A man with normal color vision marries a woman who is colorblind. What percentage of their sons would be colorblind, what percentage of their daughters would be colorblind, what percentage of their daughters would be carriers of the colorblind gene?
To determine the probabilities of their children being colorblind or carriers of the colorblind gene, we need to understand the inheritance pattern of colorblindness. In this case, let's assume that the man does not carry the colorblind gene.
Colorblindness is an X-linked recessive trait, meaning it is primarily passed down on the X chromosome. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). The colorblind gene is typically carried on the X chromosome.
Since the woman is colorblind, she must have two copies of the colorblind gene (XcXc), where 'Xc' represents the colorblind gene, whereas a man without colorblindness would have a normal gene (XCY).
Now, let's consider the possibilities for their children:
1. Sons: A man passes his Y chromosome to his sons and his X chromosome to his daughters. Since the woman is colorblind (XcXc), all of her sons will receive her X chromosome with the colorblind gene (XcY). Therefore, all their sons will be colorblind (100%).
2. Daughters: The daughters will receive one X chromosome from the father (X) and one from the mother (Xc). The presence of a normal X chromosome from the father will prevent the daughters from being colorblind. Therefore, none of their daughters will be colorblind (0%).
3. Daughters as carriers: As stated earlier, the daughters received one X chromosome from the father (X) and one from the mother (Xc). Since they have one X chromosome with the colorblind gene (Xc), they will be carriers of the colorblind gene. Therefore, all of their daughters will be carriers of the colorblind gene (100%).
In summary, if a man with normal color vision marries a colorblind woman, all of their sons will be colorblind (100%), none of their daughters will be colorblind (0%), but all of their daughters will be carriers of the colorblind gene (100%).
In order to determine the percentages, let's start by understanding the inheritance pattern of colorblindness. Color blindness is an X-linked recessive disorder, which means it is primarily passed down through the X chromosome.
Since the man has normal color vision, we can assume he does not carry the colorblind gene, as the gene for color blindness is typically found on the X chromosome. However, the woman is colorblind, which means she carries the colorblind gene on one of her X chromosomes.
With this information, let's calculate the percentages:
1. Sons: Since the father does not carry the colorblind gene, the only way for a son to be colorblind is if he inherits the colorblind gene from his mother. Since the mother is colorblind, she can only pass on her X chromosome with the colorblind gene to her son. Therefore, all of their sons would be colorblind. The percentage of their sons who would be colorblind is 100%.
2. Daughters: Daughters, on the other hand, would need to inherit one colorblind gene from their mother and one normal gene from their father to be colorblind. Since the father does not carry the colorblind gene, the daughters would not be colorblind. Therefore, the percentage of their daughters who would be colorblind is 0%.
3. Carrier daughters: In this case, all daughters will be carriers of the colorblind gene because the mother carries it. Since the colorblind gene is recessive, it would be masked by the normal gene from the father, resulting in them having normal color vision. The percentage of their daughters who would be carriers of the colorblind gene is 100%.
In summary:
- 100% of their sons would be colorblind.
- 0% of their daughters would be colorblind.
- 100% of their daughters would be carriers of the colorblind gene.
to be color blind, all X genes must have the trait. A male has only one X gene, females two X.
So the mom is XcXc, and she passes one of these Xc to each daughter and son. So the sons will be colorblind...they got it from mom. The daughers will get mom's Xc, and dad's Xo, making them carriers, but not colorblind.
Man: XoY
Mom: XcXc
daughters: XcXo
sons: XcY