Let's solve the problem step-by-step, assuming that the population is in Hardy-Weinberg equilibrium.
1. Determine the frequency of the recessive allele (q):
The problem states that the recessive allele (a) occurs with a frequency of one in a million. Therefore:
\(\[
q = \frac{1}{1,000,000} = 0.000001
\]\)
2. Determine the frequency of the dominant allele (p):
According to the Hardy-Weinberg principle, the sum of the frequencies of the dominant allele (p) and the recessive allele (q) equals 1:
\(\[
p + q = 1
\]\)
So:
\(\[
p = 1 - q = 1 - 0.000001 = 0.999999
\]\)
3. Determine the frequency of heterozygous carriers (2pq):
Heterozygous carriers (Aa) have one dominant and one recessive allele. The frequency of these carriers is represented as 2pq:
\(\[
\text{Frequency of carriers} = 2pq = 2 \times 0.999999 \times 0.000001
\]\)
\(\[
2pq = 2 \times 0.999999 \times 0.000001 = 0.000001999998
\]\)
4. Calculate the expected number of carriers in the town:
The population size is given as 14,000. To find the expected number of carriers, we multiply the carrier frequency (2pq) by the population size:
\(\[
\text{Number of carriers} = \text{Carrier frequency} \times \text{Population size}
\]\)
\(\[
\text{Number of carriers} = 0.000001999998 \times 14,000
\]\)
\(\[
\text{Number of carriers} \approx 0.027999972
\]\)
Since the number of carriers must be a whole number, we round the result to the nearest integer. Therefore, the expected number of individuals carrying the recessive allele in the town of 14,000 people is approximately:
\(\[
0
\]\)
Given that the exact value is so low, it rounds down to 0, meaning it is statistically unlikely that there are any carriers in this particular town size under the given conditions.
