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

answer: oceans, duh

the bald eagle got 50 because everything gets 10% of what they consume and since 50 is ten % of 500 hundred that means the   grasshopper had to have consumed a plant with 5,000 calories of energy