Question

c) Identify the recognition sites in the given sequences
at which E.coli will be cut and make sticky ends.
S'-GAATTC-3'
3-CTTAAG-5 plz help me​

Answer 1

Answer:

sticky end after the cutting is :-

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5'- G A A T T C

C T T AA (and). G - 3'

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Important points:-

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• sticky ends are formed by the action of restriction endonuclease enzyme E.cori
• it always formed a palindromic sequence.

Answer 2

Explanation:

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A restriction enzyme (or restriction endonuclease) is an enzyme that cuts DNA at or near specific recognition nucleotide sequences known as restriction sites. Restriction enzymes recognize a specific sequence of nucleotides and produce a double-stranded cut in the DNA. Many of them are palindromic, meaning the base sequence reads the same backwards and forwards. Recognition sequences in DNA differ for each restriction enzyme, producing differences in the length, sequence and strand orientation (5' end or the 3' end) of a sticky-end of an enzyme restriction. If we cut the DNA with EcoR1, the DNA would be cut right in the middle. All the pieces would be the same size, which would be 15 kb long. Hence 5' GAATTC 3' ; 3' CTTAAG 5' palindrome sequence can be easily cut at about the middle by EcoR1 enzyme.