Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. 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 this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.