1. frequency of A(initial) = 0.20 = p

frequency of a(initial) = 1 – 0.20 = 0.80 = q

Since the probability of allele fixation is equal to the initial allele frequency in the population, the expected fixaqtion frequencies are:

p(fix) = 0.20

q(fix) = 0.80

There is a total of 100 populations, so the expected numbers of populations fixed for the two alleles are:

A(expected) = p(100)

= (0.20) (100)

= 20

a(expected) = q(100)

= (0.80)(100)

= 80

The observed numbers of populations fixed for these alleles are:

A(observed) = 32 a(observed) = 68

The chi-square test is used to see if the expected deviates from the observed numbers of populations:

X² = (32 – 20)² + (68 – 80)²
                    20         80

= 7.2 + 1.8

= 9.0 degrees of freedom = 2 – 1 = 1

We can reject the hypothesis that genetic drift is the only force acting upon these populations, as the c ² value at the 0.05 level of significance with 1 degree of freedom is 3.841. Our calculated value of X² = 9.0 exceeds this critical value, and can therefore be rejected.

This problem could also be constructed to calculate the number of individuals in the populations that were fixed for the A and a alleles. Since there are 10 people in each population, we would expect (0.20)(1000) = 200 individuals to become fixed for the A allele. The expected number of individuals for fixation at the a allele would be 1000 – 200, or 800 individuals. Likewise, we would multiply the observed number of fixed populations by 10 to obtain the number of individuals fixed for each allele.

X² = (320 – 200)² + (680 – 800)²
            200            800

= 7.2 + 1.8 = 9.0

2. In the villages, N(females) = 50, and N(males) = 200. Since each woman can only have 3 husbands (darn!), the effective breeding population of MALES is limited to 150 individuals. If Ne equals the total effective breeding population,

Ne = 4[N(females)][N(males)]
      N(females) + N(males)

= 4 (50) (150)
      50 + 150

3. In the haplodiploid system of bees, each parent donates one allele to their daughters, but the mother gives half of her genetic information (either A1 or A2) to each of her daughters, while the father passes on 100% of his genetic information to each of his daughters. The kinship diagram for the relationship can be drawn as follows, with Z representing a hypothetical mating between mother and daughter to created a closed loop on which the kinship coefficient can be calculated:

The kinship coefficient between mothers and daughters is (1/2)(1/2)(1) = ¼ or 0.25. Between fathers and daughters, the kinship coefficient equals 1.0.

Brothers and sisters each received one allele from their mother (either A1 or A2), and they have a 50% chance that both received the same allele. Additionally, sons receive NO genetic information from their father, so the kinship diagram can be constructed as follows, where Z now represents a hypothetical mating of son and daughter:

The kinship coefficient between sons and daughters is (1/2)(1/2)(1) = ¼ or 0.25.

Finally, the kinship coefficient of two sisters can be calculated by a hypothetical mating of two daughters. The kinship diagram becomes more complicated for this mating, since there are two paths (through both the mother and father). Remember that the mother passes ½ of her genetic information onto each daughter, who in  turn each pass ½ of their genetic information onto Z. Again, Z represents the hypothetical offspring of the mating of the two daughters:

The loop with the mother as the common ancestor can be calculated as (1/2)(1/2)(1/2) = 1/8. This is then added to the loop with the father, which is (1/2)(1/2)(1) = ¼. Thus, the total kinship coefficient for sisters is 3/8, or 0.375.

Clearly, sisters have more genetic material in common (0.375) than a mother shares with her daughter (0.25). Therefore, a female bee in this haplodiploid society would promote the survival of her sisters over that of her daughters, as sisters leave more of her genetic material in the population.

4. The villages’ frequencies for the T allele at time – initial (0) are:

A = 0.4 B = 0.2 C = 0.1 D = 0.9

With a 10% migration rate to IMMEDIATELY ADJACENT villages, the migration matrix is:

If p represents the frequency of allele T, let q = the frequency of allele t. The allele frequency matrix for these alleles at the initial generation is:

Multiply the above two matrices to obtain the frequencies of the T and t alleles (p and q, respectively) in Generation 1: (if you need to see the matrix multiplication for this or any of the following steps, please email Ann)

The Generation 1 matrix can then be multiplied by the original migration matrix to obtain the allele frequencies in Generation 2:

The Generation 2 matrix can, in turn, be multiplied by the original migration matrix to obtain the allele frequencies in Generation 3:

5. The initial allele frequencies for the army and village populations are:

Using the equation 1 – M = (p_t – p_bar) / (p₀ – p_bar)

1 – M = (0.29 – 0.01) / (0.40 – 0.01)

