Textbook Question
Define and distinguish epistasis and pleiotropy.
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Ch. 4 - Gene Interaction
Chapter 4, Problem 4
In a cross of two pure-breeding lines of tomatoes producing different fruit sizes, the variance in grams (g) of fruit weight in the F₁ is 2.25 g and the variance among the F₂ is 5.40 g. Determine the genetic and environmental variance (VG and VE) for the trait and the broad sense heritability of the trait.
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Identify the given variances: F₁ variance (V_F1) is 2.25 g and F₂ variance (V_F2) is 5.40 g.
Understand that the variance in the F₁ generation (V_F1) is due to environmental variance (VE) because F₁ individuals are genetically identical.
Set VE equal to V_F1, so VE = 2.25 g.
Use the formula for F₂ variance: V_F2 = VG + VE, where VG is the genetic variance.
Solve for VG by rearranging the formula: VG = V_F2 - VE, substituting the known values to find VG.
Calculate the broad sense heritability (H^2) using the formula: H^2 = VG / V_F2.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Genetic Variance (VG)
Genetic variance (VG) refers to the portion of phenotypic variance in a trait that can be attributed to genetic differences among individuals. In the context of the tomato cross, VG can be calculated by considering the differences in fruit weight between the F₁ and F₂ generations, as these generations reflect the genetic contributions from the parent lines.
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Analyzing Trait Variance
Environmental Variance (VE)
Environmental variance (VE) is the component of phenotypic variance that arises from environmental factors affecting individuals, rather than genetic differences. In this tomato example, VE can be determined by subtracting VG from the total phenotypic variance observed in the F₂ generation, highlighting how much of the variance in fruit weight is due to external conditions.
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08:34Analyzing Trait Variance
Broad-Sense Heritability (H²)
Broad-sense heritability (H²) is a measure of the proportion of total phenotypic variance that is attributable to genetic variance. It is calculated as the ratio of VG to the total phenotypic variance (VP). In this case, understanding H² helps assess the extent to which genetic factors influence fruit weight in tomatoes, guiding breeding decisions.
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07:04Calculating Heritability
Related Practice
Textbook Question
In a test of his chromosome theory of heredity, Morgan crossed a female Drosophila with red eyes to a male with white eyes. The females were produced from Cross A shown in Figure 3.19. Predict the offspring Morgan would have expected under his hypothesis that the gene for eye color is on the X chromosome in fruit flies.
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Textbook Question
Compare and contrast broad sense heritability and narrow sense heritability, giving an example of each measurement and identifying how the measurement is used.
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Textbook Question
Describe the difference between continuous phenotypic variation and discontinuous variation. Explain how polygenic inheritance could be the basis of a trait showing continuous phenotypic variation. Explain how polygenic inheritance can be the basis of a threshold trait.
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Textbook Question
In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.
In a cross involving these three X-linked genes, do you expect any gene pair(s) to show genetic linkage? Explain your reasoning.
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Textbook Question
In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.
Do you expect any of these gene pair(s) to assort independently? Explain your reasoning.
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