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A circular plasmid of 10,000 base pairs (bp) is digested with two restriction enzymes, A and B. This produces a 3000 bp and a 2000 bp bands when visualized on an agarose gel. When digested with one enzyme at a time, only one band is visible at 5000 bp.

If the first site for enzyme A (A1) is present at the 100th base, the order in which the remaining sites (A2, B1 and B2) are present is -

(A) 3100, 5100, 8100

(B) 8100, 3100, 5100

(C) 5100, 3100, 8100

(D) 8100, 5100,. 3100

NOTE: This is a Kishore Vaigyanik Protsahan Yojana 2014 exam (stream SX) Question. I really couldn't understand the diagram shown in the solution, and my teachers were of no help either. The diagram in the solution is: solution

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Examine the effects of Enzyme A on its own.

Enzyme A performs two cuts that result in 2 fragments that are 5000bp long. We know that the first cut is made at the 100th bp. Therefore the second cut is made at the 5100th bp (A2).

You can stop there and determine that the answer is C. C is the only answer that states A2 = 5100.

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