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

A patient without symptoms has a TSH of 7.5 mIU/L on two occasions with a normal free T4. Which term fits best?

  1. AOvert primary hypothyroidism
  2. BSubclinical hypothyroidism
  3. CSecondary hypothyroidism
  4. DSubclinical hyperthyroidism
3.Arterial blood gas interpretationMedium
pH7.357.457.44PaCO₂mmHg354523HCO₃⁻mmol/L222615uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.44, PaCO₂ 23 mmHg (3.1 kPa) and HCO₃⁻ 15 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.44PaCO₂ 23 mmHg (3.1 kPa)HCO₃⁻ 15 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.31PaCO₂mmHg354531HCO₃⁻mmol/L222615appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.31, PaCO₂ 31 mmHg (4.1 kPa), HCO₃⁻ 15 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.31PaCO₂ 31 mmHg (4.1 kPa)HCO₃⁻ 15 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
10100t (h)C (mg/L)1 h40 mg/L11 h18.5 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 40 mg/L at 1 h and 18.5 mg/L at 11 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 40 mg/L (1 h)C2 = 18.5 mg/L (11 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
0.1110t (h)C (mg/L)Cpeak​ = 8τ = ?Ctrough​ = 1

Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 2 h. The target is a peak of 8 mg/L and a trough of 1 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 2 hCpeak = 8 mg/LCtrough = 1 mg/L
τ = h
7.Clinical nursingMedium

A spontaneously breathing patient is critically ill and hypoxic. Which oxygen delivery gives the highest concentration?

  1. AVenturi mask set to 24%
  2. BSimple face mask at 5 L/min with the patient sitting up
  3. CNon-rebreathe mask with reservoir bag at 15 L/min
  4. DNasal cannula at 2 L/min
8.Clinical nursingHard

A day after surgery, a patient's haemoglobin has fallen from 125 g/L to 85 g/L (12.5 to 8.5 g/dL) and their heart rate has risen to 118/min. What is the priority?

  1. ARecheck the haemoglobin next week
  2. BEncourage the patient to eat iron-rich foods
  3. CEscalate urgently, as the patient may be bleeding
  4. DDocument the result and take no further action

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: B · Subclinical hypothyroidism

Why this is the answer

A raised TSH with a normal free T4 is subclinical hypothyroidism; the pituitary is working harder to keep T4 normal. Overt hypothyroidism has a low free T4, and subclinical hyperthyroidism has a low TSH.Checked against: TSH (Thyroid-stimulating hormone) Test · MedlinePlus (NIH) (2024)

Answer

B · Subclinical hypothyroidism
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 7.44: within 7.35–7.45
PaCO₂ 23 mmHg: low (35–45)
HCO₃⁻ 15 mmol/L: low (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: appropriate compensation

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

1.5·15 + 8 = 30.5 → 28.5–32.5 mmHg

Compare

measured: 31 mmHg → appropriate compensation

Answer

appropriate compensation
5.Clinical pharmacokinetics

Answer: k = 0.07711 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(40/18.5)/10 = 0.7711/10 = 0.07711 h⁻¹

Answer

k = 0.07711 h⁻¹
6.Clinical pharmacokinetics

Answer: τ = 6 h

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.

Rate constant

k = 0.693/2 = 0.3466 h⁻¹

Dosing interval

τ = ln(8/1)/0.3466 = 2.079/0.3466 = 6 h

Answer

τ = 6 h

In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.

7.Clinical nursing

Answer: C · Non-rebreathe mask with reservoir bag at 15 L/min

Why this is the answer

A non-rebreathe mask with a filled reservoir at 15 L/min delivers a high oxygen concentration and is used first in critical illness. The dose is then titrated down to the target saturation.

Answer

C · Non-rebreathe mask with reservoir bag at 15 L/min
8.Clinical nursing

Answer: C · Escalate urgently, as the patient may be bleeding

Why this is the answer

A large fall in haemoglobin with tachycardia suggests ongoing blood loss. The nurse assesses the patient with ABCDE, checks wounds and drains and escalates for urgent review.

Answer

C · Escalate urgently, as the patient may be bleeding

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