Problem Set 7, Fall 2015
Name:
8 points total
1. You suspect that two genes, glo (required for deepwater jellyfish fluorescence) and
stn (required for their potent sting). You cross a heterozygote with homozygous
recessive and get the results shown below. In both cases, the mutant allele is recessive to
the wild type allele.
371
non-fluorescent
772
(772 + 652)/2 = 712
401
non-stinging
337
wild type
652
315
non-fluorescent, non-stinging
A. What is the probability that glo and stn are linked?
(772 – 712)2/712 + (712 – 652)2/712 = 5.1 + 5.1 = 10.2
p nearly 0.001 so probability that the genes are linked is 0.999 or 99.9%
2. The following sequence of DNA was recovered from fossil remains in Nevada. Consider each
of the following cases separately. Use the chart below to determine which of the amino acids
corresponds to each codon.
1
2
3
T C A
Ser
4
5
First
Position
U
C
A
G
6
G T A
Val
U
Phe
Phe
Leu
Leu
Leu
Leu
Leu
Leu
Ile
Ile
Ile
Met
Val
Val
Val
Val
7
8
9
A A C
Asn
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25 26 27
G A C
Asp
A T T
Ile
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
T A T
Tyr
T G C
Cys
C A C
His
T C C
Ser
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
A. What type of mutations results from a deletion of nucleotide 9? What is the sequence of the
resulting protein?
frameshift, Ser-Val-Lys-Thr-Ile-Ala-Thr-Pro-Phe/Leu
B. What would happen if a transversion occurred at position 18? What is the term for this sort of
mutation? What is the sequence of the resulting protein?
Two changes are possible.
C=>G: Cys => Trp missense
C=>A: nonsense
C. Explain the results of a transition at position 12. What type of mutation is produced? What is
the sequence of the resulting protein?
The mutation is silent. The protein sequence remains the same.
D. 5-Bromouracil is introduced at position 5. Diagram how the original GTA codon would
change as a results. GTA => GBuA => GCA
1
Problem Set 7, Fall 2015
Name:
8 points total
3. In analyzing a new compound as a suspected mutagen, you conduct the Ames test. Using
strains 1 and 2, which contain a frameshift mutation or a base substitution, respectively, in a gene
required for histidine biosynthesis, you find the results pictured below (all plates contain only
enough histidine to permit a few cycles of cell division. A spot indicates formation of a colony of
cells.)
a. Why did you include strains 1 and 2 incubated without the suspected mutagen? Why
are there a few colonies on each of those plates?
This control gives the spontaneous mutation rate.
b. Does your suspected mutagen actually function as a mutagen? If so, what type of
mutagen is it? If not, why not?
Yes, it appears to produce frameshift mutations.
c. If you found only two or three colonies on each of the 4 plates using suspected
mutagen that was not incubated with liver extract, what would you conclude?
The liver extract appears to convert a nonmutagenic compound into a mutagen.
d. Why did you add enough histidine to permit only a few rounds of cell division?
Some mutagens require replication.
4. A diploid strain of the unusual filamentous fungus A. farlowae is heterozygous for several
genes as diagrammed.
+
+
bt
+
c
rd
+
ura
gdin
Mutations in c causes the colonies to be "clumpy" in appearance. Mutation in ura results
colonies that can't grow without uracil added to the culture medium. bt results in colonies that
cannot grow without added biotin and rd results in a reddish color. The black dot indicates the
location of the centromere. The original colony was white, not clumpy and grew well on minimal
plates lacking biotin or uracil. You treat cells with a mutagen and plate on minimal medium. You
find colonies that display a number of phenotypes. What is the most likely explanation for each?
a. Most colonies are still white and not clumpy.
They have acquired no mutations that produce phenotypes.
b. Several colonies are now red. These can be divided into two categories.
i. Most of these colonies still are not clumpy.
These have acquired mutations in the previously wild-type rd gene.
ii. Some of these colonies are now clumpy.
These have lost both c and rd function. A single event that would produce
this is a deletion that affects both genes.
c. Many cells simply didn't grow much. These were transferred to rich medium
containing all nutrients required for growth.
i. Some now grew to form colonies. Mutations in bt or ura, or perhaps
dominant mutations in some metabolic gene.
ii. Some still did not grow. Dominant mutations in an essential gene.
