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

To determine the genotype of individual 5, we need to analyze the pedigree provided in the question.

1. Identify Affected Individuals: From the pedigree, individual 5 is a male (indicated by a square) and does not show symptoms of haemophilia; therefore, he must have the normal allele.

2. Determine the Parents' Genotype: Individual 5's parents are individuals 4 and 7. Since individual 4 is a normal female and individual 7 is affected (haemophiliac), individual 4 must have at least one normal allele ($X^n$) and individual 7 must have the genotype $X^hY$.

3. Inferring Genotype of Individual 5: Since individual 5 inherits his Y chromosome from his father (individual 7 doesn’t contribute an allele for clotting), the genotype of individual 5 must be $X^hX^n$ or simply $X^nY$ as he cannot inherit the $X^h$ allele from his mother if she is not a carrier.

4. Confirming Correct Option: Thus, the only option that accurately represents individual 5's genotype as the normal male with a normal clotting factor is (D) $X^nY$.