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 and her husband are both carriers (heterozygous). Based on this information, what is the probability of each of the following events? You must provide information/calculations to indicate your approach to answering the problem. If no supporting work is shown (i.e. just a number), then no credit will be given.] 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) E. their children occur in the following order: normal female, affected male, normal male. (2 points) F. their first child is heterozygous (2 points)

Explanation / Answer

Phenylketonuria (PKU) is an autosomal recessive disorder which increases the levels of phenyl alanine in blood.

A) Here both the parents are carriers which means their genotype is Aa (heterozygous condition). ( In an autosomal recessive disorder AA- Normal, Aa- carrier, aa- affected). So, for a child to get affected, the genotype should be 'aa'.

The probabilility of getting a single child affected ie., with genotype 'aa' from both parents of heterozygous condition is as follows.

Aa X Aa

aa

= 1/2 * 1/2 = 1/4.

therefore, probability that three of thier children effected = 1/4*1/4*1/4

= 1/64

B) For a child to be a male or female, there is 50% chance ie, 1/2.

For a child to be normal, the genotype should be either AA OR Aa OR aA ( carriers are considered as normal).

Probability of getting AA or Aa or aA from heterozygous parents is = (1/2*1/2) + (1/2*1/2) + (1/2*1/2)

= 1/4+1/4+1/4

= 3/4.

So, probability for their first child will be normal female= probality of a child to be normal AND it should be a female

= 3/4*1/2

= 3/8

AND RULE - Multiplication

OR RULE - Addition

C) From the above problem, probability for their child to be a normal female is 3/8.

For a child to become male or female, the probability is 50% ie.,1/2.

Probability for a child to get affected (aa) = 1/2*1/2

= 1/4.

therefore probability thta a male child is affected = probability that the child is male AND he is affected

= 1/2 *1/4

= 1/8

Final answer: The probability that their second child will be either a normal female OR an affected male= 3/8 + 1/8

= 4/8

= 1/2

D) The probability that 2 children will be normal and 1 will be affected can be by any of the following three orders.

(Normal, Normal ,Affected) OR (Normal, Affected, Normal) OR (Affected, Normal, Normal)

A normal child can have genotype either AA OR Aa OR aA.

Probability of getting AA OR Aa OR aA (Normal child)= (1/2*1/2) + (1/2*1/2) + (1/2*1/2)

= 1/4 +1/4 +1/4

= 3/4

Probability of getting affected ie., (aa) = 1/2*1/2 =1//4.

Final answer: Probability that 2 children will be normal and 1 affected is = P (Normal, Normal, Affected) + P ( Normal, Affected, Normal) + P ( Affected, Normal, Normal)

= (3/4*3/4*1/4) + (3/4*1/4*3/4) + (1/4*3/4*3/4)

= 9/64+9/64+9/64

= 27/64

E) From the abobe question B, Probability of getting normal female is 3/8.

From question C, Probability of getting affected male is 1/8

Probability of getting normal male is= P(getting male) AND P( normal ie, with genotypes AA OR Aa OR aA)

=1/2* 3/4

=3/8

Final answer: Probability that their children will occur in the following order of Normal Female, Affected male, Normal male is = 3/8*1/8*3/8

= 9/512

F) Probability that their first child is hetrozygous ie, it can be either Aa OR aA

= (1/2*1/2) + (1/2*1/2)

=1/4+1/4

=2/4

=1/2

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