2
Problem Set 7, Fall 2015
Name:
8 points total
5. Several Loch Ness Monsters have been captured recently, and have been subjected to careful
genetic analyses. Three linked, recessive mutations were identified:
tn mutants have tan body color
scl mutants have large scales
fla mutants have flat fins
In a cross of triple heterozygous individuals (parent 1)with homozygous recessive (parent 2), you
find:
tan, large scales, flat fins
tan, flat fins
tan, large scales
large scales, flat fins
tan
flat fins
large scales
all wild type
5
tn scl fla/tn scl fla
422
tn + fla/tn scl fla
36
tn scl +/tn scl fla
33
+ scl fla/tn scl fla
26
tn + +/tn scl fla
43
+ + fla/tn scl fla
449
+ scl +/tn scl fla
6
+ + +/tn scl fla
1020 total
A. What is the order of the three linked genes?
scl is in the middle
B. Calculate the map distances between all three pair-wise combinations of genes.
tn to scl: (36 + 43 + 5 + 6)/1020 X 100% = 8.8 cM
scl to fla: (26 + 33 + 5 + 6)/1020 X 100 = 6.9 cM
fla to tn: 8.8 + 6.9 = 15.7 cM
C. What is the genotype of the two parents used in the cross?
tn + fla/ + scl + X tn scl fla/tn scl fla
3
Problem Set 7, Fall 2015
Name:
8 points total
6. Please compare the sequence at the top to each DNA sequence that appears below, and
identify the type of mutation present. If it’s a substitution mutation, please identity as to whether
it is a transition mutation or a transversion mutation. Starting with the ATG on the top strand, left
hand side, please give the sequence of the resulting protein.
5’ATGGCATGGATAAGCTACGATCG3’
3’TACCGTACCTATTCGATGCTAGC5’
5’ATGGCATGAATAAGCTACGATCG3’
3’TACCGTACTTATTCGATGCTAGC5’
5’ATGGATGGATAAGCTACGATCG3’
3’TACCTACCTATTCGATGCTAGC5’
5’ATGGCATGCATAAGCTACGATCG3’
3’TACCGTACGTATTCGATGCTAGC5’
Met-Ala-Trp-Ile-Ser-Tyr-Asp-Arg
nonsense (and transition) produces
Met-Ala-stop
frameshift produces Met-Asp-Glystop
Transversion produces a missense
changing Met-Ala-Cys-Ile-Ser-TyrAsp-Arg
7. Provide the name of the mutagen, a brief description of its mode of action in the table below
and the type of mutation that results. See the first line for an example.
Name of compound
Mechanism of action
example: ultraviolet light
causes pyrimidine dimers
Type of mutation
produced
frameshift (other answers
possible)
ethylates G
G-C to A-T trnasitions
intercalator
induces frameshifts
Incorporated in place of
A but can base pair with
C
A-T to G-C transitions
name: EMS or ethylmethane
sulfonate
name: acridine orange
name: 2-aminopurine
8. In your studies of three recessive corn genes, su, v and gl, you find that the three genes are
linked. Recessive mutations in each produce: variable amounts of sugar in kernels (su),
prominent leaf veins (v) and glossy leaves (gl). Two homozygous corn plants are crossed. The
4
Problem Set 7, Fall 2015
Name:
8 points total
resulting F1 progeny all have normal kernels, veins and leaves. Some of the F1 plants were test
crossed to produce:
270
7
variable sugar, glossy leaves, prominent veins
glossy leaves
48
glossy leaves, prominent veins
60
prominent veins
4
40
235
62
variable sugar, prominent veins
variable sugar
wild type
variable sugar, glossy leaves
a. What were the genoytpes and phenotypes of the original homozygous parents used to
produce the F1s?
su gl v/ su gl v X + + +/+ + +
b. Calculate the map distance between:
su and gl: (40+48+7+4)/726 X 100% = 13.6 cM
gl and v: (60+62+7+4)/726 X 100% = 18.3 cM
su and v: 13.6 + 18.3 = 31.9 cM
c. Another recessive mutation in corn is w, which gives wrinkled kernels. This gene is
known to be 5.9 mu from v. A heterozygous plant with mutations in the trans orientation
is test crossed to produce:
440
436
63
61
wrinkled
glossy
wrinkled, glossy
wild type
What is the genetic distance between gl and w?
(61 + 63)/1000 X 100% = 12.4 cM
d. Draw a genetic map of the 4 linked genes. Show all distances.
su
13.6
12.4
gl
5
w 5.9
v