FCH 532 Homework 2
Answer key
FCH 532 Homework 2
Answer key
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1. The base sequence of one of the strands of a 20-bp duplex DNA is:
5’-GTACCGTTCGACGGTACATC-3’
What is the base sequence of its complementary strand?
5’-GATGTACCGTCGAACGGTAC-3’
Draw the standard Watson-Crick base pair structures showing hydrogen bonds. Where is
the minor groove? Where is the major groove? Explain how this structure is important
for the structure of DNA, Chargaff’s rules and heredity.
The base pairing as shown here is
responsible for the double helix
form. The distance is the same
between A and T base pairs and
G and C base pairs. The base
pairing explains Chargaff’s rules
(A=T, G=C). Hydrogen bonding
between these specific base pairs
explains how this occurs as shown
in the figure. The two strands
are complimentary of one
another so each can act as a
template for the other. Semiconservative replication
determines heredity.
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FCH 532 Homework 2
Answer key
2. Non-Watson-Crick base pairs are of biological importance. For example: (a)
Hypoxanthine (6-oxopurine) is often one of the bases of the anticodon of tRNA . With
what base on mRNA is hypoxanthine likely to pair? Draw the structure of this base pair.
(b) The third position of the codon-anticodon interaction between tRNA and mRNA is
often a G-U base pair. Draw a plausible structure for such a base pair.
(c) many of species of tRNA contain a hydrogen bonded U-A-U assembly. Draw two
plausible structures for this assembly in which each U forms at least two hydrogen bonds
with the A.
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FCH 532 Homework 2
Answer key
(d) Mutations may arise during DNA replication when mispairing occurs as a result of a
rare tautomeric form of a base. Draw the structure of a base pair with proper WatsonCrick geometry that contains a rare tautomeric form of adenine. What base sequence
change would be caused by such mispairing?
3. (a) What is the molecular mass and contour length of a segment of B-DNA that
specifies a 40-kD protein?
Average molecular weight of amino acid = 105.2 dalton
Assume that the 40 KDa protein is composed of 380 amino acids.
Number of nucleotides= 380 X 3=1140 nt
Average molecular weight per nucleotide = 499.5
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FCH 532 Homework 2
Answer key
Molecular mass of B-DNA is 5.69 X 105 da
The contour length of B-DNA =1140 nt/10 nt X 34 Å = 3876 Å.
(b) How many helical turns does this DNA have and what is its axial ratio (length to
width ratio)?
Helical turns = 1140 nt/10 nt/turn =114 turns
The width of B-DNA = 20 Å
Axial ratio = 3876Å/20Å=193.8
4. What are the differences between RNA polymerase and DNA polymerase?
1. RNA polymerase can work without a primer. DNA polymerase requires an
oligonucleotide primer base-paired to the template strand.
2. DNA polymerase has proofreading mechanisms =3’-5’ exonuclease for most DNA
polymerases and 5’-3’ exonuclease activity for DNA pol I. RNA polymerase does
not have the proofreading exonuclease.
5. Describe continuous and discontinuous replication.
DNA strands are simultaneously replicated in a continuous and discontinuous
strand. The leading strand is continuously replicated 5’ to 3’ and the discontinuous
replication occurs on the lagging strand in short segments from the 5’ to 3’ parental
strand. Primase introduces primers on the discontinuous strand that are removed
by DNA pol I. Gaps are sealed by DNA ligase.
6. The following duplex DNA is transcribed from right to left as printed:
5’-TC TGA CTA TTC AGC TCT CTG GCA CAT AGCA-3’
3’-AG ACT GAT AAG TCG AGA GAC CGT GTA TCGT-5’
(a) Identify the template strand. Clue is the wording of this question. Transcribed
from right to left!
5’-TC TGA CTA TTC AGC TCT CTG GCA CAT AGCA-3’
(b) What is the amino acid sequence of the polypeptide that this DNA sequence
encodes? Assume that translation starts at the first initiation codon.
Met-Cys-Gln-Arg-Ala-Asp-Stop
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FCH 532 Homework 2
Answer key
(c) Why doesn’t the UGA sequence in the mRNA transcript cause transcription to
terminate?
It is out of reading frame.
7. After undergoing splicing, a mature mRNA has the following sequence, where the
vertical line indicates the position of the splice junction (the nucleotides from between
which an intron had been removed).
5’-CUAGAUGGUAG|
GUACGGUUAUGGGAUAACUCUG-3’
What is the sequence of the polypeptide specified by this mRNA? Assume that
translation starts at the first initiation codon.
(a) 5’-CUAG AUG GUA GGU ACG GUU AUG GGA UAA CUCUG-3’
MVGTVMG-STOP
(b) What would the polypeptide sequence be if the splicing system had erroneously
deleted the GU on the 3’ side of the splice junction?
MVDGYGITL
(c) What would the polypeptide sequence be if the splicing system had erroneously
failed to excise a G at the splice junction?
MVGYGYGITL
(c) is there any relationship between the polypeptides specified in b and c and, if so,
why?
The two polypeptides are identical except that polypeptide b has an added residue with
respect to polypeptide c (Asp at third position). This is due to deleting two nucleotides in
the protein encoding portion of an mRNA shifts the reading frame in the same way as
adding one nucleotide. The deletion and addition occur at the same position and so in b
mRNA is one codon longer and polypeptide is one residue longer than c.
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