Question Video: Calculating the Probability of Being a Phenylketonuria Carrier with Homozygous and Heterozygous Affected Parents | Nagwa Question Video: Calculating the Probability of Being a Phenylketonuria Carrier with Homozygous and Heterozygous Affected Parents | Nagwa

# Question Video: Calculating the Probability of Being a Phenylketonuria Carrier with Homozygous and Heterozygous Affected Parents Biology • First Year of Secondary School

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Phenylketonuria (PKU) is an inherited disorder caused by a recessive allele (r) located on chromosome 12 in humans. If a male who is homozygous for the disorder with the genotype rr reproduces with a heterozygous female, what is the probability that a child of theirs will be a healthy carrier of the disorder PKU?

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### 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.

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