Exam practice

The instructor’s Lecture 7 exam questions, then drills and written questions sorted into written exam, exam status unclear, and seminar topics

This page collects everything you can practise rather than read. It starts with the instructor’s own example exam questions for Lecture 7. The rest is sorted by what the lecturer said about the exam (lecture notes, 29 September):

For the 20 concepts on Martin’s key-concepts list (GWAS, MR, proteomics), see Martin’s key concepts: one module with a quick check per concept.

Drills draw new numbers each time you press New problem ↻: work on paper, type your answers, press Check, then open the full solution. For written questions, answer out loud or on paper before you open a model answer. Specific medicines, mutations and the course’s running examples are left out, since the lecturer said not to focus on them.

Instructor’s questions: the 11 questions in the first section are from “Lecture 7 - questions.pdf”, described on Canvas as “examples (but not an exhaustive list) of exercises and possible exam questions”. The questions are the instructor’s; the model answers are written for this app and are not an official answer key.

Drills and written questions: the drills cover every numerical worked example in the course. The written questions are written for this app. Their grouping follows the lecturer’s guidance: 🎯 topics the instructor tagged EXAM in Lecture 7, 🎓 the professor’s word and concept list, 🗣️ seminar content (computer exercises 1–2), ❔ unclear, 📌 AI-inferred. Q17 (a running-example question) was removed, and Q19 now uses a made-up variant; the other numbers are unchanged.

🎯 Instructor’s example questions: Lecture 7

These come straight from “Lecture 7 - questions.pdf”, the closest thing to real exam questions in the course. Answer each one on paper first.

Model answer. Every value measured on a high-throughput platform is the sum of the true biological signal we want and variation added by the measurement process. Differences between samples can therefore come from biology or from the technology, and we must remove as much technical variation as possible (preprocessing) before comparing groups. Sources of technical variation named in the lecture (any two): RNA extraction, labelling, hybridization, scanner/detection noise, batch effects. → 7.4

Model answer. The fluorescence at each spot contains background noise on top of the true signal, and this matters most for weakly expressed miRNAs. Simple subtraction (foreground − background) is too naive because the background estimate is itself noisy: for low-intensity spots the result is often zero or negative, which is biologically meaningless and cannot be log-transformed, and the low-intensity values become very unstable. normexp models observed intensity = true signal + background, with the signal exponentially distributed and the background normally distributed, and replaces each value by the expected true signal given the observed intensity. The corrected values are always positive, and low-intensity measurements become more reliable. → 7.5

Model answer. It makes the distribution of intensities identical on every array: rank the values within each array, compute the mean across arrays at each rank, and replace every value by the mean for its rank. Each array keeps its internal order (the highest miRNA stays highest), but differences in overall level and spread between arrays disappear. It is needed because arrays differ systematically for technical reasons (amount of RNA, labelling efficiency, scanner settings, batch). The assumption is that the overall distribution of expression should be similar across samples, so a whole-array shift is technical. Without normalization a miRNA could look “up” in a sample simply because that whole array was brighter. → 7.5

\[t = \frac{\bar x_1 - \bar x_2}{s\sqrt{1/n_1 + 1/n_2}}\]

Model answer. \(s\) is estimated separately for every miRNA from only \(n_1 + n_2 - 2 = 4\) degrees of freedom. With so few observations, the sampling variability of \(s\) is huge: a feature’s estimate can easily be far below or far above its true value just by chance. Across hundreds or thousands of features, many extreme estimates are guaranteed.

