Question

A population of flowers is in Hardy-Weinberg equilibrium with an allele frequency for white flowers (w) of 40%. What percentage of the flowers will have the colored or dominant phenotype?

A. 16%
B. 25%
C. 40%
D. 60%
E. 84%

Answer 1

In a population at Hardy-Weinberg equilibrium with a 40% frequency of allele for white flowers, 84% of the flowers will have the colored (dominant) phenotype.

Step-by-step explanation:

When a population of flowers is in Hardy-Weinberg equilibrium with a 40% allele frequency for the white flowers (w), and assuming complete dominance for the colored (non-white) flowers allele, we can calculate the percentage of flowers that have the colored phenotype. The frequency of the white allele (w) is 0.4 (or 40%), which means the frequency of the colored allele (let's call it W) is 0.6 (or 60%). Using the Hardy-Weinberg equation (p² + 2pq + q² = 1), we can determine the proportion of the population with the dominant phenotype.

The dominant phenotype includes both the homozygous dominant (WW) and the heterozygous (Ww) genotypes, which is calculated as p² + 2pq. Plugging in our known values gives us (0.6²) + 2(0.6)(0.4), which equals 0.36 + 0.48, totaling 0.84, or 84%. Therefore, 84% of the flowers will have the colored phenotype.