name:_______________________
student ID:_____________________
Genetics L311 final exam
December 19, 2014
Directions: Please read each question carefully. Answer questions as concisely as possible. Excessively
long answers, particularly if they include any inaccuracies, may result in deduction of points. You may
use the back of the pages as work sheets, but please write your answer in the space allotted. However,
you must show all your work. Clearly define your genetic symbols. We will not make guesses as to
what a particular symbol is intended to mean. Also, don’t assume that strains are true-breeding unless
this is stated in the question. Finally, show all your work. Good luck.
page 2
(30 points possible)
page 3
(36 points possible)
page 4
(22 points possible)
page 5
(24 points possible)
page 6
(19 points possible)
page 7
(29 points possible)
page 8
(28 points possible)
page 9
(12 points possible)
total
(of 200 points possible)
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name:_______________________
student ID:_____________________
1. Short answers (2 pts each, for total of 30 pts)
A. Activation of a cell death program involves activation of cellular proteases
called caspases .
B. The mammalian gene found on the Y chromosome that specifies male development is
called sry .
C. Traits determined by the interactions of many genes, often in combination with environmental
factors, are said to be multifactorial .
D. One of the hallmarks of cancer, metastasis , is the spread of cancerous cells to new sites.
E. The pair rule genes specify the development of alternating segments during Drosophila
development. (I’m looking for the class of genes, not examples of members of this class.)
F. Conservative transposon is a DNA element that can excise itself from a chromosome and
move to a new position without changing the number of elements within the genome.
G. A cell or individual that deviates from a whole multiple of the haploid or monoploid
chromosome complement (eg. having one extra chromosome or missing one chromosome) is
said to be aneuploid .
H. Heritability (H2) provides an estimate of the genetic contribution to a trait.
I. A distinct breeding group within a larger interbreeding group of individuals is referred to as
a(n) subpopulation .
J. The tendency of two genes located on the same chromosome to co-segregate is
called linkage .
K. The failure of chromosomes to separate during mitosis or meiosis is nondisjunction .
L. A plant or animal with four sets of chromosomes, regardless of their origin, is said to
be tetraploid .
Please provide a brief definition of each of the following:
M. oncogene: A mutated gene that contributes to the development of cancer.
N. gene pool: The complete set of genetic information present in all the individuals in a
population.
O. reciprocal translocation: Exchange of DNA between non-homologous chromosomes.
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name:_______________________
student ID:_____________________
2. A recent study of autosomal dominant polycystic kidney disease in the Seychelles Islands estimates
the frequency of the disease at 0.0006. Assume that this population meets Hardy-Weinberg criteria.
A. What are the frequencies of the mutant and wild type alleles (6 points)?
2pq = 0.0006, p~1
or p2 = 1 – 0.0006 = 0.9994
q = 0.0003
p = 0.9997
p = 0.9997
q = 0.0003
B. What is the frequency of homozygous mutant individuals in this population (7 points)?
(0.0003)2 = 0.00000009 or 9 X 10-8
or (3 X 10-4)2 = 9 X 10-8
C. In order for Hardy-Weinberg to apply, several criteria must be met. List any three of these criteria (6
points). 1. random mating
5. no new mutation
2. allele frequencies same in male and female
6. large enough population that random
3. no selection
events are not significant
4. no migration
3. In beetles there are three mutants, q, r, s where each mutant is recessive to its wildtype allele. A cross between
a heterozygous female for all three loci and a homozygous mutant male yielded the following results:
+
+
q
q
q
+
q
+
r +
+ +
+ +
+ s
r s
+ s
r +
r s
431
47
5
420
52
22
17
6
A. What is the order of the genes on the chromosome (4 points)?
qsr
B. What is the map distance between the three linked genes (9 points)?
q-r: 16 cM
r-s: (47 + 52 + 5 + 6)/1000X100 = 11 cM
s-q: (22 + 17 + 5 + 6)/1000 X 100 = 5 cM
Assume for the rest of the problem (warning this does not necessarily represent the correct answer for a and
b) that the order and distance are the following:
r 15mu q 30mu s
C. What is the probability of obtaining a phenotypic + q + offspring from a cross between a
heterozygous (r q s/+ + +) flies and r q s homozygous recessive flies (4 points)?
