OD600 GUIDE

OD600 during bacterial growth: growth curves and log phase

How to read lag, exponential, and stationary phases from an OD600 growth curve, calculate doubling time, and know when OD stops tracking viability.

Repeated OD600 readings over time produce a growth curve. Read against the culture’s physiology, the curve tells you where the culture is, how fast it is growing, and when to act on it.

The phases

Calculating doubling time

Within the log phase, doubling time follows from any two points:

doubling time (h) = (t₂ − t₁) × log₁₀2 ÷ (log₁₀OD₂ − log₁₀OD₁)

Worked example: OD 0.10 at 2 h and OD 0.40 at 4 h. log₁₀(0.40) − log₁₀(0.10) = 0.602; (4 − 2) × 0.301 ÷ 0.602 = 1.0 h. Points must be genuinely inside the log phase — including a lag-phase point flattens the estimate, and including a stationary-phase point deflates it. When in doubt, fit a line to several log-phase points on a log-transformed plot instead of trusting two.

Inoculating to a target starting OD

To start a culture at OD 0.05 from a dense starter, apply C₁V₁ = C₂V₂: the volume of starter = (target OD × target volume) ÷ starter OD. A 50 mL culture at OD 0.05 from a starter at OD 2.0 needs 0.05 × 50 ÷ 2.0 = 1.25 mL. The OD600 calculator includes an inoculation helper that does this arithmetic and shows the substituted values.

Where OD600 misleads

From curve to specific growth rate

Doubling time is the intuitive form; the formal quantity is the specific growth rate μ:

μ (per hour) = (ln OD₂ − ln OD₁) ÷ (t₂ − t₁)
doubling time = ln 2 ÷ μ ≈ 0.693 ÷ μ

For the example above — OD 0.10 at 2 h and OD 0.40 at 4 h: μ = (ln 0.40 − ln 0.10) ÷ 2 = 1.386 ÷ 2 = 0.693 h⁻¹, and the doubling time is 0.693 ÷ 0.693 = 1.0 h, matching the log₁₀ calculation. μ is the quantity to report when comparing growth across conditions, because it composes cleanly: in a mixture of organisms, the observed μ reflects the fastest growing population, which is why a curve alone cannot establish purity.

Include a medium-only control well or tube alongside the experiment. Its reading over time exposes evaporation drift, condensation on lids, and instrument warm-up — artefacts that otherwise appear as slow “growth” in every sample.

Practical experimental design

Blank with the medium used in the experiment, keep cuvette or plate layout consistent, and include biological replicates — technical replicates catch pipetting errors, biological replicates catch the variation that actually matters. When comparing growth rates between conditions, compute μ per replicate first and compare the replicate values, rather than fitting one averaged curve; the spread between biological replicates is the honest error bar. Sample frequently through the expected log phase (every 15–30 minutes for fast growers) and less often elsewhere; missing log phase is the most common design error. Estimate cells/mL from each reading with the OD600 calculator, using a calibration from building an OD600 to cells/mL calibration — and remember that in stationary and decline phases, OD-based estimates diverge ever further from viable counts.

Frequently asked questions

How do I calculate doubling time from OD600?

Take two readings inside the exponential phase and apply: doubling time = (t₂ − t₁) × log₁₀2 ÷ (log₁₀OD₂ − log₁₀OD₁). For OD 0.10 at 2 h and OD 0.40 at 4 h, that is 2 × 0.301 ÷ 0.602 = 1.0 hour. The calculation is only valid while growth is exponential and the readings are inside the instrument's linear range.

Why does OD600 stay high when cells die?

Dead cells still scatter light. In the decline phase, CFU can fall by orders of magnitude while OD600 barely moves, because lysis — not death — is what removes scattering material. Pair plate counts with OD whenever viability matters more than biomass.

What starting OD should I use for a growth experiment?

Common starting OD600 values are 0.01–0.05 for routine growth curves, low enough that the full lag and log phases are captured. The inoculation helper in the OD600 calculator computes the volume of starter culture needed to reach a chosen starting OD via C₁V₁ = C₂V₂.

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