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?
ACerebellum
BMidbrain
CPons
DMedulla oblongata
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?
ALoss of pain and temperature on both sides only
BLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
CLoss of all sensation and movement on the left only
DRight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
3.Human physiologyMedium
How is the ejection fraction calculated?
AEnd-systolic volume ÷ stroke volume
BCardiac output ÷ heart rate
CEnd-diastolic volume ÷ end-systolic volume
DStroke volume ÷ end-diastolic volume
4.Human physiologyHard
Which spirometry finding is typical of chronic obstructive pulmonary disease (COPD)?
ANormal FEV1 with low FVC
BIncreased peak expiratory flow
CFEV1/FVC ratio above 0.9 with low FVC
DFEV1/FVC ratio below 0.7
5.MicrobiologyMedium
How is DNA transferred between bacteria during conjugation?
ABy uptake of free DNA from the environment
BBy a bacteriophage carrying bacterial DNA
CBy fusion of two bacterial cells into one
DThrough direct contact using a sex pilus
6.MicrobiologyHard
Why is Chlamydia trachomatis usually diagnosed with nucleic acid amplification tests instead of routine culture on agar?
AIt is an obligate intracellular bacterium that cannot grow on ordinary media
BIt is too large to grow on agar plates
CIt is a virus and has no DNA
DIt dies immediately in the presence of oxygen
7.PharmacologyMedium
Which rule applies when giving phenytoin intravenously?
AGive it as a rapid push to stop seizures faster
BMix or flush it only with normal saline and give it slowly
CGive it in the same line as other drugs without flushing
DMix it with glucose 5% to prevent low blood sugar
8.PharmacologyHard
A pregnant woman develops a deep vein thrombosis. Which anticoagulant is usually preferred?
AAspirin alone
BWarfarin
CLow-molecular-weight heparin, such as enoxaparin
DRivaroxaban, a direct oral factor Xa inhibitor
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 29 mmHg (3.9 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⁻.
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
1normal sinus rhythm
2atrial fibrillation
3atrial flutter, 2:1 conduction
4sinus tachycardia
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
2paced rhythm, failure to capture
3idioventricular rhythm
4dual-chamber (AV) paced rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 930 patients received a new treatment and 300 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the absolute risk reduction ARR as a percentage. Round to one decimal place.
treatment: 208/930control: 96/300
ARR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 95% and specificity 98%. The prevalence of the disease is 25%. How many true-positive results TP are expected?
N = 10 000p = 25%Se = 95%Sp = 98%
TP =
3Solutions
1.Human anatomy
Answer: D · Medulla oblongata
Why this is the answer
The medulla oblongata contains the vital centres that regulate heart rate, blood pressure and breathing rhythm.
Answer
D · Medulla oblongata
2.Human anatomy
Answer: B · 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
B · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: D · Stroke volume ÷ end-diastolic volume
Why this is the answer
Ejection fraction is the proportion of blood in the full ventricle that is ejected per beat. A normal value is about 55–70%.
Answer
D · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: D · FEV1/FVC ratio below 0.7
Why this is the answer
Narrowed airways slow expiration, so FEV1 falls more than FVC and the ratio drops below 0.7. In COPD this is not fully reversed by bronchodilators.
Answer
D · FEV1/FVC ratio below 0.7
5.Microbiology
Answer: D · Through direct contact using a sex pilus
Why this is the answer
In conjugation a donor cell forms a pilus to a recipient and transfers DNA, often a plasmid. It is an important way for antibiotic resistance genes to spread.
Answer
D · Through direct contact using a sex pilus
6.Microbiology
Answer: A · It is an obligate intracellular bacterium that cannot grow on ordinary media
Why this is the answer
Chlamydia replicates only inside host cells, so it cannot be grown on agar. NAATs such as PCR on urine or swabs are sensitive and are the standard diagnostic test.
Answer
A · It is an obligate intracellular bacterium that cannot grow on ordinary media
7.Pharmacology
Answer: B · Mix or flush it only with normal saline and give it slowly
Why this is the answer
Phenytoin precipitates in glucose solutions, so only normal saline is used. Rapid infusion can cause hypotension and arrhythmias, so the rate is limited and the patient is monitored.
Answer
B · Mix or flush it only with normal saline and give it slowly
8.Pharmacology
Answer: C · Low-molecular-weight heparin, such as enoxaparin
Why this is the answer
Heparins are large molecules that do not cross the placenta. Warfarin is teratogenic, and direct oral anticoagulants are not recommended in pregnancy.
Answer
C · Low-molecular-weight heparin, such as enoxaparin
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 21.8 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
|29 − 40| = 11 mmHg → 1.1 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.1 = 21.8 mmol/L
Answer
HCO₃⁻ = 21.8 mmol/L
10.Arterial blood gas interpretation
Answer: AGc = 18.25 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 = 143 − (112 + 15) = 16
Correction for albumin
16 + 0.25·(40 − 31) = 18.25 mmol/L
Answer
AGc = 18.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: sinus tachycardia
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 = 14 small boxes, 1500 ÷ 14 ≈ 107 per minute.
2. Rhythm: regular, every R–R is 0.56 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus tachycardia
Answer
sinus tachycardia
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: 6 QRS complexes in 6 s, × 10 = 60 per minute.
2. Rhythm: the pacing spikes are regular (every 0.88 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: 0.16 s and wide after the captured spikes; one spike is 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: ARR = 9.6 %
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 = 208/930 = 0.2237
CER = 96/300 = 0.32
Absolute risk reduction
ARR = 0.32 − 0.2237 = 0.09634 ≈ 9.6%
Answer
ARR = 9.6%
14.Evidence-based practice statistics
Answer: TP = 2375
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.