(0.15)(0.30) = 0.45 but only half of these are the + q + chromosome, so 0.045/2 = 0.0225 or 2.25%
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name:_______________________
student ID:_____________________
4. Below is a signaling pathway that regulates wing development in Drosophila, among many other
things. For the purposes of this question we will ignore any other functions of this signaling pathway.
wnt
frz
gsk3
cat
tcf
wing
What phenotypes do you expect mutations in the following to produce? Note: All mutations are loss-offunction unless otherwise indicated (gf = gain of function, assume this produces the equivalent of
continuously active protein, 8 points).
wnt (gf)
wing or maybe excess wing
gsk3
wing or maybe excess wing
tcf
no wing
wnt (gf); tcf
no wing
5. A. We looked at the genetic pathway that determines gender in C. elegans. The pathways is shown
below. What phenotype would we expect to see result from the following mutations (8 points)?
loss of xol-1
if X:A=0.5
hermaphrodite
if X:A=1.0
hermaphrodite
loss of fem-2
hermaphrodite
hermaphrodite
loss of tra-2
male
male
loss of tra-2 and loss of
fem-2
hermaphrodite
hermaphrodite
B. Most developmental pathways are conserved from simpler to more complex animals. An exception to
this general rule is provided by sex determination. Please list three ways in which sex determination is
not conserved between C. elegans, Drosophila and mammals (6 points).
worms
flies
mammals
determined at cellular level determined at cellular level
determined at level of whole
animal
xol-1 pathway
sxl pathway
sry pathway
X:A ratio
X:A ratio
Y chromosome
4
name:_______________________
student ID:_____________________
6. Progression through the cell cycle is tightly regulated. In particular, we looked at how the cell
regulates entry into S phase.
A. Please illustrate the steps involved in a cell’s decision to enter S phase. Include the relevant proteins.
You may use a diagram if you wish (6 points).
B. Cells use several different checkpoints to verify that essential steps are completed before proceeding
to the next event during the cell cycle. Please show how the cell arrests in the cell cycle prior to entry
into S phase if DNA damage is detected. Include the relevant proteins (6 points).
C. Using the pathway discussed in lecture, first determine whether the following mutations would
increase or decrease the probability of cancer, or have no effect. Please briefly state your rationale (12
points).
Mutation
Probability
Rationale
Loss of Rb
increase
E2F is free to drive cell cycle progression
Loss of Ras
decrease
Ras promotes proliferation
Loss of Bcl2
decrease or
no change
Makes it easier to kill cells
Loss of Rb plus
overexpression of Bcl2
increase
Each of these mutations increase the
risk.
5
name:_______________________
student ID:_____________________
7. The rare deepwater jellyfish K. hunterae is a diploid with two pairs of chromosomes, one long and
one short. Mutations in the lng gene cause animals to develop extremely long tentacles, mutations in blu
cause them to lose their normal blue spots and mutations in bky cause them to develop enlarged beaks. A
particular individual, named Fred, is heterozygous for lng and blu, which are on the long chromosome,
and homozygous mutant for bky. You demonstrate that the lng and blu mutant alleles are in the cis
configuration. bky is not linked to lng or blu.
A. Diagram all of the possible gametes that can result from meiotic division of Fred’s cells, including
gametes that would be formed. Include genes. Assume a single cross over between lng and blu (5 pts).
lng–
lng–
blu–
bky–
bky–
blu+
lng+
bky–
blu–
lng+
bky–
blu+
C. You find an unusual K. hunterae individual who is triploid. Please diagram this individual’s
chromosomes in the most likely arrangement in metaphase of meiosis I. You need not include genes.
Please draw a circle around one set of homologues and a box around a single pair of sister chromatids (6
points).
8. When traveling in a cave near Bloomington, you came across a species of bat, C. smithae. Two
recessive mutations are known, one causing glow-in-the-dark wings (g) and the other causing hairy feet
(h). When trying to create a monster bat for a reality TV show, you cross heterozygous parents, both
with the mutations in trans, hoping to produce bats with hairy feet and glowing wings. Since the
mutations are only 4 mu apart you assume no double crossovers. What percentage of offspring will have
glowing wings and hairy feet (8 points)?
g +/+ h X g +/+ h
Recombinant gametes are 4% from each parent, of which half
g+
will give the desired genotype offspring so
+h
0.02(0.02) = 0.0004 or 0.04%
gh
++
9. You have cloned the soc gene from mouse, a gene that you hypothesize promotes social behavior.
6
name:_______________________
student ID:_____________________
A. What would be the best test of your hypothesis? Please include the steps involved and a diagram of
the vector used in the experimental procedure (8 points).
1. Infect ES cells from black mouse using vector shown above. 2. Select for recombinants in
neomycin and ganciclovir. 3. Verify that recombinants had deletion within soc by Southern or
PCR. 4. Mix engineered ES cells with embryo derived from white mouse. 5. Implant into
pseudopregnant mouse to produce chimeric mice. 6. Cross chimeric X white to find those chimeric
mice that produce all black offspring. Use Southern or PCR to ID hets. 7. Cross hets to generate
homozygotes.
