POPH90111 · Genetic Epidemiology
Heritability
Familial aggregation told us a trait runs in families; heritability asks the sharper, quantitative question: of all the variation in a trait across a population, what proportion is due to genetic differences between people? The answer is a number between 0 and 1, and this is the most calculation-heavy stretch of POPH90111. You partition the phenotypic variance (VP = VG + VE; narrow-sense h² = VA/VP), then turn to twins — the natural experiment that holds environment roughly constant while varying genetic sharing — to run Falconer’s formula h² = 2(rMZ − rDZ), solve the ACE model for additive-genetic, common-environment and unique-environment components, and map a yes/no disease onto a hidden continuous liability. Every estimate is wrapped in the same caution: heritability is a property of a population in an environment, never of an individual, and it leans on the equal-environments assumption.
What this chapter covers
- 012.1 Heritability is a proportion of variance (VP = VG + VE; h² = VA/VP ≤ H²)
- 022.2 SD → variance (VP = SD²); estimate separately by sex and zygosity
- 032.3 Why twins? Holding the environment still (MZ ~100% vs DZ ~50% genes)
- 042.4 Falconer’s formula h² = 2(rMZ − rDZ)
- 052.5 The ACE model: A = 2(rMZ−rDZ), C = 2rDZ−rMZ, E = 1−rMZ
- 062.6 The liability-threshold model for binary traits
- 072.7 Critical appraisal: the equal-environments assumption and what h² does NOT mean
Worked example: the full ACE decomposition
- +1(a) Falconer. h² = 2(rMZ − rDZ) = 2(0.78 − 0.46) = 2(0.32) = 0.64.
- +1(b) A (additive genetic). A = 2(rMZ − rDZ) = 0.64 — ACE contains Falconer exactly.
- +1(b) C (shared environment). C = 2rDZ − rMZ = 2(0.46) − 0.78 = 0.92 − 0.78 = 0.14.
- +1(b) E (unique environment + error). E = 1 − rMZ = 1 − 0.78 = 0.22.
- +1(b) Sanity check. A + C + E = 0.64 + 0.14 + 0.22 = 1.00 ✓ — read off as 64% genes, 14% shared environment, 22% unique.
- +1(c) A negative C (when rDZ < ½rMZ) is impossible as a variance, so you drop it and fit an AE model; it usually flags non-additive genetic effects (dominance / epistasis) being mistaken for the additive split.
Key terms
- Narrow-sense heritability (h²)
- The proportion of phenotypic variance due to additive genetic variance, h² = VA/VP. Only the additive part is transmitted predictably parent-to-child and drives relatives’ resemblance, so h² is the quantity twin studies measure. Broad-sense H² = VG/VP includes dominance and epistasis, and always h² ≤ H².
- Falconer’s formula
- The twin estimate of narrow-sense heritability, h² = 2(rMZ − rDZ) for a continuous trait (or 2(concordanceMZ − concordanceDZ) for a binary one). The factor of 2 reflects that MZ twins share twice the additive genes of DZ twins, so doubling the gap in their correlations recovers the additive heritability.
- ACE model
- The variance-components model that partitions standardised phenotypic variance into A (additive genetic), C (common/shared environment) and E (unique environment + error), with A + C + E = 1. Solved from twin correlations as A = 2(rMZ−rDZ), C = 2rDZ−rMZ, E = 1−rMZ; it contains Falconer and recovers C and E for free.
- Liability-threshold model
- The device that puts a binary disease back onto a continuous scale: assume an unobserved, normally-distributed liability (genes + environment), and a person is affected only when liability exceeds a threshold T. Prevalence fixes T (disease is as common as the right-tail area beyond T), so the heritability of liability can be estimated — but it is heritability of the latent liability, not the observed yes/no status, and is sensitive to the assumed prevalence.
- Equal-environments assumption (EEA)
- The assumption underlying Falconer and ACE that MZ and DZ twin pairs share their environment to the same degree. If MZ twins are in fact treated more alike, their extra resemblance is partly shared environment masquerading as genes, and h² is over-estimated. It is the number-one critical-appraisal target for any twin estimate.
Heritability FAQ
Why is variance, not the trait value, the thing being explained?
Heritability is the proportion of the variance — the spread of differences between people — that is genetic. It says nothing about why any one person has the trait value they do; it explains why people differ from each other in one population and one environment. That is why a highly heritable trait (height is ~80% heritable) can still rise with better nutrition, and why h² tells you nothing about an individual.
Why use twins instead of ordinary relatives?
Because ordinary relatives confound genes with environment: closer kin share more of both. Twins break the tie — MZ and DZ pairs are the same age and reared together, so both share environment to roughly the same degree, but MZ twins share ~100% of their genes and DZ twins only ~50%. Hold environment constant, vary only the genetic sharing, and any extra resemblance in MZ pairs must be genetic. That difference is exactly what Falconer’s factor of 2 exploits.
Can Falconer’s formula give an impossible answer?
Yes — it is unbounded. If rDZ is less than half rMZ, h² can exceed 1, which is impossible and signals non-additive genetic effects (dominance/epistasis) or MZ-specific environment. If rDZ is more than half rMZ, the leftover implies shared environment C > 0. Falconer alone cannot separate these, which is the job of the ACE model — and when ACE returns a negative C you fit an AE model instead.
What does heritability NOT mean?
Three things examiners reward you for stating. (1) It does not mean the trait is unmodifiable — a high h² is compatible with large environmental change. (2) It says nothing about differences between groups or populations — within-group variance is not a between-group cause. (3) It tells you nothing about an individual — it is a population-in-an-environment statistic, and change the environment and h² itself changes. Add ‘missing heritability’: GWAS SNPs explain far less variance than twin studies imply.
Exam move
This is the chapter to drill arithmetically. Never drop the most-lost mark: variance = SD², and estimate h² separately by sex and zygosity because the raw spread differs across groups. Memorise the one-page route: continuous trait → VP = SD² → h² = VA/VP; twins given → check rMZ > rDZ, then Falconer h² = 2(rMZ−rDZ), then the full ACE solve with the A + C + E = 1 check; binary trait → use concordances or the liability threshold and quote the assumed prevalence. Then bank the marks-rich appraisal: name the equal-environments assumption, the ‘evidence-for-not-proof-of’ line, and the three things heritability does not mean.
