Biology 114
Name:_________________
Lab Section_____________
Prelab questions for Lab 8
1.
For each of these genetic traits which is the dominant allele and which is
the recessive allele:
•being a tongue roller
•having attached earlobes
•having a Widow’s peak
•interlocking fingers left
over right hand
•having dimples
•having mid digital hair
•having freckles
being a non tongue roller
having unattached earlobes
not having a Widow’s peak
interlocking fingers right over
left hand
not having dimples
not having mid digital hair
not having freckles
2.
Huntington’s Disease in humans is a lethal disease caused by inheriting a
dominant allele. If one of your parents carried a single allele for
Huntington’s what is the probability that you would develop the disease?
3.
If Huntington’s Disease was the result of inheriting a recessive allele instead
of a dominant allele and one of your parents carried a single allele for the
disease, what would be the probability that you would develop the
disease?
4.
What is a restriction enzyme?
5.
The process of gel electrophoresis can be used to separate DNA fragments
into bands on the gel. How is the DNA in each band different from the
other bands?
Lab 8-9-114.DOC
1
Biology 114
Lab 8
Genetics
Objectives
1.
2.
3.
Gather data on heritable human traits and use the Hardy Weinberg Principle to
estimate allele frequencies in the human population. (Part I)
Use information provided by molecular biology techniques to determine which
alleles an individual possess. (Part II)
Use agarose gel electrophoresis to separate fragments from various samples of
DNA and use this information to establish possible paternity.
Safety Precautions
1.
2.
In Part III use gloves and goggles when handling the agarose gel or loading
samples.
Keep UV light box cover closed when the UV light is on.
Materials
Table for collecting class data on 7 phenotypes
Colored sheets containing DNA sequences (a and b) representing Sam’s chromosome 4
Scissors
Scotch tape
Micropipetters, 10µl
Electrophoresis chamber and power source
agarose
gel tray and comb
10X TAE buffer (dilute to 1X for running buffer)
DNA samples (Bste II, EcoR I, Pst I, Hind III)
Introduction (Part I)
Before coming to lab today you were asked to gather data on seven specific
human heritable traits. Today you will pool these data with your classmates data and
estimate the frequencies of the various alleles present in the human population.
When we speak of the frequency of an allele we are not referring to how many
individuals show the dominant phenotype and the recessive phenotype. Rather, we are
referring to the relative abundance in the population of each of the kinds of genes
diploid individuals possess for a trait.
Consider the dominant tongue rolling gene (T) and its recessive non tongue rolling
allele (t) in a very small hypothetical population of fifty people. In this population there
are:
2 individuals with the genotype tt who are non tongue rollers,
Lab 8-9-114.DOC
2
32 with the genotype TT who are tongue rollers,
16 with the genotype Tt who are also tongue rollers.
The frequencies of these alleles are the ratios of the total number of each type of
allele (T or t) to the total number of genes for the trait. Since there are 50 people and
each person is diploid, having two genes for each trait, that totals to 100 genes. In this
population the number of T genes is 80 (32 + 32 + 16). The frequency of T is 80/100
(80% or 0.8). The number of t genes is 20
( 2 + 2 + 16) and the frequency of t is 20/100 (20% or 0.2).
If we are considering a trait with a single dominant allele and a single recessive
allele, then the sum of the frequencies of each of these alleles should total 100% or 1
depending on how we express the frequencies. In discussing the Hardy Weinberg
Principle typically the letter p is used to represent the dominant allele and q is used to
represent the recessive allele. So we can write this equation: p + q = 1 and for the
hypothetical population .8 + .2 = 1.
According to the Hardy Weinberg Principle if certain criteria are met in a
population, then the frequency of the alleles for a given trait will remain the same over
time. The Hardy Weinberg Principle applies to a biological population which is defined
as a group of individuals of the same species who live in the same region and can
reproduce with each other. The criteria that must be met are 1) the population must be
large and randomly mating 2) none of the alleles can have an adaptive advantage over
the other alleles and 3) there cannot be any immigration or emigration to and from the
population.
