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Biochemistry · Molecular techniques: RFLP in sickle cell disease
Sample itemIn prenatal diagnosis, fetal DNA is cut with the restriction enzyme MstII and probed for the beta-globin gene. The sickle mutation (GAG to GTG) destroys an MstII site. What is seen in a fetus homozygous for HbS?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
B. A single larger fragment because the restriction site is lost
Restriction enzymes recognise short palindromic sequences. The sickle point mutation abolishes one MstII site, so fragments that would be cut remain joined and a longer fragment appears on the Southern blot. A normal fetus shows the smaller fragments, and a carrier shows both patterns. Today PCR-based methods are more usual.
Option by option
A. No band, because the gene is deleted
Sickle cell disease is caused by a point mutation, and the gene is present.
B. A single larger fragment because the restriction site is lost (correct)
Correct. Loss of a cutting site yields one longer fragment.
C. Additional small fragments because a new site is created
The mutation removes a site; it does not create a new one.
D. Two small fragments identical to normal
This pattern is that of HbA, in which the site is present and cutting occurs.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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