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

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

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

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

due to the fact that it skipped a generation we can tell it is definitely not a dominant trait, therefore it must be recessive. i think this is right let me know!

explanation: