliquot 10 ml of 0.9% saline solution into a 15ml
polvpropylene culture tube: 0.9g NaCl per 100 ml distilled or deionized water.

112

V PV92 Primer/loading dye mix may collect in tube cap during shipping. To have full
volume available for student use, pool reagents by spinning tubes briefly in a
microcentrifuge or tapping tube end on desktop.

V Set up one ”boiling water bath per 12 students. This water bath should consist of a
beaker and test tube rack to allow 1.5m] centrifiage tubes to be suspended with the
reaction in the boiling water. The entire tube should not be submerged.

V Prepare student stations, each with the following materials:
l-l .5ml polypropylene tube with 500111 of 10% Chelex suspension
1-1 .Sml polypropylene tube (empty)

Instructions, student

Laboratory marker-permanent

Micropipet tips

Micropipet, 1-10ul or l-20ul

Micropipet, loo-1000111 (or l-ml transfer pipet)

Paper cup

Ready-to-Go PCR Beadm and PCR tube

1-15ml polypropylene tube with 0.9% Saline solution
Styrofoam plate turned upside down to hold (1 .Sml tubes)

V Students will share the following materials:
Beakers containing ice

DNA thermal cycler

Forceps

PV92 primer/loading dye mix

Water bath, boiling

Agarose gels w/ Ethidium bromide
Camera (optional)

Centrifirge, micro

PBR3221 BstN I size markers
Electrophoresis apparatus
Electrophoresis buffer

Micropipet tips

Micropipet, l-lOul or l-20ul

Test tube rack (Styrofoam plate upside down)
Test tubes, 1.5 ml

V Prepare 1.5% agarose solution sufiicient for the number of gels needed to hold all
student samples, by adding 5 g agarose to 333 ml of l X TBE. Each gel will need to
be approximately 8 rmn in depth to accommodate samples of to 25p].

Program thermal cycler with a step file: (30 cycles link to soak file at 4*C)
o 94*C — 30 sec
0 58"C - 30 sec
0 72*C — 30 sec

113

Sources of material:
0 Human Alu Insertion Polymorphism Kit (#21-1232)
Carolina Biological Supply Company,
2700 York Rd, Burlington,
N. Carolina 27215
1-800-334-5551
0 Deiger catalog

Problems encountered:
I only ordered the PCR kit the first time I tried this lab. Fortunately, I had some left over
materials from one of Carolina’s electrophoresis labs. Carolina Biological was nice
enough to give me the conversions for my extra Carolina BLU final concentrate stain to
use for this lab.

0 Final Stain = 50ml of final concentrate + 320ml deO

0 Gel Stain = 1.3g azure blue stain to 1L odegO
One of the bands did not show because there wasn’t enough DNA (due to pippetting
error)

Reference:

0 Human Alu Insertion Polymorphism Kit (#21-1232)
Carolina Biological Supply Company,

0 Roger Herr, Graduate Student, MSU

0 Detection of Alu by PCR,
http://www.accessexcellence.org/AE/AEPC/DNA/detection.html
John Gerlich, Assistant Professor, MSU
Dr. Merle Heideman, Professor, MSU

o Voet & Voet . 2000. Biochemistry

114

ALL YOU HAVE ALU LABORATORY

Background Information

In this experiment, the polymerase chain reaction (PCR) is used to amplify a short
nucleotide sequence from human chromosome 16. The object is to create a personal DNA
fingerprint that shows the presence (+) or absence (-) of the Alu transposable element on
each chromosome at the PV92 locus. Although DNA from various individuals is more
alike than different, many regions of human chromosomes exhibit a great deal of
diversity. Such variable sequences are termed ”polymorphic" (meaning many forms) and
are used for diagnosis of genetic disease, forensic identification, and paternity testing.
Many polymorphisms are located in the estimated 95% of the human genome that does
not encode proteins.

Alu elements are a component of the non-coding DNA of primate genomes. Alu
elements are approximately 300bp in length and derive their name from a single
recognition site for the endonuclease Alu I located near the middle of the Alu element.
Alu elements are thought to be derived from a gene that encodes the RNA component of
the signal recognition particle, which labels proteins for export from the cell.

Alus are transposable DNA sequences that ”reproduce" by copying themselves
and inserting into new chromosome locations. Alu is an example of "selfish DNA" -it
encodes no protein and appears to exist only for its own replication. The human
chromosomes contain an estimated 500,000 Alu copies, equaling 5% of the total genome.
However, it is thought that, at any point in time, only one or several Alu "masters" are
capable of transposing Each Alu insertion is the "fossil" of a unique transposition event
that occurred only once in primate evolution. Thus, all primates sharing an Alu allele are
descended from a common ancestor in whom the transposition first occurred.

An estimated 500-2,000 Alu elements are restricted to the human genome. Most
Alu mutations are "fixed," meaning that both of the paired chromosomes have an
insertion at the same locus (position). However, a number of human-specific Alus
recently have been discovered that are dimorphic -an insertion may be present or absent
on each of the paired chromosomes of different people. These dimorphic Alus inserted
within the last million years, during the evolution and dispersion of modern humans. PCR
can be used to screen individuals for the presence or absence of the Alu transposable
element.

