In a line of cherry tomatoes, the average fruit weight is 16 g. A plant producing tomatoes with an average weight of 12 g is used in one self-fertilization cross to produce a line of smaller tomatoes, and a plant producing tomatoes of 24 g is used in a second cross to produce larger tomatoes. If narrow sense heritability (h²) for this trait is 0.80, what are the expected responses to selection (R) for fruit weight in the crosses?
Table of contents
- 1. Introduction to Genetics51m
- 2. Mendel's Laws of Inheritance3h 37m
- 3. Extensions to Mendelian Inheritance2h 41m
- 4. Genetic Mapping and Linkage2h 28m
- 5. Genetics of Bacteria and Viruses1h 21m
- 6. Chromosomal Variation1h 48m
- 7. DNA and Chromosome Structure56m
- 8. DNA Replication1h 10m
- 9. Mitosis and Meiosis1h 34m
- 10. Transcription1h 0m
- 11. Translation58m
- 12. Gene Regulation in Prokaryotes1h 19m
- 13. Gene Regulation in Eukaryotes44m
- 14. Genetic Control of Development44m
- 15. Genomes and Genomics1h 50m
- 16. Transposable Elements47m
- 17. Mutation, Repair, and Recombination1h 6m
- 18. Molecular Genetic Tools19m
- 19. Cancer Genetics29m
- 20. Quantitative Genetics1h 26m
- 21. Population Genetics50m
- 22. Evolutionary Genetics29m
20. Quantitative Genetics
Heritability
Problem 19
Textbook Question
In a population of 100 inbred, genotypically identical rice plants, variance for grain yield is 4.67. What is the heritability for yield? Would you advise a rice breeder to improve yield in this strain of rice plants by selection?
Verified step by step guidance1
Understand the concept of heritability in the broad sense, which is the proportion of the total phenotypic variance that is due to genetic variance. It is given by the formula: \(H^2 = \frac{V_G}{V_P}\), where \(V_G\) is the genetic variance and \(V_P\) is the total phenotypic variance.
Note that the population consists of inbred, genotypically identical rice plants. Since all plants are genetically identical, the genetic variance \(V_G\) is effectively zero because there is no genetic variation among individuals.
Recognize that the observed variance in grain yield (4.67) must therefore be due to environmental variance \(V_E\), since \(V_P = V_G + V_E\) and \(V_G = 0\) in this case.
Calculate heritability using the formula \(H^2 = \frac{V_G}{V_P}\). Since \(V_G = 0\), heritability \(H^2\) will be zero, indicating no genetic contribution to the observed variance in yield.
Interpret the result: with zero heritability, selection for improved yield in this strain will not be effective because the variation in yield is not due to genetic differences. Advise the breeder accordingly.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Heritability in the Broad Sense
Heritability in the broad sense (H²) measures the proportion of total phenotypic variance in a trait that is due to genetic variance. It is calculated as the ratio of genetic variance to total phenotypic variance. In inbred, genotypically identical populations, genetic variance is minimal or zero, affecting heritability estimates.
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Phenotypic Variance and Genetic Variance
Phenotypic variance (Vp) is the total variation observed in a trait, arising from both genetic (Vg) and environmental (Ve) factors. In genetically identical populations, Vg is near zero, so Vp mainly reflects environmental variance. Understanding these components is essential to interpret heritability and predict response to selection.
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Selection and Breeding Implications
Selection effectiveness depends on heritability; high heritability means genetic factors strongly influence the trait, so selection can improve it. In populations with low or zero genetic variance, selection is unlikely to yield improvement, as observed variation is mostly environmental and not heritable.
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