Question

Tay Sachs is a recessive disease (t) that causes an inability to process certain nutrients which then build up in the brain, resulting in death (at a very young age). A man and a woman had a child born with this disease, but neither have the disease (otherwise they would have died as young children!). What must their genotype be in order for them to have a child with the disease? Explain your reasoning.

Answer 1

Tt (both heterozygous)

Step-by-step explanation:

According to this question, Tay Sachs is a recessive disease (t) that causes an inability to process certain nutrients which then build up in the brain, resulting in death (at a very young age). This means that this disease can only be phenotypically expressed when the recessive alleles are both present in the gene i.e. tt.

If a man and a woman had a child born with this disease, but neither have the disease, it means that both of the parents are dominant but heterozygous (carriers) for the disease. This means that they both have a genotype "Tt". So when they both mate i.e. Tt × Tt, they produce the following offsprings: TT, Tt, Tt, and tt.

The "tt" offspring is the child with the Tay Sachs disease.