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

Why is a blood β-hydroxybutyrate level often preferred to a urine ketone dipstick in DKA?

  1. AThe urine dipstick only detects ketones in people without diabetes
  2. BIt is the only test that can measure glucose and ketones together
  3. CIt measures the main ketone in DKA, and the dipstick may miss it
  4. DIt shows the average ketone level over the past three months
2.Lab values & electrolytesHard

A patient with thalassemia major has received regular red cell transfusions for years. The ferritin is now very high. What does this most likely show, and what treatment is used?

  1. AIron overload, treated with iron chelation therapy
  2. BAcute infection, treated with broad-spectrum antibiotics
  3. CLiver failure, treated with a liver transplant at once
  4. DIron deficiency, treated with IV iron supplements
3.Arterial blood gas interpretationMedium
pH7.357.457.43PaCO₂mmHg354525HCO₃⁻mmol/L222616uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.43, PaCO₂ 25 mmHg (3.3 kPa) and HCO₃⁻ 16 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.43PaCO₂ 25 mmHg (3.3 kPa)HCO₃⁻ 16 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.07PaCO₂mmHg354536HCO₃⁻mmol/L222610appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.07, PaCO₂ 36 mmHg (4.8 kPa), HCO₃⁻ 10 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.07PaCO₂ 36 mmHg (4.8 kPa)HCO₃⁻ 10 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
110t (h)C (mg/L)2 h8 mg/L10 h1.3 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 8 mg/L at 2 h and 1.3 mg/L at 10 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 8 mg/L (2 h)C2 = 1.3 mg/L (10 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
VCss,avg​ = 15 mg/LD = ? mg · q8hCL = 0.035 L/kg/h56 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 56 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.035 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 0.035 L/kg/hτ = 8 hS = 0.8F = 156 kgCss,avg = 15 mg/L
D = mg
7.Clinical nursingMedium

A patient taking a statin reports unexplained muscle pain and weakness. What should the nurse do?

  1. AAdvise more vigorous exercise to work through it
  2. BReassure them that this is a sign the drug is working
  3. CReport it, as statins can cause muscle damage
  4. DTell them to double the dose
8.Clinical nursingHard

Which patient should the nurse assess first?

  1. AA patient with diabetes whose glucose is 11 mmol/L (198 mg/dL) before lunch
  2. BA postoperative patient who rates their pain as 6/10
  3. CA patient with COPD whose SpO2 is 89% and within their target range
  4. DA patient with asthma who can only speak in single words

3Solutions

1.Lab values & electrolytes

Answer: C · It measures the main ketone in DKA, and the dipstick may miss it

Why this is the answer

In DKA, β-hydroxybutyrate is the main ketone, but most urine dipsticks detect acetoacetate, so they can underestimate ketosis. Blood β-hydroxybutyrate gives a current value that is also useful to monitor recovery.Checked against: Diabetic ketoacidosis - Medical Encyclopedia · MedlinePlus (NIH) (2025)

Answer

C · It measures the main ketone in DKA, and the dipstick may miss it
2.Lab values & electrolytes

Answer: A · Iron overload, treated with iron chelation therapy

Why this is the answer

The body has no active way to excrete iron, so repeated transfusions cause iron overload, which can damage the heart, liver and endocrine glands. A very high ferritin reflects this, and iron chelators are used to remove the excess.Checked against: Thalassemia · MedlinePlus (NIH) (2026); Ferritin Blood Test · MedlinePlus (NIH) (2025)

Answer

A · Iron overload, treated with iron chelation therapy
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 7.43: within 7.35–7.45
PaCO₂ 25 mmHg: low (35–45)
HCO₃⁻ 16 mmol/L: low (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: added respiratory acidosis

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

1.5·10 + 8 = 23 → 21–25 mmHg

Compare

measured: 36 mmHg → added respiratory acidosis

Answer

added respiratory acidosis
5.Clinical pharmacokinetics

Answer: k = 0.2271 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(8/1.3)/8 = 1.817/8 = 0.2271 h⁻¹

Answer

k = 0.2271 h⁻¹
6.Clinical pharmacokinetics

Answer: D = 294 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.

Clearance

CL = 0.035·56 = 1.96 L/h

Dose

D = Css,avg·CL·τ/(S·F) = 15·1.96·8/(0.8·1) = 294 mg

Answer

D = 294 mg
7.Clinical nursing

Answer: C · Report it, as statins can cause muscle damage

Why this is the answer

Statins can cause myopathy and, rarely, rhabdomyolysis. Unexplained muscle symptoms should be reported so creatine kinase can be checked and the drug reviewed.

Answer

C · Report it, as statins can cause muscle damage
8.Clinical nursing

Answer: D · A patient with asthma who can only speak in single words

Why this is the answer

Being unable to complete sentences is a sign of acute severe or life-threatening asthma, an airway and breathing emergency. The other patients need attention but are not immediately life-threatened.

Answer

D · A patient with asthma who can only speak in single words

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