name:_______________________
student ID:_____________________
Genetics L311 exam 2
March 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
_______
(20 points possible)
page 3
_______
(25 points possible)
page 4
_______
(23 points possible)
page 5
_______
(23 points possible)
page 6
total
(9 points possible)
_______ (of 100 points possible)
1
name:_______________________
student ID:_____________________
1. Short answers (2 points each, 20 points total)
A. Base analogs are compounds that can be incorporated into the DNA during replication
and can cause mis-pairing, thereby inducing mutation.
B. A mutation where the codon encodes the same amino acid is called silent or
synonymous .
C. Epistasis is when the phenotype expected from one locus is dependent on the genotype
of a second locus.
D. A nucleosome is composed of a histone core around which DNA is wrapped.
E. Recombinational repair uses homologous DNA to repair DNA damage.
F. A mutation in which the gene acquires some new activity is called a gain-offunction mutation.
G. A linear representation of the arrangement of genes on a chromosome is called a genetic
map (aka chromosome or linkage map) .
For the following, please provide a brief definition of the term given:
H. conditional mutation: Mutation that produces a phenotype only under certain
environmental conditions.
I. linkage: The tendency of genes located on the same chromosome to co-segregate.
J. alkyltransferase: An enzyme that removes alkyl groups from DNA such as the ethyl
group from O-6 ethyl guanine.
2
name:_______________________
student ID:_____________________
2. In nematodes three linked genes produce animals that are said to be dumpy (= Dpy, produced by
dpy-17), uncoordinated (= Unc, produced by unc-32) and variably abnormal (= Vab, produced by vab7) when mutant. You cross triply heterozygous animals with homozygous recessive and find the
following results.
wild type
Unc
Vab
Dpy
Dpy Vab
Dpy Unc
Unc Vab
Unc Vab Dpy
6
4677
157
163
4670
152
170
5
10000
A. What is the order of the genes (3 points)?
dpy-17 unc-32 vab-7
B. What is the map distance between the three linked genes (9 points)?
unc-32 – vab-7: (163 + 170 + 6 + 5)/10000 X 100 = 3.4 cM
vab-7 – dpy-17: 3.2 + 3.4 = 6.6 cM
unc-32 – dpy-17: (152 + 157 + 6 + 5)/10000 X 100 = 3.2 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:
unc-32 10mu vab-7 20mu dpy-17
What is the probability of obtaining a phenotypic unc-32 vab-7 + offspring from a cross between a
heterozygous fly(unc-32 vab-7 dpy-17/+ + +) and the triple recessive nematode (4 points)?
0.20 = freq. SCO + freq. DCO; freq. DCO = 0.1(0.2) = 0.02
0.20 = freq. SCO + 0.02 so freq. SCO = 0.18
Half of the SCO’s between vab-7 and dpy-17 produce the requisite chromosome so answer is 0.09
3. A man is homozygous mutant for a maternally silenced gene. If he marries a woman who is
homozygous wild type (Assume all who marry in are wild type, 9 points):
A. What is the probability that his sons will show the trait? 1
B. What is the probability that his daughters will show the trait? 1
C. What is the probability that his daughter's sons will show the trait? 0
D. What is the probability that his daughter's daughter will show the trait? 0
E. What is the probability that his son's sons will show the trait? 1/2
F. What is the probability that his son's daughters will show the trait? 1/2
3
name:_______________________
student ID:_____________________
4. While on an African safari, you find two true breeding strains of a unique giraffe. One strain has
bright purple spots and the other has turquoise spots. When the two strains are crossed, all the F1
individuals have bright purple spots. When F1 individuals are crossed, you find:
595 purple spots P–T– and ppT–
153 purple and turquoise spots P–tt
52 turquoise spots pptt
A. Please give the genotypes of the F2s (8 points).
B. If you cross a purple and turquoise spotted F2 with a turquoise spotted F2, what is the probability
that the offspring will have purple and turquoise spots (6 points)?
P–tt X pptt
1/3 PPtt X pptt => Pptt
2/3 Pptt X pptt => ½ Pptt and ½ pptt
Probability of purple and turquoise = 1/3(1) + 2/3(1/2) = 2/3
5. Two mutations are observed in E. qinae, a brilliant
tropical dart frog. One, b, turn the frog’s spots blue instead
of the normal red. The second, t, causes the frog’s toes to
be shorter than normal. You cross heterozygous females by
homozygous recessive (each mutation is recessive) males.
