Question

table 43.2 relationship between allele and genotype frequencies under the hardy - weinberg equilibrium
alleles a a
allele frequencies p q
genotypes aa aa aa
genotype frequencies p² 2pq q²
part 2 - questions: after completing the reading, answer the following questions. any data, calculations, or models you use to support your thinking should be explained.

1. how does the multiplication rule of probability relate to calculating the frequencies of homozygous genotypes in a population?
2. in a population in hardy - weinberg equilibrium, the frequency of homozygous recessive individuals is 0.3. what is the frequency of the dominant and recessive alleles in the population?

Explanation:

1.

The multiplication rule of probability states that the probability of two independent events occurring together is the product of their individual probabilities. In the context of Hardy - Weinberg equilibrium, for a population to have a homozygous genotype (AA or aa), two identical alleles must come together. If the frequency of allele A is \(p\) and the frequency of allele a is \(q\), the probability of getting two A alleles (AA genotype) is \(p\times p=p^{2}\) and the probability of getting two a alleles (aa genotype) is \(q\times q = q^{2}\), which is based on the multiplication rule of probability.

Step1: Identify the formula for homozygous recessive genotype frequency

The frequency of homozygous recessive individuals (\(aa\)) in Hardy - Weinberg equilibrium is \(q^{2}\). Given \(q^{2}=0.2\).

Step2: Calculate the frequency of the recessive allele \(q\)

Take the square - root of \(q^{2}\). So \(q=\sqrt{q^{2}}=\sqrt{0.2}\approx0.447\).

Step3: Calculate the frequency of the dominant allele \(p\)

Since \(p + q=1\) in Hardy - Weinberg equilibrium, then \(p = 1 - q\). Substitute \(q\approx0.447\) into the equation, so \(p=1 - 0.447 = 0.553\).

Answer:

The multiplication rule of probability is used to calculate the frequencies of homozygous genotypes in a population. For a homozygous dominant genotype (AA), if the frequency of the dominant allele \(A\) is \(p\), the frequency of AA is \(p\times p = p^{2}\). For a homozygous recessive genotype (aa), if the frequency of the recessive allele \(a\) is \(q\), the frequency of aa is \(q\times q=q^{2}\), as getting two identical alleles is an independent - like event in a large, randomly - mating population under Hardy - Weinberg equilibrium.

2.