The source of template DNA for this procedure is a sample of several thousand
squamous cells obtained from cheek cells. Cheek cells are obtained by a saline
mouthwash, collected by centrifirgation, and resuspended in Chelex. Then, the samples
are boiled to lyse the squamous cells and liberate the chromosomal DNA. The Chelex
binds metal ions, released fi'om the cells that inhibit the PCR reaction. A sample of the
clear supernatant, containing chromosomal DNA, is combined with a buffered solution of
heat-stable Taq polymerase, oligonucleotide primers, the four deoxynucleotide (dNTP)
building blocks of DNA, and the cofactor magnesium chloride (MgClz). The PCR
mixture is placed in a DNA thermal cycler and taken through 30 cycles consisting of:

115

l 30 seconds at 94°C, while the chromosomal DNA is denatured into single strands,
l 30 seconds at 58°C, while the primers form hydrogen bonds with their
complementary sequences on either side of the PV92 locus, and
l 30 seconds at 72°C, while the Taq polymerase makes complementary DNA strands
that begin with each primer.
The primers used in the experiment bracket the PV92 locus and result in selective
amplification, or copying of that region of chromosome 16. PV92 has only two alleles
indicating the presence (+) or absence (-) of the Alu transposable element. Amplification
of the (-)allele produces a DNA fiagment of 550 bp, while amplification of the (+) allele
produces an 850-bp fragment. Alleles of this size can be readily separated by agarose gel
electrophoresis. Student amplification products are loaded side-by-side in a 1.5% agarose
gel, along with size markers and electrophoresed. After staining with a visible dye, each
student can be scored as one of three Alu genotypes (++, +-, or -)

The PV92 insertion polymorphism was specifiwa selected for use in this
laboratory because it is phenotypically neutral. The Alu insertion element does not
encode protein. PV92 alleles have no known relationship to disease states, sex
determination or any other phenotype. Even though there is no chance of disclosing
phenotypic information about the experimenters, the confidentiality of your PV92
genotypes can be maintained by identifying student samples only by numbers.

Visit the DNA Learning Center WWW site
Wshockanmtml) to view animations on PCR and DNA
fingerprinting. This site will be helpful with answering the pre-lab questions.

W

G’ understand DNA amplification by polymerase chain reaction

0' practice gel electrophoresis technique

0' create a personal DNA fingerprint that shows the presence (+) or absence (-) of the
Alu transposable element on each chromosome at the PV92 locus.

PROCEDURE 1]: ISOLATE CHEEK CELL DNA

E

TERIAL

Laboratory marker

2 -1.5 ml culture tubes

15ml test tube w/ Saline solution, 0.9%
Paper cup

Centrifirge (demo desk)

1000111 Micropipet (for 500g] of 10% Chelex)
10% Chelex

Floating tube rack

Boiling water bath

Forceps

Beakers containing ice

200111 Micropipet

\\\\\\\\\\\\

116

METHQD/ PRxEDURE

1) Use a permanent marker to place your name on two clean 1.5ml tubes and on the
15m] test tube containing 10ml saline (0.9% NaCl) solution.

2) Pour the saline solution into your mouth and vigorously rinse your mouth for 10
seconds. Save test tube for later use.

3) Expel saline solution into the paper cup.

4) Carefirlly pour saline solution from the paper cup back into the original test tube and
close cap tightly. Save paper cup for later use.

5) Place your sample tube, together with other student samples, in a balanced
configuration in a clinical centrifuge and spin for 10 minutes.

6) Carefully pour ofi‘ supernatant into the paper cup. Be carefirl not to disturb the cell
pellet at the bottom of the test tube.

7) Pour supernatant from Step 6 into the sink and rinse down with water.

8) Set micropipet to 500111. Draw 10% Chelex suspension in and out of the pipet tip
several times to suspend the resin beads. Before resin settles, rapidly transfer 500111 of
Chelex suspension to the test tube containing your cell pellet.

9) Re—suspend cells by pipetting in and out several times. Examine against light to
confirm that no visible clump of cells remain.

10) Transfer 500p] of the resuspended cell pellet and Chelex into a clean 1.5ml reaction
tube labeled with your name (on the top).

11)Place your sample in a floating tube rack in the boiling water bath for 10 minutes.
Do not submerge or drop the tube into the water.

12) Use forceps to remove your tube from the boiling water bath and allow cooling for
two minutes. The tube may be placed on ice for faster cooling.

13) Place your sample tube, with others, in a balanced configuration in the
microcentrifuge and spin for 30 seconds to pellet Chelex beads at the bottom of the
tube. Alternately, let tube set for 5 minutes to allow debris to settle.

14) Use a flesh tip to transfer 200111 of the clear supernatant to a clean 1.5m] tube labeled
with your name. Avoid disturbing or transferring any of the Chelex pellet at the
bottom of the tube.

15) Store your sample on ice or in the freezer until ready to begin Procedure B.

117

PROCEDURE 2: SET UP PCR REACTIONS

MATERIAL

J 2.51.11 Micropipet

J PV92 primer/loading dye mix

I PCR tube w/ Ready-to-Go PCR Beadm
J Laboratory marker

J Beakers containing ice

M D D
1) Ask Teacher to add 22.5 it] of PV92 primer/loading buffer mix to a PCR tube
containing a Ready- To-Go PCR Bead. Flick tube with finger to dissolve bead.

2) Use fi’esh tip to add 2.5111 of your DNA to reaction tube, and tap to mix. Pool reagents
by pulsing in a microcentrifuge or by sharply tapping tube bottom on lab bench.

3) Label the cap of your tube and side (Thermal cycler has heated lids) with a number,
as assigned by your teacher. This is to ensure your results will anonymous.

4) Store all samples on ice or in the freezer until ready to amplify according to the
following profile.
I 94*C - 30 sec
1 58*C — 30 sec
1 72*C — 30 sec
(30 cycles link to soak file at 4*C)

PROCEDURE 3: LOAD AND ELECTROPHORESE PCR PRODUCTS

E
51
E

Agarose gels

250ml 1 rig/ml ethidum bromide added to the agarose
Camera (optional)

Centrifirge, micro

PBR3221 BstN I size markers
Electrophoresis apparatus

150 ml TBA Electrophoresis buffer
Loading dye

Micropipet tips

Micropipet, 1-10pl or 1-20u1

Test tube rack

Test tubes, 1.5 ml

\KKKKKKKKK'xK

NOTE: The cresol red and sucrose in the primer mix functions as loading dye, so that
amplified samples can be loaded directly into gels.

1) Your instructor will load 20111 of the pBR322-Bst N size markers into one lane of gel.

118

2) Use a micropipet with a flesh tip to add the entire PCR sample/loading dye mixture

(20111) into your assigned well of a 1.5% agarose gel. Expel any air from the tip
before loading, and be cm'eful not to push the tip of the pipet through the bottom of
the sample well.

