Question

Fill in the blank.

Hemophilia is an X-linked trait. In a population of 20,000 people in HW equilibrium, 10,000 women and 10,000 men were tested for this trait, and 15 men were found to have the disorder.

The value that you calculated for the Hemophilia allele was (to 4 decimal places) _______ and there would be _______ carrier females in this population.

Answer 1

Since Hemophilia is an X linked trait and men carry only one X chromosome.

Now, Frequency of Hemophilia causing allele in Men

= 15/10,000 = 0.0015

Since the population is in Hardy-Weinberg Equilibrium, it would be expected that the frequency of the Hemophilia causing allele in the whole population is the same as the frequency of the Hemophilia causing allele in Men.

In other words;

Fr(X^hemophilia) in Men = Fr(X^hemophilia) in total population = Fr(X^hemophilia) in women.

Now, since the frequency of the Hemophilia allele (X^hemophilia) in women is 0.0015, thus;

Frequency of Normal X = 1 - Fr(X^hemophilia) = 1 - 0.0015 = 0.9985

Thus, per Hardy-Weinberg equations, Frequency of carrier (XX^hemophilia) women

= 2 * Fr(X) * Fr(X^hemophilia)
= 2 * 0.9985 * 0.0015 = 0.00299 = 0.003

Therefore, number of carrier (XX^hemophilia) women

= Frequency of carrier (XX^hemophilia) women * Number of women
= 0.003 * 10,000 = 30