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?
ATeres major
BLatissimus dorsi
CDeltoid
DSubscapularis
2.Human anatomyHard
When inserting a chest drain or needle into the pleural space, why is it placed just above the upper border of a rib?
AThe intercostal muscles are absent above each rib
BThe intercostal vein, artery and nerve run along the lower border of each rib
CThe pleura is thickest just above each rib
DThe lung does not reach the area above each rib
3.Human physiologyMedium
Which ion enters the cell to cause the depolarisation phase of a typical neuronal action potential?
ACalcium (Ca²⁺)
BChloride (Cl⁻)
CSodium (Na⁺)
DPotassium (K⁺)
4.Human physiologyHard
In primary adrenal insufficiency (Addison's disease), which combination is expected?
ALow sodium, high potassium and high ACTH
BLow sodium, low potassium and low ACTH
CHigh sodium, high potassium and high ACTH
DHigh sodium, low potassium and low ACTH
5.MicrobiologyMedium
Which cells does HIV mainly infect and destroy?
ACD4 T helper cells
BNeutrophils
CRed blood cells
DPlatelets
6.MicrobiologyHard
In 1854 John Snow traced a cholera outbreak in London to the Broad Street water pump. Why is this study important?
AIt showed that cholera spreads through "bad air" (miasma)
BIt proved that cholera is caused by a virus
CIt showed that cholera spreads through contaminated water, founding modern epidemiology
DIt led directly to the first cholera vaccine
7.PharmacologyMedium
Which adverse effect is a well-known risk of zidovudine?
ABone marrow suppression, especially anaemia
BRed–green colour blindness
CGum overgrowth
DOrange-red colouring of the urine and tears
8.PharmacologyHard
A drug is ordered at 5 micrograms/kg/min for an 80 kg patient. The infusion contains 1,600 micrograms per mL. What rate in mL/h is needed?
A15 mL/h
B150 mL/h
C25 mL/h
D4 mL/h
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 79 mmHg (10.5 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 87 mmHg (11.6 kPa) and PaCO₂ 34 mmHg (4.5 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
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
1atrial flutter, 2:1 conduction
2ventricular tachycardia
3ventricular fibrillation
4torsades de pointes
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 380 patients with a disease (cases) and 690 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 = 299b = 331c = 81d = 359
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 73% and a specificity of 99%. It is used in a population where the prevalence is 20%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 73%Sp = 99%p = 20%
PPV = %
3Solutions
1.Human anatomy
Answer: D · 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
D · Subscapularis
2.Human anatomy
Answer: B · The intercostal vein, artery and nerve run along the lower border of each rib
Why this is the answer
The intercostal neurovascular bundle (vein, artery, nerve from above down) lies in the costal groove under each rib. Going just above the rib below avoids it.
Answer
B · The intercostal vein, artery and nerve run along the lower border of each rib
3.Human physiology
Answer: C · Sodium (Na⁺)
Why this is the answer
Voltage-gated Na⁺ channels open and Na⁺ rushes in, depolarising the membrane. K⁺ outflow then repolarises it.
Answer
C · Sodium (Na⁺)
4.Human physiology
Answer: A · Low sodium, high potassium and high ACTH
Why this is the answer
Loss of aldosterone causes sodium loss and potassium retention. Low cortisol removes negative feedback, so ACTH rises and can cause skin pigmentation.
Answer
A · Low sodium, high potassium and high ACTH
5.Microbiology
Answer: A · CD4 T helper cells
Why this is the answer
HIV binds CD4 and a co-receptor on T helper cells. Loss of these cells weakens both antibody and cell-mediated immunity, leading to AIDS.
Answer
A · CD4 T helper cells
6.Microbiology
Answer: C · It showed that cholera spreads through contaminated water, founding modern epidemiology
Why this is the answer
Snow mapped cases around the pump and linked them to its water. Removing the pump handle is a classic example of using epidemiology to find and stop a source, before the bacterium was identified.
Answer
C · It showed that cholera spreads through contaminated water, founding modern epidemiology
7.Pharmacology
Answer: A · Bone marrow suppression, especially anaemia
Why this is the answer
Zidovudine, a nucleoside reverse transcriptase inhibitor, can cause anaemia and neutropenia. Blood counts are checked regularly during treatment.
Answer
A · Bone marrow suppression, especially anaemia
8.Pharmacology
Answer: A · 15 mL/h
Why this is the answer
5 × 80 = 400 micrograms/min, × 60 = 24,000 micrograms/h, ÷ 1,600 micrograms/mL = 15 mL/h. Weight-based infusions are double-checked because errors are easy to make.
Answer
A · 15 mL/h
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 37.65 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
|79 − 40| = 39 mmHg → 3.9 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·3.9 = 37.65 mmol/L
Answer
HCO₃⁻ = 37.65 mmol/L
10.Arterial blood gas interpretation
Answer: A–a gradient = 84.4 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 = 10 × 0.04 = 0.40 s
RR = 18 × 0.04 = 0.72 s
Bazett’s formula
QTc = 0.40/√0.72 = 0.40/0.8485 = 0.471 s
Answer
QTc = 0.471 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (471 ms)
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: 0.16 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: OR = 4
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: 299/81 = 3.691
controls: 331/359 = 0.922
Odds ratio
OR = (299·359)/(331·81) = 4.004 ≈ 4
Answer
OR = 4
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 94.8 %
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.
True and false positives (per person)
Se·p = 0.73·0.2 = 0.146
(1 − Sp)·(1 − p) = 0.01·0.8 = 0.008
Positive predictive value
PPV = 0.146/(0.146 + 0.008) = 0.9481 ≈ 94.8%
Answer
PPV = 94.8%
The lower the prevalence, the lower the PPV of the same test.