3) Electrophorese at 130 volts for about 90 minutes. Adequate separation will have
occurred when the cresol red dye fiont has moved at least 50 mm from the wells.

4) Photograph your class results

RESUME
Take a picture of your groups gel and attach it here

119

INTERPRETATIQN AND CONCLUSIONS

Graphs

1. Observe the photograph of the stained gel containing your sample and those from
other students. Orient the photograph with the sample wells at the top. Interpret the
band(s) in each lane of the gel:

a)

b)

d)

Scan across the photograph to get an impression of what you see in each lane.
You should notice that virtually all student lanes contain one to three
prominent bands.

Now locate the lane containing the pBR322/ BstN I markers. Working from
the well, locate the bands corresponding to each restriction fragment: 1,857bp,
1,058 bp, 929 bp, 383 bp, and 121 bp (may be faint).

If One hamd is visib_le. Compare its migration to that of the 929-bp and 383-hp
bands in the pBR322/ BstN I lane. Ifthe PCR product migrates slightly ahead
of the 929-bp band, then the person is homozygous for the PV92 Alu insertion
(+/+). If the PCR product migrates well behind the 383-bp band, then the
person is homozygous for the absence of the PV92 Alu insertion (-/-).

If Two bands are visible. Compare migration of each PCR product to that of
the 929-bp and 383-bp bands in the pBR322/ BstN] lane. Confirm that one
PCR product corresponds to a size of about 850-bp and that the other PCR
product corresponds to a size of about 550-bp. The person is heterozygous for
the PV92 Alu insertion (+/-).

It is common to see an additional band lower on the gel. This diffuse (fuzzy)
band is "primer dimer,” an artifact of the PCR reaction that results fiom the
primers overlapping one another and amplifying themselves. Primer dimer is
approximately 50 bp, and should be in a position ahead of the 121 bp marker.
(Note: the presence of primer dimer, in the absence of other bands, confirms
that the reaction contained all components necessary for amplification, but
that there was insufficient template (DNA) to amplify the PV92 locus.)

Additional faint bands, at other positions, occur when the primers bind to
chromosomal loci other than PV92 and give rise to "nonspecific"
amplification products.

Determine the genotype distribution for the class by counting the number of
students of each genotype (+/+, +/- and -/-).

 

genotype Number of classmates
+/+
+/.
-/-

 

 

 

 

 

 

 

120

APPENDIX AIX.

ZEBRAFISH EMBRYO LABORATORY

Ins ru r’ ui

Target Group: Advanced Placement Biology students

Time Frame: 3 days

 

 

Day Time Activity

One or more 30 min. Set up tank and add fish
days before

lab

 

Lab period 1 30 min. Pre-lab: Set-up work stations
10 min. Isolate eggs into petri dishes
40 min Observe eggs and record

5 min label and return eggs

 

Lab period 2 50 min 85 work on interpretation & conclusion

 

Lab period 3 55 min h_ttp://visembryo.ucsf edu/
http://chickscope.beckmanuiuc.edu/explore/embggology/

 

 

 

 

Materials:

10 gallon fish tank
fish (males and females)
siphon

marbles

thermostat

fish net

disposable pipettes
petri dishes
microscope slides
cover slips

disecting microscopes
light microscopes

\\\\'\\\\\\\\

References:

Finer, Kim R. 1998. Internet Activities for General Biology. Belmont, CA: Wadsworth
Publishing Company.

Helm, Biology lab manual 3rd Edition.
hup://www.neuro.uoreggnW

httpz/lvisembryoucsf edu/
http://chickscope.beckman.uiuc.edu/explore/embryology/

121

 

RESULT

Answers to Questions in the Student Laboratory:

Table 1: Early Zebrafish Development Observations and Inferences

 

 

 

 

 

 

 

 

 

Illustration Labeled features Predicted age (in
hours)
Cleavage a. yolk 1 hour
b. blastodisc
c. vegetal-pole- nutrients
are located (yolk)
Blastula a. blastodenn 2.5 hours
b. animal pole
c.
Gastrula a. epiblast 5 hours
b. hypoblast
c. yolk plug
Your Choice a. somites 24 hours
b. heart and circulation
c. pigmentation in tail
\

 

122

 

INTERPRETATION AND CONCLUSIONS

1. How would you characterize the division pattern of the early cleavages?

1» holoblastic

2. How long is a typical cleavage cell cycle?

'3 15-18 minutes

3. Fill in the followingtable to summarize the major events of early development.

 

 

 

 

 

 

 

 

 

 

Formation of Early Gastrula Late Gastrula Neurula

the Blastula
Processes cleavage (cell gastrulation gastrulation neurulation
occurring proliferation) (migration & (determination (induction &

invagination) & cell organ
difierentiatiflt) diflerentiation

Type of blastocoel blastopore & gastrula brain and
structure archenteron spinal column
formed
Characteristic hollow sphere opening of archenteron neural tube
3 of structures archenteron forms
Significance cell positions difi'erentiation cells are tissues cm
of stage of germ layers determined form organs

 

4. Compare and contrast major developmental events in the sea urchin, zebrafish, and
human by completing the following table.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sea Urchin F rog Chicken
T of egg isolecithal mesolecithal telolecithal
Pattern of cleavage holoblastic holoblastic (unequal) meroblastic
(equal)
Distinguishing uniform, hollow blastocoel towards the small blastodisc on
characteristics of sphere animal pole a large quantity of
blastula yolk
How gastrulation invagination migration migration
occurs
Events of no neurulation indirect development direct development
neurulation (tadpole develops into (embryo develops
adult) directly into adult)
Distinguishing circular gray crescent, dorsal blastodisc
Characteristics of blastopore lip, circular cleavage,
later development blastopore primitive streak,
elongated
g blastopore
123

 

10.

11.

What are cytoplasmic determinants?
ix proteins made from stored mRN A that affect early development (up to gastrula),
which are produced by maternal genome.

How do cytoplasmic determinants affect early development, including cleavage of the
zygote?

is determine the orientation of early clavage planes

Q2» Ex. controls whether spiral cleavage is to the lefi or right.

