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 valve lies between the left atrium and the left ventricle?
AMitral (bicuspid) valve
BAortic valve
CTricuspid valve
DPulmonary valve
2.Human anatomyHard
Injury to the common fibular (peroneal) nerve at the neck of the fibula typically causes:
AWeak hip abduction
BFoot drop
CLoss of sensation on the sole of the foot
DInability to flex the knee
3.Human physiologyMedium
What causes the first heart sound (S1)?
AClosure of the mitral and tricuspid valves
BOpening of the aortic valve
CBlood filling the atria
DClosure of the aortic and pulmonary valves
4.Human physiologyHard
Which substance's clearance is used as the gold standard to measure glomerular filtration rate?
ACreatinine
BUrea
CInulin
DGlucose
5.MicrobiologyMedium
Which type of immunity is given by breast milk antibodies?
ANatural active immunity
BArtificial active immunity
CArtificial passive immunity
DNatural passive immunity
6.MicrobiologyHard
Why are penicillins ineffective against Mycoplasma pneumoniae?
AIt has no cell wall for penicillin to act on
BIt lives only inside red blood cells
CIt produces large amounts of beta-lactamase
DIt has a thick waxy mycolic acid layer that blocks entry
7.PharmacologyMedium
A patient wants to stop paroxetine suddenly. What should the nurse explain?
AIt can be stopped at any time, as antidepressants cause no withdrawal effects
CStopping suddenly causes the drug to work better for a while
DIt can cause dizziness, "electric shock" feelings and irritability
8.PharmacologyHard
A patient who has taken lorazepam daily for many months stops it suddenly. Which serious complication may follow?
APinpoint pupils and coma
BProfound respiratory depression
CSevere bradycardia
DWithdrawal seizures
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.52, PaCO₂ 47 mmHg (6.3 kPa) and HCO₃⁻ 37 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.52PaCO₂ 47 mmHg (6.3 kPa)HCO₃⁻ 37 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.33, PaCO₂ 65 mmHg (8.7 kPa), HCO₃⁻ 33 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 7.33PaCO₂ 65 mmHg (8.7 kPa)HCO₃⁻ 33 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 310 patients received a new treatment and 600 received the control treatment. The table shows how many in each group had a side effect. Find the number needed to harm NNH: how many patients must be treated for one extra patient to have the side effect? Round up to a whole number of patients.
treatment: 225/310control: 174/600
NNH =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 12% in relative terms (RRR = 12%). In a patient group the risk without treatment (CER) is 29% 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 = 12%CER = 29%
NNT =
3Solutions
1.Human anatomy
Answer: A · Mitral (bicuspid) valve
Why this is the answer
The mitral (bicuspid) valve separates the left atrium and left ventricle; the tricuspid valve does the same on the right side.
Answer
A · Mitral (bicuspid) valve
2.Human anatomy
Answer: B · Foot drop
Why this is the answer
The common fibular nerve supplies the dorsiflexors and evertors. Damage at the fibular neck causes foot drop and a high-stepping gait. The sole is supplied by the tibial nerve.
Answer
B · Foot drop
3.Human physiology
Answer: A · Closure of the mitral and tricuspid valves
Why this is the answer
S1 ("lub") is caused by closure of the atrioventricular valves at the start of systole. S2 ("dub") is closure of the semilunar valves.
Answer
A · Closure of the mitral and tricuspid valves
4.Human physiology
Answer: C · Inulin
Why this is the answer
Inulin is filtered freely and is not reabsorbed or secreted, so its clearance equals GFR exactly. Creatinine is slightly secreted, so it overestimates GFR a little.
Answer
C · Inulin
5.Microbiology
Answer: D · Natural passive immunity
Why this is the answer
Antibodies (mainly IgA) passed in breast milk give passive immunity acquired naturally. It is short-lived because no memory cells form.
Answer
D · Natural passive immunity
6.Microbiology
Answer: A · 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
A · It has no cell wall for penicillin to act on
7.Pharmacology
Answer: D · It can cause dizziness, "electric shock" feelings and irritability
Why this is the answer
Antidepressant discontinuation syndrome is common with short half-life drugs such as paroxetine and venlafaxine. The dose is tapered gradually under supervision.
Answer
D · It can cause dizziness, "electric shock" feelings and irritability
8.Pharmacology
Answer: D · Withdrawal seizures
Why this is the answer
Abrupt withdrawal after long-term benzodiazepine use causes anxiety, insomnia, tremor and sometimes seizures. Long-term users should have the dose tapered gradually.
Answer
D · Withdrawal seizures
9.Arterial blood gas interpretation
Answer: partially 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 7.52: alkalaemia (above 7.45)
PaCO₂ 47 mmHg: high (35–45)
HCO₃⁻ 37 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
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 + 1·2.5 = 26.5 mmol/L
chronic: 24 + 3.5·2.5 = 32.75 mmol/L
Compare
measured: 33 mmol/L → chronic
Answer
chronic
11.ECG rhythm strip interpretation
Answer: QTc = 0.507 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 12.5 × 0.04 = 0.50 s
RR = 24 × 0.04 = 0.96 s
Fridericia’s formula
QTc = 0.50/∛0.96 = 0.50/0.9865 = 0.507 s
Answer
QTc = 0.507 s
Above 0.50 s: markedly prolonged. (507 ms)
12.ECG rhythm strip interpretation
Answer: atrial rate = 79 bpm
AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.
Atrial rate
P–P = 19 small boxes = 0.76 s → 1500 ÷ 19 = 78.95 ≈ 79 per minute
Ventricular rate
R–R = 28 small boxes = 1.12 s → 1500 ÷ 28 = 53.57 ≈ 54 per minute
Answer
atrial rate = 79 bpm
Answer
atrial rate = 79 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
13.Evidence-based practice statistics
Answer: NNH = 3
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.
Event rates
EER = 225/310 = 0.7258
CER = 174/600 = 0.29
Number needed to harm
ARI = 0.7258 − 0.29 = 0.4358
NNH = 1/0.4358 = 2.295 → 3
Answer
NNH = 3
About one in every 3 treated patients has the side effect because of the treatment.
14.Evidence-based practice statistics
Answer: NNT = 29
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 = 0.29·0.12 = 0.0348
Number needed to treat
NNT = 1/0.0348 = 28.74 → 29
Answer
NNT = 29
The same relative effect gives a smaller NNT when the baseline risk is higher.