PROBLEM SET
Chapter 20 -- Population Genetics: Processes...

P. 567, #1. Consider a locus with alleles A and a in a large, randomly mating population under the influence of mutation. a. If the mutation rate of A to a is 6 x 10^-5 and the back-mutation rate to A is 7 x 10^-7, what is the equilibrium frequency of a?
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
b. If q = 0.9 in generation n, what would it be one generation later, only under the influence of mutation? In this case we need to use the equation for q in the next generation that was derived on p. 552,
             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
P. 567, #2. The following data refer to the Ro allele in the Rh blood system:
frequency in western Europe = 0.62
frequency in eastern Europe = 0.45
frequency in Mongols = 0.03
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.
qc = (1 - m)qn + m(qm)
where qc = the allele frequency in the conglomerate population
qn = the allele frequency in the native population
qm = the allele frequency in the migrant population
m = the proportion of migrants.
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
m = (qc - qn)/(qm - qn)
m = (0.45 - 0.62)/(0.03 - 0.62) = -0.17/-0.59
m = 0.29

P. 567, #7. Derive a model of selection in which the fitness of the heterozygote is half that of one of the homozygotes and twice the fitness of the other. Give expressions for the following:
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)
a. Mean population fitness
The mean population fitness (wbar) is the sum of the frequencies after selection.
wbar = p^2(1) + 2pq(.5) + q^2(.25)
b. Equilibrium allelic frequency (stable?)
The equilibrium frequency of a (qhat) is 0.0. In this model of selection, the a allele is being removed from the population and selection will proceed until it is completely removed. The frequency is stable because if the a allele is added back to the population through mutation or migration, selection will act to remove it.

P. 567, #10. Given a locus with alleles A and a in a sexually reproducing, diploid population in Hardy-Weinberg equilibrium, set up a model and the initial formula for the frequency of the dominant allele after one generation (pn+1) if selection acts against the dominant phenotype.
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)
P. 568, #12. If a locus has alleles A1 and A2, what will be the equilibrium frequency of A1 if both homozygotes are lethal?
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 A1 (qhat) is given by the following formula.
             s2       1      1
   qhat = ------- = ----- = --- = 0.5
          s1 + s2   1 + 1    2
P. 568, #13. The following data were collected from a population of Drosophila segregating sepia (s) and wild-type (s+) eye colors. A sample was taken when the eggs were deposited and later among adults. Reconstruct the model of selection.
               s+/s+          s+/s           s/s
   Egg          25             50             25
   Adult        30             60             10
Notice that the ratio of eggs to adults is the same for the s+/s+ genotype and the s+/s genotype (egg/adult = 30/25 = 60/50 = 1.2). However, the ratio of eggs to adults is much smaller for the s/s genotype (10/25 = 0.4), which indicates that fewer eggs of the s/s genotype survived and thus there must be selection against the s/s genotype.
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.

P. 568, #15. In a population of nine hundred butterflies, the frequency (p) of the fast allele of the enzyme phosphoenol pyruvate is 0.6, and the frequency of the slow form (q) is 0.4. Ninety butterflies migrate to this population and, among the migrants, the frequency of the slow allele is 0.8. Calculate the allelic frequencies of the new population.
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.
qc = (1 - m)qn + m(qm)
where qc = the allele frequency in the conglomerate (or new) population
qn = the allele frequency in the native population (0.40)
qm = the allele frequency in the migrant population (0.80)
m = the proportion of migrants (0.0909).
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.
qc = (1 - m)qn + m(qm)
qc = (1 - 0.0909) * 0.4 + 0.0909 * 0.8 = 0.36364 + 0.7272
qc = 0.43636

P. 568, #16. In a particular population with two alleles at a locus, the frequency of AA individuals = 0.25, Aa = 0.5, and aa = 0.25. If the AA genotype has a fitness = 1, Aa = 0.8, and aa = 0.6, what will the frequencies of A and a be in the next generation? Assume mutations are nonexistent.
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
freq(a) = qn+1 = freq(aa) + 0.5(freq(Aa))
qn+1 = 0.1875 + 0.2500 = 0.4375
pn+1 = 1.0 - qn+1 = 1 - 0.4375 = 0.5625

P. 568, #17. If the frequency of the N allele in a native population is 0.25, 0.32 in a conglomerate population, and 0.4 in a migrant population, what percent of the N alleles in the conglomerate population were derived from the migrant population?
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.
qc = (1 - m)qn + m(qm)
where qc = the allele frequency in the conglomerate population
qn = the allele frequency in the native population
qm = the allele frequency in the migrant population
m = the proportion of migrants.
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
m = (qc - qn)/(qm - qn)
m = (0.32 - 0.25)/(0.40 - 0.25) = 0.07/0.15
m = 0.467

P. 568, #18. Consider a population in which p = 0.9 and q = 0.1. If the forward mutation rate, A --> a is 5 x 10^-5, and the reverse mutation rate a --> A, is 2 x 10^-5, calculate the equilibrium frequency q, of the a allele. This problem is just like problem 1a above.
                           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
P. 568, #19. If the forward mutation rate, A --> a, is 5 times the reverse mutation rate, what will be the equilibrium frequency of the a allele be?
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
P. 568, #20. Calculate the frequency of the recessive b allele in a population one generation after selection if in the original population q = f(b) = 0.7 and the relative fitness of bb homozygotes is 0.4.
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
freq(b) = qn+1 = freq(bb) + 0.5(freq(Bb))
qn+1 = 0.2276 + 0.2975 = 0.5251

Last updated on 22 August 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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