1. What distinguishes genomic selection (GS) from traditional marker‑assisted selection (MAS)?
A. GS uses only a few markers linked to major QTLs
B. GS estimates effects of all genome-wide markers simultaneously
C. GS requires no genotype data
D. GS ignores linkage disequilibrium
Answer: B
Rationale: Unlike MAS, GS models estimate all marker/haplotype effects across the genome to calculate genomic breeding values (GEBVs), capturing small‑effect loci too (Wikipedia)
2. The key output of a genomic selection model is:
A. Estimated SNP allele frequency
B. Genomic estimated breeding value (GEBV)
C. Traditional EBV from pedigree
D. Genome‑wide linkage maps
Answer: B
Rationale: GEBV is the predicted genetic merit based on the genomic model that integrates marker effects (IntechOpen, Wikipedia)
3. A major advantage of GS is its ability to:
A. Eliminate phenotyping completely
B. Predict performance of un‑phenotyped lines from genotypes
C. Operate without training data
D. Only work for simple Mendelian traits
Answer: B
Rationale: GS allows selection of candidates based solely on genotypes if they are not phenotyped using a trained model (IntechOpen, Frontiers)
4. What factor strongly increases GEBV accuracy when marker density is low (<25 K SNPs), particularly in GBLUP models?
A. Effective population size
B. Marker density itself
C. Environmental variance
D. Trait dominance effects
Answer: B
Rationale: Accuracy rises sharply with increasing marker density under GBLUP methods until a plateau, especially when starting from low densities (Frontiers, pmc.ncbi.nlm.nih.gov)
5. Increasing marker density beyond a certain point may reduce prediction accuracy for traits controlled by a single major gene. This is because:
A. GBLUP fails to estimate variance
B. Overfitting reduces accuracy when the QTN is in the genotype data
C. MAF filtering removes causal SNPs
D. LD decays too fast
Answer: B
Rationale: When the major QTL allele is included, excess markers dilute its effect and can reduce accuracy (Frontiers, pmc.ncbi.nlm.nih.gov)
6. Which statistical model is often equivalent to GBLUP in marker‑effect estimation?
A. BayesB
B. Random Forest
C. Ridge‑Regression BLUP (RR‑BLUP)
D. Bayesian Variable Selection
Answer: C
Rationale: RR‑BLUP and GBLUP are mathematically equivalent in assuming all markers contribute equally to variance (Frontiers)
7. BayesR outperforms GBLUP when:
A. Predicting low‑heritability traits controlled by many small QTLs
B. Predicting high‑heritability traits controlled by one or few major genes
C. Marker density is very low (<1 K)
D. Non-additive dominance effects are the only variance source
Answer: B
Rationale: BayesR models mixture of marker effects and suits traits with strong major-QTL effects (Frontiers)
8. What is the relationship between heritability and prediction accuracy in GS?
A. Higher heritability → lower accuracy
B. Only depends on marker density
C. Higher heritability generally yields higher accuracy
D. Inverse correlation always
Answer: C
Rationale: Traits with higher h² result in more reliable GEBVs under most GS models (pmc.ncbi.nlm.nih.gov, Frontiers)
9. What is the purpose of cross‑validation (e.g. five‑fold CV) in genomic selection?
A. Genotype filtering
B. Estimate GEBV accuracy (e.g. r[GEBV, EBV] or r[GEBV, TBV])
C. Genotype imputation
D. Increase marker density artificially
Answer: B
Rationale: Cross‑validation partitions TP to test model accuracy before applying it to BP (IntechOpen)
10. According to Schaeffer (cited in GS literature), a GEBV accuracy of 0.75 in cattle could:
A. Halve genetic gain
B. Cut costs by 92% and double genetic gain
C. Double inbreeding rate
D. Replace all phenotyping permanently
Answer: B
Rationale: High GEBV accuracy greatly reduces sampling cost and speeds up selection response (IntechOpen)
11. The first official USDA genomic evaluations for Holstein dairy cattle were released in:
A. 2001
B. January 2009
C. 2013
D. 2016
Answer: B
Rationale: USDA issued first genomic EBVs for Holsteins and Jerseys in January 2009 (pubmed.ncbi.nlm.nih.gov, annualreviews.org)
12. The impact of genomic selection in dairy cattle includes all EXCEPT:
A. Doubling rate of genetic progress
B. Reducing generation interval
C. Identifying recessive lethals
D. Eliminating the need for SNP chips
Answer: D
Rationale: GS relies on SNP chips; it reduces need for progeny testing not markers (pubmed.ncbi.nlm.nih.gov)
