The chance that IV-2 is heterozygous for Tay Sachs disease is A - VCE - SSCE Biology - Question 7 - 2003 - Paper 1

To determine the likelihood that individual IV-2 is heterozygous for Tay Sachs disease, we must consider the genetics of the trait.

Tay Sachs disease is an autosomal recessive disorder, which means an individual must inherit two copies of the mutant allele to express the disease. The alleles can be represented as follows:

• T (normal allele)
• t (allele for Tay Sachs)

Assuming we have information about the genotypes of the parents and can apply a Punnett square to track the inheritance of the alleles:

• If both parents are carriers (Tt), the possible genotypes for their offspring would be:
  • TT (normal)
  • Tt (carrier)
  • Tt (carrier)
  • tt (affected)

This results in a ratio of:

• 1:2:1 for TT, Tt, and tt.

Since we only need to consider the heterozygous (Tt) chance:

• The probability that IV-2 is heterozygous (Tt) is therefore:

P(Tt) = rac{2}{4} = rac{1}{2}

Thus, the chance that IV-2 is heterozygous for Tay Sachs disease is 1/2, which corresponds to option B.