a. Assume the human genome is a random sequence with equal frequencies of A, G,

Question

a. Assume the human genome is a random sequence with equal frequencies of A, G, C, and T. If you were to choose a particular, randomly selected 6 bp piece to examine, what is the probability that one piece would have the sequence GTAGAC?

b. How many times would you expect that sequence GTAGAC to appear in the human genome by chance? (Note that the sequence could appear on either of the two strands of the double stranded DNA)

c. Considering the calculation in part b, what length would a DNA sequence need to be for you to expect it to occur no more than once within the genome?

Explanation / Answer

(a) The probability is 1/4 per bp. Since there are 6 bp, (1/4)6=1/4096. (b) (1/4096)(3.1x109bp/6)=126139 times (c) 4096*6=24576bp

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