What are homeotic genes?
is genes that control the development of segmentation in organisms

How do homeotic genes control the development of segmented organisms?
\ Each homeotic gene contains a conserved homeobox sequence that codes for a
DNA regulatory protein that can turn on other needed genes.

Describe the function of each of the following extraembryonic membranes in the
embryo:
a. Amnion- 9: filled with amnionic fluid to protect the embryo
b. Chorion- m protection & gas exchange w/ vessels of allantois
c. Allantois- 1» waste storage & and gas exchange.
(1. Yolk sac- \ vitelline vessels associated w/ the yolk sac to deliver
...... nutrients to the embryo.

What is the firnction of the vitelline blood vessels in the embryo?

is Vitelline vessels in the chick embryo are found on the surface of the yolk sac. In
early development, these vessels assist w/ oxygenation of the blood, picking up
oxygen from what has diffused into the shell. Later, the vessels pick up nutrients
from the yolk sac. The nutrient laden blood is returned to the heart.

In development of the human embryo, the trophoblast is like the chorion, surrounding
the embryo in its allantois. The trophoblast becomes the placenta during
development of the fetus. How does the firnction of the human placenta compare to
the chorion.

Q» The human placenta is responsible for oxygenation of fetal blood & for the
removal of wastes and carbon dioxide. In the chick, the chorion/ allantois
membrane system carries out the same function. In both the chick and human, the
embryo is surrounded by a fluid filled protective amnion.

124

 

 

1577.?!

c)-

ZEBRAFISH EMBRYO LABORATORY

INTRODUQIIIQN

Background Information

The heredity material within fertilized eggs, or zygotes, of animal organisms holds the
key to both structure and function. DNA guides, through time, the structural and
functional development of a single cell into an embryo and its morphogenesis into an
adult. Even aging is a result of genetic programming.

Biologists are beginning to understand the genetic control of development. Advanced
technologies as well as molecular recombinant DNA techniques have made it possible to
discover more and more and more about the genomes of organisms and to map the entire
genomes of some. Common patterns of gene function are beginning to emerge,
suggesting that some day, perhaps, we will be able to explain the mysteries of animal
development.

During this lab you will have the opportunity to explore the development of a simple
vertebrate. The zebrafish, Danio rerio, is a common aquarium fish that breeds very
easily. If fish are maintained in breeding condition, spawning occurs each day at dawn,
and with fertilization external, all stages are easily accessible. The embryo and its outer
shell are very clear optically, allowing for easy observation throughout its early
development. The small size of the embryo allows you to section very deeply into the
embryo with slight adjustments of the microscope to reveal much of the internal
structural development. Development is also very rapid. The following table gives
approximate times (afier fertilization) of key developmental stages:

Time
1st 0.7 hrs
10th

4.3 hrs

5 hrs

first movements 18 hrs
heartbeat and ' ' 24 hrs
° ' 48 hrs
and ' 72 hrs

 

Part I: Cleavage Period

In order for a fertilized egg or zygote to become a multi-cellular organism, the
zygote must divide by mitosis. During early development, this process of division is
known as cleavage.

The fertilized egg is not a uniform sphere. Differential concentrations of
cytoplasm and yolk (if present) can affect the cleavage process. The upper portion of the

125

 

slitt'tlk

 

egg, usually richest in cytoplasm, is known as the animal pole, and the lower portion of
the egg, containing more yolk, the vegetal pole. The first plane of cleavage is vertical,
bisecting both the animal and vegetal poles.

Depending upon the amount of yolk in the egg, the planes of cleavage will differ.
Ifthe plane of cleavage passes all the way through the zygote it is referred to as
hololblastic cleavage, which is typical of cells with small to medium amounts of yolk
(sea urchin and human). On the other hand, if the plane of cleavage only passes a part of
the zygote it is called meroblastic cleavage, which is typical of cells with large amounts
of yolk (zebrafish and chicken).

A second cleavage division typically occurs at a right angle to the first, producing
four cells. The third cleavage division cuts horizontally to form eight cells, four on the
top and four on the bottom .The cells produced during these cleavage divisions are
known as blastomeres. Ifthe blastomeres in the top tier lie directly above those in the
bottom tier, the pattern is said to be radial, a pattern characteristics of echinoderms and
chordates.

OEECTIVE:
0' Identify the animal pole and vegetal pole

"' Describe how the amount and distribution of cytoplasm within a fertilized egg
influence the patterns of cleavage

9' Determine the cleavage stage of various embryos

5" Describe the important developmental features at each stage

MATERIALS:

Dissecting microscope
Colored pencils
Dissecting needle
Petri dish

7 coverslips
disposable pipette
beaker of eggs

METH D/PR ED :

During this period, the first 6 cleavages occur. The cells or blastomeres, divide
synchronously at about 15 minutes. Usually, the first five cleavages are all vertical and
occur at right angles to one another, as shown in the figure below.

\\\\\\K

The sample of water in your beaker was obtained by siphoning the bottom of the fish tank
in class. In addition to extra food and waste, it should have some zebrafish embryos in
early cleavage.

1. With the pipetter, transfer several eggs into the petri dish. Try not to collect a lot of

the extra debris. Cover them entirely with the clear water at the top of your beaker to
keep them alive.

126

10.

11.

12.

Carefirlly observe your embryos with the dissecting microscope. IMPORTANT
NOTE: Try slight adjustments of your lighting to find the best viewing conditions.
Sometimes slightly less lighting allows you to see more detail.

Determine the stage of development.

Find and transfer a few 2-4 cell embryos to a small petri dish. It will be necessary to
change to a monocular scope with greater magnification to see the nuclei. To make
this change, you will need to mount the embryo, still in its shell, on a slide in a
generous drop of water beneath a coverslip bridge. You do this by stacking 3
coverslips on each side of the embryo and one above it (7 in all). This will prevent the
embryo fiom being squashed. Observe with any objectives except the 100x oil
immersion lens.

