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 artery supplies most of the anterior wall of the left ventricle and the front of the interventricular septum?
APosterior descending artery
BRight coronary artery
CLeft anterior descending artery
DCircumflex artery
2.Human anatomyHard
A large central disc prolapse at L4–L5 causes saddle anaesthesia, urinary retention and bilateral leg weakness. Which structure is compressed?
AFemoral nerve
BSciatic nerve in the buttock
CThoracic spinal cord
DCauda equina
3.Human physiologyMedium
Mean arterial pressure is approximately equal to:
AHeart rate × stroke volume
BStroke volume × total peripheral vascular resistance
CCardiac output × total peripheral resistance
DCardiac output ÷ heart rate
4.Human physiologyHard
A pulmonary embolus blocks blood flow to a well-ventilated region of lung. How does the ventilation–perfusion (V/Q) ratio of that region change?
AIt becomes low, like in pneumonia
BIt becomes very high, acting as dead space
CIt stays normal because ventilation falls too
DIt becomes zero, acting as a shunt
5.MicrobiologyMedium
Which laboratory test separates Staphylococcus aureus from most other staphylococci?
AThe Gram stain
BThe catalase test
CThe coagulase test
DThe acid-fast stain
6.MicrobiologyHard
Which feature distinguishes most exotoxins from endotoxin?
AExotoxins are only released when bacteria die
BExotoxins are proteins and can be turned into toxoid vaccines
CExotoxins are lipopolysaccharides of the Gram-negative outer membrane
DExotoxins are heat-stable lipids
7.PharmacologyMedium
A patient starting niacin (nicotinic acid) complains of flushing and warmth after doses. Which advice may reduce this?
ADouble the dose to build tolerance faster
BTake the niacin with a hot drink so that it dissolves and acts faster
CTake it on an empty stomach in a hot bath
DTake aspirin 30 minutes before the niacin, if the prescriber agrees
8.PharmacologyHard
A patient with epilepsy needs pain relief. Why is tramadol used with caution?
AIt causes a sharp rise in blood glucose
BIt always causes respiratory arrest in epilepsy
CIt lowers the seizure threshold
DIt speeds up the metabolism of all antiepileptic drugs
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 78 mmHg (10.4 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 metabolic acidosis: pH 7.24, PaCO₂ 43 mmHg (5.7 kPa), HCO₃⁻ 18 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.24PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 18 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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 Bazett’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
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
1ventricular fibrillation
2third-degree (complete) AV block
3junctional rhythm
4asystole
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 320 patients with a disease (cases) and 380 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 = 207b = 144c = 113d = 236
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 99% and a specificity of 84%. It is used in a population where the prevalence is 1%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 99%Sp = 84%p = 1%
PPV = %
3Solutions
1.Human anatomy
Answer: C · Left anterior descending artery
Why this is the answer
The left anterior descending (anterior interventricular) artery supplies the anterior left ventricle and the anterior two thirds of the septum. Its blockage causes an anterior infarct.
Answer
C · Left anterior descending artery
2.Human anatomy
Answer: D · Cauda equina
Why this is the answer
Below L1–L2 the canal holds the cauda equina. Central compression of these lumbar and sacral roots causes cauda equina syndrome, a surgical emergency.
Answer
D · Cauda equina
3.Human physiology
Answer: C · Cardiac output × total peripheral resistance
Why this is the answer
Pressure = flow × resistance, so MAP ≈ CO × TPR. Changes in either cardiac output or vascular resistance change blood pressure.
Answer
C · Cardiac output × total peripheral resistance
4.Human physiology
Answer: B · It becomes very high, acting as dead space
Why this is the answer
Ventilation without perfusion gives a very high V/Q ratio (dead space). A shunt, such as in pneumonia, is the opposite: perfusion without ventilation.
Answer
B · It becomes very high, acting as dead space
5.Microbiology
Answer: C · The coagulase test
Why this is the answer
S. aureus produces coagulase, which converts fibrinogen to fibrin. All staphylococci are Gram-positive and catalase-positive, so those tests do not separate them.
Answer
C · The coagulase test
6.Microbiology
Answer: B · Exotoxins are proteins and can be turned into toxoid vaccines
Why this is the answer
Exotoxins are secreted proteins with specific actions and are often very potent; many can be inactivated into toxoids. Endotoxin is lipopolysaccharide, more heat-stable and cannot be made into a toxoid.
Answer
B · Exotoxins are proteins and can be turned into toxoid vaccines
7.Pharmacology
Answer: D · Take aspirin 30 minutes before the niacin, if the prescriber agrees
Why this is the answer
Niacin causes prostaglandin-mediated vasodilation and flushing. Aspirin beforehand, taking it with food and avoiding hot drinks and alcohol reduce this effect.
Answer
D · Take aspirin 30 minutes before the niacin, if the prescriber agrees
8.Pharmacology
Answer: C · It lowers the seizure threshold
Why this is the answer
Tramadol can cause seizures, especially at high doses or with other drugs that lower the seizure threshold. Its serotonergic action also raises the risk of serotonin syndrome.
Answer
C · It lowers the seizure threshold
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.8 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
|78 − 40| = 38 mmHg → 3.8 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.8 = 27.8 mmol/L
Answer
HCO₃⁻ = 27.8 mmol/L
10.Arterial blood gas interpretation
Answer: added respiratory 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.
Expected PaCO₂
1.5·18 + 8 = 35 → 33–37 mmHg
Compare
measured: 43 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: QTc = 0.375 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 = 9 × 0.04 = 0.36 s
RR = 23 × 0.04 = 0.92 s
Bazett’s formula
QTc = 0.36/√0.92 = 0.36/0.9592 = 0.375 s
Answer
QTc = 0.375 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (375 ms)
12.ECG rhythm strip interpretation
Answer: asystole
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: no countable ventricular rate.
2. Rhythm: none; an almost flat line.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: no QRS complexes.
Interpretation
asystole
Answer
asystole
A flat line is always checked: look at the patient, the leads and the gain, and view a second lead, because a loose electrode or fine VF can look like asystole.
13.Evidence-based practice statistics
Answer: OR = 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.
Odds of exposure
cases: 207/113 = 1.832
controls: 144/236 = 0.6102
Odds ratio
OR = (207·236)/(144·113) = 3.002 ≈ 3
Answer
OR = 3
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 5.9 %
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.
True and false positives (per person)
Se·p = 0.99·0.01 = 0.0099
(1 − Sp)·(1 − p) = 0.16·0.99 = 0.1584
Positive predictive value
PPV = 0.0099/(0.0099 + 0.1584) = 0.05882 ≈ 5.9%
Answer
PPV = 5.9%
The lower the prevalence, the lower the PPV of the same test.