Often students of genetics not familiar with this principle expect recessive genes in
a population to decrease in frequency over time. The mathematics from which the
Hardy Weinberg Principle is derived demonstrates that this expectation is incorrect. We
will not study this mathematical problem now but use the equations involved in
demonstrating the Hardy Weinberg Principle for another purpose. We will use these
equations to determine what proportion of a population is homozygous dominant and
heterozygous.
In the hypothetical population described above you were told how many
individuals were TT and Tt. Both of these different genotypes have the same phenotype
and they cannot be distinguished by observation. If an individual is a tongue roller, you
cannot tell by how they do it whether they are TT or Tt. The data you have gathered on
tongue rolling tells us how many tongue rollers and non tongue rollers there are. You
will use it to calculate how many of the tongue rollers are TT and how many are Tt. You
will not be able to tell who is TT or Tt however.
In order to determine how many individuals in a population are homozygous
dominant and how many are heterozygous, we use the frequencies of the alleles.
These frequencies can be determined from the number of individuals who are
homozygous recessive.
In another hypothetical population there are 9 homozygous recessives in a group
of 100. This is a frequency of .09 for the homozygous recessive phenotype. The
Lab 8-9-114.DOC
3
frequency of the recessive allele is the square root of this number √.09 = 0.3. Why? If
the frequency of the recessive allele is any number q then the probability of possessing
two of those alleles or being homozygous recessive would be q X q or q2. This comes
from the law of probability which states that the probability of 2 events occurring
together (inheriting a recessive allele from both parents) is the product of the probability
of each event. The frequency of the homozygous recessive phenotype in the
population (0.09) then is q2 and the frequency of the recessive allele (0.3) is the square
root. Since p + q = 1 we can also calculate p to be 0.7.
Knowing these frequencies allows us to determine the frequencies of the
homozygous dominant and heterozygous individuals. The homozygous dominant
would be p x p or p2 which in this population would be 0.49. If there are 100 people and
we now know 9 are homozygous recessive and 49 are homozygous dominant that
means the remaining 42 people must be heterozygous. Notice that 2pq or 2 X 0.7 X 0.3
= 0.42. The p X q represents a heterozygous individual. Can you figure out why the
formula for determining the number of heterozygous individuals is 2pq?
Procedure (Part I)
1.
Enter your data on the seven traits in the table provided in class and calculate the
class totals for each trait. Enter these numbers in Table 1 of the assignment sheet.
This must be completed during the lab period but the remainder of Part I can be
completed outside of the scheduled lab time.
2.
Calculate the frequency of the recessive phenotype in the population of individuals
polled by the class. Enter these numbers in Table 1.
3.
Using the frequency of the recessive phenotype (=q2) calculate the frequency of
the recessive allele (q) and the frequency of the dominant allele (p). Enter these
numbers in Table 1.
4.
From the values of p and q determine the frequency of homozygous dominant and
heterozygous individuals in the population. Enter these numbers in Table 1.
5.
Using these frequencies determine the actual number of individuals who were
polled that are homozygous dominant and heterozygous. Enter these numbers in
Table 1. You can check your work by adding these numbers and the number
individuals with the recessive phenotype. They should total to the number of
individuals polled.
Introduction (Part II)
In the previous section we worked with a mathematical process that allowed us to
determine how many people possessed each of the different genotypes but we could
not tell who had a homozygous dominant genotype and a heterozygous genotype. In
the next section we will use some more recently developed molecular techniques that
allow us to analyze the DNA of individuals and determine if they carry specific alleles.
The trait we will be working with is Huntington’s Disease.
Lab 8-9-114.DOC
4
Huntington’s disease is caused by an autosomal dominant allele. The disease is
characterized by involuntary muscular movement and mental changes. It is a
progressive illness; walking, swallowing, speaking, reasoning become increasingly
impaired. The effect on families can be devastating because no cure has been
developed and it cannot be treated effectively. Because the onset is typically late in life,
individuals bear offspring and pass this gene on before they realize they carry it.
