4
$\begingroup$

In terms of genetics, which type of reproduction is better against adverse mutations? I understand a bit of Muller's Ratchet—in which asexual organisms accumulate a number of deleterious mutations in their genomes. However, doesn't the crossover process and corresponding genetic diversity in sexual-reproducing organisms make them more susceptible to these mutations?

$\endgroup$
2
$\begingroup$

Mutations can be caused by much more than just recombinatory events--polymerase fidelity, DNA damage repair, even environmental factors can have a huge impact on DNA damage. While most DNA polymerases have an error rate of around one mutation per billion, there are different levels of fidelity among different polymerases (McCulloch and Kunkel, 2008).

Regarding the fidelity of crossover/recombination events, it is true that these events increase the likelihood of a lethal or detrimental mutation. However, recombination is not a random process. Recombination hotspots are chromosomal locations where the rate of recombination can be hundreds of times higher than other locations. These hotspots are generally in "safe" locations, not in the middle of a crucial gene, etc. There are also recombinase enzymes specific to certain sites of recombination.

As to "which is better", I don't think the answer is as straightforward as you might think. You might think a sexually-reproducing eukaryote with a very high-fidelity DNA polymerase and safe, specific recombination hotspots would be ideal. That situation is generally what we find in nature, so you're (probably) correct. However, at the other extreme, think of a virus*. Viruses have a much, much, much higher rate of mutation than eukaryotes--orders of magnitude higher. This is how a virus survives, basically out-evolving the host's defenses. A high rate of mutation works for a virus, and a low rate of mutation works for a multicellular eukaryote.

*Of course, there is always debate about whether or not viruses are alive. However, they definitely reproduce.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.