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 starting dabigatran, a direct oral anticoagulant. Which laboratory test is most important to review before the first dose?

  1. AINR, to set the dose of the anticoagulant
  2. BLipase, to rule out an inflamed pancreas first
  3. CTSH, to rule out an overactive thyroid gland
  4. DKidney function (creatinine clearance)
2.Lab values & electrolytesHard

A patient with a long-standing lung condition has a serum bicarbonate of 32 mEq/L, an arterial PaCO₂ that is always high and a pH of 7.36. Which condition fits best?

  1. ACOPD with chronic CO₂ retention and metabolic compensation
  2. BAcute diabetic ketoacidosis with severe dehydration and vomiting
  3. CAcute aspirin overdose with rapid breathing
  4. DAcute severe diarrhea with large bicarbonate losses
3.Arterial blood gas interpretationMedium
pH7.357.457.42PaCO₂mmHg354529HCO₃⁻mmol/L222618uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.42, PaCO₂ 29 mmHg (3.9 kPa) and HCO₃⁻ 18 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.42PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 18 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 145, Cl⁻ 101 and HCO₃⁻ 9 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.

Na⁺ 145 mmol/LCl⁻ 101 mmol/LHCO₃⁻ 9 mmol/L
delta ratio =
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg51 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.

51 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
86 kg73 · manSCr 2 mg/dLfe​ = 0.66D = ? mg

Practice exercise (made-up values, not clinical guidance). A 73-year-old man weighs 86 kg and has a serum creatinine of 2 mg/dL (177 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 levetiracetam dose is 1000 mg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.66 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.

73 years86 kgmanSCr = 2 mg/dL1000 mgfe = 0.66CrCl normal 100 mL/min
D = mg
7.Clinical nursingMedium

Which tool is commonly used at the bedside to screen for delirium?

  1. AThe Glasgow-Blatchford score
  2. BThe Braden scale
  3. CThe Confusion Assessment Method (CAM)
  4. DThe Malnutrition Universal Screening Tool
8.Clinical nursingHard

A nurse teaches a patient with a new ileostomy. Which statement shows a need for further teaching?

  1. A"I will empty the bag when it is about a third to half full."
  2. B"My output will be firm and formed like before."
  3. C"I will drink extra fluids when it is hot."
  4. D"My stoma should look pink or red and moist."

3Solutions

1.Lab values & electrolytes

Answer: D · Kidney function (creatinine clearance)

Why this is the answer

Dabigatran is excreted mainly by the kidneys, so poor kidney function raises the drug level and bleeding risk. Kidney function is checked before starting and regularly afterwards. Routine INR monitoring is not used for direct oral anticoagulants.Checked against: Dabigatran · MedlinePlus (NIH) (2025)

Answer

D · Kidney function (creatinine clearance)
2.Lab values & electrolytes

Answer: A · COPD with chronic CO₂ retention and metabolic compensation

Why this is the answer

In chronic CO₂ retention, as in severe COPD, the kidneys retain bicarbonate over days and bring the pH back towards normal, often to the low end of the range. DKA and diarrhea lower bicarbonate, and aspirin overdose causes rapid breathing that lowers the PaCO₂.Checked against: Respiratory Acidosis · StatPearls (NCBI Bookshelf)

Answer

A · COPD with chronic CO₂ retention and metabolic compensation
3.Arterial blood gas interpretation

Answer: fully 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 7.42: within 7.35–7.45
PaCO₂ 29 mmHg: low (35–45)
HCO₃⁻ 18 mmol/L: low (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: delta ratio = 1.533

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 = 145 − (101 + 9) = 35

Delta ratio

(35 − 12)/(24 − 9) = 23/15 = 1.533

Answer

delta ratio = 1.533

Between 1 and 2: a pure high anion gap acidosis.

5.Clinical pharmacokinetics

Answer: LD = 582.1 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 = 0.7·51 = 35.7 L

Amount of active drug needed

Vd·C = 35.7·15 = 535.5 mg

Loading dose

LD = 535.5/(0.92·1) = 582.1 mg

Answer

LD = 582.1 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: D = 604 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 − 73)·86/(72·2) = 40.01 mL/min

Dose fraction (Dettli)

Q = 1 − 0.66·(1 − 40.01/100) = 0.6041

Adjusted dose

D = 0.6041·1000 = 604.1 ≈ 604 mg

Answer

D = 604 mg

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

7.Clinical nursing

Answer: C · The Confusion Assessment Method (CAM)

Why this is the answer

The CAM, and similar tools such as the 4AT, check for an acute onset or fluctuating course, inattention and disorganised thinking or altered consciousness. Early detection allows causes to be treated.

Answer

C · The Confusion Assessment Method (CAM)
8.Clinical nursing

Answer: B · "My output will be firm and formed like before."

Why this is the answer

Ileostomy output is liquid to porridge-like because it bypasses the colon. The other statements describe correct care: good hydration, emptying the bag before it is too heavy and recognising a healthy stoma.

Answer

B · "My output will be firm and formed like before."

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