Download These practice questions are from prior LS4 finals and are courtesy

These practice questions are from prior LS4 finals and are courtesy of Drs. Laski, Chen
and Sagasti.
A few short questions – no short answer questions on the exam but good practice
1. If a grandfather has a Y-linked trait, what is the probability that his grandson (his daughter's
son) will have this trait?
Zero. He does not pass on his Y chromosome to his daughter.
2. A male is affected with the X-linked recessive condition hemophilia. He marries a carrier
woman. What percentage of their male children will be affected?
50%. The father’s status is unimportant because he does not pass his X-chromosome on to
his male offspring.
3. A woman is the offspring of a father with the X-linked recessive condition hemophilia. What
is the likelihood that her first child will be affected?
25%. The woman must be a carrier so there is a 50% chance she will pass on the mutant
allele. There is also a 50% chance her child will be male.
4. The ability to roll the tongue is controlled by a single gene with two alleles. The allele for
being able to roll is dominant over the allele for not being able to roll. In a given population of
10,000 individuals there are 7400 tongue rollers and 2600 non-rollers. Assuming the population
is at Hardy-Weinberg equilibrium, how many heterozygous tongue rollers are there in the
population?
q2 = .26, q = .51, p = .49, 2pq = .4998 = 4,998 people out of 10,000
5. Phenylketonuria (PKU) is a disease in which people who are homozygous for the recessive
PKU allele lack the ability to properly breakdown the amino acid phenylalanine. Build up of
phenylalanine causes mental retardation in the children born with PKU. Given that PKU disease
occurs in 1 out of 10,000 births and that the U.S. population is so large that Hardy-Weinberg
equilibrium has been achieved for this gene, 1) what is the PKU allele frequency, and 2) what is
the proportion of carriers in the population?
q2 = 1/10,000, q = 1/100; 2pq = 2/100 = 1/50
6. The wild type zebrafish has short tail. From a genetic screen, you identified three recessive
mutations that cause the long-tail phenotype (A, B and C). You cross the true breeding mutant
A fish to the true breeding mutant B fish, and found that all progeny have normal tail. However,
all progeny from the cross of true breeding mutant A fish to the true breeding mutant C fish have
long tails. Provide a genetic explanation for this phenomenon. Circle the correct answer(s).
1. A and B are allelic.
2. A and C are allelic.
3. B and C are allelic.
4. The allele strength is A > C > B.
7. E. coli strain B is doubly infected with two rII mutants of phage T4. 0.1 ml of a 106 dilution
of the progeny is plated on E. coli B and 0.1 ml of a 104 dilution of the progeny is plated on E.
coli K. 50 plaques appeared on strain B, 6 on strain K. Calculate the recombination frequency
between these two mutations.
2 X 6 / 5000 = 0.0024 = 0.24%
8. Which termination codon(s) can be suppressed by a point mutation in the anticodon of trp
tRNA? UAG and UGA
Longer problems
1. Manx cats are tailless and when crossed with one another produce on average 1 long-tailed
(wild type) cat for every 2 Manx. The M (Manx) allele is lethal in homozygous condition due to
problems arising during development. Thus, a MM genotype is lethal, a Mm cat is Manx
(tailless), whereas a mm cat is wild type with a long tail.
A. A large number of Manx cats (Mm) are put on an island and reproduce. The F1 progeny are
composed of Manx and wild type cats in a 2:1 ratio, respectively. What is the allele frequency of
the M allele in this F1 population. (circle correct answer)
1 3/4 2/3 1/2 1/3 1/4 0 none of above
If none of above, the correct answer is:
B. The F1 population breeds randomly among themselves, producing an F2 population. What
fraction of the F2 population are Manx (have no tails). (circle correct answer)
1 3/4 2/3 1/2 1/3 1/4 0
If none of above, the correct answer is:
Keep in mind that the MM offspring die so you have to correct for this in the surviving
population.
2. Ten percent of the males of a large and randomly mating population are colorblind, a recessive
X-linked trait. A representative group of 1000 from this population migrates to a South Pacific
island,where there are already 1000 inhabitants and where 30 percent of the males are colorblind.
