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
The rotator cuff consists of supraspinatus, infraspinatus, teres minor and which other muscle?
ASubscapularis
BDeltoid
CLatissimus dorsi
DTeres major
2.Human anatomyHard
Which structure passes through the inguinal canal in males?
AObturator nerve
BGreat saphenous vein
CSpermatic cord
DFemoral artery
3.Human physiologyMedium
What is the main stimulus for breathing in a healthy person?
AA rise in arterial CO₂
BA rise in blood glucose
CA fall in blood pH caused by lactate only
DA fall in arterial O₂
4.Human physiologyHard
Which substance's clearance is used as the gold standard to measure glomerular filtration rate?
AGlucose
BCreatinine
CUrea
DInulin
5.MicrobiologyMedium
In the chain of infection, what is the "reservoir"?
AThe route by which the pathogen leaves the infected host's body
BThe way the pathogen enters a new host
CThe place where the pathogen normally lives and multiplies
DThe person who is most at risk of infection
6.MicrobiologyHard
Which feature distinguishes most exotoxins from endotoxin?
AExotoxins are only released when bacteria die
BExotoxins are lipopolysaccharides of the Gram-negative outer membrane
CExotoxins are proteins and can be turned into toxoid vaccines
DExotoxins are heat-stable lipids
7.PharmacologyMedium
How many micrograms are in 0.5 mg?
A50 micrograms
B5 micrograms
C500 micrograms
D5000 micrograms
8.PharmacologyHard
In aspirin overdose the urine may be made alkaline with sodium bicarbonate. Why does this increase salicylate excretion?
AAlkaline urine increases glomerular filtration
BMore salicylate is ionised in the tubule, so less is reabsorbed
CBicarbonate binds salicylate and makes it inactive
DBicarbonate speeds up the liver metabolism of salicylate to inactive forms
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.51, PaCO₂ 31 mmHg (4.1 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.51PaCO₂ 31 mmHg (4.1 kPa)HCO₃⁻ 24 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 55 mmHg (7.3 kPa) and PaCO₂ 44 mmHg (5.9 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 Bazett’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 with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1rate
2rhythm
3P waves
4PR interval
5QRS duration
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 2100 person-years with 53 new cases, and unexposed people 2400 person-years with 24 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 53/2100 PYnot exposed: 24/2400 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 20% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.99. 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 20%RR = 1.99
NNH =
3Solutions
1.Human anatomy
Answer: A · Subscapularis
Why this is the answer
SITS: supraspinatus, infraspinatus, teres minor and subscapularis. Teres major is close by but is not part of the cuff.
Answer
A · Subscapularis
2.Human anatomy
Answer: C · Spermatic cord
Why this is the answer
The inguinal canal carries the spermatic cord in males and the round ligament in females. The femoral vessels pass below the inguinal ligament instead.
Answer
C · Spermatic cord
3.Human physiology
Answer: A · A rise in arterial CO₂
Why this is the answer
Central chemoreceptors respond to CO₂ (through the pH of the cerebrospinal fluid). Low O₂ adds strongly to the drive only when PaO₂ falls well below normal, for example at high altitude.
Answer
A · A rise in arterial CO₂
4.Human physiology
Answer: D · Inulin
Why this is the answer
Inulin is filtered freely and is not reabsorbed or secreted, so its clearance equals GFR exactly. Creatinine is slightly secreted, so it overestimates GFR a little.
Answer
D · Inulin
5.Microbiology
Answer: C · The place where the pathogen normally lives and multiplies
Why this is the answer
The chain of infection links the agent, reservoir, portal of exit, mode of transmission, portal of entry and susceptible host. Breaking any link stops transmission.
Answer
C · The place where the pathogen normally lives and multiplies
6.Microbiology
Answer: C · Exotoxins are proteins and can be turned into toxoid vaccines
Why this is the answer
Exotoxins are secreted proteins with specific actions and are often very potent; many can be inactivated into toxoids. Endotoxin is lipopolysaccharide, more heat-stable and cannot be made into a toxoid.
Answer
C · Exotoxins are proteins and can be turned into toxoid vaccines
7.Pharmacology
Answer: C · 500 micrograms
Why this is the answer
0.5 mg × 1000 = 500 micrograms. Writing "micrograms" in full (or "mcg" in US practice), not "µg", helps avoid confusion with "mg".
Answer
C · 500 micrograms
8.Pharmacology
Answer: B · More salicylate is ionised in the tubule, so less is reabsorbed
Why this is the answer
Salicylate is a weak acid. In alkaline urine more of it is in the ionised form, which is "trapped" in the tubule and excreted. This is called ion trapping.
Answer
B · More salicylate is ionised in the tubule, so less is reabsorbed
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 7.51: alkalaemia (above 7.45)
PaCO₂ 31 mmHg: low (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)
Compensation
uncompensated
Answer
uncompensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 103.9 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.
Alveolar oxygen
PAO₂ = 0.3·713 − 44/0.8 = 213.9 − 55 = 158.9 mmHg
Gradient
A–a = 158.9 − 55 = 103.9 mmHg
Answer
A–a gradient = 103.9 mmHg
11.ECG rhythm strip interpretation
Answer: QTc = 0.451 s
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 = 11.5 × 0.04 = 0.46 s
RR = 26 × 0.04 = 1.04 s
Bazett’s formula
QTc = 0.46/√1.04 = 0.46/1.02 = 0.451 s
Answer
QTc = 0.451 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (451 ms)
12.ECG rhythm strip interpretation
Answer: QRS duration
Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.
Five steps
1. Rate: R–R = 23 small boxes, 1500 ÷ 23 ≈ 65 per minute.
2. Rhythm: regular, every R–R is 0.92 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: 0.12 s, wide (0.12 s or more).
First abnormal step
QRS duration → sinus rhythm with wide QRS
Answer
QRS duration
13.Evidence-based practice statistics
Answer: IRR = 2.52
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: 53/2100 = 25.24 per 1000 person-years
not exposed: 24/2400 = 10 per 1000 person-years
Rate ratio
IRR = 25.24/10 = 2.524 ≈ 2.52
Answer
IRR = 2.52
14.Evidence-based practice statistics
Answer: NNH = 6
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.