Question

EcoR1 and Rsa1 are restriction endonucleases that need 6 and 4 bp sequences respectively for cleavage. In a 10 kb DNA fragment, how many probable cleavage sites are present for these enzymes?
A.0 EcoR1 and 10 Rsa1
B.1 EcoR1 and 29 Rsa1
C.4 EcoR1 and 69 Rsa1
D.2 EcoR1 and 39 Rsa1

Answer 1

Hint:Restriction enzymes are the enzymes that cut DNA fragments at specific sites. They recognize a sequence of base pairs and cut at specific sites and are used in recombinant DNA technology.

Complete step by step answer: Restriction enzymes are of two types. They are exonucleases and endonucleases. Exonuclease cuts DNA at the end sites while endonucleases cut at specific sites within the DNA.
Restriction endonucleases inspect DNA, find a recognition sequence of 4 to 8 nucleotides palindrome sites, and cut a little away from the center of palindrome sites. It produces single-stranded ends called sticky ends. Know that the same restriction enzymes cut the vector DNA and foreign DNA which are later joined by DNA ligases.
Here, EcoR1 cuts at 6 base pair sequences and Rsa1 at 4 base pair sequences.
As there are 4 nitrogen bases in total, the total number of possibilities for cleavage will be 4096 for EcoR1 and 256 for Rsa1.
We get this by taking a permutation combination as there are 4 nitrogen bases in total, the same for both enzymes.
As the total amount of DNA is 10000, dividing the probabilities, we get 2 possible cleavage sites for EcoR1 and 40 for Rsa1.
Hence the correct option is D.

Note:Palindrome sites are the site of recognition for restriction enzymes. We get the same reading frame when they are read both forward and backward. Palindrome sequence in DNA is a sequence of base pairs that can be read in the same manner when they are read both forward and backward, in both strands.
We cut all the DNA fragments by the same restriction enzyme so that we get the same kind of sticky ends that can be joined later by ligases. If we cut vectors and DNA by different enzymes, recombinant DNA can’t be generated.