Question

Restriction enzyme Fnu4HI cuts a four nucleotide DNA sequence GCGC. Assuming that the GC content of a DNA molecule is 50%, what will be the average size of a Fnu4HI fragment?

Restriction enzyme Fnu4Hl cuts a four nucleotide DNA sequence GCGC. Assuming that the GC content of a DNA molecule is 50%, what will be the average size of a Fnu4Hl fragment? O A: 0.25 kb O'B: 0.50 kb O C: 1 kb OD: 2 kb UE: 4 kb

Answer 1

Let us assume that there is an infinite stretch of DNA with each base chosen at random such that the probability of having GC is 50% and that of having AT is 50%. Hence each base can occur at a position with a probability of 25%. Now, the restriction enzyme used here is Fnu4HI. Hence this is a high fidelity enzyme.

The site used by it for cutting is GCGC.

The probability of having a sequence GCGC is 0.25*0.25*0.25*0.25 = 0.00390625.

Hence to obtain one such site in a truly random sequence, the sequence length searched would be:

1/0.00390625 = 256bp = 0.256kb or approximately 0.25kb i.e. option A.