Haemophilia is a human disease in which the blood of an affected individual does not clot. The disease is known to be caused by a sex-linked recessive allele. The family pedigree shows the pattern of inheritance of this disease in a family. I II III If $X^h$ is the allele for haemophilia and $X^n$ is the allele for normal clotting, what is the genotype of individual 5? (A) $X^hY^h$ (B) $X^hX^n$ (C) $X^nX^h$ (D) $X^nY$

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Haemophilia is a human disease in which the blood of an affected individual does not clot - HSC - SSCE Biology - Question 9 - 2001 - Paper 1

Question 9

Haemophilia is a human disease in which the blood of an affected individual does not clot. The disease is known to be caused by a sex-linked recessive allele. The f... show full transcript

Worked Solution & Example Answer:Haemophilia is a human disease in which the blood of an affected individual does not clot - HSC - SSCE Biology - Question 9 - 2001 - Paper 1

Step 1

Identify the individuals affected by haemophilia

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Answer

From the pedigree provided, we can observe that individual 7 is a haemophiliac female and individual 8 is also identified as haemophiliac. Since haemophilia is a sex-linked recessive trait, we note the implications this has on the genotype of other individuals.

Step 2

Determine the sex and genotype of individual 5

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Answer

Individual 5 is a male, and since his mother (individual 4) is a normal female (not a carrier), he could only inherit the normal allele from his mother. Therefore, individual 5 must carry the normal cloting allele from his mother, and since males inherit one X chromosome from their mother and a Y chromosome from their father, the genotype for individual 5 is ultimately $X^nY$ .

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