Population genetics
Introduzido| English | Significado |
|---|---|
| Hardy–Weinberg equilibrium | Hardy–Weinberg equilibrium |
| heterozygote advantage | heterozygote advantage |
| founder effect/ˈfaʊndə ɪˈfekt/ | founder effect |
"Your carrier risk is 1 in 25"
- A healthy person with no family history is told that number before any test. It comes from one equation and the number of people born with the disease.
- The same arithmetic runs newborn screening, carrier panels and every risk quoted for a couple.
The Hardy–Weinberg equation
- For a gene with two alleles at frequencies p and q (p + q = 1), the genotype frequencies are $p^2 + 2pq + q^2 = 1$: this is Hardy–Weinberg equilibrium 哈迪-温伯格平衡.
- It holds if mating is random, the population is large, and there is no selection, migration or new mutation.
In a population at Hardy–Weinberg equilibrium, q = 0.1. What percentage of people are carriers (Aa)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
p = 0.9, so 2pq = 2 × 0.9 × 0.1 = 0.18 = 18%.
Why the 2?
- Each child draws one allele from each parent at random. AA has probability $p^2$ and aa has $q^2$.
- A carrier can get A from the mother and a from the father, or the other way round: two routes, so $2pq$.
The 2 in 2pq counts the two ways of being a carrier, A from the mother or A from the father.
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Aa and aA are both carriers, so their probabilities add.
From disease rate to carrier rate
- Cystic fibrosis affects about 1 in 2500 Northern European births, so $q^2 = 1/2500$ and $q = 1/50$.
- Carriers, about 4% of people: $2pq = 2 \times \tfrac{49}{50} \times \tfrac{1}{50} \approx \tfrac{1}{25}$. When q is small, the carrier rate is about $2q$.
A recessive disease affects 1 birth in 3600. The carrier rate is about 1 in how many people?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
q = √(1/3600) = 1/60, and 2pq ≈ 2 × 1/60 = 1/30.
When the equation bends
- Heterozygote advantage 杂合子优势: sickle-cell carriers resist malaria, so the allele stays common in malaria regions although homozygotes are selected against.
- The founder effect 奠基者效应: a population descended from a few ancestors can carry a rare allele at high frequency, such as Tay–Sachs carriers, about 1 in 30 in Ashkenazi Jews against about 1 in 300 generally.
Why is the sickle-cell allele common in malaria regions?
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
Heterozygote advantage balances the loss of homozygotes, a known departure from simple Hardy–Weinberg.
Match each idea to its example.
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
The equation gives the default; selection and founders move it.
A negative screen is not zero
- No carrier panel finds every variant. With carrier chance c and detection rate d, the chance of still being a carrier after a negative test is $\dfrac{c(1-d)}{1-cd}$.
- For cystic fibrosis, $c = 1/25$ and $d = 0.92$ give $\dfrac{0.04 \times 0.08}{1 - 0.0368} \approx \dfrac{1}{300}$: the residual risk.
Why is a negative carrier screen not quoted as zero risk?
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The undetected fraction leaves a small residual chance, which depends on the gene and the panel.
Worked example: a couple with no family history
- She has the population carrier chance of 1/25; her partner screened negative, leaving a residual of 1/300.
- Risk for their child: $\tfrac{1}{25} \times \tfrac{1}{300} \times \tfrac14 = \tfrac{1}{30\,000}$. Carrier screening is this one multiplication.
Worked example: a known carrier
- (Fictional) One partner is a known carrier; the other screens negative, with a residual of 1/300.
- Risk: $1 \times \tfrac{1}{300} \times \tfrac14 = \tfrac{1}{1200}$. The counsellor quotes this number and never says "no risk".
One partner is a known carrier. The other screens negative, with a residual carrier chance of 1/400. Their child's risk is 1 in how many?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
1 × 1/400 × 1/4 = 1/1600.
Three common slips
- Square-rooting a carrier rate. $\sqrt{\ }$ works on the disease rate ($q^2$); carriers are $2pq$.
- Using the equation where selection or founders distort it. Check the population first.
- "A negative screen means no risk." The residual is small but real, and it depends on the gene and the panel.
Recap
- $p^2 + 2pq + q^2 = 1$; disease rate gives $q = \sqrt{q^2}$, and carriers are $2pq \approx 2q$.
- Cystic fibrosis: 1/2500 → q = 1/50 → carriers about 1/25.
- Heterozygote advantage and founder effects bend the numbers; residual risk after a negative screen is quoted, never zero.