Download Problems 10

BS 50—Genetics and Genomics
Week of Oct 31
Additional Practice Problems for Section
Answer Key
Question 1.
Loss-of-function mutations are significantly more frequent than gain-of-function
mutations. Why is this so?
Mutations are more likely to reduce or eliminate gene function than to enhance
it. As stated in the text, “By randomly changing or removing one of the
components of a machine, it is much easier to break it (that is, loss of function)
than alter the way it works (that is, gain of function)
Question 2
a) Define “transition mutations” and “transversion mutations.” Give one example of
each.
Transitions: Change from a purine to a purine or a pyrimidine to a pyrimidine.
Examples: A to G; G to A; C to T; T to C
Transversions: Change from a purine to a pyrimidine or vice versa. Examples:
A to C or T; G to C or T; C to A or G; T to A or G.
b) The frequency of spontaneous transitions is significantly higher than the rate of
spontaneous transversions. Using everything you know about transitions and
transversions, propose an explanation for why transitions are more common.
In a normal double-stranded piece of DNA, purines are always paired with
pyrimidines and vice versa. Because purines and pyrimidines are different sized
molecules (made of two or one ring, respectively), the consistent pairing results
in a consistent distance between the two sugar-phosphate backbones. If two
purines or two pyrimidines pair, the double helix will be distorted, and thus be
easily recognized by the various repair systems.
For a transversion to occur, it is necessary for two purines or two pyrimidines to
pair (for at least one round of replication). This will result in a distortion and
repair. Thus, transversions are not likely to occur because they are easy to
detect by repair systems. Whereas transitions do require mispairing, they do not
cause the severe distortions of the double helix, and are thus more likely to
“escape” the repair systems.
Question 3
Shown below is a list of statements (a-k) and types of mutations (1-8). On the blank line
following each mutation, write the letter(s) of all statements that apply to that type of
mutation. Hint: Each statement may be used more than once and each type of mutation
may have more than one correct statement.
a) A mutation that changes UUA to UUG
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b) A mutation that gives methionine instead of leucine
c) Created by the addition of a nucleotide to a coding region
d) A stop codon is read as an amino acid
e) A chemically similar amino acid is replaced by the mutation
f) A mutation that changes CCU to ACU
g) Deleting a nucleotide in a coding region gives this type of mutation
h) Mutation does not alter the peptide
i) A mutation changing UAU to UAG
j) Premature termination codon is responsible for this mutation
k) A chemically different amino acid is replaced by the mutation
1) Missense mutation: b, e, f, k
2) Silent mutation: a, h
3) Frameshift mutation: c, g
4) Nonsense mutation: i, j
5) Synonymous mutation: a, h
6) Suppressor mutation: d
Question 4.
A partial peptide sequence for the wild type and three mutant alleles of a gene, PET1, are
shown below. Each mutant was caused by a single point mutation.
Wild Type: met
ile
arg
met asp
Mutant 1:
met ile
gln
Mutant 2:
met ile
arg met
Mutant 3:
met
ser met asp
ile
asn
lys trp…
gly …
gly
lys
trp…
lys
trp…
a) Using the amino acid sequence of the wild-type and the three mutants, deduce the
exact DNA sequence of the coding strand of the wild-type allele.
Each mutation helps determine the correct codon possible for the amino acids.
Students therefore cannot get full credit for indicating alternate codons for an
amino acid (except for the lys codon, which could be AAA or AAG).
Wild Type:
met ile
arg
met asp
lys
ATG ATC AGA ATG GAT AA(G/A)
Mutant 1:
met ile
gln
asn
gly
ATG ATC CAG AAT GGA TAA
trp……
TGG
2
Mutant 2:
met ile
arg
met gly
lys
ATG ATC AGA ATG GGT AA(G/A)
Mutant 3:
met ile
ser
ATG ATC AG(C/T)
trp……
TGG
met asp
lys
ATG GAT AA(G/A)
trp……
TGG
b) What type of mutation (transition vs. transversion vs. indel; missense vs. nonsense vs.
frameshift) has occurred in each mutant?
Mutant 1: Insertion of either a C in the first base of the third codon or a T, C or
A in the third base of the second codon; Frameshift mutation.
Mutant 2: A to G transition in the second base of the fifth codon; Missense
mutation
Mutant 3: A to C or T transversion in the third base of the third codon;
Missense mutation.
Question 5.
5-bromouracil is an unstable base analog of thymine that spontaneously converts between
an enol and keto form. In the enol form, 5-bromouracil pairs with G. In the keto form, 5bromouracil pairs with A. Use a diagram to show how this chemical can cause AT to GC
transitions during DNA replication.
Answer:
T
T
T
A
A
T
A
5BU added during first round of
replication. 5BU is incorporated in
the daughter strand across from A
5BUketo
nd
A
A
2 DNA replication.
No 5BU in the cell
enol
5BU
5BUenol
G
T
G
C
G
A
3rd DNA replication.
No 5BU in the cell
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Question 6.
A) (6 pts) Yes. Some combination of: each has only one centromere; each has a
telomere at its two termini; there is no aneuploidy.
B) (6 pts) No. Any combination of 1-2-6-7 has two centromeres; 5-3-4-7 has three
telomeres; segment 7 appears twice.
C) (8 pts) Yes. In a heterozygote, during meiosis, translocated chromosomes and
normal chromosomes pair. Adjacent I segregation (or independent assortment) results
in aneuploidy for segments 1 and 5; in each case, one is duplicated and one is missing.
(An indication of understanding of the concept of aneuploidy must be demonstrated for
full credit).
Question 7.
a) In humans, the red photoreceptor gene is 96% identical to the green photoreceptor
gene. The red and green photoreceptor genes are also next to each other on the X
chromosome. Diagram how unequal crossing over can result in red/green colorblindness
and gene duplication.
Red
Green
Non-functional hybrid protein
Gene duplication
b) Gene duplication can lead to the formation of gene families. The globin family
includes two clusters of genes, the alpha globin cluster and beta globin cluster. Why is it
important for humans to have both fetal and adult forms of hemoglobin?
Fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin. This allows
the fetus to draw oxygen from its mother’s bloodstream.
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