LS4 Final Exam—Fall 2010
Name:
University ID:
Instructions:
There are 10 problems on this exam. Each is worth 15 points.
1. Make your answers clear and concise.
2. Use the back of each sheet to work out the problems
3. Make it clear what your final answer is. Circling it helps.
4. Write in blue or black ink only.
5. Do not write in the boxes. They are for grading purposes.
Score:
Problem 1
Problem 6
Problem 2
Problem 7
Problem 3
Problem 8
Problem 4
Problem 9
Problem 5
Problem 10
Total
Name:
UID:
Problem 1
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
and show in which step each mutant is defective. (6 points)
biosynthetic pathway
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). (3 points)
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. (6 points)
1
2
3
4
5
6
7
1
2
3
4
5
6
7
-
+
+
+
+
+
+
-
+
-
+
+
-
-
+
+
-
+
-
+
+
-
-
+
+
-
+
-
Name:
UID:
Problem 2
ABO bloodtypes are controlled by three alleles: A B i. There are four phenotypic
bloodtypes: A (AA or Ai), B (BB or Bi), AB (AB) and O (ii).
A) Write out an equation that describes Hardy-Weinberg genotype frequencies when
there are three alleles in a population. Use p, q, and r to represent allele frequencies. (3
points)
p^2 + 2pq + q^2 + 2pr + r^2 + 2qr = 1
B) Nauru is the world’s smallest island nation, with a population of about 10,000 people.
All that is known about the island is that allelic frequency for A is 0.158 and that there
are 625 type O people. Assuming Hardy-Weinberg equilibrium, calculate: (6 points)
i = answer: square root of (625/10,000) = .25
I^B = answer: 1 - .158 (I^A) - .25 (i) = .592
# Bloodtypes: Part B:
total number of people:
A^2
B^2
i^2
+ 2 Ai
+ 2 Bi + 2 AB = 1
.158^2 .592^2
.25^2 2*.158*.592 2*.592*.25 + 2*.158*.592
Multiple each by 10,000 to get the number of people
Bloodtype A = AA + Ai =
1040 (or 1039.64)
AA = .158^2 * 10,000
= 250 people (249.64)
Ai = 2*.158*.25 * 10,000 = 790 people
Bloodtype B = BB + Bi = 6465
(or 6465.65)
BB = .592^2 * 10,000
= 3505 (3504.64)
Bi = 2*.592*.25 * 10,000 = 2960
C) A geneticist claims the population is not in Hardy-Weinberg equilibrium, and that
actual bloodtypes are AA: 1600, Ai: 400, BB: 4525, Bi: 1200, AB: 1650 and O: 625.
What are the actual allelic frequencies? (6 points)
I^ A = 1600 +1600 +400 +1650 / 20,000
I^ B = 4525 + 4525 +1200 +1650 / 20,000
i=
400 + 1200 + 625 + 625 / 20,000
= .2625
= .595
= .1425 or 1-.2625-.595=
Name:
UID:
Problem 3
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 non-homologously. 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. (6 points)
ROH
NeoR
ROH
GFP
B) How will you distinguish cells that integrated the targeting transgene (either
homologously or non-homologously) from cells that did not integrate the transgene? (2
points)
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 homologously from those
that integrated it non-homologously? (2 points)
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? (3 points)
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? (2 points)
50% of the pups that fluoresce red will also have the Parkin knockout allele.
Name:
UID:
Problem 4
Six recessive point mutations are known to reside in 3 phage genes required for making
plaques on R bacteria.
You decide to co-infect these point mutations pairwise to determine which mutants
belong to the same complementation groups. You start by infecting a vial full of millions
of R bacteria with about ten particles of mutant phage A and about ten particles of
mutant phage B.
A) Is this experiment likely to be informative? Why or why not? (Answer in one short
phrase or sentence!) (1 point)
No, with such small amounts of phage A and phage B it is VERY unlikely that both
phage will coinfect the same bacterium.