13. Which termite of LD (linkage disequilibrium) is crucial for genomic prediction?
A. Between different breeding programs
B. Between any random individuals
C. Between marker loci and QTLs in TP and BP
D. Only between pedigree relations
Answer: C
Rationale: GS depends on LD structure; high LD between markers and QTL across TP and BP improves prediction (IntechOpen, pmc.ncbi.nlm.nih.gov)
14. Genomic selection is especially effective for:
A. Major gene resistance traits
B. Traits with simple inheritance
C. Polygenic traits controlled by many small‑effect QTLs
D. Traits with no heritable components
Answer: C
Rationale: GS captures small‑effect loci better than MAS and is ideal for complex traits like yield or drought tolerance (IntechOpen, Frontiers)
15. Genotyping‑by‑sequencing (GBS) is favored in GS because it:
A. Requires no reference genome
B. Combines SNP discovery and genotyping at reduced cost
C. Generates data only for known QTLs
D. Cannot be used for large genome species
Answer: B
Rationale: GBS allows high-density genotyping and discovery simultaneously, often without a reference genome → cost efficiency for GS (IntechOpen)
16. The minimum marker density needed for stable prediction accuracy in a rabbit GS study was about:
A. 5 K SNPs
B. 10 K
C. 50 K
D. 250 K
Answer: C
Rationale: A meat‑rabbit study found prediction and heritability estimation stabilised at ≈50 K marker density (pmc.ncbi.nlm.nih.gov)
17. One potential way to reduce genomic prediction cost without losing much accuracy is:
A. Using fewer individuals in TP
B. Using pedigree rather than genotype data
C. Applying feature selection (e.g. Markov blankets) to drop non‑informative markers
D. Ignoring cross‑validation
Answer: C
Rationale: Feature selection can streamline GS models and reduce cost with minimal accuracy loss (arxiv.org)
18. GS can help in hybrid breeding by:
A. Obviating need for tester lines
B. Predicting hybrid performance using parental genotypes
C. Needing to test all cross combinations in field
D. Removing heterosis
Answer: B
Rationale: GS speeds up hybrid breeding by predicting combining ability and reducing full hybrid testing (IntechOpen)
19. Which of the following is NOT a limitation of genomic selection?
A. Requires continuous model retraining with changing allele frequencies
B. Needs high infrastructure for genotyping
C. Always leads to increased inbreeding without management
D. Depends on phenotype data at all cycles
Answer: D
Rationale: GS reduces phenotyping in selection cycles but still needs initial phenotypes for model training and occasional updates (IntechOpen)
20. LD decay and effective population size (Ne) influence GS because:
A. Smaller Ne requires fewer markers
B. Faster LD decay means markers must be denser to capture QTL linkages
C. LD is irrelevant if pedigree is known
D. GS works better if LD decays faster
Answer: B
Rationale: Marker density must scale with Ne and LD decay rate to maintain accuracy (pmc.ncbi.nlm.nih.gov, IntechOpen)
21. In wheat GS studies, inclusion of multiple previous breeding cohorts in training:
A. Decreased prediction accuracy
B. Increased accuracy, especially when target BP is less related
C. Made no difference
D. Prevented model convergence
Answer: B
Rationale: Using several prior cohorts improves accuracy across diverse validation sets (pmc.ncbi.nlm.nih.gov)
22. When minor allele frequency (MAF) is too low, which method suffers more accuracy loss?
A. GBLUP
B. BayesR
C. Pedigree BLUP
D. Additive index GS
Answer: B
Rationale: BayesR's convergence and accuracy are sensitive to presence of many low‑MAF SNPs (Frontiers)
23. GS helps manage inbreeding because:
A. It always reduces genetic variation
B. It avoids all crosses between relatives
C. Models can track relationships from genomic data, allowing control
D. It ignores relationships
Answer: C
Rationale: GS enables breeders to monitor and manage inbreeding using genomic relationship matrices (IntechOpen, pmc.ncbi.nlm.nih.gov)
24. The correlation r(GEBV, TBV) is typically estimated via:
A. Direct phenotypes alone
B. r(GEBV, EBV) × sqrt(h²)
C. Pedigree relationships
D. Linkage map distance
Answer: B
Rationale: r(GEBV, TBV) ≈ r(GEBV, EBV) ÷ sqrt(r(EBV, TBV)), and EBV is usually √h² of phenotype (IntechOpen)
25. In practical GS programs, TP size:
A. Should be as small as possible
B. Affects prediction accuracy significantly—larger TP increases accuracy
C. Has no relation to heritability
D. Should always contain only full-sib families
Answer: B
Rationale: Accuracy scales with TP size, particularly for low‑heritability traits (pmc.ncbi.nlm.nih.gov, IntechOpen)
