name:_______________________
student ID:_____________________
Genetics L311 final exam
May 6, 2015
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
(31 points possible)
page 4
(24 points possible)
page 5
(28 points possible)
page 6
(29 points possible)
page 7
(48 points possible)
page 8
(10 points possible)
total
(of 200 points possible)
1
name:_______________________
student ID:_____________________
1. Short answers (2 pts each, for total of 30 pts)
A. A protein that specifies different fates depending on its level is a(n) morphogen .
B. Aneuploidy is the condition where the number of chromosomes present differs from
normal by less than a full haploid chromosome set.
C. Many familial cancer syndromes are cell recessive but transmit as autosomal dominant. The
explanation for this phenomenon is that in one or more cells, wild type gene function is lost. This
occurs most commonly through the phenomenon of loss of heterozygosity , where the
chromosome, or a portion of the chromosome with the wild type gene is lost.
D. The variance or total variance provides a measure of variation, including both genetic and
environmental factors.
E. The migration of cancerous cells from one site in the body to another is called metastasis .
F. Epistasis is when the phenotype expected from one locus is dependent on the genotype of a
second locus.
G. A gene altered by dominant (usually) mutation that contributes to the development of cancer
is called a(n) oncogene .
H. During conservative transposition , the number of mobile DNA elements within the cell
remains constant.
I. The complete set of genetic information within all of the individuals in a population is
the gene pool .
J. Several cells that have the same developmental potential (i.e. can adopt the same fate)
constitute a(n) equivalence group .
K. A(n) Robertsonian translocation is a special type of rearrangement in which the long arms
of two acrocentric chromosomes fuse.
L. Cosegregation of genes on the same chromosome is
linkage .
Please provide a brief definition of each of the following:
M. dosage compensation: The normalization of expression of genes on the X chromosome
between males and females.
N. alkyltransferase: An enzyme that removes alkyl groups from DNA such as the ethyl
group from O-6 ethyl guanine.
O. linkage disequilibrium: The occurance of combinations of alleles that differ from what
would be expected by chance, based on their frequency in the population.
2
name:_______________________
student ID:_____________________
2. The frequency of Tay Sachs disease (TS) in the Ashkenazi Jewish population is approximately 1 in
3600. Tay Sachs transmits as an autosomal recessive disease.
A. What are the frequencies of the mutant and wild type Tay Sachs alleles among the Ashkenazi Jewish
population (6 pts)?
q2 = 1/3600, q = 1/60 = 0.0167
p = 0.9833
B. What is the frequency of Tay Sachs carriers (i.e. heterozygotes) in this population (6 pts)?
2(0.9833)(0.0167) = 0.033
C. A couple from this population seeks advice from their genetic counselor. With a previous husband,
the woman had a child with Tay Sachs. The two men are unrelated to each other and her current spouse
does not have any family history of Tay Sachs. What is the risk that their first child will have the disease
(6 pts)?
1(0.033)(0.25) = 0.008
3. In your studies of the exceptionally high-flying bat, S. reifeisae, in Bloomington you discovered three
recessive mutations: (b) produces brown instead of black fur, (c) causes curved instead of straight toes,
and (q) causes unusually loud squeaks. You cross triply heterozygous females with homozygous
recessive males and find the following results:
74 brown fur, curved toes, normal squeaks
567 brown fur, straight toes, loud squeaks
12 brown fur, straight toes, normal squeaks
80 brown fur, curved toes, loud squeaks
14 black fur, curved toes, loud squeaks
578 black fur, curved toes, normal squeaks
86 black fur, straight toes, loud squeaks
89 black fur, straight toes, normal squeaks
1500
A. What is the order of the three genes (3 pts)
bqc
B. What is the distance between the genes (6 pts)?
b and c: 25.4 cM
c and q: (80+89+12+14)/1500X100 = 13.0 cM
b and q: (74+86+12+14)/1500X100 = 12.4 cM
C. 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:
b 8 mu c 22mu q
What is the probability of obtaining a phenotypic + c + offspring from a cross between a heterozygous
(b c q/+ + +) bats and b c q homozygous recessive bats (4 pts)?
