Question

Eco $RI$ and Rsa $I$ restriction endonucleases require $6$ 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 Eco RI and 10 Rsa I
• B. 1 Eco RI and 29 Rsa I
• C. 4 Eco RI and 69 Rsa I
• D. 2 Eco RI and 39 Rsa I

Answer 1

Answer: (D)

The frequency by which a restriction enzyme will cut DNA is mainly a function of the length of the sequence, it is sensitive to, like Rsa I recognition sequence has $4$ base pairs. Out of sheer probability, we see that with $4$ positions and each position having potentially $4$ different values there are $4^4 = 256$ different possibilities for given $4-$ base long strand. Therefore, theoretically, this enzyme will cut $1$ in $256 4$-base pair long sites. On the other hand, Eco RI recognises a sequence of $6$ base pairs $4^6 = 4096$ possible combinations with this length. Total DNA length =$ 10\, kb = 10,000 ,bp $ So, probable cleavage would be
Eco RI$=\frac{10000}{4096}=2.44$
Rsa I $=\frac{10000}{256}=39.06$