The genetic material of the Escherichia coli cell with a diameter of about 1 ? m

Question

The genetic material of the Escherichia coli cell with a diameter of about 1 ?m and a length of about 2 ?m, 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 anucleoid 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 mmand a length of about 2 mm.
Recall that V=pr2h for a cylinder.

Express your answer to two decimal places.

Explanation / Answer

First of all we will calculate the volume of the bacterial cell having diameter of 1µm (or radius of 0.5µm) and length of 2µm.

We know the volume of cylinder with length l and radius r is given by V=?r2l.

Therefore, vol of E. coli cell will be ?(0.5)2(2) cubic. µm = 1.57 cubic. µm

Now, lets think of DNA as a long thread of length 1.36 mm and thickness (dia) of 2nm. Therefore, to fit this DNA thread inside the cell we would like to wrap it like a spool with a diameter nearly equal to the dia of cell. We will then calculate the number of turns or coils on the spool, which we will use to calculate the length of DNA spool.

Let’s assume the dia of each coil of DNA is equal to the (actually slightly less than) the dia of cell. Therefore, circumference of each coil of DNA spool will be =? d = ?. 1µm = 3.14 µm

We will now calculate the no. of turns or coils on DNA spool-

No of DNA coils = Total length of DNA/circumference of each coil

                            = 1.36 x 103µm/ 3.14 µm= 1360 µm / 3.14 µm

                         = 433 coils

If the DNA is wrapped tightly enough, the length of the DNA spool can be given by

Thickness of DNA strand x No. of coils = 2nm x 433 = 866 nm

Now the vol of DNA spool will be = ?r2l = ? (0.5µm)20.866µm

                                                             = 0.68 cubic. µm

Now we will calculate what percentage of internal vol of bacterial cell will be occupied by the coiled DNA-

% vol occupied = 100 x vol of DNA spool/ Vol of cell = 100 x 0.68/1.57 = 43.31 %

Thanks!

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