Suppose a population is carrying a condition controlled by two alleles: R (dominant) and r (recessive). Only homozygous individuals that have two copies of the recessive allele have the condition. If the r allele is found in 15 percent of the population, what's the frequency of the RR genotype? A. 2.25 percent
B. 25.5 percent
C. 72.25 percent
D. 12.75 percent

Answers

Answer 1
Answer:

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

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

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

Expressed in percentage, p² = 72.25%.

Answer 2
Answer: Simple its d 72 and quarter percent

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Good luck on that test of your and i hope you past the test