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 had a total gastrectomy. Which long-term therapy is needed to prevent anemia?

  1. AMonthly transfusions of red blood cells
  2. BWeekly doses of oral vitamin K tablets
  3. CLifelong vitamin B12 replacement
  4. DLifelong oral folic acid tablets only
2.Lab values & electrolytesHard

A young adult has fatigue, bruising and fever. The white cell count is 85 × 10³/µL with many blasts, the hemoglobin is 7.9 g/dL (79 g/L) and the platelets are 22 × 10³/µL. What is the most likely diagnosis?

  1. APolycythemia vera
  2. BIron-deficiency anemia
  3. CAcute viral infection
  4. DAcute leukemia
3.Arterial blood gas interpretationMedium

A practice gas shows PaCO₂ 55 mmHg (7.3 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.

PaCO₂ 55 mmHg (7.3 kPa)baseline: 40 mmHg, 24 mmol/L
HCO₃⁻ = mmol/L
4.Arterial blood gas interpretationHard
pH7.357.457.19PaCO₂mmHg354573HCO₃⁻mmol/L222627acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.19, PaCO₂ 73 mmHg (9.7 kPa), HCO₃⁻ 27 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 7.19PaCO₂ 73 mmHg (9.7 kPa)HCO₃⁻ 27 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
5.Clinical pharmacokineticsMedium
Vd = 7 L/kgC = 1 mcg/LLD = ? mcg76 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 76 kg patient is to be loaded with digoxin tablets to reach a level of 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.

76 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg4.5 → 15 mg/LLD = ? mg78 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 78 kg patient has a phenytoin level of 4.5 mg/L; the target is 15 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?

78 kgVd = 0.7 L/kgC = 4.5 → 15 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium

A patient started on an ACE inhibitor develops a persistent dry cough. What is the likely cause?

  1. AAsthma caused by a beta-blocker
  2. BAn allergic reaction to a recently given penicillin
  3. CA recognised side effect of the ACE inhibitor
  4. DPulmonary embolism
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 urinary retention
  3. CAcute kidney injury with no urine production
  4. DUrinary tract infection without retention

3Solutions

1.Lab values & electrolytes

Answer: C · Lifelong vitamin B12 replacement

Why this is the answer

Intrinsic factor from the stomach is needed to absorb vitamin B12 in the ileum. After a total gastrectomy, B12 must be replaced for life, usually by injection, to prevent macrocytic anemia and nerve damage.Checked against: Vitamin B12 or folate deficiency anaemia - Causes · NHS (2023); Vitamin B12 deficiency anemia · MedlinePlus (NIH) (2026)

Answer

C · Lifelong vitamin B12 replacement
2.Lab values & electrolytes

Answer: D · Acute leukemia

Why this is the answer

Many circulating blasts with anemia and thrombocytopenia suggest acute leukemia: malignant cells crowd out normal production in the marrow. Despite the high white count, the patient has few working white cells and is at high risk of infection and bleeding.Checked against: Laboratory Evaluation of Acute Leukemia · StatPearls (NCBI Bookshelf) (2025)

Answer

D · Acute leukemia
3.Arterial blood gas interpretation

Answer: HCO₃⁻ = 25.5 mmol/L

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.

Change in PaCO₂

|55 − 40| = 15 mmHg → 1.5 × 10 mmHg

Expected HCO₃⁻

HCO₃⁻ = 24 + 1·1.5 = 25.5 mmol/L

Answer

HCO₃⁻ = 25.5 mmol/L
4.Arterial blood gas interpretation

Answer: acute

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 + 1·3.3 = 27.3 mmol/L
chronic: 24 + 3.5·3.3 = 35.55 mmol/L

Compare

measured: 27 mmol/L → acute

Answer

acute
5.Clinical pharmacokinetics

Answer: LD = 760 mcg

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 = 7·76 = 532 L

Amount of active drug needed

Vd·C = 532·1 = 532 mcg

Loading dose

LD = 532/(1·0.7) = 760 mcg

Answer

LD = 760 mcg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: LD = 623.2 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 = 0.7·78 = 54.6 L

Rise needed

15 − 4.5 = 10.5 mg/L

Loading dose

LD = 54.6·10.5/(0.92·1) = 623.2 mg

Answer

LD = 623.2 mg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: C · A recognised side effect of the ACE inhibitor

Why this is the answer

Dry cough is a common side effect of ACE inhibitors such as ramipril. An angiotensin II receptor blocker is often substituted.

Answer

C · A recognised side effect of the ACE inhibitor
8.Clinical nursing

Answer: B · 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

B · Acute urinary retention

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