STEPTRACKERGENETICS + BIOSTATISTICS | POPULATION GENETICS | HIGH-YIELD REVIEW

Hardy-Weinberg & Population Genetics

Turn a disease prevalence into a carrier frequency in two lines — and know when the equation stops applying.

FOCUS ON
p + q = 1 and p² + 2pq + q² = 1. For an AR disease: prevalence = q² → take the square root for q → carrier frequency = 2pq ≈ 2q (p ≈ 1 when rare). For X-linked recessive: male prevalence = q; female prevalence = q². Assumes no mutation, no selection, random mating, no migration, large population.
QUANTITY
FORMULA
WORKED EXAMPLE
CLUE
Allele frequencies
p (dominant/normal) + q (recessive) = 1
q = 0.01 → p = 0.99
Always add to 1
Affected (AR)
q²
Disease in 1/10,000 → q² = 1/10,000
Starting point for most questions
Recessive allele
q = √prevalence
√(1/10,000) = 1/100
—
Carrier frequency
2pq ≈ 2q when rare
2 × 1/100 = 1/50
Heterozygotes
Unaffected homozygotes
p²
(0.99)² ≈ 0.98
—
X-linked recessive in males
Male prevalence = q
Hemophilia A in 1/5,000 males → q = 1/5,000
Males are hemizygous
X-linked recessive in females
Female prevalence = q² · carriers = 2pq ≈ 2q
(1/5,000)² — extremely rare
Female carriers = 2 × male prevalence
Child of two random people (AR)
(carrier freq)² × 1/4
(1/50)² × 1/4 = 1/10,000
Reconstructs prevalence
Known carrier × random partner
1 × (2q) × 1/4
1 × 1/50 × 1/4 = 1/200
Very commonly asked
Unaffected sib of affected × random partner
2/3 × 2q × 1/4
2/3 × 1/50 × 1/4 = 1/300
Combine pedigree and population risk

Mechanism — prevalence to carrier in three steps

Prevalence= q²
→
√ → q
→
Carriers = 2q(p ≈ 1)

Example, cystic fibrosis in 1/2,500 → q = 1/50 → carrier frequency ≈ 1/25.

Assumptions and what breaks them

ASSUMPTION
VIOLATED BY
EFFECT
No mutation
New mutations
Adds alleles
No selection
Heterozygote advantage (sickle cell–malaria)
Keeps allele common
Random mating
Consanguinity
↑ Homozygotes
No migration
Gene flow
Shifts frequencies
Large population
Founder effect, genetic drift
Rare allele becomes common

How it's tested

COMMON CONFUSIONFor X-linked diseases the male prevalence is q itself — don't take a square root.
THE TRAP"Probability their child is affected" needs the carrier probability of BOTH parents times 1/4 — candidates often forget the 1/4.
YOUR APPROACHWrite q² = prevalence → q → 2q. Multiply each parent's chance of carrying, then × 1/4 for an AR child.

If you see… think…

AR disease in 1/10,000 → carrier frequency ≈ 1/50
AR disease in 1/40,000 → q = 1/200, carriers ≈ 1/100
X-linked disease in 1/10,000 males → q = 1/10,000
Female carrier frequency for that X-linked disease → ≈ 1/5,000
Known carrier with a random partner (carriers 1/50) → child risk 1/200
Rare disease common in an isolated group → founder effect
Sickle trait common in malaria regions → heterozygote advantage
Consanguinity raises → homozygote frequency
Heterozygote frequency → 2pq
Assumption: random mating, large population, no selection/migration/mutation → Hardy-Weinberg equilibrium