0.08(0.22) = 0.0176 but divide by 2 because half of the doubles have the right combination of
alleles so 0.0176/2 = 0.0088
3
name:_______________________
student ID:_____________________
4. Skunks produce a putrid liquid that they spray when they feel threatened. An imaginary genetic
pathway for the formation of this liquid is shown below.
stnk
fnk
odr
stnch
rnk
putrid liquid
What phenotype would be expected for the following mutants (12 points)?
Gain of function fnk: stinky
Gain of function odr: stinky
Loss of function stnch: stinky
Gain of function odr and gain of function stnch: no smell
5. In multi-cellular organisms, specific mechanisms regulate the cell cycle and function to prevent the
proliferation of cancerous cells. What effect will each of the following mutations have on the likelihood
of developing cancer. Please briefly state your rationale (12 points).
Mutation
A non-phosphorylatable but otherwise
functional form of Rb protein
Probability
decrease
Rationale
E2F always bound by Rb
Constitutively active Ras (i.e. always active) increase
Ras promotes proliferation
eliminates Bcl2 function
decrease
will increase apoptosis
Constitutively active Abl
increase
promotes proliferation
4
name:_______________________
student ID:_____________________
6. 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 (6 points)?
constitutively active xol-1
if X:A=0.5
male
if X:A=1.0
male
loss of tra-2
male
male
constitutively active tra-1 and loss of
fem-2
hermaphrodite
hermaphrodite
B. Dosage compensation in Drosophila and C. elegans is implemented after sex determination has
occurred in the developing embryo. Why must this be true? Please explain in the context of what you
know about these processes during Drosophila development (6 points).
Sex determination relies on a difference in expression from genes on the X chromosome between
males and females, for example to turn on the sxl early promoter only in females. If dosage
compensation comes on too early this difference will be lost.
7. The exotic sea squirt N. kumarae is diploid with two pairs of chromosomes, one long and one short.
Mutations in the orn gene cause animals to develop orange coloration, mutations in opq causes them to
be opaque rather than transparent and mutations in ssp cause them to siphon water at an unusually high
rate. A particular individual, named Sharon, is heterozygous for orn and opq, and homozygous mutant
for ssp. You demonstrate that the orn and opq mutant alleles are in the trans configuration. ssp is located
on the short chromosome not linked to orn or opq.
A. Please show one of Sharon’s cells in prophase of mitosis. Please include the alleles of each gene.
Draw a circle around one pair of homologs and a square around one pair of sister chromatids (6 points).
B. Occasionally meiosis goes awry. Please show the produces of meiosis where nondisjunction of the
long chromosome has occurred during meiosis I. You do not need to show the genes (10 pts).
5
name:_______________________
student ID:_____________________
8. Drosophila anterior-posterior body axis is patterned by a series of sequentially acting classes of
genes. A. Please provide the names and function of each class of gene, the order in which they act, and
provide an example of one gene in each class (10 points).
morphogens provide A-P information (bcd, hb or nos)
gap genes begin to divide embryo into regions (knirps or kruppel)
pair rule genes establish
segments(ftz or eve)
homeotic genes specify segment
identity (antp or btx)
segment polarity genes define A-P within segment
and reinforce segment boundaries(wg or en)
B. You find a mutation in a gene that you call antenna-leg, in which the animal has antennae growing
where legs should be. What class of genes is antenna-leg likely to belong to (4 points)?
