217k views
3 votes
When conducting a restriction digest with the endonuclease Mst II, how many fragments do you expect to see from a person's PCR product who is homozygous for the normal Hb A gene?

O no fragments
O 1 fragment
O 2 fragments
O 3 fragments

1 Answer

5 votes

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.

User Peter
by
8.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.