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 receives repeated nebulized albuterol (salbutamol) for severe asthma. Which laboratory change should the nurse watch for?

  1. AA rise in serum sodium
  2. BA fall in serum potassium
  3. CA rise in serum calcium
  4. DA fall in blood glucose
2.Lab values & electrolytesHard

A patient on a heparin infusion has a platelet count that falls from 260 to 110 × 10³/µL on day 7. Which action is most appropriate?

  1. AGive a platelet transfusion and continue the heparin
  2. BContinue the infusion and increase the rate as per protocol
  3. CReport it at once: heparin-induced thrombocytopenia is possible
  4. DTake no action, because the count is still just above 100 × 10³/µL
3.Arterial blood gas interpretationMedium
pH7.357.457.25PaCO₂mmHg354574HCO₃⁻mmol/L222631uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.25, PaCO₂ 74 mmHg (9.9 kPa) and HCO₃⁻ 31 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.25PaCO₂ 74 mmHg (9.9 kPa)HCO₃⁻ 31 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 144, Cl⁻ 114, HCO₃⁻ 21 mmol/L and albumin 30 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 144 mmol/LCl⁻ 114 mmol/LHCO₃⁻ 21 mmol/Lalbumin 30 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
VCss​ = 15 mg/LR = ? mg/hCL = 0.03 L/kg/h89 kg

Practice exercise (made-up values, not clinical guidance). An 89 kg patient receives a theophylline infusion. The clearance is 0.03 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 15 mg/L?

89 kgCL = 0.03 L/kg/hCss = 15 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Vd = 7 L/kg0.3 → 0.8 mcg/LLD = ? mcg84 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). An 84 kg patient has a digoxin level of 0.3 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?

84 kgVd = 7 L/kgC = 0.3 → 0.8 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium

After a laparoscopic cholecystectomy, a patient has pain in the shoulder tip. What is the most likely cause?

  1. AA broken collarbone from being moved onto the table
  2. BGas used during the operation irritating the diaphragm
  3. CAn allergic reaction to the dressing
  4. DA heart attack caused by the stress of surgery
8.Clinical nursingHard

A patient took a large paracetamol (acetaminophen) overdose six hours ago and says they feel well. Why is treatment still urgent?

  1. AParacetamol damage only shows up in the kidneys, many weeks later
  2. BPatients who feel well after six hours never need any treatment at all
  3. CLiver damage develops later, and the antidote works best when given early
  4. DParacetamol overdose causes immediate breathing problems

3Solutions

1.Lab values & electrolytes

Answer: B · A fall in serum potassium

Why this is the answer

Beta-2 agonists push potassium into cells and can cause hypokalemia, especially at high doses; they also raise blood glucose. For this reason nebulized albuterol is sometimes used as an extra treatment for hyperkalemia.

Answer

B · A fall in serum potassium
2.Lab values & electrolytes

Answer: C · Report it at once: heparin-induced thrombocytopenia is possible

Why this is the answer

Heparin-induced thrombocytopenia causes a large fall in platelets (often more than half), usually 5–14 days after starting heparin, and carries a high risk of clots. The prescriber stops all heparin and starts a non-heparin anticoagulant if HIT is suspected.Checked against: Heparin-Induced Thrombocytopenia · StatPearls (NCBI Bookshelf) (2023)

Answer

C · Report it at once: heparin-induced thrombocytopenia is possible
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 7.25: acidaemia (below 7.35)
PaCO₂ 74 mmHg: high (35–45)
HCO₃⁻ 31 mmol/L: high (22–26)

Compensation

partially compensated

Answer

partially compensated
4.Arterial blood gas interpretation

Answer: AGc = 11.5 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.

Anion gap

AG = 144 − (114 + 21) = 9

Correction for albumin

9 + 0.25·(40 − 30) = 11.5 mmol/L

Answer

AGc = 11.5 mmol/L

Still within about 8–12 after correction.

5.Clinical pharmacokinetics

Answer: R = 40.05 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.

Clearance

CL = 0.03·89 = 2.67 L/h

Infusion rate

R = 2.67·15 = 40.05 mg/h

Answer

R = 40.05 mg/h
6.Clinical pharmacokinetics

Answer: LD = 420 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 = 7·84 = 588 L

Rise needed

0.8 − 0.3 = 0.5 mcg/L

Loading dose

LD = 588·0.5/(1·0.7) = 420 mcg

Answer

LD = 420 mcg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: B · Gas used during the operation irritating the diaphragm

Why this is the answer

Carbon dioxide used to inflate the abdomen can irritate the diaphragm, causing referred shoulder-tip pain. It usually settles within a few days with mobilisation and analgesia.

Answer

B · Gas used during the operation irritating the diaphragm
8.Clinical nursing

Answer: C · Liver damage develops later, and the antidote works best when given early

Why this is the answer

Paracetamol overdose may cause few early symptoms, but liver failure can develop over days. Acetylcysteine is most effective when started early, guided by timed paracetamol levels.

Answer

C · Liver damage develops later, and the antidote works best when given early

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