STUDY QUESTIONS: LEC 5-7
1. How does complexity of bacterial genome differ from that of eukaryotic (calf) genome?
Bacterial genome has more unique sequences while eukaryotic genomes have a lower
proportion of unique sequences, and a large amount of highly repetitive sequences. However,
calf genome is more complex as it has a greater amount of unique DNA overall.
2. Explain C value paradox.
C value is the DNA content of a haploid cell. There is no correlation between the amount of DNA
(size of genome) and the apparent complexity of organisms.
3. List and briefly explain factors that influence DNA renaturation kinetics.
ï Size of the DNA fragment (indirect) ï the larger the size, the harder it is for complementary
bases to find each other.
ï Complexity of genome (indirect) ï a more complex genome will have more unique
sequences that will renature very slowly as there are few similar copies, and so each strand
must find its complement
ï DNA concentration (direct) ï at [higher], strands will be able to find each other faster.
ï Ionic strength (direct) ï higher salt concentration will help neutralize âve charge of DNA
and therefore help annealing.
ï Time (direct)ï over time, more strands will renature
ï Temperature (indirect)ï 20-25 degrees below Tm.
ï pH ï affects H-bonds
4. You have found a new species of insects. To evaluate the complexity of the genome of this
species, you isolate genomic DNA from, fragment the DNA to uniform 500 base pair pieces,
denature the DNA and measure the rate of reassociation. Your data is represented in the
curve below (sorry for the bad drawing):
(a) How many classes of DNA (in respect to sequence complexity) are found in this organism?
3; highly repetitive, moderately repetitive, and unique
(b) What can you say about the relative complexity of each class? What fraction of the genome
falls into each class?
Highly repetitive Moderately repetitive Unique
>1000 copies >10 copies 1-10 copies
Less complex Moderately complex Very complex
25% 25% 50% 5. List three (3) differences between prokaryotic Topoisomerase I and Gyrase.
Topoisomerases are enzymes that recognize and regulate supercoiling. They play an important
role in replication and transcription.
Topoisomerase I Gyrase
Makes 1 cut Makes 2 cuts
Removes âve supercoils Introduces âve supercoils;
Relaxes +ve supercoils.
L# = +1 L#= -2
Doesnât use ATP Uses ATP
6. What are topological isomers of DNA?
They are identical molecules of DNA that differ only in their degree of supercoiling (from relaxed
to tightly supercoiled).
7. Explain the importance of DNA supercoiling for the cell survival?
The only way to get the 2.5m long genome in a nucleus if the DNA is highly coiled. It is therefore
important for packing of DNA and making it more condensed.
It also reduces the induced stress by twisting the double helix.
Coiling also prevents access of enzymes to DNA, which prevents unwanted processes from
taking place. E.g., if DNA were to be unwinding at all times, DNAP would make multiple copies of
the DNA at all times, costing the cell nutrients and energy and therefore, decreasing its viability.
8. What are the differences between primary (or secondary, or tertiary) structures of RNA and
Primary Sugar-phosphate âchainâ Similar to DNA except it can exhibit different
with purine and pyrimidine bases as side conformations
SecondaryDouble helical structure (hydrogen Areas of regular helices and discontinuous
bonding between A-T and G-C; stacking helices with stem-loops of hairpins. Frequently
interactions; phosphate backbone fold back on themselves to form base-paired
âoutsideâ) segments between short stretches of
Tertiary Double stranded DNA (both circular AND Formed through interactions of secondary
linear) makes complexes with proteins - structures: lack of constraint by long-range
supercoil regular helices means RNA has high degree of
Supercoiling = coiling of a coil. rotational freedom in backbone of its non-base-
paired regions â capable of folding into
complex tertiary structures
Also, formation of unconventional U:A:U base
triple is possible.
Pseudo-knots can be formed due to base-pairing
between sequences that arenât adjacent. 9. You have a small 4800 bp long circular DNA. It has a linking number of 450 (L=450). What are
the twist (T) and the writhe (W) of this DNA? What assumptions about the structure of the
DNA have you made in your answer? (Hint: what is the definition of the twist #?)
Ans: T = # of bp / # of bp/turn
= 4800/10 = 480 turns
L = T+ W
W = -30
The DNA has 30 negative supercoils and protein can easily get in and transcribe it
Also, the DNA has 480 complete turns assuming the DNA is in B form.
10. Thinking question 1: Many different mutations have been observed in almost all genes.
However, only few have been isolated in histones. How would you explain this finding?
Mutations that survive to be passed on to the next generations are in general favoured by
natural selection, however, histone proteins are very essential for the cell, and their function is
very well carried out by their structure as is. Therefore, natural selection is unlikely to favour
mutations in the genes coding for histones and organisms with mutated genes are not likely to
survive to pass on their genes. This could have led to the highly conserved nature of the gene
sequence of histones.
11. Thinking question 2: would it be an easy task to produce polyclonal antibodies to h