In Mendel's crosses with peas, seed shape is due to a single allele, and the allele for smooth seeds is dominant to the allele for wrinkled.If 20 percent of a population has the smooth allele, _______ percent of the plants should have wrinkled seeds at Hardy-Weinberg equilibrium.
A) 4
B) 20
C) 36
D) 80
E) 64

Question

Quizplus · Accepted Answer

At Hardy-Weinberg equilibrium, the frequency of the dominant allele (for smooth seeds) is p = 0.8 (since 20% of the population has the recessive allele, q = 0.2). 
Using the Hardy-Weinberg equation, we can calculate the frequency of the homozygous recessive genotype (wrinkled seeds): 
q^2 = (0.2)^2 = 0.04 
Converting this frequency to a percentage, we get 4%. 
Therefore, 100% - 4% = 96% of the plants have at least one copy of the dominant allele. 
To calculate the frequency of the heterozygous genotype (i.e. plants that are carriers of the recessive allele), we can use the formula: 
2pq = 2(0.8)(0.2) = 0.32 
Converting this frequency to a percentage, we get 32%. 
Therefore, the percentage of plants with wrinkled seeds (i.e. the homozygous recessive genotype) is 4%, as calculated above.

In Mendel's Crosses with Peas, Seed Shape Is Due to a Single