Download 06BIO201 Exam 2 KEY

Name _________KEY___________________
Section ______________
Biology 201 (Genetics)
Exam #2 120 points
20 October 2006
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Read the question carefully before answering. Think before you write.
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you are in the exam.
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Sign the honor pledge if applicable.
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work.
Signature: _________________________________________________________
1
12 pts.
1. In the late 1950s Herbert Taylor grew bean root-tip cells in a solution of radioactive thymidine and
allowed them to undergo one round of DNA replication and then stopped the experiment in the
middle of mitosis. His results are shown in the following figure, where the darkly shaded
chromosomes contain radioactive thymidine. Draw diagrams explaining the pattern of radioactivity
observed in each of the chromatids after the first round of DNA replication. Use dashed lines to
indicate radioactivity and box your answer when done.
This was suggested problem at the end of Ch 14.
12 pts.
12 pts.
2. This table gives the results of assays of percentages of bases from nucleic acids isolated from
different sources. For each source, indicate what type of nucleic acid (RNA or DNA) is present in the
source by writing the answer in the table. For each source, indicate whether it is likely to be mostly
single stranded or mostly double stranded if it can be ascertained from the data provided.
Source
A
G
T
C
U
(1)
(2)
(3)
20
20
20
30
30
20
0
20
30
10
30
30
40
0
0
DNA or
RNA
RNA
DNA
DNA
Single or double stranded
single
double
single
3. A biochemist isolated and purified what she thought were all of the various molecules needed for
DNA replication. She then recombined them to replicate DNA in vitro. After she allowed for
replication to occur, she isolated the DNA that was newly synthesized, heated it to break the
hydrogen bonds between complementary strands, and electrophoresed the products on a gel. She
noted that there were both some long strands of DNA and also numerous short segments of DNA a
few hundred nucleotides long. List the two most likely components of the replication system could
be missing based on her data.
DNA ligase and DNA polymerase I
2
12 pts.
4. You treat bacteria with UV light. Fill in the following two graphs with the data you expect to obtain
and answer the questions provided.
a) What type of mutations does UV light cause?
Thymine-thymine dimmers; also accepted induced mutations or base pair substitutions
b) Name two mechanisms that repair the damage caused by UV light.
Photoreactivation and excision repair
c) What happens to percent survival with increasing UV dose and why?
Survival decreased with increasing UV dose because there is an increase in mutation frequency and thus
an increase in the chance that a mutation will occur in a gene essential for life. Alternative answer was
because there are too many mutations and DNA repairs systems can not work fast enough to keep up
with the mutations.
15 pts.
5. Imagine you can eliminate any of the following components of eukaryotic transcription at will.
Match the components listed below with the ultimate result of eliminating the component by putting
the appropriate letter in the blank next to the component.
component
RNA polymerase
Basal transcription factors
polyA tail polymerase
capping
spliceosome
ultimate result on protein production
(choose from a , b, c, or d below)
a
a
b
b
c
a = Elimination of the component would ultimately result in no protein production.
b = Elimination of the component would ultimately result in less protein production.
c = Elimination of the component would ultimately result in protein production where the amino
acid sequences of the proteins are wrong.
d = Elimination of the component would ultimately have no effect on protein production.
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15 pts.
6. This sequence of bases contains a gene encoding a protein:
3’ TCTGCCCTTACGGATTGATCCGG 5’
5’ AGACGGGAATGCCTAACTAGCCC 3’
a) Write out the sequence of the mRNA that would be transcribed.
5’ AGACGGGAAUGCCUAACUAGCCC
b) Show the 5’ end of the mRNA.
See above
c) Indicate the 5’ and 3’ ends of the DNA on the DNA strand given above.
See above
d) Write out the amino acid sequence of the protein encoded by this gene.
methionine – proline - asparagine
12 pts.
7. The genetic code is redundant/degenerate but unambiguous. Clearly explain (a) how the code is
redundant and (b) why the code must be unambiguous.
The genetic code is redundant because more than one codon can specify a particular amino acid due to
(1) wobble and (2) many amino acids are carried by more than one type of tRNA, with different
anticodons.
The genetic code must be unambiguous, which means that a particular codon will only specify one
amino acid. If the code were not unambiguous, when the translation apparatus went to “decode” a
particular codon, more than one amino acid type could be put in the growing polypeptide chain. This
would cause the protein made to possible not be functional because the “wrong” amino acid could be
present.
4
Multiple choice section: (30 points total – 6 points per question)
Write your answer in the blank provided to the left. If you want to explain your answer, you can do so next to
the question.
d
1. In the domestic rabbit, the recessive allele b (brown) results in brown fur and the recessive allele f (furless)
results in no fur. A female rabbit heterozygous for both genes was testcrossed with a male rabbit and the
progeny were: 19 nonbrown, furless; 21 brown, fur; 4 brown furless; and 6 nonbrown, fur. Estimate the
distance between the fur color gene and the fur amount genes.
a. 80 map units
b. 40 map units
c. 10 map units
d. 20 map units
e. 5 map units
f. 0.2 map units
d
2. The pedigree below shows the inheritance of a biochemical disorder in humans. Affected individuals are
indicated by filled-in circles (females) and squares (males). This disease state appears to be caused by a
____________ gene. Jenny’s genotype is ____________.
a.
b.
c.
d.
e.
dominant; Aa
recessive; aa
dominant; aa
recessive; Aa
incomplete dominant; Aa
a
3. In E. coli, a particular mutation in a section of DNA results in the total absence of one protein.
Where is the mutation most likely located relative to the transcriptional start site? Minus indicates
upstream of the transcriptional start site and plus indicated downstream of the transcriptional start
site.
a.
-10 basepairs
b.
0 basepairs
c.
+10 basepairs
d.
-100 basepairs
e.
a or c
5
c
4. Below is the data set from an experiment designed to determine the pathway for synthesis of the
vitamin spiderflavin in bacteria U. richmondi. The set up was similar to the one gene – one enzyme
experiment of Beadle and Tatum. Bacterial cells from various mutant strains (1, 2, and 3) or normal
strain X were added to agar plates that contained the metabolic intermediates as indicated. The
metabolic intermediates include Ortho-Spiderin, (O), Para-Webin (P), and Trans-Westhamptonin
(T). The plates were incubated overnight, and growth was assessed. Which biochemical pathway is
most likely based on the information provided?
Normal X
Mutant 1
Mutant 2
Mutant 3
a.
b.
c.
d.
e.
None
growth
No growth
No growth
No growth
Compound added:
O
P
growth
growth
growth
No growth
growth
growth
No growth
No growth
T
growth
growth
growth
growth
precursor → O → P → T à spiderflavin
precursor → T → P→ O à spiderflavin
precursor → P → O → T à spiderflavin
precursor → O → T → P à spiderflavin
precursor → T → O → P à spiderflavin
e
5. Which mutation is/are most detrimental to the structure and function of a protein?
a.
a 1 basepair deletion
b.
a 1 basepair insertion
c.
a 3 basepair deletion
d.
a 3 basepair insertion
e.
a and b
f.
c and d
g.
All of the above are equally detrimental to the protein
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