  • False positives: features with a randomly small \(s\) get a small denominator and a huge \(t\), even when the mean difference is tiny.
  • False negatives (lost power): features with a randomly large \(s\) get a small \(t\), so real differences are missed.
  • The ranking of “top hits” is driven partly by noise in \(s\) rather than by effect size, and the unstable p-values also undermine the multiple-testing correction applied afterwards. → 7.7–7.8

Model answer. The moderated t-statistic (limma) is a t-statistic in which each feature’s variance is replaced by a moderated variance estimated with empirical Bayes, sharing information across all features:

\[\tilde s_g^2 = \frac{d_0 s_0^2 + d_g s_g^2}{d_0 + d_g}\]

“Shrinking toward a common estimate” means each gene’s own variance \(s_g^2\) is pulled toward a global variance \(s_0^2\) estimated from all genes. The result is a weighted average: weight \(d_0\) (prior degrees of freedom) on the global value and weight \(d_g\) (the gene’s own residual df) on its own estimate. Extremely small and extremely large variances are pulled toward the middle, which stabilizes the denominator of \(t\) and gives fewer false positives and more power.

  • Few replicates (small \(d_g\)): \(d_0\) dominates, and each variance is pulled strongly toward \(s_0^2\).
  • Many replicates (large \(d_g\)): the gene’s own data dominate, and \(\tilde s_g^2 \to s_g^2\), so the moderated t approaches the ordinary t. → 7.7

Model answer. The variance sits in the denominator of \(t\). A randomly tiny \(s^2\) makes \(t\) very large and the p-value very small, even if the mean difference is modest or zero, so the miRNA is called significant because of noise in \(s^2\) rather than a real effect. Moderation replaces \(s^2\) by \(\tilde s^2 = (d_0 s_0^2 + d\, s^2)/(d_0 + d)\). Because the global variance \(s_0^2\) is typical rather than tiny, \(\tilde s^2\) is pulled up toward it, the denominator grows, and \(t\) shrinks back to a realistic size. With few samples the pull is strong, which is exactly when the risk is largest. (Try the moderated t drill below.) → 7.7

Model answer.

  • FWER (family-wise error rate) \(= P(V \ge 1)\): the probability of making at least one false positive among all \(m\) tests. Here \(V\) is the number of false positives.
  • FDR (false discovery rate) \(= E\left[V/R\right]\), with \(V/R\) taken as 0 when \(R = 0\): the expected proportion of false positives among all tests called significant, where \(R\) is the number of rejections.

Key difference: FWER treats a single false positive anywhere as failure of the whole experiment, which is very strict. FDR accepts that some hits will be false and controls the fraction of false hits among the discoveries. FWER is controlled by Bonferroni (α/m); FDR is controlled by Benjamini–Hochberg. FDR control gives more power when many tests are run. → 7.11

miRNA p-value
miR-A 0.0360
miR-B 0.0008
miR-C 0.0620
miR-D 0.0140
miR-E 0.0071
miR-F 0.0810
miR-G 0.0090
miR-H 0.0330
miR-I 0.0038
miR-J 0.0420
  1. How many are significant with no correction (p < 0.05)? (b) With Bonferroni (α/m)? (c) Bonferroni-adjusted p for miR-H (p × m); is it significant? (d) Sort, give ranks 1–10, show (i/m)·Q; how many are significant with BH, and which rank sets the cutoff? (e) Compute p₍ᵢ₎·m/i for every rank. (f) BH-adjusted p for rank 5? (g) BH-adjusted p for rank 3? (Hint: not simply p₍₃₎·m/3.)

Model answer. m = 10.

(a) All except miR-C (0.062) and miR-F (0.081): 8 significant.

(b) α/m = 0.05/10 = 0.005. Only miR-B (0.0008) and miR-I (0.0038): 2 significant.

(c) p_adj = 0.0330 × 10 = 0.33 > 0.05, so miR-H is not significant after Bonferroni.

