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 on warfarin for atrial fibrillation has a target INR of 2.0–3.0. How does the usual target for a patient with a mechanical mitral valve compare?

  1. AMuch higher, about 5.0–6.0
  2. BThe same, 2.0–3.0
  3. CHigher, about 2.5–3.5
  4. DLower, about 1.0–1.5
2.Lab values & electrolytesHard

A patient with acute kidney injury has a potassium of 6.8 mEq/L despite medical treatment, a pH of 7.12 and pulmonary edema not responding to diuretics. Which treatment is most likely needed?

  1. AA potassium supplement
  2. BUrgent dialysis
  3. CMore oral fluid
  4. DAn IV magnesium dose
3.Arterial blood gas interpretationMedium
pH7.357.457.55PaCO₂mmHg354550HCO₃⁻mmol/L222642uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.55, PaCO₂ 50 mmHg (6.7 kPa) and HCO₃⁻ 42 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.55PaCO₂ 50 mmHg (6.7 kPa)HCO₃⁻ 42 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 143, Cl⁻ 115 and HCO₃⁻ 10 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.

Na⁺ 143 mmol/LCl⁻ 115 mmol/LHCO₃⁻ 10 mmol/L
delta ratio =
5.Clinical pharmacokineticsMedium
110100t (h)C (mg/L)2 h25 mg/L18 h5.1 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 25 mg/L at 2 h and 5.1 mg/L at 18 h. Assume first-order elimination. Find the half-life t½.

C1 = 25 mg/L (2 h)C2 = 5.1 mg/L (18 h)
t½ = h
6.Clinical pharmacokineticsHard
Vd = 0.25 L/kg0.5 → 7 mg/LD = ? mg74 kg

Practice exercise (made-up values, not clinical guidance). A 74 kg patient on regular gentamicin doses is at steady state. The target is a peak of 7 mg/L and a trough of 0.5 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?

74 kgVd = 0.25 L/kgCpeak = 7 mg/LCtrough = 0.5 mg/L
D = mg
7.Clinical nursingMedium

Which finding with a groin hernia needs urgent surgical review?

  1. AThe lump has become painful and tender and cannot be pushed back
  2. BThe lump gets bigger when the patient coughs
  3. CThe lump has been present for several years without pain
  4. DThe lump disappears completely when the patient lies down flat
8.Clinical nursingHard

Which route and site are recommended for the first dose of adrenaline (epinephrine) in anaphylaxis on a ward?

  1. AIntravenous bolus given by the nurse without monitoring
  2. BIntramuscular injection into the anterolateral thigh
  3. COral tablet dissolved under the tongue
  4. DSubcutaneous injection into the upper arm

3Solutions

1.Lab values & electrolytes

Answer: C · Higher, about 2.5–3.5

Why this is the answer

Mechanical heart valves, especially in the mitral position, have a high risk of thrombosis, so the INR target is usually higher, about 2.5–3.5. An INR of 5 or more carries a high bleeding risk.Checked against: Label: WARFARIN SODIUM tablet · DailyMed (NLM/FDA) (2012)

Answer

C · Higher, about 2.5–3.5
2.Lab values & electrolytes

Answer: B · Urgent dialysis

Why this is the answer

Indications for urgent dialysis in AKI include refractory hyperkalemia, severe metabolic acidosis, fluid overload not responding to diuretics, uremic complications and some poisonings. This patient has three of them.Checked against: Acute tubular necrosis · MedlinePlus (NIH) (2025)

Answer

B · Urgent dialysis
3.Arterial blood gas interpretation

Answer: partially compensated

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 7.55: alkalaemia (above 7.45)
PaCO₂ 50 mmHg: high (35–45)
HCO₃⁻ 42 mmol/L: high (22–26)

Compensation

partially compensated

Answer

partially compensated
4.Arterial blood gas interpretation

Answer: delta ratio = 0.4286

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 = 143 − (115 + 10) = 18

Delta ratio

(18 − 12)/(24 − 10) = 6/14 = 0.4286

Answer

delta ratio = 0.4286

Below 1: an added normal anion gap acidosis.

5.Clinical pharmacokinetics

Answer: t½ = 6.977 h

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(25/5.1)/16 = 1.59/16 = 0.09935 h⁻¹

Half-life

t½ = 0.693/0.09935 = 6.977 h

Answer

t½ = 6.977 h
6.Clinical pharmacokinetics

Answer: D = 120.3 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 = 0.25·74 = 18.5 L

Dose

D = 18.5·(7 − 0.5) = 120.3 mg

Answer

D = 120.3 mg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: A · The lump has become painful and tender and cannot be pushed back

Why this is the answer

An irreducible, painful, tender hernia, especially with vomiting, suggests incarceration or strangulation. This is a surgical emergency because the trapped bowel can become ischaemic.

Answer

A · The lump has become painful and tender and cannot be pushed back
8.Clinical nursing

Answer: B · Intramuscular injection into the anterolateral thigh

Why this is the answer

IM adrenaline into the outer thigh is absorbed quickly and is the first-line route. IV adrenaline is reserved for experienced specialists with full monitoring because of the risk of arrhythmias.

Answer

B · Intramuscular injection into the anterolateral thigh

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