To calculate the percentage of heterozygous individuals in a population where 98 out of 200 individuals express the recessive phenotype, we can use the Hardy-Weinberg principle.
Let’s denote:
- \( p \) = frequency of the dominant allele
- \( q \) = frequency of the recessive allele
According to Hardy-Weinberg, the frequencies of the genotypes can be represented as:
- \( p^2 \) = frequency of homozygous dominant individuals
- \( 2pq \) = frequency of heterozygous individuals
- \( q^2 \) = frequency of homozygous recessive individuals
Since 98 individuals express the recessive phenotype, we can say that: \[ q^2 = \frac{98}{200} = 0.49 \]
Now, we can find \( q \) by taking the square root of \( q^2 \): \[ q = \sqrt{0.49} = 0.7 \]
Next, we can find \( p \) since \( p + q = 1 \): \[ p = 1 - q = 1 - 0.7 = 0.3 \]
Now, we can find the frequency of heterozygous individuals (2pq): \[ 2pq = 2 \times p \times q = 2 \times 0.3 \times 0.7 = 0.42 \]
To find the percentage of the population that is heterozygous, we multiply by 100: \[ 0.42 \times 100 = 42% \]
Thus, the answer is \( c) 42% \).