Restriction Mapping - Homework question
This is not as hard, as it first seems. Lets have a look at the single enzyme digests first: The digest with enzyme A and B only leads to products which are 5kB (5000 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 is a bit more tricky, but not much. You get two products from it, with 2kb and 3kB. This means that enzyme B cuts between the restriction sites for enzyme A, resulting in these two fragments. Have a look at the sketch I made below (only a quick one and not to scale):


A1 and A2 are the cutting sites for enzyme A, B1 and B2 the sits 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 5kB (doesn't matter which) and that both A and both B sites are located 5kB from each other.

A1 is at position 100, A2 has to be at position 5100, B1 is located 2kB behind A1 (and therefore at position 2100), and B2 2kB behind position A2 (and therefore at position 7100). To get the same restriction pattern it is also possible that B1 and B2 are located 3kB behind their respective enzyme A position (so at 3100 and 8100) and you still get the same pattern of 2kB and 3kB on the gel.