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

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).

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.