CBSE Class 12 Biology PYQs

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From the CBSE Class 12 Biology Sample Paper 2027.

A stretch of an euchromatin has 200 nucleosomes, how many base pairs (bp) will be there in the stretch and what would be the length of the typical euchromatin?

1. 20,000 bp and 13,000 × 10⁻⁹ m
2. 10,000 bp and 10,000 × 10⁻⁹ m
3. 40,000 bp and 13,600 × 10⁻⁹ m
4. 40,000 bp and 13,900 × 10⁻⁹ m

Answer

C. 40,000 bp and 13,600 × 10⁻⁹ m

Explanation

A nucleosome is a DNA–histone packaging unit. The school-level average repeat is about 200 base pairs, including the DNA associated with the core and linker region. It is not exactly the core-wrapped length alone, which is smaller.

The question gives 200 nucleosomes. First convert that unit count to DNA base pairs using the assumed average: 200 nucleosomes × 200 bp per nucleosome = 40000 bp. The nucleosome units cancel.

Adjacent base pairs in the standard B-DNA length calculation are separated by about 0.34 nm, and 1 nm = 10⁻⁹ m.

DNA length = 40000 bp × 0.34 nm/bp = 13600 nm = 13600 × 10⁻⁹ m = 1.36 × 10−5 m = 13.6 μm. The bp units cancel, leaving length. Multiply by 0.34; do not divide by it or treat 200 nucleosomes as only 200 base pairs.

Final answer: C, 40000 bp and 13600 × 10⁻⁹ m.

A and B use the wrong DNA amount and length. D gives the correct base-pair count but an inconsistent length. C matches both multiplication steps.

Important wording limit: this is the contour length of the DNA if extended according to the base-pair spacing, not the actual end-to-end length of a packed euchromatin fibre. Packaging shortens its spatial extent, and the QP's phrase “length of euchromatin” should not erase that distinction.

See this sample paper