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

Which IV fluid can be given through the same line as a red cell transfusion?

  1. A0.45% sodium chloride
  2. B5% dextrose in water
  3. CLactated Ringer's (Hartmann's) solution
  4. D0.9% sodium chloride
2.Lab values & electrolytesHard

During an IV magnesium sulfate infusion, which assessment best detects toxicity early?

  1. AKnee reflexes, respiratory rate and urine output
  2. BSkin color and capillary refill every hour
  3. CBowel sounds and abdominal girth
  4. DBlood glucose and pupil size every hour
3.Arterial blood gas interpretationMedium
pH7.357.457.38PaCO₂mmHg354565HCO₃⁻mmol/L222637uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.38, PaCO₂ 65 mmHg (8.7 kPa) and HCO₃⁻ 37 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.38PaCO₂ 65 mmHg (8.7 kPa)HCO₃⁻ 37 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 138, Cl⁻ 115, HCO₃⁻ 11 mmol/L and albumin 15 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 138 mmol/LCl⁻ 115 mmol/LHCO₃⁻ 11 mmol/Lalbumin 15 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 15 mg/LLD = ? mg52 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 52 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.

52 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
95 kg74 · womanSCr 1.4 mg/dLCrCl → τ = ? h

Practice exercise (made-up values, not clinical guidance). A 74-year-old woman weighs 95 kg and has a serum creatinine of 1.4 mg/dL (124 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 gentamicin regimen is 80 mg every 8 h for a creatinine clearance of 120 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.

74 years95 kgwomanSCr = 1.4 mg/dL80 mg q8hCrCl normal 120 mL/min
τ = h
7.Clinical nursingMedium

Why are alternate sutures sometimes removed first when taking out stitches?

  1. ABecause the remaining sutures dissolve on their own within a day
  2. BTo reduce the cost of dressings
  3. CTo make the procedure faster
  4. DTo check the wound stays closed before removing the rest
8.Clinical nursingHard

A patient with an acute GI bleed has a heart rate of 120/min and blood pressure 90/60 mmHg. What is the nursing priority after calling for help?

  1. AGive a large meal to restore energy
  2. BSecure large-bore IV access and prepare for fluid or blood replacement
  3. CEncourage walking to the bathroom alone
  4. DDelay action until the stool colour is checked

3Solutions

1.Lab values & electrolytes

Answer: D · 0.9% sodium chloride

Why this is the answer

Only 0.9% sodium chloride is used with red cells. Dextrose solutions can cause hemolysis, and lactated Ringer's contains calcium, which can cause clotting in the citrated blood. Drugs are never added to blood products.

Answer

D · 0.9% sodium chloride
2.Lab values & electrolytes

Answer: A · Knee reflexes, respiratory rate and urine output

Why this is the answer

Loss of the knee reflex is an early warning sign of magnesium toxicity and appears before slowed breathing. Magnesium is cleared by the kidneys, so falling urine output increases the risk. IV calcium gluconate counteracts magnesium toxicity.Checked against: Magnesium Toxicity · StatPearls (NCBI Bookshelf) (2026); Magnesium Toxicity · StatPearls (NCBI Bookshelf)

Answer

A · Knee reflexes, respiratory rate and urine output
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 7.38: within 7.35–7.45
PaCO₂ 65 mmHg: high (35–45)
HCO₃⁻ 37 mmol/L: high (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: AGc = 18.25 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 = 138 − (115 + 11) = 12

Correction for albumin

12 + 0.25·(40 − 15) = 18.25 mmol/L

Answer

AGc = 18.25 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: LD = 487.5 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 = 0.5·52 = 26 L

Amount of active drug needed

Vd·C = 26·15 = 390 mg

Loading dose

LD = 390/(0.8·1) = 487.5 mg

Answer

LD = 487.5 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: τ = 18.2 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.

Creatinine clearance (Cockcroft–Gault)

CrCl = (140 − 74)·95/(72·1.4)·0.85 = 52.87 mL/min

New interval

τ = 8·120/52.87 = 18.16 ≈ 18.2 h

Answer

τ = 18.2 h

In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.

7.Clinical nursing

Answer: D · To check the wound stays closed before removing the rest

Why this is the answer

Removing alternate sutures allows the nurse to see whether the wound edges stay together. If they begin to separate, the remaining sutures can be left and the team informed.

Answer

D · To check the wound stays closed before removing the rest
8.Clinical nursing

Answer: B · Secure large-bore IV access and prepare for fluid or blood replacement

Why this is the answer

Tachycardia and hypotension suggest significant blood loss. ABCDE assessment, wide-bore IV access, bloods including cross-match and resuscitation come first, followed by endoscopy.

Answer

B · Secure large-bore IV access and prepare for fluid or blood replacement

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