Question

You are studying a population of geese in which there are two color phases, brown and gray. Color in this species is controlled by a single gene, with brown dominant to gray. A random sample of 250 geese shows that 210 are brown. What percentage of the brown geese are heterozygous? (Assume that the population is in Hardy-Weinberg equilibrium.)

1) 36
2) 43
3) 48
4) 57
5) 84

Answer 1

The percentage of the brown geese that are heterozygous can be calculated using the Hardy-Weinberg principle. 48% of the brown geese in the sampled population are heterozygous for the color gene.

Step-by-step explanation:

To determine what percentage of the brown geese are heterozygous, we must use the Hardy-Weinberg equilibrium principle. Since color in this species is controlled by a single gene, with brown dominant to gray, we can label the dominant allele as 'B' and the recessive allele as 'b'. According to the Hardy-Weinberg principle, the allele frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. The frequencies of the genotypes can be represented as p^2, 2pq, and q^2, where p^2 is the frequency of the homozygous dominant genotype (BB), 2pq is the frequency of the heterozygous genotype (Bb), and q^2 is the frequency of the homozygous recessive genotype (bb).

In a sample of 250 geese, 40 are gray (recessive phenotype, so q^2 = 40/250 = 0.16). To find q, we take the square root of 0.16, which gives us q = 0.4. The frequency of the dominant allele (p) is then calculated as 1 - q, so p = 1 - 0.4 = 0.6. The frequency of the heterozygous individuals (2pq) equates to 2 * 0.6 * 0.4 = 0.48 or 48%. Therefore, 48% of the brown geese are expected to be heterozygous (Bb).

Hence, the answer is option 3) 48, which means that 48% of the brown geese sampled are heterozygous for the gene controlling the color phase.