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 hemodialysis has an arteriovenous (AV) fistula in the left arm. Where should the nurse draw routine blood samples?

  1. AFrom the left hand
  2. BFrom the fistula
  3. CFrom the left wrist
  4. DFrom the right arm
2.Lab values & electrolytesHard

A healthy woman at 10 weeks of pregnancy has a slightly low TSH with a normal free T4 and no symptoms. What is the most likely explanation?

  1. AThe result shows severe primary hypothyroidism
  2. BShe has pituitary failure from the pregnancy
  3. ChCG stimulates the thyroid in early pregnancy
  4. DShe has developed Graves disease in pregnancy
3.Arterial blood gas interpretationMedium

A practice gas shows a metabolic acidosis with HCO₃⁻ 10 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.

HCO₃⁻ 10 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
pH7.357.457.10PaCO₂mmHg354543HCO₃⁻mmol/L222613appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.10, PaCO₂ 43 mmHg (5.7 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.10PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 13 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
VCss​ = 25 mg/LR = ? mg/hCL = 4.5 L/h

Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 4.5 L/h. What infusion rate R in mg/h gives a steady-state level of 25 mg/L?

CL = 4.5 L/hCss = 25 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Vd = 7 L/kg0.5 → 0.8 mcg/LLD = ? mcg66 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 66 kg patient has a digoxin level of 0.5 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?

66 kgVd = 7 L/kgC = 0.5 → 0.8 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium

Which is a major risk factor for Clostridioides difficile infection?

  1. ARegular exercise
  2. BTaking regular paracetamol
  3. CRecent broad-spectrum antibiotic use
  4. DA diet high in fibre and fresh fruit
8.Clinical nursingHard

Which feature best distinguishes delirium from dementia?

  1. ADelirium starts suddenly and fluctuates, while dementia develops gradually
  2. BDelirium is permanent, while dementia is reversible
  3. CDementia always causes visual hallucinations
  4. DDelirium only affects people under 50, while dementia affects older people

3Solutions

1.Lab values & electrolytes

Answer: D · From the right arm

Why this is the answer

Venipuncture, blood pressure readings and IV lines are avoided in the arm with an AV fistula or graft, to prevent damage, clotting and infection of the access. Only the dialysis team uses the access, for dialysis.

Answer

D · From the right arm
2.Lab values & electrolytes

Answer: C · hCG stimulates the thyroid in early pregnancy

Why this is the answer

hCG is structurally similar to TSH and stimulates the thyroid, peaking at the end of the first trimester, so TSH often falls slightly. With a normal free T4 and no symptoms, this is usually physiologic and is interpreted with pregnancy-specific ranges.Checked against: TSH (Thyroid-stimulating hormone) Test · MedlinePlus (NIH) (2024)

Answer

C · hCG stimulates the thyroid in early pregnancy
3.Arterial blood gas interpretation

Answer: PaCO₂ = 23 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.

Winter’s formula

PaCO₂ = 1.5·10 + 8 = 23 mmHg
range: 21–25 mmHg

Answer

PaCO₂ = 23 mmHg
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

1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg

Compare

measured: 43 mmHg → added respiratory acidosis

Answer

added respiratory acidosis
5.Clinical pharmacokinetics

Answer: R = 112.5 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.

Infusion rate

R = 4.5·25 = 112.5 mg/h

Answer

R = 112.5 mg/h
6.Clinical pharmacokinetics

Answer: LD = 198 mcg

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 = 7·66 = 462 L

Rise needed

0.8 − 0.5 = 0.3 mcg/L

Loading dose

LD = 462·0.3/(1·0.7) = 198 mcg

Answer

LD = 198 mcg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: C · Recent broad-spectrum antibiotic use

Why this is the answer

Broad-spectrum antibiotics disrupt normal gut flora, letting C. difficile overgrow. Older age, hospital stay and proton pump inhibitors also increase the risk.

Answer

C · Recent broad-spectrum antibiotic use
8.Clinical nursing

Answer: A · Delirium starts suddenly and fluctuates, while dementia develops gradually

Why this is the answer

Delirium develops over hours to days, with fluctuating attention and consciousness, and is often reversible when the cause is treated. Dementia is a gradual, progressive decline. Patients with dementia are at higher risk of delirium.

Answer

A · Delirium starts suddenly and fluctuates, while dementia develops gradually

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