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ICAR AICE-JRF/SRF Statistical Genomics: QTL Mapping, GWAS & Genomic Selection: practice questions

ICAR AICE-JRF/SRF - Agricultural Statistics, Computer Application, Bioinformatics (PhD admission and fellowship) · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
100written to this exam in this subject
10syllabus units

What this subject covers

Statistics for molecular-breeding data: genetic map construction, interval mapping of QTL, genome-wide association with mixed models, multiple-testing control, genomic prediction (GBLUP, ridge and Bayesian alphabet), and the analysis of high-throughput expression data.

In the exams: AICE-JRF Bioinformatics Unit 5 'Statistical Genomics' (heavy); AICE-JRF Ag Stat Unit 3 (present); AICE-JRF Genetics & Plant Breeding molecular units (present)

Where the 100 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Statistical Genomics: QTL Mapping, GWAS & Genomic Selection. Difficulty: 19 easy, 63 medium, 18 hard.

Linkage map construction and map functions11
QTL mapping: single-marker, interval and composite interval mapping11
Genomic selection: ridge regression BLUP and Bayesian methods (BayesA, BayesB)11
Genome-wide association studies with mixed models10
Genomic relationship matrices and GBLUP10
Gene expression and RNA-seq differential analysis10
Statistical genomics descriptive and inferential basics10
Molecular markers and genotype data quality9
Linkage disequilibrium and population structure9
Multiple testing in genomics: Bonferroni and FDR9

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Molecular markers and genotype data quality · hard

An SSR locus has three alleles at frequencies 0.5, 0.3 and 0.2. What is its expected heterozygosity?

  1. 0.38
  2. 0.55
  3. 0.50
  4. 0.62
Answer and working

D. 0.62

Hₑ = 1 − Σpᵢ² = 1 − (0.25 + 0.09 + 0.04) = 0.62. 0.38 is Σpᵢ² itself; 0.55 is close to the PIC, which subtracts a further term.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

2 · QTL mapping: single-marker, interval and composite interval mapping · hard

How does Haley-Knott regression approximate interval mapping?

  1. Regression on expected QTL genotypes
  2. It replaces the trait values by their ranks and runs a rank test
  3. It fits a Bayesian model with an unknown number of QTLs
  4. It tests only at marker positions and interpolates LOD between them
Answer and working

A. Regression on expected QTL genotypes

Using expected genotype values in a least-squares regression is fast and closely matches maximum likelihood except where markers are sparse or effects are large.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

3 · Linkage disequilibrium and population structure · medium

Which LD measure most directly determines the power to detect a causal variant through a nearby marker?

  1. D
  2. r²
  3. D′
  4. F_ST
Answer and working

B. r²

The sample size needed through a marker is roughly 1/r² times that needed with the causal variant itself. D′ can be 1 even when r² is small.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

4 · Linkage disequilibrium and population structure · medium

How does a biparental population avoid the confounding by structure seen in diversity panels?

  1. It contains more alleles at each locus than a diversity panel does
  2. All individuals share the same two parents
  3. It removes linkage disequilibrium between all markers
  4. It requires no phenotypes for mapping
Answer and working

B. All individuals share the same two parents

Designed crosses have no structure by construction, but they sample only two alleles and have extended LD, so resolution is coarser.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

5 · Multiple testing in genomics: Bonferroni and FDR · easy

A GWAS tests 500,000 SNPs. What is the Bonferroni per-test threshold for a family-wise error rate of 0.05?

  1. 5 × 10⁻⁸
  2. 0.05
  3. 1 × 10⁻⁷
  4. 1 × 10⁻⁵
Answer and working

C. 1 × 10⁻⁷

α/m = 0.05/500,000 = 1 × 10⁻⁷, i.e. −log₁₀ p = 7. 5 × 10⁻⁸ is the conventional threshold in human GWAS for about a million independent tests.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

6 · Genomic relationship matrices and GBLUP · medium

What does G capture that the pedigree relationship matrix A does not?

  1. Only the expected relationship from recorded parents
  2. The environmental covariance between animals
  3. Dominance effects of every gene
  4. Realised, not expected, DNA sharing
Answer and working

D. Realised, not expected, DNA sharing

A gives expectations (full sibs 0.5); G measures actual sharing, which varies around 0.5, and captures unrecorded relationships.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

7 · Genomic relationship matrices and GBLUP · medium

Using r = √[Nh²/(Nh² + M_e)], what is the expected accuracy of genomic prediction with a reference of 1,000 animals, h² = 0.5 and M_e = 1,000?

  1. 0.58
  2. 0.33
  3. 0.71
  4. 0.50
Answer and working

A. 0.58

r = √(500/1,500) = √0.333 = 0.58. 0.33 forgets the square root.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

8 · Genomic selection: ridge regression BLUP and Bayesian methods (BayesA, BayesB) · hard

A dairy programme has i = 2.0 and σ_A = 10 units. Progeny testing gives accuracy 0.85 and generation interval 6 years; genomic selection gives accuracy 0.70 and interval 2.5 years. What is the annual genetic gain under each?

  1. Progeny testing 2.83; genomic 5.60 units per year
  2. Progeny testing 17.00; genomic 14.00 units per year
  3. Progeny testing 5.60; genomic 2.83 units per year
  4. Progeny testing 2.83; genomic 2.83 units per year
Answer and working

A. Progeny testing 2.83; genomic 5.60 units per year

ΔG = i r σ_A/L: progeny testing 2 × 0.85 × 10/6 = 2.83; genomic 2 × 0.70 × 10/2.5 = 5.60. Despite lower accuracy, GS nearly doubles gain through the shorter interval. Ignoring L gives 17 and 14.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

9 · Genomic selection: ridge regression BLUP and Bayesian methods (BayesA, BayesB) · medium

Why can random k-fold cross-validation overstate the accuracy of genomic prediction for new breeding families?

  1. Random folds always contain too few markers
  2. Relatives span training and validation folds
  3. It uses the same marker effects for all individuals
  4. It ignores the heritability of the trait
Answer and working

B. Relatives span training and validation folds

Predictions of close relatives are easy; new families are less related to the training set. Forward or leave-family-out validation mimics actual use.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

10 · Statistical genomics descriptive and inferential basics · medium

What is the purpose of the bootstrap in QTL mapping?

  1. To set genome-wide significance thresholds by shuffling phenotypes
  2. To impute missing marker genotypes
  3. Confidence intervals for QTL position
  4. To order markers on a linkage group
Answer and working

C. Confidence intervals for QTL position

Resampling individuals with replacement and re-scanning gives an empirical distribution of the peak position. Permutation, not bootstrap, gives thresholds.

Further reading: Lynch and Walsh, Genetics and Analysis of Quantitative Traits; Isik, Holland and Maltecca, Genetic Data Analysis for Plant and Animal Breeding; original papers cited in the manual

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Statistical Genomics: QTL Mapping, GWAS & Genomic SelectionThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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