B. To further understand the function of soc, you wish to determine where in animals the gene is
expressed. What method would you use to obtain this information? Please include a diagram of the
vector that you would use but you do not need to include the steps involved (6 points).
Use a reporter transgene. You could fuse either the entire soc gene or just its promoter to gfp (or
lacZ or luciferase)
10. In C. elegans, cell-cell signaling specifies vulval cell fates, causing the Pn.p cells to adopt primary,
secondary or tertiary fates. Give the predicted affect of the following mutations in each of the cell fates
listed (i.e. increase, decrease or no change for each cell type, 3 pts each, 15 pts total):
primary
tertiary
rationale
A. loss of lin-3/EGF
decrease
increase
pathway can’t inhibit
transcription factors
B. gain-of-function mpk-1 (kinase
downstream of Ras)
increase
decrease
transcription factors
inhibited in all 6 cells
C. gain of mpk-1 plus
loss of EGF
increase
decrease
kinase is
downstream, will
inhibit factors
D. loss of transcription factor
(lin-1 or lin-31)
increase
decrease
all cells adopt
primary fate
E. gain of mpk-1 plus
increase
decrease
loss of transcription factor
11. While exploring the outer regions of the galaxy, Captain Kirk and his crew
7
all cells adopt
primary, trans.
factors downstream
name:_______________________
student ID:_____________________
discover a unique creature called a Tribble. Tribbles are fuzzy, round, purring creatures that can
reproduce very quickly and make great pets. So merchants at the nearby Starbase are very interested in
tribble genetics. Specifically, they want to know about the genetics of the three most common colors:
black, white and tan. A purebred black tribble is bred to a purebred white tribble. F1s are crossed to give
the F2s shown below:
453 Black
B–W–
301 Tan
B–ww & bbW–
49 White
bbww
A. Please give the genotypes of the F2s shown above (8 points).
B. White tribbles from the F2 generation are crossed to black tribbles from the F2 generation. What is the
probability of getting tan tribbles from this cross (6 points)?
(1/9) bbww X BBWW => all BbWw (black)
(2/9) bbww X BbWW => 1/2 BbWw (black) and 1/2 bbWw (tan)
(2/9) bbww X BBWw => 1/2 BbWw (black) and 1/2 Bbww (tan)
(4/9) bbww X BbWw => 1/4 BbWw (black), 1/4 bbWw (tan), 1/4 Bbww (tan), ¼ bbww (wh)
So the answer is 1/9(1) + 2/9(1/2) + 2/9(1/2) + 4/9(1/4) = 4/9
C. A few tribbles are found to have unusually short fur. Short-furred males are crossed to wild-type
females producing the F1s shown below. Short-furred F1 males are then crossed to short-furred female
F1s to produce the F2s shown below.
F1s:
½ long furred
½ short furred
F2s
2/3 short furred
1/3 long furred
How do you explain these results (6 points)?
The 2:1 ratio in the F2 suggests that there might be lethality in the homozygous mutants.
12. Imagine that you are counseling a couple who have just had a child born with Down syndrome,
which typically results from trisomy 21. The couple’s first child was also affected by Down syndrome.
When you examined the child’s karyotype you find 46 chromosomes.
A. How might you explain this result (5 points)?
Robertsonian translocation
B. The couple wish to have another child. Which of the following provides the best estimate of the
probability that the couple’s next pregnancy will be affected by Down syndrome (3 points)?
i. the couple’s next pregnancy is certain to be affected by Down syndrome (probability = 1)
ii. the couple’s next pregnancy is certain not to be affected by Down syndrome (probability = 0)
iii. the probability is the same as for the general population, which is ~1/25,000
iv. the probability is higher than the general population risk, and may be as high as 1/3
v. based on Mendelian principles, the probability is exactly 1/2
8
name:_______________________
student ID:_____________________
13. Kennedy disease is a disorder that causes muscle cramps and progressive weakness. The following
is a human pedigree of a family affected with this disorder. Kennedy disease is extremely rare in the
population.
A. What is the most likely mode of inheritance of this
disorder (2 points)?
X-linked recessive
B. What are the genotypes of the following individuals in
the pedigree (6 points)?
I-1: XAXA
II-2: XAXa
III-2: XAXA or a
?
C. What is the probability that individual IV-1 will be affected (2 points)?
½X½=¼
D. What is the probability that III-7 carries the mutation that produces Kennedy disease (2 points).
½
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