. Carefully adjust your lighting to accentuate the nuclei. They are present during the

first half of each cycle and their shapes change systematically. The nuclei are globular
in very early interphase and become spherical by late interphase. As the cells enter
mitosis, the nuclei are elliptical and then take on the shape of a rugby-style football
shortly before they disappear during prophase. The mitotic chromosomes are more
difficult to visualize. Approximately 5 minutes after the nuclei disappear, you
should see cleavage taking place.

Gently use the small probe to roll the embryo about the dish. . Take care not to
damage the chorion. By observing the cell arrangements fiom different angles you
should be able to figure out the way the cleavages occur. Early cleavages are
incomplete, or meroblastic, leaving the yolky region (vegetal pole) of the embryo
uncleaved. Orient your embryo so that you are looking down on the animal pole.

Follow what is happening during several successive divisions, trying to keep track of
more than a single embryo. Notice that divisions often seem to occur along

stereotyped "cleavage planes".

Draw sketches in Table l, but work quickly for things may happen fast while you are
not watching.

Record the times of your observations. You may find that the first 5 or so cleavages
occur in a regular, but not an invariant, pattern.

Illustrate one of your embryos and indicate the cleavage stage you think it represents
in Table 1 below. Your lab partner should draw the other embryo.

Find two embryos in the Blastula Period. Follow the same directions as above.

Refer to your textbook and other references to describe the characteristic features of
your embryo. Record them in Table l.

127

 

RESULT

Table 1: Early Zebrafish Development Observations and Inferences

 

Illustration

Labeled features

Predicted age (in hours)

 

Cleavage

 

Blastula

 

Gastrula

 

 

Your Choice

 

 

 

INTERPRETATION AND CONCLUSIONS

1.

How would you characterize the division pattern of the early cleavages?

128

 

 

2. How long is a typical cleavage cell cycle?

 

3. Fill in the following table to summarize the major events of early development.

 

Formation of Early Gastrula Late Gastrula Neurula
the Blastula

 

Processes
occurring

 

Type of
structure
formed

 

Characteristic
s of structures

 

Significance
of stage

 

 

 

 

 

 

 

4. Compare and contrast major developmental events in the sea urchin, zebrafish, and
human by completing the following table.

 

Sea Urchin Zebra fish Human

 

Iypeofess

 

Pattern of
cleavage

 

Distinguishing
characteristics of
blastula

 

How gastrulation

OCCUI'S

 

Events of
neurulation

 

fifinguishing
Characteristics of
later development

 

 

 

 

 

129

 

10.

11.

What are cyt0plasmic determinants?

 

 

How do cytoplasmic determinants affect early development, including cleavage of the
zygote?

 

 

What are homeotic genes?

 

 

How do homeotic genes control the development of segmented organisms?

 

 

Describe the firnction of each of the following extraembryonic membranes in the
embryo:

‘5 Amnion

\ Chorion

Q; Allantois

\ Yolk sac

 

What is the function of the vitelline blood vessels in the embryo?

 

 

 

 

 

 

In development of the human embryo, the trophoblast is like the chorion, surrounding
the embryo in its allantois. The trophoblast becomes the placenta during
development of the fetus. How does the firnction of the human placenta compare to
the chorion.

 

 

 

 

 

130

APPENDIX BI.

STUDENT BIOTECHNOLOGY AND DEVELOPMENTAL BIOLOGY SURVEY

1. How would you rate the laboratory experiences in this course in comparison to other
science courses?
E. coli Transformation
[excellent, very good, satisfactory, fairly 800d, unsatisfactory]

Gel Electrophoresis
[excellent, very good, satisfactory, fairly good, unsatisfactory]

All You Have ALU
[excellent, very good, satisfactory, fairly good, unsatisfactory]

Zebrafi sh Embryo
[excellent, very good, satisfactory, fairly good, unsatisfactory]

2. How well did the laboratory work relate to the ideas discussed in class?
[excellent, very good, satisfactory, fairly good, unsatisfactory]

b)

. How would you rate the classroom simulations of laboratory techniques in terms of
helping you to understand biotechnology concepts taught in this course?
Breakfast DNA-> protein
[excellent, very good, satisfactory, fairly good, unsatisfactory]

Manipulating DNA Replication
[excellent, very good, satisfactory, fairly good, unsatisfactory]

4. How would you rate the Medical Technology and Genetic Testing Field Trip to MSU
in terms of interest and usefirlness to the study of biology?
[excellent, very good, satisfactory, fairly good, unsatisfactory]

LII

. How do you rate the way you were graded on labs, tests, and essays in this course?
[excellent, very good, satisfactory, fairly good, unsatisfactory]

O‘

. How would you rate the laboratory classroom environment (how well students know,
help and are support one another) ?
[excellent, very good, satisfactory, fairly good, unsatisfactory]

\l

. How would you rate the interpersonal behavior of the teacher?
[excellent, very good, satisfactory, fairly good, unsatisfactory]

131

APPENDIX BH.

DNA STRUCTURE AND REPLICATION PRETEST

Use the figure below to answer questions 1-5.

1. A chemical group that, together with a sugar and a nitrogen base, makes up a
nucleotide

2. A hydrogen bond
3. A pyrimidine
4. A 5’ carbon of deoxyribose

5. Most likely to be broken during replication

 

(A)

 

(C)

 

132

6. DNA replication can be described as

a. semiconservative
b. conservative
c. degenerative
d. dispersive
e radical

In DNA replication, DNA polymerase catalyzes the reaction in which:

the double helix unwinds

the sugar-phosphate bonds of each strand are broken

a phosphate group is added to the 3’-carbon or 5’-carbon of ribose

a nucleotide with a base complementary to the base on the template strand is
added to the new DNA strand

e. the two nucleotide strands come together and intertwine to form a double helix

9.09"!”

Answer Key;

1
2
3.
4.
5
6
7

. b

QNGNO—O

133

DNA STRUCTURE AND REPLICATION POST-TEST

Ug the figu_re below to answer questions 1-5.

1.

