Question

4. Phenylketonuria (fen-ul-ke-toe-NU-re-uh), also called PKU, is a rare inherited disorder that causes an amino acid called phenylalanine to build up in your body. PKU is caused by a defect in the gene that helps create the enzyme needed to break down phenylalanine. Without the enzyme necessary to process phenylalanine, a dangerous buildup of this amino acid can develop when a person with PKU eats foods that are high in protein. This can eventually lead to very serious health problems. Thus, people with PKU (babies, children and adults) for the rest of their lives need to follow a diet that limits phenylalanine, which is found mostly in foods that contain protein. PKU is a disease caused by a recessive allele (p). Normal individuals have at least one dominant allele (P). Problem: If two cousins marry, one who is a carrier (heterozygous for PKU), the other with PKU, please predict the probability that they could have children with the following genotype or phenotype: a) With PKU ? b) Carriers ? c) Normal (No PKU; Not Carrier)

Answer 1

Given Phenylketonuria (PKU) is caused due to recessive allele that means only a individual who is Homozygous for recessive allele will have PKU.

We have been given Alleles for PKU that are : P and p

P = Dominant = unaffected or no PKU

p = recessive = affected only in Homozygous Recessive condition

Now question says 2 cusions marry, one is carrier for PKU ( heterozygous for PKU) and other with PKU ( affected with PKU). So, their Genotypes will be :

Carrier for PKU = Pp ( heterozygous)

With PKU = pp ( affected)

We have to tell the probabilities of having different kinds of progenies.

First we will cross the cusions :

	P	p
p	Pp	pp
p	Pp	pp

Pp = heterozygous for PKU = unaffected = carrier = 2

pp = Homozygous for PKU = affected = diseased = 2

1) Probability of having children with PKU = pp

PKU affected individuals in cross = 2

Total children = 4

Probability = 2/4

= 1/2

2) Probability of having children as carriers = Pp

Number of children with Pp = 2

Total children = 4

Probability = 2/4

= 1/2

3) Probability of having children as normal, not carrier or no PKU = PP Genotype

Number of children with PP = 0

Total children = 4

Probability = 0/4

= 0

They can't have a child (or children) with PP, Homozygous Dominant Genotype.

Thank you.?