Question

1. E. coli DNA polymerase V has the ability to bypass thymine dimers. However, Pol V tends to incorporate G rather than A opposite the dam-aged T bases. Would you expect Pol V to be more or less processive than Pol III? Explain.

2. Explain why base excision repair, nucleotide excision repair, and mismatch repair—which all require nucleases to excise damaged DNA—require DNA ligase.

3. Why are there no Pol I mutants that completely lack 5′ → 3′ exo-nuclease activity?

Answer 1

Ans- 1   E. coli DNA polymerase V has the ability to bypass thymine dimers. However, Pol V tends to incorporate G rather than A opposite the dam-aged T bases. The processivity of DNA Pol III is higher than POL V . The ability of Pol V to add and surpass few nucleotides is due to its Dna repair activity. The processivity rate is the speed by which a enzyme can add nucleotides during daughter strand synthesis. This activity is exhibited by pol III and pol V plays role in dna repair by changing the incorrect nucleotides. So we cannot use different activities to compare processivity.

Ans-2 base excision repair, nucleotide excision repair, and mismatch repair- which all require nucleases to excise damaged DNA—require DNA ligase because after these activities, Dna is not bond together and these activities break the Dna at the point where is some deformities in the structure than the normal. Dna ligase helps in formation of covalent phosphodiester bond between the 3' hydroxyl end of one nucleotide and 5' phosphate end of another nucleotide. It uses 2 ATP for this purpose.

Ans-3 There is no Pol I mutants that completely lack 5′ → 3′ exo-nuclease activity because this activity is very important for the survival of the organism in which this enzyme is present. This activity ensures that the organism is capable to live and reproduce further , and survive. So DNA POL I always have 5'-3' exonuclease activity.

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