(d)–(e)

Rank i miRNA p₍ᵢ₎ (i/m)·Q p₍ᵢ₎ ≤ (i/m)·Q? p₍ᵢ₎·m/i BH-adjusted p
1 miR-B 0.0008 0.005 ✓ 0.0080 0.0080
2 miR-I 0.0038 0.010 ✓ 0.0190 0.0190
3 miR-E 0.0071 0.015 ✓ 0.0237 0.0225
4 miR-G 0.0090 0.020 ✓ 0.0225 0.0225
5 miR-D 0.0140 0.025 ✓ (cutoff) 0.0280 0.0280
6 miR-H 0.0330 0.030 ✗ 0.0550 0.0514
7 miR-A 0.0360 0.035 ✗ 0.0514 0.0514
8 miR-J 0.0420 0.040 ✗ 0.0525 0.0525
9 miR-C 0.0620 0.045 ✗ 0.0689 0.0689
10 miR-F 0.0810 0.050 ✗ 0.0810 0.0810

The largest rank with p₍ᵢ₎ ≤ (i/m)·Q is i = 5 (0.0140 ≤ 0.025), and ranks 6–10 all fail. 5 significant with BH: miR-B, miR-I, miR-E, miR-G, miR-D. (Note how close ranks 6 and 7 are: 0.0330 vs 0.030 and 0.0360 vs 0.035.)

(f) Rank 5: the minimum over j ≥ 5 of p₍ⱼ₎·m/j = min(0.0280, 0.0550, 0.0514, 0.0525, 0.0689, 0.0810) = 0.028. Here the minimum is at rank 5 itself.

(g) Rank 3: p₍₃₎·m/3 = 0.0071 × 10/3 = 0.0237, but rank 4 gives 0.0225, which is smaller. The adjusted value is the running minimum from the bottom, so p_adj(3) = 0.0225 (the minimum occurs at rank 4). This keeps adjusted p-values non-decreasing down the sorted list.

Summary: uncorrected 8, BH 5, Bonferroni 2 significant. In R, p.adjust(p, "BH") and p.adjust(p, "bonferroni") give the same values. → 7.12–7.13b

Model answer. Any two of the four:

  1. Too many tests: with m ≈ 955, Bonferroni requires p < 0.05/955 ≈ 0.00005, so most true effects are missed and power collapses.
  2. Exploratory aim: the screen produces a shortlist of candidates, so a small, controlled fraction of false leads is acceptable in exchange for finding more true effects.
  3. Validation follows: hits are re-tested with an independent method or cohort (miR-375 was confirmed by qRT-PCR and in situ hybridization in a separate cohort).
  4. Correlated features: co-regulated miRNAs make the effective number of independent tests smaller than m, so Bonferroni is needlessly strict. → 7.14

Given: M1 had more high-grade tumours (P = 0.0011), more Ki67 > 3% (P = 0.00023), more chromosome 14 gain (P = 0.0013) and shorter overall survival than M2 (log-rank P = 0.029).

Model answer.

(a) Fisher’s exact test for categorical variables such as grade. It tests association in a contingency table (cluster × category) and is exact, so it is valid with the small cell counts you get with a few dozen tumours, where the χ² approximation is unreliable. The t-test for numerical variables such as age. It compares the mean of a continuous variable between two groups (M1 vs M2).

(b) M1 tumours are more proliferative and more aggressive: higher grade (G2/G3) and more Ki67 > 3%, both markers of cell division. They also more often carry gain of chromosome 14, fitting the idea that siNETs are driven by chromosomal alterations rather than point mutations. Clinically, M1 patients have shorter overall survival (log-rank P = 0.029). So M1 is a molecular subtype with a worse prognosis, and M2 is a less proliferative subtype with a better one. The most upregulated miRNAs in M1 (miR-1246, miR-663a) are known oncomiRs; miR-488-3p, a reported tumour suppressor, is down. → 7.16

Model answer. M1 and M2 were created from the miRNA expression data by unsupervised clustering, which by construction puts samples with similar expression together and samples with different expression apart. Testing the same miRNA data for differences between those clusters asks a question whose answer was built in by the first step. Even pure noise, clustered into two groups, would give many “significant” miRNAs.

