In a population of gerbils, long hair (h) is completely dominant over short hair (h). if 18% of the population has short hair, calculate the percentage of the population that is expected to be heterozygous (hh).
a) 29%
b) 49%
c) 58%
d) 80%

b) 49%

Explanation:

To answer this question, we must assume that the population of gerbils is in Hardy-Weinberg equilibrium.

According to this law, the frequencies of the genotypes will be:

Where p is the frequency of the H allele and q is the frequency of the h allele.

p+q=1

If 18% of the population has short hair, the frequency of the hh genotype is 0.18.

Therefore:

The expected Hh population is 2pq:

The percentage of the population expected to be heterozygous is 0.49x100=49%

the correct answer is ( hypoglycemia ); good luck

explanation: