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Bioavailability (F) Practice Problems

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Absolute bioavailability compares how much drug reaches the systemic circulation from an oral dose versus an IV dose (which is 100% by definition). It is calculated from the dose-normalized areas under the concentration-time curve (AUC).

FormulaF = (AUC_oral ÷ Dose_oral) × (Dose_IV ÷ AUC_IV)
Worked example

IV 100 mg gives AUC 50; oral 200 mg gives AUC 60. First term: 60 ÷ 200 = 0.30; second term: 100 ÷ 50 = 2; F = 0.30 × 2 = 0.60 = 60%.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

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Bioavailability (F)
A 200 mg IV dose gives an AUC of 50 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.
%

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How to set up a bioavailability problem

Absolute bioavailability is the fraction of an oral dose that reaches the systemic circulation, measured against an IV dose of the same drug, which by definition is 100% available. The measurement is the area under the concentration-time curve (AUC), because AUC is proportional to the amount of drug the body was exposed to. The one thing that makes this a calculation rather than a lookup is that the oral and IV doses in a study are usually different sizes, so the AUCs have to be normalized to their doses before they can be compared.

F = (AUC_oral ÷ dose_oral) × (dose_IV ÷ AUC_IV)   =   (AUC_oral × dose_IV) ÷ (dose_oral × AUC_IV)

Both forms are the same equation. The second one, written as a single fraction, is the one to use on a calculator, because it avoids rounding two separate factors and multiplying rounded numbers together.

Worked example

A 100 mg IV dose gives an AUC of 40 mg·hr/L. A 250 mg oral dose gives an AUC of 60 mg·hr/L.

numerator = 60 × 100 = 6,000

denominator = 250 × 40 = 10,000

F = 6,000 ÷ 10,000 = 0.6

So F is 0.6, or 60%. Report it as a fraction or a percent depending on what the question asks for; the practice problems here ask for the percent.

Now look at what happens if you skip the normalization: 60 ÷ 40 is 1.5, which would mean the oral dose was 150% bioavailable, more drug absorbed than was given. That's impossible, and it's the built-in check. If F comes out above 1, the doses weren't accounted for.

Why it's dimensionless

The AUC units cancel top and bottom, and so do the dose units. F is a pure ratio. Don't write mg or mg·hr/L after it. That also means the AUC can be in any unit as long as both AUCs are in the same one, and the same for the doses.

Relative bioavailability

The same setup compares two non-IV formulations, for example a generic tablet against the brand: F_relative = (AUC_test ÷ dose_test) × (dose_reference ÷ AUC_reference). Nothing changes except that the reference isn't 100% by definition, so a relative F above 1 is possible and just means the test product was absorbed better than the reference.

Using F in a dose calculation

Once F is known it turns up as a divisor in oral dosing: if an IV dose of 200 mg is needed and F is 0.6, the equivalent oral dose is 200 ÷ 0.6, about 333 mg. And in the loading dose equation from pharmacokinetics, LD = C_target × Vd ÷ (S × F), the F in the denominator is this number. A lower F means a larger oral dose to get the same amount into the body.

Where students lose the points

10 Bioavailability (F) practice problems with worked solutions

Every problem below is worked start to finish. Try it yourself first, then check the steps. For a fresh problem every time, use the practice box above. It generates unlimited variations of these same question types.

Problem 1: Bioavailability (F)

A 50 mg IV dose gives an AUC of 60 mg·hr/L. A 400 mg oral dose gives an AUC of 240 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 50 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 240 mg·hr/L × Dose_iv 50 mg = 12000.
  4. Denominator: Dose_po 400 mg × AUC_iv 60 mg·hr/L = 24000.
  5. F = 12000 ÷ 24000 = 0.5, i.e. 50%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 50 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 2: Bioavailability (F)

A 200 mg IV dose gives an AUC of 40 mg·hr/L. A 100 mg oral dose gives an AUC of 18 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 90 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 18 mg·hr/L × Dose_iv 200 mg = 3600.
  4. Denominator: Dose_po 100 mg × AUC_iv 40 mg·hr/L = 4000.
  5. F = 3600 ÷ 4000 = 0.9, i.e. 90%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 90 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 3: Bioavailability (F)

A 100 mg IV dose gives an AUC of 80 mg·hr/L. A 100 mg oral dose gives an AUC of 72 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 90 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 72 mg·hr/L × Dose_iv 100 mg = 7200.
  4. Denominator: Dose_po 100 mg × AUC_iv 80 mg·hr/L = 8000.
  5. F = 7200 ÷ 8000 = 0.9, i.e. 90%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 90 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 4: Bioavailability (F)

A 50 mg IV dose gives an AUC of 30 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 22.5 mg·hr/L × Dose_iv 50 mg = 1125.
  4. Denominator: Dose_po 150 mg × AUC_iv 30 mg·hr/L = 4500.
  5. F = 1125 ÷ 4500 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 5: Bioavailability (F)

A 200 mg IV dose gives an AUC of 60 mg·hr/L. A 150 mg oral dose gives an AUC of 31.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 70 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 31.5 mg·hr/L × Dose_iv 200 mg = 6300.
  4. Denominator: Dose_po 150 mg × AUC_iv 60 mg·hr/L = 9000.
  5. F = 6300 ÷ 9000 = 0.7, i.e. 70%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 70 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 6: Bioavailability (F)

A 50 mg IV dose gives an AUC of 40 mg·hr/L. A 200 mg oral dose gives an AUC of 40 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 40 mg·hr/L × Dose_iv 50 mg = 2000.
  4. Denominator: Dose_po 200 mg × AUC_iv 40 mg·hr/L = 8000.
  5. F = 2000 ÷ 8000 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 7: Bioavailability (F)

A 100 mg IV dose gives an AUC of 50 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 30 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 22.5 mg·hr/L × Dose_iv 100 mg = 2250.
  4. Denominator: Dose_po 150 mg × AUC_iv 50 mg·hr/L = 7500.
  5. F = 2250 ÷ 7500 = 0.3, i.e. 30%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 30 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 8: Bioavailability (F)

A 100 mg IV dose gives an AUC of 25 mg·hr/L. A 250 mg oral dose gives an AUC of 15.625 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 15.625 mg·hr/L × Dose_iv 100 mg = 1562.5.
  4. Denominator: Dose_po 250 mg × AUC_iv 25 mg·hr/L = 6250.
  5. F = 1562.5 ÷ 6250 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 9: Bioavailability (F)

A 100 mg IV dose gives an AUC of 40 mg·hr/L. A 100 mg oral dose gives an AUC of 12 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 30 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 12 mg·hr/L × Dose_iv 100 mg = 1200.
  4. Denominator: Dose_po 100 mg × AUC_iv 40 mg·hr/L = 4000.
  5. F = 1200 ÷ 4000 = 0.3, i.e. 30%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 30 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 10: Bioavailability (F)

A 50 mg IV dose gives an AUC of 25 mg·hr/L. A 300 mg oral dose gives an AUC of 37.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 37.5 mg·hr/L × Dose_iv 50 mg = 1875.
  4. Denominator: Dose_po 300 mg × AUC_iv 25 mg·hr/L = 7500.
  5. F = 1875 ÷ 7500 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

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