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 cranial nerve controls movement of the tongue?
AHypoglossal nerve (XII)
BFacial nerve (VII)
CVagus nerve (X)
DGlossopharyngeal nerve (IX)
2.Human anatomyHard
Where does the thoracic duct usually empty its lymph into the blood?
ARight atrium directly
BJunction of the right subclavian and right internal jugular veins
CJunction of the left subclavian and left internal jugular veins
DInferior vena cava at the level of the kidneys
3.Human physiologyMedium
Which hormonal event triggers ovulation?
AA surge of luteinising hormone (LH)
BA surge of prolactin
CA rise in progesterone from the corpus luteum
DA fall in oestrogen
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
BIt blocks the binding of calcium to troponin
CIt hyperpolarises cardiac cells, so they can never fire
DIt speeds up the SA node until tachycardia causes fainting
5.MicrobiologyMedium
What is the incubation period?
AThe time between infection and the first symptoms
BThe time an antibiotic takes to work
CThe time during which a patient remains infectious to others
DThe time a culture needs in the laboratory
6.MicrobiologyHard
In 1854 John Snow traced a cholera outbreak in London to the Broad Street water pump. Why is this study important?
AIt showed that cholera spreads through contaminated water, founding modern epidemiology
BIt led directly to the first cholera vaccine
CIt proved that cholera is caused by a virus
DIt showed that cholera spreads through "bad air" (miasma)
7.PharmacologyMedium
A patient with suspected sepsis has blood cultures ordered and a first antibiotic dose due. What should the nurse do?
AGive the antibiotic only if the temperature is above 40 °C (104 °F)
BGive the antibiotic first, then take blood cultures the next day
CWait for the culture results before giving any antibiotic
DTake the blood cultures first, then give the antibiotic at once
8.PharmacologyHard
A drug has a volume of distribution of 40 L and complete bioavailability. Which loading dose gives a target plasma concentration of 2 mg/L?
A80 mg
B160 mg
C42 mg
D20 mg
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 22 mmHg (2.9 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⁻.
A practice gas shows a metabolic alkalosis with HCO₃⁻ 39 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 39 mmol/L
PaCO₂ = mmHg
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
3sinus rhythm with PVCs
4atrial flutter, 4:1 conduction
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, 170 patients received a new treatment and 290 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the absolute risk reduction ARR as a percentage. Round to one decimal place.
treatment: 27/170control: 93/290
ARR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 35% in relative terms (RRR = 35%). In a patient group the risk without treatment (CER) is 18% 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 = 35%CER = 18%
NNT =
3Solutions
1.Human anatomy
Answer: A · Hypoglossal nerve (XII)
Why this is the answer
The hypoglossal nerve supplies the intrinsic and most extrinsic tongue muscles. A lesion makes the protruded tongue deviate towards the affected side.
Answer
A · Hypoglossal nerve (XII)
2.Human anatomy
Answer: C · Junction of the left subclavian and left internal jugular veins
Why this is the answer
The thoracic duct drains lymph from both legs, the abdomen and the left side of the upper body into the left venous angle. The right lymphatic duct drains the rest into the right venous angle.
Answer
C · Junction of the left subclavian and left internal jugular veins
3.Human physiology
Answer: A · 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
A · A surge of luteinising hormone (LH)
4.Human physiology
Answer: A · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
Why this is the answer
High extracellular potassium makes the resting potential less negative. Many fast sodium channels then become inactivated, slowing conduction and risking fatal arrhythmias.
Answer
A · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: A · The time between infection and the first symptoms
Why this is the answer
The incubation period runs from exposure to the onset of symptoms. People can sometimes already transmit the infection during it.
Answer
A · The time between infection and the first symptoms
6.Microbiology
Answer: A · It showed that cholera spreads through contaminated water, founding modern epidemiology
Why this is the answer
Snow mapped cases around the pump and linked them to its water. Removing the pump handle is a classic example of using epidemiology to find and stop a source, before the bacterium was identified.
Answer
A · It showed that cholera spreads through contaminated water, founding modern epidemiology
7.Pharmacology
Answer: D · Take the blood cultures first, then give the antibiotic at once
Why this is the answer
Cultures are taken before antibiotics so the organism can be identified, but antibiotics must not be delayed in sepsis. Taking cultures quickly and then giving the antibiotic meets both aims.
Answer
D · Take the blood cultures first, then give the antibiotic at once
8.Pharmacology
Answer: A · 80 mg
Why this is the answer
Loading dose = 40 L × 2 mg/L = 80 mg. The loading dose depends on Vd, while the maintenance dose depends on clearance.
Answer
A · 80 mg
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20.4 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
|22 − 40| = 18 mmHg → 1.8 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.8 = 20.4 mmol/L
Answer
HCO₃⁻ = 20.4 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 48.3 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
0.7·39 + 21 = 48.3 mmHg
Answer
PaCO₂ = 48.3 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
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 0.98 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
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: irregularly irregular, R–R from 0.62 to 0.96 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rhythm → atrial fibrillation
Answer
rhythm
13.Evidence-based practice statistics
Answer: ARR = 16.2 %
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 = 27/170 = 0.1588
CER = 93/290 = 0.3207
Absolute risk reduction
ARR = 0.3207 − 0.1588 = 0.1619 ≈ 16.2%
Answer
ARR = 16.2%
14.Evidence-based practice statistics
Answer: NNT = 16
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 = 0.18·0.35 = 0.063
Number needed to treat
NNT = 1/0.063 = 15.87 → 16
Answer
NNT = 16
The same relative effect gives a smaller NNT when the baseline risk is higher.