The hemophilia gene is located 12 map units from the red-green colorblindness gene on the X chromosome. Red-green colorblindness is recessive to normal color vision. A woman with normal blood clotting and normal color vision, but whose father was a hemophiliac and whose mother was red-green colorblind, marries a man who is not a hemophiliac and is not colorblind.

What percent of their children will be colorblind and have hemophilia? Show work ?

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The hemophilia gene is located 12 map units from the red-green colorblindness gene on the X chromosome. Red-green colorblindness is recessive to normal color vision. A woman with normal blood clotting and normal color vision, but whose father was a hemophiliac and whose mother was red-green colorblind, marries a man who is not a hemophiliac and is not colorblind.


What percent of their children will be colorblind and have hemophilia? Show work

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To determine the percent of their children who will be colorblind and have hemophilia, we need to understand the inheritance pattern of these two traits and use the concept of genetic recombination.

First, let's define the genotypes of the parents. The woman has the genotype X^N X^n, where X^N represents the normal allele for color vision and blood clotting, and X^n represents the recessive allele for red-green colorblindness. Her father was a hemophiliac, so he must have had the genotype X^h Y, where X^h represents the recessive allele for hemophilia, and Y represents the male sex chromosome. Her mother was red-green colorblind, which means she had the genotype X^n X^n.

The man, on the other hand, is not a hemophiliac and is not colorblind. Hence, he must have the genotype X^N Y.

To determine the percent of their children who will be colorblind and have hemophilia, we need to consider the inheritance pattern and recombination between the X chromosome alleles.

Since the genes for color vision and hemophilia are located 12 map units apart on the X chromosome, there is a 12% chance of genetic recombination occurring between these two genes during gamete formation.

Now, let's consider the possible combinations of alleles from each parent:

1. X^N X^h (Normal color vision, hemophilia)
2. X^N X^n (Normal color vision, red-green colorblindness)
3. X^h Y (Hemophilia, normal color vision)
4. X^n Y (Red-green colorblindness, normal color vision)

However, the recombination event allows for new combinations:
5. X^N X^n X^h (Normal color vision, red-green colorblindness, and hemophilia)

So, we have five possible genotypes for the children.

Now, let's calculate the percentages:

Out of the five possible genotypes, there are two that result in children having both colorblindness and hemophilia: (1) X^N X^h and (5) X^N X^n X^h.

Since each possible genotype has an equal chance of being inherited, there is a 2 out of 5 chance (or 2/5) that a child will be colorblind and have hemophilia.

To convert this fraction to a percentage, we multiply 2/5 by 100:

(2/5) * 100 = 40%

Therefore, 40% of their children will be colorblind and have hemophilia.

Note: This is a simplified explanation, assuming no other genetic factors or modifiers are involved. In reality, the inheritance of these traits can be more complex, and the actual percentages may vary depending on various other factors.