Problem 1. Part A.

The generations of each population in this case are discrete, so only persons within each age category are reproducing. Since time is not important, the Net Reproductive Rate equation:

R₀ = S (l_x) (m_x)
^x

can be considered a measure of fitness.

R_{0 POP A} = (0.7)(0) + (0.5)(1) + (0.3)(1) + (0.2)(1) (0.1)(1)

R_{0 POP A} = 1.1

R_{0 POP B} = (0.8)(0) + (0.7)(1) + (0.5)(0.6) + (0.4)(0.1) + (0.01)(0.1)

R_{0 POP B} = 1.041

Part B.

Euler’s equation can be rewritten to reflect the possibility of cross-cohort reproduction. Time is a factor in this case, but each generation retains its distinct mx value. Since e^-rx » rx when r has a small value,

1 = S (1 – rx) (l_x) (m_x)

Population A:

1 = [(1 – 0r)(0.70)(0)] + [(1 – 1r)(0.50)(1)] + [(1 – 2r)(0.30)(1)] +

[(1 – 3r)(0.20)(1)] + [(1 – 4r)(0.10)(1)]

1 = 0.5 – 0.5r + 0.3 – 0.6r + 0.2 – 0.6r + 0.1 – 0.4r

1 = 1.1 – 2.1r

solve for r, and R = e^r = e^0.0476

R_A = 1.048771047

Population B:

1 = [(1 – 0r)(0.80)(0)] + [(1 – 1r)(0.70)(1)] + [(1 – 2r)(0.50)(0.6)] +

[(1 – 3r)(0.40)(0.1)] + [(1 – 4r)(0.01)(0.1)]

1 = 0.7 – 0.7r + 0.3 – 0.6r + 0.04 – 0.12r + 0.001 – 0.004r

1 = 1.041 – 1.424r

solve for r, and R = e^r = e^0.02879

R_B = 1.029210635

The fitnesses of these phenotypes seem to be only slightly affected by the different demographic assumptions; those calculated under the assumption of overlapping generations are only a bit below those seen in the discrete generation scenario. The relatively low R values (net reproductive rate, which are also a measure of fitness) suggest that the populations are reproducing under a K-selection model. Individuals are able to survive for several generations (especially in Population B, where 40% survive into the fourth age cohort), but their fecundity rates are not exceptionally high. No offspring are produced in age cohort = 0 in either population, and after individuals reach viability (age cohort = 1) in Population B, their fecundity drops off. Population A does demonstrate a continued stable fecundity rate of 1.0 across all generations, but the frequency of those viable is less than that of Population B. In both cases, individuals seem to be reproducing to replace themselves, not to increase the overall size of the population.

Problem 2.

Summary of Allele Frequencies and Fitness Levels

In Environment 1, the heterozygote genotypes are favoured over both homozygous types, but neither allele is a deleterious recessive, so the environment is not experiencing stable polymorphism. The relative fitness level in Environment 1’s BB homozygote genotype is greater than the AA homozygote. According to Falconer and MacKay (19__, pg. 38), this relationship of homozygotes is important in the establishment of an intermediate polymorphic equilibrium. Environment 2 shows disruptive selection; the heterozygotes are at a disadvantage when compared to the AA and BB homozygote genotypes.

To calculate the relative fitness values for the next generation, and thereby establish the existence of a polymorphic equilibrium point, the initial frequency of the A allele, p0 must be established. For both environments, p can be calculated by counting the number of A alleles:

p₀ = 2(160) + 1(480) / 2(1000) = 0.4

The initial relative fitnesses for the three genotypes are:

Environment 1        Environment 2

w₁₁ = 0.598879        w₁₁ = 1.129176

w₁₂ = 1.0                 w₁₂ = 1.0

w₂₂ = 0.888057         w₂₂ = 1.134371

The average relative frequencies are:

Environment 1

w = p² (w₁₁) + 2pq (w₁₂) + q² (w₂₂)

w = (0.16)(0.598879) + 2(0.4)(0.6)(1) + (0.36)(0.89)

w = 0.8964

Environment 2

w = p² (w₁₁) + 2pq (w₁₂) + q² (w₂₂)

w = (0.16)(1.13) + 2(0.4)(0.6)(1) + (0.36)(1.13)

w = 1.0676

p’ value after 1 generation:

Environment 1

p’ = [p²(w₁₁) + pq(w₁₂)] / w

p’ = [(0.16)(0.6) + (0.24)(1)] / 0.08964

p’ = 0.3748

Environment 2

p’ = [p²(w₁₁) + pq(w₁₂)] / w

p’ = [(0.16)(1.13) + (0.24)(1)] / 1.0676

p’ = 0.3942

D p value:

Environment 1                      Environment 2

D p = p’ – p₀                              D p = p’ – p₀

D p = 0.3748 – 0.4                      D p = 0.3942 – 0.4

D p = -0.0252                           D p = -0.0058

w₁, the average relative fitness in the next generation:

Environment 1

w₁ = p’² (w₁₁) + 2p’q’ (w₁₂) + q’²(w₂₂)

w₁ = (0.1405) (0.6) + 2 (0.3748) (0.6252) (1) + (0.3909) (0.89)

w₁ = 0.9009

Environment 2

w₁ = p’² (w₁₁) + 2p’q’ (w₁₂) + q’²(w₂₂)

w₁ = (0.1405) (1.13) + 2 (0.5947) (0.4053) (1) + (0.1643) (1.13)

w₁ = 0.8265

Equilibrium value of AA:

Environment 1

p = (w₁₂ – w₂₂) / (2w₁₂ – w₁₁ – w₂₂)

p = (1.0 – 0.89) / [2(1.0) – 0.6 – 0.89]

p = 0.2157 q = 0.7843

Environment 2

p = (w₁₂ – w₂₂) / (2w₁₂ – w₁₁ – w₂₂)

p = (1.0 – 1.13) / [2(1.0) – 1.13 – 1.13]

p = 0.4845 q = 0.5155

Population 1 is headed toward a stable equilibrium point due to the relationshipsof the allele frequencies. The frequency of starting heterozygote individuals exceeds the values of their homozygote counterparts, and this domination continues into the adult generation of Population 1. No matter how close or disparate the relationship between homozygote genotype frequencies, they will always be exceeded by heterozygotes. Population 2, however, shows a rapid decrease in the number of heterozygous individuals between the starting and adult generations. The homozygote genotypes are very close in frequency, and dominate the heterozygotes. This pattern of disruptive selection leaves Population 2 without a stable polymorphic equilibrium point, and if it is headed anywhere, it is toward fixation.

Problem 3. The following table presents the allele frequencies, values for H_S, H_I, H_T, and F-statistics for the three ethnic/linguistic subpopulations of Sardinia. As some of you probably noticed, even the smallest deviations and rounding errors can lead to huge differences in the final answers. A few hints: p and q CANNOT be calculated simply by taking the square root of the p² or q² frequencies you found by dividing the observed number of individuals by the total per subpopulation and locus. Instead, you need to consider the p (or M alleles) that are in the heterozygotes, as well. For the H_S of each subpopulation, use the equation [1 – (M² + N²)] (obviously, for the MN blood group). This takes into account the frequency of the different alleles per subpopulation.

Across the three subpopulations, the values for the average F-statistics are:

F_ST = 0.004209374

F_IS = 0.055060078

F_IT = 0.058974396

These values indicate that marriage within language groups, rather than the Wahlund Effect, has contributed more to the increase in heterozygotes in these subpopulations.

If you need help or have questions about the NTSYS portion of the assignment, please email Ann and set up an appointment.