Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

Sagot :

batolisis

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

answer:

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

View image batolisis
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.