So this a question about recombinant DNA, the questions are answered but can som

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So this a question about recombinant DNA, the questions are answered but can someone explain how to solve part c (1 &2) please?

Restriction enzymes are extensively used in molecular biology. Below are the recognition sites of two of these enzymes, BamHl and Bcll. a) BamHl, cleaves after the first G. Does cleavage by BamHl result in a 5' or 3 overhang? What is the sequence of this overhang? C 3' 5' 3 CCTA Cleavage by BamHI results in a 5' overhang as seen here: 2 cCTAG b) Bcll cleaves after the first T. Does cleavage by Bcll result in a 5' or 3 overhang? What is the sequence of this overhang? 5' T 3 ACTA 5' 5' T Cleavage by Bcll results in a 5' overhang as seen here c) You are given the DNA shown below 5' ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG 3' 3' TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC 5' i) If this DNA was cut with BamHl, how many DNA fragment would you expect? Write out the sequence of these double-stranded DNA fragments 5 ATTGAG 3' 3 TAACTCCTAG 5 5 GATCCGTAATGTGTCCTGATCACGCTCCACG 3' GCATTACACAGGACTAGTGCGAGGTGC 5 i) If the DNA shown above was cut with Bcll, how many DNA fragment would you expect? Write out the sequence of these double-stranded DNA fragments 5 ATTGAGGATCCGTAATGTGTCCT 3 3 TAACTCCTAGGCATTACACAGGACTAG 5 5' GA TGCGAGGTGC 5

Explanation / Answer

part c.

i) So recognition site for BamHI is- 5'-GGATCC-3' and the enzyme cuts after the first G. In the 5' end of the given sequence, from 6th base to 11th base, the recognition sequence stretch (GGATCC) is present. So the enzyme cuts after 6th base at forward strand and after 10th base at the complementary strand. As a result of this double-stranded break, the given two double-stranded fragments are produced, one with 5' overhang and one with 3' overhang.

ii) Recognition site for BcII is- 5'-TGATCA-3' and the enzyme cuts after the first T. In the 5' end of the given sequence, from 23rd base to 28th base, the recognition sequence stretch (TGATCA) is present. So the enzyme cuts after 23rd base at forward strand and after 27th base at the complementary strand. As a result of which, the given two fragments are generated as similar as the first case.

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