26. Which platform allows genotype imputation and cost savings in species with no SNP chip?
A. Phenotyping arrays
B. GBS or low-coverage sequencing with imputation
C. QTL‑specific markers
D. No genotype data needed
Answer: B
Rationale: Low‑coverage sequencing plus imputation enables high-density genotype estimates affordably (pmc.ncbi.nlm.nih.gov)
27. Implementation of GS in honey bees showed:
A. No improvement over pedigree-based methods
B. GEBV improved predictive ability for honey yield and work traits compared to pedigree-only models
C. Genomic data reduced predictive ability
D. GS isn’t feasible in invertebrates
Answer: B
Rationale: A honey‑bee study found increased predictive ability using GS for several traits versus pedigree models (arxiv.org)
28. Feature‑selection approaches like Markov blanket allow GS to:
A. Increase marker density arbitrarily
B. Drop uninformative markers and maintain accuracy
C. Replace phenotypes
D. Remove dominance effects
Answer: B
Rationale: Reducing marker set saves cost and simplifies models without harming prediction (arxiv.org)
29. Which challenge of GS becomes acute as generations pass since TP was trained?
A. Generation interval increases
B. Predictor markers lose LD with QTL—lower accuracy
C. Cost of genotyping decreases
D. Heritability becomes infinite
Answer: B
Rationale: Recombination and allele-frequency changes erode prediction accuracy over generations if not refreshed (IntechOpen)
30. GS adoption in maize/wheat breeding helps primarily to:
A. Avoid field trials entirely
B. Skip hybrid testing
C. Shorten breeding cycles and increase rate of gain per year
D. Identify only major QTL
Answer: C
Rationale: GS greatly accelerates selection by reducing time between cycles without phenotyping every candidate (IntechOpen, Frontiers)
31. Multi-trait GS indices are beneficial because:
A. Each trait must be predicted by separate models
B. They integrate GEBVs across traits in selection decisions
C. They do not require marker data
D. Only usable in animals, not plants
Answer: B
Rationale: Selection indices can combine multiple trait GEBVs for optimizing overall breeding goals (IntechOpen)
32. GS is less likely to be useful when:
A. Phenotyping is expensive and slow
B. Trait heritability is extremely low (<0.1) unless TP is large
C. Genotyping cost is minimal
D. LD structure is strong
Answer: B
Rationale: Very low heritability traits require large TP for meaningful GEBV accuracy (pmc.ncbi.nlm.nih.gov, IntechOpen)
33. In hybrid maize breeding, GS can:
A. Select elite hybrid directly without parents
B. Predict combining ability to identify promising parental crosses
C. Replace genetic diversity
D. Ignore inbreeding depression
Answer: B
Rationale: GS models can screen multiple cross-combinations and eliminate many lines early (IntechOpen)
34. A common way to build the genomic relationship matrix (GRM) for GBLUP is:
A. Mead’s algorithm
B. VanRaden method using centered SNP matrix
C. Pedigree-based IBS calculations
D. QTL linkage-only models
Answer: B
Rationale: VanRaden’s method calculates GRM from marker genotypes and allele frequencies (pmc.ncbi.nlm.nih.gov, Frontiers)
35. Which machine‑learning method can capture non-additive interactions in GS?
A. Ridge‑regression
B. Linear GBLUP
C. Support‑Vector Machines or Random Forests
D. Pedigree BLUP
Answer: C
Rationale: Kernel and ML methods (linking epistasis, dominance, non-linear patterns) can improve accuracy for complex architectures (Frontiers, IntechOpen)
36. For which scenario would BayesR converging be more problematic?
A. Low marker density (e.g., <50 K)
B. High marker density (e.g., >200 K)
C. Traits with moderate heritability controlled by many small QTLs
D. GBLUP equivalent situations
Answer: B
Rationale: BayesR requires more MCMC iterations and may fail to converge with very high SNP counts (Frontiers)
