Homework 8 - KEY
BIO 2
1a.
Protein synthesis occurs with relatively high fidelity. In prokaryotes, incorrect amino acids are
inserted at the rate of approximately 5 x 10-4 codons. What is the probability that an averagesized polypeptide of 300 amino acids has exactly the amino acid sequence specified in the
mRNA? Show your calculation and express your final answer as a percent.
(5 x 10-4 mutations/codon) x (300 codons/polypeptide) = 0.15 mutations/polypeptide
This means that 15% of all polypeptides translated from the mRNA will carry a
different amino acid than was coded by the mRNA and 85% will have exactly the
same sequence as was coded by the mRNA.
1a.
The error rate of prokaryotic DNA replication is much lower than the error rate of prokaryotic
protein synthesis. Why do you think the replication enzymes have evolved to have so much
greater a level of fidelity than the enzymes involved in translation?
If a mistake happens at the DNA level, ALL polypeptides produced from the
transcription and translation of that gene from that point on will carry the error.
However, if the mistake happens at the level of translation, only the one polypeptide
will be affected. If that polypeptide fails to fold or function properly, it probably
won’t matter because so many other normal polypeptides will be translated and able
to compensate for the error.
2.
What amino acid would a tRNA with the anticodon 5’-UAG-3’ carry? Explain your logic.
The anticodon 5’-UAG-3’ will base pair to the codon 5’-CUA-3’. Therefore, the tRNA
will carry leucine.
3.
During translation, RNA:RNA base-pairing is extremely important. Describe three different
stages during prokaryotic translation in which RNA:RNA base-pairing plays an important role
in the process. In each case, identify the type of RNA(s) involved and the specific role that the
base-pairing plays in ensuring that translation takes place properly.
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a.
During prokaryotic translation, the ribosome binding site in the 5’ UTR of the
mRNA base-pairs with a ribosomal RNA sequence in the small ribosomal
subunit. This enables the small ribosomal subunit to recognize the mRNA and
initiate translation.
b.
During elongation of translation, anticodons in the tRNAs must base-pair with
codons in the mRNA. This ensures that the proper amino acids are brought to
the ribosome in the right order so that the correct polypeptide is made.
c.
tRNAs must fold back upon themselves to form double-stranded RNA regions
that stabilize their secondary and tertiary structures. The 3-D structures of
these molecules are vital for their proper functions during translation. Each
tRNA must be recognized by its corresponding amino-acyl tRNA synthetase
and “charged” with the correct amino acid. Each tRNA fits into the active site
of its corresponding amino-acyl tRNA synthetase because of its unique 3-D
configuration. Similarly, tRNAs fit into the A,P, and E sites on the ribosome
because of their specific 3-D conformations.
4.
Among Betazoids, the ability to read minds is under the control of a gene called “mindreader”
(abbreviated mr). Most Betazoids can read minds, but rare mutations in the mr gene result in
two alternative phenotypes: delayed-receivers and insensitives. Delayed-receivers have some
mind-reading ability but perform the task much more slowly than normal Betazoids and often
make mistakes. However, insensitives cannot read minds at all and are often placed in
institutions at a young age, since they are unable to survive in normal Betazoid society.
Betazoid genes do not have introns, 5’ UTRs, or 3’ UTRS, so the gene only contains coding
DNA. It is 2,331 nucleotides in length and Betazoids, like humans, use a three-letter genetic
code.
The table below shows some data from a Star Fleet study in which the mutations carried by
five unrelated mindreader mutants were analyzed.
Description of mutation
Bio 2
Phenotype
Nonsense mutation in codon 774
delayed-receiver
Missense mutation in codon 52
delayed-receiver
Deletion of nucleotides 934-939
delayed-receiver
Missense mutation in codon 192
insensitive
Deletion of nucleotides 83-93
insensitive
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For each mutation, provide a plausible explanation for why it gives rise to its associated
phenotype and not to the other phenotype. For example, hypothesize why the nonsense
mutation in codon 774 gives rise to the milder delayed-receiver phenotype rather than the
more severe insensitive phenotype, and then repeat this type of analysis for the other
mutations. (NOTE: More than one explanation is possible in each case.)
a.
Nonsense mutation in codon 774
The polypeptide is 776 amino acids in length (2,331/3 = 777, minus one for the stop
codon). A nonsense mutation changes an amino acid codon to a stop codon. Usually,
such mutations result in a non-functional protein and a more severe phenotype
because the polypeptide is terminated prematurely. However, in this particular case,
the mutation happens so late in the polypeptide that only a few amino acids are
affected near the C terminus. This provides a good explanation for why the less
severe phenotype results.
b.
Missense mutation in codon 52
A missense mutation switches one amino acid for another. Sometimes, missense
mutations can cause a protein to fold incorrectly or disrupt an active site in the
protein such that it is completely non-functional. In other cases, however, the
protein may be only mildly affected or its function may not be altered at all. In this
case, the switch apparently changes the function of the protein a moderate amount
but only enough to cause the delayed-receiver phenotype.
c.
Deletion of nucleotides 934-939
This mutation deletes exactly 6 nucleotides from the mRNA. Although this will cause
a localized disruption in the polypeptide sequence, the reading frame of the mRNA
will recover immediately after the mutation and the rest of the polypeptide
downstream of the mutation site will be made normally. As long as the
missing/changed amino acids are not vital for the protein’s folding or function, a
mild phenotype can result, which is what has probably happened in this case.
d.
Amino acid substitution mutation in codon 192
This mutation apparently affected an amino acid that is vital to the proper
folding/function of the protein. The affected amino acid may be in the active site of
an enzyme or in some other area of the protein that needs a particular amino acid at
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that site in order to function normally. This mutation renders the protein nonfunctional, leading to the more severe phenotype.
e.
Deletion of nucleotides 83-93
This mutation deletes 11 nucleotides from the gene and, because 11 is not a multiple
of 3, will cause a reading frame shift in the mRNA. All of the codons downstream of
the mutation site will be misread and “garbage” amino acids will be added.
Moreover, a out-of-frame stop codon is likely to be encountered fairly soon, leading
to early termination of translation. Since the change is so early in the gene
sequence, it is highly unlikely that such a change would leave the polypeptide with
any residual function, thus explaining the severe phenotype.
5-6. More practice with unit conversions and common lab calculations. Try to work these
problems without looking at your notes. Remember to show your entire calculation and to
strike through units to receive full credit!
5.
The following primer set is used during a PCR reaction. What annealing temperature would you
choose to run the reaction? Show your calculations and explain your logic.
forward primer: 5’ – aatgcgttatggaagagct – 3’
reverse primer: 5’ – ccacggcgattagttattga – 3’
Tm of forward primer = (11 x 2) + (8 x 4) = 54 deg C
Tm of reverse primer = (11 x 2) + (9 x 4) = 58 deg C
The annealing temperature should be about 52 deg C, a couple of degrees below the
lower of the two primer Tms.
6.
You cut a viral genome that contains 12,344 bp with the restriction enzyme Eco RI,
producing 5 bands, the shortest of which is 422 bp. What is the smallest amount of viral DNA
(in ng) that you would need to cut in order to ensure that you would see the 422 bp band on
a gel, given that the lower limit of detection of ethidium bromide staining is 10 ng?
(422/12,344)(X) = 10 ng;
X = (10 ng)/(0.0342) = 292 ng
You would need to cut at least 292 ng of the viral DNA in order to see the 422 bp
band on an ethidium bromide-stained gel.
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