Read the given passage and answer the questions that follow:
In prokaryotes such as E. coli, though they do not have a defined nucleus the DNA is not scattered throughout the cell. The DNA is held with some proteins in a specific region. In eukaryotes this organisation is much more complex. There is a set of positively charged basic proteins called histones. Histones are organised to form a unit of 8 molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer.
If the length of E. coli DNA is 1.36 mm, what will be the number of base pairs in E. coli?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 x 106
In a DNA double helix the distance between two neighbouring base pairs is 0.34 nm. So the number of base pairs equals the length of the DNA divided by 0.34 nm.
- Length of E. coli DNA = 1.36 mm = 1.36 × 10−3 m
- Distance per base pair = 0.34 nm = 0.34 × 10−9 m
- Number of base pairs = (1.36 × 10−3) ÷ (0.34 × 10−9) = 4 × 106 bp
The value 6.6 × 109 bp in Option (4) is the book’s figure for a typical mammalian cell, not for E. coli. Options (1) and (2) do not come out of the division. The book lists E. coli as about 4.6 × 106 bp; with the 1.36 mm length given here, the division gives 4 × 106 bp.
NCERT Extract:
Packaging of DNA Helix – “Taken the distance between two consecutive base pairs as 0.34 nm (0.34×10–9 m), if the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with distance between two consecutive bp, that is, 6.6 × 109 bp × 0.34 × 10-9m/bp), it comes out to be approximately 2.2 metres.”