37. Which policy reduces risk of reduced long-term response in GS?
A. Stop using GS after first cycle
B. Use only major gene markers
C. Re-estimate model every cycle and infuse new germplasm
D. Avoid updating the TP
Answer: C
Rationale: Update GS models each generation and incorporate new lines to maintain LD‑QTL relationships (IntechOpen)
38. In GS, the breeder's equation is modified by:
A. Excluding heritability
B. Replacing phenotypic accuracy with GEBV accuracy
C. Using only phenotype data
D. Ignoring selection intensity
Answer: B
Rationale: Response per cycle depends on selection intensity, genetic variance, and accuracy – replaced by r(GEBV, TBV) in GS (IntechOpen)
39. Estimating dominance effects in GS is attempted with:
A. GBLUP‑AD models including dominance GRM
B. BayesA only models
C. MAS
D. No models support dominance
Answer: A
Rationale: Some GS variants include additive‑dominance relationship matrices to predict non-additive genetic variance (pmc.ncbi.nlm.nih.gov, Frontiers)
40. Annual genetic gain per year improves under GS mainly because:
A. Marker density increases annually
B. Phenotyping costs are negligible
C. Generation intervals are shortened
D. Pedigree information is eliminated
Answer: C
Rationale: GS reduces time per selection cycle, accelerating genetic progress per unit time (IntechOpen, Frontiers)
41. A training population that is genetically unrepresentative of the breeding population will:
A. Increase GEBV accuracy
B. Not affect predictions
C. Lead to reduced prediction accuracy
D. Only benefit low heritability traits
Answer: C
Rationale: Relatedness between TP and BP strongly affects accuracy – less relation → lower accuracy (pmc.ncbi.nlm.nih.gov, IntechOpen)
42. What happens if TP and BP share low LD structure?
A. GS still works well
B. Accuracy drops if marker‑QTL LD differs between them
C. Heritability increases
D. Cross-validation accuracy always remains high
Answer: B
Rationale: Mismatched LD patterns across TP and BP reduce GEBV reliability (IntechOpen)
43. How does GS aid in germplasm pre‑breeding?
A. Replaces all phenotyping of landraces
B. Filters diverse accessions by predicting their GEBVs for traits of interest
C. Eliminates hybrids testing
D. Reduces marker density requirement
Answer: B
Rationale: GS allows selecting valuable accessions from gene banks without phenotyping every line (IntechOpen)
44. Reliability of GEBV in new environments is best improved by:
A. Using a TP phenotyped only in one environment
B. Ignoring genotype×environment interactions
C. Including G×E in model via multi‑environment phenotyping
D. Training only on parent lines
Answer: C
Rationale: Multi‑environment TP phenotypes help account for G×E and improve GEBV portability (IntechOpen)
45. Which model remains widely used in practice due to efficiency and robust performance?
A. BayesR
B. GBLUP
C. SVM
D. Single-step BLUP with no marker data
Answer: B
Rationale: GBLUP remains the most popular due to its speed, simplicity, and decent performance across architectures (Frontiers, pmc.ncbi.nlm.nih.gov)
46. Why might breeders choose moderate (≈25 K) SNP density for GS?
A. It’s the minimum needed for any trait
B. It balances prediction accuracy with computation cost
C. It eliminates Markov blanket necessity
D. It always gives highest accuracy
Answer: B
Rationale: ~25 K SNP density often offers stable accuracy with manageable computing and cost (Frontiers, pmc.ncbi.nlm.nih.gov)
47. GS allows shifting phenotyping emphasis to:
A. Early generation nurseries for candidate line selection
B. Phenotyping only final varieties
C. Removing field testing entirely
D. Phenotyping after selection decisions
Answer: A
Rationale: Phenotyping resources are reallocated toward training TPs and validation sets rather than screening every candidate (IntechOpen)
48. One reason why GS is less common in public sector breeding (especially in developing countries) is:
A. Lack of complex traits
B. High cost of genotyping infrastructure
C. Overwhelming marker density requirements
D. Pedigree records are mandatory
Answer: B
Rationale: GS requires computing infrastructure and genotyping capacity, often beyond reach of smaller programs (IntechOpen)
49. To mitigate biased GEBV due to inflated marker effect estimates, breeders can:
A. Use only low‑density SNPs
B. Update the model in each cycle and adequately size TP
C. Ignore GEBV and continue phenotypes only
D. Use MAS instead
Answer: B
Rationale: Regular retraining and sufficient TP helps avoid overestimation and maintain accuracy (IntechOpen)
50. A refinery key reason why GS has revolutionized crop and animal breeding is:
A. It removed need for any breeding expertise
B. It allows breeders to skip field research entirely
C. It combines high-coverage markers with prediction models to shorten cycles and capture polygenic effects
D. It works only for disease resistance traits
Answer: C
Rationale: GS leverages dense genomic data with statistical models to drive faster and more accurate breeding, especially for complex traits (Wikipedia, Frontiers)
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