A. You suspect that the genes are linked. Therefore you use
chi square to text this possibility. What is the probability
that the genes are linked? What can you conclude (9 points)?
(421 – 400)2/400 + (379 – 400)2/400 = 1.1 + 1.1
= 2.2 This means the two genes have a 10%
probability of being unlinked. There is a ~90%
probability that
the genes are linked. We conclude that the
genes may be unlinked because 90% does not
reach the threshold of significance.
4
blue spots, short toes
blue spots, normal toes
red spots, short toes
red spots, normal toes
Total
214
191
188
207
800
name:_______________________
student ID:_____________________
6. In studies of the rare Pham’s field mouse, you find a most unusual individual animal. Recessive
mutation in g causes green fur. Recessive mutations in the linked gene r causes the fur to appear
ruffled rather than smooth. The remarkable individual has smooth brown fur everywhere, except for a
single large patch of green, ruffled fur.
A. Name a process that might produce such an unusual individual (4 points).
mitotic recombination (could also be mitotic nondisjunction or chromosome loss)
B. Diagram the process, showing relevant chromosomes and genes (10 points).
g
g + +
r
r
g
+
r
+
g
+
r
+
+ +
+
g
r
g
+
r
+
+
7. In class we discussed five different epigenetic phenomena. Please name any three of these and
briefly explain the mechanisms by which each functions (9 points).
1. Transcriptional control of cell fate: activation of a transcription factor such as MyoD causes
continued MyoD synthesis, maintaining cell fate
2. Genomic imprinting: silencing of gene inherited from mother or father by methylation of DNA
3. Position effect variegation: Gene is silenced when placed in a new chromosomal context by
chromosome rearrangement. Expansion or retraction of heterochromatin:euchromatin
boundary leads to repression in some cells and expression in others.
4. X chromosome inactivation: One of the two X chromosomes is largely silenced by combined
effects of Xist RNA, DNA methylation and histone modification
5. Prion: trait is transmitted by misfolded protein that interacts with native, endogenous protein
promoting its misfolding.
5
name:_______________________
student ID:_____________________
8. The giant west coast banana slug grows 5 – 6 inches long and is normally found in shades of yellow
or brown. Suppose you have identified the gene that controls pigmentation in banana slugs. The wildtype sequence includes the sequence:
ATG TCT TAT TGT ATT GGG GTG
Met Ser Tyr Cys Ile Gly Val
You sequence the gene from several strains and find the results given below. In each case please name
the type of mutation, give the sequence of the resulting protein and give an example of a mutagen other
than ionizing radiation that is likely to produce the change listed. An asterisk indicates the approximate
location of the mutation. Do not use the same mutagen more than once (3 points each).
*
A.
ATG TCT TAA TGT ATT GGG GTG
Protein:
Met Ser stop
Name of mutation: nonsense or transversion
Mutagen:
UV light
*
B.
GTG TCT TAT TGT ATT GGG GTG
Protein:
Val Ser Tyr Cys Ile Gly Val
Name of mutation: transition or missense
Mutagen:
5 BU or 2AP
*
C.
ATG TCT TTA TTG TAT TGG GGT G
Protein:
Met Ser Leu Leu Tyr Trp Gly
Name of mutation: frameshift
Mutagen:
intercalator
First
Position
U
C
A
G
U
Phe
Phe
Leu
Leu
Leu
Leu
Leu
Leu
Ile
Ile
Ile
Met
Val
Val
Val
Val
Second Position
C
A
Ser
Tyr
Ser
Tyr
Ser
Stop
Ser
Stop
Pro
His
Pro
His
Pro
Gln
Pro
Gln
Thr
Asn
Thr
Asn
Thr
Lys
Thr
Lys
Ala
Asp
Ala
Asp
Ala
Glu
Ala
Glu
Third
Position
G
Cys
Cys
Stop
Trp
Arg
Arg
Arg
Arg
Ser
Ser
Arg
Arg
Gly
Gly
Gly
Gly
U
C
A
G
U
C
A
G
U
C
A
G
U
C
A
G
6