Please use the "Talk to Ed and Mike" forum in Moodle if you wish to discuss the assignment further.
Note that I have used "P" to represent the dominant allele and "p" to represent the recessive allele for the PAH gene that is associated with Phenylketonuria (PKU).
The 4 types of sperm = PAH, X -- pah, Y -- PAH, Y -- pah, X
The woman is a carrier (heterozygous) for Phenylketonuria (PKU) (PAH, pah). She has two X chromosomes. All her eggs will contain an X chromosome. Half her eggs will contain a dominant allele (PAH) and half will carry a recessive allele (pah). The answer is 50% of her eggs will contain the recessive allele (pah) and an X chromosome.
| Woman's eggs (from Question #3) | ||
| Man's sperm (from question #2) |
PAH, X | pah, X |
| PAH, X |
|
|
| PAH, Y |
|
|
| pah, X |
|
|
| pah, Y |
|
|
There is a 2/8, or 25% chance that this man and woman will have a boy who is heterozygous (PAH, pah) for the PAH gene locus.
Another way to figure this is to multiply probabilities:
1/2 chance that a child will be heterozygous (PAH, pah)
1/2 chance that a child will be a boy.
1/2 X 1/2 = 1/4 chance that a child will be a boy who is is
heterozygous (PAH, pah) for the PAH gene locus.