Question

The pedigree below concerns the autosomal recessive disease phenylketonuria (PKU). The couple marked A and B are contemplating having a baby but are concerned about the baby having PKU. What is the probability of the first child having PKU

Answer 1

Answer: Hello your question some missing data attached below is the complete question

answer:

P( first child having PKU ) 0.25 ≤ x ≤ 0.5

Step-by-step explanation:

The pedigree father has PKU ( pp ) ( From the top right )

pedigree mother = PP

The possible resultant progeny = Pp

Resultant progeny marries non-carrier ( Pp x PP ) = PP , PP, pP, pP

Hence B is either ; PP ( non carrier ) or Pp ( carrier )

From left

one of the Resultant progeny = pp ( affected ). this simply means pedigree parents where both carriers or sufferers i.e. Pp or pp

Hence A is either ; Pp or pp

The probability of their first child having PKU

= PP x Pp = PP, Pp, PP, Pp ( probability = 0 in this case )

= Pp x Pp = PP, Pp, pP, pp ( probability = 1/4 * 100 = 25% )

= Pp x pp = Pp, Pp, pp, pp ( probability = 2/4 * 100 = 50% )

P( first child having PKU ) 0.25 ≤ x ≤ 0.5

lets denote dominant Gene = PP, recessive Gene = pp