B) What may be a better way to do this experiment? (Answer in one short phrase or
sentence!) (1 point)
Coinfect with an excess of both mutant phage A and mutant phage B, OR
Coinfect with an excess of one mutant phage and a smaller amount of the other mutant
phage.
You decide that your initial approach had problems and perform several
complementation tests using optimal methods. The results are reported below.
a
b
c
d
a
b
c
d
e
F
0
+
+
+
+
+
0
+
+
0
+
0
0
+
0
0
+
0
0
+
e
f
0
+ = complementation; 0 = non-complementation; and blank = not-tested.
D) From the results of the complementation tests, list below which mutants are in the
same genes (2 points) and complete the chart above (7 points).
a
b,e
c,d,f
1 point for each correct answer on the chart.
Problem continues on the next page
Name:
UID:
E) You find another 3 phage mutants, g, h, and i, that have mutations in the same gene
and decide to determine whether recombination can occur between each pair. In order
to do this experiment, with what relative amounts of each phage will you infect bacteria?
(1 point)
Equal amounts (and enough to saturate bacteria).
F) You obtain the following results from your coinfections:
g + h = rare plaques
h + i = no plaques
g + i = no plaques
Which of these mutations is most likely to be a deletion? Explain very briefly. (3 points)
Mutation i is most likely a deletion (1 point). For the extra 2 points, you need explain
your answer. Restating the information given does not count (e.g. “because i doesn’t
overlap with h or with g” is not worth any points). Acceptable answers include:
-i is a deletion that overlaps g and h. This may also be in the form of a drawing.
-Mentioning that g and h make rare plaques, they are point mutations that cannot
recombine with i.
-i is either a point mutation that is in the same nucleotide position as g and h or a
deletion, but because we know g and h are in different positions (because they
recombine), then i must be a deletion.
Name:
UID:
Problem 5
You are studying a recessive condition in a strain of mice that causes blue eyes instead
of red eyes. You find two microsatellites, C and G that may be useful for mapping this
locus. Each microsatellite has 2 alleles – C1 is 1kb, C2 is 2kb, G4 is 4 kb and G5 is 5
kb. You cross a female mouse that is rr (blue-eyed) C1/C1 and G4/G4 with a male
mouse that is RR C2/C2 and G5/G5. You cross your F1 mice back to the rr C1/C1
G4/G4 mother. You perform PCR for the microsatellites of 16 blue-eyed progeny and
run the results on the gel below.
Blue
Kb
1
2
3
4
5
6
7
8
9
Blue
10
11
12
13
14
15
Red
16 Mom Dad F1
5
4
3
2
1
A) Fill in the bands for the F1 mouse on the gel. (1 point)
See gel
B) What color eyes will the F1 have? (1 point)
Red
C) Which microsatellite is linked to the eye color gene (3 points) and how many map
units away from the gene is it (4 points)?
G. 4/16 = 25% recombination = 25mu
D) If you were to genotype red-eyed progeny, what percentage of each genotype would
you expect at the C and G microsatellites? List proportions of each genotype expected.
C microsatellite (3 points): 50% C1/C1 50% C1/C2
G microsatellite (3 points): 75% G5/G4 25% G4/G4
Name:
UID:
Problem 6
You have collected a set of deletion and point mutations in a bacteriophage gene
required for the formation of plaques.
A) You coinfected the deletion mutants(I-III) with the haploid point mutants (1-5) and
assessed their ability to make plaques. The results are shown below (+ means plaques
were formed). Map out the relative position of the eight mutations. (9 points)
1
2
3
4
5
I
+
-
-
+
-
II
-
+
+
+
-
III
+
-
+
-
+
I:
----------------(
)----------------II:
------(
)----------------------------III:
----------------------------(
)---------Point:
1
5 3
2
4
You discover that this gene has three different “domains.” Each domain has a different
function in the protein. A diagram of the gene is shown below.
