The restriction enzyme HindIII cleaves double-stranded DNA atthe sequence 5’-AAG

Question

The restriction enzyme HindIII cleaves double-stranded DNA atthe sequence 5’-AAGCTT-3’ (a six-nucleotidesequence).

a. In a random stretch of DNA with equal proportions of A, T, Cand G, what will be the average distance (in base pairs) betweeneach HindIII site ?

b. Assuming that there are equal proportions of A, T, C and G inthe human genome, what would be the average number of HindIIIrestriction sites in the human genome (human genome size is3.4x109 bp) ?

c. In fact, human DNA contains about 30% of A nucleotides. Basedupon this, recalculate your answers for a and b.

Explanation / Answer

a) .25 probability of each base pair .25^6 (6 is from the number of bases in the restriction site) isthe probability of finding the restriction site in randomDNA. (probability x distance) = 1 so  1/probability = distance ==> distance = 4096 bp b) 3.4x10^9 bp / 4096 bp = 8.301x10^5 c) 30% A, 30% T since A is complimentary to T 100-60 = 40 20% C, 20%G Our restriction sequence is AAGCTT so.. probability: (.3)(.3)(.2)(.2)(.3).(.3) = 3.24x10^-4 1 / 3.24x10^-4 = 3086 bp in human genome: 3.4x10^9 bp / 3086 bp = 1.102x10^6

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