calculation-past exam - (Nov/09/2007 )
this is from a past exam and not sure how to do it...
You are working with a bacterium in which the genome consists of 60% GC base pairs. The frequency of finding recognition sites for the enzyme MseII, which cleaves at the sequence AATT, is expected to be
is it sumthing like (1/4x1/4x1/4x1/4)x0.20??
Hmm.. I never learned about probabilities (probably should go back and try to learn it), but if you are using the 1/4 x 1/4 x 1/4 x 1/4 to get the probablility of the AATT based on the 1 in 4 possibility at each position - then I would say rather than using 25% in each case you should use 20% (100-60=40% AT, divided in half =20%). So instead of 25% x 25% x 25% x 25%, shouldn't it be 20% x 20% x 20% x 20%?
generic 4 base cutter cuts at a frequency of 1/4^4 i.e 1/256 bp in a normal genome i.e. 50%GC
hope this helps