a mother carried the sex-linked gene for color blindness and the father had normal color vision, what would be the probability of having a colorblind son?
*using a Punnett Square
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To determine the probability of having a colorblind son when a mother carries the sex-linked gene for color blindness and the father has normal color vision, we can use a Punnett Square.
First, let's understand the inheritance pattern of sex-linked genes. In humans, the gene for color blindness is located on the X chromosome, and it is recessive. Males have one X and one Y chromosome, while females have two X chromosomes. Since males only have one X chromosome, they have a higher chance of expressing sex-linked recessive traits.
Let's represent the sex chromosomes of the parents using letters. For this explanation, we will use "X" to represent the normal color vision allele and "x" for the colorblindness allele.
The mother's genotype is Xx (carrying the colorblindness gene) since she is a carrier. The father's genotype is XY (since he has normal color vision).
Now, let's create a Punnett Square to determine the probabilities of their children's genotypes:
| X | x |
______|________|________|
X | XX | Xx |
x | Xx | Xx |
Now let's analyze the Punnett Square:
- There are 4 possible combinations of genotypes:
- XX: Represents a daughter with normal color vision
- Xx: Represents a daughter who is a carrier but has normal color vision
- Xx: Represents a son who has normal color vision
- xx: Represents a son with color blindness
Out of the four possible combinations, two represent a son with normal color vision, and one represents a son with color blindness.
Therefore, the probability of having a colorblind son in this scenario is 1 out of 4, or 25%.