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 carpal bone is most commonly fractured, typically after a fall on an outstretched hand?
ALunate
BScaphoid
CPisiform
DHamate
2.Human anatomyHard
A patient with a large aneurysm of the aortic arch develops hoarseness. Which nerve is most likely compressed?
ALeft recurrent laryngeal nerve
BRight recurrent laryngeal nerve
CExternal laryngeal nerve
DLeft phrenic nerve
3.Human physiologyMedium
In skeletal muscle, where does most of the calcium for contraction come from?
AThe sarcoplasmic reticulum
BThe mitochondria
CThe nucleus
DThe extracellular fluid
4.Human physiologyHard
Which mechanism mainly creates the concentration gradient in the medulla of the kidney?
AADH acting on the descending limb
BThe countercurrent multiplier of the loop of Henle
CSecretion of urea into the proximal tubule
DGlomerular filtration
5.MicrobiologyMedium
What is the main difference between innate and adaptive immunity?
AInnate immunity uses antibodies and forms memory; adaptive immunity does not
BAdaptive immunity acts within minutes of infection
CInnate immunity develops only after vaccination
DAdaptive immunity is specific and forms memory; innate immunity does not
6.MicrobiologyHard
Why is Chlamydia trachomatis usually diagnosed with nucleic acid amplification tests instead of routine culture on agar?
AIt dies immediately in the presence of oxygen
BIt is an obligate intracellular bacterium that cannot grow on ordinary media
CIt is a virus and has no DNA
DIt is too large to grow on agar plates
7.PharmacologyMedium
Why should doxycycline not be taken at the same time as milk or antacids?
AAntacids turn it into a toxic compound
BCalcium and other metal ions bind it and reduce absorption
CThe combination causes severe constipation
DMilk speeds up its breakdown in the liver by inducing enzymes
8.PharmacologyHard
A patient with Parkinson's disease has nausea. Why should metoclopramide be avoided?
AIt increases dopamine levels in the brain too much and causes psychosis
BIt blocks dopamine receptors in the brain and can worsen the parkinsonism
CIt inhibits the absorption of all oral drugs
DIt causes severe diarrhoea in Parkinson's disease
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 149 mmol/L, glucose 17.2 mmol/L and urea 19.3 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 149 mmol/Lglucose 17.2 mmol/Lurea 19.3 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 88 mmHg (11.7 kPa) and PaCO₂ 54 mmHg (7.2 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
Interpret this lead II strip. Which interpretation can you defend from this single 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
12:1 AV block (type undetermined)
2first-degree AV block
3sinus bradycardia
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 100 patients with a disease (cases) and 570 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 = 68b = 234c = 32d = 336
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 89% and specificity 81%. The prevalence of the disease is 23%. How many positive results n₊ are expected in total?
N = 10 000p = 23%Se = 89%Sp = 81%
n₊ =
3Solutions
1.Human anatomy
Answer: B · Scaphoid
Why this is the answer
The scaphoid is the most often fractured carpal bone. Tenderness in the anatomical snuffbox is a key clinical sign.
Answer
B · Scaphoid
2.Human anatomy
Answer: A · Left recurrent laryngeal nerve
Why this is the answer
The left recurrent laryngeal nerve loops under the aortic arch beside the ligamentum arteriosum. It supplies most laryngeal muscles, so compression causes hoarseness.
Answer
A · Left recurrent laryngeal nerve
3.Human physiology
Answer: A · The sarcoplasmic reticulum
Why this is the answer
Depolarisation of the T-tubules opens ryanodine receptors in the sarcoplasmic reticulum, releasing stored calcium into the cytoplasm.
Answer
A · The sarcoplasmic reticulum
4.Human physiology
Answer: B · The countercurrent multiplier of the loop of Henle
Why this is the answer
The ascending limb actively transports NaCl into the interstitium without letting water follow, and the countercurrent flow multiplies this into a steep gradient that ADH then uses to concentrate urine.
Answer
B · The countercurrent multiplier of the loop of Henle
5.Microbiology
Answer: D · 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
D · Adaptive immunity is specific and forms memory; innate immunity does not
6.Microbiology
Answer: B · 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
B · It is an obligate intracellular bacterium that cannot grow on ordinary media
7.Pharmacology
Answer: B · Calcium and other metal ions bind it and reduce absorption
Why this is the answer
Tetracyclines form complexes with calcium, magnesium, aluminium and iron in the gut, which reduces absorption. Doses are separated from these products.
Answer
B · Calcium and other metal ions bind it and reduce absorption
8.Pharmacology
Answer: B · It blocks dopamine receptors in the brain and can worsen the parkinsonism
Why this is the answer
Metoclopramide crosses the blood–brain barrier and antagonises D2 receptors. In Parkinson's disease this worsens rigidity and bradykinesia, so antiemetics that do not block central dopamine receptors are preferred.
Answer
B · It blocks dopamine receptors in the brain and can worsen the parkinsonism
9.Arterial blood gas interpretation
Answer: Osm = 335 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 = 2·149 + 17.2 + 19.3 = 334.5 ≈ 335 mOsm/kg
Answer
Osm = 335 mOsm/kg
10.Arterial blood gas interpretation
Answer: A–a gradient = 15.62 mmHg
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.
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 13 × 0.04 = 0.52 s
RR = 25 × 0.04 = 1.00 s
Fridericia’s formula
QTc = 0.52/∛1.00 = 0.52/1 = 0.52 s
Answer
QTc = 0.52 s
Above 0.50 s: markedly prolonged. (520 ms)
12.ECG rhythm strip interpretation
Answer: 2:1 AV block (type undetermined)
2:1 AV block: with two P waves for every QRS, no two conducted P waves follow each other, so you cannot see whether the PR would lengthen (type I) or stay constant (type II) before a non-conducted P wave. From one strip it is called a 2:1 AV block; a longer strip or a change in the conduction ratio decides the type. A narrow QRS makes type I more likely and a wide QRS type II, but neither proves it.
Five steps
1. Rate: R–R = 36 small boxes, 1500 ÷ 36 ≈ 42 per minute.
2. Rhythm: regular, every R–R is 1.44 s.
3. P waves: upright and regular (P–P 0.72 s, 83 per minute), two for every QRS.
4. PR interval: constant at 0.18 s in the conducted beats.
5. QRS: 0.14 s, wide (0.12 s or more).
Why the type stays open
Every second P wave is not conducted, so no two conducted beats follow each other and a lengthening PR could never show.
2:1 AV block (type undetermined)
Answer
2:1 AV block (type undetermined)
A narrow QRS makes type I more likely and a wide QRS type II, but only a longer strip or a change in the conduction ratio decides.
13.Evidence-based practice statistics
Answer: OR = 3.05
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: 68/32 = 2.125
controls: 234/336 = 0.6964
Odds ratio
OR = (68·336)/(234·32) = 3.051 ≈ 3.05
Answer
OR = 3.05
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: n₊ = 3510
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.
With and without the disease
10 000·0.23 = 2300
10 000 − 2300 = 7700
All positive results
TP = 2300·0.89 = 2047
FP = 7700·0.19 = 1463
n₊ = 2047 + 1463 = 3510
Answer
n₊ = 3510
Of the 3510 positives only 2047 have the disease (PPV 58.3%).