Q.In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive bases increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix (which has 2×10^9 bp) in the presence of saturating amount of this compound.

When an intercalating compound inserts itself between every pair of stacked nitrogenous bases, it increases the distance between consecutive base pairs. Normally, in the B-form DNA double helix, two consecutive base pairs are separated by 0.34 nm. Here, the compound stretches that distance to 0.44 nm.

The DNA molecule in question contains 2×10⁹ base pairs. Since the compound is present in saturating amounts, it intercalates between every consecutive pair, effectively making the pitch per base pair 0.44 nm instead of the standard 0.34 nm.

To find the total length of the double helix under these conditions:

• Number of base pairs = 2×10⁹
• Distance per base pair (with compound) = 0.44 nm
• Total length = 2×10⁹ × 0.44 nm = 0.88×10⁹ nm = 8.8×10⁸ nm …

The question describes a classic molecular biology experiment involving intercalating agents — compounds that slip between the stacked nitrogenous bases of the DNA double helix like cards sliding into a deck. These agents don't break bonds; they simply wedge themselves into the narrow gaps between consecutive base pairs, forcing the helix to stretch.

Under normal circumstances, the DNA double helix maintains a remarkably regular geometry. Each base pair sits 0.34 nm above the next, a spacing determined by the van der Waals forces and π-π stacking interactions between the flat aromatic rings of the bases. This 0.34 nm rise per base pair is one of the defining structural features discovered through X-ray crystallography and is central to understanding DNA's compactness.

When the intercalating compound saturates the DNA — meaning it occupies every available space between base pairs — it pushes each pair further apart. The distance increases from 0.34 nm to 0.44 nm, an expansion of 0.10 nm per base pair.

Now we calculate the total length in both states.

Normal DNA (without intercalator):

The DNA molecule contains 2×10⁹ base pairs. Since each pair contributes 0.34 nm to the length:

Length = (number of base pairs) × (distance per base pair)

Length = 2×10⁹ × 0.34 nm

Length = 0.68×10⁹ nm

Length = 6.8×10⁸ nm

Converting to more convenient units (1 nm = 10⁻⁹ m):

Length = 6.8×10⁸ × 10⁻⁹ m = 0.68 m = 68 cm …

Instead of computing both lengths independently, use a ratio: the intercalator changes only the rise-per-base-pair, from 0.34 nm to 0.44 nm, a scale factor of 0.44/0.34 ≈ 1.29 -- so the treated length is simply about 1.29 times the untre …