Answer :
The frequency of the recessive allele f in a population where 84 out of 400 individuals cannot tolerate caffeine is approximately 0.459.
To find the frequency of the f allele (autosomal recessive trait for inability to tolerate caffeine) in the population, we first need to determine the genotype frequencies for the individuals who are unable to tolerate caffeine (ff) and those who are able to tolerate caffeine (FF and Ff).
Given:
- Total population size = 400 individuals
- Number of individuals unable to tolerate caffeine (ff) = 84
Let's denote:
- Frequency of the f allele as q
- Frequency of the F allele as p
We know that in a population:
[tex]\[ p^2 + 2pq + q^2 = 1 \][/tex]
We are given that inability to tolerate caffeine (f) is an autosomal recessive trait, so ff individuals express the trait.
From the information given:
- Number of ff individuals = 84
- Total population size = 400 individuals
[tex]\[ q^2 = \frac{\text{Number of ff individuals}}{\text{Total population size}} = \frac{84}{400} = 0.21 \][/tex]
Now, we can find the frequency of the f allele (q):
[tex]\[ q = \sqrt{0.21} \approx 0.459 \][/tex]
Therefore, the frequency of the f allele in this population is approximately 0.459 or 45.9%.
