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Sagot :
To determine the value of [tex]\( q \)[/tex], we need to follow the steps for calculating allele frequencies under Hardy-Weinberg equilibrium conditions.
1. Determine the frequency of the recessive phenotype (green eye color in this case):
- Given that 20 out of 100 organisms are green, the frequency of the recessive phenotype ([tex]\( q^2 \)[/tex]) is:
[tex]\[ q^2 = \frac{20}{100} = 0.20 \][/tex]
2. Calculate the value of [tex]\( q \)[/tex] (the frequency of the recessive allele) by taking the square root of [tex]\( q^2 \)[/tex]:
- Solving for [tex]\( q \)[/tex], we get:
[tex]\[ q = \sqrt{0.20} \approx 0.4472 \][/tex]
So, the value of [tex]\( q \)[/tex] is approximately [tex]\( 0.45 \)[/tex]. Thus, the correct answer is:
C. 0.45
1. Determine the frequency of the recessive phenotype (green eye color in this case):
- Given that 20 out of 100 organisms are green, the frequency of the recessive phenotype ([tex]\( q^2 \)[/tex]) is:
[tex]\[ q^2 = \frac{20}{100} = 0.20 \][/tex]
2. Calculate the value of [tex]\( q \)[/tex] (the frequency of the recessive allele) by taking the square root of [tex]\( q^2 \)[/tex]:
- Solving for [tex]\( q \)[/tex], we get:
[tex]\[ q = \sqrt{0.20} \approx 0.4472 \][/tex]
So, the value of [tex]\( q \)[/tex] is approximately [tex]\( 0.45 \)[/tex]. Thus, the correct answer is:
C. 0.45
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