Hardy-Weinberg Equilibrium

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Hardy-Weinberg Equilibrium

Using the Hardy-Weinberg Equilibrium, it is possible to establish the ratio of dominant to recessive alleles.

P represents the dominant allele.

q represents the recessive allele.

Example:

Parents: Aa x Aa
Gametes: A or a A or a
F1 genotypes: AA Aa Aa aa

Applying the P / q

P2 + 2Pq + q2

100% of the population will be AA, Aa or aa

Therefore: P2 + 2Pq + q2 = 1

Also,

The total number of dominant alleles + The total number of recessive alleles = 100%

Therefore:

P + q = 1

Example:

A genetic disease is caused by a recessive allele. The frequency of the disease in the population is 1 in 2,000. Calculate the frequency of the carrier genotype.

Let P = normal, q = disease allele.

Therefore:

NN = normal (P2),

Nn = carrier (2Pq),

nn = sufferer (q2).

So,

q2 = 1 in 2,000 = 0.0005

q = √0.0005 = 0.0224

Since P + q = 1

P = 1 - q = 1 - 0.0224

P = 0.9776

So,

2Pq = 2 (0.9776 x 0.0224) = 0.044

Or,

In 23 people (roughly 5% of the population are carriers).

S-cool exclusive!!