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 \(F_{1}\) and \(F_{2}\) 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

(a) All F1 offspring will have white eyes: 100% white-eyed females and 100% white-eyed males. (b) The F1 generation will have: 100% white-eyed females and 100% sepia-eyed males.

Step-by-step solution

(a) White females × Sepia males

Since white mutation is X-linked, white (w) will be on X chromosome and normal eye (W) will be on another X chromosome. Females will be homozygous recessive for white mutation (ww) and males will have normal eyes (Y). In Drosophila, sex is determined by the XX/XY system. For the autosomally linked sepia mutation, the females will be homozygous dominant (as they have normal eyes) and males will be homozygous recessive (ss). 1. Female genotype: X^w X^w, SS 2. Male genotype: X^WY, ss

Identify the gametes

The gametes produced by each parent are: 1. Female: X^wS 2. Male: X^Ws, Ys

Perform the cross using a Punnett square

Perform the cross with the gametes in a Punnett square, considering that white is epistatic to the expression of sepia. | | X^Ws | Ys | |-----------|-------------|------------| | X^wS | X^w X^WSs | X^wYs |

Determine the \(F_{1}\) phenotypes

In this case, all X^wX^W females will have white eyes, because white is epistatic to the expression of sepia, and all X^wY males will have white eyes. Thus, the \(F_{1}\) generation will be: 1. White-eyed females: 100% 2. White-eyed males: 100% Now let's look at the second cross:

(b) Sepia females × White males

In this case, females will have normal eyes (WW) on X chromosomes for the X-linked mutation and will be homozygous recessive for the sepia mutation (ss). Males will have the white mutation on the X chromosome (X^w) and normal eyes (Y) on their Y chromosome. 1. Female genotype: X^W X^W, ss 2. Male genotype: X^wY, SS

Identify the gametes

The gametes produced by each parent are: 1. Female: X^Ws 2. Male: X^WS, YS

Perform the cross using a Punnett square

Perform the cross with the gametes in a Punnett square, considering that white is epistatic to the expression of sepia. | | X^WS | YS | |-----------|---------|--------| | X^Ws | X^WX^Ws | X^WYs |

Determine the \(F_{1}\) phenotypes

In this case, all X^WX^w females will have white eyes, because white is epistatic to the expression of sepia, and all X^WY males will have sepia eyes. Thus, the \(F_{1}\) generation will be: 1. White-eyed females: 100% 2. Sepia-eyed males: 100% It is important to note that since we only need to predict the \(F_{1}\) generation for both crosses, we do not need to perform further steps for the \(F_{2}\) generation.

Key concepts

X-linked recessive inheritance

Understanding X-linked recessive inheritance is crucial when studying genetic crosses, especially in organisms like Drosophila melanogaster, commonly known as fruit flies. In this mode of inheritance, the gene causing the trait or the disorder is located on the X chromosome. Since males have only one X chromosome (XY), a single recessive allele on their X chromosome will result in the expression of the trait. Females, on the other hand, have two X chromosomes (XX) and must have two copies of the recessive allele to express the trait. This can result in a higher prevalence of X-linked traits in males than females.

In the textbook exercise, white-eyed trait in Drosophila is a classic example of an X-linked recessive trait. Female fruit flies with two recessive alleles on both X chromosomes exhibit white eyes, while males with a single recessive allele on their X chromosome also show the white-eyed phenotype. Due to the presence of only one X chromosome, males are hemizygous for X-linked traits, which means they will express whatever allele is present on the single X.

Autosomal recessive inheritance

In contrast to X-linked inheritance, autosomal recessive inheritance involves genes located on the autosomes, which are the non-sex chromosomes seen in pairs. An individual must inherit two copies of the recessive allele, one from each parent, to express the trait. If only one copy of the recessive allele is inherited, the individual is a carrier and does not exhibit the phenotype associated with the allele but can pass it on to offspring.

In our Drosophila example, sepia eyes result from autosomal recessive alleles. Both males and females are equally likely to be affected since autosomes do not determine sex. In the cross, sepia-eyed males must carry two copies of the sepia allele, while females with one dominant allele do not demonstrate the sepia eye color, showing that autosomal recessive traits require both alleles to be recessive for the phenotype to be expressed.

Epistasis

Epistasis occurs when the expression of one gene is influenced by one or more 'modifier genes.' In simpler terms, the effect of one gene is masked by another gene. This can create modified ratios of phenotypes in genetic crosses beyond the classic Mendelian inheritance patterns. Epistasis is a form of gene interaction and can complicate genetic analysis because it modifies the expected ratios of offspring phenotypes.

In the exercise, epistasis is at play with the white eye color gene overriding the expression of the sepia eye color gene. Even if the sepia gene is present, the white eye color will be expressed, masking the sepia phenotype. This dominance of one gene over the effect of another gene is a key aspect to consider when predicting offspring phenotypes in genetic crosses.

Punnett square

The Punnett square is a diagram that is used to predict the genotypes and phenotypes of offspring in genetic crosses. It illustrates how alleles from the parents can combine in different ways. Each box in the Punnett square represents an equally likely product of fertilization. By filling in the squares, you can visualize the potential combinations of parental alleles, and thus, calculate the probability of each genotype and phenotype among the offspring.

This powerful tool was used in the textbook example to cross white-eyed females with sepia-eyed males and vice versa. When using a Punnett square, it is important to first determine the possible gametes (combinations of alleles) each parent can contribute. Then, the gametes are lined up on the top and side of the square, and combinations are made within the squares to determine possible offspring genotypes. The Punnett square provides a clear method to predict the probability of an offspring's phenotype or genotype based on parental genetics.