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 patient starts lisinopril (an ACE inhibitor). Which laboratory values should the nurse expect to be monitored?

  1. AALT and bilirubin
  2. BCalcium and phosphate
  3. CTSH and free T4
  4. DPotassium and creatinine
2.Lab values & electrolytesHard

A patient with long-standing rheumatoid arthritis has a hemoglobin of 10.5 g/dL (105 g/L), a low serum iron, a low total iron-binding capacity (TIBC) and a high-normal ferritin. Which condition fits best?

  1. AAcute blood loss anemia
  2. BVitamin B12 deficiency anemia
  3. CIron-deficiency anemia
  4. DAnemia of chronic disease (inflammation)
3.Arterial blood gas interpretationMedium

Practice values: Na⁺ 128 mmol/L, glucose 9.5 mmol/L and urea 21.8 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 128 mmol/Lglucose 9.5 mmol/Lurea 21.8 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
pH7.357.457.32PaCO₂mmHg354528HCO₃⁻mmol/L222614appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.32, PaCO₂ 28 mmHg (3.7 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.32PaCO₂ 28 mmHg (3.7 kPa)HCO₃⁻ 14 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
110t (h)C (mg/L)1 h6 mg/L7 h1.5 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 1 h and 1.5 mg/L at 7 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 6 mg/L (1 h)C2 = 1.5 mg/L (7 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
Vd = 0.5 L/kg4.5 → 15 mg/LLD = ? mg51 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 kg patient has a theophylline level of 4.5 mg/L; the target is 15 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?

51 kgVd = 0.5 L/kgC = 4.5 → 15 mg/LS = 0.8
LD = mg
7.Clinical nursingMedium

What is one advantage of a direct oral anticoagulant (DOAC) such as apixaban compared with warfarin?

  1. AIt is safe in all patients with severe kidney failure
  2. BIt does not need routine INR monitoring
  3. CIt never interacts with other drugs
  4. DIt has no risk of bleeding
8.Clinical nursingHard

A patient with severe asthma becomes quiet, with a "silent chest", exhaustion and a falling respiratory rate. What does this indicate?

  1. ALife-threatening asthma with impending respiratory arrest
  2. BAnxiety only
  3. CImprovement of the attack as the bronchospasm settles and the airways relax
  4. DNormal recovery after a nebuliser

3Solutions

1.Lab values & electrolytes

Answer: D · Potassium and creatinine

Why this is the answer

ACE inhibitors reduce aldosterone, so potassium can rise, and they lower the filtration pressure in the kidney, so creatinine can rise. Both are checked before and soon after starting or increasing the dose.Checked against: Hyperkalemia · StatPearls (NCBI Bookshelf) (2025)

Answer

D · Potassium and creatinine
2.Lab values & electrolytes

Answer: D · Anemia of chronic disease (inflammation)

Why this is the answer

In chronic inflammation, iron is trapped in stores, so serum iron is low but ferritin is normal or high and TIBC is low. In iron deficiency, ferritin is low and TIBC is high. Iron supplements help iron deficiency much more than anemia of chronic disease.Checked against: Ferritin Blood Test · MedlinePlus (NIH) (2025); Serum iron test · MedlinePlus (NIH) (2026)

Answer

D · Anemia of chronic disease (inflammation)
3.Arterial blood gas interpretation

Answer: Osm = 287 mOsm/kg

Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.

Calculated osmolality

Osm = 2·128 + 9.5 + 21.8 = 287.3 ≈ 287 mOsm/kg

Answer

Osm = 287 mOsm/kg
4.Arterial blood gas interpretation

Answer: appropriate compensation

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.

Expected PaCO₂

1.5·14 + 8 = 29 → 27–31 mmHg

Compare

measured: 28 mmHg → appropriate compensation

Answer

appropriate compensation
5.Clinical pharmacokinetics

Answer: k = 0.231 h⁻¹

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.

Elimination rate constant

k = ln(6/1.5)/6 = 1.386/6 = 0.231 h⁻¹

Answer

k = 0.231 h⁻¹
6.Clinical pharmacokinetics

Answer: LD = 334.7 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 = 0.5·51 = 25.5 L

Rise needed

15 − 4.5 = 10.5 mg/L

Loading dose

LD = 25.5·10.5/(0.8·1) = 334.7 mg

Answer

LD = 334.7 mg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: B · It does not need routine INR monitoring

Why this is the answer

DOACs have predictable effects so INR monitoring is not needed. They still cause bleeding, have some interactions and need kidney function checked.

Answer

B · It does not need routine INR monitoring
8.Clinical nursing

Answer: A · Life-threatening asthma with impending respiratory arrest

Why this is the answer

A silent chest, exhaustion, cyanosis and confusion are life-threatening signs. The patient needs immediate senior and critical care input.

Answer

A · Life-threatening asthma with impending respiratory arrest

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