Question

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

Part 1. calculate the volume of the e. coli cell and express the answer in um^3.

Part 2. using the same cylindrical model, calculate the surface area of an e. coli cell in um^2. (round to 2 decimal places.

Answer 1

The volume of an E. coli cell modeled as a cylinder is approximately 1.57 um³, and the surface area is approximately 7.85 um².

Step-by-step explanation:

Part 1: Volume of an E. coli Cell

To calculate the volume of the E. coli cell, we use the formula for the volume of a cylinder: V = πr²h, where V is the volume, r is the radius, and h is the height. Given the diameter is 1 micrometer, the radius is 0.5 micrometers. The length, which is the height h in this case, is 2 micrometers. Plugging in the values, we get:

V = π(0.5²)(2) = π(0.25)(2) = π(0.5) ≈ 1.57 um³.

Part 2: Surface Area of an E. coli Cell

The surface area A of a cylinder is calculated using the formula A = 2πrh + 2πr². Plugging in the radius and height:

A = 2π(0.5)(2) + 2π(0.5²) = 2π(1) + 2π(0.25) = 2π + 0.5π ≈ 7.85 um².

The volume of the E. coli cell is approximately 1.57 um³, and the surface area is approximately 7.85 um², with the surface area being rounded to two decimal places as requested.

Answer 2

`A=6.28\ \mu m^2`

Part 1

`V=1.57 \ \mu m^3`

Part 2

`A=6.28\ \mu m^2`

Step-by-step explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

`V=\pi r^2l`

`V=\pi * 0.5^2* 2 \ \mu m^3`

`V=1.57 \ \mu m^3`

Surface area,A

A=π d l

`A=\pi * 1 * 2\ \mu m^2`

`A=6.28\ \mu m^2`

Part 1

`V=1.57 \ \mu m^3`

Part 2

`A=6.28\ \mu m^2`