How to calculate cell doubling time (with a worked example)

Work out the doubling time of a cell culture from two or more counts, using the log-linear fit, and spot the lag phase that throws it off.

Doubling time is the time a growing culture takes to double its cell number: with two counts it is the time between them times ln 2, divided by the natural log of how many times the count grew. The doubling time calculator does this for two counts, or fits a straight line through three or more on a log scale and gives an interval.

What you need

  • Two or more cell counts from the same culture, each with the time it was taken. Four to six counts, one a day, work much better than two.
  • Counts measured the same way every time: cells per mL, cells per well or total cells. Mixing units breaks the maths.
  • A rough idea of when the cells were growing steadily. Counts taken just after seeding (the lag) or after the dish filled up (the plateau) should be left out.

Step by step

  1. Get the counts. Count each sample with a hemocytometer or a cell counter. The hemocytometer cell counter gives cells per mL from a square-by-square tally.
  2. Type them in. On the doubling time page, choose hours or days, then add a row for each count with its time. You can also open Paste a table and paste two columns (time and count) from a spreadsheet.
  3. Read the headline. With two counts the tool uses the exact two-point formula. With three or more it fits ln(count) against time by least squares and shows the doubling time, the growth rate, a 95 % interval, R² (how close the points lie to a straight line) and the number of generations.
  4. Look at each interval. The table “Between each pair of counts” gives a doubling time for every step. A step much slower than the rest, at the start or the end, is usually lag or plateau.
  5. Untick the odd counts. Clear the box next to a count to leave it out of the fit, and watch the doubling time and R² update.

The formulas are:

Two counts: Td = (t₂ − t₁) × ln 2 ÷ ln(N₂ ÷ N₁)

Several counts: fit ln N = ln N₀ + r × t, where r is the growth rate per hour; then Td = ln 2 ÷ r.

Generations over the fitted span = r × (t_end − t_start) ÷ ln 2.

The same page then has a passage planner that uses your doubling time to say how many cells to seed for a target confluence on a chosen day. For bacteria measured by optical density instead of counts, use the OD600 growth curve tool.

A worked example

A made-up culture, counted once a day in cells/mL:

Time (h) Count (cells/mL)
0 50,000
24 82,000
48 160,000
72 330,000
96 610,000

With all five counts ticked, the fit gives a doubling time of 26.01 h, a 95 % interval of 23.47 to 29.18 h and R² = 0.9965.

The step table tells a clearer story. From 0 to 24 h the doubling time is 33.6 h; after that it is 24.9 h, 23.0 h and 27.1 h. The first day looks slow, which is typical of freshly split cells settling in.

Untick the 0 h count and fit the last four:

  • Growth rate r = 0.02810 per hour (0.6744 per day)
  • Doubling time = ln 2 ÷ 0.02810 = 24.67 h
  • 95 % interval 22.58 to 27.17 h, R² = 0.9991
  • Generations over the 72 h from 24 to 96 h: 0.02810 × 72 ÷ 0.6931 = 2.919, a 7.563-fold rise

As a hand check with just two counts, 24 h and 96 h: Td = 72 × 0.6931 ÷ ln(610,000 ÷ 82,000) = 49.91 ÷ 2.007 = 24.87 h. That is close to the fit, as it should be.

Check your answer

  • Count the doublings by eye. From 82,000 to 610,000 is a bit under 8-fold, which is just under 3 doublings (2³ = 8). Three doublings in 72 hours is 24 h each, so an answer near 25 h makes sense.
  • Run it forwards: 82,000 × 2^(72 ÷ 24.67) ≈ 620,000, close to the 610,000 you counted.
  • If R² is below about 0.95, or the tool warns that growth slows towards the end, some counts are from the lag or plateau. Untick them and fit only the steady part.

Questions people ask

Is doubling time the same as cell cycle time? Not always. Doubling time is for the whole population, so cells that die or stop dividing make it longer than the time one cell takes to divide.

Why log the counts? Steady growth multiplies the count by the same factor each hour, so on a log scale it becomes a straight line. A straight line is easy to fit, and its slope is the growth rate.

How many counts do I need? Two give an answer but no idea of its error. Three or more steady-growth counts give an interval, and five or six give a narrow one.

Sources

  • Exponential growth N = N₀ × 2^(t / Td), the two-count formula Td = (t₂ − t₁) × ln 2 ÷ ln(N₂ ÷ N₁) and the least-squares fit of ln(count) against time: general science knowledge.
  • The counts in the example are made up.

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