Dr. Ruth Ballard
BIO 184
Page 1
02/06/16
Fall 2006
HOMEWORK 1
KEY
1. (6 points)
The coding region of an mRNA begins with the following sequence:
5’ – AUGCCCAUUGGGCGCAACAAGCAUAUGGCC – 3’
a.
Draw the template strand of the DNA molecule coding for this region of the mRNA. Label
its 5’ and 3’ ends.
ANSWER:
b.
Draw the coding strand of the DNA molecule coding for this region of the mRNA. Label its
5’ and 3’ ends.
ANSWER:
c.
3’ – TACGGGTAACCCGCGTTGTTCGTATACCGG – 5’
5’ – ATGCCCATTGGGCGCAACAAGCATATGGCC – 3’
Use the Genetic Code (in your textbook) to translate the mRNA into a polypeptide. Label
the N and C termini of the polypeptide.
ANSWER:
N- met-pro-ile-gly-arg-asn-lys-his-met-ala- C
2. (4 points)
The following sequence is from the middle portion of exon 4 of the human APP gene:
5’- TCTCTCTCGGTGCTTGGCCTC – 3’
3’- AGAGAGAGCCACGAACCGGAG – 5’
Assuming that the top strand is the template strand, what is the mRNA sequence that will be
transcribed from this region of the gene? Label its 5’ and 3’ ends.
ANSWER:
5’ – GAGGCCAAGCACCGAGAGAGA – 3’
3. (15 points)
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 recessive 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.
Dr. Ruth Ballard
BIO 184
Page 2
02/06/16
Fall 2006
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 where the mutations carried by five
unrelated mindreader mutants were analyzed.
Description of mutation
Nonsense mutation in codon 774
Missense mutation in codon 52
Deletion of nucleotides 934-939
Missense mutation in codon 192
Deletion of nucleotides 83-93
Phenotype
delayed-receiver
delayed-receiver
delayed-receiver
insensitive
insensitive
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
ANSWER: A nonsense mutation replaces an amino acid-coding codon for a
stop codon. Usually, this is associated with a severe phenotype because it
causes a polypeptide to be terminated prematurely. However, if the
mutation occurs very late in the coding region of the gene and only a few
amino acids are lost, the polypeptide might have some residual function. In
this case, the mutation occurred in codon 774, which corresponds to
nucleotides 2,320-2,322 (774 x 3 = 2,322). Since the entire gene is comprised
of coding DNA and is 2,331 nucleotides long, only 3 amino acids are actually
deleted from the carboxyl terminal. Therefore, the milder phenotype is
probably explained by the protein having some function since most of it is
intact.
b.
Missense mutation in codon 52
ANSWER: A missense mutation causes one amino acid to be replaced by
another but does not add or delete amino acids. A missense mutation can
cause a mild phenotype if the amino acid being replaced is not in a domain
of the protein that is vital for its function, but does cause some disruption in
protein folding and/or substrate recognition. A milder phenotype is also
often associated with a situation in which the new amino acid has a similar
“R” group chemistry to the original amino acid it has replaced. This often
allows the protein to have some level of function.
c.
Deletion of nucleotides 934-939
Dr. Ruth Ballard
BIO 184
Page 3
02/06/16
Fall 2006
ANSWER: This deletion mutation eliminates 2 amino acids in the
polypeptide (a total of 6 bases) but does not affect the downstream reading
frame of the mRNA because the total number of base-pairs in the deletion is
divisible by three. Therefore, the 5’ and 3’ ends of the protein will be made
correctly and the only disruption is the loss of the amino acids involved in
the deletion. As long as these amino acids are not vital to the protein’s
function, a mild phenotype could result.
d.
Amino acid substitution mutation in codon 192
ANSWER: This missense mutation must involve an amino acid that is vital
to the proper folding and/or function of the protein. This can happen when
the amino acid is exchanged for one with a very different “R” group
chemistry or if the only amino acid that can function in that site is the one
that was in the wild-type form of the protein.
e.
Deletion of nucleotides 83-93
ANSWER: This deletion involves a total of 11 base-pairs, a number that is
not divisible by three. As a result, this mutation will result in a frame-shift
as the mRNA is being read by the ribosome, and the entire polypeptide
sequence C-terminal to the mutation site will be completely different from
that of the wild-type form. The polypeptide will probably also be
considerably shortened since an off-frame stop codon is predicted to occur
about once in every twenty-one codons. This particular reading frame
mutation is particularly bad because it occurs so early in the gene sequence,
and very little of the polypeptide will be made correctly. Such a mutation
almost always results in a non-functional protein and a severe phenotype.
3. (5 points)
During translation, the ribosome must be able to recognize and bind to an mRNA. Explain how this
process occurs in prokaryotes, using a diagram to support your answer.
ANSWER: In prokaryotes, a portion of the 16S ribosomal RNA
(which is part of the small ribosomal subunit) recognizes the Shine
Delgarno sequence in the 5’ UTR of the mRNA through WatsonCrick base-pairing. This 7 base-pair interaction is strong enough to
stabilize the small ribosomal subunit on the mRNA until the other
units of the translation machinery can bind.
Dr. Ruth Ballard
BIO 184
Page 4
02/06/16
Fall 2006
4. (4 points)
Nitrogen and carbon are more abundant in proteins than sulfur. Why did Hershey and Chase use
radioactive sulfur instead of nitrogen or carbon to label the protein portion of their bacteriophages
in their experiments to determine whether parental protein or parental DNA is necessary for
progeny phage production?
ANSWER: Selective labeling of the protein as different from the DNA would have
been impossible because DNA also contains carbon and nitrogen. The
bacteriophages would have had both biomolecules labeled and their movement
during the experiment could not have been monitored separately.
5. (6 points)
The diagram below shows the structure of a eukaryotic pre-mRNA. Below the diagram, draw your
own picture of how this mRNA would look after processing into its mature form. Include as much
detail as you can and be sure to label your drawing thoroughly.
exon 1
exon 2
exon 3
5’--------AUG ---------------------------------------------------------------UGA-------AAUAAA --------3’
ANSWER:
exon 1
exon 2
exon 3
--------AUG---------------------------------------UGA------AAUAA ----AAAAAAAAAAAAAAAA…3’
5’ CAP (5’-5’ linkage with
methylated G)
Introns have been
removed by the
spliceosome and
exons have been
connected
The 3’UTR downstream
from the AAUAA
poyladenylation signal is
cleaved and a polyA tail is
added (tails vary in length
but can contain as many as
250 A residues)