Question

small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of this population, 10 of the individuals have albino skin, which is due to the homozygous recessive condition (aa). Determine the frequency of the dominant (p) and recessive (q) allele? How many individuals would you expect to be homozygous dominant (AA), heterozygous (Aa) and homozygous recessive (aa)?

Answer 1

Answer:

Frequency of p $= 0.684$

Frequency of p $= 0.316$

Number of individuals with homozygous dominant (AA) $= 47$

Number of individuals with heterozygous (Aa)$= 43$

Number of individuals with homozygous recessive (aa) $= 10$

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals $(q^{2})$ is $\frac{10}{100} = 0.1$

Now, q will be

$= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316$

As per Hardy Weinberg's equation -

$p + q = 1$

Substituting the value of q in above equation, we get -

$p + 0.316 = 1p = 1 -0.316\\p = 0.684$

Now the frequency of homozygous dominant (AA) will be

$p^{2} = 0.684^{2} \\= 0.467$

Hence, out of 100 people $0.467 * 100 = 46.7 or 47$ people are homozygous dominant (AA)

Like wise out of 100 people $0.1 * 100 = 10$ people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

$p^{2} + q^{2} + 2pq = 1$

Substituting the values in above equation, we get -

$0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433$

So, out of 100 people $0.433 * 100 = 43.3 or 43$ people are heterozygous (Aa)