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

Question

Biology

Tianna Briggs

15 April, 22:57

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

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Alison Levy · Answer 1

C) 2.0 kb

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.