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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

How should a blood sample for serum ammonia be handled?

  1. ALeft at room temperature for several hours first
  2. BShaken firmly to mix it before it is sent away
  3. CPlaced on ice and sent to the laboratory at once
  4. DKept warm and sent with the next routine batch
2.Lab values & electrolytesHard

After a crush injury, a patient's urine dipstick is strongly positive for blood, but microscopy shows almost no red cells. CK is very high. What best explains this?

  1. AA kidney stone is scraping the ureter lining
  2. BThe sample was contaminated with menstrual blood
  3. CBleeding is coming from the bladder lining
  4. DMyoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretationMedium

Practice values: Na⁺ 137 mmol/L, glucose 21.9 mmol/L and urea 14.9 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 137 mmol/Lglucose 21.9 mmol/Lurea 14.9 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
pH7.357.457.27PaCO₂mmHg354525HCO₃⁻mmol/L222611appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.27, PaCO₂ 25 mmHg (3.3 kPa), HCO₃⁻ 11 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.27PaCO₂ 25 mmHg (3.3 kPa)HCO₃⁻ 11 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg93 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 93 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 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.

93 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
VCss,avg​ = ?300 mg · q12hCL = 0.04 L/kg/h70 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 70 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.04 L/kg/h. Each dose is 300 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.04 L/kg/hτ = 12 hS = 1F = 170 kgD = 300 mg
Css,avg = mg/L
7.Clinical nursingMedium

Which is a first-line nursing measure to help prevent constipation in hospital?

  1. ARestricting fluids to reduce urine output
  2. BGiving an enema to every patient on admission
  3. CKeeping the patient in bed
  4. DEncouraging fluids, fibre and mobility
8.Clinical nursingHard

Which feature best distinguishes delirium from dementia?

  1. ADelirium is permanent, while dementia is reversible
  2. BDelirium only affects people under 50, while dementia affects older people
  3. CDementia always causes visual hallucinations
  4. DDelirium starts suddenly and fluctuates, while dementia develops gradually

3Solutions

1.Lab values & electrolytes

Answer: C · Placed on ice and sent to the laboratory at once

Why this is the answer

Ammonia can rise in a standing sample as red cells and proteins break down. The sample is usually placed on ice and analyzed promptly to avoid a falsely high result.

Answer

C · Placed on ice and sent to the laboratory at once
2.Lab values & electrolytes

Answer: D · Myoglobin from damaged muscle is in the urine

Why this is the answer

The dipstick blood test detects heme, which is present in both hemoglobin and myoglobin. Positive "blood" without red cells, together with a very high CK, indicates myoglobinuria from rhabdomyolysis, which can damage the kidneys. IV fluids and close monitoring of urine output and potassium are needed.

Answer

D · Myoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretation

Answer: Osm = 311 mOsm/kg

Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.

Calculated osmolality

Osm = 2·137 + 21.9 + 14.9 = 310.8 ≈ 311 mOsm/kg

Answer

Osm = 311 mOsm/kg
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

1.5·11 + 8 = 24.5 → 22.5–26.5 mmHg

Compare

measured: 25 mmHg → appropriate compensation

Answer

appropriate compensation
5.Clinical pharmacokinetics

Answer: LD = 1415 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 = 0.7·93 = 65.1 L

Amount of active drug needed

Vd·C = 65.1·20 = 1302 mg

Loading dose

LD = 1302/(0.92·1) = 1415 mg

Answer

LD = 1415 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: Css,avg = 8.929 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 = 0.04·70 = 2.8 L/h

Average steady-state level

Css,avg = S·F·D/(CL·τ) = 1·1·300/(2.8·12) = 8.929 mg/L

Answer

Css,avg = 8.929 mg/L
7.Clinical nursing

Answer: D · Encouraging fluids, fibre and mobility

Why this is the answer

Adequate fluids, a fibre-rich diet, mobility and privacy for toileting help prevent constipation. Laxatives are added if needed, especially with opioids.

Answer

D · Encouraging fluids, fibre and mobility
8.Clinical nursing

Answer: D · Delirium starts suddenly and fluctuates, while dementia develops gradually

Why this is the answer

Delirium develops over hours to days, with fluctuating attention and consciousness, and is often reversible when the cause is treated. Dementia is a gradual, progressive decline. Patients with dementia are at higher risk of delirium.

Answer

D · Delirium starts suddenly and fluctuates, while dementia develops gradually

ExamProblems · examproblems.com/courses/lab-values-for-nurses