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Ch. 4 - Extensions of Mendelian Genetics
Klug - Concepts of Genetics 12th Edition
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Tutti i libri di testoKlug 12th EditionCh. 4 - Extensions of Mendelian GeneticsProblema 29a
Capitolo 4, Problema 29a

In a cross in Drosophila involving the X-linked recessive eye mutation white and the autosomally linked recessive eye mutation sepia (resulting in a dark eye), predict the F₁ and F₂ results of crossing true-breeding parents of the following phenotypes: Note that white is epistatic to the expression of sepia.
white females x sepia males

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Step 1: Understand the genetic basis of the problem. The white eye mutation is X-linked and recessive, meaning it is located on the X chromosome and only expressed in the absence of a dominant allele. The sepia eye mutation is autosomal and recessive, meaning it is located on a non-sex chromosome and only expressed when both alleles are recessive. Additionally, white is epistatic to sepia, meaning the white mutation will mask the expression of sepia if present.
Step 2: Assign genotypes to the true-breeding parents. For the white-eyed female, her genotype is XᵂXᵂ (homozygous recessive for the white mutation). For the sepia-eyed male, his genotype is XᵂY (normal X chromosome and Y chromosome) and ss (homozygous recessive for the sepia mutation).
Step 3: Determine the F₁ generation. Perform a Punnett square for the X-linked white mutation and the autosomal sepia mutation. For the X-linked mutation, the female contributes Xᵂ, and the male contributes either X or Y. For the autosomal sepia mutation, the female contributes S (dominant allele), and the male contributes s (recessive allele). Combine these contributions to determine the genotypes and phenotypes of the F₁ offspring.
Step 4: Predict the phenotypes of the F₁ generation. All F₁ females will have the genotype XᵂX (heterozygous for white) and Ss (heterozygous for sepia), resulting in wild-type eyes because the dominant alleles mask the recessive mutations. All F₁ males will have the genotype XᵂY (hemizygous for white) and Ss (heterozygous for sepia), resulting in white eyes because the X-linked white mutation is expressed in males.
Step 5: Determine the F₂ generation. Cross the F₁ individuals (XᵂX Ss females with XᵂY Ss males). Use a Punnett square to calculate the probabilities of all possible combinations of X-linked and autosomal alleles. Consider the epistatic relationship where the white mutation masks the sepia mutation. Analyze the resulting genotypes to predict the phenotypic ratios in the F₂ generation.

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X-linked Inheritance

X-linked inheritance refers to genes located on the X chromosome. In Drosophila, males have one X and one Y chromosome, while females have two X chromosomes. This means that X-linked recessive traits, like the white eye mutation, will manifest in males if they inherit the affected X, while females require two copies of the recessive allele to express the trait.
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X-Inactivation

Epistasis

Epistasis is a genetic interaction where the expression of one gene is affected by another gene. In this case, the white eye mutation is epistatic to the sepia mutation, meaning that the presence of the white allele will mask the expression of the sepia allele, regardless of its genotype. This concept is crucial for predicting phenotypic outcomes in genetic crosses.
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Epistatic Genes

Punnett Square

A Punnett square is a diagram used to predict the genotypes and phenotypes of offspring from a genetic cross. By organizing the alleles of the parents, it allows for a visual representation of possible combinations in the F₁ and F₂ generations. This tool is essential for understanding inheritance patterns, especially when dealing with multiple alleles and epistatic interactions.
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Chi Square Analysis
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