Lab Values for Nurses: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Lab values & electrolytes

Method

  1. Compare the value with the laboratory’s own reference range, then with the patient’s baseline and trend.
  2. Check the sample: hemolysis, a delay, or blood drawn near an IV can give false results.
  3. Link the value to its typical signs and to the drugs and conditions that cause it.
  4. Look for interactions: low potassium or magnesium increases digoxin toxicity; dehydration raises lithium.
  5. Act on the patient: assess, hold a drug if the level is unsafe, read back critical values and inform the prescriber.

Arterial blood gas interpretation

Normal ranges
pH 7.35–7.45, PaCO₂ 35–45 mmHg (4.7–6.0 kPa), HCO₃⁻ 22–26 mmol/L
ROME
Respiratory: PaCO₂ opposite to pH; Metabolic: HCO₃⁻ equal to pH
Winter’s formula
PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2
Metabolic alkalosis
PaCO₂ ≈ 0.7·HCO₃⁻ + 21
Anion gap
AG = Na⁺ − (Cl⁻ + HCO₃⁻) ≈ 8–12
Albumin-corrected gap
AG + 2.5·(4.0 − alb g/dL) = AG + 0.25·(40 − alb g/L)
Delta ratio
(AG − 12)/(24 − HCO₃⁻)
P/F ratio
PaO₂/FiO₂
Alveolar gas equation
PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8
Serum osmolality
2·Na⁺ + glucose + urea (2·Na⁺ + glucose/18 + BUN/2.8)

Method

  1. Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
  2. Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
  3. Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
  4. In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
  5. Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.

Clinical pharmacokinetics

Rate constant and half-life
k = ln 2/t½ = 0.693/t½
First-order decay
C = C₀·e−kt = C₀·(1/2)t/t½
Two levels
k = ln(C₁/C₂)/Δt
Volume and clearance
Vd = dose/C₀, CL = k·Vd
Loading dose
LD = Vd·C/(S·F)
Maintenance rate
R = CL·Css/(S·F)
Average steady state
Css,avg = S·F·D/(CL·τ)
Steady state
1 − (1/2)ⁿ after n half-lives; ≈ 97% after 5
Infusion
C = (R/CL)(1 − e−kt)
Interval and dose from peak and trough
τ = ln(Cpeak/Ctrough)/k, D = Vd·(Cpeak − Ctrough)
Cockcroft–Gault
CrCl = (140 − age) × kg/(72 × SCr mg/dL), × 0.85 for women
Renal adjustment
dose × CrCl/normal; Dettli Q = 1 − fe·(1 − CrCl/normal)

Method

  1. Convert the half-life to a rate constant: k = 0.693/t½.
  2. Turn per-kg values into totals: Vd (L) = L/kg × weight, CL (L/h) = L/kg/h × weight.
  3. Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
  4. For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
  5. Check the result: levels fall by half every half-life, and steady state takes about five half-lives.

2Practice problems

1.Lab values & electrolytesMedium

A woman who is 30 weeks pregnant has a new blood pressure of 152/98 mm Hg. Which urine finding would most support preeclampsia?

  1. AGlucose in the urine
  2. BNitrites in the urine
  3. CKetones in the urine
  4. DProtein in the urine
2.Lab values & electrolytesHard

A patient on stable digoxin starts amiodarone. A week later, the patient has nausea and visual changes, and the digoxin level is above the range. What is the most likely explanation?

  1. AThe patient has started a viral illness
  2. BDigoxin has raised the amiodarone level
  3. CAmiodarone has raised the digoxin level
  4. DAmiodarone has lowered potassium only
3.Arterial blood gas interpretationMedium
pH7.357.457.49PaCO₂mmHg354541HCO₃⁻mmol/L222630uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.49, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 30 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.49PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 30 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 136, Cl⁻ 98, HCO₃⁻ 21 mmol/L and albumin 33 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 136 mmol/LCl⁻ 98 mmol/LHCO₃⁻ 21 mmol/Lalbumin 33 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 12 mg/LLD = ? mg84 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). An 84 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 12 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.

