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 has IV fluids running into the left forearm. Where should the nurse take a blood sample?

  1. AFrom the opposite arm
  2. BFrom any vein, since the fluid does not affect results
  3. CFrom the IV cannula while the fluid is running
  4. DFrom just above the IV site in the same arm
2.Lab values & electrolytesHard

A patient receiving chemotherapy has a white cell count of 1.8 × 10³/µL with 20% neutrophils and 5% bands. What is the absolute neutrophil count (ANC)?

  1. A4,500/µL
  2. B450/µL
  3. C1,800/µL
  4. D360/µL
3.Arterial blood gas interpretationMedium
pH7.357.457.37PaCO₂mmHg354518HCO₃⁻mmol/L222610uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.37, PaCO₂ 18 mmHg (2.4 kPa) and HCO₃⁻ 10 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.

pH 7.37PaCO₂ 18 mmHg (2.4 kPa)HCO₃⁻ 10 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.32PaCO₂mmHg354534HCO₃⁻mmol/L222617added normal-gap acidosispure high-gap acidosisadded metabolic alkalosis

A practice gas shows pH 7.32, PaCO₂ 34 mmHg (4.5 kPa) and HCO₃⁻ 17 mmol/L, with Na⁺ 133 and Cl⁻ 99 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.

pH 7.32PaCO₂ 34 mmHg (4.5 kPa)HCO₃⁻ 17 mmol/LNa⁺ 133 mmol/LCl⁻ 99 mmol/L
  1. 1added normal-gap acidosis
  2. 2pure high-gap acidosis
  3. 3added metabolic alkalosis
5.Clinical pharmacokineticsMedium
VCss​ = 2.5 mg/LR = ? mg/minCL = 850 mL/min

Practice exercise (made-up values, not clinical guidance). A patient has a lidocaine clearance of 850 mL/min. What infusion rate R in mg/min keeps the steady-state level at 2.5 mg/L?

CL = 850 mL/minCss = 2.5 mg/L
R = mg/min
6.Clinical pharmacokineticsHard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 1

Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 2 h. The target is a peak of 6 mg/L and a trough of 1 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 2 hCpeak = 6 mg/LCtrough = 1 mg/L
τ = h
7.Clinical nursingMedium

Which signs suggest a blood transfusion reaction in the first minutes?

  1. AFever, chills, back pain and low blood pressure
  2. BImproved colour and energy
  3. CSlow, steady rise in haemoglobin over the next day
  4. DMild hunger
8.Clinical nursingHard

A patient who has had a kidney stone is taught about prevention. Which statement shows a need for further teaching?

  1. A"I will try to pass pale urine."
  2. B"I will drink less water so I don't need to pass urine as often."
  3. C"I will aim to drink plenty of fluids spread through every day."
  4. D"I will follow the diet advice for my stone type."

3Solutions

1.Lab values & electrolytes

Answer: A · From the opposite arm

Why this is the answer

Blood taken near a running infusion is mixed with the IV fluid, causing false results, such as a high glucose or a low hemoglobin. The sample is taken from the other arm, or according to policy if that is not possible.

Answer

A · From the opposite arm
2.Lab values & electrolytes

Answer: B · 450/µL

Why this is the answer

ANC = 1,800 × (0.20 + 0.05) = 450/µL. A value below 500/µL is severe neutropenia with a high risk of serious infection, so neutropenic precautions and early reporting of fever are essential.

Answer

B · 450/µL
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 7.37: within 7.35–7.45
PaCO₂ 18 mmHg: low (35–45)
HCO₃⁻ 10 mmol/L: low (22–26)

Compensation

Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 21–25 mmHg, so the measured 18 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: added normal-gap acidosis

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 = 133 − (99 + 17) = 17

Delta ratio

(17 − 12)/(24 − 17) = 0.7143
added normal-gap acidosis

Answer

added normal-gap acidosis
5.Clinical pharmacokinetics

Answer: R = 2.125 mg/min

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 in L/min

850 mL/min = 0.85 L/min

Infusion rate

R = 0.85·2.5 = 2.125 mg/min

Answer

R = 2.125 mg/min
6.Clinical pharmacokinetics

Answer: τ = 5.2 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.

Rate constant

k = 0.693/2 = 0.3466 h⁻¹

Dosing interval

τ = ln(6/1)/0.3466 = 1.792/0.3466 = 5.17 ≈ 5.2 h

Answer

τ = 5.2 h

In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.

7.Clinical nursing

Answer: A · Fever, chills, back pain and low blood pressure

Why this is the answer

An acute haemolytic or febrile reaction can cause fever, rigors, loin or back pain, breathlessness and hypotension. Stop the transfusion and call for help.

Answer

A · Fever, chills, back pain and low blood pressure
8.Clinical nursing

Answer: B · "I will drink less water so I don't need to pass urine as often."

Why this is the answer

A high fluid intake dilutes urine and is the most important general measure to prevent stones. Dietary advice depends on the stone's composition.

Answer

B · "I will drink less water so I don't need to pass urine as often."

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