Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
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.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which nerve supplies the diaphragm?
AVagus nerve
BPhrenic nerve
CAccessory nerve
DIntercostal nerves
2.Human anatomyHard
After a stroke, a patient speaks fluently but uses meaningless words and cannot understand speech. Where is the lesion most likely to be?
AInferior frontal gyrus of the dominant hemisphere
BPosterior superior temporal gyrus of the dominant hemisphere
CCerebellar hemisphere
DPrecentral gyrus of the non-dominant hemisphere
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
AEjection fraction
BStroke volume
CPeripheral resistance
DBlood pressure
4.Human physiologyHard
Why does a deep sea diver risk decompression sickness when surfacing quickly?
AOxygen becomes toxic at the surface
BThe lungs collapse when the pressure falls
CNitrogen dissolved at high pressure forms bubbles as the pressure falls
DCarbon dioxide cannot leave the lungs fast enough
5.MicrobiologyMedium
Pneumocystis jirovecii pneumonia in a person with advanced HIV is an example of:
AAn opportunistic infection
BA food-borne infection
CA vector-borne infection
DA zoonotic infection
6.MicrobiologyHard
Why are penicillins ineffective against Mycoplasma pneumoniae?
AIt produces large amounts of beta-lactamase
BIt has no cell wall for penicillin to act on
CIt lives only inside red blood cells
DIt has a thick waxy mycolic acid layer that blocks entry
7.PharmacologyMedium
Which patient should usually not use the combined oral contraceptive pill?
AA 25-year-old woman with heavy periods
BA 22-year-old non-smoker with acne
CA 38-year-old woman who smokes 20 cigarettes a day
DA 30-year-old non-smoker with no medical problems
8.PharmacologyHard
A patient receives high-dose IV aciclovir for herpes encephalitis. Which nursing action reduces the main risk to the kidneys?
AKeep the patient well hydrated and give the infusion slowly
BRestrict oral and IV fluids to prevent brain swelling
CGive the infusion as a rapid push
DGive a loop diuretic before each dose
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 145 mmol/L, glucose 15.3 mmol/L and urea 3.3 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 145 mmol/Lglucose 15.3 mmol/Lurea 3.3 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.31, PaCO₂ 79 mmHg (10.5 kPa), HCO₃⁻ 38 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.31PaCO₂ 79 mmHg (10.5 kPa)HCO₃⁻ 38 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
This lead II strip shows sinus rhythm with one pause. Measure the pause, the R–R interval across it, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
pause = s
12.ECG rhythm strip interpretationHard
Identify the rhythm on this lead II strip. Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1sinus tachycardia
2supraventricular tachycardia (SVT)
3ventricular tachycardia
4torsades de pointes
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 280 patients with a disease (cases) and 230 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 40b = 58c = 240d = 172
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 31% in relative terms (RRR = 31%). In a patient group the risk without treatment (CER) is 10% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 31%CER = 10%
NNT =
3Solutions
1.Human anatomy
Answer: B · Phrenic nerve
Why this is the answer
The phrenic nerve (roots C3–C5) is the motor supply to the diaphragm. High cervical cord injury above C3 therefore stops spontaneous breathing.
Answer
B · Phrenic nerve
2.Human anatomy
Answer: B · Posterior superior temporal gyrus of the dominant hemisphere
Why this is the answer
This is Wernicke's (receptive) aphasia, caused by damage to Wernicke's area in the posterior superior temporal gyrus. Broca's area in the inferior frontal gyrus causes non-fluent aphasia.
Answer
B · Posterior superior temporal gyrus of the dominant hemisphere
3.Human physiology
Answer: B · Stroke volume
Why this is the answer
CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.
Answer
B · Stroke volume
4.Human physiology
Answer: C · Nitrogen dissolved at high pressure forms bubbles as the pressure falls
Why this is the answer
At depth more nitrogen dissolves in the tissues. If pressure falls too fast it comes out of solution as bubbles in tissues and blood, causing "the bends".
Answer
C · Nitrogen dissolved at high pressure forms bubbles as the pressure falls
5.Microbiology
Answer: A · An opportunistic infection
Why this is the answer
Opportunistic infections take advantage of a weakened immune system. In HIV they become common when the CD4 count falls to low levels.
Answer
A · An opportunistic infection
6.Microbiology
Answer: B · It has no cell wall for penicillin to act on
Why this is the answer
Mycoplasma lacks a cell wall, so beta-lactams have no target. Macrolides or tetracyclines, which act on the ribosome, are used instead.
Answer
B · It has no cell wall for penicillin to act on
7.Pharmacology
Answer: C · A 38-year-old woman who smokes 20 cigarettes a day
Why this is the answer
Smoking over the age of 35 greatly increases the risk of myocardial infarction and stroke with combined hormonal contraception. Progestogen-only or non-hormonal methods are preferred.
Answer
C · A 38-year-old woman who smokes 20 cigarettes a day
8.Pharmacology
Answer: A · Keep the patient well hydrated and give the infusion slowly
Why this is the answer
Aciclovir can crystallise in the tubules and cause acute kidney injury. Good hydration, slow infusion and monitoring of kidney function reduce this risk.
Answer
A · Keep the patient well hydrated and give the infusion slowly
9.Arterial blood gas interpretation
Answer: Osm = 309 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 = 2·145 + 15.3 + 3.3 = 308.6 ≈ 309 mOsm/kg
Answer
Osm = 309 mOsm/kg
10.Arterial blood gas interpretation
Answer: chronic
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 HCO₃⁻
acute: 24 + 1·3.9 = 27.9 mmol/L
chronic: 24 + 3.5·3.9 = 37.65 mmol/L
Compare
measured: 38 mmol/L → chronic
Answer
chronic
11.ECG rhythm strip interpretation
Answer: pause = 1.84 s
Narrow-complex tachycardia and pauses: supraventricular tachycardia (SVT) is a regular tachycardia at about 150–250/min with narrow QRS complexes; the P waves are usually not visible because they are hidden in the QRS complex or the T wave. Sinus tachycardia in an adult at rest is usually below about 150/min and has a visible P wave before every QRS; atrial flutter with 2:1 conduction often runs at about 150/min and shows sawtooth waves. In sinus arrest the sinus node stops firing for a moment: a pause with no P wave and no QRS whose length is not a whole number of P–P cycles (a pause of exactly two or three cycles points to sinus exit block instead).
Measure
46 small boxes from R to R across the pause
Convert
pause = 46 × 0.04 s = 1.84 s
Compare with P–P
1.84 ÷ 0.80 ≈ 2.3 P–P cycles, not a whole number
Answer
pause = 1.84 s
No P wave appears in the pause, and the pause is not a whole number of P–P cycles: sinus arrest. A pause of exactly two or three cycles would point to sinus exit block.
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: OR = 0.49
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Odds of exposure
cases: 40/240 = 0.1667
controls: 58/172 = 0.3372
Odds ratio
OR = (40·172)/(58·240) = 0.4943 ≈ 0.49
Answer
OR = 0.49
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: NNT = 33
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.
Absolute risk reduction
ARR = 0.1·0.31 = 0.031
Number needed to treat
NNT = 1/0.031 = 32.26 → 33
Answer
NNT = 33
The same relative effect gives a smaller NNT when the baseline risk is higher.