Implications: the p-values are not valid. They are biased toward zero because the null distribution assumed by the test does not account for the groups having been selected from the data, and the multiple-testing correction does not fix this. The list of differentially expressed miRNAs (76 up, 65 down) mainly describes what separates the clusters; it is not independent evidence of biological differences. It should be treated as descriptive or hypothesis-generating, and validated in an independent cohort (or analysed with methods that account for the selection). The associations with clinical variables (grade, Ki67, survival) are more informative, because those variables were not used to build the clusters. → 7.16b

Part 1: ✍️ Written exam

Everything in this part is on the written exam: 🎯 instructor EXAM slides (Lecture 7), 🎓 words and concepts on the professor’s list (Lectures 1–3), and 📌 AI-inferred topics (Lectures 1, 7, 8).

Calculation drills

🎯 Multiple testing (Lecture 7)

Benjamini–Hochberg and Bonferroni side by side. Do several: the “largest k” rule and the running minimum for adjusted p-values are where mistakes happen.

  1. Sort the p-values from smallest to largest; write the rank \(i\) next to each.
  2. Compute the threshold \(\frac{i}{m} Q\) for every rank.
  3. Find the largest \(i\) with \(p_{(i)} \le \frac{i}{m} Q\). Call it \(k\). Do not stop at the first failure.
  4. Everything with rank \(\le k\) is significant.
  5. Adjusted p: compute \(p_{(i)} \cdot m / i\), then go from the bottom up taking the running minimum. Cap at 1.
  6. Bonferroni for comparison: one threshold \(\alpha/m\) (or \(p \times m\)), same for all.
  7. State the interpretation: BH controls the FDR (expected share of false hits among hits); Bonferroni controls the FWER (probability of any false hit).

🎯 Preprocessing and moderated statistics (Lecture 7)

🎓 Genetics calculations (Lecture 3)

HWE and recombination frequency are on the professor’s list (“how to check for HWE” is under “uncertain about use”).

Written questions

Each question links to the section it comes from. Open the model answer only after you have tried.

🎯 Lecture 7: high-dimensional data

Model answer. In omics data we measure thousands of features on few replicates, e.g. 3 vs 3 samples, giving only \(n_1 + n_2 - 2 = 4\) degrees of freedom per feature. Each feature’s variance \(s^2\) is estimated separately and is therefore very unstable. By chance some features get a tiny \(s^2\); since \(s^2\) is in the denominator of \(t\), they get huge t-values although the mean difference is small → false positives. Others get a randomly large \(s^2\), masking real effects → lost power. This instability affects every p-value and so also undermines the multiple-testing correction.

The moderated t (limma, empirical Bayes) shares information across all features: each variance is shrunk toward a common variance \(s_0^2\) estimated from the whole dataset, \(\tilde s^2 = (d_0 s_0^2 + d\, s^2)/(d_0 + d)\). With few replicates the prior weight \(d_0\) dominates and the shrinkage is strong; with many replicates the feature’s own variance dominates. Result: fewer false positives, more power, more reliable FDR-adjusted results. → 7.7–7.8

Model answer. FWER = P(at least one false positive among all tests), controlled by Bonferroni (\(p \le \alpha/m\)). FDR = the expected proportion of false positives among all tests called significant, E[FP / positives], controlled by Benjamini–Hochberg.

For omics screens FDR/BH is preferred: (1) with thousands of tests Bonferroni’s threshold is so small that most true effects are missed (0.05/955 ≈ 0.00005); (2) the analysis is exploratory, producing a shortlist where a controlled fraction of false leads is acceptable; (3) hits are validated afterwards in an independent cohort/method (miR-375 by qRT-PCR and in situ hybridization); (4) features are correlated, so the effective number of tests is smaller than \(m\) and Bonferroni is needlessly strict. → 7.11–7.14