1 – M = 0.28 / 0.39

so, the initial frequency of non-migrants (1 – M) = 0.717948717

migrants (M) = 0.282051282

Now, we can add time depth to the problem:

a) after 5 generation, the estimated migration rate into the villages is:

m₅ = 1 – exp{ln(0.717948717)/5}

m₅ = 0.064123192

b) after 10 generations, the estimated migration rate into the villages is:

m₁₀ = 1 – exp{ln(0.717948717)/10}

m₁₀ = 0.032592739

6. Part A:

Total number of individuals = 10000

Observed # SS individuals = 2875

Observed # SF individuals = 4250

Observed # FF individuals = 2875

Expected SS frequency = p² Expected # SS individuals = 10000(p²)

Expected SF frequency = 2pq Expected # SF individuals = 10000(2pq)

Expected FF frequency = q² Expected # FF individuals = 10000(q²)

Expected genotypic frequency of S = 2(2875 + (4250) = 0.5
                                                                    2(10000)

Expected genotypic frequency of F = 1 - p = 0.5

Since the expected genotypic frequencies of S and F include those alleles in both homozygotes and heterozygotes, we need to calculate the expected number of phenotypic individuals in the population.

Expected # SS individuals = 10000 (0.5)² = 2500

Expected # SF individuals = 10000(2)(0.5)(0.5) = 5000

Expected # FF individuals = 10000 (0.5)² = 2500

Compare the expected and observed numbers of individuals using the Chi-square test:

X² = (2875 – 2500)² + (4250 – 5000)² + (2875 - 2500) ²
                  2500                 5000                  2500

X² = 56.25 + 112.50 + 56.25

X² = 225 degrees of freedom = 3 - 1 - 1 = 1

c ² at a 0.05 level of significance is 3.841. Since X² = 225 is much larger than the critical value for the significance level of 0.05, we can reject the null hypothesis that this population is in Hardy-Weinberg equilibrium.

Part B:

Flare sub-population total # flarers = 3750

Observed # SS flarers = 234

Observed # SF flarers = 1406

Observed # FF flarers = 2110

Expected SS frequency = p² Expected # SS individuals = 3750(p²)

Expected SF frequency = 2pq Expected # SF individuals = 3750(2pq)

Expected FF frequency = q² Expected # FF individuals = 3750(q²)

Expected genotypic frequency of S = 2(234 + 1406) = 0.249866666
                                                                        2(1406)

Expected genotypic frequency of F = 1 - p = 0.750133333

Expected # SS flarers = (0.249866666)²(3750) = 234.1250654

Expected # SF flarers = (0.249866666) (0.750133333) (3750) = 1405.749863

Expected # FF flarers = (0.750133333)²(3750) = 2110.125065

X² = (234 – 234.1250654)² + (1406 – 1405.749863)² + (2110 - 2110.125065)²
                    234.1250654                1405.749863                2110.125065

X² = (6.680768781 x 10^-5) + (4.450899866 x 10^-5) + (7.412477338 x 10^-6)

X² = 1.187291638 x 10^-4

Non-flare sub-population total # flarers = 6250

Observed # SS flarers = 2641

Observed # SF flarers = 2844

Observed # FF flarers = 765

Expected SS frequency = p² Expected # SS individuals = 6250(p²)

Expected SF frequency = 2pq Expected # SF individuals = 6250(2pq)

Expected FF frequency = q² Expected # FF individuals = 6250(q²)

Expected genotypic frequency of S = 2(2641 + 2844) = 0.65008
                                                                            2(6250)

Expected genotypic frequency of F = 1 - p = 0.34992

Expected # SS non-flarers = (0.65008)²(6250) = 2641.27504

Expected # SF non-flarers = (0.65008) (0.34992) (6250) = 2843.44992

Expected # FF non-flarers = (0.34995)²(6250) = 765.27504

X² = (2641 - 2641.27504)² + (2844 – 2843.44992)² + (765 - 765.27504)²
                  2641.27504              2843.44992              765.27504

X² = (2.864033486 x 10^-5) + (1.06415803 x 10^-4) + (9.884943013 x 10^-5)

X² = 2.339055679 x 10^-4

Both flaring and non-flaring sub-opulations have 3 - 1 - 1 = 1 degree of freedom. At the 0.05 significance level, c ² has a critical value of 3.841 for 1 degree of freedom. The two sub-populations have X² vales less than this critical value, so their genotypic frequencies are in Hardy-Weinberg equilibrium. The decrease in homozygotes and increase in heterozygotes in both sub-populations suggest that the flarers and non-flarers had not been previously breeding. The pooling of the sub-populations resulted in the Wahlund effect for the villages.

