My teacher told me that when DNA polymerase makes an error (roughly every 10 million nucleotides?) that if, for example, it matches an A with a G that the error remains and is the main cause of point mutations. How do the two bases that aren't complementary remain bonded?


1 Answer 1


How do the two bases that aren't complementary remain bonded?

Non complementary bases can only form non-Watson-Crick bonds, which are very unstable and noticeably different.
Therefore any mismatch in the DNA of the genome will form a recognisable disruption of the DNA helix. This disruption is usually removed very quickly by the cells, since it makes DNA much more unstable.

Most cells have multiple mechanisms to detect and repair these sites, with one system specifically dedicated to DNA mismatch repair.
This system acts on mismatches directly after DNA replication, since at this point specific markers (sometimes DNA methylation, sometimes not fully understood) can differentiate between the newly synthesised DNA strand (which may contain an error) and the old one.

The repair mechanism will remove the mismatch and replace it with a properly bonded base pair, so that mismatches in the DNA are always short term issues - it is still possible though that 'wrong' mismatching base is replaced, which then (finally) leads to a mutation.

The error rate of DNA polymerase (1 in 10^7 sounds about right, this is the combination of the initial error rate and the DNA polymerase proof-reading mechansim) therefore has to multiplied with the error rate of this repair mechansim, which is around 10^3 (I can't find a source for this right now), to get a final error rate of DNA replication in cells.

  • 1
    $\begingroup$ "They do not" is incorrect. Non complementary base pairs do form, and remain until they are removed. They are less stable, and produce a detectable change in the 3-d structure of the duplex. The rest of your answer is correct. $\endgroup$
    – De Novo
    Commented Mar 22, 2019 at 16:29
  • $\begingroup$ @DeNovo edited the answer accordingly $\endgroup$
    – Nicolai
    Commented Mar 24, 2019 at 12:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .