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

When should a blood sample for a digoxin level be drawn to be interpreted correctly?

  1. AAt any time, because digoxin levels do not change during the day
  2. BAt least 6–8 hours after a dose, often just before the next one
  3. CWithin 30 minutes after an oral dose to show the peak level
  4. DOnly after a meal, because food is needed for absorption
2.Lab values & electrolytesHard

A patient with diabetic ketoacidosis is receiving an IV insulin infusion. Which laboratory value must be monitored especially closely?

  1. APotassium, because insulin moves it into the cells and it can fall quickly
  2. BSodium, because insulin causes the kidneys to excrete very large amounts of it
  3. CHemoglobin, because insulin breaks down red blood cells
  4. DAlbumin, because insulin makes the liver stop producing it
3.Arterial blood gas interpretationMedium
pH7.357.457.54PaCO₂mmHg354522HCO₃⁻mmol/L222618uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.54, PaCO₂ 22 mmHg (2.9 kPa) and HCO₃⁻ 18 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.54PaCO₂ 22 mmHg (2.9 kPa)HCO₃⁻ 18 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

A practice gas shows a metabolic alkalosis with HCO₃⁻ 36 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.

HCO₃⁻ 36 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
VCss​ = 15 mg/LR = ? mg/hCL = 0.035 L/kg/h86 kg

Practice exercise (made-up values, not clinical guidance). An 86 kg patient receives a theophylline infusion. The clearance is 0.035 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 15 mg/L?

86 kgCL = 0.035 L/kg/hCss = 15 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
54 kg63 · manSCr 2.3 mg/dLfe​ = 0.6D = ? mg

Practice exercise (made-up values, not clinical guidance). A 63-year-old man weighs 54 kg and has a serum creatinine of 2.3 mg/dL (203 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual ciprofloxacin dose is 500 mg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.6 of the drug is excreted unchanged by the kidneys. Using the Dettli dose fraction Q = 1 − fe·(1 − CrCl/normal CrCl), find the adjusted dose D. Round to the nearest whole number.

63 years54 kgmanSCr = 2.3 mg/dL500 mgfe = 0.6CrCl normal 100 mL/min
D = mg
7.Clinical nursingMedium

Which examination findings suggest a large pleural effusion on one side?

  1. AWheeze heard over both lungs
  2. BA hyper-resonant chest with the trachea pulled towards that side
  3. CDull percussion and reduced breath sounds over the lung base
  4. DNormal breath sounds and normal percussion
8.Clinical nursingHard

A patient after abdominal surgery has not passed urine for eight hours, has lower abdominal discomfort and a palpable bladder. What is the most likely problem?

  1. ADehydration only
  2. BAcute kidney injury with no urine production
  3. CUrinary tract infection without retention
  4. DAcute urinary retention

3Solutions

1.Lab values & electrolytes

Answer: B · At least 6–8 hours after a dose, often just before the next one

Why this is the answer

Digoxin distributes slowly from blood to tissues, so a sample taken too soon after a dose is falsely high. Levels are drawn at least 6–8 hours after a dose, commonly as a trough before the next dose.

Answer

B · At least 6–8 hours after a dose, often just before the next one
2.Lab values & electrolytes

Answer: A · Potassium, because insulin moves it into the cells and it can fall quickly

Why this is the answer

In DKA the body has lost potassium in the urine, but the serum level may be normal or high because acidosis and lack of insulin shift potassium out of cells. Insulin drives it back in, so the level can fall fast. Potassium is checked often and replaced as prescribed.

Answer

A · Potassium, because insulin moves it into the cells and it can fall quickly
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 7.54: alkalaemia (above 7.45)
PaCO₂ 22 mmHg: low (35–45)
HCO₃⁻ 18 mmol/L: low (22–26)

Compensation

partially compensated

Answer

partially compensated
4.Arterial blood gas interpretation

Answer: PaCO₂ = 46.2 mmHg

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

0.7·36 + 21 = 46.2 mmHg

Answer

PaCO₂ = 46.2 mmHg

The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.

5.Clinical pharmacokinetics

Answer: R = 45.15 mg/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.

Clearance

CL = 0.035·86 = 3.01 L/h

Infusion rate

R = 3.01·15 = 45.15 mg/h

Answer

R = 45.15 mg/h
6.Clinical pharmacokinetics

Answer: D = 275 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.

Creatinine clearance (Cockcroft–Gault)

CrCl = (140 − 63)·54/(72·2.3) = 25.11 mL/min

Dose fraction (Dettli)

Q = 1 − 0.6·(1 − 25.11/100) = 0.5507

Adjusted dose

D = 0.5507·500 = 275.3 ≈ 275 mg

Answer

D = 275 mg

In practice the dose is rounded to a convenient amount and checked against the product information.

7.Clinical nursing

Answer: C · Dull percussion and reduced breath sounds over the lung base

Why this is the answer

Fluid in the pleural space muffles breath sounds and makes percussion stony dull. A large effusion may push the trachea away from the affected side.

Answer

C · Dull percussion and reduced breath sounds over the lung base
8.Clinical nursing

Answer: D · Acute urinary retention

Why this is the answer

Anaesthesia, opioids and pain can cause retention. A bladder scan confirms residual volume; catheterisation is often needed.

Answer

D · Acute urinary retention

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