Why recessive genes do not die out
Tonight's idea
A gene comes in two versions, the alleles \(A\) and \(a\). Everyone carries two copies, giving three genotypes: \(AA\), \(Aa\), \(aa\). Suppose a fraction \(p\) of all copies in a population are \(A\) and \(q = 1 - p\) are \(a\). If mating is random, a child's two copies are independent draws from the pool, so
\(P(AA) = p^2,\quad P(Aa) = 2pq,\quad P(aa) = q^2,\) and \(p^2 + 2pq + q^2 = (p + q)^2 = 1\).
Now count the \(a\) copies among the children. Each \(aa\) child carries two, each \(Aa\) child one, so the new fraction of \(a\) is
\[q' = q^2 + \tfrac12(2pq) = q(q + p) = q.\]
Nothing moved. Whatever mix you start with, one generation of random mating puts the genotypes at \(p^2, 2pq, q^2\), and they stay there: the Hardy–Weinberg equilibrium (Hardy and Weinberg, 1908).
The consequence that matters concerns a recessive allele, whose effect shows only in \(aa\). The affected fraction is \(q^2\); the carriers \(Aa\), who show nothing, are \(2pq \approx 2q\) when \(q\) is small. Carriers outnumber the affected by \(2p/q\), and almost every \(a\) copy sits in one of them, out of selection's reach. So a harmful recessive allele lingers for thousands of years.
Watch one: \(q = 0.01\)
A recessive condition with \(q = 0.01\) affects \(q^2 = 0.0001\) of births, one in 10,000. Carriers are \(2pq = 2(0.99)(0.01) = 0.0198\), about one in 50. Carriers outnumber the affected by \(2p/q = 198\) to 1.
Where are the \(a\) copies? Carriers hold \(pq\) of all copies as \(a\); the affected hold \(q^2\). The share in carriers is \(pq/(pq + q^2) = p = 0.99\). Ninety-nine of every hundred recessive alleles sit in healthy people. Selection against the condition touches one copy in a hundred.
Random draws from a pool, as in a Grade 10 probability sheet, but with the pool refilled every generation.
Your turn
- \(p = 0.7\), \(q = 0.3\). (a) Give the three genotype fractions and the counts in 1,000 people. (b) Count the \(a\) copies among the 2,000 copies in those people and check that their share is still 0.3.
- Two isolated villages of 1,000 people merge: one is entirely \(AA\), the other entirely \(aa\). Mating is now random. (a) Find \(p\) and \(q\). (b) Give the genotype fractions among the children. (c) Show the grandchildren have the same fractions.
- A recessive condition affects 1 birth in 2,500. (a) Find \(q\). (b) What fraction of people are carriers, as "one in how many"? (c) How many carriers are there per affected person?
- In a sample of 200 people, 120 are \(AA\), 60 are \(Aa\) and 20 are \(aa\). (a) Find \(p\) by counting copies, not people. (b) How many of each genotype would Hardy–Weinberg predict? (c) The sample has too few \(Aa\). Suggest what about the population could cause that.
Stretch: a dose of selection
Suppose \(aa\) individuals never reproduce. Show that the next generation's frequency is \(q' = q/(1 + q)\), so \(1/q\) rises by exactly 1 each generation. Starting from \(q = 0.01\), how many generations does it take to halve \(q\)? Compare with an allele whose effect is dominant and equally harsh.
Answers
- (a) \(AA\): \(0.49\) (490), \(Aa\): \(2(0.7)(0.3) = 0.42\) (420), \(aa\): \(0.09\) (90). (b) \(a\) copies: \(2 \times 90 + 420 = 600\) out of 2,000, so \(q = 0.3\) again.
- (a) 2,000 \(A\) copies and 2,000 \(a\) copies out of 4,000: \(p = q = \tfrac12\). (b) \(\tfrac14, \tfrac12, \tfrac14\). (c) Among the children, \(q = \tfrac14 + \tfrac12 \cdot \tfrac12 = \tfrac12\), unchanged, so the grandchildren are again \(\tfrac14, \tfrac12, \tfrac14\). The parents were far from equilibrium (no \(Aa\) at all); one generation fixed it.
- (a) \(q^2 = 1/2500\), so \(q = 1/50 = 0.02\). (b) \(2pq = 2(0.98)(0.02) = 0.0392\), about one person in 26. (c) \(2p/q = 98\) carriers per affected person.
- (a) \(A\) copies: \(2 \times 120 + 60 = 300\) of 400, so \(p = 0.75\), \(q = 0.25\). (b) \(200 \times 0.5625 = 112.5\) \(AA\), \(200 \times 0.375 = 75\) \(Aa\), \(200 \times 0.0625 = 12.5\) \(aa\). (c) Observed 120, 60, 20: both homozygotes are over-represented. Mating is not random with respect to this gene: people tend to pair with partners of their own genotype. Relatives marrying does this, and so does a population made of two groups that rarely intermarry, like the villages of question 2 before the merger.
Stretch. Only \(AA\) (fraction \(p^2\)) and \(Aa\) (fraction \(2pq\)) reproduce. Among their copies the share of \(a\) is \(\dfrac{pq}{p^2 + 2pq} = \dfrac{q}{p + 2q} = \dfrac{q}{1 + q}\), using \(p + q = 1\). Then \(1/q' = (1 + q)/q = 1/q + 1\). From \(q = 0.01\), \(1/q\) goes from 100 to 200 in exactly 100 generations, roughly 2,500 years. (Check: \(0.1 \to 0.0909 \to 0.0833 \to 0.0769\), that is \(1/10, 1/11, 1/12, 1/13\).) A dominant allele that stopped its carriers reproducing would vanish in a single generation, because every copy is exposed. Recessive alleles survive by hiding.