If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?

The Hardy-Weinberg equilibrium principle states that in a large, randomly mating population with no factors such as mutation, migration, or natural selection affecting it, the frequencies of alleles and genotypes will remain constant from generation to generation. The formula for Hardy-Weinberg equilibrium is: p^2 + 2pq + q^2 = 1 Where p is the frequency of the dominant allele (A), q is the frequency of the recessive allele (a), p^2 represents the frequency of the homozygous dominant genotype (AA), 2pq represents the frequency of the heterozygous genotype (Aa), and q^2 represents the frequency of the homozygous recessive genotype (aa).

We are given that 4% of the population, or 0.04, expresses the recessive trait. This means that 0.04 is the frequency of the homozygous recessive genotype (aa) in the population. Therefore, q^2 = 0.04.

To find the frequency of the recessive allele (a), we take the square root of q^2: q = sqrt(q^2) = sqrt(0.04) = 0.2

Since p and q must sum to 1, we can find the frequency of the dominant allele (A) by subtracting the frequency of the recessive allele (q) from 1: p = 1 - q = 1 - 0.2 = 0.8

If two individuals do not express the recessive trait, they must have either the AA or Aa genotype. To find the probability that their offspring will express the recessive trait, we will perform a Punnett square analysis. The probabilities of the parental genotypes are: P(AA) = p^2 = (0.8)^2 = 0.64 P(Aa) = 2pq = 2(0.8)(0.2) = 0.32 P(aa) = q^2 = 0.04 Since the parents do not express the recessive trait, they must have either AA or Aa genotypes. Their combined frequency is 0.64 + 0.32 = 0.96. Now we will calculate the probability of their offspring being aa (expressing the trait) using the Punnett square: Aa x Aa --> 1/4 AA, 1/2 Aa, 1/4 aa As aa is the only genotype that expresses the trait, the offspring's probability is 1/4. However, we must account for the probability that both parents are carriers (Aa) based on the combined frequency of AA and Aa parents (0.96): Probability(Aa parent) = 0.32 / 0.96 = 1/3 Thus, the probability that both parents are carriers (Aa) and that their offspring will express the recessive trait (aa) is: (1/3) * (1/3) * (1/4) = 1/36 ≈ 0.0278 Therefore, the probability that the offspring of two non-trait-expressing individuals will express the recessive trait is approximately 2.78%.