looks like a homeotic mutation
9. In C. elegans, cell-cell signaling specifies vulval cell fates. These fates are categorized as either
primary, secondary or tertiary. 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 EGF
decrease
increase
can’t activate kinase to inhibit
transcription factors
B. gain-of-function mpk-1 (kinase
increase
decrease
transcription factors inhibited
in all 6 cells
increase
decrease
Ras downstream, activates
kinases
D. transcription factors (lin-1 or lin-31) decrease
that can’t be inactivated
increase
no cells will adopt primary fate
E. loss of transcription factors
plus loss of Ras
decrease
transcription factors
downstream; can’t inhibit
primary fate
downstream of Ras)
C. loss of EGF plus
activating mutation in Ras
increase
6
name:_______________________
student ID:_____________________
10. You discover two true-breeding strains of the brilliant tropical dart frog, E. qinae. Strain A has a
loud trill (=song) and long legs. Strain B has a soft trill and short legs. You cross strain A males with
strain B females and find that the F1 offspring all have a loud trill and long legs. From a cross of F1
animals, you obtain the following:
301
medium trill, medium legs
T–ll
904
loud trill, long legs
T–L–
403
soft trill, short legs
ttL– and ttll
A. Please fill in the genotypes of the F2 progeny on the lines above (8 points).
B. From a cross of medium trill, medium leg F2 X Strain B, what fraction of the progeny will have soft
trill and short legs (8 points)?
T–ll X ttll
2/3(1/2) = 1/3
TTll
=> Ttll
Ttll
=> Ttll and ttll
C. You discover two different strains of the same species of frog that differ in skin color. Strain 1 has
blue skin whereas strain 2 has green. A cross of strain 1 female with strain 2 male gives all greenskinned F1s. Half of the progeny of a cross of F1s are blue and the other half are green. How can you
explain this result? Please name the mode of inheritance spot color displays (8 points).
Looks like maternally imprinted gene. F1s are green because mom’s wild type allele is silenced.
The F2’s are half and half because half inherit the mutant allele from dad and half inherit the wild
type allele.
11. In your studies of the unusual flying fish species D. phamae you find two strains. Strain 1 has light
blue scales and long fins. The second strain has purple scales and short fins. A cross of strain 1 females
with strain 2 males produces F1 that all have lavender scales. Half of the F1’s have long fins and half
have short fins. You cross the lavender scaled, short finned F1s and find:
BBll 1/12 purple scales, long fins
Note: Short fins are dominant but
Bbll 1/6 lavender scales, long fins
recessive lethal. Color is
bbll 1/12 light blue scales, long fins
semidominant.
BBLl 1/6 purple scales, short fins
BbLl 1/3 lavender scales, short fins
bbLl 1/6 light blue scales, short fins
A. Please give the genotypes of the F2s on the lines above (12 points).
B. Purple scale, short finned F2 X light blue scaled, long finned F2. What fraction of the offspring will
have lavender scales and short fins (4 points)?
BBLl X bbll => BbLl and Bbll, so ½ have lavender scales and short fins
C. In another group of the same species of fish you find mutants that produce yellow spots rather than
the usual orange. When you cross true-breeding yellow females with true-breeding orange males, the
7
name:_______________________
student ID:_____________________
progeny all have yellow spots. When you cross these the next generation all have orange spots. How can
you explain this result? Please name the mode of inheritance spot color displays (8 points).
This looks like maternal effect. Mom’s nuclear genotype determines the offspring’s phenotype.
12. Alkaptonuria (aka black urine disease) is a rare inherited genetic disorder that causes the urine of
affected individuals to turn brown or black after prolonged exposure to air. It results from a defect in
tyrosine metabolism. Affected individuals show the trait from birth. The pedigree to the right shows an
extended family including two individuals affected by the trait.
A. What mode of inheritance does alkaptonuria show
(4 points)?
autosomal recessive
B. Obligate carriers are individuals who must carry
the mutation that produces the trait. Please list the
obligate carriers in the pedigree at right (4 points).
I-1, I-2, II-4, III-2, III-3, III-5, IV-2, IV-3
C. What is the probability that II-3 carries the
mutation (i.e. is heterozygous, 3 points)?
2/3
D. What is the probability that IV-2 and IV-3’s next
child will be a boy who is affected by the disease (3 points)?
½ X ½ X ½ = 1/8
8