GATE 2022 BT Q.64: solution and answer

Official answer: 2Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

1. Given: the plasmid is circular and has exactly one unique restriction site each for enzymes a, b and c (3 sites total, one copy of each).

2. Key rule: on a CIRCULAR DNA molecule, the number of linear fragments produced equals the number of cut sites used. (This differs from linear DNA, where n internal cuts give n+1 fragments — on a circle there are no free ends to begin with, so each additional cut simply adds one more piece to the ring you have already opened.)

3. Digesting with a + b (2 cuts on the circle) gives 2 fragments; the stem states these are of equal size, so let each fragment have size $X$ (plasmid total size $=2X$).

4. Since c has only ONE restriction site in total, and the stem says c produces equal-size fragments from only ONE of the two a+b fragments, c's single site must lie inside that one $X$-sized fragment (not the other).

5. Digesting with a + b + c together: 3 cuts on the circular plasmid $\Rightarrow$ 3 linear fragments in total.

6. The fragment that c does not cut stays intact at its original size, $X$.

7. The fragment that c does cut is split by c's one internal cut into two pieces; since the stem specifies these are of equal size, each piece is $X/2$.

8. So the three fragments produced are $X$, $X/2$, $X/2$ — three fragments but only two distinct sizes.

9. On agarose gel electrophoresis, fragments of identical size co-migrate and appear as a single band, so the two $X/2$ fragments merge into one band.

10. Total distinct bands observed $=2$ (one band at $X$, one band at $X/2$).