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 type of epithelium lines most of the trachea?
ASimple cuboidal epithelium
BStratified squamous epithelium
CTransitional epithelium
DPseudostratified ciliated columnar epithelium
2.Human anatomyHard
Why is the proximal pole of the scaphoid at risk of avascular necrosis after a waist fracture?
AIt has no blood supply at all
BIts blood supply enters mainly from the distal end
CIts only artery runs through the lunate
DIt is drained by a single vein that clots easily
3.Human physiologyMedium
Which hormonal event triggers ovulation?
AA surge of prolactin
BA rise in progesterone from the corpus luteum
CA fall in oestrogen
DA surge of luteinising hormone (LH)
4.Human physiologyHard
Which combination of signs is expected in a patient in early hypovolaemic shock after major bleeding?
ABradycardia, cool pale skin and high urine output
BBradycardia, warm flushed skin and high urine output
CTachycardia, warm flushed skin and high urine output
DTachycardia, cool pale skin and reduced urine output
5.MicrobiologyMedium
How is DNA transferred between bacteria during conjugation?
AThrough direct contact using a sex pilus
BBy a bacteriophage carrying bacterial DNA
CBy fusion of two bacterial cells into one
DBy uptake of free DNA from the environment
6.MicrobiologyHard
In the 1840s Ignaz Semmelweis lowered deaths from puerperal fever in a Vienna maternity clinic. What did he introduce?
AHandwashing with chlorinated lime before examining women
BTreating mothers with penicillin
CKeeping mothers in separate isolation rooms after delivery
DVaccination of all mothers before birth
7.PharmacologyMedium
How does clopidogrel reduce platelet aggregation?
AIt activates antithrombin
BIt inhibits platelet COX-1
CIt blocks the glycoprotein IIb/IIIa receptor directly
DIt blocks the platelet P2Y12 receptor for ADP
8.PharmacologyHard
A woman with epilepsy who takes an antiepileptic drug plans to become pregnant. Which advice is standard?
AStop all antiepileptic drugs straight away as soon as she stops contraception
BNo change is needed, as antiepileptics carry no risk in pregnancy
CAvoid folic acid because it lowers drug levels
DTake folic acid before conception and seek specialist review of her medication
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.56, PaCO₂ 38 mmHg (5.1 kPa) and HCO₃⁻ 33 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.56PaCO₂ 38 mmHg (5.1 kPa)HCO₃⁻ 33 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows pH 7.22, PaCO₂ 23 mmHg (3.1 kPa) and HCO₃⁻ 9 mmol/L, with Na⁺ 133 and Cl⁻ 101 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
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
2junctional rhythm
3atrial flutter, 4:1 conduction
4atrial fibrillation
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 tachycardia
2ventricular fibrillation
3sinus tachycardia
4atrial flutter, 2:1 conduction
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 5500 person-years with 138 new cases, and unexposed people 4500 person-years with 45 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 138/5500 PYnot exposed: 45/4500 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 41% in relative terms (RRR = 41%). In a patient group the risk without treatment (CER) is 7% 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 = 41%CER = 7%
NNT =
3Solutions
1.Human anatomy
Answer: D · Pseudostratified ciliated columnar epithelium
Why this is the answer
Respiratory epithelium is pseudostratified ciliated columnar with goblet cells. Mucus traps particles and cilia move it up towards the pharynx.
Answer
D · Pseudostratified ciliated columnar epithelium
2.Human anatomy
Answer: B · Its blood supply enters mainly from the distal end
Why this is the answer
Branches of the radial artery enter the scaphoid mainly distally and flow backwards. A fracture through the waist can cut off the proximal pole, causing avascular necrosis.
Answer
B · Its blood supply enters mainly from the distal end
3.Human physiology
Answer: D · A surge of luteinising hormone (LH)
Why this is the answer
Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.
Answer
D · A surge of luteinising hormone (LH)
4.Human physiology
Answer: D · Tachycardia, cool pale skin and reduced urine output
Why this is the answer
Baroreceptor-driven sympathetic activation raises heart rate and constricts skin vessels. Reduced renal perfusion, ADH and aldosterone lower urine output.
Answer
D · Tachycardia, cool pale skin and reduced urine output
5.Microbiology
Answer: A · 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
A · Through direct contact using a sex pilus
6.Microbiology
Answer: A · Handwashing with chlorinated lime before examining women
Why this is the answer
Semmelweis noticed higher mortality in the ward run by doctors who also did autopsies. Hand disinfection with chlorinated lime sharply reduced deaths, long before germ theory was accepted.
Answer
A · Handwashing with chlorinated lime before examining women
7.Pharmacology
Answer: D · It blocks the platelet P2Y12 receptor for ADP
Why this is the answer
Clopidogrel is activated in the liver and irreversibly blocks the P2Y12 ADP receptor on platelets. It is often combined with aspirin after a heart attack or stent.
Answer
D · It blocks the platelet P2Y12 receptor for ADP
8.Pharmacology
Answer: D · Take folic acid before conception and seek specialist review of her medication
Why this is the answer
Folic acid before and in early pregnancy reduces the risk of neural tube defects. Medication should be reviewed before conception, but drugs should not be stopped suddenly because uncontrolled seizures are also dangerous.
Answer
D · Take folic acid before conception and seek specialist review of her medication
9.Arterial blood gas interpretation
Answer: uncompensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.56: alkalaemia (above 7.45)
PaCO₂ 38 mmHg: normal (35–45)
HCO₃⁻ 33 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 42.1–46.1 mmHg, so the measured 38 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
10.Arterial blood gas interpretation
Answer: added normal-gap 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.
Anion gap
AG = 133 − (101 + 9) = 23
Delta ratio
(23 − 12)/(24 − 9) = 0.7333
added normal-gap acidosis
Answer
added normal-gap 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: 9 QRS complexes in 6 s, × 10 = 90 per minute.
2. Rhythm: irregularly irregular, R–R from 0.42 to 0.78 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
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: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: IRR = 2.51
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.
Incidence rates
exposed: 138/5500 = 25.09 per 1000 person-years
not exposed: 45/4500 = 10 per 1000 person-years
Rate ratio
IRR = 25.09/10 = 2.509 ≈ 2.51
Answer
IRR = 2.51
14.Evidence-based practice statistics
Answer: NNT = 35
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 = 0.07·0.41 = 0.0287
Number needed to treat
NNT = 1/0.0287 = 34.84 → 35
Answer
NNT = 35
The same relative effect gives a smaller NNT when the baseline risk is higher.