# Option D.4 – The Hardy-Weinberg Principle

Option D.4 – The Hardy-Weinberg Principle

D.4.1 – Explain how the Hardy-Weinberg equation is derived Evolution involves changes in allele frequency. The Hardy-Weinberg equation is often

Evolution involves changes in allele frequency. The Hardy-Weinberg equation is often used for doing calculations of allele frequency. The Hardy-Weinberg equation is a formula that gives proportions of the diploid

The Hardy-Weinberg equation is a formula that gives proportions of the diploid genotypes formed by random union of haploid gametes, or random mating between diploids.

If there are two alleles of a gene in a population, the frequency of the alleles is represented by p and q. The total frequency of the alleles is 1: When there is random mating, the chance of inheriting two copies of the first allele is p x p. The chance of inheriting two copies of the second allele q x q. Thus, the expected frequency of the homozygous genotypes is p2 and q2, while the heterozygous genotype is 2pq. The sum of all the frequencies is 1.  D.4.2 – Calculate allele, genotype and phenotype frequencies for two alleles of a gene, using the Hardy-Weinberg equation

Allele Frequencies

The gene for tasting phenylthiocarbamide (PTC) has two alleles: ability to taste is dominant (T) and no tasting is the recessive (t).

In a survey of 3200 people, 922 could not taste PTC: a frequency of 0.288.

The genotype homozygous recessive (t t) is called q.

Thus, q2 = 0.288
q = 0.537
p = T allele, so p = (1 – q) = 0.463

Genotype Frequencies

Cystic fibrosis is caused by recessive alleles of a chloride channel gene. A screening of 27000 people who did not have cystic fibrosis tested for carriers of

A screening of 27000 people who did not have cystic fibrosis tested for carriers of the recessive gene. The frequency of the normal allele = p = 0.9776

The frequency of the normal allele = p = 0.9776

Frequency of the cystic fibrosis allele = q = 0.0224

When the tested people have children, the chance of the child having cystic fibrosis: q2 = (0.0224)2 = 0.000502

So, one child in 1900 would have cystic fibrosis. Meanwhile, the chance of a child being a carrier is:

2pq = 2(0.9776 x 0.0224) = 0.0438

So, one child in 23 would be a carrier.

Phenotype Frequencies

Pea plants have dwarf and tall varieties, in a ratio of three tall to one dwarf. The plants are then allowed to disperse seeds naturally.

Frequency of T (p) = 0.75
Frequency of t (q) = 0.25

Thus, the frequency of (t t) = q2 = 0.252 = 0.0625

The frequency of tall plants = 1 – q2 = 1 – 0.0625 = 0.0375

D.4.3 – State the assumptions made when the Hardy-Weinberg equation is used