Download Biol 207 Workshop #4 New Problems:

1/30/04
Biol 207 Workshop #4
New Problems:
1. You are studying an organism capable of surviving high UV doses.
Describe 3 possible mechanisms could allow this organism to survive?
The organism could up-regulate the repair pathways that fix the DNA
damage caused by UV light. One repair pathway could involve phytolase
that splits the pyrimidine dimers. Another pathway excises the
pyrimidine dimer and repairs the sequence using the complementary
sequence.
The organism could be polyploid. Having many copies of each gene would
provide backups in case one allele is mutated.
The organism could produce a pigment that would absorb UV before it
could cause damage.
The organism could have a UV opaque shell or covering that protects the
inner cells and tussues.
2. Draw a diagram of a chromosome during the following stages of the cell cycle: G1 and G2.
Indicate the following structures at least once in one of the diagrams: 5’-DNA molecule-3’, DNA
duplex, centromere, telomere, and sister chromatid.
Answers
1. G1
5’
3’
centromere
1 chromatid=2 paired ssDNA
G2
telomere
telomere
3’
5’
DNA
molecule
DNA
molecule
Sister
chromatids
3. The following sequence of DNA undergoes a spontaneous mutation, resulting in the
deamination of cytosine. The deamination of Cytosine results in uracil, which pairs with adenine.
5'-A-T-A-T-T-C*-T-A-3' (C* represents the deaminated cytosine)
3'-T-A-T-A-A-G -A-T-5'
a) Draw a series of diagrams that follow the DNA, and its replication products through 2 rounds of
replication.
b) Label the second generation of products as wild type or mutant.
c) What kind of mutation is this?
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(a+b)
5'-A-T-A-T-T-C*-T-A-3'
3'-T-A-T-A-A-G -A-T-5'
5'-A-T-A-T-T-C*-T-A-3'
3'-T-A-T-A-A-A -A-T-5'
5'-A-T-A-T-T-C-T-A-3'
3'-T-A-T-A-A-G-A-T-5'
Mutant
5'-A-T-A-T-T-C*-T-A-3'
3'-T-A-T-A-A-A -A-T-5'
Mutant
5'-A-T-A-T-T-T-T-A-3
3'-T-A-T-A-A-A-A-T-5'
Wild type
5'-A-T-A-T-T-C-T-A-3'
3'-T-A-T-A-A-G-A-T-5'
Wild type
5'-A-T-A-T-T-C-T-A-3'
3'-T-A-T-A-A-G-A-T-5'
c) Transition
4. Using a diagram, show how “slippage” of the template strand during DNA replication can cause a
deletion mutation.
ACC
ACCAACCAACCAACCACCACCAACCA (Template strand)
TGGTTGGTTGGTTGG TGGTGGTTGGT (replicated strand misses the repeat that
slips out)
When the ACC repeat slips out the newly replicated strand using it as a
template loses one TGG repeat.
5. Why are auxotrophic mutants easier to find in haploid organisms like Neurospora than in diploid
organisms?
It is more likely to knockout the one functional gene copy in a haploid than the two functional
copies in a diploid.
6. A spontaneous mutation can occur in mice that causes a white coat color rather than the normal
black coat color. The mouse with this spontaneous mutation has a mosaic coat with black and
white patches of hair. This mutant phenotype is frequently observed in mice that have a small
deletion on one chromosome compared to mice that do not have a deletion in this region.
a) Provide an genetic explanation for the mosaic phenotype?
The mouse has a mosaic phenotype because it is a mixture of wild type cells and mutant cells. The
mutant cells have the spontaneous mutation and produce the white coat color. If the mutation is
recessive, both alleles for the gene must be dysfunctional for the cells to produce a white color
coat.
b) Will this spontaneous mutation be passed on to the next generation? Explain
It depends on when the mutation took place. If the mutation occurred very early during
development it will occur in both somatic and germ line tissues. However, if it occurred later in
development, after the somatic and germ line tissues differentiated, then the mutation will only
occur in the somatic tissues. The mutation must occur in the germ cells to be passed on to the next
generation.
c) Why would the mutant phenotype frequently occur in mice that have a deletion on one
chromosome? The homologue to this chromosome is wild type.
A deletion will remove many genes within a region on a chromosome. Possibly, one of these
genes is necessary to produce the black coat color. Since one homologue has a deletion that
could include this gene, only the other homologue has a functional copy of this gene. A
spontaneous mutation affecting the functional gene will interrupt the function of the single
functional allele and cause the cells to express the mutant phenotype. In comparison, wild type
mice that have no deletions will have two functional alleles of this gene. A spontaneous mutation
must knockout both of the genes for the cells to express the mutant phenotype. The chances of
two spontaneous mutations occurring are much less likely than one.
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7. Determine what type of Muller morph best describes the following mutants. The mutant allele is
designated with an “m.” A wild type allele is indicated by a “+” sign. A deficiency that includes the
mutation is indicated by “def.”
a) A new type of dark pea is discovered. Different alleles combinations and the phenotype
produced by these combinations are described in the following table. The phenotype is
ranked from using a scale from 1-4 with 4 being white, 2 light green, and 4 dark green.
Genotype
Phenotype
+/+
2
+/m
3
def/m
2
m/m
4
Muller Morph?
Hypermorphic
b) A mutation in rats is discovered that results in obesity. Different allele combinations and the
average rat weight produced by these combinations are described in the following table.
Genotype
Weight (g)
m/+
~300
m/def
~600
m/m
~450
+/+
~200
Muller’s Morph? Hypomorphic
8.The following table lists mutations affecting the daf-2 gene. Mutations in daf-2 can either reduce
or increase a C. elegans’ average life span. A wild type C. elegans lives for an average of 14
days. Genomic deletions (DEL) that include daf-2 act recessively. Determine if the mutations are
amorphic, antimorphic, neomorphic, hypomorphic, or hypermorphic.
Allele (m)
m/+
m/DEL
m/m
Mutation
+
14(+/+)
14(+/DEL)
14(+/+)
wild type
e949
14 (e49/+)
30 (e49/DEL)
25 (e49/e49)
hypomorph
m41
14
37
37
amorph
m65
12
14
10
hypermorph
m577
*
*
*
neomorph
sa223
22
37
37
antimorph
h474
14
25
16
hypomorph
s54
14
36
32
hypomorph
ed27
6
8
dies as larvae hypermorph
*Both XO and XX progeny develop as males. Normally, XX develop as hermaphodites and XO
develop as males.
(Deletions or deficiencies that include a gene act as amorphic mutations.)
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