Genetics: A Conceptual Approach

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Last Time Restriction enzymes (or restriction endonucleases) make double- stranded cuts in DNA at specific nucleotide sequences Recognition sequences: - are usually 4 to 8 base pairs long - are palindromic: they read the same forward and backward 5’-CAGTCGATCGTACGTTCGA AAGCTT AGCTGATGACGATAGC-3’ 3’-GTCAGCTAGCATGCAAGCT TTCGAA TCGACTACTGCTATCG-5’ Hind III site

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Hind III: 5’- AAGCTT-3’ Probability that Hind III site occurs in a random DNA sequence : In random sequence: A = 0.25 T = 0.25 G = 0.25 C = 0.25 P( Hind III) = P(5’-AAGCTT-3’) = (0.25) 6 = 0.000244 (or 1/4098 base pairs) Probability that Hind III site occurs in a genome with 60% A-T base pairs : In this genome: A = 0.30 T = 0.30 G = 0.20 C = 0.20 P( Hind III) = P(5’-AAGCTT-3’) = (0.3)(0.3)(0.2)(0.2)(0.3)(0.3) =
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Unformatted text preview: 0.000324 (or 1/3086 base pairs) Hind III: 5’- AAGCTT-3’ Probability that Hind III site occurs in a genome with 60% A-T base pairs: Random sequence: A = 0.30 T = 0.30 G = 0.20 C = 0.20 P( Hind III) = P(5’-AAGCTT-3’) = (0.3)(0.3)(0.2)(0.2)(0.3)(0.3) = 0.000324 (or 1/3086 base pairs) Number of Hind III sites in genome of 3,000,000,000 base pairs: # sites = genome size x probability of restriction site = 3,000,000,000 x 0.000324 = 972,000 Hind III sites in the genome Average size of DNA fragments produced by digestion with Hind III: average size = genome size / number of Hind III sites = 3,000,000,000 / 972,000 = 3086 base pairs...
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This note was uploaded on 05/06/2010 for the course BIO 311 taught by Professor Otto during the Spring '10 term at George Mason.

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Additional slides lecture 15 - 0.000324 (or 1/3086 base...

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