Question

4. What is an Okazaki fragment and explain why it is found only in the lagging strand in a replication fork, but not the leading strand.

5. What is the role of a telomerase in eukaryotes and why are they not needed in bacterial cells?

6. Using a translation table, there is only one possible sequence of nucleotides in the template strand of DNA listed below that would code for the polypeptide sequence Phe-Leu-Ile-Val. Identify the correct one and explain for all other choices why they cannot be correct.

A) 5' UUG-CUA-CAG-UAG 3'.

B) 3' TTT-GTT-TAG-CAG 5'.

C) 5' AAG-GAA-TAA-CAA 3'.

D) 3' AAA-AAT-ATA-ACA 5'.

E) 3' AAA-GAA-TAA-CAA 5'.

Answer 1

4. In replication fork DNA polymerase III moves in only one direction that is 5' - 3' direction because, it can only add the new bases at the 3' end. But, the DNA is antiparellel in nature which runs in both the direction. So, DNA polymerase has to polymerise both the strands at a time though, it runs in only one direction. Therefore to make this thing possible it makes a clamp which take a particular length of DNA in a bubble. This particular length will be polymerised along the continuous strand. By this way, the whole length of the DNA will be polymerised in a discontinuous manner. This is called as lagging strand or Okazaki fragment.

5. Telomerase maintains the length of the DNA. Okazaki fragment causes the shortening of the DNA fragment at each replication cycle. Bacetria contain a circular genome. So, unlike our linear DNA, there is no shortening of DNA in bacterial replication.

6. E. 3' AAA- GAA-TAA-CAA 5' mRNA for the DNA will be 5' UUU- CUU- AUU- GUU 3'. This codes phe- Leu- Ile- Val.

First option contains UAG stop codon, Second option's first codon is AAA which does not code for Phe. Third option's first codon UUG does not code for Phe. Fourth option's third codon UAU does not code for Ile.