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 warfarin for atrial fibrillation has an INR of 1.4. What does this mean?

  1. AThe dose is exactly right and needs no change
  2. BThe patient is at a high risk of bleeding and needs vitamin K at once
  3. CAnticoagulation is below target, so clots and stroke are more likely
  4. DThe patient no longer needs to take warfarin
2.Lab values & electrolytesHard

A patient on clozapine for 2 years has been stable. Which statement shows a need for further teaching?

  1. A"I can stop my blood tests since I feel well"
  2. B"I'll report a sore throat or fever at once"
  3. C"I'll keep all my scheduled blood test visits"
  4. D"I'll tell other doctors that I take clozapine"
3.Arterial blood gas interpretationMedium
pH7.357.457.10PaCO₂mmHg354540HCO₃⁻mmol/L222612uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.10, PaCO₂ 40 mmHg (5.3 kPa) and HCO₃⁻ 12 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.

pH 7.10PaCO₂ 40 mmHg (5.3 kPa)HCO₃⁻ 12 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 146, Cl⁻ 114, HCO₃⁻ 16 mmol/L and albumin 19 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 146 mmol/LCl⁻ 114 mmol/LHCO₃⁻ 16 mmol/Lalbumin 19 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg54 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 54 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.

54 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
61 kg72 · womanSCr 1.5 mg/dLCrCl → D = ? mg

Practice exercise (made-up values, not clinical guidance). A 72-year-old woman weighs 61 kg and has a serum creatinine of 1.5 mg/dL (133 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual cefazolin dose is 1000 mg every 8 h for a creatinine clearance of 120 mL/min. Keeping the interval, the dose is reduced in proportion to the creatinine clearance. Find the adjusted dose D. Round to the nearest whole number.

72 years61 kgwomanSCr = 1.5 mg/dL1000 mg q8hCrCl normal 120 mL/min
D = mg
7.Clinical nursingMedium

Which adverse effects should be monitored for in a patient receiving IV gentamicin?

  1. AHair loss and weight gain
  2. BHyperglycaemia and tremor
  3. CConstipation and urinary retention
  4. DKidney damage and hearing loss
8.Clinical nursingHard

A postoperative patient has a heart rate of 115/min, temperature 38.6 °C, respiratory rate 26/min and blood pressure 85/50 mmHg despite a fluid bolus. What does this most likely indicate?

  1. AA normal inflammatory response in the first days after surgery
  2. BMild dehydration only
  3. CSeptic shock requiring urgent critical care review
  4. DAnxiety about discharge

3Solutions

1.Lab values & electrolytes

Answer: C · Anticoagulation is below target, so clots and stroke are more likely

Why this is the answer

An INR below the target range means the blood clots too readily for this indication. The prescriber is informed, and the dose and possible causes, such as missed doses or interactions, are reviewed.Checked against: Prothrombin Time Test and INR (PT/INR) · MedlinePlus (NIH) (2024)

Answer

C · Anticoagulation is below target, so clots and stroke are more likely
2.Lab values & electrolytes

Answer: A · "I can stop my blood tests since I feel well"

Why this is the answer

Clozapine requires ongoing ANC monitoring according to the schedule, because severe neutropenia can occur without warning. The patient must keep blood test appointments and report signs of infection at once.

Answer

A · "I can stop my blood tests since I feel well"
3.Arterial blood gas interpretation

Answer: uncompensated

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 7.10: acidaemia (below 7.35)
PaCO₂ 40 mmHg: normal (35–45)
HCO₃⁻ 12 mmol/L: low (22–26)

Compensation

Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 24–28 mmHg, so the measured 40 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.

uncompensated

Answer

uncompensated
4.Arterial blood gas interpretation

Answer: AGc = 21.25 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 = 146 − (114 + 16) = 16

Correction for albumin

16 + 0.25·(40 − 19) = 21.25 mmol/L

Answer

AGc = 21.25 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: LD = 616.3 mg

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 = 0.7·54 = 37.8 L

Amount of active drug needed

Vd·C = 37.8·15 = 567 mg

Loading dose

LD = 567/(0.92·1) = 616.3 mg

Answer

LD = 616.3 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: D = 272 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.

Creatinine clearance (Cockcroft–Gault)

CrCl = (140 − 72)·61/(72·1.5)·0.85 = 32.65 mL/min

Adjusted dose

D = 1000·32.65/120 = 272.1 ≈ 272 mg

Answer

D = 272 mg

In practice the dose is rounded to a convenient amount and checked against the product information.

7.Clinical nursing

Answer: D · Kidney damage and hearing loss

Why this is the answer

Gentamicin is nephrotoxic and ototoxic. Blood levels and renal function are monitored, and any tinnitus, dizziness or hearing change is reported.

Answer

D · Kidney damage and hearing loss
8.Clinical nursing

Answer: C · Septic shock requiring urgent critical care review

Why this is the answer

Infection signs with hypotension not responding to fluid suggest septic shock. Vasopressors and critical care input are likely needed, along with antibiotics and source control.

Answer

C · Septic shock requiring urgent critical care review

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