Question

Restriction enzyme HinfI cleaves a five nucleotide DNA sequence GA(A/T)TC. The ambiguity in the central position - (A/T) - means that either A or T can occur in the cleavage site. Assuming that each of four nucleotides is equally likely to occur at any position on a DNA molecule, an average HinfI cleavage fragment is about A) 0.5 kb B) 1.0 kb C) 2.0 kb D) 4.0 kb E) 8.0 kb

Answer 1

C) 2.0 kb

Step-by-step explanation:

It is given that out of the 4 nucleotides A, T, C & G each one has equal probability to occur at any position on the DNA molecule which simply means that the probability of occurrence of any nucleotide at a position is 1/4.

Also, it is given that probability of occurrence of either A or T at 3rd position is equal which means that the probability at that particular position will be 2/4 = 1/2.

Now, GA(A/T)TC is the DNA sequence where Restriction enzyme HinfI cleaves so the total probability of an average HinfI cleavage fragment will be = 1/4 x 1/4 x 1/2 x 1/4 x 1/4 = 0.00195 = 0.2 i.e. 2 kb.