Consequently this disease is easily transmitted to later generations.
A diagnostic test has been developed using DNA analysis that allows individuals
carrying an allele for Huntington’s to be recognized. It involves using enzymes to cut
the DNA into fragments called restriction fragments and then determining the length of
these fragments using a process called gel electrophoresis. In gel electrophoresis the
DNA fragments are pulled through a gel matrix by an electric current. DNA is negatively
charged and the fragments then move toward the positive pole created by the current.
The gel matrix impedes the larger fragments more than the smaller fragments so
different sized fragments travel through the gel at different rates and separate into bands
according to their size. The allele for Huntington’s disease has a repetition of a CAG
base sequence that is longer than the non-Huntington’s disease allele. This causes the
Huntington’s disease allele fragment to migrate more slowly than the non-Huntington’s
disease allele in the gel matrix and it can be recognized.
Individuals who have a history of this disease in their family have an agonizing
decision to make—whether to be tested for this incurable disease or not. Testing bears
consequences because insurance companies and HMO’s have in the past disqualified
or reduced benefits to individuals known to carry this disease. However, those
individuals who are tested do add to the overall knowledge of the disease and hope for
a cure or effective treatment becomes more concrete. Whether to be tested or not is an
individual decision. The following case study is true, however, the names have been
changed to protect “Sam” and his family.
Sam received a registered letter from Columbia University requesting that he
submit a blood sample for their Huntington’s disease study. His mother was healthy
and his father died when he was a young child, so he thought the letter was meant for
someone else, not himself. However, when the second registered letter from Columbia
University arrived, he began to wonder. So he called his mother.
His mother told him that it was not a mistake, Sam’s paternal grandfather had died
of complications of the disease at age 58. When Sam’s father, Dave, died his mother
had moved back to the Midwest and severed nearly all contact with Dave’s family. Sam
knew that his father had died in a car accident. Road conditions were slippery and
Sam’s father had lost control of his car and slammed into a bridge embankment. Sam’s
mother, Cheryl, told Sam that she always wondered if Dave purposefully did not try to
compensate when the car began to skid. He had been depressed. Dave knew his own
father had Huntington’s disease and he thought that he was displaying some of the
early symptoms. Sam’s mother thought that Dave had, in effect, committed suicide.
Sam later learned that one of his uncles, Mike, had Huntington’s disease. Each of his
cousins, like himself, had a 50-50 chance of inheriting the disease.
Lab 8-9-114.DOC
5
Sam felt tremendous conflict and irrational guilt. He and his wife, Cindy, had two
children with a third on the way. He wondered if he had the allele for Huntington’s
disease. He worried that he had passed it on to his children. And he wondered whether
he really wanted to know if increasing physical disability and dementia were his fate.
Put yourself in Sam’s shoes and think about what you would have decided: would you
have had the DNA analysis performed?
In this exercise, you will model the DNA testing that would diagnose whether Sam
inherited the allele for Huntington’s disease.
Procedure (Part II)
1. Work in groups of two. Obtain the colored sheets containing information about
Sam’s DNA. These are print outs of the base sequence in a short segment of
human chromosome 4 that includes part of the Huntington gene. There are 2
different print outs (a and b) each representing one of the chromosomes of his
diploid chromosome 4 pair. Take a copy of each.
2. Number the columns on each print out and assemble each into a long chains of DNA
bases by taping the columns together in sequential order.
3. The restriction enzyme used to cut this DNA segment recognizes the sequence
GGTGT ACACC and cuts between the last T and the first A. Mark the cuts with a
pencil. Count the resulting number of base pairs in each fragment and record the
number for each fragment in Table 2 of the assignment sheet.
4. Figure 1 is a diagram of a gel showing the migration pattern of the DNA of members
of Sam’s family who have been tested for Huntington’s. Using the information you
have recorded in Table 2, diagram Sam’s DNA. Use a thin line to represent unique
fragments and a thick line to represent fragments that both chromosomes have in
common.
