Aprospective mother and a prospective father are hemophiliacs. (enter your answers as fractions.) (a) find the probability that their child would be a hemophiliac male. (b) find the probability that their child would be a hemophiliac female. (c) find the probability that their child would be a carrier. (d) find the probability that their child would be symptom-free.

So remember that Hemophilia is a recessive, x linked trait.

For a woman to have hemophilia, she must have the trait linked to both x chromosomes. For a man to have hemophilia, he must have the trait linked to his singular x chromosome.

For this, I will be making a Punnett Square to determine the Possibilities:

Since there is a 1/2 chance for their offspring to be a boy, and of that 1/2 both would be hemophiliacs, there is a 1/2 chance for the child to be a hemophiliac male.

As previously mentioned, for a woman to have hemophilia, they would need to have that trait attached to both x chromosomes. Since there is a 1/2 chance for their offspring to be a girl, and of that 1/2 both would be hemophiliacs (since they both carry the trait on both chromosomes), there is a 1/2 chance that they will have a hemophiliac female.

So a carrier is someone who carries a recessive trait, but it isn't displayed due to the dominant trait masking it. With x-linked traits, only women can be carriers since they carry more than one x chromosome. What this asks is the probability of an offspring having the trait attached to only 1 of the x chromosomes. Looking at the girls, since both carry the traits on both x chromosomes, there is 0 chance of a carrier.

So symptom free is as it seems, without the hemophilia trait. Looking at the table, since all the offspring contain the hemophilia trait, there is 0 chance that any of their offspring will go symptom free.

the forst and 3rd

step-by-step explanation: