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 muscle is the main extensor of the elbow?
ACoracobrachialis
BBiceps brachii
CBrachialis
DTriceps brachii
2.Human anatomyHard
A tumour at the apex of the lung causes drooping of the eyelid, a small pupil and loss of sweating on the same side of the face. Which pathway is disrupted?
ACervical sympathetic chain
BOculomotor nerve
CFacial nerve
DParasympathetic fibres of the vagus
3.Human physiologyMedium
Why is the oxygen–haemoglobin dissociation curve S-shaped?
ABinding of one oxygen makes the next bind more easily
BRed cells change size as they take up oxygen
COxygen dissolves better in plasma at high partial pressure
DHaemoglobin can bind only one oxygen molecule
4.Human physiologyHard
Constriction of the efferent arteriole of a glomerulus, with the afferent arteriole unchanged, tends to:
AHave no effect on GFR at all
BStop filtration completely
CRaise glomerular capillary pressure and GFR
DLower glomerular capillary pressure and GFR
5.MicrobiologyMedium
Which antibody class is the first to rise in a new infection?
AIgA
BIgM
CIgE
DIgG
6.MicrobiologyHard
Joseph Lister reduced infection after surgery in the 1860s. What was his main innovation?
AUsing carbolic acid (phenol) as an antiseptic on wounds and instruments
BVaccinating surgeons against staphylococci
CUsing steam autoclaves to sterilise all surgical instruments before each operation
DGiving sulfonamide antibiotics after every operation
7.PharmacologyMedium
When does oseltamivir give the most benefit in influenza?
AOnly after a bacterial pneumonia has developed
BWhen started within 48 hours of the first symptoms
CAt any time, as the timing makes no difference
DOnly after symptoms have lasted at least one week
8.PharmacologyHard
A malnourished patient with low serum albumin shows nystagmus and unsteady gait. Her total phenytoin level is within range. What is the most likely explanation?
ALow albumin speeds up phenytoin metabolism
BMore phenytoin is unbound when albumin is low, so free levels rise
CThe total level is high, but the laboratory is wrong
DLow albumin makes phenytoin bind more tightly, so the drug is less active
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 75 mmHg (10.0 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 7.10, PaCO₂ 43 mmHg (5.7 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.10PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 13 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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
1sinus arrhythmia
2atrial fibrillation
3atrial flutter, 4:1 conduction
4atrial flutter, 2:1 conduction
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. The QRS complexes are regular: find the ventricular 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
ventricular rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 850 patients received a new treatment and 920 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 24/850control: 64/920
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 40% in relative terms (RRR = 40%). In a patient group the risk without treatment (CER) is 14% 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 = 40%CER = 14%
NNT =
3Solutions
1.Human anatomy
Answer: D · Triceps brachii
Why this is the answer
Triceps brachii inserts on the olecranon of the ulna and extends the elbow. Biceps and brachialis are elbow flexors.
Answer
D · Triceps brachii
2.Human anatomy
Answer: A · Cervical sympathetic chain
Why this is the answer
This is Horner's syndrome: ptosis, miosis and anhidrosis from interruption of sympathetic fibres, here by an apical (Pancoast) tumour pressing on the sympathetic chain.
Answer
A · Cervical sympathetic chain
3.Human physiology
Answer: A · Binding of one oxygen makes the next bind more easily
Why this is the answer
Haemoglobin shows cooperative binding: each bound O₂ increases the affinity of the remaining sites, giving a sigmoid curve.
Answer
A · Binding of one oxygen makes the next bind more easily
4.Human physiology
Answer: C · Raise glomerular capillary pressure and GFR
Why this is the answer
Efferent constriction holds blood in the glomerulus, raising its pressure and GFR. Angiotensin II acts this way, which is why ACE inhibitors can lower GFR.
Answer
C · Raise glomerular capillary pressure and GFR
5.Microbiology
Answer: B · IgM
Why this is the answer
IgM appears first in a primary response; IgG follows and provides long-term protection. A positive IgM therefore suggests recent infection.
Answer
B · IgM
6.Microbiology
Answer: A · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
Why this is the answer
Lister applied carbolic acid to wounds, dressings and instruments, which greatly reduced post-operative infection. This started antiseptic surgery.
Answer
A · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
7.Pharmacology
Answer: B · When started within 48 hours of the first symptoms
Why this is the answer
Neuraminidase inhibitors work best early, while the virus is multiplying. Started within 48 hours, they shorten illness; in high-risk or hospitalised patients they may still be given later.
Answer
B · When started within 48 hours of the first symptoms
8.Pharmacology
Answer: B · More phenytoin is unbound when albumin is low, so free levels rise
Why this is the answer
Phenytoin is highly protein bound. When albumin is low, the free fraction rises, so toxicity can occur with a "normal" total level; a free phenytoin level or corrected level is more useful.
Answer
B · More phenytoin is unbound when albumin is low, so free levels rise
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.5 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
|75 − 40| = 35 mmHg → 3.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.5 = 27.5 mmol/L
Answer
HCO₃⁻ = 27.5 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
1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg
Compare
measured: 43 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: atrial fibrillation
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: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.44 to 1.00 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: ventricular rate = 33 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 = 20 small boxes = 0.80 s → 1500 ÷ 20 = 75 ≈ 75 per minute
Ventricular rate
R–R = 46 small boxes = 1.84 s → 1500 ÷ 46 = 32.61 ≈ 33 per minute
Answer
ventricular rate = 33 bpm
Answer
ventricular rate = 33 bpm
Third-degree block with a wide-complex ventricular escape (usually 20–40 per minute).
13.Evidence-based practice statistics
Answer: RR = 0.41
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 = 24/850 = 0.02824
CER = 64/920 = 0.06957
Relative risk
RR = 0.02824/0.06957 = 0.4059 ≈ 0.41
Answer
RR = 0.41
14.Evidence-based practice statistics
Answer: NNT = 18
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 = 0.14·0.4 = 0.056
Number needed to treat
NNT = 1/0.056 = 17.86 → 18
Answer
NNT = 18
The same relative effect gives a smaller NNT when the baseline risk is higher.