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 vessels carry oxygenated blood from the lungs to the heart?
APulmonary arteries
BVenae cavae
CBronchial arteries
DPulmonary veins
2.Human anatomyHard
A stab wound cuts the left half of the spinal cord at T10. Which pattern of loss is expected below the lesion?
ALeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
BRight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
CLoss of all sensation and movement on the left only
DLoss of pain and temperature on both sides only
3.Human physiologyMedium
According to Poiseuille's law, if the radius of an arteriole halves, its resistance to flow:
AIncreases about 16-fold
BIncreases 4-fold
CDoubles
DHalves
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AConcentric hypertrophy with a thicker wall
BThinning of the wall with dilation
CNo change because the valve is not part of the ventricle
DAtrophy of the muscle due to less work
5.MicrobiologyMedium
Cholera causes profuse "rice-water" diarrhoea. What is the most important treatment?
AAntiviral drugs
BAntifungal drugs
CStopping all oral fluid intake so that the bowel can rest and recover
DReplacing fluid and salts by oral or intravenous rehydration
6.MicrobiologyHard
Why must a negative-sense RNA virus such as influenza carry its own RNA-dependent RNA polymerase into the host cell?
AThe polymerase protects the virus from antibodies
BHost ribosomes cannot read any viral genes
CThe polymerase is needed to integrate the genome into host DNA
DHost cells cannot make mRNA from an RNA template
7.PharmacologyMedium
How long does it take, approximately, for a drug given at regular intervals to reach a steady state?
A10 half-lives
B4 to 5 half-lives
CIt depends only on the dose size
D1 half-life
8.PharmacologyHard
Drug X gives a maximum response of 100% with an EC50 of 1 mg. Drug Y gives a maximum response of 100% with an EC50 of 10 mg. Which statement is correct?
AX has higher efficacy than Y
BY has higher efficacy than X
CX is more potent than Y, and both have the same efficacy
DY is more potent than X, and both have the same efficacy
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 49 mmHg (6.5 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a respiratory acidosis: pH 7.26, PaCO₂ 74 mmHg (9.9 kPa), HCO₃⁻ 32 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 7.26PaCO₂ 74 mmHg (9.9 kPa)HCO₃⁻ 32 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
Interpret this lead II rhythm strip. Which rhythm is it? 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
1atrial flutter, 4:1 conduction
2atrial flutter, 3:1 conduction
3normal sinus rhythm
4atrial flutter, 2:1 conduction
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the 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
1second-degree AV block type II
2third-degree (complete) AV block
3dual-chamber (AV) paced rhythm
4paced rhythm, failure to capture
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 150 patients with a disease (cases) and 520 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 = 86b = 182c = 64d = 338
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 21% in relative terms (RRR = 21%). In a patient group the risk without treatment (CER) is 21% 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 = 21%CER = 21%
NNT =
3Solutions
1.Human anatomy
Answer: D · Pulmonary veins
Why this is the answer
Four pulmonary veins drain oxygenated blood into the left atrium. They are the main exception to the rule that veins carry deoxygenated blood.
Answer
D · Pulmonary veins
2.Human anatomy
Answer: A · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
Why this is the answer
In Brown-Séquard syndrome the corticospinal tract and dorsal columns are lost on the same side, while the spinothalamic tract, which crosses near entry, causes contralateral pain and temperature loss.
Answer
A · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: A · Increases about 16-fold
Why this is the answer
Resistance is proportional to 1/r⁴, so halving the radius raises resistance by 2⁴ = 16. This is why small changes in arteriole diameter control blood flow so strongly.
Answer
A · Increases about 16-fold
4.Human physiology
Answer: A · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.
Answer
A · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: D · Replacing fluid and salts by oral or intravenous rehydration
Why this is the answer
Cholera toxin makes gut cells secrete large amounts of water and salt. Rapid rehydration saves most lives; antibiotics only shorten the illness.
Answer
D · Replacing fluid and salts by oral or intravenous rehydration
6.Microbiology
Answer: D · Host cells cannot make mRNA from an RNA template
Why this is the answer
A negative-sense genome must first be copied into positive-sense mRNA. Human cells have no enzyme that copies RNA from RNA, so the virus brings its own polymerase.
Answer
D · Host cells cannot make mRNA from an RNA template
7.Pharmacology
Answer: B · 4 to 5 half-lives
Why this is the answer
With repeated dosing, the level approaches steady state after about 4–5 half-lives, whatever the dose. A loading dose can be used to reach it faster.
Answer
B · 4 to 5 half-lives
8.Pharmacology
Answer: C · X is more potent than Y, and both have the same efficacy
Why this is the answer
A lower EC50 means less drug is needed for half the effect, so X is more potent. Both reach the same maximum, so their efficacy is equal.
Answer
C · X is more potent than Y, and both have the same efficacy
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 24.9 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.
Change in PaCO₂
|49 − 40| = 9 mmHg → 0.9 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·0.9 = 24.9 mmol/L
Answer
HCO₃⁻ = 24.9 mmol/L
10.Arterial blood gas interpretation
Answer: acute on 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 + 1·3.4 = 27.4 mmol/L
chronic: 24 + 3.5·3.4 = 35.9 mmol/L
Compare
measured: 32 mmol/L → acute on chronic
Answer
acute on chronic
11.ECG rhythm strip interpretation
Answer: atrial flutter, 4:1 conduction
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Five steps
1. Rate: R–R = 24 small boxes, 1500 ÷ 24 ≈ 63 per minute.
2. Rhythm: regular, every R–R is 0.96 s.
3. P waves: none; regular sawtooth flutter waves every 0.24 s (250 per minute), 4 for each QRS.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 4:1 conduction
Answer
atrial flutter, 4:1 conduction
12.ECG rhythm strip interpretation
Answer: paced rhythm, failure to capture
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: 4 QRS complexes in 6 s, × 10 = 40 per minute.
2. Rhythm: the pacing spikes are regular (every 0.96 s), but the R–R is not, because a QRS is missing after some spikes.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.14 s and wide after the captured spikes; 2 spikes are not followed by a QRS (no capture).
Interpretation
paced rhythm, failure to capture
Answer
paced rhythm, failure to capture
Failure to capture: the pacemaker fires but the heart does not respond. Textbooks treat it as urgent to report, because the patient may depend on the pacemaker.
13.Evidence-based practice statistics
Answer: OR = 2.5
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: 86/64 = 1.344
controls: 182/338 = 0.5385
Odds ratio
OR = (86·338)/(182·64) = 2.496 ≈ 2.5
Answer
OR = 2.5
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NNT = 23
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 = 0.21·0.21 = 0.0441
Number needed to treat
NNT = 1/0.0441 = 22.68 → 23
Answer
NNT = 23
The same relative effect gives a smaller NNT when the baseline risk is higher.