2
3.
4

VI

A chemical group that, together with a sugar and a nitrogen base, makes up a
nucleotide

A hydrogen bond
A pyrimidine
A 5’ carbon of deoxyribose

Most likely to be broken during replication

 

/(D)
crr,
\
o o-
\P/
o/ \\o
o

 

 

134

6. DNA replication can be described as
semiconservative

conservative

degenerative

dispersive

radical

sop-ova

7. In DNA replication, DNA polymerase catalyzes the reaction in which:

the double helix unwinds

the sugar-phosphate bonds of each strand are broken

a phosphate group is added to the 3’-carbon or 5’-carbon of ribose

a nucleotide with a base complementary to the base on the template strand is
added to the new DNA strand

e. the two nucleotide strands come together and intertwine to form a double helix

9'." 9‘!”

8. DNA contains all of the following molecules EXCEPT:
adenine

guanine

deoxyribose

phosphate

uricil

 

sup-99‘s»

9. “Primers” that initiate DNA replication consist of
a. RNA nucleotides
b. DNA nucleotides
c. Okazaki fragments
(1. DNA polymerase
e nucleosomes

10. The two strands of a DNA molecule are connected by

a. hydrogen bonds between the codons and anticodons

b. hydrogen bonds between the bases of one strand and the bases of the second
strand

c. hydrogen bonds between deoxyribose sugar molecules of one strand and
deoxyribose molecules of the second strand

d. covalent bonds between phosphate groups

e. covalent bonds between the nitrogen bases

11. All of the following combinations of nucleotides are examples of normal base pairing
EXCEPT
a. an adenine DNA nucleotide to a thymine DNA nucleotide

b. a guanine DNA nucleotide to a cytosineDNA nucleotide

c. a thymine DNA nucleotide to an adenine DNA nucleotide

d. a cytosine DNA nucleotide to a guanineDNA nucleotide

e a uricil RNA nucleotide to a thymine DNA nucleotide

135

12. All of the following enzymes are involved in DNA replication EXCEPT
helicase

DNA ligase

DNA polymerase

RNA polymerase

primase

{DP-OFT!”

uestions 13-16 refer to the dia am of DNA on the first e.

 

13. Label adenine
14. Label cytosine
15. Label guanine
16. Label thymine

W 13. adenine

. b 16. thymine

 

O-OU‘NGD-NGNO—G

 

(C)

136

APPENDIX BIII.

BACTERIAL TRANSFORMATION PRE AND POST-TEST

. Once a plasmid has incorporated specific genes such as the gene coding for the
antibiotic ampicillin, into its genome, the plasmid may be cloned by

a. inserting it into a virus to generate multiple copies
b. treating it with a restriction enzymes in order to cut the molecule into small pieces
c. inserting it into a suitable bacterium in order to produce multiple copies
(1. running it on a gel electrophoresis in order to determine the size of the gene of
interest
e. infecting it with a mutant cell in order to incorporate the gene of interest
. A bacteriophage is a
a. bacterium that atttacks viruses
b. virus that attacks bacteria
c. bacterium that attacks eukaryotic cells
d. parasitic bacterium
e. parasitic eukaryotic cell

. Which of the following correctly describes plasmids?

oer-99‘s»

They are composed only of RNA

They are composed of RNA and protein

They are DNA segments in the chromosomes of bacteria
They are the DNA cores of viruses

They can be transferred between bacteria during conjugation

. In bacteria, a small circle of DNA found outside the main chromosome is called a

99.0w»

Plasmid

cDNA

RFLP

PCR

Genetic fingerprint

. All of the following cell components are found in prokaryotic cells EXCEPT

cup-99's»

DNA

Ribosomes

Cell membranes
Nuclear envelope
Enzymes

137

6. Which of the following statements about plasmids is correct?
They are synthesized in the endoplasmic reticulum

They are found only in eukaryotic cells

They are composed of RNA

They are larger in size than bacterial chromosomes

They are self-replicating.

990.6.»

A plasmid

a. can be used as a DNA vector
b. is a type of bacteriophage

c is a type of cDNA

d. is a retrovirus

e both b and c

 

8. DNA molecules with complementary “sticky ends” associate by

a. covalent bonds

b. hydrogen bonds

c. ionic bonds

d. disulfide bonds

e. phosphodiester linkages

Answer Kgy:

”$9999.“?
c‘moamocro

138

APPENDIX BIV.

GEL ELECTROPHORESIS PRE AND POST-TEST

Questions 1-3 refers to the following semi-log graph of results from the standard
restriction enzyme used in a gel electrophoresis procedure. Phage lambda DNA
molecules digested with HindIII are used as the standard.

1 0.000

1000

Number of Base Pairs in
a DNA Fragment

 

100

0 2 4 6 8
Migration Distance in cm.

1. A fragment of phage lambda DNA produced by EcoRI endonuclease migrates
6 cm. If this fragment is produced during the same electrophoresis procedure
as the standard shown in the graph above, how large is the fragment?

a. 150 base pairs

b 600 base pairs
c. 800 base pairs
d. 1000 base pairs
e 6000 base pairs

2. What is the relationship between migration

distance and DNA fragment size?

a. migration distance is independent of DNA
fragment size

b. longer DNA fragments travel a greater
distance than shorter fragments

c. migration distance is inversely proportional to
the fragment size.

d. Migration distance is directly proportional to
the fragment size.

e. The heavier the fragments size the greater the
migration distance.

139

3. According to the graph, approximately how far
will a DNA fragment with 3000 base pairs travel
on the gel.

1.0 cm

2.0 cm

2.3 cm

2.6 cm

5.0 cm

sup-99's»

4. Human DNA and a particular plasmid both have sites that can be cut by the
restriction enzymes HindIII and EcoRI. To make recombinant DNA, one should

cut the plasmid with EcoRI and the human DNA with Hind III

use 15le to cut both the plasmid and the human DNA

use Hind III to out both the plasmid and the human DNA

a or b

b or c

era-99:1»

5. Which of the following sequences is not palindromic?

a. 5’-AAGCTT-3’ & 3’-"I'1‘CGAA-5’
b. 5’-GATC-3’ & 3’-CTAG-5’
c. 5’-GAATTC-3’ & 3’-CTTAAG-5’
d. 5’-CTAA-3’ & 3’-GATT-5’
6. Gel electrophoresis separates nucleic acids on the basis of differences in
length
charge

nucleotide sequence
relative proportions of adenine and guanine
relative proportions of thymine and cytosine

{DP-9.0".”

