International A Level Biology / Microbiology

Calculating exponential growth and the growth rate constant revision notes

Edexcel International A Level Biology (YBI11), chapter 17: Microbiology, specification point 6.3, A.0.5, A.2.5. Free lesson.

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Bacteria double once per generation during exponential growth, so you can calculate their number using powers of 2. The growth rate constant, k, measures generations per hour and can be used to find the mean generation time.

How to calculate the number after several generations

Each generation doubles the population. After n generations, the starting number has therefore been multiplied by 2ⁿ. The number of generations is not the same as the time: divide the time by the generation time, using the same units for both. N is the number of bacteria at the end, and N₀ is the starting number. n is the number of generations. It is a count, so it has no unit.

How to calculate the growth rate constant

The growth rate constant, k, tells you how many generations occur per hour. Use two counts from the exponential phase. The difference between their base-10 logarithms measures the increase in population. Each doubling adds about 0.301 to the log, so dividing by 0.301 × time converts that increase into generations per hour. N₁ is the earlier count and N₂ is the later count. t is the time between the counts, in hours. k = (log₁₀ N₂ − log₁₀ N₁) ÷ (0.301 × t), in generations per hour.

How to find mean generation time

Once you know k, the mean generation time is the average time for the population to double. It is the reciprocal of k. If k is in generations per hour, the answer from 1 ÷ k is in hours. Mean generation time = 1 ÷ k. Convert hours to minutes when needed.

How to use values from a log graph

Check what the graph's axis shows. If the values are already labelled as log₁₀ N, they are logarithms, so use them directly in the subtraction. Taking their logs again gives the wrong result. For readings of 7.5 and 4.5 over 5 hours, k = (7.5 − 4.5) ÷ (0.301 × 5) = 3.0 ÷ 1.505 = 1.99 generations per hour. For 3.0 × 10³ rising to 4.8 × 10⁴ cells per cm³ in 2 hours, k = (4.681 − 3.477) ÷ (0.301 × 2) = 1.204 ÷ 0.602 = 2.0 generations per hour.

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