ANSWER KEY
FCH 532 Homework 5
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1. What are the differences and similarities between FASTA, CLUSTAL, and
BLAST algorithms?
FASTA
CLUSTAL
BLAST
1. Matrix: PAM or
1. multiple sequence
1. matrix: BLOSUM62
BLOSUM
alignment
2. The higher the E
2. Gap penalty
2. different matrix and
value, the less
parameters
gap penalties.
significant the
3. Ktup = 1 or 2
3. Start with the highest
alignement.
4. The smaller the ktup
scoring pairwise
3. Used for pairwise
value, the more
alignment.
alignments
sensitive the alignment.
5. Used for pairwise
alignments.
2. Why is it preferred to use polypeptide sequence rather than nucleic acid sequence
to determine relationships between genes?
The proteins derived from the polypeptide sequence specify the function of the
encoding genes. Because different triplet codons can determine the same protein
sequences, and protein sequences determine the function, there may be higher
variability at the nucleic acid sequence compared to the polypeptide sequence.
Therefore, it is move valuable to evaluate the relationships based on the
polypeptide sequence. DNA sequence alignments also do not use substitution
matrices which can account for similarities in function between amino acid side
chains.
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3. Outline the differences and similarities between B-DNA, A-DNA, and Z-DNA.
(i.e. handedness, pitch, etc.)
4. What are the differences of hydrogen bonding, hydrophobic forces, and
electrostatic interactions with regards to nucleic acid structures?
Hydrogen bonds are required for proper base pairing during DNA replication.
Contribute little to the stabilization of the double helix.
Hydrophobic forces account for most the stabilizing interactions of DNA.
Stacking interactions in aqueous solutions are derived from hydrophobic
interactions.
Electrostatic interactions
5. A-T base pairs in DNA exhibit greater variability in their propeller twisting than
do G-C base pairs. Suggest the structural basis for this phenomenon.
A-T basepairs have 2 hydrogen bonds and are less stable than the 3 H-bonds for GC basepairs. This accounts for the differences between the two different structures’
range of flexibility and thus propeller twisting.
6. Calculate the Tm of a double stranded DNA molecule with 0.5 mol% G+C content
in a 0.1M NaCl solution.
Tm = 41.1XGC + 16.6 log[Na+] + 81.5
= 41.1 (0.5 mol%) + 16.6 log[0.1] +81.5 = 85.4 C
7. At Na+ concentrations >5M, the Tm of DNA decreases with increasing [Na+].
Explain this behavior. (Hint: consider the solvation requirements of Na+)
ANSWER KEY
At high concentrations of Na+ , there is little water available to promote
hydrophobic bonding since most water molecules are involved in solvating the
ions in the solution. Consequently, the structure of the double-helical DNA,
which is largly stabilized by hydrophobic forces, is destabilized by increasing
[Na+] as is indicated by its decreasing melting temperature.
8. When the helix axis of a closed circular duplex DNA of 2340 bp is constrained to
lie in a plane, the DNA has a twist (T) of 212. When released, the DNA takes up
its normal twist of 10.4 bp per turn. Indicate the values of the linking number (L),
writhing number (W), and twist for both the constrained and unconstrained
conformational states of this DNA circle. What is the super helix density, , of
both the constrained and unconstrained DNA circles?
L=T+W. For the constrained DNA circle, W=0 so that L=T=212. For the
unconstrained DNA circle, L=212 since this quantity is invariant, T=2340
bp/10.4 bp/turn =225 and W=L-T=212-225= -13.
For the constrained DNA circle, superhelical density = W/T =0/212 =0. For the
unconstrained DNA circle, the superhelical density = -13/225 = -0.058 (a value
that is typical of naturally occurring DNA circles in vivo).
9. A closed circular duplex DNA has a 100-bp segment of alternating C and G
residues. On transfer to a solution containing a high salt concentration, this
segment undergoes a transition from the B conformation to the Z conformation.
What is the accompanying change in its linking number, writing number, and
twist?
In B-DNA to Z-DNA conversion, a right-handed helix with one turn per 10.4
base pairs converts to a left-handed helix with one turn per 12 base pairs. Since
a right handed duplex has a positive twist, the twist decreases. Thus,
T = -(100/10.4) + (-100/12) = -17.9 turns
The linking number must remain constant (L = 0) since no covalent bonds are
broken. Hence, the change in W is
W = -T= 17.9 turns
10. Type IA and type II topoisomerases exhibit no significant sequence similarity,
but have been suggested that they are related based on their enzymatic
mechanisms. What are the similarities in the mechanisms?
Types IA and II topoisomerases both transiently cleave DNA strands by the
nucleophilic attack of an active site Tyr OH group on the scissile phosphate to form
a phosphoTyr linkage to the DNA strand’s newly liberated 5’ end. In addition, both
types of enzyme probably function via a strand passage mechanism (cleaving a
single- or double-stranded DNA, passing a similar strand through a break and
resealing the break) rather than a controlled rotation mechanism such as appears to
be mediated by type IB topoisomerases (cleaving one strand of a duplex DNA,
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letting one portion of the opposite strand rotate about its now unconstrained sugar
phosphate bonds opposite the break, and then resealing the break).