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haldane_1937

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Haldane, J. B. (1937). The effect of variation of fitness. The American Naturalist, 71(735), 337-349.

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Abstract: In a species in equilibrium variation is mainly due to two causes. Some deleterious genes are being weeded out by selection at the same rate as they are produced by mutation. Others are preserved because the heterozygous form is fitter than either homozygote. In the former case the loss of fitness in the species is roughly equal to the sum of all mutation rates and is probably of the order of 5 per cent. It is suggested that this loss of fitness is the price paid by a species for its capacity for further evolution. N - Nf + N\mu

Notes

Symbols
  • $\mu$ is the mutation rate per locus per generation.
  • $N$ is the population size.
  • $x$ is the frequency of individuals carrying a copy of the mutant allele.
  • $f$ is the relative average fitness of individuals carrying the mutant allele.
  • $y$ is half the number of heterozygotes, which are $2y$.
  • Here, $p$ is the mutant allele frequency.
  • Here, $s$ is the reduction in fitness of the mutation. $s = 1-f$.
Dominant Fitness Effect

Pages 341–342 deal with a dominant fitness effect and the assumptions that the species is diploid, deleterious allele is rare (so most individuals with a copy only have one copy), and the unmutated fitness is one.

A fraction of the unmutated loci are expected to mutate each generation.

$$N\mu (2-x)$$

Mutated copies are removed by selection each generation ($1-f$ survive).

$$N x (1-f) $$

At equilibrium the rate of removal is equal to the rate of input.

$$N x (1-f) = N\mu (2-x)$$

$$Nx - Nxf = 2N\mu - N\mu x$$ $$Nx - Nxf + N\mu x = 2N\mu$$ $$x (N - Nf + N\mu) = 2N\mu$$ $$x = \frac{2N\mu}{N - Nf + N\mu} = \frac{2\mu}{1 - f + \mu}$$

Assuming $\mu$ is much smaller than one and that the mutant alleles are rare gives

$$ x = 2 p (1-p) \approx 2 p \approx \frac{2\mu}{s}\mbox{.}$$

$$ p \approx \frac{\mu}{s}\mbox{.}$$

The loss of fitness (from one) to the species is

$$x (1-f) = x - fx = \frac{2\mu}{1 - f + \mu} - \frac{2\mu f}{1 - f + \mu} = 2\mu \left(\frac{1-f}{1-f+\mu}\right) \approx 2\mu\mbox{.}$$

This is approximately $2\mu$ if $\mu$ is small relative to $1-f$. The interesting thing about this is that, at equilibrium, the loss of fitness is only a function of the mutation rate and is independent of the average fitness of individuals carrying the mutant alleles. The reason for this is the inverse relationship between the loss of fitness and the equilibrium frequency of the allele. Mutant alleles with a higher fitness attain a higher frequency in the population. So even though the loss of fitness is small more individuals are affected. And vice versa, Mutants with a large loss of fitness are maintained at a low frequency and fewer individuals are affected by the large loss. On average over an entire population these factors cancel out and the average loss of fitness is only twice the mutation rate (for dominant effects).

Recessive Fitness Effect

Page 344 treats the case of recessive loss of fitness.

A fraction of the unmutated loci are expected to mutate each generation. (Only half of the heterozygous loci are unmutated.)

$$2N\mu (1-x-y)$$

Mutated copies are removed by selection each generation. (Each loss removes two copies of the mutant alleles so $2x$.)

$$N 2 x (1-f) $$

At equilibrium

$$2N\mu (1-x-y) = N 2 x (1-f)$$ $$2N\mu -2N\mu x-2N\mu y = N 2 x - N 2 x f$$ $$2N\mu -2N\mu y = N 2 x - N 2 x f + 2N\mu x$$ $$2N\mu -2N\mu y = x (N 2 - N 2 f + 2N\mu )$$ $$x =\frac{2N\mu -2N\mu y}{ N 2 - N 2 f + 2N\mu }$$ $$x =\frac{\mu -\mu y}{ 1 - f + \mu } = \frac{\mu (1 - y)}{ 1 - f + \mu }$$

If $y$ and $\mu$ are small

$$x \approx \frac{\mu}{ 1 - f }$$

$$p^2 = x \approx \frac{\mu}{ 1 - f } = \frac{\mu}{s}$$ $$p \approx \sqrt{\frac{\mu}{s}}$$

The average loss of fitness in the population is

$$x (1-f) = x - fx \approx \frac{\mu}{1 - f} - \frac{\mu f}{1 - f} = \mu \left(\frac{1-f}{1-f}\right) = \mu\mbox{.}$$

haldane_1937.1568516892.txt.gz · Last modified: 2019/09/15 03:08 by floyd