Assuming that Hardy Weinberg equilibrium applies throughout (in the two original populations
before emigration and in the mixed population immediately following the arrival of the
immigration), what fraction of males and females can be expected to be colorblind in the
generation immediately following the arrival of the immigrants?
percent females colorblind: 4%
percent males colorblind: 20%
3. Below is shown the RNA sequence from the imaginary protein coding region of the rII gene of
bacteriophage T4. As you can see the gene encodes a protein that is 7 amino acids long. Previous
analysis shows that the Val-Val-Val amino acids at the C terminus of the protein (underlined
below) are all that is required for rII+ activity. You isolate 4 mutations in the gene. Mutation #1
and #2 are both positive frameshift mutations, the base inserted is shown in bold. Mutations #3
and #4 are both minus frameshift mutations, the base missing is shown as a gap in the sequence.
rII phenotype
Wild Type 5'UUAUGCCUGGUAAAGUCGUCGUCUGAUACUAA
rII+
MetProGlyLysValValValStop
#1
#2
#3
#4
5'UUAGUGCCUGGUAAAGUCGUCGUCUGAUACUAA3'
5'UUAUGCGCUGGUAAAGUCGUCGUCUGAUACUAA3'
5'UUAUGCCU GUAAAGUCGUCGUCUGAUACUAA3'
5'UUAUGCCUGGUAA GUCGUCGUCUGAUACUAA3'
rIIrIIrIIrII-
A. The following double mutants are made. For each double mutant write out the sequence of
the rII protein it will make. Predict (circle) whether the phage will be rII+ or rII-.
Double mutant
#1, #3
rII+ or rII-
Sequence
5’UUAGUGCCUGUAAAGUCGUCGUCUGAUACUAA3’
No Met codon so no protein
#2, #3
rII+ or rII5’UUAUGCGCUGUAAAGUCGUCGUCUGAUACUAA3’
MetArgCysLysValValValStop (could be rIIif ArgCys can’t functionally substitute for ProGly
#2, #4
rII+ or rII5’UUAUGCGCUGGUAAGUCGUCGUCUGAUACUAA3’
MetArgTrpStop
4. The human gene for hemophilia is on the X chromosome. Below is a pedigree from a family
affected with hemophilia. Blackened symbols indicate that the person has hemophilia. To help
with genetic diagnosis, a probe that detects an RFLP (restriction fragment length polymorphism)
on the X chromosome is used. This probe detects either a 7 kb restriction enzyme fragment or
3 kb and 4 kb restriction enzyme fragments. The RFLP pattern for all the members of the
pedigree is shown. The recombination distance between the RFLP and the hemophilia locus is
10%. What information could be given to the woman designated with the arrow as to the
likelihood of her first son having hemophilia?
The disorder segregates with the 3+4 kb allele. She inherited the 7 kb allele from her
carrier mother. There is a 10% chance she is a carrier.
5. You isolated 7 haploid mutant yeast strains that are unable to make cholesterol, and therefore
die in minimal media. You culture them separately and add in each of 4 intermediates. The
results are shown below (+ means ability to grow).
C
J
P
S
Cholesterol
1
+
+
-
+
+
2
-
-
-
+
+
3
-
+
-
+
+
4
-
-
-
+
+
5
-
-
-
-
+
6
-
+
-
+
+
7
-
-
-
+
+
A) Draw out the biosynthetic pathway and show in which step each mutant is defective.
P ---(1)---> C ---(3, 6) ---> J ---(2, 4, 7) ---> S ---(5)---> Cholesterol
B) For the following double mutants, indicate which compound accumulates (i.e. the last
compound in the pathway that can be produced).
1, 3
____P______
2, 4
____J______
1, 5
____P______
C) If you assume that each step in this pathway requires just one enzyme, indicate the expected
results of a complementation test among the 7 mutants. Assume the mutations are recessive. Use
+ to indicate complementation.