Model answer. \(955 \times 0.05 \approx 48\) false positives. Bonferroni: \(0.05/955 \approx 5.2 \times 10^{-5}\) (≈ 0.00005). With 955 independent null tests the chance of at least one false positive is \(1 - 0.95^{955} \approx 1\), so the cutoff itself must change. → 7.10

Model answer. Observed signal = biological signal + technical noise (RNA extraction, labelling, hybridization, scanner, batch). Steps: (1) quality control; (2) background correction within each array with normexp, which models observed = signal (exponential) + background (normal) and estimates the true signal, giving positive values and better low-intensity estimates; (3) normalization between arrays with quantile normalization: rank the values in each array, average across arrays at each rank, replace each value by its rank mean, so all arrays get an identical distribution while keeping each array’s internal order; (4) average replicate probes (4–8 spots) per miRNA. Assumption of quantile normalization: the overall expression distribution should be the same in every sample, so distribution differences are technical. → 7.4–7.5

Model answer. This is circular reasoning: the clusters were defined from the same miRNA data that was then tested. Clustering produces groups that differ in the clustering variables almost by construction, so significant differences are expected even in pure noise and the p-values are inflated. The differences are not independent evidence. Valid alternatives: test the cluster–clinical associations (grade, Ki67, chromosome 14 gain, which used other data), validate the clusters in an independent cohort, or use methods that account for the selection. → 7.16b

Model answer. \(2^{1.5} \approx 2.83\): expression is about 2.8 times higher in the first group. log₂FC = 0 means no change, +1 twice as high, −1 half as high. Log₂ transformation makes the distribution more symmetric and reduces the influence of very large raw values; on the log scale, fold changes become differences. → 7.9

Model answer. Two samples from the same patient are not independent. Including both would double-weight those patients and violate the independence assumption of the test, biasing variance estimates and p-values. → 7.9b

🎓 Lectures 1–3: biology and genetics

Model answer. Germline: arise in egg/sperm, present in every cell, heritable, detected in blood/saliva, cause inherited disorders and hereditary cancer syndromes. Somatic: acquired during life in non-reproductive cells, confined to the tissue/tumour, not heritable, need a biopsy of the tissue, drive cancers and benign tumours (e.g. AML, detected by targeted deep sequencing and VAF). → 3.4

Model answer. In the reference transcript NM_000999, version 3 (NM_ = mRNA/coding reference), c. position 451 of the coding DNA has C replaced by T. At the protein level (p.) arginine 151 becomes tryptophan: a single-base substitution that changes the amino acid, i.e. a missense variant (codon 151 spans c.451–453, so c.451 is its first base). rs123456789 would be its dbSNP identifier, the same across genome builds, whereas genomic positions (g., on an NC_ chromosome reference) differ between GRCh37 and GRCh38. p.Arg151Ter (also written p.Arg151*): the codon becomes a stop codon, a nonsense variant. p.Arg151=: the amino acid is unchanged, a silent variant. c.451del: one base deleted, which shifts the reading frame, a frameshift (written e.g. p.Arg151fs). → 2.4, 3.3b–3.3c

Model answer. miRNA: ~22 nt, genome-encoded, binds targets imperfectly, inhibits translation or promotes degradation, regulates many mRNAs (why it is widely dysregulated in cancer). siRNA: ~21–23 nt, from double-stranded RNA, binds with near-perfect complementarity, cleaves one specific mRNA, used as designed drugs. → 3.2

Model answer. Linkage: loci close on the same chromosome tend to be inherited together. Recombination frequency \(r\) = recombinant offspring / total, 0% (complete linkage) to 50% (unlinked). LD: non-random association of alleles at different loci in a population. Detected by comparing observed haplotype frequency with the expected product of allele frequencies (e.g. expected AB = 0.5 × 0.4 = 0.20); a difference indicates LD. Quantified with \(r^2\) (> 0.8 very strong; < 0.01 treated as independent). → 3.7

