Answer :
Let’s analyze the inheritance of the bar eye trait in fruit flies. We are given that bar eye is a dominant X-linked trait, denoted by B, and non-bar eye is the recessive condition, denoted by b.
A dominant X-linked trait means that the gene causing the trait is located on the X chromosome, and only one copy of the dominant allele is required to express the trait. We denote the X chromosomes carrying the dominant allele as X^B and those carrying the recessive allele as X^b.
Here's the step-by-step explanation:
1. Determine the Genotypes of the Parents:
- The heterozygous bar-eyed female fly will have the genotype X^BX^b.
- The non-bar-eyed male fly will have the genotype X^bY.
2. Set Up the Punnett Square:
Create a Punnett square to predict the possible genotypes of the offspring from the cross of X^BX^b (female) with X^bY (male).
The configuration of the Punnett square will be:
```
X^B X^b
X^b | X^BX^b | X^bX^b
Y | X^BY | X^bY
```
3. Analyze the Punnett Square:
- First row:
- X^BX^b: Offspring will be a bar-eyed female, having one X^B from the mother and one X^b from the father.
- X^bX^b: Offspring will be a non-bar-eyed female, having two X^b, one from each parent.
- Second row:
- X^BY: Offspring will be a bar-eyed male, having one X^B from the mother and Y from the father.
- X^bY: Offspring will be a non-bar-eyed male, having one X^b from the mother and Y from the father.
4. Determine the Phenotypic Ratio:
- One female offspring will have the genotype X^BX^b (bar-eyed).
- One female offspring will have the genotype X^bX^b (non-bar-eyed).
- One male offspring will have the genotype X^BY (bar-eyed).
- One male offspring will have the genotype X^bY (non-bar-eyed).
5. Expected Ratio:
Given the analysis of the Punnett square, the expected ratio of phenotypes among the four offspring is:
- One bar-eyed female.
- One non-bar-eyed female.
- One bar-eyed male.
- One non-bar-eyed male.
Therefore, the correct answer is:
- One bar-eyed and one non-bar-eyed female, one bar-eyed and one non-bar-eyed male
A dominant X-linked trait means that the gene causing the trait is located on the X chromosome, and only one copy of the dominant allele is required to express the trait. We denote the X chromosomes carrying the dominant allele as X^B and those carrying the recessive allele as X^b.
Here's the step-by-step explanation:
1. Determine the Genotypes of the Parents:
- The heterozygous bar-eyed female fly will have the genotype X^BX^b.
- The non-bar-eyed male fly will have the genotype X^bY.
2. Set Up the Punnett Square:
Create a Punnett square to predict the possible genotypes of the offspring from the cross of X^BX^b (female) with X^bY (male).
The configuration of the Punnett square will be:
```
X^B X^b
X^b | X^BX^b | X^bX^b
Y | X^BY | X^bY
```
3. Analyze the Punnett Square:
- First row:
- X^BX^b: Offspring will be a bar-eyed female, having one X^B from the mother and one X^b from the father.
- X^bX^b: Offspring will be a non-bar-eyed female, having two X^b, one from each parent.
- Second row:
- X^BY: Offspring will be a bar-eyed male, having one X^B from the mother and Y from the father.
- X^bY: Offspring will be a non-bar-eyed male, having one X^b from the mother and Y from the father.
4. Determine the Phenotypic Ratio:
- One female offspring will have the genotype X^BX^b (bar-eyed).
- One female offspring will have the genotype X^bX^b (non-bar-eyed).
- One male offspring will have the genotype X^BY (bar-eyed).
- One male offspring will have the genotype X^bY (non-bar-eyed).
5. Expected Ratio:
Given the analysis of the Punnett square, the expected ratio of phenotypes among the four offspring is:
- One bar-eyed female.
- One non-bar-eyed female.
- One bar-eyed male.
- One non-bar-eyed male.
Therefore, the correct answer is:
- One bar-eyed and one non-bar-eyed female, one bar-eyed and one non-bar-eyed male