5. Compare Sam's DNA with the band patterns of other members of his family and his
family's pedigree (Figure 2) and determine if Sam inherited the disease.
Lab 8-9-114.DOC
6
Figure 1
Lab 8-9-114.DOC
7
Figure 2
Lab 8-9-114.DOC
8
Introduction (Part III )
In the previous section you used information about alleles of genes obtained
through molecular techniques such as DNA sequencing, producing restriction
fragments and gel electrophoresis. You did not perform the actual techniques. In this
section you will have an opportunity to perform one of the techniques, gel
electrophoresis, and use the information obtained to do some genetic analysis. We
have prepared samples of DNA isolated from cells. We also have prepared gels made
of a substance called agarose that are used to separate DNA fragments cut with
enzymes. You will load these gels with the DNA samples and expose the gels to an
electric current that will cause the fragments to move through the gel and separate into
bands according to the fragment lengths. The samples provided will represent DNA
taken from a child, its mother and several people who are the possible fathers of the
child. You will use the results of the gel electrophoresis to determine which of the men
are most likely to be the father.
Procedure (Part III)
Your instructor will explain how the electrophoresis process works and show you
how to use the apparatus. The gel has been prepared ahead of time as well as
the buffer solution.
1.
Obtain the five samples of DNA (child, mother and 3 possible fathers) that are
contained in microcentrifuge tubes. In addition to DNA these samples contain a
purple dye that will allow us to monitor how fast the electrophoresis is progressing
and a stain for DNA that is visible under UV light. This stain is toxic and you must
wear gloves!!!!!
2.
Prepare the gel for loading with the DNA. Fill the chamber with the running buffer.
Both ends of the chamber should be filled and the surface of the gel should be
covered with buffer. Remove any large air bubbles under the tray by sliding the
tray back and forth.
3.
Pipette 10µl of each sample into a different well on the gel. Keep track of the
location (lane number) of each sample.
4.
Run the electrophoresis apparatus at 100V for 60 minutes. Periodically check the
progress of the dye in the samples. We do not want the dye to run off the bottom
of the gel.
5.
After 60 minutes turn off the power and unplug the electrodes from the power
supply.
6.
Wearing gloves and goggles, remove the tray from the chamber. Allow the gel on
the tray to cool for several minutes. Gently slide the gel from the tray onto the
surface of the UV light box. Close the lid and observe your gel. Be sure not to lift
the lid of the light box when the light is on or to look directly at the UV light. It can
damage your eyes.
Lab 8-9-114.DOC
9
7.
Make a diagram of the bands that have formed in each lane in the space provided
on your assignment sheet and use this information to determine which man is
most likely to be the father of this child.
Lab 8-9-114.DOC
10
Assignment
Table 1
Total #
polled
# dominant
phenotype
# recessive
phenotype
Frequency
recessive
phenotype
Frequency
recessive
allele
Frequency
dominant
allele
Frequency
of homozygous
dominant
Tongue
rolling
Ear
lobes
Widow’s
peak
Finger
locking
Dimples
Digital
hair
Freckles
Lab 8-9-114.DOC
11
Frequency
heterozygous
# homozygous
dominant
# heterozygous
1.
Are dominant alleles always more common (have a higher frequency in a
population) than recessive alleles? Support your answer with information
from the data collected.
2.
In order to have the heterozygotes be the largest group in a population
what criteria is necessary in the gene frequencies of the
alleles?
3.
If the frequency of the dominant allele in a population is some number p
and the frequency of the recessive allele is some number q, then why is the
probability of being a heterozygote in the population 2pq rather than just p
X q?
Table 2
Fragment lengths (number of bases)
Chromosome (a)
Chromosome (b)
4.
Did Sam inherit the Huntington allele?
Lab 8-9-114.DOC
12
Drawing of DNA restriction fragment bands after electrophoresis.
Mother
5.
Child
Candidate 1
Candidate 2
Candidate 3
Which of the men (indicate lane number from your drawing) is most likely
the father of the child?
Lab 8-9-114.DOC
13