Question

In four o'clock plants, there are three phenotypes for flower color but only two alleles. In one population, q2 = 0.169, and the population is assumed to be in Hardy-Weinberg equilibrium. The table below shows the number of plants of each phenotype that are expected in a population of 1,000 plants in the next generation. In the next generation, what p value is expected?

A. 0.105
B. 0.411
C. 0.516
D. 0.589

Answer 1

In a population of 500 plants, with a dominant violet (V) and recessive white (v) flower color alleles, if p = 0.8 and q = 0.2, we can use the Hardy-Weinberg equation to calculate the expected number of individuals with each genotype and the expected number of plants with each flower color.

Step-by-step explanation:

In a population of 500 plants with dominant violet (V) and recessive white (v) flower color alleles, if p = 0.8 and q = 0.2, we can calculate the expected number of individuals with each genotype. The genotype frequencies can be calculated using the Hardy-Weinberg equation: p² + 2pq + q² = 1.

For homozygous dominant (VV): p² = (0.8)² = 0.64, so the expected number would be 0.64 x 500 = 320 plants.

For heterozygous (Vv): 2pq = 2 x 0.8 x 0.2 = 0.32, so the expected number would be 0.32 x 500 = 160 plants.

For homozygous recessive (vv): q² = (0.2)² = 0.04, so the expected number would be 0.04 x 500 = 20 plants.

Therefore, the expected number of plants with violet flowers (homozygous dominant and heterozygous) would be 320 + 160 = 480 plants, and the expected number of plants with white flowers (homozygous recessive) would be 20 plants.