1
2
3
4
5
6
7
1
2
3
4
5
6
7
-
+
+
+
+
+
+
-
+
-
+
+
-
-
+
+
-
+
-
+
+
-
-
+
+
-
+
-
6. You decide to make a mouse model of Parkinson’s disease by knocking out the gene Parkin.
Unfortunately, there is a world-wide shortage of ganciclovir, but you realize that you can use
GFP in place of the typical negative selection to screen out ES cells in which the targeting
transgene has integrated into the incorrect location by non-homologous recombination. One
copy of GFP is sufficient to obtain green fluorescent cells.
A) Draw the DNA targeting transgene for knocking out the Parkin gene, including the location
of any genes used for selection and regions of homology.
GFP
5’-promoter
Parkin Exon(s)
NeoR
Parkin Exon(s)
terminator -3’
B) How will you distinguish cells that integrated the targeting transgene (by either homologous
or non-homologous recombination) from cells that did not integrate the transgene?
Add neomycin to the ES cell culture medium: all cells which did not integrate the Neomycin resistance
gene will die.
C) How will you distinguish cells that integrated the transgene by homologous recombination
from those that integrated it in the incorrect place by non-homologous recombination?
ES cells in which homologous recombination occurred will have not have GFP, whereas cells in which
non-homologous integration occurred will have the whole transgene, including GFP, and will thus
fluoresce green. Pick the non-green cells to obtain your knock out line!
To help differentiate between the host blastocyst genome and the ES cell genome, you performed
your integration in ES cells that were hemizygous (have one integrated copy) for a transgene that
causes Red Fluorescent Protein (RFP) expression in all cells. One copy of RFP is enough to
make cells fluoresce red. You inject the knockout cells into a wildtype blastocyst and cross the
resulting chimeras to wildtype mice.
D) One of the chimera x wildtype pairings produces 40 pups, 5 of which fluoresce red. What
percentage of this chimera’s germline derives from ES cells?
25% of the chimera’s germline is derived from the ES
E) Of the red pups, what percent will also have a Parkin knockout allele?
50% of the pups that fluoresce red will also have the Parkin knockout allele.
7. You performed a mutagenesis screen in zebrafish. You first mutagenized male fish with the
chemical ENU, then crossed them to wildtype females to create a set of F1 animals.
A) If you crossed two of these F1 animals to each other, would you be likely to see the
phenotype of a recessive mutant in the F2? Why or why not? Answer in one sentence or
phrase. (Note that zebrafish don’t have sex chromosomes).
No. The mutagenesis (of the male’s germline) was random and the chance of two F1
progeny having the exact same gene mutated is very low.
B) If instead you mutagenized both males and females and crossed these together to make the F1
generation, would you be likely to see a recessive phenotype in the F2? Why or why not?
Answer in one sentence or phrase.
No. The mutagenesis (of the male’s and female’s germlines) was random and the chance
of two F1 progeny having the exact same gene mutated is still very low.
An F1 male carried two recessive mutations on different chromosomes that could cause
phenotypes—inability to swim and small fins. You crossed this fish to a wild type female to
create an F2 family.
C) If you performed random sibling crosses within the F2 generation of this family, what
percentage of crosses would yield fish with both kinds of mutants (e.g. fish that can’t swim and
fish with small fins)?
1/16
D) If you crossed F2 females back to their father, what percentage of crosses would yield
clutches with both kinds of mutants?
1/4
E) If a cross between between the F1 father and an F2 daughter yielded both kinds of mutants,
what percentage of the offspring would be unable to swim and have small fins?
1/16
8. A single nucleotide polymorphism is very closely linked to a gene for which there is a
recessive disease allele. This disease is not sex-linked. You realize that you can assay this SNP
with a RFLP. You genotype a family with this RFLP and obtain the following gel.
A) If there was no recombination between the RFLP and the gene, what were the parental
haplotypes in the father and mother (i.e. what alleles of the RFLP and the disease gene were on
the same chromosome). Indicate both chromosomes for both parents. Use + and – to indicate
the wildtype and disease alleles, and N (not cutting) and C (cutting) to indicate the RFLP alleles.
