This can be solved either by using a Punnet Square or you can use the formula (n(n+1))/2, where n is the number of alleles. However, using either method will give you the number of possible genotypes regardless if it's homozygous or heterozygous. If we use the second formula, that will give you 10 possible genotypes. Since you only want to know the number of heterozygous genotypes, you should subtract the number of possible homozygous genotypes. In this case, there are four. That is (A1,A1), (A2,A2), (A3,A3), and (A4,A4). Knowing this, you'd be able to figure out that there are six heterozygous genotypes.
(A1,A2), (A2,A3), (A3,A4), (A1,A4), (A1,A3), (A2,A4)
Answer: The possible genotype for their child is Hh. All the children will be heterozygous for Huntington (Hh).
The possible phenotype for their child is Huntington disease.
Explanation: If H represents the trait for Huntington and h represents normal trait; and if Huntington trait (H) is dominant over normal trait (h), therefore the genotype of the mother who is homozygous for Huntington disease is HH and the genotype of the man who is homozygous normal is hh.
A cross between the man and woman will produce offsprings who are all heterozygous for Huntington disease Hh. Phenotypically, the offsprings will manifest as Huntington disease.
See the attached punnet square for more information
Enzymes onto the end of substrates
Example
preoxidase
telemorase
polymerase