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This paper uses two equations to explain the different conditions under which sexually antagonistic genetic variance is maintained in a population, while allowing for unequal dominance, for autosomally linked loci:

$h_f / 1-h_m + h_fs_f <$ $ $ $ s_m/s_f < 1-h_f/h_m(1-s_f)$

And X-chromosome linked (X-linked) loci:

$2 h_f / 1+h_f s_f < $ $ $ $ s_m/s_f < $ $ $ $2(1-h_f)/1-h_f s_f$

where $s_m$ and $s_f$ are the selection coffecients against the less fit homozygote (or hemizygote) in males and females respectively where the most fit genotype has a relative fitness of 1. Similarly $h_m$ and $h_f$ represent the dominance of the less fit allele in males and females, and $h_m$ is not applicable in the X-linked equation because there is no dominance (due to hemizygosity it can only be homozygote).

I'm having a little trouble understanding how $s$ and $h$ are defined here.

For $s$, is it the fitness of the deleterious homozygote or the difference in relative fitness between the two homozygotes? Such that either $s$ = the relative fitness of the homozygote, or, $s$ = 1 - relative fitness of deleterious homozygote, where one represents the fitness of the fittest homozygote thus giving the fitness differential.

For $h$, is the dominance defined as the fitness difference between the less-fit homozygote and the heterozygote, or the most-fit homozygote and the heterozygote? Or is it the deviation from the average fitness of the two homozygotes?

Going from the graph (see below for values of fitnesses), what are the values of $s_m$, $s_f$, $h_m$, and $h_f$ where the dashed line is Females and solid line is Males. I see possible values of either $s_m$ = 0.2 or 0.8 (fittest genotype - fitness of weakest), likewise $s_f$ is either 0.3 or 0.7, $h_m$ is either 0.1 (deviation from most-fit genotype), 0.7 (deviation from least-fit) , or 0.3 (deviation from average) and $h_f$ is either 0.3, 0.4, or 0.05.

enter image description here

The values here are, as fitnesses for genotypes A1A1, A1A2, and A2A2, for males, 1.0, 0.9, 0.2, and for females 0.3, 0.7, 1.0.

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Fry (2010) borrowed his variables from Kidwell et al. (1977). Kidwell defines the fitness of each genotype as,

$w_{m1}$, $w_{f1}$ = male and female fitness of the A$_1$A$_1$ genotype.

$w_{m2}$, $w_{f2}$ = male and female fitness of the A$_1$A$_2$ genotype.

$w_{m3}$, $w_{f3}$ = male and female fitness of the A$_2$A$_2$ genotype.

Kidwell then establishes the following parameters for each fitness variable for opposing additive selection,

$w_{m1} = w_{f3} = 1$,

$w_{m2} = 1-.5s_m$,

$w_{f2} = 1-.5s_f$,

$w_{m3} = 1-s_m$, and

$w_{f1} = 1-s_f$.

Therefore,

$s_m = 1-w_{m3}$, and

$s_f = 1-w_{f1}$.

$s_m$ and $s_f$ are the relative fitness differences between the two homozygous genotypes of each gender. The heterozygotes are assumed to have half of that differential. This is what Fry refers to on page 2 (page 1511 of the publication):

In Kidwell et al.’s notation ... the most fit genotype in a given sex has a relative fitness of 1, and $s_m$ and $s_f$ are the selection coefficients against the less-fit homozygote (or hemizygote) in males and females, respectively.

For opposing selection with arbitrary dominance, Kidwell parameterizes the heterozygous fitness components as

$w_{m2} = 1-h_ms_m$, and

$w_{f2} = 1-h_fs_f$.

Notice that $h$ is modifying the fitness difference $s$ in the heterozygotes. This is due to incomplete dominance of one allele over the other. If both alleles in a heterozygote contribute equally to the phenotype, then $h = 0.5$, and you have the additive fitness described above ($w = 1-.5s$).

If one allele is incompletely dominant over the other, the selective difference $s$ will be modified by some amount, $h$. If A$_1$ is only slightly dominant over A$_2$, (say, $h=0.6$), the effect on $w$ won't be much different than if $h=0.5$. Both alleles will affect fitness but the more dominant allele will have a slightly greater affect on fitness than the other allele. If one allele is much more dominant, then $h$ becomes larger and the more dominant allele contributes more to the overall fitness than the other allele.

Finally, if $h=1$ then one allele is completely dominant so the other allele will not contribute at all to fitness. When $h=1$, the equations reduce to $w = 1 - s$ (for the respective genders), remembering that $h$ is based on the dominance of the allele with maximum fitness in the other gender.

Consider a female. Her maximum fitness is the A$_2$A$_2$ genotype. If A$_1$ is completely dominant over A$_2$, then a heterozygous female would have fitness equal to the A$_1$A$_1$ genotype which, for females is defined as $w = 1-s_f$. On the other hand, if A$_1$ is not completely dominant over A$_2$, then the heterozygote will have some reduction in fitness but not as much as for complete dominance.

To answer your final question about your figure, given the following values for A$_1$A$_1$, A$_1$A$_2$, and $_2$A$_2$, respectively:

$w_m = 1, 0.9, 0.2$, and

$w_f = 0.3, 0.7, 1$, then

$s_m = 1 - 0.2 = 0.8$, and

$s_f = 1 - 0.7 = 0.7$.

For the heterozygosity effect ($h$), given that

$w = 1 - hs$, then

$h = \frac{1-w}{s}$, therefore

$h_{m2} = \frac{1 - 0.9}{0.8} = 0.125$.

A$_1$ is nearly completely dominant over A$_2$ in males, so the presence of A$_2$ is the male is not reducing his fitness by very much. For females,

$h_{f2} = \frac{1-0.7}{0.7} = 0.429$.

The presence of the dominant A$_1$ allele in females has stronger heterozygous effect so her fitness is reduced more, compared to heterozygous males.

Citations

Fry, J.D. 2010. The genomic location of sexually antagonistic variation: some cautionary comments. Evolution 64: 1510-1516.

Kidwell, J.F. et al. 1977. Regions of stable equilibria for models of differential selection in the two sexes under random mating. Genetics 85: 171-183.

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  • $\begingroup$ Hi @MikeTaylor Thanks for the answer, I've just edited a couple of the annotations which I think were incorrect, could you check the edit history to proof them for me please $\endgroup$ – rg255 Sep 18 '14 at 8:26
  • $\begingroup$ also should "Therefore, sm=1−wm1, and sf=1−wf3." read as "Therefore, sm=1−wm 3 , and sf=1−wf 1 ." $\endgroup$ – rg255 Sep 18 '14 at 8:30
  • $\begingroup$ @GriffinEvo Thanks for your careful proofing. I also added a bit more to answer your question about your figure. $\endgroup$ – Michael S Taylor Sep 18 '14 at 12:41
  • $\begingroup$ Thanks again, I had gotten myself tangled - I knew how to define a selection coefficient but got spiralled out by self doubt when reading the paper and getting tangled in the math of it all! It's all cleared up for me now (and I've managed to recreate fry's results, using dummy data in R I have managed to duplicate figure 3a and 3b) $\endgroup$ – rg255 Sep 19 '14 at 9:28

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