Chapter 20 -- Population Genetics: Processes...

In this case, we will use the formula for the equilibrium value of q that was derived on page 541,

u q at equilibrium = ------- u + v where u equals the forward mutation rate and v equals the back-mutation rate. Thus, 6 x 10^-5 q at equilibrium = ----------------------- = 6 x 10^-5 + 7 x 10^-7 .00006 .00006 q at equilibrium = ------------------- = --------- .00006 + .0000007 .0000607 q at equilibrium = 0.988

q(n+1) = q(n) + u*p(n) - v*q(n) where q(n+1) is q in the next generation and q(n) is q in the current generation p(n) is p in the current generation u is the forward mutation rate v is the back-mutation rate. q(n+1) = 0.9 + (0.1 * 0.00006) - (.9 * .0000007) q(n+1) = 0.9 + 0.000006 - 0.00000063 q(n+1) = 0.90000537

What is the total proportion of alleles that have entered the eastern European population?

This is a straight forward example of the continent-island model of migration, in which the western European population is the island, the eastern European population is the conglomerate population (presumably, it was identical to the western European population before migration), and the Mongols represent the migrant or continent population. The relationship between these three populations and migration is defined by this equation.

where qc = the allele frequency in the conglomerate population

In this case the three allele frequencies are given and we need to find m (the rate of migration). To do this, we rearrange the equation to yield

The most fit genotype will have a relative fitness of 1.0 and this will be one of the homozygotes. The heterozygote will then have a fitness of 0.5 and the other homozygote will have a fitness that is half of the heterozygote and thus 0.25. We can then make the following chart.

genotypes AA Aa aa relative fitness 1.00 0.50 0.25 freq after selection p^2(1) 2pq(.5) q^2(.25)

The mean population fitness (wbar) is the sum of the frequencies after selection.

The equilibrium frequency of

We can then make the following chart.

genotypes AA Aa aa relative fitness 1-s 1 1 freq after selection p^2(1-s) 2pq(1) q^2(1) wbar = p^2(1-s) + 2pq + q^2 = 1-s*p^2 frequency p^2(1-s)/wbar 2pq/wbar q^2/wbar pn+1 = freq(AA) + 0.5*freq(Aa) = (p^2(1-s) + pq)/wbar pn+1 = (p^2(1-s) + pq)/(1-s*p^2)

The selection model can be represented by the following chart.

genotypes A1A1 A1A2 A2A2 relative fitness 1-s1 1 1-s2 where s1 = 1 and s2 = 1 relative fitness 0 1 0 freq after selection p^2(0) 2pq(1) q^2(0)The equilibrium frequency of

s2 1 1 qhat = ------- = ----- = --- = 0.5 s1 + s2 1 + 1 2

s+/s+ s+/s s/s Egg 25 50 25 Adult 30 60 10

The model of selection is thus selection against one of the homozygotes, the s/s genotype. In this model of selection, the frequency of the "s" allele will decrease in the population until it reaches zero (qhat = 0.0).

To actually estimate the selection coefficient, we need to determine the relative fitness. The first step is to scale 1.2 to 1 as this is the most fit genotype. To do this we divide the fitness of each genotype by 1.2 (the fitness of the most fit genotype), which converts all fitness values to relative fitness, with the most fit genotype being 1. Next divide 0.4 by 1.2, which yields the relative fitness of the s/s genotype, which is 0.33.

This is a straight forward example of the continent-island model of migration, in which the 900 butterflies represent the island, the 90 butterflies represent the migrant or continent population, and the proportion of migrants is 90/(90 + 900) = 0.0909. The relationship between these two populations and migration is defined by this equation.

where qc = the allele frequency in the conglomerate (or new) population

In this case we are given sufficient information to calculate the allele frequency of the slow allele in the conglomerate population. We need only to substitute the appropriate values in the equation above and solve for qc.

We can then make the following chart.

genotypes AA Aa aa freq before selection 0.25 0.50 0.25 relative fitness 1.00 0.80 0.60 freq after selection 0.25 * 1 0.5 * 0.8 0.25 * 0.6 wbar = 0.25 + 0.40 + 0.15 = 0.80 frequency 0.25/0.8 0.4/0.8 0.15/0.8

This is a straight forward example of the continent-island model of migration, in which the native population had an allele frequency of 0.25, the conglomerate population has an allele frequency of 0.32, and the migrant (or continent) population has an allele frequency of 0.40. The relationship between these three populations and migration is defined by this equation.

where qc = the allele frequency in the conglomerate population

In this case the three allele frequencies are given and we need to find m (the rate of migration). To do this, we rearrange the equation to yield

u q at equilibrium = ------- u + v where u equals the forward mutation rate and v equals the back-mutation rate. Thus, 5 x 10^-5 q at equilibrium = ----------------------- = 5 x 10^-5 + 2 x 10^-5 .00005 .00005 q at equilibrium = ------------------- = --------- .00005 + .00002 .00007 q at equilibrium = 0.714

This problem is just like problem 1a above except we have the forward mutation rate defined in terms of the back-mutation rate (u = 5*v).

u q at equilibrium = ------- u + v where u equals the forward mutation rate and v equals the back-mutation rate. Thus, 5v q at equilibrium = -------- = 5v + v 5v 5 q at equilibrium = ---- = --- = 0.833 6v 6

We can then make the following chart, where the genotype frequencies before selection are determined from Hardy-Weinberg proportions.

genotypes BB Bb bb freq before selection 0.09 0.42 0.49 relative fitness 1.00 1.00 0.40 freq after selection 0.09 * 1 0.42 * 1 0.49 * 0.4 wbar = 0.09 + 0.42 + 0.196 = 0.706 frequency .09/.706 .42/.706 .196/.706

Last updated on 22 August 1996.