Question

Heterochromia iridium is a recessive autosomal condition in hurnons in which the colour of the iris of one eye is different from the colour of the other eye. Consider that in a human population, in a survey of 5000 people, two people were found to have heterochromia. Also, consider that the population is at Hardy-Weinberg equilibrium at this locus. Based on this information, answer the question:
(a) Assume that mating occurs at random with respect to this character. What frequency of the matings in this population will be between heterozygous individuals and homozygous dominant individuals?

Answer 1

THE HARDY-WEINBERG LAW GIVES US THIS EQUATION;

p² + 2pq + q² = 1 and p + q = 1

Where;

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

Having that in mind; The first thing to do is to determine p and q.

So since Heterochromia iridium is a recessive autosomal condition, it will be represented by hh, and therefore the gene for normal eyes is H.

Percentage of Heterochromia individuals $=\dfrac{2}{5000} x 100$ = 0.04%

Thus,q²(percentage of homozygous recessive individuals) = 0.04% = 0.0004

To determine q, simply take the square root of q²

q is therefore square root of 0.0004 = 0.02

Since p+q = 1, then p must be 1-0.02 = 0.98

Now then, to answer the questions; What frequency of matings will be;

(a) Heterozygous individuals (Hh)

= 2pq

= 2 (0.98)(0.02)

= 0.0397

(b) Homozygous dominant (HH)

= p²

= (0.98)²

= 0.9604

I think I deserve a bonus for this ($1 is a shame)