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 part of the brainstem contains the main cardiovascular and respiratory centres?
AMidbrain
BPons
CMedulla oblongata
DCerebellum
2.Human anatomyHard
A lesion of the left optic tract causes which visual field defect?
ABlindness in the left eye only
BRight homonymous hemianopia
CLeft homonymous hemianopia
DBitemporal hemianopia
3.Human physiologyMedium
Which receptor is stimulated when the patellar tendon is tapped in the knee-jerk reflex?
AMuscle spindle
BNociceptor
CPacinian corpuscle
DGolgi tendon organ
4.Human physiologyHard
Why does breathing 100% oxygen correct hypoxaemia from a true right-to-left shunt only poorly?
AThe extra oxygen is all exhaled as CO₂
BShunted blood is already fully saturated
CPure oxygen damages haemoglobin
DThe shunted blood never comes into contact with alveolar gas
5.MicrobiologyMedium
What is the main difference between innate and adaptive immunity?
AAdaptive immunity acts within minutes of infection
BInnate immunity uses antibodies and forms memory; adaptive immunity does not
CAdaptive immunity is specific and forms memory; innate immunity does not
DInnate immunity develops only after vaccination
6.MicrobiologyHard
A traveller returns from sub-Saharan Africa with fever. Malaria is suspected. Which species is most likely to cause severe and cerebral malaria?
APlasmodium malariae
BPlasmodium vivax
CPlasmodium ovale
DPlasmodium falciparum
7.PharmacologyMedium
A patient with insulin-treated diabetes is found unconscious with very low blood glucose. There is no IV access. Which drug can a trained helper give?
AOral glucose gel poured into the mouth
BGlucagon by injection or nasal spray
CIntramuscular adrenaline
DRapid-acting insulin under the skin
8.PharmacologyHard
Before starting long-term bisphosphonate treatment, a patient is advised to have a dental check-up. Why?
ABisphosphonates stain the teeth black if taken for many years
BThey are linked to osteonecrosis of the jaw after dental surgery
CBisphosphonates dissolve tooth enamel within weeks
DBisphosphonates cause severe gum overgrowth
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 31 mmHg (4.1 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
This lead II strip shows atrial flutter with a fixed conduction ratio. Find the ventricular rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
ventricular rate = bpm
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). A case-control study compares 60 patients with a disease (cases) and 410 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 = 37b = 119c = 23d = 291
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 38% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.41. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.
control 38%RR = 1.41
NNH =
3Solutions
1.Human anatomy
Answer: C · Medulla oblongata
Why this is the answer
The medulla oblongata contains the vital centres that regulate heart rate, blood pressure and breathing rhythm.
Answer
C · Medulla oblongata
2.Human anatomy
Answer: B · Right homonymous hemianopia
Why this is the answer
The left optic tract carries fibres from the left half of each retina, which see the right visual field. Its lesion causes loss of the right half of the field in both eyes.
Answer
B · Right homonymous hemianopia
3.Human physiology
Answer: A · Muscle spindle
Why this is the answer
Tapping the tendon stretches quadriceps muscle spindles. Their afferents synapse directly on motor neurons in a monosynaptic reflex that contracts the muscle.
Answer
A · Muscle spindle
4.Human physiology
Answer: D · The shunted blood never comes into contact with alveolar gas
Why this is the answer
Shunted blood bypasses ventilated alveoli and mixes deoxygenated blood into arterial blood. Extra inspired oxygen cannot reach it, unlike in V/Q mismatch or hypoventilation.
Answer
D · The shunted blood never comes into contact with alveolar gas
5.Microbiology
Answer: C · Adaptive immunity is specific and forms memory; innate immunity does not
Why this is the answer
Innate defences (barriers, phagocytes, complement) act fast and in the same way each time. Adaptive immunity (B and T cells) takes days, is specific to each antigen and forms memory.
Answer
C · Adaptive immunity is specific and forms memory; innate immunity does not
6.Microbiology
Answer: D · Plasmodium falciparum
Why this is the answer
P. falciparum causes very high parasite levels and makes infected red cells stick in small vessels, including in the brain. It is common in sub-Saharan Africa and is a medical emergency.
Answer
D · Plasmodium falciparum
7.Pharmacology
Answer: B · Glucagon by injection or nasal spray
Why this is the answer
Glucagon breaks down liver glycogen and raises blood glucose. Nothing is given by mouth to an unconscious patient because of the risk of choking and aspiration.
Answer
B · Glucagon by injection or nasal spray
8.Pharmacology
Answer: B · They are linked to osteonecrosis of the jaw after dental surgery
Why this is the answer
Osteonecrosis of the jaw is a rare but serious adverse effect, mostly after invasive dental work. Dental problems are treated before starting, and good oral hygiene is maintained.
Answer
B · They are linked to osteonecrosis of the jaw after dental surgery
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 22.2 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
|31 − 40| = 9 mmHg → 0.9 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·0.9 = 22.2 mmol/L
Answer
HCO₃⁻ = 22.2 mmol/L
10.Arterial blood gas interpretation
Answer: AGc = 12.25 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.
Anion gap
AG = 140 − (111 + 20) = 9
Correction for albumin
9 + 0.25·(40 − 27) = 12.25 mmol/L
Answer
AGc = 12.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: ventricular rate = 100 bpm
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.
Flutter waves
F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute
QRS complexes
R–R = 15 small boxes = 0.60 s → 1500 ÷ 15 = 100 per minute
conduction ratio = 300 ÷ 100 = 3:1
Answer
ventricular rate = 100 bpm
12.ECG rhythm strip interpretation
Answer: atrial rate = 68 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 = 22 small boxes = 0.88 s → 1500 ÷ 22 = 68.18 ≈ 68 per minute
Ventricular rate
5 QRS complexes in 6 s × 10 = 50 per minute
Answer
atrial rate = 68 bpm
Answer
atrial rate = 68 bpm
Second-degree type II: 3 P waves for every 2 QRS complexes, with a constant PR.
13.Evidence-based practice statistics
Answer: OR = 3.93
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: 37/23 = 1.609
controls: 119/291 = 0.4089
Odds ratio
OR = (37·291)/(119·23) = 3.934 ≈ 3.93
Answer
OR = 3.93
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NNH = 7
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.