The diagram below shows a segment of DNA with a total length of 4,900 base pairs. The
arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).

 

 

 

Enz Enz Enz Enz

X Y X X
DNA Segment I—--V V V V 1
Length (base pairs) 400 500 1200 1300 1500

7. Explain how the principles of gel electrophoresis allow for the separation of DNA
fragments

140

 

8.

Describe the results you would expect from electrophoretic separation of fragments
fiom the following treatments of the DNA segment above. Assume that the digestion
occurred under appropriate conditions and went to completion

a. DNA digested with only enzyme X

b. DNA digested with enzyme X and Y

Answer Kg:

>1

8.

99:99:19?

”9.00.00:

lpt each with a 4 points max.

E Electricity - Electrical potential (charge, field) moves fragments

E Charge - Negatively charged fragments move toward (+) anode through gel (-)
charge due to phosphate groups

Rate/size - Smaller fragments move faster (farther) relative to larger fragments.
Describe logarithmic relationship

Calibraton - DNAs of known molecular weights are used as markers/standards
Apparatus - DNA is stained for visualization of bands/explains use of wells, gel
material, tracking dye, buffers.

55!

lpt each with a 4 points max.

a. DNA digested with only enzyme X
E Describes 400, 1300, 1500, l700,bp fragments
E correct 4 band diagram

b. DNA digested with enzyme X and Y

E Describe 400, 500, 1200, 1300, 1500 bp fragments
E correct 5 band diagram

141

APPENDIX BV.

POLYMERASE CHAIN REACTION PRE AND POST-TEST

Question 1 refers to the following diagram representing the bands produced by an
electrophoresis procedure using DNA from four human individuals. Each DNA sample

is treated with the same restriction enzyme.

1. Which of the following is a correct

1m
1

 

interpretation of the gel electrophoresis

data?
a.

b.

C.

Individual 1 could be an offspring of
individuals 3 & 4.

Individual 1 could be an offspring of
individuals 2 & 3.

Individual 2 could be an offspring of
individuals 1 & 3.

Individual 3 could be an offspring of
individuals 2 & 4.

All four individuals could be the
same people.

I'm

1W
3

 

Question 2 refers to the following diagram of a DNA segment

1
5'-rcrcoAcr-3'
s-AcaoctroA-sr

W
4

 

 

2. Arrows in the above diagram show the cleavage points for the restriction enzyme
T an. Which of the following describes the correct DNA segments produced after
treatment with the restriction enzyme?

a.

5’-TCT
3 ’ -AGAGC

two fi'agments: and

5 ’ -TCTCGACT-3 ’
3 ’-AGAGCTGA—5 ’

two fragments:

5 ’-TCTCGTGA—3 ’
3’-AGAGCACT-5’

two fragments:

four fragments:
four fragments:

142

CGACT-3 ’
TGA-S’

5’-TCT, 3’-AGAGC, CGACT-3’, and TGA-S’
5’-TCTCG, ACT-3’, 3’-AGAGC, and TGA-S’

Question 3 refers to the following diagram. DNA material from each of four individuals

was treated with a restriction enzyme. The products were separated using gel
electrophoresis. The results, which show relative migration distances for DNA fragments

from each individual, are given below.

3. Which of the following is a correct

Indv‘flu-l lndvldud lndiv'dud Individud
2 3 4

 

interpretation
of the gel electrophoresis data?
a. all four individuals could be the

same person

b. Individual 3 could be an offspring

of individuals 2 and 4. ‘—— -——-
c. Individual 2 could be an offspring __

of individuals 1 and 3. -—-—— __
d. Individual 1 could be an offspring _— ——-

of individuals 2 and 3. ___
e. Individual 1 could be an offspring __

of individuals 3 and 4.

4. The PCR technique uses

sop-9.0:.»

heat-resistant DNA polymerase
reverse transcriptase

DNA ligase

A nd b

B and c

5. In restriction fragment length polymorphism (RFLP) analysis to determine parentage

a.

b.

C.

d.
e.

every band present in a child would be expressed to be present in both of the true
parents

every band present in a child would be expected to be present in at least one of the
true parents

every band present in a true parent would be expected to be present in all of the
children

a and b

b and c

6. Why is the PCR technique valuable?

 

 

7. Describe the major steps of PCR.

E

g

¥

¥

143

Answgr Key:

.V‘PP’NT"

.9

5555-"

O‘NOND-

Polymerase chain reaction (PCR) is a method of greatly increasing the quantity of a
specific DNA or RNA sequence.

lpt each with a 3 points max.

The double-stranded DNA to be amplified is first heated to separate the two strands.
Synthetic primers are bound to separated strands. The primers define the ends of the
sequences to be amplified.

The two new strands are synthesized with DNA polymerase extending the strands
from the primers giving complementary strands.

Each cycle is repeated 30 times, doubling the amount of DNA each time.

144

Answer Key:

.V‘PP’Nf"

.O‘

5555.‘I

700930-

Polymerase chain reaction (PCR) is a method of greatly increasing the quantity of a
specific DNA or RNA sequence.

lpt each with a 3 points max.

The double-stranded DNA to be amplified is first heated to separate the two strands.
Synthetic primers are bound to separated strands. The primers define the ends of the
sequences to be amplified.

The two new strands are synthesized with DNA polymerase extending the strands
from the primers giving complementary strands.

Each cycle is repeated 30 times, doubling the amount of DNA each time.

144

APPENDIX BVI.

