5. Tay-Sachs disease is caused by a recessive allele. The frequency of this allele is 0.1 in a population of 3600 people. What is the frequency of the dominant allele, and how many of the 3600 people will be heterozygous for the condition

Answer: Frequency of dominant allele is 0.9 and people who are heterozygous for the condition are 648.

Explanation:

According to Hardy-Weinberg equation, p represents frequency of dominant allele, q represent frequency of recessive allele.

And also p + q = 1.

Given frequency of recessive allele, q as 0.1.

p, frequency of dominant allele is

1 - 0.1 = 0.9

Therefore, frequency of dominant allele = 0.9.

Number of people that will be heterozygous will gotten from the percentage of heterozygous of the total population.

Heterozygous is represented as 2pq, that is the product of 2, p and q.

Therefore we have 2 multiply by p by q.

2 x 0.9 x 0.1 = 0.18 or 18%.

Since we are to find the number and not only percentage:

So from the population given as 3600, number of heterozygous is 18% of 3600.

18% of 3600 = 648.

The frequency of the dominant allele is 0.9 and there will be 648 heterozygous people for the condition.

Explanation:

According to Hardy Weinberg's equilibrium p2 + 2pq + q2 = 1

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p2 = percentage of homozygous dominant individuals

q2 = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals = frequency of the dominant allele

Therefore (1 - 0.1) = 0.9. (frequency of dominant allele)

(0.18x3600) = 648 (number of peopleheterozygous for Tay Sachs, 18% of the population).