Q.46 A restriction endonuclease has a recognition site of 3 bases. Assuming random arrangement of nucleotides, the probability that this endonuclease will cut a piece of DNA is _______ (rounded off to three decimal places).

Q.46 A restriction endonuclease has a recognition site of 3 bases. Assuming random
arrangement of nucleotides, the probability that this endonuclease will cut a piece
of DNA is _______ (rounded off to three decimal places).

The probability that a restriction endonuclease with a 3-base recognition site cuts DNA at any specific position, assuming random nucleotide arrangement, is 0.016.

Calculation Method

DNA consists of four nucleotides (A, T, C, G), each with equal probability of 1/4 at any position in random sequence. For a 3-base recognition site, the enzyme requires a specific sequence of three consecutive nucleotides, so the probability is (1/4) × (1/4) × (1/4) = 1/64. This equals 0.015625, which rounds to 0.016 when taken to three decimal places.

CSIR NET Context

This question appears in CSIR NET Life Sciences exams, testing probability in molecular biology and restriction enzyme frequency. Some sources list 0.015, but standard rounding of 0.015625 yields 0.016 (nearest thousandth). No options are provided in the query, but the fill-in answer aligns with 1/64 for 3-base cutters versus rarer 6-base cutters (1/4096).

Article:

Restriction Endonuclease 3 Base Recognition Site: Probability Calculation Guide

In molecular biology, understanding how a restriction endonuclease 3 base recognition site determines the probability that this endonuclease will cut a piece of DNA is crucial for CSIR NET Life Sciences aspirants. Assuming random arrangement of nucleotides, this probability reveals cutting frequency in DNA cloning and genetic engineering.

The key formula stems from four equally likely bases (A, C, G, T). For a specific 3-base sequence, each position has 1/4 chance, yielding (1/4)^3 = 1/64 or 0.015625, rounded to 0.016. This means cuts occur every 64 bases on average in random DNA.

Step-by-Step Solution for CSIR NET Q.46

  1. Identify recognition length: 3 bases.

  2. Probability per position: (1/4)^3 = 1/64.

  3. Decimal: 1 ÷ 64 = 0.015625.

  4. Round to three decimals: 0.016 (0.015625 > 0.0155 threshold).

No multiple-choice options exist here—it’s numerical response. Common traps include confusing with 4-base (1/256) or 6-base (1/4096) enzymes.

Applications in Biotechnology

Restriction endonuclease enzymes like 3-base cutters fragment DNA frequently, ideal for small inserts in vectors. In random DNA, frequency scales as 4^n where n = bases, aiding restriction mapping.

Recognition Bases Probability Average Cut Frequency
3 0.016 (1/64) Every 64 bp 
4 0.004 (1/256) Every 256 bp 
6 0.000244 (1/4096) Every 4 kb 

Master this for CSIR NET genetics, molecular biology sections—practice with varying lengths.

Leave a Reply

Your email address will not be published. Required fields are marked *

Latest Courses