Question

The recognition sequence of restriction enzyme `A` is GGCC and that of `B` is GAATAC. Which statement among the following is true regarding the probable frequency of cutting sites in a genome?
(a) Both `A` and `B` cut equally frequently.
(b) `A` cuts once every 676 nucleotides and `B` 1024 nucleotides.
(c) `A` cuts every 256 nucleotides and `B` 4096 nucleotides.
(d) `A` cuts every 256 nucleotides and `B` 1024 nucleotides.

Answer 1

Hint: The frequency of cutting in a DNA sequence for a particular restriction enzyme is once every ${4^n}$, where n is the number of bases in the recognition sequence of the restriction enzyme and 4 is derived from the fact that there are 4 different possible nucleotides (G, A, T or C) that may be inserted in a position.

Complete answer:
 As the cutting frequency is ${4^n}$ therefore for restriction enzyme `A` it is ${4}^{4}$=256 and for `B` it is ${4}^{6}$=4096. Hence, restriction enzyme `A` cuts after every 256 base pairs, and restriction enzyme `B` cuts after every 4096 base pairs.
-A restriction enzyme cleaves DNA into fragments within a molecule at restriction sites.
-Every restriction enzyme has a recognition sequence through which it binds to the gene. Recognition sequence may be palindromic also (i.e., when read from backward or front it is the same).
-These enzymes are found in bacteria and provide a defense mechanism against viruses.
-The restriction enzymes identify the recognition sequence and cleave the DNA strand over there and as the recognition site is encountered again it cleaves the strand again.
-There are two types of restriction enzymes: Endonuclease which make cuts on DNA at specific positions.
Exonucleases are the ones that cut nucleotides from the ends of DNA.

So, the correct answer is `A` cuts every 256 nucleotides, and `B` cuts every 4096 nucleotides`.

Note:
Every restriction enzyme has its own recognition sequence through which it binds and then cleaves the DNA strand from there. So as the restriction enzyme encounters its recognition sequence it cleaves the DNA strand from there. So as many times the restriction enzyme encounters its recognition sequence it makes that many cuts in the DNA strand.