PhDP Posted May 18, 2008 Posted May 18, 2008 I just want to answer to something SkepticLance said about vitamin C, first because he made a mistakes, but also because it leads to a simple explanation of the molecular clock (the idea that we can use DNA to date events). The other thing that relates to loss of vitamin C manufacturing ability' date=' in addition to the fruit diet mentioned by ecol, is a very low population number. The loss came about because of a mutation in a key gene, that spread to all humans. That spread would not have come as a result of natural selection, since there was no selective advantage. It would have come as a result of genetic drift, which is something that does NOT happen in large populations. Thus, the loss happened at a time when our forebears were very low in population size. We can also conclude that a similar event CANNOT happen today, with 6.5 billion people interbreeding.[/quote'] Effective population size is fundamental to, well, pretty much everything in population genetics. But in this case, it doesn't matter. For diploids, the probability of fixation 'f' of a new mutant with selection coefficient 's' and population size 'N' is given by; [math] f(s) = \frac{2s}{1-e^{-4Ns}} [/math] (We get this equation from the diffusion approx.) Now, losing our capacity to produce vitamin C doesn't matter if we already get this vitamin in our diet, so a mutation that would destroy our capacity to produce vitamin C will have a selection coefficient of 0 (in short, it's neutral). With s = 0 and l'Hôpital's rule we find that; [math] f(0) = \frac{d(2s)/ds}{d\left(1-e^{-4Ns}\right)/ds} = \frac{2}{4Ne^{-4Ns}}=\frac{1}{2N} [/math] If the mutation rate for this gene is [math]\mu[/math], then the total number of mutations appearing in a population is simply; [math] \mu 2N [/math] ...because a mutation can appear in every single individual. High population = more mutations. As a last step, we combine this with the probability of fixation to get the rate of substitution 'k', that is, how many mutants will reach fixation in a generation; [math] k = \mu 2N \times P(s) [/math] With this equation we can see that the rate of substitution depend largely on the effective population size. However, in the neutral case; [math] k = \mu 2N \times P(0) = \mu 2N\frac{1}{2N} = \mu [/math] This is simple, but it's also an extraordinary result; the rate of substitution of neutral mutations is NOT affected by time or population size, it depends only on the mutation rate... Molecular clock If we can find a "neutral region" shared by two species, that is, a region of DNA where mutations are neutral, then, only by looking at how many substitutions occured, we can estimate the time of divergence of these two species.
SkepticLance Posted May 18, 2008 Posted May 18, 2008 To PhDP What if the mutation rate is once per million years? As far as I know, this mutation has not happened more than the one single time. I could easily be wrong. It is possible that, in some other taxonomic group, it happens on a regular basis. If I am wrong, please let me know the paper or article where this is described, since I would be most interested.
PhDP Posted May 18, 2008 Author Posted May 18, 2008 What if the mutation rate is once per million years? No. Frakking. Way. It HAS to have happened several times. Look at the equation, the probability of fixation of a neutral mutation is 1/2N. It's very low. The mutation rate for humans is about 200 mutations/genome/generations, and there are several ways to break a gene, from simple mutations like a point mutation to catastrophic mutations like a frameshift. More to the point, since it has lost its function, the gene that was responsible for the synthesis of vitamin C has accumulated several mutations; Biochimica et Biophysica Acta 1472:408-411.
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