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Annabeth and Percy are concerned about having a child with Canavan disease, which causes brain tissue degeneration

Biology May 03, 2021

Annabeth and Percy are concerned about having a child with Canavan disease, which causes brain tissue degeneration. This condition, which is very rare, has affected Annebeth's uncle (her mother's brother) and Percy's sister. No one else in either family has the condition.

a) What is the most likely mode of inheritance of Canavan disease? 

b)Based on their family history, what is the probability Annabeth and Percy's first child will be affected by Canavan disease?

c)What is the probability their first child will be a carrier?

d)What is the probability their first child will be affected by both diseases?

 

Expert Solution

Answer:

From the given data :

a.)

Autosomal Recessive:

  • This condition is inherited in an autosomal recessive pattern, which means both copies of the gene in each cell have mutations.
  • The parents of an individual with an autosomal recessive condition each carry one copy of the mutated gene, but they typically do not show signs and symptoms of the condition.
  • When both parents are found to carry the Canavan gene mutation, there is a one in four (25%) chance with each pregnancy that the resulting child will be affected with Canavan disease.

b.)

  • Since the uncle (Mother’s brother) of Annabeth and the sister of Percy are affected, the Annabeth and the Percy are carrier for this disease therefore;

P (child with canavan disease) = P (Annabeth is a carrier) x P (Percy is a carrier)

  • Since both parents (Father and Mother) are found to carry the Canavan gene mutation, there is a probability of one in four (25%) with each pregnancy that the resulting child will be affected with Canavan disease.

Therefore ¼ = (2/3 X ½ ) X (2/3) X ¼ = 1/18 = 0.0556 = 5.56 %

2 children are carrier 1 child is affected and 1 child is normal

c.)

P (child with Aa)

= 1/3*2/3* ½ + 2/3*2/3* ½ + 1/3*1/3* ½

= 7/18

= 0.389 = 38.9%

d.)

P (child with Canavan disease) X P (child with hemophilia)

1/18 X P (Annabeth is XHXh) X P (Xh from Annabeth X P ( Y from percy)

=1/18 X ½ X ½ X ½ = 1/44 = 0.0069 = 0.69 %.

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