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Different domains of DNA polymerase contain different activity, like 5'->3' polymerisation and 3'->5' proof reading activity (a general case), and these domains can be exploited separately to make them perform single activity at a time. For example if we are conducting detection of some sort of mutation on a strand of DNA, we need to suppress the activity of DNA polymerase proof reading, ie, stopping it to correct the error it makes.

Since the proof reading activity increases the fidelity of replication to some 99.9%, I want to know what would be the efficiency if there is no proof reading activity.

To simplify my question- If DNA repair system commits 1 mistakes out of 10^8 base pairs it added in nucleotide chain, how many errors would occur if we remove it's proof reading mechanism?

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  • $\begingroup$ What research have you done to find the answer to this question yourself? Have you looked at related questions on this list? Try the right-hand side bar. There were questions recently on this topic and you will find the answer there. $\endgroup$ – David Sep 9 '18 at 18:49
  • $\begingroup$ Actually I don't understand the dna sequencing tag $\endgroup$ – Shred Sep 10 '18 at 8:24
  • $\begingroup$ @David do I need to show you my research papers or my notes to be able to ask question here? $\endgroup$ – Sahu V Kumar Sep 10 '18 at 17:41
  • $\begingroup$ @Shred DNA sequencing tag is for those people who mostly use DNA exonucleases for sequencing which is also related to my question. $\endgroup$ – Sahu V Kumar Sep 10 '18 at 17:42
  • $\begingroup$ The objective of this site is "to build a library of detailed answers to every question about biology". So, obviously, the first thing you should do is check that there isn't already an answer to your question here. The Help to this site provides tips on asking good questions, in particular : "Have you thoroughly searched for an answer before asking your question? … This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers." $\endgroup$ – David Sep 10 '18 at 19:57
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Klenow Fragment could be used to have a polymerase which lacks the exonuclease activity: not just the 5' -> 3' but also the 3' -> 5' .

Lacking the exonucleases activity, the only kind of control is given by the biochemical property of the enzyme: remember that the DNA polymerase has an active site similar to a right hand, which make possible to have correct couple between A,G,C,T. Then, according to this study, the error rate is something around 1e-4.

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