Q.42 A 152 nm long Watson–Crick double helical DNA (B–DNA) will contain _____ turns.
A 152 nm long Watson-Crick double helical B-DNA contains 45 turns. This calculation uses standard structural parameters for B-DNA, where the helical rise is 0.34 nm per base pair and approximately 10 base pairs occur per helical turn.
Calculation Steps
First, determine the total number of base pairs (bp) in the DNA:
Length = 152 nm, rise per bp = 0.34 nm.
Number of bp = 152 / 0.34 ≈ 447 bp.
Next, divide by base pairs per turn (10 bp/turn for B-DNA):
Number of turns = 447 / 10 ≈ 44.7, which rounds to 45 turns for exam contexts.
This matches CSIR NET-level precision, as competitive exams often use 10 bp/turn exactly.
B-DNA Parameters
B-DNA, the predominant physiological form, features:
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Helical rise: 0.34 nm (3.4 Å) per bp.
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Base pairs per turn: ~10 (precisely 10–10.5).
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Pitch (full turn length): ~3.4 nm.
Variations like 10.5 bp/turn yield ~42.6 turns, but 10 bp/turn is standard for such problems.
Common Exam Options Explained
MCQ options often test these values:
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45 turns: Correct, using 10 bp/turn (447 bp / 10).
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44 turns: If using ~10.4 bp/turn (447 / 10.4 ≈ 43, rounds down).
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50 turns: Error assuming 0.3 nm rise (507 bp / 10).
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42–43 turns: Precise 10.5 bp/turn (447 / 10.5 ≈ 42.6).
Introduction to 152 nm Watson-Crick B-DNA Turns
Discover how many helical turns exist in a 152 nm long Watson-Crick double helical B-DNA segment. This key CSIR NET calculation uses B-DNA’s 0.34 nm rise per base pair and 10 base pairs per turn, yielding exactly 45 turns for precise problem-solving in molecular biology exams.
Core Structural Features of B-DNA
Watson and Crick’s 1953 model defines B-DNA as the right-handed double helix dominant in vivo. Key metrics include:
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Axial rise: 0.34 nm per base pair.
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Base pairs per helical turn: 10 (idealized for calculations).
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Pitch: 3.4 nm (full 360° turn).
These parameters enable straightforward length-to-turns conversion, vital for competitive exams like CSIR NET Life Sciences.
Step-by-Step Calculation: 152 nm B-DNA Turns
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Compute base pairs: Total length ÷ rise per bp = 152 nm ÷ 0.34 nm/bp ≈ 447 bp.
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Compute turns: Base pairs ÷ bp per turn = 447 ÷ 10 = 44.7 ≈ 45 turns.
Using 10.5 bp/turn gives ~42.6 (less common in MCQs). Verify with Python: confirms 44.7 rounds to 45.
| Parameter | Value | Source |
|---|---|---|
| Length | 152 nm | Query |
| Rise per bp | 0.34 nm | |
| bp per turn | 10 | |
| Total turns | 45 | |
Exam Tips and Common Pitfalls
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Always assume B-DNA means 10 bp/turn unless specified.
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Convert units carefully: 1 nm = 10 Å; rise is 3.4 Å = 0.34 nm.
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Options like 44 arise from 10.4 bp/turn—stick to 10 for integers.
For CSIR NET, practice variants: e.g., 45,000 bp DNA has 4,500 turns.
Comparison: B-DNA vs. Other Forms
| DNA Form | bp/Turn | Rise/bp | Pitch |
|---|---|---|---|
| B-DNA | 10 | 0.34 nm | 3.4 nm |
| A-DNA | 11 | 0.26 nm | 2.9 nm |
| Z-DNA | 12 | 0.37 nm | 4.4 nm |
B-DNA’s parameters make it ideal for such queries.
