Q.64 The possible number of SalI restriction sites in a 9 kb double-stranded DNA,
with all four bases occurring in equal proportion (rounded off to the nearest integer)
is _________.
Calculating SalI Restriction Sites in 9 kb DNA
SalI recognizes the palindromic sequence G^TCGAC, a 6-base pair (bp) site where cleavage occurs between G and T. In DNA with equal base proportions (25% each A, T, C, G), the expected number of sites follows a probabilistic model based on random sequence distribution. For a 9 kb (9000 bp) double-stranded DNA, the calculation yields 2 sites when rounded to the nearest integer.
✅ Correct Answer: 2
Key Calculation: 9000 bp ÷ 4096 ≈ 2.20 → Rounded to 2
Recognition Sequence
SalI cuts at G/TCGAC (5′-3′) and its complement C/GAGTC, producing sticky ends. This 6-bp specificity means each position has a 1/4 probability of matching (since p(G)=p(T)=p(C)=p(A)=0.25). The full site probability is (1/4)⁶ = 1/4096 per possible starting position.
Probability Calculation
The average number of sites equals total possible positions times site probability. In 9000 bp dsDNA, SalI sites can start at ~8995 positions (accounting for overlap negligible at 6 bp). Thus, expected sites = 8995 / 4096 ≈ 2.196. Rounded to nearest integer: 2.
• Probability per site:
(1/4)⁶ = 1/4096• Positions: 9000 – 6 + 1 = 8995
• Expected value:
8995 × (1/4096) ≈ 2.196• Nearest integer: 2 (since 0.196 < 0.5)
Common approximation uses 9000 / 4096 ≈ 2.20, still rounding to 2.
Why This Matters in Research
Such calculations predict fragment sizes for cloning or mapping, vital in genetics and biotech experiments. Fewer sites (e.g., for 6-cutters like SalI) suit large inserts compared to frequent cutters like 4-bp enzymes.
