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 vessels carry oxygenated blood from the lungs to the heart?
APulmonary arteries
BVenae cavae
CPulmonary veins
DBronchial arteries
2.Human anatomyHard
A stab wound cuts the left half of the spinal cord at T10. Which pattern of loss is expected below the lesion?
ARight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
BLoss of all sensation and movement on the left only
CLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
DLoss of pain and temperature on both sides only
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
AEjection fraction
BPeripheral resistance
CStroke volume
DBlood pressure
4.Human physiologyHard
After a head injury, a patient passes large volumes of very dilute urine and plasma sodium rises. What is the most likely cause?
AUncontrolled diabetes mellitus
BExcess aldosterone
CExcess ADH release (SIADH)
DLack of ADH release (cranial diabetes insipidus)
5.MicrobiologyMedium
Which type of immunity is given by breast milk antibodies?
ANatural active immunity
BArtificial passive immunity
CArtificial active immunity
DNatural passive immunity
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe bacteria made a heat-stable toxin in the food before it was eaten
BThe bacteria multiply very fast inside the gut and invade the intestinal wall
CThe immune system reacts to the bacteria within minutes
DThe bacteria invade the bloodstream immediately
7.PharmacologyMedium
Apixaban and rivaroxaban are anticoagulants that directly inhibit:
AThe platelet P2Y12 receptor
BThrombin
CFactor Xa
DVitamin K epoxide reductase
8.PharmacologyHard
An older woman started on an SSRI two weeks ago is now confused, and her plasma sodium is low. What is the most likely mechanism?
ASSRI-induced inappropriate ADH secretion causing water retention
BDirect damage to the adrenal glands
CLoss of sodium through severe diarrhoea caused by the SSRI
DIncreased aldosterone secretion
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 23 mmHg (3.1 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
Practice values: Na⁺ 144, Cl⁻ 113 and HCO₃⁻ 11 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 144 mmol/LCl⁻ 113 mmol/LHCO₃⁻ 11 mmol/L
delta ratio =
11.ECG rhythm strip interpretationMedium
This lead II strip shows atrial flutter. Measure the flutter-wave interval and find the atrial rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
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
1supraventricular tachycardia (SVT)
2ventricular fibrillation
3idioventricular rhythm
4torsades de pointes
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 570 patients received a new treatment and 440 received the control treatment. The table shows how many in each group had a side effect. Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 51/570control: 22/440
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 29% in relative terms (RRR = 29%). In a patient group the risk without treatment (CER) is 24% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 29%CER = 24%
NNT =
3Solutions
1.Human anatomy
Answer: C · Pulmonary veins
Why this is the answer
Four pulmonary veins drain oxygenated blood into the left atrium. They are the main exception to the rule that veins carry deoxygenated blood.
Answer
C · Pulmonary veins
2.Human anatomy
Answer: C · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
Why this is the answer
In Brown-Séquard syndrome the corticospinal tract and dorsal columns are lost on the same side, while the spinothalamic tract, which crosses near entry, causes contralateral pain and temperature loss.
Answer
C · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: C · Stroke volume
Why this is the answer
CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.
Answer
C · Stroke volume
4.Human physiology
Answer: D · Lack of ADH release (cranial diabetes insipidus)
Why this is the answer
Without ADH the collecting ducts stay impermeable to water, so dilute urine is produced and the patient loses free water, raising plasma sodium.
Answer
D · Lack of ADH release (cranial diabetes insipidus)
5.Microbiology
Answer: D · Natural passive immunity
Why this is the answer
Antibodies (mainly IgA) passed in breast milk give passive immunity acquired naturally. It is short-lived because no memory cells form.
Answer
D · Natural passive immunity
6.Microbiology
Answer: A · 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
A · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: C · Factor Xa
Why this is the answer
Apixaban and rivaroxaban are direct factor Xa inhibitors. Dabigatran directly inhibits thrombin, warfarin blocks vitamin K recycling and clopidogrel blocks P2Y12.
Answer
C · Factor Xa
8.Pharmacology
Answer: A · SSRI-induced inappropriate ADH secretion causing water retention
Why this is the answer
SSRIs can cause the syndrome of inappropriate antidiuretic hormone secretion (SIADH), especially in older people. Water retention dilutes plasma sodium, causing hyponatraemia and confusion.
Answer
A · SSRI-induced inappropriate ADH secretion causing water retention
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20.6 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
|23 − 40| = 17 mmHg → 1.7 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.7 = 20.6 mmol/L
Answer
HCO₃⁻ = 20.6 mmol/L
10.Arterial blood gas interpretation
Answer: delta ratio = 0.6154
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.
Anion gap
AG = 144 − (113 + 11) = 20
Delta ratio
(20 − 12)/(24 − 11) = 8/13 = 0.6154
Answer
delta ratio = 0.6154
Below 1: an added normal anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: atrial rate = 300 bpm
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.
Flutter waves
F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute
QRS complexes
R–R = 20 small boxes = 0.80 s → 1500 ÷ 20 = 75 per minute
conduction ratio = 300 ÷ 75 = 4:1
Answer
atrial rate = 300 bpm
12.ECG rhythm strip interpretation
Answer: torsades de pointes
More ventricular rhythms: torsades de pointes is a polymorphic ventricular tachycardia, usually 150–250/min, in which the wide complexes change in height and twist around the baseline, pointing up for a few beats and then down. Textbooks link it to a prolonged QT interval, and magnesium sulfate is the drug resuscitation guidelines name for it. An idioventricular rhythm is a ventricular escape rhythm: regular, wide QRS (0.12 s or more), no P waves, at 20–40/min (an accelerated idioventricular rhythm runs at 40–100/min). Monomorphic VT has identical wide complexes at more than 100/min. VF is chaotic, without any repeating complex.
Five steps
1. Rate: very fast, the complexes are about 6 small boxes apart (about 250 per minute).
2. Rhythm: fast and roughly regular, but the height and direction of the complexes change in a repeating spindle pattern.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: wide and polymorphic; the complexes twist around the baseline, pointing up for a few beats and then down.
Interpretation
torsades de pointes
Answer
torsades de pointes
Torsades de pointes is a polymorphic VT linked to a prolonged QT interval. This is identification practice only.
13.Evidence-based practice statistics
Answer: RR = 1.79
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 = 51/570 = 0.08947
CER = 22/440 = 0.05
Relative risk
RR = 0.08947/0.05 = 1.789 ≈ 1.79
Answer
RR = 1.79
14.Evidence-based practice statistics
Answer: NNT = 15
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.
Absolute risk reduction
ARR = 0.24·0.29 = 0.0696
Number needed to treat
NNT = 1/0.0696 = 14.37 → 15
Answer
NNT = 15
The same relative effect gives a smaller NNT when the baseline risk is higher.