a.  Because the yellow-black allele (a) is recessive, only homozygotes of this allele are yellow-black.  Assuming
         no selection, and therefore that the genotype frequencies are in Hardy-Weinberg proportions, the proportion
         of individuals that are yellow-black is the square of the gene frequency of the yellow-black allele, q² :

              q² = proportion yellow-black = 120/(360+120) = 0.25.

                Consequently, q = 0.5 .  The frequency, p, of the blue-black allele (A) is then p = 1 - q = 0.5 .

    b.  Again, assuming Hardy-Weinberg frequencies,

               p_AA = p² = (0.5 x 0.5) = 0.25

               p_Aa = 2pq = (2 x 0.5 x 0.5) = 0.5

               p_aa = q² = (0.5 x 0.5) = 0.25

    c.  The standard deviation of the magnitude of gene frequency change due to genetic drift is

          s = (pq/2N)^1/2 = (0.5 x 0.5/2x480)^1/2 = 0.016.

         Consequently, 95% of the time the change in gene frequency will be less than 2s, or 0.032.

    d.  You needed to assume that no selection was acting on this locus and that genotypes were in
         Hardy-Weinberg equilibrium.

Problem 2.

    a.   p_A  = (6400   +   0.5 x 3200)/(6400 + 3200 + 400) = 0.8

    b.  Fitnesses are given by W = ml.  Hence for the three genotypes, the fitnesses are

            Genotype                    W

              AA                     2 x 0.3 = 0.6
              Aa                     2 x 0.5 = 1.0
              aa                      2 x 0.2 = 0.4

    c.   Use the equation for gene frequency change to predict the gene frequencies in the next generation:

          p'  = p(pW_AA + qW_Aa)/(p²W_AA + 2pqW_Aa + q²W_aa)

                 = (0.8) [(0.8 x 0.6) + (0.2 x 1.0)]/[(0.8² x 0.6) + (2 x 0.8 x 0.2 x 1.0) + (0.2² x 0.4)]

                 = 0.755

    d.  The fitnesses indicate heterozygote superiority.  Consequently, the equilibrium gene frequency of allele A is given by

              p* = (W_Aa - W_aa)/ (W_Aa - W_AA + W_Aa - W_aa)

                     =  (1-.4)/(1 - .6 + 1 - .4)

                     =  0.6

Problem 3.

    a.  Results (i - iii) indicate that short-spined individuals are homozygous (say genotype aa), long-spined individuals
         are homozygous (genotype AA), and medium-spined individuals are heterozygous (genotype Aa).  One may
         therefore set up the following table with the data provided:

                                                                            genotype
                                                              AA                Aa                aa
            No. adults (time t)                    2000            1000               500
            gametes/individual                 25,000          40,000          20,000
            No. adults (time t+1)                4900            2100               450

         The value of m for each genotype is given (gametes/individual).  To find l for each individual, we first need to know
         how many zygotes are formed of each genotype.  To do this, we find the gene frequency of the A allele among
         the zygotes, which in turn is equal to the gene frequency among the gametes.  To find the gamete gene frequency,
         we first find the total number of gametes produced by each genotype:

            Genotype                Total Number of Gametes Produced (= No. adults x gametes/individual)

          AA                            2000 x 25,000 = 50 x 10⁶
          Aa                            1000 x 40,000 = 40 x 10⁶
          aa                              500  x 20,000 = 10 x 10⁶

         The gene frequency of A in the gamete pool is thus

          p^g = (No. gametes with allele A)/(Total No. gametes)

               =  (50x10⁶  + (1/2)x40x10⁶ )/(50x10⁶ + 40x10⁶ + 10x10⁶ )

               = 0.7

         Since random mating does not change gene frequencies, the gene frequencies among the zygotes must be

          p = 0.7   and q = 1 - p = 0.3

         We also know that zygotes will exhibit Hardy-Weinberg freuqencies, so that the genotype frequencies in the
         zygotes are given by

            Genotype   Frequency
              AA        (0.7)² = 0.49
              Aa         2 x 0.7 x 0.3 = 0.42
              aa         (0.3)² = 0.09

         But the total number of zygotes is simply 1/2 the total number of gametes, or 1/2  x 100 x 10⁶ , or 50 x 10⁶ .
         Thus, the number of zygotes of each type is simply the total number of zygotes times the proportion that are
         of that type, or

             Genotype   No. zygotes
              AA          0.49 x 50 x 10⁶ = 24.5 x 10⁶
              Aa          0.42  x 50 x 10⁶ = 21 x 10⁶
              aa           0.09 x 50 x 10⁶  = 4.5 x 10⁶

         Finally, l for a given genotype is simply (No. surviving adults)/(No. zygotes), so we have

             Genotype             l
              AA         4900/24.5 x 10⁶ = 0.0002
              Aa          2100/21 x 10⁶    = 0.0001
              aa           450/4.5 x 10⁶    = 0.0001

         Now that estimates of m and l are available for each genotype, fitness can be calculated as W = ml :

             Genotype              Fitness
              AA          0.0002 x 25,000 = 5
              Aa          0.0001 x 40,000 = 4
              aa           0.0001 x 20,000 = 2

    c.  Gene frequencies among the zygotes were calculate in part b.

    d.  Since W_AA > W_Aa > W_aa , natural selection will eliminate the a allele and the equilibrium gene frequency
         will be p = 1.
 

Problem 4.

    a.  Since all individuals are Bb initially, one half of the alleles are B and one half are b.  Hence,
       p_B = p_b = 0.5.

    b.  First calculate the fitnesses of each genotype:

             Genotype              Fitness
              BB                    0 x 0 = 0
              Bb                100 x 0.01 = 1
              bb                100 x 0.01 = 1

         From these values it is evident that allele B is a recessive lethal.  For a recessive lethal, the equation for
         gene frequency change reduces to

              p' = p/(1+p).

        After 1 generation, the gene frequency is p' = 0.5/(1+0.5) = 0.33.

        Iterating this 4 more times gives the gene frequency after 5 generations as p' = 0.143.
        Iterating an additional 5 times gives the gene frequency after 10 generations as p' = 0.083.

        Note:  You can easily verify that after n generations, the gene frequency will be p' = p/(1+ np).

    c.  Either by trial and error, or by using the formula p' = p/(1+ np), one can calculate that 8 generations
         are needed to reduce the gene frequency from an initial value of 0.5 to 0.1 or less.

    d.  Since W_BB < W_Bb = W_bb , natural selection will eliminate the B allele and the equilibrium gene frequency
         will be p_b = 1.
 
 

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