POPH90111 · Genetic Epidemiology
The Complete Exam Bible · S1 2026Genetic Epidemiology
Genetic Epidemiology teaches how to measure the genetic contribution to disease and to appraise the study designs that produce the evidence — familial aggregation, twin heritability, genetic association and GWAS, Mendelian randomisation, penetrance, gene–environment interaction and genetic screening. The whole subject is a calculation + critical-appraisal toolkit: every assessment item is the same chain — calculate a measure, interpret it, then critically appraise the design that produced it. Because all three assessments are open-book and take-home (an MCQ and two written assignments, with no closed-book exam), no fact is worth a mark on its own — the marks live in the method, the interpretation and the appraisal. This guide teaches each to assignment standard.
What POPH90111 covers
Seven chapters → one assignment-ready map of the genetic-epidemiology pipeline. Each links to its free chapter guide.
How POPH90111 is assessed
| Component | Weight | Format |
|---|---|---|
| Online MCQ | 10% | Open-book · ~30-min LMS quiz of ~10 questions · Introduction + Module 1 · completable any time within a one-week window |
| Assignment 1 (written, take-home) | 40% | Modules 1–3 · submitted online over a multi-week window |
| Assignment 2 (written, take-home) | 50% | Modules 4–8 · submitted online over a multi-week window — confirm the exact weights and dates in your subject guide |
Twin heritability by Falconer’s formula — the signature calculation, mark by mark
- +1Check the direction first. rMZ = 0.78 > rDZ = 0.46, so MZ twins resemble each other more than DZ twins — consistent with a genetic contribution. (If rMZ ≤ rDZ, h² would be about 0 and you would stop here.)
- +2(a) Apply Falconer. h² = 2(rMZ − rDZ) = 2(0.78 − 0.46) = 2 × 0.32 = 0.64. About 64% of the variation in female height in this population is additively genetic.
- +1(b) A (additive genetic). A = 2(rMZ − rDZ) = 0.64 — identical to Falconer, because ACE contains it.
- +1(b) C and E. C = 2rDZ − rMZ = 2(0.46) − 0.78 = 0.14 (shared environment); E = 1 − rMZ = 1 − 0.78 = 0.22 (unique environment + error). Check: 0.64 + 0.14 + 0.22 = 1.00 ✓.
- +1(c) The caveat. A higher MZ-than-DZ correlation is evidence for, but not proof of, an inherited genetic aetiology; heritability is a property of a population in an environment, not of an individual, and the estimate leans on the equal-environments assumption.
Key terms
- Familial aggregation
- The clustering of a disease in families more than chance would allow, measured by an odds ratio, relative risk, standardised morbidity ratio or recurrence-risk ratio λR. It is evidence for — never proof of — a genetic role, because relatives also share an environment.
- Heritability (h²)
- The proportion of the phenotypic variance in a population that is due to additive genetic variance, h² = VA/VP. It describes differences between people in one population and environment, not the trait of any individual. Falconer’s twin estimate is h² = 2(rMZ − rDZ).
- Genome-wide significance
- The strict p-value threshold a genetic association must clear in a GWAS, p < 5×10⁻⁸ (about −log10 = 7.3 on a Manhattan plot). It is the Bonferroni correction for roughly one million independent common-variant tests, and a genuine hit must also replicate.
- Mendelian randomisation
- An instrumental-variable method that uses a genetic variant as a proxy for a modifiable exposure. Because alleles are allocated randomly at conception, the variant is largely free of confounding and reverse causation, so it behaves like the randomisation arm of a trial — subject to its three IV assumptions (relevance, independence, exclusion restriction).
- Penetrance
- The probability that a person with a given genotype develops the disease by a specified age, Pr(disease | genotype, age) — a conditional, age-specific risk. It is usually incomplete (below 1), and clinic-recruited carrier families over-estimate it (ascertainment bias).
POPH90111 FAQ
Is POPH90111 hard?
It is conceptually rich but calculation-light: most marks come from applying a standard measure (an odds ratio, Falconer’s h², a Wald ratio, NNT/NNS) and then interpreting and critically appraising the study design. Because every assessment is open-book and take-home, the difficulty is not memorisation — it is reasoning precisely about bias, scale and what a number does and does not mean.
How is POPH90111 assessed?
There is no closed-book exam. Your grade is three take-home pieces: an open-book online MCQ (about 10%, covering the Introduction and Module 1), Assignment 1 (about 40%, Modules 1–3) and Assignment 2 (about 50%, Modules 4–8). Confirm this year’s exact weights and dates in your subject guide.
What does POPH90111 cover?
The genetic-epidemiology pipeline end to end: foundations and familial aggregation (Hardy–Weinberg, linkage disequilibrium, pedigrees, OR/RR/SMR), twin heritability (Falconer and the ACE model), genetic association and GWAS (allelic χ², the 5×10⁻⁸ threshold, Manhattan and QQ plots), Mendelian randomisation, penetrance and expressivity, gene–environment interaction, and genetic screening (sensitivity/specificity/PPV, ROC/AUC, NNT and NNS).
Do I need a strong statistics or genetics background for POPH90111?
No. The subject builds the genetics vocabulary from scratch and the statistics it needs are arithmetic-level — ratios, a 2×2 table, a variance decomposition, a number-needed-to-treat. The harder skill it trains is critical appraisal: naming the signature bias of each study design and stating the right caveat.
Is using AskSia for POPH90111 cheating?
No. AskSia is a study reference written in our own words — we host none of your lecturer’s files, and Sia teaches you the method to earn the marks; it does not complete or sit your assessments.
How to study for the exam
Treat the subject as a pipeline of paired skills — for every module, learn the signature calculation and the signature critical-appraisal point together, because every assignment item is “calculate → interpret → appraise.” Drill the recurring calculations until they are automatic (build a 2×2 then OR = ad/bc; Falconer h² = 2(rMZ−rDZ) and the ACE solve; the Wald ratio; ARR → NNT → NNS), and rehearse the matching appraisal for each design (recall and selection bias in case-control aggregation, the equal-environments assumption in twin studies, population stratification in GWAS, horizontal pleiotropy in MR, ascertainment bias in penetrance, the independence assumption in case-only G×E, and PPV-at-prevalence in screening). Because the assessments are open-book, treat the formula boxes and design-appraisal grids as a lookup — the marks are in the interpretation you write, and the one line that recurs everywhere is ‘evidence for, but not proof of, a genetic aetiology.’