📌 Lecture 1: biomarkers

Model answer. Exposure (cotinine for tobacco), biological process (CRP for inflammation), disease/tissue injury (troponin for myocardial injury), future risk (LDL-C for atherosclerosis), treatment response (LDL-C change on lipid-lowering therapy). Being a biomarker does not imply causality. → 1.3b

📌 Lecture 8: proteomics

Model answer. Batch = plate, reagent lot, run, day, instrument, operator. Prevent (design): randomize samples across plates, balance cases and controls, technical replicates across plates, common reference samples on every plate. Detect: boxplots by plate, PCA, clustering, controls, replicates. Correct (analysis): normalization across plates/batches (incl. reference-sample normalization), adjust for plate/run or include batch as a covariate. Also consider pre-analytical variation (fasting, time to centrifugation, freeze–thaw, hemolysis) that may differ systematically between cases and controls. → 8.6

Model answer. Antibody-based (ELISA, immunoassays, Western blot): targeted, quantitative with standards, clinical routine. Mass spectrometry: trypsin digestion → LC → electrospray → masses; detects isoforms, allows discovery; complex, dynamic-range problems, labelled peptides for calibration. High-throughput affinity (Olink, SomaScan): thousands of proteins in large cohorts (UKBB: ~50,000 people, ~3,000 proteins), relative abundance in NPX (log₂; 1 NPX = doubling). → 8.4–8.5

Part 2: ❔ Exam status unclear

GWAS and MR theory that is not part of computer exercises 1–2, plus polygenic risk scores and protein MR. Several of these questions touch concepts on Martin’s key-concepts list (🔑): those concepts are exam-relevant; see Martin’s key concepts.

Calculation drills

❔ GWAS effect sizes (Lecture 4)

“How to check for association between SNP and phenotype” is on the professor’s list under “uncertain about use”.

❔ Polygenic risk scores (Lecture 9)

Written questions

❔ Lectures 4–6: GWAS and MR theory

Model answer. Vertical: the SNP affects the outcome through the exposure (SNP → PCSK9 → LDLR → LDL-C → CHD); this is the causal chain MR measures, not a problem. Horizontal: the SNP also affects the outcome through a pathway independent of the exposure (SNP → inflammation → CHD); this violates the exclusion restriction and biases the MR estimate, because part of the SNP–outcome association is wrongly attributed to the exposure. → 6.6

Model answer. Genotype is fixed at conception, so it precedes the disease: no reverse causation. Alleles are allocated randomly at meiosis, like treatment in an RCT, so they are (under the independence assumption) unrelated to confounders such as smoking. An allele that lowers LDL-C is a lifelong “natural trial” of lower LDL-C. MR is cheaper and possible where RCTs are not, but relies on the three instrument assumptions. → 6.3–6.4

Model answer. If allele frequencies and the phenotype both differ between ancestry groups, pooling them creates a spurious SNP–phenotype association even without any biological effect. Handling: restrict to genetically homogeneous ancestry groups; adjust for genetic principal components; use mixed models that account for relatedness (e.g. regenie); check the QQ plot and genomic inflation. → 4.9

Model answer. Strength, Consistency, Specificity, Temporality, Biological gradient (dose–response), Plausibility, Coherence, Experiment, Analogy. → 6.2

❔ Lecture 8: protein MR

Model answer. A pQTL is a SNP associated with a protein’s abundance. cis: near the encoding gene (mechanisms: transcription, splicing, stability, secretion). trans: far away, > 500 kb or another chromosome (mechanisms: transcription factors, receptors, signalling). cis-pQTLs are preferred because the link to the protein is direct and there is less risk of horizontal pleiotropy; they still need all three MR assumptions. Protein MR: \(\beta_{MR} = \beta_{GO}/\beta_{GP}\). → 8.9–8.10

