Sunday 23 May 2010

The Hardy Weinberg Distribution

The Hardy Weinberg distribution allows you to find out the prevalence of homozygotes and heterozygotes for given alleles if you know the overall prevalence of the disease.

Thus, say a trait has two phenotypes, a and b. "a" has an allele called p and "b" has an allele called q.

The prevalence of homozygotes for p or q would be p^2 and q^2 respectively. The prevalence of heterozygotes would be 2pq.

This is easy to understand. Say if you dip into a bag containing two types of marbles- a & b, whose proportions are p & q respectively. The chances that you pick out "a" first time is of course, p. The chance of picking "a" twice in a row is p*p, and similarly the chances of picking "b" back to back is q*q.

If you picked "a" first, the chance that you'd pick "b" next is of course p*q. If you pick "b" first the chance that you'd pick "a" next q*p. Thus the chance that you'd pick a combination of a & b in two attempts is 2pq.

Let's look at two applications of this.

An African boy has sickle cell anaemia, an autosomal recessive disease with a prevalence of 1 in 500 among Blacks. His sister is getting married to a man of same ethnic origin, and wants to know the odds that her partner, who is unrelated to her, is a carrier.

For an autosomal recessive disorder, cases are obligate homozygotes, and therefore will have their distribution given by p^2. Thus, p^2= 1/500, or p= sq root 1/500= 1/22 (approx).

Thus, the likelihood that her husband would be a heterozygote would be 2pq or 2*1/22*21/22= 1/11 (approx).

Let's take another example. Haemophilia has a population prevalence of 1 in 5000. What's the likelihood that a random female would be a carrier?

Since haemophilia is an X linked trait, the prevalence of the disease is the same as that of the defective allele (since affected boys carry the defective allele on their only X chromosome). Thus, p= 1/5000.

Therefore the chance that an unselected female would be a carrier is 2pq= 2*1/5000*4999/5000= 1/2500.

The Hardy Weinberg distribution assumes that there is no inbreeding among relatives. In practice, many rare disorders express themselves more frequently in communities who inbreed. I'll deal with ways to get around this in another blog entry.

No comments:

Post a Comment