Question

# The genetic material of the Escherichia coli cell with a diameter of about 1 μm and...

The genetic material of the Escherichia coli cell with a diameter of about 1 μm and a length of about 2 μm, consists of a DNA molecule with a diameter of 2 nm and a total length of 1.36 mm. (The molecule is actually circular, with a circumference of 1.36 mm.) To be accommodated in a cell that is only a few micrometers long, this large DNA molecule is tightly coiled and folded into a nucleoid that occupies a small proportion of the cell's internal volume. Calculate the smallest possible volume the DNA molecule could fit into, and express it as a percentage of the internal volume of the cylindrical bacterial cell with a diameter of about 1 μm and a length of about 2 μm. Recall that V=pr2h for a cylinder.

Smallest possible volume the DNA molecule could fit into
= Cross-section Area of DNA * Length of DNA
= Pi * (Diameter of DNA/2)2 * Length of DNA
= 22/7 * (2*10-9 m / 2)2 * 1.36*10-3 m
= 4.27 * 10-21 m3

Volume of cylindrical bacterial cell
= Cross-section Area of Cell * Length of Cell
= Pi * (Diameter of Cell/2)2 * Length of Cell
= 22/7 * (1*10-6 m / 2)2 * 2*10-6 m
= 3.14 * 10-18 m3

Percentage of the internal volume
= 4.27 * 10-21 m3 / 3.14 * 10-18 m3
= 0.135%

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