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Dilution factors and patient results

12 min

Skills you'll build

  • Calculate a dilution factor and specimen fraction from prepared volumes
  • Calculate the original concentration from a diluted measurement

At the bench: Prepare a dilution using the required sample and diluent volumes

Read the full reference

Try first

Try first

A specimen is diluted by adding 100 µL of serum to 400 µL of saline. What is the dilution factor?

Keep going to build the reasoning.

Right. Keep going to follow the reasoning.

Keep going to build the reasoning.

Keep going to build the reasoning.

Get the idea

Use the explanation to check your first decision.

Find the factor

A dilution factor says how many times the finished mixture spreads out the specimen. Divide the whole mixture by the part of it that came from the specimen:

dilution factor = (specimen volume + diluent volume) ÷ specimen volume

Written as a ratio, 1:5 means one part specimen in five parts of finished mixture, so the factor is 5.1 Either of these makes a 1:5 dilution:

  • Make up 200 µL of specimen to 1,000 µL.
  • Mix 100 µL of specimen with 400 µL of diluent.

The diluent is only part of the mixture. Dividing the diluent volume by the specimen volume gives 4 for that same 1:5 mixture, and every result corrected with 4 comes out 20% low.

Carry the reading back to the specimen

The analyzer measured the diluted mixture. In a 1:5 dilution that mixture holds a fifth of the specimen's concentration, so multiply the reading by the factor:

patient result = measured diluted result × dilution factor

A diluted creatine kinase (CK) that reads 365 U/L in a 1:10 mixture came from a specimen at 3,650 U/L.1

Check the direction before you go on:

  • The patient result is always larger than the diluted reading.
  • A patient result smaller than the reading means the reading was divided by the factor.

Use the mixture as it was made

The factor comes from the mixture as it was made. If a planned 1:5 dilution ends at 1,100 µL because extra diluent went in, the factor is 1,100 ÷ 200 = 5.5, and the result is multiplied by 5.5.2

Before the multiplied result is reported, the diluted reading itself has to lie inside the method's analytical measuring interval.3 That decision has its own step.

Follow the bench demonstration

Sources for this step
References
  1. Bishop ML, Fody EP, Van Siclen C, Mistler JM, Moy M. Clinical Chemistry: Principles, Techniques, and Correlations. 9th ed. Jones & Bartlett Learning; 2023.
  2. Flowers P, Theopold K, Langley R, Robinson WR. Molarity. In: Chemistry 2e. OpenStax; 2019. Accessed September 23, 2026. https://openstax.org/books/chemistry-2e/pages/3-3-molarity
  3. Clinical and Laboratory Standards Institute. Establishing and Verifying an Extended Measuring Interval Through Specimen Dilution and Spiking. 1st ed. CLSI guideline EP34. Clinical and Laboratory Standards Institute; 2018. Reaffirmed March 2023. Accessed September 23, 2026. https://clsi.org/media/sj3d0lue/ep34ed1e_reaffirmed_sample.pdf

Watch one

Follow each move and the reason beside it.

A urine creatinine reads above the analyzer's measuring interval. You mix 50 µL of urine with 450 µL of diluent, rerun the mixture and get 38.2 mg/dL, which is inside the interval. What creatinine do you report for the specimen?

  1. Add the volumes to get the finished mixture: 50 µL + 450 µL = 500 µL.

    The factor compares the whole mixture with the specimen in it, so the diluent volume alone is not enough.

  2. Divide the mixture by the specimen volume: 500 ÷ 50 = 10. The dilution factor is 10.

    The specimen is one part in ten of the mixture.

  3. Multiply the diluted reading by the factor: 38.2 mg/dL × 10 = 382 mg/dL.

    The analyzer measured a mixture ten times weaker than the specimen, so the specimen's concentration is ten times the reading.

  4. Check the direction: 382 mg/dL is larger than 38.2 mg/dL, as it must be.

    A corrected result smaller than the reading would mean the factor was divided in.

Report 382 mg/dL.

Practice this skill

Your turn

Problem 1 of 3 · Hints available

You pipette 200 µL of serum into 1,000 µL of diluent. What is the dilution factor?

Hint
  1. Start with the volume of the finished mixture.
  2. The mixture is 200 µL + 1,000 µL = 1,200 µL.
  3. Divide the mixture by the part of it that is serum.
Show the answer

6

The mixture is 200 + 1,000 = 1,200 µL, and 1,200 ÷ 200 = 6. Adding serum to a stated volume of diluent makes a mixture larger than that volume.

Review Parts and dilution factor

Problem 2 of 3 · Hints available

A serum is diluted 1:5 and the diluted mixture reads 112 U/L for lipase, inside the measuring interval. Which result do you report?

Dividing the reading by 5 gives 22.4 U/L. The specimen is stronger than the mixture made from it, so the reading is multiplied.

Divided the measured result by the dilution factor

The diluted mixture reads lower than the specimen, so the specimen's result is higher than the reading. Dividing a 365 U/L reading by a factor of 10 gives 36.5 U/L, 100 times lower than the true 3,650 U/L. The reading is multiplied by the factor of the mixture that was prepared.

That is the diluted mixture's lipase. The specimen holds five times as much.

The factor is 5, so the result is 112 U/L × 5 = 560 U/L, larger than the reading, as a corrected result must be.

Hint
  1. The analyzer measured a mixture five times weaker than the specimen.
  2. The reported result must be larger than the reading.

Review Concentration and patient-result calculation

Problem 3 of 3 · On your own

A triglyceride reads above the measuring interval. You dilute 100 µL of serum with 500 µL of diluent, and the mixture reads 492 mg/dL, inside the interval. What triglyceride do you report?

Show the answer

2,952 mg/dL

The mixture is 100 + 500 = 600 µL, so the factor is 600 ÷ 100 = 6, and 492 mg/dL × 6 = 2,952 mg/dL.

Review Concentration and patient-result calculation

Use it

Use the specimen and readings to make the next bench decision. Finish the case to complete this stage.

Try this at the Dilution stationCK · U/L>1000UNDILUTED · ABOVE AMR1:5 to 500 µLDilute specimen

For reference