Generation Time Calculator
During exponential growth a bacterial population doubles at a fixed interval called the generation or doubling time. Enter two cell counts and the elapsed time to find the generation time, growth rate, and specific growth rate, or switch to reverse mode to predict how large a culture will be after a set number of doublings.
Formula
Worked example
A culture grows from N₀ = 1,000 to N = 64,000 cells in t = 120 min. Fold increase = 64. Generations n = log₂(64) = 6. Generation time g = 120 ÷ 6 = 20 min. Specific growth rate μ = ln(2) ÷ (20/60 hr) = 2.079 /hr. Growth rate constant k = ln(64) ÷ 120 = 0.0347 /min.
What generation time measures
Generation time, also called doubling time, is the average interval a bacterial population needs to double in number while it is growing exponentially. Because cells divide by binary fission, one cell becomes two, two become four, and the count climbs as a power of two rather than in a straight line. The number of doublings that occurred over a measured period is n = log₂(N ÷ N₀), and dividing the elapsed time t by that number gives g, the time for a single division. This calculator accepts time in minutes, hours, or days so you can enter measurements exactly as recorded without manual conversion.
Growth rate constant k and specific growth rate mu
Microbiologists use two related kinetic parameters alongside generation time. The growth rate constant k = ln(N ÷ N₀) ÷ t (units: per minute or per hour) describes how quickly the natural logarithm of the count rises, and is the slope of a semi-log plot of population versus time. The specific growth rate mu = ln(2) ÷ g converts generation time to a per-hour rate and is the standard input to Monod kinetics, yield calculations, and fed-batch models used in biotechnology and fermentation. A shorter generation time means a larger mu and a steeper growth curve.
Reverse mode: predicting a future population
If you already know the generation time of an organism (from the literature or a previous experiment) you can flip the calculator into reverse mode to predict how large the culture will be after a chosen period. The formula is N = N₀ × 2^(t/g). This is useful for planning harvest times, estimating contamination growth, or scaling fermentation volumes. Toggle "Show generation-by-generation projection" to see the population at every doubling, which helps identify the best sampling windows.
How to take valid measurements
The calculation only holds during the exponential, or log, phase of the growth curve, when nutrients are plentiful and division proceeds at a constant maximum rate. After inoculation a culture first sits in lag phase while cells adjust, and it later enters stationary phase as nutrients run out and waste accumulates. Sampling across those phases mixes growing and non-growing intervals and overstates the true generation time. To get a clean value, take both counts well inside log phase, use the same measurement method and units for N₀ and N, and keep temperature and medium constant between the two readings.
Typical generation times of common bacteria and yeasts
| Organism | Generation time | Specific growth rate (μ) | Notes |
|---|---|---|---|
| Escherichia coli | ~20 min | ~2.08 /hr | Rich broth, 37 °C |
| Staphylococcus aureus | ~28-30 min | ~1.39 /hr | Optimal growth |
| Bacillus subtilis | ~26 min | ~1.60 /hr | Log phase, LB |
| Pseudomonas aeruginosa | ~35 min | ~1.19 /hr | Nutrient broth |
| Lactobacillus acidophilus | ~66-87 min | ~0.48 /hr | MRS broth, 37 °C |
| Saccharomyces cerevisiae | ~90 min | ~0.46 /hr | YPD medium, 30 °C |
| Mycobacterium tuberculosis | ~15-20 hr | ~0.04 /hr | Slow grower |
| Treponema pallidum | ~30 hr | ~0.02 /hr | Very slow grower |
Approximate doubling times under favourable laboratory conditions; actual values vary with medium, temperature, and inoculum history.
Frequently asked questions
What is the difference between generation time and specific growth rate?
Generation time (g) is the time for one doubling, measured in minutes or hours. The specific growth rate mu = ln(2) ÷ g converts that to a per-hour rate and is the standard kinetic parameter in microbiology. A generation time of 20 min (0.333 hr) gives mu = 0.693 ÷ 0.333 = 2.08 /hr. Smaller g means larger mu and faster growth.
What is the difference between k and mu?
The growth rate constant k = ln(N ÷ N₀) ÷ t is the slope of a semi-log growth plot and has units of per-time. The specific growth rate mu = ln(2) ÷ g is equivalent to k only when the base used for generations is e (natural). In the classic binary-fission framework, k = mu = ln(2)/g, so they are numerically identical. Some textbooks use k with base-10 logarithms (k = log₁₀(N/N₀) / t), which gives a slightly different value; this calculator uses natural logarithms for both.
Why does generation time use log base 2 but mu uses natural log?
Each binary-fission event doubles the population, so counting doublings is most natural in base 2: n = log₂(N ÷ N₀). The specific growth rate mu, however, is defined through the continuous exponential differential equation dN/dt = mu × N, whose solution is N = N₀ × e^(mu × t), making natural logarithm the correct choice. The two connect through mu = ln(2) ÷ g, because ln(2) = log₂(e)^(-1).
Does this work outside the exponential growth phase?
No. The formula assumes a constant doubling rate, which only happens during the exponential or log phase. During lag phase cells are adapting and barely dividing, and during stationary phase growth stops as resources are exhausted. Measuring across those phases blends growing and non-growing time and produces a generation time that is too long to be meaningful. Take both readings while the culture is actively doubling.
How do I use optical density (OD600) instead of cell counts?
OD₆₀₀ is proportional to cell concentration in the linear range (roughly 0.05 to 0.7 absorbance units), so you can enter raw OD values directly as N₀ and N in this calculator as long as both readings are within the linear range and the same cuvette path length was used. The resulting generation time is identical to what you would get from direct counts because the fold increase N/N₀ is the same in both units.
How accurate is the reverse-solve prediction?
The prediction assumes strictly exponential growth with no lag or stationary phase, no nutrient limitation, and constant temperature. In practice, cultures start to slow as density rises and nutrients deplete. Use the reverse-solve result as an upper-bound estimate and verify with a real OD or count measurement before making critical decisions (e.g., before a transformation, induction, or harvest step).