Q.46 Given below are two statements : One is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) :Restriction enzymes its particular recognition sequence and cut the
ss DNA / ds DNA to produce cohesive and blunt fragments.
Reasons (R) :Restriction enzyme with A — base recognition will cut, on average. every
256 base pairs if all four nucleotides are present in equal proportions.
In the light of the above statements, choose the most appropriate answer from the options given
below :
(1)Both (A) and (R) are correct and (R) is the correct explanation of (A)
(2)Both (A) and (R) are correct but (R) is NOT the correct explanation of (A)
(3)(A) is correct but (R) is not correct
(4)(A) is not correct but (R) is correct
Both Assertion (A) and Reason (R) are correct, but (R) is NOT the correct explanation of (A). Option (2) is correct.
Option Analysis
(1) Both correct; (R) explains (A): Incorrect. While both statements are true, (R)’s frequency calculation (4^4 = 256 bp for a 4-base site) doesn’t explain how enzymes produce cohesive/blunt ends.
(2) Both correct; (R) NOT explanation: Correct. (A) describes cutting mechanism (recognition → cohesive 5’/3′ overhangs or blunt ends on dsDNA; ssDNA typically not cut). (R) gives statistical cut frequency assuming equal A/C/G/T, but unrelated to end types.
(3) (A) correct, (R) incorrect: Wrong. (R) is accurate—4-base recognition (e.g., A-only misread; actually typical like GATC) cuts ~every 256 bp (1/4^4).
(4) (A) incorrect, (R) correct: False. (A) is true for Type II enzymes (dsDNA target; cohesive e.g., EcoRI sticky ends; blunt e.g., SmaI).
Restriction enzymes recognition sequence questions dominate GATE Life Sciences molecular biology sections. This assertion-reason tests cutting mechanisms vs. statistical frequency.
Cutting Mechanism (A)
Type II enzymes bind palindromic sites (4-8 bp, e.g., EcoRI: GAATTC), hydrolyzing phosphodiester bonds. Cohesive ends (sticky, overhangs); blunt (straight cut). Primarily dsDNA; ssDNA rarely/inefficiently.
Frequency Calculation (R)
For n-base site, random equal nucleotides: cut every 4^n bp. 4-base (e.g., ~TaqI TCGA): 4^4 = 256 bp average. Explains fragment size in digests, not end types.
| Aspect | Assertion (A) Details | Reason (R) Details |
|---|---|---|
| Focus | Recognition → Cut type | Statistical spacing (4^n bp) |
| Examples | EcoRI (cohesive); EcoRV (blunt) | 6-base: every 4096 bp |
| Relevance | Cloning (sticky ends ligate easy) | Digest prediction (gel patterns) |
Exam Strategy
GATE BT favors (2) for unrelated true statements. Note: Real sites mixed bases (not “A-base”); equal proportions assumed. Practice with NEB enzyme tables. Ideal for recombinant DNA prep.
