Suppose a population is carrying a condition controlled by two alleles: p (dominant) and p (recessive). only homozygous individuals that have two copies of the recessive allele have the condition. if the p allele is found in 15 percent of the population, what's the frequency of the pp genotype?

a. 25.5 percent
b. 72.25 percent
c. 2.25 percent
d. 12.75 percent

The answer is B. 72.25 percent.

The Hardy-Weinberg principle can be used:

p² + 2pq + q² = 1 and p + q = 1

where p and q are the frequencies of the alleles, and p², q² and 2pq are the frequencies of the genotypes.

The p allele (q) is found in 15% of the population:
q = 15% = 15/100
Thus, q = 0.15

To calculate the P allele frequency (p), the formula p + q = 1 can be used:
If p + q = 1, then p = 1 - q
p = 1 - 0.15
Thus, p = 0.85

Knowing the frequency of the P allele (p), it is easy to determine the frequency of the PP genotype (p²):
p² = 0.85² = 0.7225

Expressed in percentage, p² = 72.25%.

i believe the answer would be b. hope it : )

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