Sample Solution from
Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
3rd Edition
ISBN: 9780840064189
Chapter 1
Problem 1SP
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Textbook Problem

Suppose you have a culture of bacteria, where the density of each bacterium is 2.0 g / cm 3 .

  1. If each bacterium is 5 μ m × 5 μ m × 20 μ m in size, find the number of bacteria if their total mass is 30 grams. Recall that 1 μ m = 10 − 6 meters.
  2. Suppose that you learn that the sizes of bacteria range from 4 μ m × 5 μ m × 15 μ m to 5 μ m × 6 μ m × 25 μ m What is the range of the possible number of bacteria making up the total mass of 30 grams?

Expert Solution

a.

To determine

To find: The number of bacteria when total mass of culture of bacteria is given.

Answer

The number of bacteria in 30 gram is 3×1010 .

Explanation of Solution

Given:

The size of each bacteria is 5μm×5μm×20μm total mass of bacteria is 30 grams and density of bacterium is 2.0g/cm3, and 1μm=10-6m .

Formula used:

Mass of a bacteria = Size of bacterium × density of bacteria

Calculation:

The size of bacterium

  V=5μm×5μm×20μm=(5×10−4cm)×(5×10−4cm)×(20×10−4cm)=500×10−12cm3

  ( ∵1μm=10−6m=10−4cm )

Mass of bacterium mB=V× density of bacterium =500×10−12cm3×2.0gcm3=1000×10−12g=10−9g .

Therefore, number of bacteria in 30 g mass =30mB=3010−9=30×109=3×1010.

Conclusion:

The number of bacteria in 30 gram is 3×1010 .

Expert Solution

b.

To determine

To find: The range of possible number of bacteria of different size in 30 gram mass.

Answer

The range of number of bacteria of size 5μm×6μm×25μm and 4μm×5μm×15μm in 30 gram mass is 2×1010  −  5×1010 .

Explanation of Solution

Given:

The density of bacteria given in part (a)

Formula used:

The same as used in part (a).

Calculation:

The mass of bacteria of size 4μm×5μm×15μm

  m1B=(4×10-4cm)×(5×10-4cm)×(15×10-4cm)×2.0 gcm3=300×10−12×2=6×10−10g

Therefore, the number of bacteria of this kind in 30 gram mass =30m1B=306×10−10=5×1010.

The mass of bacteria of size 5μm×6μm×25μm

  m2B=(5×10-4cm)×(6×10-4cm)×(25×10-4cm)×2.0 gcm3=750×10−12×2=1.5×10−9g

Therefore, the number of bacteria of this kind in 30 gram mass =30m2B=301.5×10−9=20×109=2×1010

Thus, the range of number of bacteria of size 5μm×6μm×25μm and 4μm×5μm×15μm in 30 gram mass is 2×1010  −  5×1010 .

Conclusion:

The range of number of bacteria of size 5μm×6μm×25μm and 4μm×5μm×15μm in 30 gram mass is 2×1010  −  5×1010 .

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