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
Where does fertilisation most commonly take place?
AOvary
BBody of the uterus
CCervix
DAmpulla of the uterine tube
2.Human anatomyHard
An ECG shows an infarct in leads II, III and aVF (inferior wall). Which coronary artery is most commonly blocked?
ALeft main coronary artery
BLeft anterior descending artery
CRight coronary artery
DFirst diagonal branch
3.Human physiologyMedium
How does glucagon raise blood glucose?
AIt slows glucose absorption from the gut
BIt stimulates glucose uptake by muscle
CIt stimulates glycogen breakdown and gluconeogenesis in the liver
DIt blocks insulin receptors on fat cells
4.Human physiologyHard
In Graves' disease, what thyroid function results are expected and why?
ALow T4 and low TSH, because the hypothalamus fails
BLow T4 and high TSH, because antibodies destroy the gland
CHigh T4 and low TSH, because antibodies stimulate the TSH receptor
DHigh T4 and high TSH, because the anterior pituitary gland is overactive
5.MicrobiologyMedium
In the lysogenic cycle of a bacteriophage, what happens to the phage DNA?
AIt is destroyed by the bacterium
BIt is immediately used to make new phages that burst the cell
CIt is integrated into the bacterial chromosome as a prophage
DIt leaves the cell through the pilus
6.MicrobiologyHard
What is the key difference between bactericidal and bacteriostatic antibiotics?
ABactericidal drugs work only outside the body
BBactericidal drugs work only on viruses; bacteriostatic drugs work on bacteria
CBactericidal drugs kill bacteria; bacteriostatic drugs stop growth and rely on the immune system to clear them
DBacteriostatic drugs always work faster than bactericidal drugs
7.PharmacologyMedium
Intravenous vancomycin is commonly used to treat infections caused by:
APseudomonas aeruginosa
BEscherichia coli
CMRSA
DCandida albicans
8.PharmacologyHard
Combining several drugs that prolong the QT interval raises the risk of which life-threatening arrhythmia?
ATorsades de pointes
BFirst-degree heart block
CSinus tachycardia
DAtrial fibrillation
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 140 mmol/L, glucose 27.5 mmol/L and urea 17.3 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 140 mmol/Lglucose 27.5 mmol/Lurea 17.3 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 135, Cl⁻ 105 and HCO₃⁻ 14 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 135 mmol/LCl⁻ 105 mmol/LHCO₃⁻ 14 mmol/L
delta ratio =
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
1sinus bradycardia
2first-degree AV block
3normal sinus rhythm
4junctional rhythm
12.ECG rhythm strip interpretationHard
This lead II strip shows a conduction problem between the atria and the ventricles. 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
1junctional rhythm
2first-degree AV block
3third-degree (complete) AV block
4second-degree AV block type I (Wenckebach)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 440 patients received a new treatment and 410 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.
treatment: 106/440control: 131/410
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 77% and specificity 78%. The prevalence of the disease is 18%. How many people with the disease are missed by the test (false negatives, FN)?
N = 10 000p = 18%Se = 77%Sp = 78%
FN =
3Solutions
1.Human anatomy
Answer: D · Ampulla of the uterine tube
Why this is the answer
Fertilisation usually occurs in the ampulla of the uterine (fallopian) tube. The embryo then travels to the uterus to implant.
Answer
D · Ampulla of the uterine tube
2.Human anatomy
Answer: C · Right coronary artery
Why this is the answer
The inferior wall is usually supplied by the right coronary artery through the posterior descending artery (right-dominant circulation), so inferior infarcts are mostly RCA occlusions.
Answer
C · Right coronary artery
3.Human physiology
Answer: C · It stimulates glycogen breakdown and gluconeogenesis in the liver
Why this is the answer
Glucagon from alpha cells acts on the liver to break down glycogen and make new glucose from non-carbohydrate sources.
Answer
C · It stimulates glycogen breakdown and gluconeogenesis in the liver
4.Human physiology
Answer: C · High T4 and low TSH, because antibodies stimulate the TSH receptor
Why this is the answer
TSH-receptor antibodies stimulate the thyroid independently of TSH. The resulting high T4 and T3 suppress pituitary TSH by negative feedback.
Answer
C · High T4 and low TSH, because antibodies stimulate the TSH receptor
5.Microbiology
Answer: C · It is integrated into the bacterial chromosome as a prophage
Why this is the answer
In lysogeny the phage genome integrates into the host chromosome and is copied with it. Later it may switch to the lytic cycle, making new phages and lysing the cell.
Answer
C · It is integrated into the bacterial chromosome as a prophage
6.Microbiology
Answer: C · Bactericidal drugs kill bacteria; bacteriostatic drugs stop growth and rely on the immune system to clear them
Why this is the answer
Bacteriostatic drugs halt multiplication so host defences can eliminate the bacteria. Bactericidal drugs are preferred when defences are weak or hard to reach, such as in endocarditis or neutropenia.
Answer
C · Bactericidal drugs kill bacteria; bacteriostatic drugs stop growth and rely on the immune system to clear them
7.Pharmacology
Answer: C · MRSA
Why this is the answer
Vancomycin is a glycopeptide that blocks cell wall synthesis in Gram-positive bacteria. It is a standard choice for MRSA. It cannot pass the outer membrane of Gram-negative bacteria.
Answer
C · MRSA
8.Pharmacology
Answer: A · Torsades de pointes
Why this is the answer
A long QT interval delays repolarisation and can trigger torsades de pointes, which may turn into ventricular fibrillation. Drugs such as some macrolides, antipsychotics and antiarrhythmics add to this risk.
Answer
A · Torsades de pointes
9.Arterial blood gas interpretation
Answer: Osm = 325 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 = 2·140 + 27.5 + 17.3 = 324.8 ≈ 325 mOsm/kg
Answer
Osm = 325 mOsm/kg
10.Arterial blood gas interpretation
Answer: delta ratio = 0.4
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 = 135 − (105 + 14) = 16
Delta ratio
(16 − 12)/(24 − 14) = 4/10 = 0.4
Answer
delta ratio = 0.4
Below 1: an added normal anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: sinus bradycardia
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: R–R = 36 small boxes, 1500 ÷ 36 ≈ 42 per minute.
2. Rhythm: regular, every R–R is 1.44 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.20 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus bradycardia
Answer
sinus bradycardia
12.ECG rhythm strip interpretation
Answer: third-degree (complete) AV block
AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.
Five steps
1. Rate: R–R = 55 small boxes, 1500 ÷ 55 ≈ 27 per minute.
2. Rhythm: R–R regular (2.20 s) and P–P regular (0.68 s), but the two are independent.
3. P waves: upright and regular at 88 per minute, but not linked to the QRS complexes; some hide in a QRS or a T wave.
4. PR interval: varies from beat to beat; there is no fixed relation.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
third-degree (complete) AV block
Answer
third-degree (complete) AV block
Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.
13.Evidence-based practice statistics
Answer: NNT = 13
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 = 106/440 = 0.2409
CER = 131/410 = 0.3195
Number needed to treat
ARR = 0.3195 − 0.2409 = 0.0786
NNT = 1/0.0786 = 12.72 → 13
Answer
NNT = 13
Treating 13 patients prevents about one event.
14.Evidence-based practice statistics
Answer: FN = 414
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.