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 artery is most commonly used to feel the pulse at the wrist?
AAxillary artery
BUlnar artery
CRadial artery
DBrachial artery
2.Human anatomyHard
A blow to the side of the head at the pterion causes an extradural haematoma. Which vessel is usually torn?
AMiddle cerebral artery
BSuperficial temporal artery
CMiddle meningeal artery
DBridging cerebral veins
3.Human physiologyMedium
Besides lowering glucose, insulin moves which ion into cells, an effect used to treat hyperkalaemia?
AChloride
BCalcium
CPotassium
DSodium
4.Human physiologyHard
In Graves' disease, what thyroid function results are expected and why?
ALow T4 and low TSH, because the hypothalamus fails
BHigh T4 and low TSH, because antibodies stimulate the TSH receptor
CHigh T4 and high TSH, because the anterior pituitary gland is overactive
DLow T4 and high TSH, because antibodies destroy the gland
5.MicrobiologyMedium
What is the main virulence factor of Streptococcus pneumoniae, and the target of its vaccines?
AIts cholera-like toxin
BIts flagella
CIts polysaccharide capsule
DIts endospores
6.MicrobiologyHard
In Griffith's 1928 experiment, mice died after injection of heat-killed smooth (S) pneumococci mixed with live rough (R) pneumococci. What did this show?
AViruses carry genes between bacteria
BA "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
CRough pneumococci are always more virulent than smooth ones
DHeat does not kill pneumococci
7.PharmacologyMedium
How does metformin mainly lower blood glucose?
AIt replaces missing insulin
BIt stimulates insulin release from the pancreas
CIt blocks glucose absorption in the kidney
DIt reduces glucose production by the liver
8.PharmacologyHard
Zero-order elimination, as seen with alcohol at normal intake, means that:
AElimination stops at high concentrations
BA constant amount is eliminated per unit of time
CThe half-life stays constant whatever the level
DA constant fraction is eliminated per unit of time
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 138 mmol/L, glucose 7.6 mmol/L and urea 2.9 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 138 mmol/Lglucose 7.6 mmol/Lurea 2.9 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.23, PaCO₂ 22 mmHg (2.9 kPa), HCO₃⁻ 9 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.23PaCO₂ 22 mmHg (2.9 kPa)HCO₃⁻ 9 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
1atrial fibrillation
2first-degree AV block
3junctional rhythm
4sinus bradycardia
12.ECG rhythm strip interpretationHard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? 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
1pulseless electrical activity (PEA)
2ventricular fibrillation
3asystole
4normal sinus rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 950 patients received a new treatment and 930 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: 157/950control: 205/930
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 23% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 2.24. 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 23%RR = 2.24
NNH =
3Solutions
1.Human anatomy
Answer: C · Radial artery
Why this is the answer
The radial artery lies superficially on the thumb side of the wrist, against the radius, which makes the pulse easy to feel.
Answer
C · Radial artery
2.Human anatomy
Answer: C · Middle meningeal artery
Why this is the answer
The middle meningeal artery lies under the pterion, where the skull is thin. Its rupture bleeds between bone and dura. Bridging veins cause subdural haematomas.
Answer
C · Middle meningeal artery
3.Human physiology
Answer: C · Potassium
Why this is the answer
Insulin stimulates the Na⁺/K⁺-ATPase and shifts potassium into cells. Insulin with glucose is therefore used to lower dangerously high plasma potassium.
Answer
C · Potassium
4.Human physiology
Answer: B · High T4 and low TSH, because antibodies stimulate the TSH receptor
Why this is the answer
TSH-receptor antibodies stimulate the thyroid independently of TSH. The resulting high T4 and T3 suppress pituitary TSH by negative feedback.
Answer
B · High T4 and low TSH, because antibodies stimulate the TSH receptor
5.Microbiology
Answer: C · Its polysaccharide capsule
Why this is the answer
The capsule protects S. pneumoniae from phagocytosis. There are many capsular types, and pneumococcal vaccines contain capsular polysaccharides from the most common ones.
Answer
C · Its polysaccharide capsule
6.Microbiology
Answer: B · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
Why this is the answer
Something from the dead S strain turned live R bacteria into capsule-forming, virulent S bacteria. Avery, MacLeod and McCarty later showed that this transforming principle was DNA.
Answer
B · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
7.Pharmacology
Answer: D · It reduces glucose production by the liver
Why this is the answer
Metformin mainly reduces hepatic gluconeogenesis and improves insulin sensitivity. Because it does not stimulate insulin release, hypoglycaemia is rare when used alone.
Answer
D · It reduces glucose production by the liver
8.Pharmacology
Answer: B · A constant amount is eliminated per unit of time
Why this is the answer
When the metabolising enzymes are saturated, a fixed amount is removed per hour regardless of concentration, so the half-life gets longer as the level rises. Phenytoin and high-dose aspirin can behave the same way.
Answer
B · A constant amount is eliminated per unit of time
9.Arterial blood gas interpretation
Answer: Osm = 287 mOsm/kg
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.
Calculated osmolality
Osm = 2·138 + 7.6 + 2.9 = 286.5 ≈ 287 mOsm/kg
Answer
Osm = 287 mOsm/kg
10.Arterial blood gas interpretation
Answer: appropriate compensation
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
1.5·9 + 8 = 21.5 → 19.5–23.5 mmHg
Compare
measured: 22 mmHg → appropriate compensation
Answer
appropriate compensation
11.ECG rhythm strip interpretation
Answer: junctional rhythm
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 = 27 small boxes, 1500 ÷ 27 ≈ 56 per minute.
2. Rhythm: regular, every R–R is 1.08 s.
3. P waves: inverted, just before each QRS.
4. PR interval: 0.10 s, shorter than 0.12 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
junctional rhythm
Answer
junctional rhythm
12.ECG rhythm strip interpretation
Answer: pulseless electrical activity (PEA)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 16 small boxes, 1500 ÷ 16 ≈ 94 per minute.
2. Rhythm: regular, every R–R is 0.64 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.12 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
normal sinus rhythm
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
pulseless electrical activity (PEA)
Answer
pulseless electrical activity (PEA)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: RR = 0.75
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 = 157/950 = 0.1653
CER = 205/930 = 0.2204
Relative risk
RR = 0.1653/0.2204 = 0.7497 ≈ 0.75
Answer
RR = 0.75
14.Evidence-based practice statistics
Answer: NNH = 4
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.