To solve this problem, we need to follow a logical series of steps derived from the given Hardy-Weinberg equilibrium principles. Here's the detailed breakdown:
1. Identify the given quantities:
- Number of organisms with green eyes: 44
- Total number of organisms: 100
2. Determine the frequency of the recessive phenotype:
- The frequency of organisms with green eyes (recessive phenotype) is [tex]\( q^2 \)[/tex].
- Since 44 out of 100 organisms have green eyes, the frequency of the recessive phenotype is [tex]\( q^2 = \frac{44}{100} = 0.44 \)[/tex].
3. Calculate the allele frequency for the recessive allele (q):
- The frequency of the recessive allele [tex]\( q \)[/tex] is obtained by taking the square root of [tex]\( q^2 \)[/tex].
- Thus, [tex]\( q = \sqrt{0.44} \approx 0.6633 \)[/tex].
4. Calculate the allele frequency for the dominant allele (p):
- The sum of the frequencies of both alleles must equal 1 (i.e., [tex]\( p + q = 1 \)[/tex]).
- Therefore, [tex]\( p = 1 - q \)[/tex].
- Substituting the value of [tex]\( q \)[/tex] we found, [tex]\( p = 1 - 0.6633 \approx 0.3367 \)[/tex].
So, the value of [tex]\( p \)[/tex] is approximately 0.34.
With these calculations in place, the answer to the problem is:
B. 0.34
