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Genetics Worksheet

0350 Genetics Fall 2020
Assignment Problems: Cumulative
(1) Analysis of vulval development in C. elegans has been important in understanding some key signaling
pathways, some that play roles in human cancer. The vulva is the organ through which fertilized eggs leave the
mother. Wild-type worms only have one vulva, but mutants have been identified that either have more than one
vulva or no vulva. These include mutations in the mpk-1, lin-1, lin-39, let-23 and lin-3 genes. Below is a table
that lists single or double mutants and their phenotype.
Mutant (all are homozygous
mutants in the indicated gene(s))
phenotype
mpk-1
no vulva
lin-1
multiple vulvas
lin-3
no vulva
lin-39
no vulva
let-23
no vulva
mpk-1 lin-1
multiple vulvas
lin-3 lin-1
multiple vulvas
lin-39 lin-1
no vulva
let-23 lin-1
multiple vulvas
1 (1a) Who originally chose C. elegans as a model system to study genetics
A: Sydney Brenner
B: Francis Crick
C: Seymour Benzer
D: Barbara McCintock
2 (1b) What other major contribution did he make to genetics?
A: Uncovered the secrets of the lac operon
B: Used T4 phage rII gene frameshift mutants to show that the Genetic Code is composed of triplet codons
with no commas
C: Used recombination between thousands of T4 phage rII gene mutants to demonstrate that a gene is likely
just a linear sequence of nucleotides
D: Identified the first gene linked to a chromosome, the white gene of Drosophila
 
 
Assignment Problems: Cumulative 0350 Genetics Fall 2020
1
3 (1c) Which of the pathways below is consistent with the data (if you cannot position some genes
unambiguously then they must be placed at the same position in the pathway).
4(1d) Which of the following genes is most likely to encode for a transcription factor?
A: lin-3
B: let-23
C: lin-39
D: mpk-1
The following table shows where wild-type protein encoded by the genes is expressed during vulval
development.
Gene
Cells in which the wild-type gene
is expressed
mpk-1
Vulva and cells outside
lin-1
Cells outside of the vulva only
lin-3
Cells outside of the vulva only
lin-39
Vulva and cells outside
let-23
Vulva and cells outside
5 (1e) Which gene is most likely to encode for a secreted signaling protein?
A: lin-3
B: let-23
C: lin-39
D: lin-1
6 (1f) Which gene is most likely to encode for a receptor for the signaling protein you identified in (1e)?
A: lin-3
B: let-23
C: lin-39
D: lin-1
Assignment Problems: Cumulative 0350 Genetics Fall 2020
2
(2) A pure breeding population of reindeers at the North Pole have red noses. Those in Finland all have black
noses while those in Russia all have tan noses. To investigate the genetics of this trait, crosses were done
between these pure breeding populations; the most revealing of these was a cross between the Finnish and
Russian with the F1 progeny all being black, but a cross between these yielding the following:
Black: 89
Tan: 41
Red: 30
7 (2a) Explain these results
A: Single gene, two alleles, one incompletely dominant over the other
B: Single gene, two alleles, each codominant
C: Single gene, three alleles, allelic series
D: Two genes, each with two alleles, one completely dominant over the other, and one gene recessively
epistatic to the other
8(2b) Would all the tan offspring be pure-breeding like the ones from Russia?
A: Yes
B: No
9 (2c) What color would the offspring be from a cross between reindeer from the North Pole and Russia
A: red
B: tan
C: black
D: pink
10 (2d) Reindeers at the North Pole can fly while those from anywhere else cannot; the ability to fly is
controlled by a single gene. A cross between reindeers from the North Pole and Finland produced F1 progeny
could not fly and had black noses; these F1s were backcrossed reindeers from the North Pole resulted in
following F2 progeny:
Black flightless
61
Red flier
59
Black flier
39
Red flightless
41
Is the gene controlling black or red noses in Finnish/North Pole, likely to be linked to that controlling the ability
to fly? (Table on last page if needed)
A: Yes
B: No
Assignment Problems: Cumulative 0350 Genetics Fall 2020
3
(3) A new mutation that results in eight-legged flies was isolated and was found to be recessive, the gene
name was designated extra-legs, el, and the wild-type allele, el
+
and the mutant, el
R
. The mutation was
mapped approximately to a region on the 2
nd
chromosome that included the following known mutant markers
(the mutant marker alleles are all recessive and their map position on the chromosome is indicated in the table
below).