DEVELOPMENTAL BIOLOGY PRE AND POST-TEST

. In human females fertilization normally occurs in the

sup-99‘s»

ovary
uterus

vagina
oviduct
cervix

. In mammalian embryo, the somites give rise to the notocord

9'.“ 9".”

the central nervous system
the lining of the mouth
the eyes

muscles and vertebrae

. Gastrulation results in three primary tissue layers that give rise to all the organs and
tissues of the body. Which of the following statements is true.

sup-99's»

Endoderrn gives rise to muscle.
Epiderm gives rise to skin
Mesoderm gives rise to bone
Ectoderm gives rise to the gut lining
Periderm gives rise to skin

. Embryonic induction is best illustrated by which of the following:

9.9-9.7!"

F orrnation of the lens in the ectoderrn after contact with the underlying optic cup
Replacement of cartilage by bone

Lysosomal action on the degenerating tail of a tadpole

Development of the amnion surrounding the embryo

Development of the mesoderm into the notochord

Which of the following describes the correct sequence of stages during

embryogenesis:

a. Cleavage, blastula formation, gastrulation
b. Cleavage, gastnrlation, blastula formation
c. Blastula formation, gastrulation, cleavage
d. Blastula formation, cleavage, gastrulation
e. Gastrulation, cleavage, blastula formation

145

 

Questions 6-9 refer to the following choices

9.

9999‘s»

Cleavage
Organogenesis
Gastrulation
Neurulation
Fertilization

This process establishes the primary germ layers.

Two haploid cells fuse to form a diploid cell.

The number of cells increases, but there is no increase in total cell mass and there is
little or no differentiation

Cells migrate over the dorsal lip of the blastopore.

10. In animals, all of the following are associated with embryonic development EXCEPT

sup-99's»

migration of cells to specific areas
formation of germ layers

activation of all the genes in each cell
inductive tissue interactions

cell division at a relatively rapid rate

11. In humans primary oocytes are located in the

9.0-9.0“?

cervix

uterus

corpus luteum
oviduct

ovary

12. All of the following statements about the placenta are correct EXCEPT

a. It permits an interchange of C02 and 02 between maternal and fetal blood.
b. It forms fi'om tissues of both the embryo and the uterus.
c. It permits the mixing of maternal and fetal blood.
d. It firnctions as an endocrine gland.
e. It provides the embryo with a way to dispose of its nitrogenous waste products.
Answgr Kg:
1. d 7. e
2. e 8. a
3. c 9. d
4. a 10. c
5. a l 1. e
6. c 12. c

146

 

APPENDIX CII.

Table 2: INDIVIDUAL STUDENT8' PRE AND POST-TEST RESULTS

 

148

APPENDIX CIII.

Table 3: T-TEST RESULTS FOR ALL PRE AND POST-TESTS

Pre-test Post-test t- Test Inverse of
Mean Mean ' tdistribution

0.90 8.00 1.62E-13
NA Structure 0.80 3.50 1.66E-15

NA 0.75 2.40 4.39E-10

ransforrnation 1 .75 6. 75 8.23E- l 5
' 2.25 5.40 4.67E-12 5000000

El ' 0.10 3.24 4.36E-O9

C 1.10 4.60 1.67E-20

0.00 2.83 3.25E-10

   

149

APPENDIX CIV.

Table 4: AVERAGE PRE AND POST-TEST PERCENTAGES

est Pre Ave % Post Ave %

7.50
NA Structure 20.00
NA ' ' 25.00
ransformation 21.88

37.50
1.43
22.00
0.00
16.91

 

150

APPENDIX CV.

Table 5: DNA STRUCTURE AND REPLICATION PRE AND POST-TEST

 

151

 

APPENDIX CV1.

Table 6: BACTERIAL TRANSFORMATION PRE AND POST-TEST

# Pretest Post-test Difference
T=20 T=20
l 3
1 5
1 2
1 l
16
12
10
1 1
l 00
12.5

1
2
3
4
5
6
7
8

 

152

APPENDIX CVII.

Table 7: GEL ELECTROPHORESIS PRE AND POST-TEST

# PreTest PostTest Difference
T=20 T=20

19
19
17
16
18
19
108
5.4

 

153

APPENDIX CVHI.

Table 8: POLYMERASE CHAIN REACTION PRE AND POST-TEST

# Pretest Post-test Difference
T=20 T=20

20
19
20
13
20
92

 

154

APPENDIX CIX.

Table 9: DEVELOPMENTAL BIOLOGY PRE AND POST-TEST

Girls Post Girls Difference
1 3 1

11

fl

lowQO‘VlAWNr—I
mxoqu—wx—Ncom

\IQOOMNOO—‘NWAN-d

4
0
l
1
0
0
6
2
0
2
4
l

HW—‘OOOOOHt—I
Or-‘t-‘ONOKOOO

 

155

APPENDIX CX.

Table 10: STUDENT'S GENETICS AND DEVELOPMENTAL BIOLOGY SURVEY

 

Percentages of Student Responses

Question

 

 

Excellent Very Good Satisfactory Fairly Good

  

1. How would you rate the laboratory
experiences in this course in comparison
to other science courses?

 

 

a Zebrafish Development 11.10 50.00 16.70 22.20
b. E. coli Transformation 44.40 55.60
c. Gel Electrophoresis 27.80 55.60 1 1.10 5.60
d. All you have ALU 72.20 22.20 5.60
2. How well did the laboratory work relate
to the ideas discussed in class? 16.70 77.80 5.60
3 . How would you rate the classroom
simulations of laboratory techniques in terms
of helping you to understand biotechnology
concepts taught in this course? U
a Breakfast DNA--> protein 55.60 16.70 22.20 5.60
b. Manipulating DNA Replicaiton 33.30 66.70
4. How would you rate the Medical
Technology and Genetic Testing Field
Trip to MSU in terms of interest and 72.20 22.20 5.60
usefulness to the study of biology?
5 . How well did the essays relate to
the ideas discussed in class? 11.10 38.90 44.40 5.60L
6. How do you rate the way you were
graded on labs, and tests in this course? 44.40 44.40 11.10
7. How would you rate the laboratory
Classroom environment (how well students
know, help and support of one another). 33.3 44.4 22.2
8. How would you rate the interpersonal
behavior of the teacher? 44.40 33.30

 

 

156

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