84 kgVd = 0.5 L/kgC = 12 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Vd = 0.5 L/kg6 → 10 mg/LLD = ? mg73 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 73 kg patient has a theophylline level of 6 mg/L; the target is 10 mg/L. The drug is given intravenously as aminophylline (S = 0.8), and the volume of distribution is 0.5 L/kg. What extra loading dose LD raises the level to the target?

73 kgVd = 0.5 L/kgC = 6 → 10 mg/LS = 0.8
LD = mg
7.Clinical nursingMedium

A patient is admitted with cellulitis of the leg. Why might the nurse mark the edge of the redness on the skin?

  1. ATo monitor whether the infection is spreading or improving
  2. BTo help the patient find the sore area
  3. CTo mark the site for an injection
  4. DTo show the surgeon exactly where to make the incision
8.Clinical nursingHard

A patient has new facial weakness. Which feature suggests a stroke rather than Bell's palsy?

  1. AThe patient has pain behind the ear
  2. BThe patient has had a recent cold
  3. CThe forehead can still be wrinkled on the affected side
  4. DThe whole of one side of the face, including the forehead, is weak

3Solutions

1.Lab values & electrolytes

Answer: D · Protein in the urine

Why this is the answer

Preeclampsia is new hypertension after 20 weeks of pregnancy with proteinuria or other signs of organ involvement, such as low platelets, raised liver enzymes or rising creatinine. Proteinuria reflects damage to the glomerular filter.Checked against: Preeclampsia · MedlinePlus (NIH) (2026)

Answer

D · Protein in the urine
2.Lab values & electrolytes

Answer: C · Amiodarone has raised the digoxin level

Why this is the answer

Amiodarone reduces digoxin clearance, so the level can rise significantly, and the digoxin dose is often reduced when amiodarone is started. Nausea and visual changes are classic signs of digoxin toxicity.

Answer

C · Amiodarone has raised the digoxin level
3.Arterial blood gas interpretation

Answer: uncompensated

Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.

ROME, value by value

pH 7.49: alkalaemia (above 7.45)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 30 mmol/L: high (22–26)

Compensation

uncompensated

Answer

uncompensated
4.Arterial blood gas interpretation

Answer: AGc = 18.75 mmol/L

Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.

Anion gap

AG = 136 − (98 + 21) = 17

Correction for albumin

17 + 0.25·(40 − 33) = 18.75 mmol/L

Answer

AGc = 18.75 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: LD = 630 mg

Clearance and volume: at steady state the dosing rate equals the elimination rate, so the maintenance rate = CL × Css. A loading dose is LD = Vd × C/(S × F), with S the salt factor (fraction of active drug in the salt) and F the bioavailability (1 for IV). From two levels, k = ln(C₁/C₂)/Δt; for an IV bolus Vd = dose/C₀ and CL = k × Vd. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.

Volume of distribution

Vd = 0.5·84 = 42 L

Amount of active drug needed

Vd·C = 42·12 = 504 mg

Loading dose

LD = 504/(0.8·1) = 630 mg

Answer

LD = 630 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: LD = 182.5 mg

Designing a regimen: with rapid IV doses the level falls from peak to trough during the interval, so τ = ln(Cpeak/Ctrough)/k and the dose that restores the peak is Vd × (Cpeak − Ctrough). The average steady-state level is Css,avg = S·F·dose/(CL·τ). An infusion rises as C = (R/CL)(1 − e−kt). For a drug cleared by the kidneys, the dose is scaled with the creatinine clearance. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.

Volume of distribution

Vd = 0.5·73 = 36.5 L

Rise needed

10 − 6 = 4 mg/L

Loading dose

LD = 36.5·4/(0.8·1) = 182.5 mg

Answer

LD = 182.5 mg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: A · To monitor whether the infection is spreading or improving

Why this is the answer

Marking the margin helps staff see whether the redness is spreading despite treatment, which may mean antibiotics need review. The limb is also elevated and observations monitored.

Answer

A · To monitor whether the infection is spreading or improving
8.Clinical nursing

Answer: C · The forehead can still be wrinkled on the affected side

Why this is the answer

In a stroke (upper motor neurone lesion), the forehead is usually spared because it has bilateral cortical supply. Bell's palsy (lower motor neurone) weakens the whole side, including the forehead.

Answer

C · The forehead can still be wrinkled on the affected side

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