Video Transcript
Phenylketonuria, PKU, is an
inherited disorder caused by a recessive allele, lowercase r, located on chromosome
12 in humans. If a male who is homozygous for the
disorder with the genotype lowercase r lowercase r reproduces with a heterozygous
female, what is the probability that a child of theirs will be a healthy carrier of
the disorder PKU? (A) Zero percent, (B) 25 percent,
(C) 50 percent, (D) 75 percent, or (E) 100 percent.
Every human is the product of
fertilization, which is the process of fusion of a sperm cell containing half the
genetic material of the father with an egg cell containing half the genetic material
of the mother. So each child will receive one copy
of each gene from their father, located on the chromosomes they receive from him,
and the other copy of each gene from their mother, located on the chromosomes they
receive from her.
We can use a Punnett square to find
all of the possible genetic combinations that the child can have of the gene causing
PKU. As the father is homozygous for the
disorder, he has two of the same recessive lowercase r alleles of the gene. The mother is heterozygous for the
disorder. This means she has two different
alleles, one lowercase r allele and one uppercase R allele. She may pass on either of these to
her offspring. As PKU is caused by the recessive
allele lowercase r, two copies of the allele lowercase r are needed for the disorder
to be expressed. As long as the child has one copy
of uppercase R, it will be healthy in terms of not expressing the disorder PKU. To be a carrier of the disorder,
the child needs to have one lowercase r allele. In other words, a child that is a
healthy carrier of the disorder PKU has the genotype uppercase R lowercase r.
If we return to our Punnett square,
we can see that a child of these particular parents can have two different
genotypes: lowercase r lowercase r or uppercase R lowercase r. Since two out of the four, or half,
of the possibilities in the Punnett square are the uppercase R lowercase r genotype
that we are looking for, that means that the probability that a child of these
parents will be a healthy carrier of PKU is about half. Half expressed as a percentage is
50 percent, so the correct answer is (C). The probability that a child of
these parents will be a healthy carrier of PKU is 50 percent.
