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4.4.2Genetics, Variation and Interdependence

Using Logarithms When Investigating Bacteria — AQA A Level Biology 7402 Revision Notes & Topical Questions

5 min read 9 parts Lesson 23 of 46 1 quiz

📚 AQA A Level Biology 7402 Hub / Topical Revision
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4.4.2 Using Logarithms When Investigating Bacteria — AQA A Level Biology 7402 notes: Use logarithms and exponential notation to describe and compare numbers of bacteria and calculate population growth.Bacteria can reproduce very rapidly by binary fission, so populations grow by h…

  • standard form
  • generation time
  • Log (exponential) phase
  • Stationary phase
  • Death (decline) phase
  • Exam-style question 1

Section 1 of 9Why use logarithms?

Bacteria can reproduce very rapidly by binary fission, so populations grow by huge amounts, from a few cells to many millions in hours.

Plotting such numbers on a normal linear scale is difficult because the large values make the early part of the graph seem flat.
Standard form
Biologists therefore use standard form and logarithms to handle numbers across many orders of magnitude.
A logarithm to base 10 is the power to which 10 is raised to give the number: log10(1000) = 3, because 103 = 1000.
  • A change of 1 in the log value is a ten-fold change in the number of cells.

Section 2 of 9Exponential growth and the doubling time

In ideal conditions, each bacterium divides into two at regular intervals, so the number doubles each generation: 1, 2, 4, 8, 16...

The number of bacteria after n divisions is N0 x 2n, where N0 is the starting number.
Generation time
The time taken to double is the generation time (doubling time).
For example, if one cell with a generation time of 20 minutes divides for 3 hours, there are 9 generations and the number of cells is 1 x 29 = 512.

For a starting population of 500 cells, with 6 divisions: 500 x 26 = 500 x 64 = 32,000 cells, which in standard form is 3.2 x 104.

Section 3 of 9Plotting a growth curve with log values

A typical bacterial growth curve in a closed culture has four phases, and the numbers of cells are best plotted as log10 of the number of cells against time.

1Lag phase
Lag phase cells adapt to the medium and synthesise enzymes; little division, so the log line is flat or rises slowly.
2Log (exponential) phase
Log (exponential) phase cells divide at a constant rate; on a log scale this is a straight line with a positive slope.
3Stationary phase
Stationary phase the rate of cell division equals the rate of cell death as nutrients become limiting and toxic waste accumulates,
  • so the line levels off.
4Death (decline) phase
Death (decline) phase death exceeds division, and the line falls.

The straight line in the log phase makes it easy to find the growth rate. Because a log scale compresses large numbers, you can see all four phases on one graph.

Section 4 of 9Calculating with logarithms

Calculators have a log button for base 10. To convert a count to a log value, press log and the number: log10(2.5 x 106) = 6.40. To convert back, calculate 10x: 106.40 = 2.5 x 106. If the log value rises from 4.0 to 6.0 in 8 hours, the numbers have increased 100-fold (102).

The specific growth rate or the number of generations can also be found.

Calculating with logarithms
The number of generations n = (log N - log N0) / 0.301, as log102 = 0.301.
  • For N0 = 103 and N = 106, n = (6 - 3) / 0.301 = 9.97, or about 10 generations.
If this took 5 hours, the generation time = 300 minutes / 10 = 30 minutes.

Section 5 of 9Estimating numbers of bacteria and links

Bacterial numbers can be estimated by a viable count: a sample of culture is serially diluted (often in ten-fold steps), spread on agar, incubated and the colonies counted.

1
The original concentration = number of colonies x dilution factor.
2
For example, 45 colonies from 0.1 cm3 of a 10-4 dilution means 45 x 104 / 0.1 = 4.5 x 106 cells per cm3.
3
Total counts using a haemocytometer include dead cells, whereas viable counts include only living cells that can form colonies.
4
This topic links to aseptic technique, growth of populations, antibiotic effects and the mathematics of the practical skills section.

Section 6 of 9Practice with bacterial growth data

Suppose a culture has 4.0 x 104 cells per cm3 at 2 hours and 6.4 x 105 cells per cm3 at 6 hours, within the log phase.

Practice with bacterial growth data
The increase is 16-fold, which is 24, so there were 4 generations in 4 hours and the generation time is 1 hour.
Using logs, log 6.4 x 105 is 5.81 and log 4.0 x 104 is 4.60, a difference of 1.21, and 1.21 / 0.301 = 4.0 generations, which agrees.
Checking an answer by two methods is a good habit.

Section 7 of 9Exam-style questions

  1. A culture starts with 2 x 103 bacteria. The generation time is 30 minutes. Calculate the number of bacteria after 4 hours, giving your answer in standard form. [3]3 marks

  2. Explain why the number of bacteria in a growth curve is often plotted as a logarithm. [2]2 marks

  3. A sample is diluted 10-3 and 0.1 cm3 is spread on agar. 36 colonies grow. Calculate the number of bacteria per cm3 in the original culture. [3]3 marks

Section 8 of 9Key terms

TermMeaning
LogarithmThe power to which a base must be raised to give a number.
Generation timeTime taken for a population of cells to double.
Binary fissionDivision of a bacterial cell into two.
Lag phaseEarly phase with little division as cells adapt.
Exponential phasePhase of constant, rapid doubling of the population.
Stationary phasePhase where division rate equals death rate.
Serial dilutionRepeated dilution of a sample by a constant factor.
Viable countA count of living cells able to form colonies.

Section 9 of 9Exam technique

1
Using 2n instead of 2 to the power n for the number of cells.
2
Forgetting that a rise of 1 on a log10 scale means a ten-fold rise.
3
Not converting time to the same units as the generation time.
4
Reading the stationary phase as the point where cells stop dividing; division equals death.
5
Confusing the dilution factor with the amount plated.

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