Question

A circular plasmid of $10,000$ base pairs $(bp)$ is digested with two restriction enzymes, $A$ and $B$, to produce a $3,000 \,bp$ and a $2,000\, bp$ bands when visualised on an agarose gel. When digested with one enzyme at a time, only one band is visible at $5,000\, bp$. If the first site for enzyme $A\, (A1)$ is present at the $100 \text{th}$ base, the order in which the remaining sites $(A2, B1 $ and $B2)$ are present is

• A. 3,100, 5,100, 8,100
• B. 8,100, 3,100, 5,100
• C. 5,100, 3,100, 8,100
• D. 8,100, 5,100, 3,100

Answer 1

Answer: (C)

Lets have a look at the single enzyme digests first. The digest with enzyme $A $ and $B$ only leads to products which are $5\,kb\, (5,000 \,bp)$ away from each other. Since, they are of the same size, both equally sized restriction fragments appear as one band. So, each enzyme cuts the plasmid exactly in half. The double digest will form two products from it, with $2 \,kb$ and $3 \,kb$. This means that enzyme $B$ cuts between the restriction sites for enzyme $A$, resulting in these two fragments. See the sketch below

$A1 $ and $A2$ are the cutting sites for enzyme $A, B1$ and $B1$ are the sites for enzyme $B$. The numbers on the outside are the positions on the DNA sequence. You can see that a single digest leads to two fragments of $5\,kb$ and that both $A$ and both $B$ sites are located $5\, kb$ from each other.
$A1$ is a t position $100, A2$ has to be at position $5,100, B1 $ is located $2\,kb $ behind $A1 $ (and therefore at position $2,100$) and $B2\, 2\,kb$ behind position $A2$ (and therefore at position $7,100$ ). To get the same restriction pattern it is also possible that $B1$ and $B2$ are located $3\,kb$ behind their respective enzyme A position (so at $3,100$ and $8,100$) and you still get the same pattern of $2\,kb$ and $3\,kb$ on the gel.