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NEETZoologyPrinciples of Inheritance and Variation

A man and a woman, both carriers for sickle-cell anaemia, plan to have a child. What is the probability that their child will be affected by the disease, and what is the specific mRNA codon at the sixth position of the beta globin gene that will be present in the homozygous condition in the affected child?

Options

  1. A1 4 and GUG
  2. B1 2 and GUG
  3. C1 4 and GAG
  4. D1 4 and CTC

Correct answer

A. 1 4 and GUG

Step-by-step solution

Sickle-cell anaemia is an autosome-linked recessive trait. The parents are carriers, meaning their genotype is Hb^A Hb^S . When two carriers are crossed ( Hb^A Hb^S Hb^A Hb^S ), the probability of having an affected child ( Hb^S Hb^S ) is 1 4 . The disease is caused by a point mutation where the normal mRNA codon GAG is replaced by GUG at the sixth position of the beta globin chain. Therefore, the affected child will have the mutant GUG codon in the homozygous state. Answer: 1 4 and GUG

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