Domain A
Q
R
Domain B
S
Domain C
T
B) If there was a single nucleotide insertion at position S, will the reading frame of
domain C be changed? Explain your answer in one sentence or phrase. (2 points)
Yes, because domain C is still in the same gene as domain B.
C) If there was a single nucleotide insertion at position Q and a single nucleotide
deletion at position R, and the phage can still make plaques, what does this tell you
about the function of domain A? Explain in one sentence or phrase. (2 points)
Domain A is not important for the protein’s ability to make plaques.
D) You found a single nucleotide substitution at position R that made the protein
completely non-functional. Given your conclusion above about the function of domain A,
what kind of alteration of the gene could this substitution mutation have caused?
Explain in one sentence or phrase. (2 points)
There was a nonsense mutation that resulted in a premature stop codon.
Name:
UID:
Problem 7
Goltz's syndrome is a dominant X-linked trait. In a population that is in HWE, 1 of every
100 males has this syndrome.
A) What is the allele frequency for Goltz's syndrome allele in the male population? (2
points)
1/100
B) What is the frequency of the standard (non-Goltz) allele in the population as a whole?
(allele frequencies are the same in males and females) (3 points)
99/100
C) What is the percentage of females who suffer from Goltz's syndrome in this
population? (5 points)
1-0.99*0.99=0.0199
D) Joe and Sally both have Goltz’s syndrome and want to know the chances that their
first-born child will be healthy. Both of Joe’s parents have Goltz's syndrome, but his
brother is healthy. If Joe wants to marry Sally, who was adopted and therefore knew
nothing about her parents’ genetic history, what can you advise them about the chance
that their first-born child will be healthy? (To receive any partial credit, you must show
you work concisely and clearly!) (5 points)
2pq/(2pq+p^2)*0.5*0.5=198/199*0.5*0.5=0.25
Name:
UID:
Problem 8
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). (2 points)
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. (2 points)
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 wildtype 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 clutches with both kinds of mutants (e.g. fish that
can’t swim and fish with small fins)? (5 points)
1/16
D) If you crossed F2 females back to their father, what percentage of crosses would
yield clutches with both kinds of mutants? (4 points)
1/4
E) If a cross between the F1 father and an F2 daughter yielded both kinds of mutants,
what percentage of the clutch would be unable to swim and have small fins? (2 points)
1/16
Name:
UID:
Problem 9
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. (6 points)
0 points if you did not draw haplotypes.
Father:
Mother:
+
N
-
N
+
N
-
C
Father: 2 points for exactly as shown. No partial credit.
Mother: 2 points for each correct chromosome (4 total).
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. (5 points)
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. (4 points)
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.
Name:
UID:
Problem 10
You wish to map a zebrafish mutation with respect to the centromere. To do so, you
decide to create fish with no male genetic contribution and double the female genetic
contribution. You destroy the male contribution by UV-irradiating sperm and, in theory,
you could double the female contribution by: 1) Blocking meiosis 1 so that homologous
chromosomes don’t separate, or 2) Blocking meiosis 2 so that sister chromatids don’t
separate.
A) Using the chart below, indicate whether animals in the next generation (write Yes or
No) will have more homozygous loci than their mother, whether they will have
completely non-recombinant chromosomes (i.e. chromosomes identical to their
mother’s), and whether this procedure will be useful for mapping the mutation with
respect to the centromere. (12 points)
Block meiosis 1
Block meiosis 2
More homozygous loci?
No
Yes
Non-recombinant chromosomes?
Yes
No
Useful for mapping?
No
Yes
B) You realize that another way to double the contribution from the mother is to make
haploid embryos and then prevent the first mitotic division of the embryo. Indicate in
the chart below what the consequences of this procedure will be. (3 points)
Block first mitosis
More homozygous loci?
Yes
Non-recombinant chromosomes?
No
Useful for mapping?
Yes
This last 1 point question may have been the trickiest on
the exam. Although you can’t map with respect to the
centromere as easily as with blocking meisois 2, all will
have recombinant chromosomes so SNPs can be used to
find loci associated with a phenotype.