Question

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 \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of crossing true-breeding parents of the following phenotypes: (a) white females \(\times\) sepia males (b) sepia females \(\times\) white males Note that white is epistatic to the expression of sepia.

Short answer

Question: Predict the F1 and F2 generation results of a cross between Drosophila with X-linked recessive white eye mutation and autosomally linked recessive sepia eye mutation, considering that white eye mutation is epistatic to sepia eye mutation. Answer: In cross (a), the F1 generation will have all white-eyed females but no F2 results can be predicted. In cross (b), the F1 generation consists of white-eyed females and sepia-eyed males, while the F2 generation will have 50% white-eyed and 50% sepia-eyed offspring for both males and females.

Step-by-step solution

(a) Cross white females × sepia males

Step 1: Determine the genotypes of the parents White female = XwXw (homozygous recessive white) Sepia male = XsY, ss (hemizygous sepia, homozygous recessive sepia) Step 2: Perform the cross and find F1 results Each offspring will inherit one X-linked gene from the mother and one autosomal gene from both parents. Offspring's genotypes: XwXs, sS Phenotypes for F1: All females will have white eyes, as white mutation is epistatic.

(b) Cross sepia females × white males

Step 1: Determine the genotype of the parents Sepia female = XX, ss (homozygous recessive sepia) White male = XwY (hemizygous recessive white) Step 2: Perform the cross and find F1 results Each offspring will inherit one X-linked gene from both parents and one autosomal gene from the mother. Offspring's genotypes: - Female: XsXw, sS (sepia and white double heterozygous) - Male: XsY, sS (hemizygous sepia) Phenotypes for F1: - All females will have white eyes, as white mutation is epistatic. - All males will have sepia eyes.

Predict F2 results for both crosses

For (a), all offspring in the F1 generation are XwXs, sS. Since no males are available for mating, it's not possible to predict the F2 generation for cross (a). For (b), the F1 generation consists of XsXw, sS females and XsY, sS males. By crossing F1 females with F1 males, we can analyze their combinations of genes: Offspring's genotypes: - Female: XsXs, sS (sepia heterozygous) - Female: XsXw, sS (white heterozygous) - Male: XsY, sS (sepia hemizygous) - Male: XwY, sS (white hemizygous) Phenotypes for F2: - Females: 50% white eyes (as white mutation is epistatic), 50% sepia eyes - Males: 50% white eyes (as white mutation is epistatic), 50% sepia eyes In conclusion, for cross (a), the F1 generation will have all white-eyed females but no F2 results can be predicted. For cross (b), the F1 generation consists of white-eyed females and sepia-eyed males, and the F2 generation will have 50% white-eyed and 50% sepia-eyed offspring for both males and females.

Key concepts

X-linked inheritance

X-linked inheritance is a genetic trait that is associated with genes located on the X chromosome. In Drosophila, or fruit flies, this form of inheritance is crucial for determining several phenotypes, such as eye color.

Males have only one X chromosome (XY), while females have two (XX). Therefore, males are more likely to express recessive traits linked to the X chromosome as they do not have a second X chromosome to counter a recessive gene's effects. This is seen in the exercise where a white-eyed trait in Drosophila is X-linked.

When we talk about white-eyed females in the exercise, they have the genotype XwXw, being homozygous recessive. On the other hand, sepia-eyed males are only hemizygous since they have the genotype XsY.

In a cross like (a), where white females are crossed with sepia males, the presence of an X-linked gene from the mother's side greatly influences the offspring's eye color phenotype.

Epistasis

Epistasis occurs when one gene affects or "masks" the expression of another gene. It's an important concept when discussing the inheritance of traits like eye color in Drosophila. In our exercise, the white mutation is epistatic to the sepia mutation. This means that if the white gene is present, it masks the expression of the sepia gene, regardless of the sepia gene's genotype.

For instance, in both cross (a) and (b), white eyes show up in some offspring even if they inherit a sepia gene because the white gene is epistatic. This masking effect is crucial for predicting phenotypes because it shows that not every gene directly translates to a trait due to the genetic interactions they may have.

Understanding epistasis helps in predicting outcomes in genetic crosses, making it evident why certain expected traits don't always appear, even when inherited.

Genotype prediction

Predicting genotypes requires understanding both the parental genotypes and the genetic principles such as X-linked inheritance and epistasis. This process often involves determining what alleles the offspring are likely to inherit from their parents.

In cross (a) from the exercise, we used the genotypes XwXw for white-eyed females and XsY, ss for sepia-eyed males. This combination lets us predict that all \(F_1\) females inherit the Xw gene, resulting in white eyes due to epistasis, while all males inherit Xs and show sepia eyes because they lack the second X-chromosome for the epistatic effect to take place.

For cross (b), the offspring range from white to sepia eyes in the F1 and F2 generations. By applying principles like segregation and independent assortment, we identify all potential genetic combinations in the offspring. Ultimately, predicting outcomes in genetic crosses involves translating genetic notation into phenotypic probabilities.