Final answer:
A person who is homozygous for the normal Hb A gene would expect to see 3 fragments when conducting a restriction digest with the endonuclease Mst II.
Step-by-step explanation:
When conducting a restriction digest with the endonuclease Mst II, a person who is homozygous for the normal Hb A gene would expect to see 3 fragments in their PCR product.
Restriction digest is a process where DNA is cut into smaller fragments using restriction enzymes. Mst II is an endonuclease that recognizes a specific DNA sequence and cleaves the DNA at that site.
In the case of a homozygous individual with the normal Hb A gene, Mst II would recognize and cut the DNA in two places, resulting in three fragments.