Model answer. Horizontal pleiotropy (especially trans-pQTLs); assay binding artifacts (coding variants change the epitope, fake lower abundance); amount ≠ function (variants changing activity without abundance are invisible); LD creating false appearance of causality (different SNPs in LD for protein and disease), checked by colocalization; one pQTL affecting several proteins. → 8.11

❔ Lecture 9: two-sample MR theory and PRS

Model answer. The SNP → exposure effects come from one GWAS and the SNP → outcome effects from another, in different people. It needs only summary statistics per instrument: rsID, effect allele (and other allele, EAF for harmonisation), β and SE for the exposure, β and SE for the outcome; no individual-level data. All three MR assumptions still apply (relevance, independence, exclusion restriction); two-sample MR adds practical requirements: the SNPs must be found in both GWAS, the alleles harmonised, and the two samples comparable (ancestry, sex, age) but not overlapping. → 9.1–9.2

Model answer. Weak instruments with non-overlapping samples → bias toward the null. Sample overlap (people in both GWAS) → bias toward the confounded observational estimate. Winner’s curse (SNP effects overestimated in the discovery GWAS that selected them) → β_GX too large → estimate too small (toward the null); corrected by estimating effects in an independent replication sample. Different populations → associations may not transport; direction unpredictable. Also: horizontal pleiotropy is still possible, so report sensitivity analyses. → 9.8

Model answer. Genotype acts from conception, so MR estimates the effect of a lifelong difference in exposure; lifetime exposure builds risk. A treatment in adulthood changes the exposure for a shorter period, so its effect is usually smaller. MR is most reliable about the direction of a causal effect. → 9.7

Model answer. MR asks whether X causes Y (population level, a few valid instruments). A PRS asks who is at higher risk (individual level, many SNPs, weak and pleiotropic SNPs allowed). \(PRS_i = \sum_j G_{ij}\hat\beta_j\): allele dosage (0/1/2) times GWAS effect size, summed over SNPs. It needs a discovery (base) sample giving the β’s and an independent target sample with individual genotypes, phenotype and covariates; overlap between them inflates the apparent accuracy. → 9.9–9.11

Model answer. Classical clumping + thresholding (C+T, “pruning and thresholding”): keep the lead SNP per locus, include SNPs with P below a threshold, trying several thresholds and validating the choice out of sample. Alternatives from statistical learning: model the LD between SNPs instead of clumping, and use LASSO/ridge or Bayesian shrinkage with a train/test split, which pulls noisy large effects toward zero (counters winner’s curse). Weak SNPs are allowed because a PRS only needs to predict: each adds a little signal, like weak learners; validity as a causal instrument is not required. → 9.12

Model answer. A PRS has no unit; it is a relative, roughly normally distributed measure and does not translate directly into a probability of disease. It is usually less predictive than age, sex or SBP. Evaluate: discrimination (AUC; does it rise when the PRS is added?), calibration (do predicted risks match observed?), and added prediction: compare age + sex + smoking + SBP with the same model + PRS (incremental R² or AUC). A strong OR per SD does not guarantee good screening; reclassification measures are more clinically relevant, and accuracy drops in ancestries unlike the discovery GWAS. → 9.13

Part 3: 🗣️ Seminar topics (computer exercises 1–2)

Examined at the seminar, and seminar content can also come on the written exam (corrected 2 October). The lecturer’s own seminar questions, with model answers, are on Seminar questions. These cover what computer exercises 1 and 2 do: GWAS output, lead SNPs and clumping, matching summary statistics, harmonising alleles, and IVW, MR-Egger and weighted median. See also Computer exercises: concepts.

