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 spinal nerve roots form the brachial plexus?
AC3 to C7
BT1 to T5
CC1 to C4
DC5 to T1
2.Human anatomyHard
In the femoral triangle, what is the order of the main structures from lateral to medial?
AFemoral artery, femoral nerve, femoral vein
BFemoral nerve, femoral vein, femoral artery
CFemoral nerve, femoral artery, femoral vein
DFemoral vein, femoral artery, femoral nerve
3.Human physiologyMedium
Which hormonal event triggers ovulation?
AA fall in oestrogen
BA surge of luteinising hormone (LH)
CA surge of prolactin
DA rise in progesterone from the corpus luteum
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 is a common cause of infection in burns and in people with cystic fibrosis, and can produce a green pigment?
AClostridium tetani
BHelicobacter pylori
CPseudomonas aeruginosa
DStreptococcus pyogenes
6.MicrobiologyHard
What does an R0 (basic reproduction number) of 3 mean?
AThree doses of vaccine are needed
BOn average one case infects three others in a fully susceptible population
CThree percent of infected people die
DThe infection lasts three days
7.PharmacologyMedium
A man taking tamsulosin is planning cataract surgery. What should the nurse advise?
AStop the drug for one day and no further action is needed
BThe drug has no link with eye surgery
CTell the eye surgeon, as it can cause floppy iris syndrome
DTake a double dose on the day of surgery
8.PharmacologyHard
Why is the TPMT enzyme often tested before azathioprine is started?
ATPMT is needed to absorb azathioprine from the gut
BTPMT testing shows whether the patient has tuberculosis
CHigh TPMT activity causes an allergic reaction to azathioprine
DLow TPMT activity causes toxic levels and marrow suppression
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.52, PaCO₂ 48 mmHg (6.4 kPa) and HCO₃⁻ 38 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.52PaCO₂ 48 mmHg (6.4 kPa)HCO₃⁻ 38 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows pH 7.32, PaCO₂ 34 mmHg (4.5 kPa) and HCO₃⁻ 17 mmol/L, with Na⁺ 134 and Cl⁻ 96 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
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 arrhythmia
2normal sinus rhythm
3sinus bradycardia
4atrial fibrillation
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). In a randomised trial, 430 patients received a new treatment and 810 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: 18/430control: 57/810
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 14% in relative terms (RRR = 14%). In a patient group the risk without treatment (CER) is 26% 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 = 14%CER = 26%
NNT =
3Solutions
1.Human anatomy
Answer: D · C5 to T1
Why this is the answer
The brachial plexus is formed by the ventral rami of C5, C6, C7, C8 and T1. It gives rise to the main nerves of the arm.
Answer
D · C5 to T1
2.Human anatomy
Answer: C · Femoral nerve, femoral artery, femoral vein
Why this is the answer
From lateral to medial: nerve, artery, vein, then empty space and lymphatics (NAVEL). This is why the femoral artery pulse lies between the nerve and vein.
Answer
C · Femoral nerve, femoral artery, femoral vein
3.Human physiology
Answer: B · A surge of luteinising hormone (LH)
Why this is the answer
Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.
Answer
B · A surge of luteinising hormone (LH)
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: C · Pseudomonas aeruginosa
Why this is the answer
Pseudomonas aeruginosa is an opportunistic pathogen of burns, ventilated patients and cystic fibrosis lungs. It is naturally resistant to many antibiotics.
Answer
C · Pseudomonas aeruginosa
6.Microbiology
Answer: B · On average one case infects three others in a fully susceptible population
Why this is the answer
R0 is the average number of secondary cases from one case when everyone is susceptible. Herd immunity needs roughly 1 − 1/R0 of the population immune: about 67% for R0 = 3.
Answer
B · On average one case infects three others in a fully susceptible population
7.Pharmacology
Answer: C · Tell the eye surgeon, as it can cause floppy iris syndrome
Why this is the answer
Tamsulosin can cause intraoperative floppy iris syndrome, which complicates cataract surgery. The surgeon needs to know so that the technique can be adapted.
Answer
C · Tell the eye surgeon, as it can cause floppy iris syndrome
8.Pharmacology
Answer: D · Low TPMT activity causes toxic levels and marrow suppression
Why this is the answer
Thiopurine methyltransferase helps inactivate azathioprine and mercaptopurine. People with low or absent TPMT activity need much lower doses or another drug to avoid life-threatening myelosuppression.
Answer
D · Low TPMT activity causes toxic levels and marrow suppression
9.Arterial blood gas interpretation
Answer: partially compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.52: alkalaemia (above 7.45)
PaCO₂ 48 mmHg: high (35–45)
HCO₃⁻ 38 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: pure high-gap 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.
Anion gap
AG = 134 − (96 + 17) = 21
Delta ratio
(21 − 12)/(24 − 17) = 1.286
pure high-gap acidosis
Answer
pure high-gap acidosis
11.ECG rhythm strip interpretation
Answer: sinus arrhythmia
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: 7 QRS complexes in 6 s, × 10 = 70 per minute.
2. Rhythm: irregular, R–R from 0.74 to 0.98 s, lengthening and shortening gradually.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.16 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus arrhythmia
Answer
sinus arrhythmia
12.ECG rhythm strip interpretation
Answer: rhythm
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: 7 QRS complexes in 6 s, × 10 = 70 per minute.
2. Rhythm: irregular, R–R from 0.70 to 1.02 s, lengthening and shortening gradually.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.16 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rhythm → sinus arrhythmia
Answer
rhythm
13.Evidence-based practice statistics
Answer: NNT = 36
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 = 18/430 = 0.04186
CER = 57/810 = 0.07037
Number needed to treat
ARR = 0.07037 − 0.04186 = 0.02851
NNT = 1/0.02851 = 35.08 → 36
Answer
NNT = 36
Treating 36 patients prevents about one event.
14.Evidence-based practice statistics
Answer: NNT = 28
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 = 0.26·0.14 = 0.0364
Number needed to treat
NNT = 1/0.0364 = 27.47 → 28
Answer
NNT = 28
The same relative effect gives a smaller NNT when the baseline risk is higher.