Lessons · Medicine foundations · the Hardy-Weinberg equation
Hardy-Weinberg: two frequencies that add up to one
If p and q are the frequencies of the two forms of a gene, then p + q = 1 and p² + 2pq + q² = 1, where p² and q² are the two homozygous groups and 2pq is the heterozygotes.
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
What it is for
The classic exam question gives you how common the recessive form is and asks how common carriers are. It looks like population genetics and it is a square root, a subtraction and a multiplication, in that order, every single time.
How to think about it
The number you are usually given is q², because that is the group you can actually count. Take its square root to get q. Subtract from one to get p. Then 2pq is the carrier frequency. Check at the end that p² + 2pq + q² comes to one; if it does not, one of the three steps slipped.
Worked example
q² = 0.01, given by the problemOne in a hundred has the recessive form.
q = sqrt(0.01) = 0.1, the allele frequencyThe square root of the group frequency, not the group frequency itself.
p = 1 − 0.1 = 0.9, because the two frequencies add to oneThere are only two forms, so they must fill the whole of one.
2pq = 2 × 0.9 × 0.1 = 0.18, the carrier frequencyEighteen in a hundred, far more than the one in a hundred you started from.
Your turn
A problem gives q² = 0.04, so q = 0.2 and p = 0.8. Write the line that gives the carrier frequency.
2 × 0.8 × 0.2 =
Solve one, graded on the server
The trap
Answering with q when the question asked for carriers. q is the allele frequency, one letter in the pool; 2pq is the share of people carrying one of each, and the factor of two is there because the pairing can happen either way round.