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?
AUlnar artery
BBrachial artery
CRadial artery
DAxillary artery
2.Human anatomyHard
After a fracture of the midshaft of the humerus, a patient cannot extend the wrist and fingers ("wrist drop"). Which nerve is most likely injured?
AMedian nerve
BUlnar nerve
CRadial nerve
DAxillary nerve
3.Human physiologyMedium
What is the main effect of aldosterone on the kidney?
AIncreased filtration in the glomerulus
BSodium reabsorption and potassium secretion in the distal nephron
CGlucose reabsorption in the proximal tubule
DWater reabsorption by inserting aquaporins only
4.Human physiologyHard
Spirometry shows reduced FEV1 and reduced FVC, with a normal or high FEV1/FVC ratio. What pattern is this?
AA normal result in an older adult
BAn obstructive pattern, as in emphysema or chronic bronchitis
CA restrictive pattern, as in pulmonary fibrosis
DAn obstructive pattern, as in asthma
5.MicrobiologyMedium
Which statement is one of Koch's postulates?
AThe organism must be found in healthy people as well
BThe organism must be visible with the naked eye
CThe organism must cause the same disease when introduced into a healthy susceptible host
DThe disease must respond to antibiotics
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe bacteria invade the bloodstream immediately
BThe bacteria made a heat-stable toxin in the food before it was eaten
CThe bacteria multiply very fast inside the gut and invade the intestinal wall
DThe immune system reacts to the bacteria within minutes
7.PharmacologyMedium
A malnourished patient with alcohol dependence needs IV glucose. Which vitamin should be given before or with it?
AVitamin C
BVitamin D
CThiamine (vitamin B1)
DVitamin K (phytomenadione)
8.PharmacologyHard
A drug with a short half-life is given at a dosing interval much longer than its half-life. What is the result at steady state?
ASteady state is never reached
BVery stable levels with almost no fluctuation
CLarge swings between peak and trough levels
DGradual accumulation to toxic levels
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 29 mmHg (3.9 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.20, PaCO₂ 48 mmHg (6.4 kPa), HCO₃⁻ 18 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.20PaCO₂ 48 mmHg (6.4 kPa)HCO₃⁻ 18 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
2junctional rhythm
3atrial flutter, 4:1 conduction
4sinus rhythm with PVCs
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). A case-control study compares 390 patients with a disease (cases) and 140 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 = 285b = 73c = 105d = 67
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 60% for a disease. A test with sensitivity 86% and specificity 92% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 60%Se = 86%Sp = 92%positive
p′ = %
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 · Radial nerve
Why this is the answer
The radial nerve runs in the radial (spiral) groove of the humeral shaft. It supplies the wrist and finger extensors, so injury causes wrist drop.
Answer
C · Radial nerve
3.Human physiology
Answer: B · Sodium reabsorption and potassium secretion in the distal nephron
Why this is the answer
Aldosterone acts on the distal tubule and collecting duct to increase Na⁺ reabsorption and K⁺ secretion. Water follows the sodium, raising blood volume.
Answer
B · Sodium reabsorption and potassium secretion in the distal nephron
4.Human physiology
Answer: C · A restrictive pattern, as in pulmonary fibrosis
Why this is the answer
In restrictive disease the lungs are small or stiff, so both FEV1 and FVC fall but the ratio is preserved. Obstruction lowers FEV1 more than FVC.
Answer
C · A restrictive pattern, as in pulmonary fibrosis
5.Microbiology
Answer: C · The organism must cause the same disease when introduced into a healthy susceptible host
Why this is the answer
Koch's postulates say the microbe must be found in every case, grown in pure culture, cause the disease in a healthy host and be recovered again from that host. Some pathogens cannot be cultured, so the postulates have limits.
Answer
C · The organism must cause the same disease when introduced into a healthy susceptible host
6.Microbiology
Answer: B · The bacteria made a heat-stable toxin in the food before it was eaten
Why this is the answer
S. aureus produces enterotoxins in food left at warm temperatures. The toxin is already there when the food is eaten and survives reheating, so symptoms start quickly.
Answer
B · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: C · Thiamine (vitamin B1)
Why this is the answer
Giving glucose without thiamine to a thiamine-deficient patient can trigger or worsen Wernicke's encephalopathy. Thiamine is therefore given before or alongside glucose.
Answer
C · Thiamine (vitamin B1)
8.Pharmacology
Answer: C · Large swings between peak and trough levels
Why this is the answer
When the interval is long compared with the half-life, most of each dose is eliminated before the next one. Levels rise high after each dose and fall low before the next, so there may be toxic peaks or ineffective troughs.
Answer
C · Large swings between peak and trough levels
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 18.5 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
|29 − 40| = 11 mmHg → 1.1 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.1 = 18.5 mmol/L
Answer
HCO₃⁻ = 18.5 mmol/L
10.Arterial blood gas interpretation
Answer: added respiratory 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.
Expected PaCO₂
1.5·18 + 8 = 35 → 33–37 mmHg
Compare
measured: 48 mmHg → added respiratory acidosis
Answer
added respiratory 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: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.56 to 1.00 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: P waves
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 = 21 small boxes, 1500 ÷ 21 ≈ 71 per minute.
2. Rhythm: regular, every R–R is 0.84 s.
3. P waves: inverted, just after each QRS.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
First abnormal step
P waves → accelerated junctional rhythm
Answer
P waves
13.Evidence-based practice statistics
Answer: OR = 2.49
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: 285/105 = 2.714
controls: 73/67 = 1.09
Odds ratio
OR = (285·67)/(73·105) = 2.491 ≈ 2.49
Answer
OR = 2.49
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: p′ = 94.2 %
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.