Microbiology

Colony Count & CFU Workbench

Reconcile plate counts across volumes and dilutions, normalize liquid or extracted samples, and keep zero and TNTC observations honest.

Biology · experimental measurements

Trace each plate back to the volume of original sample it represents.

Private calculations in your browser · explicit inputs and model boundaries
Example preview · Two dilution levelsColonies versus original-equivalent volume
00360.0025720.0051080.00751440.01A: 0.01, 144B: 0.01, 118C: 0.001, 26D: 0.001, 20Original-equivalent volume (mL)Colonies

Points are exact plate counts. The line shows the pooled concentration multiplied by each original-equivalent volume. It passes through zero; departures can reflect sampling or method differences, not a diagnosis.

  1. 1EnterProvide the known values
  2. 2CalculateResults update automatically
  3. 3VerifyReview the details and units
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Plate records
1 row
Row 1

Empty rows are ignored until edited. Keep commas and tabs out of individual entries; use the paste view for comma- or tab-separated records.

One row per plate: label, colonies (whole count or TNTC), plated volume, cumulative reciprocal dilution. Enter 100 for 1:100, including every prior dilution.

Calculation result

Enter valid values to see the result.

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Understand the relationship

The reasoning behind the result

A dilution changes the volume of original sample represented

Ei = Vi / Di; expected coloniesi = C Ei

Vi is the plated volume in mL, Di is cumulative reciprocal dilution and Ei is original-equivalent volume in mL. A 0.1 mL plate from a 1:100 dilution represents 0.001 mL of original suspension. The reciprocal dilution is 100, not 0.01.

Include all dilution steps between the stated original suspension and the plate. In extract modes the original suspension is the final extract; its mass or surface normalization happens afterward and must not be counted as a second dilution.

Pool counts against exposure, not against the number of plates

C = (N1 + N2 + …) / (E1 + E2 + …)

Ni is an exact colony count. Dividing total colonies by total original-equivalent volume produces an exposure-weighted concentration. An arithmetic average of individual CFU/mL estimates gives small and large exposure volumes equal influence and can produce a different answer.

The plot compares observed colonies with C × Ei. A discrepancy can reflect random counts, clumping, crowding, recovery or an incorrect dilution. The line visualizes the pooled estimate; it does not certify that the records meet a method's eligibility rules.

An extract needs its own sample denominator

Recovered CFU per g or cm² = C × Vextract / Qsample

Vextract is the final recovered extract volume, and Qsample is the original sampled mass in grams or area in cm². The product C × Vextract estimates the colony-forming units represented by that extract. Dividing by Qsample changes the reporting basis.

The volume of diluent added may differ from the final extract volume. Recovery can also be incomplete or variable. This calculation reports an entered measurement model and never invents an extraction-efficiency correction.

Zero counts and a detection limit are different information

Zero recorded colonies gives a zero point estimate. Under an ideal common-rate Poisson model, the probability of zero counts is exp(−C × E), where E is total exposure. Setting that probability to 0.05 gives the one-sided 95% upper bound −ln(0.05)/E.

The optional central 95% interval instead assigns 2.5% to each tail. With zero counts its upper bound is −ln(0.025)/E, so it differs from the one-sided bound. For positive counts the central interval inverts Poisson tail probabilities. All limits are normalized to the selected sample basis and exclude uncertainty in volumes, dilutions and recovery.

One count divided by exposure is a resolution step, not a validated detection limit. TNTC is censored information, not zero or a known maximum; it prevents this uncensored estimate and its interval from being reported.

Follow the numbers

Pooling two dilution levels

  1. The first two plates each represent 1/100 = 0.01 mL of original suspension. The next two each represent 1/1000 = 0.001 mL.
  2. The four plates contain 144 + 118 + 26 + 20 = 308 colonies in 0.022 mL of original-equivalent volume.
  3. The pooled concentration is 308/0.022 = 14,000 CFU/mL. It predicts 140 colonies on each first-level plate and 14 on each second-level plate.

The estimate reconciles all exact counts against the volume actually represented, rather than averaging differently diluted counts.

Quick guide

How to use this calculator

  1. Choose liquid, extracted mass or sampled surface as the reporting basis.
  2. Enter each comparable plate with its own volume and cumulative dilution; retain zero and TNTC observations.
  3. For extracts, provide the final recovered volume and the original mass or surface area.
  4. Inspect the volume ledger, per-plate estimates and pooled result. Enable the probability interval only if its sampling assumptions apply.

Calculation method

Calculation and interpretation

Trace each plate back to the volume of original sample it represents.

Ei = plated volumei / reciprocal dilutioni; C = Σcolonies / ΣEi; CFU per sample unit = C × final extract volume / original sampled quantity

Worked example

Pooling two dilution levels

The estimate reconciles all exact counts against the volume actually represented, rather than averaging differently diluted counts.

Ei = plated volumei / reciprocal dilutioni; C = Σcolonies / ΣEi; CFU per sample unit = C × final extract volume / original sampled quantity

Supported inputs

Precision and limits

Method-specific measurement

No organism, medium, incubation condition, countable range, recovery correction, laboratory approval or product-safety criterion is selected here.

Censored observations

TNTC is not an exact count. A censored-count analysis must follow the actual method; this tool retains these records without inventing a pooled value.

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