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 is scheduled for a liver biopsy. Which laboratory results should the nurse check before the procedure?

  1. ATSH and free T4
  2. BHbA1c and lipid panel
  3. CPlatelet count and PT/INR
  4. DSerum sodium and chloride
2.Lab values & electrolytesHard

A patient has a hematocrit of 60% and a raised red cell mass. Which result suggests polycythemia vera rather than secondary polycythemia from chronic hypoxia?

  1. AA low oxygen saturation
  2. BA high carboxyhemoglobin
  3. CA low erythropoietin level
  4. DA high erythropoietin level
3.Arterial blood gas interpretationMedium
pH7.357.457.47PaCO₂mmHg354543HCO₃⁻mmol/L222630uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.47, PaCO₂ 43 mmHg (5.7 kPa) and HCO₃⁻ 30 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.47PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 30 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.31PaCO₂mmHg354572HCO₃⁻mmol/L222635acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.31, PaCO₂ 72 mmHg (9.6 kPa), HCO₃⁻ 35 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.

pH 7.31PaCO₂ 72 mmHg (9.6 kPa)HCO₃⁻ 35 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
5.Clinical pharmacokineticsMedium
110100t (h)C (mg/L)C₀ = 12 mg/Lt½ = 2.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 320 mg of tobramycin. The level back-extrapolated to time zero is 12 mg/L and the half-life is 2.5 h. Find the volume of distribution Vd.

D = 320 mgC₀ = 12 mg/Lt½ = 2.5 h
Vd = L
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​15 hC = ?

Practice exercise (made-up values, not clinical guidance). A theophylline infusion runs at 30 mg/h with no loading dose. The patient's clearance is 2.5 L/h and the half-life is 6 h. What is the level C after 15 h?

R = 30 mg/hCL = 2.5 L/ht½ = 6 ht = 15 h
C = mg/L
7.Clinical nursingMedium

Why must levodopa for Parkinson's disease be given at the exact times prescribed in hospital?

  1. ALate doses cause dangerous high potassium
  2. BEarly doses cause severe low blood glucose and sweating attacks
  3. CDelays can quickly worsen rigidity, immobility and swallowing
  4. DThe drug only works if taken at midnight
8.Clinical nursingHard

Why are patients with nephrotic syndrome at increased risk of blood clots?

  1. AAnticoagulant proteins are lost in the urine and clotting factors rise
  2. BThey lose all their red cells in the urine, which thickens the remaining blood
  3. CTheir blood pressure is always very low
  4. DThey lose most of their platelets in the urine every day

3Solutions

1.Lab values & electrolytes

Answer: C · Platelet count and PT/INR

Why this is the answer

Liver biopsy carries a risk of bleeding, and liver disease often lowers platelets and clotting factors. Platelets and PT/INR are checked beforehand, and abnormal results are reported so they can be corrected or the procedure postponed.Checked against: Liver biopsy · MedlinePlus (NIH) (2025)

Answer

C · Platelet count and PT/INR
2.Lab values & electrolytes

Answer: C · A low erythropoietin level

Why this is the answer

In polycythemia vera, a marrow mutation (usually JAK2) drives red cell production, so erythropoietin is low. In secondary polycythemia from hypoxia (COPD, high altitude, smoking), erythropoietin is high. Often white cells and platelets are also raised in polycythemia vera.Checked against: Erythropoietin test · MedlinePlus (NIH) (2025); Polycythemia vera · MedlinePlus (NIH) (2025)

Answer

C · A low erythropoietin 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 7.47: alkalaemia (above 7.45)
PaCO₂ 43 mmHg: normal (35–45)
HCO₃⁻ 30 mmol/L: high (22–26)

Compensation

uncompensated

Answer

uncompensated
4.Arterial blood gas interpretation

Answer: chronic

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 HCO₃⁻

acute: 24 + 1·3.2 = 27.2 mmol/L
chronic: 24 + 3.5·3.2 = 35.2 mmol/L

Compare

measured: 35 mmol/L → chronic

Answer

chronic
5.Clinical pharmacokinetics

Answer: Vd = 26.67 L

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 = 320/12 = 26.67 L

Answer

Vd = 26.67 L
6.Clinical pharmacokinetics

Answer: C = 9.879 mg/L

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.

Steady-state level

Css = 30/2.5 = 12 mg/L

Rate constant

k = 0.693/6 = 0.1155 h⁻¹

Level after the given time

C = 12·(1 − e−0.1155·15) = 9.879 mg/L

Answer

C = 9.879 mg/L
7.Clinical nursing

Answer: C · Delays can quickly worsen rigidity, immobility and swallowing

Why this is the answer

Levodopa has a short effect, so missed or late doses can lead to severe stiffness, falls, swallowing problems and, rarely, a dangerous hyperpyrexia-like state. Parkinson's medicines are treated as time-critical.

Answer

C · Delays can quickly worsen rigidity, immobility and swallowing
8.Clinical nursing

Answer: A · Anticoagulant proteins are lost in the urine and clotting factors rise

Why this is the answer

Loss of proteins such as antithrombin in the urine, together with increased hepatic production of clotting factors, creates a hypercoagulable state. Some patients need anticoagulation.

Answer

A · Anticoagulant proteins are lost in the urine and clotting factors rise

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