Phenylketonuria?

Question

Phenylketonuria is an inherited disease caused by a recessive allele. If a woman has this disease, and her husband is heterozygous for this disease, what is the probability that their first child will be a phenotypically normal one?
A) 1/4
B) 1/2
C) 1/16
D) 1

Answer 1

Do a punnet (sp?) square. If the mom has it, that's pp (let P = dominant normal and p = recessive disease)

Dad is heterozygous, meaning he's Pp

So pp x Pp has four results:
___p___ p
P__Pp__Pp
p__pp__pp

So the 4 results are two Pp and two pp. Phenotypically normal will be any presence of P, since the disease is recessive.

Ergo, 1/2 chance. B is correct.

Answer 2

1/2. Phenotypically Normal.

Answer 3

Pp= heterzygous
pp= recessive
P p
p Pp pp
p Pp pp

Thus 1/2 chance for phenotypically normal child

Answer 4

The answer is B. You can do a simple Punett square for this one. Since it's a recessive trait, the woman must be homozygous for that allele. We'll say she's rr, and he's Rr. So after the cross, you get rR, rR, rr, and rr. The R is dominant, and half of the squares have R in them.

Answer 5

Sometimes. There are several mutations that can cause a defective allele at that locus (phenylalanine dehydroxylase gene), and at least one is known to be a base substitution. In general, human genetic diseases can be caused by many different kinds mutations at the same locus. Sickle-cell anemia is odd because we know EXACTLY what mutation causes it.

Answer 6

B) one in two
The p.k.u. gene is recessive.
you need 2 recessive defective genes to express the disease. (which is the inability to breakdown the amino acid phenylalanine.)
D.