1. Calculating the risk of an inherited disorder by pedigree analysis. An autoso
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1. Calculating the risk of an inherited disorder by pedigree analysis. An autosomal recessive gene (p) in humans results in the absence of the enzyme phenylalanine hydroxylase and causes and inherited disease called phenylketonuria (PKU). Individuals with PKU (pp) develop severe mental retardation if untreated because of the accumulation of compounds such as phenylpyruvic acid that are toxic to the central nervous system. If detected early in life and placed on a diet with low concentrations of phenylalanine, individuals with PKU develop normal abilities. In the following pedigree, circles represent females, squares represent males, open symbols represent individuals with normal phenylalanine hydrolase activity (P_) and shaded symbols represent individuals with PKU (pp). Unless there is evidence to the contrary, assume that individuals brought into the family by marriage do not carry the allele (p) for PKU. Calculate the probability of PKU occurring in a child produced by the mating of individuals IV-1 and IV-2. 10 12 IVExplanation / Answer
PKU is an autosomal recessive disorder.
The parents in generation I must be heterozygous for the PKU allele.
WT dominant allele = P
Mutant recessive allele = p
Pp X Pp ----> PP Pp Pp pp
The probability for II-2 to be heterozygous = 1/2
II-1 must be heterozygous.
The probability for III-4 to be heterozygous = 1/2 X 1/2 = 1/4
The probability for II-3 to be heterozygous = 1/2
II-4 must be homozygous WT.
The probability for III-5 to be heterozygous = 1/2 X 1/2 = 1/4
The probability for IV-1 to be heterozygous = 1/4 X 1/4 X 1/2 = 1/32
The probability for III-9 to be heterozygous = 1
The probability for II-7 to be heterozygous = 1/2
The probability for III-10 to be heterozygous = 1/2
The probability for IV-2 to be heterozygous = 1 X 1/2 = 1/2
The probability for the child of IV-1 X IV-2 to be affected = 1/4 X 1/32 X 1/2 = 1/256
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