For each of the following restriction endonucleases, calculate theaverage distan

Question

For each of the following restriction endonucleases, calculate theaverage distance between restriction sites in an organismwith a random DNA sequence and equal proportions of all 4nucleotides. The symbol "Z" means any purine (G or A), "Q"means any pyrimidine (C or T), and "N" stands for anynucleotide. Show work.

a)    5'-CTAG-3'

Here's how I did it:

The probability of a restriction site is 1/4n,where n is the number of nucleotides in the site, and theaverage distance between sites is the reciprocal of this.So....

44= 256 base pairs



b)      5'-GGANTCC-3'

47= 16384 base pairs



c) 5'-ZCGQ-3'

44= 256 base pairs.

Explanation / Answer

Okay, you have part a of the problem correct. However, you used aprobability of 1/4 for Z, Q, and N which is not true in this case.When you have Z or Q you have two choices so getting one of thechoices gives you a probability of 1/2. For N which is theprobability of getting any of the nucleotides equals 1 because youwill always get a nucleotide. The probabilities are as follows: A = 1/4 C = 1/4 T = 1/4 G = 1/4 N (any nucleotide) = 1 Q (C or T) = 1/2 Z (G or A) = 1/2 a) The probability of the sequence for CTAG equals 1/4 x 1/4 x 1/4 x 1/4 = 1/256 Distance 256. b) The probability of N (any nucleotide) is 1 so the probability ofthe sequence for GGANTCC equals 1/4 x 1/4 x 1/4 x 1 x 1/4 x 1/4 x 1/4 = 1/4096 Distance is 4096. c) The probability of Z (either G or A) is 1/2 and the probabilityof Q (either C or T) is 1/2 so the probability of the sequence forZCGQ equals 1/2 x 1/4 x 1/4 x 1/2 = 1/64 Distance is 64

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