Question

if 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population are homozygous dominant?

Explanation:

Step1: Calculate frequency of recessive phenotype

The frequency of the recessive phenotype ($q^{2}$) is the number of recessive - expressing individuals divided by the total number of individuals. So, $q^{2}=\frac{98}{200}=0.49$.

Step2: Calculate frequency of recessive allele

Take the square - root of $q^{2}$ to find the frequency of the recessive allele ($q$). So, $q = \sqrt{0.49}=0.7$.

Step3: Calculate frequency of dominant allele

Since $p + q=1$ (where $p$ is the frequency of the dominant allele), then $p = 1 - q$. Substituting $q = 0.7$, we get $p=1 - 0.7 = 0.3$.

Step4: Calculate frequency of homozygous dominant

The frequency of homozygous dominant individuals is $p^{2}$. So, $p^{2}=(0.3)^{2}=0.09$.

Step5: Convert to percentage

To convert the frequency to a percentage, multiply by 100. So, $0.09\times100 = 9\%$.

Answer:

9%