Father:
Mother:
B) If the unaffected daughter in the pedigree has a child with a man who is a heterozygous
carrier of the disease allele, what are the chances that their first child will be affected by this
disease? Again, assume no recombination.
The unaffected daughter is a carrier. The probability of having an affected child when both parents
are carriers is ¼.
C) If the unaffected son in the pedigree has a child with a woman who is a heterozygous carrier
of the disease allele, what are the chances that their first child will be affected by this disease?
Again, assume no recombination.
There is a 50% chance that the unaffected son is a +/-. The probability of two carriers having an
affected child is ¼. So the answer is: (½) (¼) = 1/8.
9. Whether the earlobe is free or attached is inherited in a simple Mendelian manner. Hanging
earlobes are dominant to attached earlobes. A group of scientists is trying to identify the gene
responsible for this trait by positional cloning with microsatellites. They have located a family
where the mother has attached earlobes, the father has hanging earlobes, and of 12 children, 6
have attached earlobes. The results of the microsatellite analysis of the mom, dad, and 12
children are shown below. Microsatellites A and B come in two forms, A1 and A2, B1 and B2.
Ee designates people with hanging earlobes, and ee designates people with attached earlobes.
ee
Ee
ee
ee
Ee
Ee
Ee
Ee
Ee
ee
Ee
ee
ee
Ee
Mom
Dad
1
2
3
4
5
6
7
8
9
10
11
12
B2
B1
A2
A1
A) Which microsatellite is linked to the gene that causes attached earlobes?
Marker A is linked to the gene for the phenotype because Dad gave allele A2 to 6 Ee
(hanging) and only 1 ee (attached). The latter child is a recombinant. He also gave allele
A1 to 4 ee offspring and only 1 Ee child, the latter being a recombinant. Marker B is
unlinked because Dad gave allele B1 to 3 Ee and 3 ee children, which is random
inheritance.
B) Which microsatellite allele is linked to the “e” allele in the father? The Dad’s “attached” (e)
allele is A1.
C) How far apart are the attached earlobes gene and the microsatellite listed in part A? (Hint:
Because the mother is homozygous at both loci, you cannot consider her when calculating
recombination frequency. Just consider the paternal chromosomes!) Since mom is homozygous
at both markers, the only alleles we can use for recombination assessment are Dad’s. 2
recombinants in 12 measureable events = 16.6%.
10. You discover a population of flies living in your coffee maker and notice that some of them
have slightly rough eyes. You notice that a separate population, living in your trash bin, also
includes some flies with rough eyes. Upon further investigation, you realize that both
populations (M1 and M2) have a recessive lethal mutation that causes the rough eye phenotype
in a heterozygote. In other words, when homozygous, the mutation leads to death, when
heterozygous, it leads to rough eyes. You counted the genotypes in both populations and found
that M1 has 100 rough-eyed flies and 500 wildtype-eyed flies, whereas M2 has 300 rough-eyed
flies and 300 flies with wildtype eyes.
A) Calculate the allele frequencies of both populations (M1 and M2).
M1
M2
p=
.91
p=
.75
q=
.09
q=
.25
B) You’ve realized that both populations have a mutation in the same gene. You put the flies
together to make a single population. Calculate the new allele frequencies of the single
population.
Populations are of equal size so average the allele frequencies.
p=
.83
q=
.17
C) Assuming Hardy-Weinberg Equilibruim, if you count 3000 flies in the next generation, how
many flies of each genotype do you expect? (write out all genotypes and number of flies for
each)
WT = p2 = (.83)(.83) = .6889 = .69
Rough = 2pq = 2(.83)(.17) = .2822 = .28
Lethal = q2 = (.17)(.17) = .0289 = .03
After eliminating the lethals, the normalized phenotypic frequencies are:
WT = .69/.69+.28 = .71 of the total = 2130 flies
Rough = .28/.69+.28 = .29 of the total = 870 flies