Question

in the cross aabbcc × aabbcc, what is the probability of producing the genotype aabbcc?
1/4
1/8
1/16
1/64
question 8
1 pts
phenylketonuria (pku) is a disease in humans that results from the abnormal metabolism of the amino acid phenylalanine. if untreated, it can lead to severe brain damage in infants. two normal parents have a child who has pku. what can you determine about the inheritance of the pku allele from this information?
it is recessive.
it is dominant.
it is pleiotropic
it is epistatic.

Explanation:

Step1: Analyze each gene pair separately

For the \(Aa\times Aa\) cross, the probability of getting \(AA\) is \(\frac{1}{4}\) (since \(AA:Aa:aa = 1:2:1\)).
For the \(Bb\times Bb\) cross, the probability of getting \(BB\) is \(\frac{1}{4}\) (since \(BB:Bb:bb=1:2:1\)).
For the \(Cc\times Cc\) cross, the probability of getting \(CC\) is \(\frac{1}{4}\) (since \(CC:Cc:cc = 1:2:1\)).

Step2: Use the multiplication rule

Since the inheritance of each gene pair is independent, we multiply the probabilities of getting the desired genotype for each pair.
The probability of \(AABBCC=( \frac{1}{4})\times(\frac{1}{4})\times(\frac{1}{4})\)

According to Mendelian genetics, when two normal - phenotype parents (who are carriers, i.e., heterozygous) have a child with a recessive disorder (PKU in this case), it follows the recessive inheritance pattern. In recessive inheritance, an individual must inherit two recessive alleles (one from each parent) to express the disorder. If it were dominant, at least one of the parents would have to show the PKU phenotype. Pleiotropic refers to a single gene affecting multiple traits, and epistatic refers to one gene affecting the expression of another gene, which is not indicated here.

Answer:

\(\frac{1}{64}\)