Gene
Wild-type
allele
(dominant)
Wild-type
phenotype
Mutant marker
allele
(recessive)
Mutant
phenotype
Map position
in map units
wingless
+
wg
Full sized wing
wg
T
No wing
25
orange
+
or
Red eyes
5
or
Orange eyes
37
short bristles
+
sb
Long bristles
S
sb
Short bristles
52
Three crosses were set up:
T
5
(i) wg
R
or
T
el
5
/ wg
R
or
el
x wg
+
+
+
or
+
el
+
/ wg
+
or
el
(ii) or
5
S
R
sb
5
el
S
/ or
R
sb
el
x wg
+
+
+
or
+
el
+
/ wg
+
or
el
(iii) wg
T
S
R
sb
T
el
S
/ wg
R
sb
el
x wg
+
+
+
or
+
el
+
/ wg
+
or
el
The F1 progeny from these crosses were then crossed separately to: wg
T
5
S
or
R
sb
T
el
5
/ wg
S
or
R
sb
el
(i) F1 heterozygous parent: wg
T
5
R
or
+
el
+
/ wg
+
or
el
Full wing, red eyes, 6 legs
884
Full wing, orange eyes, 6 legs
46
No wing, red eyes, 6 legs
66
No wing, orange eyes, 6 legs
4
Full wing, red eyes, 8 legs
2
Full wing, orange eyes, 8 legs
23
No wing, red eyes, 8 legs
29
No wing, orange eyes, 8 legs
455
95/1509=0.629
(ii) F1 heterozygous parent: or
5
S
R
sb
+
el
+
/ wg
+
or
el
Red eyes, long bristles, 6 legs
872
Red eyes, short bristles, 6 legs
70
Orange eyes, long bristles, 6 legs 8
Orange eyes, short bristles, 6 legs 50
Red eyes, long bristles, 8 legs
26
Red eyes, short bristles, 8 legs
5
Orange eyes, long bristles, 8 legs 39
Orange eyes, short bristles, 8 legs 433
(iii) F1 heterozygous parent: wg
T
S
R
sb
+
el
+
/ wg
+
or
el
Full wing, long bristles, 6 legs
736
Full wing, short bristles, 6 legs
200
No wing, long bristles, 6 legs
61
No wing, short bristles, 6 legs
14
Full wing, long bristles, 8 legs
8
Assignment Problems: Cumulative 0350 Genetics Fall 2020
4
Full wing, short bristles, 8 legs
23
No wing, long bristles, 8 legs
103
No wing, short bristles, 8 legs
322
11 (3a) Between which two marker genes is the el gene located
A. wg and or
B. wg and sb
C.
or and sb
12 (3b) What is the distance between wg and el
A.
7 m.u
B.
9 m.u
C.
13 m.u.
D.
16 m.u.
13 (3c) Is there anything unusual about the numbers of any of the classes of phenotype in the three crosses?
A There are a lot more parentals that would be expected
B There are a lot fewer double recombinants that would be expected
C Both classes of parentals should be approximately the same but are not
D The total number of recombinants is too high compared to the parentals
 
Assignment Problems: Cumulative 0350 Genetics Fall 2020
5
(4) If you were to generate mutations for the his gene in Salmonella gene with X-rays and with the chemicals
EMS and profavine (i.e. separately) and then sequenced the gene, what might you expect to find in each
case?
14 (4a) X-rays
A: No sequence because the whole gene is deleted
B: Point mutations
C: Small insertions and deletions
D: Several pyrimidine dimers
15 (4b) EMS
A: No sequence because the whole gene is deleted
B: Point mutations
C: Small insertions and deletions
D: Several pyrimidine dimers
16 (4c) Proflavine
A: No sequence because the whole gene is deleted
B: Point mutations
C: Small insertions and deletions
D: Several pyrimidine dimers
17 (4d) Which of these mutations might be useful to use in the Ames test?
A: All three
B: X-ray only
C: EMS only
D: EMS and proflavine
18 (4e) As well as a mutation in the his gene, the strain of Salmonella used in the Ames test also usually
carries a mutation in what other gene?
A: Antibiotic resistance
B: uvrA or uvrB
C: DNA polI
D: RND efflux pump
Assignment Problems: Cumulative 0350 Genetics Fall 2020
6
(5) Different temperature sensitive mutants of E. coli and yeast in genes encoding proteins involved in
transcription were grown at the permissive temperature for several generations and while still actively growing
the temperature was raised to the restrictive temperature. After raising temperature, the synthesis of new RNA
was assessed using radioactively labeled nucleotides. Assume the mutations completely inactivate the
proteins at the restrictive temperature and that function of the mutant protein was abolished instantly
temperature was raised.
What would happen to the level of new RNA synthesized (i.e. radioactive incorporation into RNA) in the
following mutants after raising the temperature?)
A Transcription would stop instantly
B Some RNA would still be synthesized but will be reduced and eventually stop
C Levels would probably increase
D No effect on the level of transcription
19 (5a) Sigma factor (E. coli)
20 (5b) Subunit ß of RNA polymerase (E. coli)
21 (5c) RNA Pol II (yeast)
22 (5d) Rho protein (E. coli)
Assignment Problems: Cumulative 0350 Genetics Fall 2020
7
(6) Hedgehog (Hh) is a secreted signaling protein that activates the Hh signaling pathway in responding cells
resulting in the activation of Hh-target genes that can control cell activities such as proliferation. The pathway is
inappropriately activated in some cancers.
Key proteins in the Hh pathway are as follows:
Hedgehog (Hh): secreted signaling protein found outside cells
Patched (Ptc): transmembrane receptor, externally portions of the protein bind Hh, other regions bind to Smo.
Note, Ptc is the Hh receptor, but unlike most receptors its activity is inactivated by its ligand, Hh.
