Question

Calculate the surface area and volume for a rod-shaped bacterium; consider it a perfect cylindrical microbe with 1,300 nm X 4,000 nm dimension.

Answer 1

The shape of the bacterium is cylindrical (rod-shaped)

Diameter of the bacterium = 1300 nm

Therefore radius (r) = 1300/2 = 650 nm

Height or length of bacterial cell (l) = 4000 nm

Surface area of bacterial cell = 2πrl + 2πr²

= 2πr (l + r)

= 2 x 3.14 x 650 ( 4000 + 650)

= 4082 (4650)

= 18981300 nm²

= 18.98 um² (square micrometers)

Volume of the bacterial cell = πr²l

= 3.14 x (650)² x 4000

= 5306600000 nm³

= 5.306 um³ (cubic micrometers)