Clinical GeneticsMarch 21, 20264 min read

Q-Bank Breakdown: Hardy-Weinberg equilibrium — Why Every Answer Choice Matters

Clinical vignette on Hardy-Weinberg equilibrium. Explain correct answer, then systematically address each distractor. Tag: Genetics > Clinical Genetics.

Q-Bank Breakdown: Hardy-Weinberg equilibrium — Why Every Answer Choice Matters

Tag: Genetics > Clinical Genetics

Hardy-Weinberg equilibrium (HWE) questions look like pure math—until a vignette forces you to decide which allele is common, what “carrier frequency” actually means, and what assumptions you’re allowed to make. This post walks through a classic USMLE-style vignette, nails the correct answer, and then dissects each distractor so you don’t lose points to wording traps.

The Clinical Vignette (USMLE-Style)

A 26-year-old pregnant woman receives carrier screening. She is heterozygous for an autosomal recessive disorder. The father is from the same population. The disorder occurs in 1 in 10,000 live births in this population. Assume the population is in Hardy-Weinberg equilibrium. What is the approximate probability that this couple will have an affected child?

Answer choices: A. 1/400
B. 1/200
C. 1/100
D. 1/50
E. 1/10,000

Step-by-Step: Solve Using Hardy-Weinberg

Step 1: Translate incidence into allele frequency

For an autosomal recessive disease, affected frequency = q².

  • Given incidence = 1/10,000
  • So q² = 1/10,000
  • Therefore q = 1/100 = 0.01

Step 2: Find carrier frequency in the population

Carrier frequency = 2pq.

  • If q is small, p ≈ 1
  • So 2pq ≈ 2q ≈ 2(0.01) = 0.02 = 1/50

This means the father has about a 1/50 chance of being a carrier.

Step 3: Combine probabilities for the couple

Mother is a known carrier (probability = 1). If father is a carrier, chance of affected child from two carriers is 1/4.

P(affected child)=P(father is carrier)×14=150×14=1200P(\text{affected child}) = P(\text{father is carrier}) \times \frac{1}{4} = \frac{1}{50} \times \frac{1}{4} = \frac{1}{200}

✅ Correct answer: B. 1/200

Why Every Answer Choice Matters (Distractor Autopsy)

A. 1/400 — Half-right math, wrong carrier frequency

How you get it:

  • You might correctly apply the 1/4 Mendelian risk, but underestimate paternal carrier frequency as 1/100 instead of 1/50.

Common mistake:

  • Confusing q with carrier frequency.
    • Here, q = 1/100, but carriers ≈ 2q = 1/50.

High-yield takeaway: For rare AR disease, carrier frequency ≈ 2q, not q.

B. 1/200 — Correct

This reflects:

  • q² = 1/10,000 → q = 1/100
  • carrier frequency ≈ 2q = 1/50
  • affected given two carriers = 1/4
  • (1/50)(1/4) = 1/200

C. 1/100 — Forgetting the 1/4

How you get it:

  • You compute father’s carrier risk correctly as 1/50, but then assume that if dad is a carrier, the child is automatically affected.

Reality:

  • Two carriers produce:
    • 1/4 affected
    • 1/2 carriers
    • 1/4 unaffected non-carriers

High-yield takeaway: In AR inheritance with two carriers, affected risk is always 25% per pregnancy.

D. 1/50 — This is the father’s carrier probability, not the baby’s disease risk

How you get it:

  • You stop after finding 2pq ≈ 1/50.

Why it’s wrong:

  • The question asks the probability of an affected child, which requires multiplying by 1/4 after confirming paternal carrier status.

High-yield takeaway: When one parent is a known carrier, the workflow is:

  1. Use HWE to get partner carrier frequency
  2. Multiply by 1/4

E. 1/10,000 — Mixing up population incidence with this couple’s risk

How you get it:

  • You assume the child’s risk equals the population disease frequency.

Why it’s wrong:

  • This couple is higher risk than the general population because mom is a known carrier.

High-yield takeaway: Once you’re told a parent is a carrier, you’re no longer dealing with a “random newborn”—you’re doing conditional probability.

High-Yield Hardy-Weinberg Facts (USMLE Gold)

The core equations

  • p + q = 1
  • p² + 2pq + q² = 1
    • p²: homozygous normal
    • 2pq: heterozygous carriers
    • q²: homozygous affected (for AR disease)

Classic AR approximation for rare diseases

If the disease is rare, q is small, so p ≈ 1:

  • 2pq ≈ 2q
  • carrier frequency ≈ 2 × √(incidence)

When Hardy-Weinberg assumptions matter

HWE assumes:

  • Large population
  • Random mating
  • No mutation
  • No migration
  • No natural selection

USMLE loves scenarios that violate “random mating,” e.g.:

  • Consanguinity → increases homozygosity → increases AR disease risk beyond HWE expectations
  • Founder effect (e.g., Ashkenazi Jewish populations with Tay-Sachs) → changes allele frequencies
  • Selection (e.g., sickle cell trait advantage in malaria regions) → alters expected genotype distribution

Common Step pitfalls

  • Treating q as carrier frequency
  • Forgetting the 1/4 Mendelian risk
  • Using population incidence even when given a known carrier in the family
  • Forgetting that HWE applies to populations, not individual families—until you use it to estimate a partner’s carrier status

Quick “Exam-Day” Template

If vignette says: AR disease incidence = q², and one partner is a known carrier:

  1. Compute q = √(q²)
  2. Estimate carrier frequency ≈ 2q
  3. Multiply: P(affected child)≈(2q)×14=q2P(\text{affected child}) \approx (2q)\times\frac{1}{4} = \frac{q}{2}

For this question: q = 0.01 → risk ≈ 0.005 = 1/200.

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