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

Two sets of blood cultures are ordered for a patient with suspected sepsis. How are they best collected?

  1. AFrom two separate venipuncture sites, using strict aseptic technique
  2. BAfter the first antibiotic dose, so the infusion is not delayed
  3. CFrom the existing IV line while the infusion continues to run
  4. DBoth from the same site, one minute apart, to save time for the patient
2.Lab values & electrolytesHard

Why can an HbA1c result be misleading in a patient who received a large blood transfusion two weeks ago?

  1. ATransfusions always raise HbA1c to falsely high levels
  2. BHbA1c measures sugar on white cells, not red cells
  3. CHbA1c is accurate in every situation
  4. DMany red cells are donor cells, not exposed to the patient’s glucose
3.Arterial blood gas interpretationMedium
pH7.357.457.51PaCO₂mmHg354547HCO₃⁻mmol/L222636uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.51, PaCO₂ 47 mmHg (6.3 kPa) and HCO₃⁻ 36 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.51PaCO₂ 47 mmHg (6.3 kPa)HCO₃⁻ 36 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.32PaCO₂mmHg354532HCO₃⁻mmol/L222616appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.32, PaCO₂ 32 mmHg (4.3 kPa), HCO₃⁻ 16 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.32PaCO₂ 32 mmHg (4.3 kPa)HCO₃⁻ 16 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
0.1110t (h)C (mg/L)2 h6 mg/L10 h0.9 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 2 h and 0.9 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.

C1 = 6 mg/L (2 h)C2 = 0.9 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
VCss,avg​ = ?450 mg · q12hCL = 0.03 L/kg/h98 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 98 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.03 L/kg/h. Each dose is 450 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.03 L/kg/hτ = 12 hS = 1F = 198 kgD = 450 mg
Css,avg = mg/L
7.Clinical nursingMedium

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

  1. APulmonary embolism
  2. BAn allergic reaction to a recently given penicillin
  3. CAsthma caused by a beta-blocker
  4. DA recognised side effect of the ACE inhibitor
8.Clinical nursingHard

A patient has new facial weakness. Which feature suggests a stroke rather than Bell's palsy?

  1. AThe forehead can still be wrinkled on the affected side
  2. BThe whole of one side of the face, including the forehead, is weak
  3. CThe patient has had a recent cold
  4. DThe patient has pain behind the ear

3Solutions

1.Lab values & electrolytes

Answer: A · From two separate venipuncture sites, using strict aseptic technique

Why this is the answer

Taking cultures from two separate sites helps distinguish true bacteremia (growth in both sets) from skin contamination (growth in only one). The skin is cleaned carefully, the bottle tops are disinfected, and cultures are taken before antibiotics.

Answer

A · From two separate venipuncture sites, using strict aseptic technique
2.Lab values & electrolytes

Answer: D · Many red cells are donor cells, not exposed to the patient’s glucose

Why this is the answer

HbA1c forms as glucose attaches to hemoglobin over the life of a red cell. Transfused donor cells and conditions that shorten red cell life give falsely low values, so a glucose log or other markers are used instead.

Answer

D · Many red cells are donor cells, not exposed to the patient’s glucose
3.Arterial blood gas interpretation

Answer: partially 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 7.51: alkalaemia (above 7.45)
PaCO₂ 47 mmHg: high (35–45)
HCO₃⁻ 36 mmol/L: high (22–26)

Compensation

partially compensated

Answer

partially 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

1.5·16 + 8 = 32 → 30–34 mmHg

Compare

measured: 32 mmHg → appropriate compensation

Answer

appropriate compensation
5.Clinical pharmacokinetics

Answer: t½ = 2.923 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.

Elimination rate constant

k = ln(6/0.9)/8 = 1.897/8 = 0.2371 h⁻¹

Half-life

t½ = 0.693/0.2371 = 2.923 h

Answer

t½ = 2.923 h
6.Clinical pharmacokinetics

Answer: Css,avg = 12.76 mg/L

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 = 0.03·98 = 2.94 L/h

Average steady-state level

Css,avg = S·F·D/(CL·τ) = 1·1·450/(2.94·12) = 12.76 mg/L

Answer

Css,avg = 12.76 mg/L
7.Clinical nursing

Answer: D · 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

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

Answer: A · The forehead can still be wrinkled on the affected side

Why this is the answer

In a stroke (upper motor neurone lesion), the forehead is usually spared because it has bilateral cortical supply. Bell's palsy (lower motor neurone) weakens the whole side, including the forehead.

Answer

A · The forehead can still be wrinkled on the affected side

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