7.

Part A. Assume random mating of two homozyous individuals with genotypes SN/SN and Fn/Fn. In this parental generation (G0), we have no heterozygous individuals, and the frequencies of their gametes are:

P₁₁ = SN = 0.5

P₁₂ = Sn = 0.0 (remember, no hets in the parental generation!)

P₂₁ = FN = 0.0

P₂₂ = Fn = 0.5

We can therefore figure out the disequilibrium value for the parental generation (G0):

D = P₁₁P₂₂ - P₁₂P₂₁

D = (0.5)(0.5) - (0)(0)

D = 0.25

The genotypes in the next generation (G1) under random mating are:

1/4 SN/SN 1/2 SN/Fn 1/4 Fn/Fn

The frequencies of the gametes in G1 can be calculated:

P₁₁’ = SN = P₁₁ - rD = 0.4375

P₁₂’ = Sn = P₁₂ - rD = 0.0625

P₂₁’ = FN = P₂₁ - rD = 0.0625

P₂₂’ = Fn = P₂₂ - rD = 0.4375

These gametic frequencies allow us to calculate the disequilibrium value for G1:

D’ = P₁₁’ P₂₂’ - P₁₂’ P₂₁’

D’ = (0.4375)² - (0.0625)²

D’ = 0.1875 [alternatively, you could use the equation: D’ = (1 - r) D]

The genotypes in the next generation (G2) under random mating would be:

1/4 SS 1/2 SF 1/4 FF

1/4 NN 1/2Nn 1/4 nn

The frequencies of the gametes in G2 can be calculated:

P₁₁’’ = SN = P₁₁’ - rD = 0.3906

P₁₂’’ = Sn = P₁₂’ - rD = 0.1094

P₂₁’’ = FN = P₂₁’ - rD = 0.1094

P₂₂’’ = Fn = P₂₂’ - rD = 0.3906

These gametic frequencies allow us to calculate the disequilibrium value for G1:

D’’= P₁₁’’P₂₂’’- P₁₂’’P₂₁’’

D’ = (0.3906)² - (0.1094)²

D’ = 0.1406

Alternatively, you could use the equation: D^t = (1 - r)^t D to get an answer of D² = 0.1406 in the second generation.

Part B. Assuming 100% positive assortative mating in the parental (G0) generation, we can only have mating between SN/SN and SN/SN individuals, or between Fn/Fn and Fn/Fn individuals. Again, each of these genotypes starts with a frequency of 0.5. We can figure out the disequilibrium value for either; it will be the same.

SN/SN x SN/SN                Fn/Fn x Fn/Fn

D = (0.5)(0.5) - (0)(0)     D = (0.5)(0.5) - (0)(0)

D = 0.25                                  D = 0.25

Even though a Punnett square will give you heterozygous individuals, you need to force these to have a frequency of 0.0 in the next generation (G1); only homozygotes will be in the G1 population:

The disequilibrium value for the G1 generation is:

D’ = (0.5)(0.5) - (0)(0)

D’ = 0.25

Again in the G2 generation, you will only have homozygotes, each with a frequency of 0.5. Therefore, the disequilibrium value for G2 will be the same as it was for G0 and G1:

D’’ = (0.5)(0.5) - (0)(0)

D’’ = 0.25

Under a model of 100% positive assortative mating, linkage disequilibrium does not break down. The population is not in Hardy-Weinberg equilibrium, because the population consists of only homozygous individuals, no hets.

8. Assuming a model of 100% negative assortative mating, we can only breed SN/SN and Fn/Fn homozygous individuals. Each has a frequency of 0.5 in the parental (G0) generation, so, like the previous two problems, the disequilibrium value in G0 is:

D = P₁₁ P₂₂ - P₁₂ P₂₁

D = (0.5)(0.5) - (0)(0)

D = 0.25

Because this is a NAM situation, you need to calculate the gametes of G1 using slightly tweaked equations:

P₁₁’ = SN = 1/2(1 - r) = 0.375

P₁₂’ = Sn = r/2 = 0.125

P₂₁’ = FN = r/2 = 0.125

P₂₂’ = Fn = 1/2(1 - r) = 0.375

Now, use these gamete frequencies in the disequilibrium equation to get the disequilibrium value for G1:

D’ = P₁₁’ P₂₂’ - P₁₂’ P₂₁’

D’ = (0.375) (0.375) - (0.125) (0.125)

D’ = 0.125

And for the second generation (G2):

D’’ = (1 - r) D’

D’’ = (1 - 0.25) (0.125)

D’’ = 0.09375

So, NAM is the most effective of the three methods for breaking down linkage disequilibrium, followed by random mating.