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
The sympathetic nervous system leaves the central nervous system from which spinal segments?
ASacral only (S2–S4)
BCervical only
CCranial and sacral
DThoracic and upper lumbar (T1–L2)
2.Human anatomyHard
Which ligament of the knee prevents the tibia from sliding forward relative to the femur?
APatellar ligament
BMedial collateral ligament
CAnterior cruciate ligament
DPosterior cruciate ligament
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
APeripheral resistance
BStroke volume
CBlood pressure
DEjection fraction
4.Human physiologyHard
Why does a person with lactase deficiency get diarrhoea and bloating after drinking milk?
ALactose is absorbed too quickly and raises blood glucose
BMilk protein triggers an allergic reaction
CLactose damages the villi directly
DUndigested lactose draws water into the gut and is fermented by colonic bacteria
5.MicrobiologyMedium
Why are Gram-negative bacteria often harder to treat with antibiotics?
AThey divide much more slowly
BThey have no ribosomes
CThey have a thicker peptidoglycan layer
DTheir outer membrane blocks many drugs
6.MicrobiologyHard
Which hepatitis viruses are mainly spread through contaminated food and water?
AHepatitis B and D
BHepatitis C and E
CHepatitis A and E
DHepatitis B and C
7.PharmacologyMedium
A patient is starting olanzapine. Which monitoring is most important over the following months?
AWeight, blood glucose and blood lipids
BINR and platelet count
CPeak expiratory flow
DHearing tests and vision tests every month
8.PharmacologyHard
Some receptors show activity even without an agonist. How does an inverse agonist differ from a neutral antagonist at such receptors?
AIt destroys the receptor permanently
BIt has no effect when given alone
CIt reduces activity below the baseline level
DIt increases activity above the baseline
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 14 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 14 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 62 mmHg (8.3 kPa) and PaCO₂ 40 mmHg (5.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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 fibrillation
2atrial flutter, 2:1 conduction
3atrial flutter, 4:1 conduction
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
Interpret this lead II strip. Which interpretation can you defend from this single 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
2second-degree AV block type I (Wenckebach)
3first-degree AV block
42:1 AV block (type undetermined)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 340 patients with a disease (cases) and 500 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 = 24b = 65c = 316d = 435
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 91% and specificity 99%. The prevalence of the disease is 29%. How many false-positive results FP are expected?
N = 10 000p = 29%Se = 91%Sp = 99%
FP =
3Solutions
1.Human anatomy
Answer: D · Thoracic and upper lumbar (T1–L2)
Why this is the answer
Sympathetic preganglionic neurons lie in the lateral horn from T1 to L2. The parasympathetic system has a craniosacral outflow.
Answer
D · Thoracic and upper lumbar (T1–L2)
2.Human anatomy
Answer: C · Anterior cruciate ligament
Why this is the answer
The anterior cruciate ligament limits anterior translation of the tibia; a positive anterior drawer or Lachman test suggests an ACL tear.
Answer
C · Anterior cruciate ligament
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: D · Undigested lactose draws water into the gut and is fermented by colonic bacteria
Why this is the answer
Unabsorbed lactose acts osmotically, pulling water into the lumen. Colonic bacteria ferment it, producing gas and short-chain fatty acids.
Answer
D · Undigested lactose draws water into the gut and is fermented by colonic bacteria
5.Microbiology
Answer: D · Their outer membrane blocks many drugs
Why this is the answer
The outer membrane of Gram-negative bacteria, with lipopolysaccharide and porins, keeps many antibiotics out and adds efflux and enzyme defences.
Answer
D · Their outer membrane blocks many drugs
6.Microbiology
Answer: C · Hepatitis A and E
Why this is the answer
Hepatitis A and E spread by the faecal–oral route. B, C and D spread through blood and body fluids and can cause chronic infection.
Answer
C · Hepatitis A and E
7.Pharmacology
Answer: A · Weight, blood glucose and blood lipids
Why this is the answer
Olanzapine and several other atypical antipsychotics cause weight gain, insulin resistance and dyslipidaemia. Weight, waist size, glucose and lipids are checked regularly.
Answer
A · Weight, blood glucose and blood lipids
8.Pharmacology
Answer: C · It reduces activity below the baseline level
Why this is the answer
A neutral antagonist simply blocks agonists and leaves baseline activity unchanged. An inverse agonist stabilises the inactive receptor state, so constitutive activity falls below baseline.
Answer
C · It reduces activity below the baseline level
9.Arterial blood gas interpretation
Answer: PaCO₂ = 29 mmHg
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.
Winter’s formula
PaCO₂ = 1.5·14 + 8 = 29 mmHg
range: 27–31 mmHg
Answer
PaCO₂ = 29 mmHg
10.Arterial blood gas interpretation
Answer: A–a gradient = 37.73 mmHg
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.
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: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 4:1 conduction
Answer
atrial flutter, 4:1 conduction
12.ECG rhythm strip interpretation
Answer: 2:1 AV block (type undetermined)
2:1 AV block: with two P waves for every QRS, no two conducted P waves follow each other, so you cannot see whether the PR would lengthen (type I) or stay constant (type II) before a non-conducted P wave. From one strip it is called a 2:1 AV block; a longer strip or a change in the conduction ratio decides the type. A narrow QRS makes type I more likely and a wide QRS type II, but neither proves it.
Five steps
1. Rate: R–R = 42 small boxes, 1500 ÷ 42 ≈ 36 per minute.
2. Rhythm: regular, every R–R is 1.68 s.
3. P waves: upright and regular (P–P 0.84 s, 71 per minute), two for every QRS.
4. PR interval: constant at 0.16 s in the conducted beats.
5. QRS: 0.08 s, narrow (below 0.12 s).
Why the type stays open
Every second P wave is not conducted, so no two conducted beats follow each other and a lengthening PR could never show.
2:1 AV block (type undetermined)
Answer
2:1 AV block (type undetermined)
A narrow QRS makes type I more likely and a wide QRS type II, but only a longer strip or a change in the conduction ratio decides.
13.Evidence-based practice statistics
Answer: OR = 0.51
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: 24/316 = 0.07595
controls: 65/435 = 0.1494
Odds ratio
OR = (24·435)/(65·316) = 0.5083 ≈ 0.51
Answer
OR = 0.51
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: FP = 71
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.
With and without the disease
10 000·0.29 = 2900
10 000 − 2900 = 7100
False positives
FP = 7100·0.01 = 71
Answer
FP = 71
Of the 2710 positives only 2639 have the disease (PPV 97.4%).