Answer:
51.84%
Explanation:
p² + 2pq + q² = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
p = 0.72 => p² = 0.72² = 0.5184
=> Percentage of the rats are normal sized: 51.84%
