4. Phenylketonuria (PKU) is an inherited genetic disorder caused by a recessive

Question

4. Phenylketonuria (PKU) is an inherited genetic disorder caused by a recessive allele. A woman an husband are both carriers (heterozygous the following events? [You must provide information/calculations to indicate your approach t answe given.] d her ). Based on this information, what is the probability of each of ring the problem. If no supporting work is shown (i.e. just a number), then no credit will be A. all three of their children will be have PKU (2 points) B. their first child will be a normal female (2 points) C. their second child with be either a normal female or an affected male (2 points) D. two children will be normal and one affected (2 points)

Explanation / Answer

From the question,following information is given;

PKU is an autosomal recessive condition, that is would affect the person where both alleles are recessive and affects male and females equally.

Normal herein would include both,a person being a carrier but not affected by the recessibve allele and also person who doesnt have the affected allele.

For the problem, considering "P" as dominant allele and "p" as recessive, the disease PKU being expressed with the genotype "pp"."PP" and "Pp" would be normal. The parents are both "Pp"

A. probability of one child having PKU would be same as the other two since each crossingover would be an independent event. The cross between "Pp" and "Pp" would yield four possible genotypes i.e (using Punnett square)

1 "PP", 2 "Pp" , 1"pp"

There is 1/4 the probability that a child is affected by the PKU. For all three children to be affected by PKU, the probability would be 1/4 * 1/4 * 1/4 = 1/64 (all probaility multiplied since independent of each other)

B. Considering similar allele segregation as part A. which is 1 "PP", 2 "Pp" , 1"pp" , probability of first child being normal is 3/4 and being a female would be 50% or 1/2 , considering both as independent events, 1/2 * 3/4 = 3/8

C. Probability of second child being normal and female is 3/8 (from part B) ;

probaility of second child being affected and male is 1/4 * 1/2 = 1/8

(considered the event of second child being male or female / normal or affected as an independent event)

so second child being either normal female or affected male is 3/8 + 1/8 = 1/2

D. Probability of being normal is 3/4 and being affected is 1/4, considering each child birth/crossing over as an independent event, probability is 3/4 *3/4 * 1/4 = 9/64

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