Smoothened (Smo): transmembrane protein that positively regulates the Hh pathway by binding and
inactivating Cos2. Smo activity is inhibited by Ptc.
Costal-2 (Cos2): cytoplasmic protein that has 3 separate binding domains. It binds to Smo and that blocks its
activity. It binds to GSK/PKA and that forms an active complex. It binds to Gli when the latter active complex
forms and that results the processing of Gli to the repressor form.
GSK3 and PKA: kinases that phosphorylate Gli to result in its partial degradation
Glioblastoma (Gli) Transcription factor that can act as a repressor or activator of transcription. In the absence
of Hh, it is partially degraded to a repressor form. In the presence of Hh this does not happen and the full
length form is an activator.
Absence of Hh: Ptc inactivates Smo by binding to it and preventing it interacting with Cos2, which then binds to
Gli, and GSK3 and PKA. The latter 2 phosphorylate Gli and this targets it for partial degradation. The smaller
from of Gli enters the nucleus, binds to Hh-target genes and represses their expression.
Presence of Hh: Hh binds to its receptor which is Ptc and this prevents Ptc from inactivating Smo, which can
now bind Cos2 and preventing GSK3 and PKA from phosphorylating Gli. Gli then remains full length and can
enter the nucleus and bind to and activate Hh-target genes.
(N.B. some liberties have been taken with pathway to simply the question so don’t take it all completely
literally)
Assignment Problems: Cumulative 0350 Genetics Fall 2020
8
23 (6a) Which of the simple genetic pathways below is correct based on the information above.
The table below lists mutations in different regions of proteins in Ptc, Smo and Cos2 in the Hh pathway.
(Determine whether the pathway will be on or off in each mutant, i.e. is the gene G is expressed in a
homozygous mutant.
(c) Determine if the mutation will be a simple loss of function (SLF), possibly a dominant negative (DN) or
probably a gain of function (GOF). (0.33 points each)
To work out answers to (c) you need to first determine the loss of function phenotype for the gene in question
(one mutation of each is obviously a SLF), then determine if each of the other mutations in the same gene
would result in the same phenotype or not – if the same then the mutation is either SLF or DN, if the opposite
then it would be a GOF. To work out if it is SLF or DN you have to consider what would happen in a
heterozygote: can the mutant protein interfere with wild-type protein in some way? Could it block the wild-type
protein or would it compete with the wild-type for a protein the wild-type needs to interact with?
Note, a mutation in one part of a protein could be DN, while mutation in a different part could be GOF.
Hint…only one of the mutants is predicted to act as a DN
Protein Mutation (region inactivated)
(b) Is pathway always on or
off in homozygous mutant
(i.e. is gene G expressed)
(c) Is the mutation
SLF, possibly DN
or probably GOF?
Ptc
Deletion of gene
24 6b
32 6j
Assignment Problems: Cumulative 0350 Genetics Fall 2020
9
Smo
Deletion of gene
25 6c
33 6k
Cos2
Deletion of gene
26 6d
34 6l
Ptc
Hh binding domain
27 6e
35 6m
Ptc
Smo binding domain
28 6f
36 6n
Smo
Ptc binding domain
29 6g
37 6o
Cos2
Smo binding domain
30 6h
38 6p
Cos2
Gli binding domain
31 6i
39 6q
24-32 (6b- 6i) Determine whether the pathway will be on or off in each mutant, i.e. is the gene G is expressed
in a homozygous mutant.
A = On
B = Off
33-41 (6j- 6q) Determine if the mutation will be a simple loss of function (SLF), possibly a dominant negative
(DN) or probably a gain of function (GOF).
A = SLF
B = DN
C = GOF
To work out answers you need to first determine the loss of function phenotype for the gene in question (one
mutation of each is obviously a SLF), then determine if each of the other mutations in the same gene would
result in the same phenotype or not – if the same then the mutation is either SLF or DN, if the opposite then it
would be a GOF.
Remember, a GOF mutation is only in relation to the function of the protein itself, not as to whether the
pathway is activated by the mutation. So for a gene that functions to inhibit the pathway a SLF mutation will
result in the pathway being on all the time, but a GOF would result in the pathway being off all the time.
To work out if it is SLF or DN you have to consider what would happen in a heterozygote: can the mutant
protein interfere with wild-type protein in some way? Could it block the wild-type protein or would it compete
with the wild-type for a protein the wild-type needs to interact with?
Note, a mutation in one part of a protein could result in a DN mutation, while mutation in a different part could
be GOF.
Hint…only one of the mutants is predicted to act as a DN
40 (6r) Based on what you know about other signaling pathways, why might a GSK3 mutant have more
complicated phenotypes than would be predicted based on its role in Hh signaling? (2 points)
A. It is redundant
Assignment Problems: Cumulative 0350 Genetics Fall 2020
10
B. It is also involved another pathway – in the Wnt pathway – so loss of GSK3 would affect both pathways
C. It is haploinsufficient
D. Mutations are always dominant negative
Assignment Problems: Cumulative 0350 Genetics Fall 2020
11

 
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