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 superficial vein in the front of the elbow is most often used for venepuncture?
AMedian cubital vein
BAxillary vein
CBrachial vein
DGreat saphenous vein
2.Human anatomyHard
Which muscle is the main flexor of the hip joint?
AIliopsoas
BAdductor magnus
CGluteus maximus
DBiceps femoris
3.Human physiologyMedium
Since action potentials are all-or-none, how does a neuron signal a stronger stimulus?
ABy changing the neurotransmitter it releases
BBy firing action potentials more often
CBy making each action potential larger
DBy making each action potential longer
4.Human physiologyHard
A patient has taken high-dose prednisolone for a year. What is the state of the hypothalamic–pituitary–adrenal axis?
AACTH is suppressed and the adrenal cortex has atrophied
BACTH is normal and cortisol production is increased
CACTH is high and the adrenal cortex is enlarged
DThe adrenal medulla has stopped making adrenaline and noradrenaline
5.MicrobiologyMedium
Which organism causes typhoid fever?
AVibrio cholerae
BSalmonella Typhi
CCampylobacter jejuni
DShigella dysenteriae
6.MicrobiologyHard
How do retroviruses such as HIV copy their genome into the host DNA?
AHost DNA polymerase copies the RNA without help
BThe viral DNA is copied directly by host RNA polymerase
CThe viral RNA is translated into DNA by ribosomes
DReverse transcriptase makes DNA from viral RNA, and integrase inserts it
7.PharmacologyMedium
A patient with heart failure starts furosemide. Which self-monitoring instruction is most useful?
ACount the respiratory rate each evening
BCheck the blood glucose before every meal
CWeigh daily at the same time on the same scales
DMeasure the temperature twice a day and record it in a diary
8.PharmacologyHard
A 500 mg intravenous dose gives an initial plasma concentration of 10 mg/L. What is the volume of distribution?
A50 L
B5000 L
C5 L
D500 L
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 151 mmol/L, glucose 10.7 mmol/L and urea 10.5 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 151 mmol/Lglucose 10.7 mmol/Lurea 10.5 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 42 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 42 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = 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 fibrillation
2supraventricular tachycardia (SVT)
3ventricular tachycardia
4sinus tachycardia
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 810 patients received a new treatment and 960 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk reduction RRR as a percentage. Round to one decimal place.
treatment: 131/810control: 259/960
RRR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 15% for a disease. A test with sensitivity 64% and specificity 65% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 15%Se = 64%Sp = 65%positive
p′ = %
3Solutions
1.Human anatomy
Answer: A · Median cubital vein
Why this is the answer
The median cubital vein crosses the cubital fossa between the cephalic and basilic veins. It is superficial, large and fairly fixed, so it is easy to puncture.
Answer
A · Median cubital vein
2.Human anatomy
Answer: A · Iliopsoas
Why this is the answer
Iliopsoas (psoas major and iliacus) is the strongest hip flexor. Gluteus maximus extends the hip.
Answer
A · Iliopsoas
3.Human physiology
Answer: B · By firing action potentials more often
Why this is the answer
Stimulus strength is coded by the frequency of action potentials and by how many neurons are recruited, not by their amplitude.
Answer
B · By firing action potentials more often
4.Human physiology
Answer: A · ACTH is suppressed and the adrenal cortex has atrophied
Why this is the answer
Exogenous glucocorticoid suppresses CRH and ACTH, so the adrenal cortex atrophies. Stopping the drug suddenly can cause an adrenal crisis, so it is tapered.
Answer
A · ACTH is suppressed and the adrenal cortex has atrophied
5.Microbiology
Answer: B · Salmonella Typhi
Why this is the answer
Salmonella enterica serotype Typhi causes typhoid fever, a systemic illness with prolonged fever. Some people become chronic carriers, often in the gallbladder.
Answer
B · Salmonella Typhi
6.Microbiology
Answer: D · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
Why this is the answer
Reverse transcriptase converts the RNA genome into DNA, which integrase inserts into the host genome as a provirus. Both enzymes are drug targets.
Answer
D · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
7.Pharmacology
Answer: C · Weigh daily at the same time on the same scales
Why this is the answer
Daily weight is the best home measure of fluid balance. A rapid gain suggests fluid retention, and a rapid loss suggests overdiuresis and dehydration, both of which should be reported.
Answer
C · Weigh daily at the same time on the same scales
8.Pharmacology
Answer: A · 50 L
Why this is the answer
Vd = 500 mg ÷ 10 mg/L = 50 L. This is close to total body water, so the drug spreads beyond the plasma but is not extensively stored in tissues.
Answer
A · 50 L
9.Arterial blood gas interpretation
Answer: Osm = 323 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 = 2·151 + 10.7 + 10.5 = 323.2 ≈ 323 mOsm/kg
Answer
Osm = 323 mOsm/kg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 50.4 mmHg
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
0.7·42 + 21 = 50.4 mmHg
Answer
PaCO₂ = 50.4 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QT = 0.36 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
9 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 9 × 0.04 s = 0.36 s
Answer
QT = 0.36 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.84 s).
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: RRR = 40.1 %
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 = 131/810 = 0.1617
CER = 259/960 = 0.2698
Relative risk reduction
ARR = 0.2698 − 0.1617 = 0.1081
RRR = 0.1081/0.2698 = 0.4005 ≈ 40.1%
Answer
RRR = 40.1%
14.Evidence-based practice statistics
Answer: p′ = 24.4 %
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.