Calculation drills

Written questions

Model answer. 1. Relevance: the variant is robustly associated with the exposure (genome-wide significant). Threat: weak instruments, which bias and destabilize the estimate. 2. Independence: the variant is not associated with confounders of the exposure–outcome relationship. Threats: population stratification, assortative mating, selection effects. 3. Exclusion restriction: the variant affects the outcome only through the exposure. Threat: horizontal pleiotropy (another pathway, e.g. SNP → inflammation → CHD). → 6.5

Model answer. Wald ratio (one instrument): \(\beta_{\text{outcome}}/\beta_{\text{exposure}}\), change in outcome per genetically predicted unit of exposure (log-odds for binary outcomes). IVW: weighted regression of β-outcome on β-exposure through the origin = precision-weighted average of the Wald ratios; assumes all instruments valid. Weighted median: consistent if at least ~50% of the instrument weight is valid, so robust to a minority of bad instruments. MR-Egger: free intercept; a non-zero intercept indicates directional horizontal pleiotropy; less precise and more sensitive to weak instruments, relies on the InSIDE assumption. Trust grows when all methods point in the same direction, since their weaknesses differ. Always state the exposure unit and the direction. → 6.7–6.9

Model answer. Nearby SNPs are correlated (in LD) because they are inherited together, so one causal variant makes all SNPs in LD with it look associated. The locus is the region containing the associated signal; the lead SNP is the one with the smallest p-value there. It is not necessarily causal: it may be a proxy, and the causal variant may not even be on the array. Ways forward: conditional analysis, credible sets, functional annotation, protein folding models, pQTL evidence, experiments (knock-in mice). → 4.8, 5.3–5.4

Model answer. About 1 million independent tests: \(0.05/10^6 = 5 \times 10^{-8}\) (Bonferroni); at 0.05 you would get ~50,000 false positives. Significance means statistical robustness only. It says nothing about effect size (huge samples make tiny effects significant), causality or public-health relevance (rs11591147 has a big effect but is rare, 1.5%). → 4.7, 4.10

Model answer. Which allele is the effect allele (A1/EA); whether β is an OR or log(OR) for binary outcomes; the scale of the phenotype (log? SD units?); the genome build (GRCh37/hg19 vs GRCh38). When combining datasets also check SNP coverage, allele/strand alignment and ancestry of the LD reference panel. rsIDs help avoid mismatches. → 4.6, 5.6

Model answer. (1) Select instruments for X: P < 5×10⁻⁸, sufficiently strong (F = (β/SE)² > 10), independent (LD clumping). (2) Look up the same SNPs in the outcome GWAS (use a proxy SNP in high LD if one is missing). (3) Harmonise alleles so both β’s refer to the same effect allele; flip the sign when swapped, handle strand, exclude ambiguous palindromic SNPs. (4) Estimate: Wald ratio per SNP, IVW over SNPs. (5) Sensitivity analyses: weighted median, MR-Egger (intercept), leave-one-out, heterogeneity. (6) Interpret, stating the exposure unit. The slides stress that the difficulty lies in steps 1–3. → 9.3–9.6

Model answer. The outcome β is per G; per A it is +0.03. Wald ratio = 0.03 / 0.08 = 0.375: Y increases 0.375 units per unit of genetically predicted X. Without harmonising: −0.03 / 0.08 = −0.375, the opposite conclusion (X looks protective). → 9.5

Extra material

Not required reading; see Section 9.14.

Model answer. Each is the OR for CHD per 1 mmol/L genetically higher level, adjusted for the other exposure (multivariable MR, IVW). Per mmol/L remnant-C looks much more harmful, but the SDs differ (0.30 vs 0.82 mmol/L): per SD the ORs are 1.36 vs 1.29, so always compare on a stated unit. Per apoB particle, the lipolysis SNP cluster gave OR 1.76 vs 1.33 for the receptor cluster, suggesting remnants are more atherogenic per particle. Limitations: exposure and outcome largely from the same cohort (sample overlap, winner’s curse; addressed by a two-sample replication in CARDIoGRAMplusC4D); IVW assumes all 1125 SNPs valid (checked with MR-Egger, robust IVW, MR-Lasso); mainly European ancestry; remnant-C computed indirectly and remnant apoB not measured; MR reflects lifelong differences, not a drug effect. → 9.14