Question

DNA polymerase III builds new DNA strands in the 5' to 3' direction...always adding the new nucleotide to the 3' end of the existing strand. As it adds new nucleotides, it proofreads its work. If a mistake is detected, DNA polymerase will act as an exonuclease and excise the incorrect nucleotide. A. What supplies the energy needed to add the new nucleotide to the existing chain? B. Please explain why evolution has favored 5’ to 3’ polymerases, as opposed to 3’ to 5’ ones (hint...what happens if DNA Polymerase removes a nucleotide?).

Answer 1

A. answer- The energy needed to add the new nucleotide to the existing chain during replication is supplied by nucleoside triphosphates - ATP, GTP, TTP and CTP, which is high energy molecule like ATP. Nucleotide triphosphates serves both as a souce of energy and source of nucleotide to drive polymerization.

The source of energy  comes from breaking high energy phosphate bonds on the nucleotide-triphosphate's , the energy released is used to form the phosphodiester bond between the new nucleotide and the existing chain of DNA.

B. answer- Evolution has favoured 5' to 3' polymerases as opposed to 3' to 5' ones may be because DNA polymerase add new nucleotides only to the 3'-OH group of growing chain of DNA during polymerization.

During proofreading mechanism, when DNA polymerase removes nucleotide then the energy required for addition of new nucleotide  will be always from the triphosphate of the incoming base in case 5' to 3' replication direction of polymerase.

Whereas  in case 3' to 5' replication direction of polymerase, it would not be possible to add new nucleotide after removal of incorrect base from 5' end because there will be no energy at 5' end of resulting monophosphate in existing DNA and 3'-OH group of incoming nucleotide.

This is the reason why evolution has favoured 5’ to 3’ polymerases, as opposed to 3’ to 5’ ones.