The table below shows genotype frequencies in each of four d

The table below shows genotype frequencies in each of four diploid populations. Which appear to be in Hardy-Weinberg equilibrium? Which have more heterozygotes than predicted by HW equilibrium? Which have fewer heterozygotes? Show your work. Two geneticists (we\'ll call them \"Brad\" and \"Angelina\") attempt to estimate the frequency of a recessive allele in a population of beach slugs. Brad observes that 1 percentage of the population has the recessive phenotype: he therefore estimates the allele frequency as q = Squareroot (0.01) = 0.10. Angelina, using a molecular test, determines that 71 percentage of the population is homozygous for the dominant allele, and 28 percentage is heterozygous. Using Angelina\'s data, what would be the best estimate of the frequency of the recessive allele? Why do you think this differs from Brad\'s estimate? With three alleles, there will be six possible genotypes in a diploid organism (see below). If the frequencies of A_1, A_2, and A_3 are p, q, and r, respectively, what will the genotype frequencies be at Hardy-Weinberg equilibrium? Fill in below and briefly explain your logic. The first one is already done for you)

Solution

1).

(i). The frequency of A1 = frequency of A1A1 + ½ (frequency of A1A2)

A1 = 0.64 + ½ (0.32) = 0.8

The frequency of A2 = frequency of A2A2 + ½ (frequency of A1A2)

A2 = 0.04 + ½ (0.32) = 0.2

Expected frequency of A1A1 = 0.64

Expected frequency of A2A2 = 0.04

A1+A2 = 0.8+0.2 = 1. Thus, the population is in H-W equilibrium.

(ii). The frequency of A1 = frequency of A1A1 + ½ (frequency of A1A2)

A1 = 0.60 + ½ (0.40) = 0.8

The frequency of A2 = frequency of A2A2 + ½ (frequency of A1A2)

A2 = 0.00 + ½ (0.40) = 0.2

Expected frequency of A1A1 = 0.64

Expected frequency of heterozygotes is = 2*0.8*0.2 = 0.32

A1+A2 = 0.8+0.2 = 1. Thus, the population is in H-W equilibrium.

But the population has more number of heterozygotes than expected by the H-W equilibrium.

(iii). The frequency of A1 = frequency of A1A1 + ½ (frequency of A1A2)

A1 = 0.16 + ½ (0.48) = 0.4

The frequency of A2 = frequency of A2A2 + ½ (frequency of A1A2)

A2 = 0.36 + ½ (0.48) = 0.6

Expected frequency of A1A1 = 0.16

Expected frequency of A2A2 = 0.36

A1+A2 = 0.4+0.6 = 1. Thus, the population is in H-W equilibrium.

(iv). The frequency of A1 = frequency of A1A1 + ½ (frequency of A1A2)

A1 = 0.34 + ½ (0.12) = 0.4

The frequency of A2 = frequency of A2A2 + ½ (frequency of A1A2)

A2 = 0.54 + ½ (0.32) = 0.6

Expected frequency of heterozygotes is = 2*0.4*0.6 = 0.48

A1+A2 = 0.4+0.6 = 1. Thus, the population is in H-W equilibrium.

But the population has less number of heterozygotes than expected by the H-W equilibrium.