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
The circle of Willis at the base of the brain connects which two arterial supplies?
AExternal carotid and subclavian arterial systems
BLeft and right external carotid arteries
CInternal carotid and vertebrobasilar systems
DJugular and vertebral systems
2.Human anatomyHard
A stroke affecting the left middle cerebral artery in a right-handed person most likely causes:
ALeft-sided weakness and aphasia
BIsolated loss of balance
CRight-sided weakness and left visual neglect
DRight-sided weakness and aphasia
3.Human physiologyMedium
Why is the oxygen–haemoglobin dissociation curve S-shaped?
AOxygen dissolves better in plasma at high partial pressure
BRed cells change size as they take up oxygen
CHaemoglobin can bind only one oxygen molecule
DBinding of one oxygen makes the next bind more easily
4.Human physiologyHard
What is the main reason the ventricles do not contract at the same moment as the atria?
AThe SA node fires twice per beat
BThe ventricles have a longer refractory period only
CThe Purkinje fibres conduct slowly
DConduction is delayed at the atrioventricular node
5.MicrobiologyMedium
Pneumocystis jirovecii pneumonia in a person with advanced HIV is an example of:
AAn opportunistic infection
BA zoonotic infection
CA vector-borne infection
DA food-borne infection
6.MicrobiologyHard
In Griffith's 1928 experiment, mice died after injection of heat-killed smooth (S) pneumococci mixed with live rough (R) pneumococci. What did this show?
ARough pneumococci are always more virulent than smooth ones
BA "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
CHeat does not kill pneumococci
DViruses carry genes between bacteria
7.PharmacologyMedium
An adult patient is due to receive digoxin. The apical pulse, counted for a full minute, is 52 beats per minute. What should the nurse do?
AWithhold the dose and notify the prescriber
BGive half the dose and document the pulse
CGive the dose and recheck the pulse in one hour
DGive the dose with food to slow absorption
8.PharmacologyHard
A patient with Parkinson's disease has nausea. Why should metoclopramide be avoided?
AIt causes severe diarrhoea in Parkinson's disease
BIt blocks dopamine receptors in the brain and can worsen the parkinsonism
CIt increases dopamine levels in the brain too much and causes psychosis
DIt inhibits the absorption of all oral drugs
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 24 mmHg (3.2 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 respiratory acidosis: pH 7.30, PaCO₂ 75 mmHg (10.0 kPa), HCO₃⁻ 36 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.30PaCO₂ 75 mmHg (10.0 kPa)HCO₃⁻ 36 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
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). A test has a sensitivity of 97% and a specificity of 74%. Find the positive likelihood ratio LR+. Round to two decimal places.
Se = 97%Sp = 74%
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 87% and specificity 75%. The prevalence of the disease is 4%. How many false-positive results FP are expected?
N = 10 000p = 4%Se = 87%Sp = 75%
FP =
3Solutions
1.Human anatomy
Answer: C · Internal carotid and vertebrobasilar systems
Why this is the answer
The circle of Willis links the internal carotid arteries with the basilar artery (formed by the vertebral arteries), allowing some collateral flow.
Answer
C · Internal carotid and vertebrobasilar systems
2.Human anatomy
Answer: D · Right-sided weakness and aphasia
Why this is the answer
The left MCA supplies the left motor cortex (face and arm of the right side) and Broca's and Wernicke's areas, so right hemiparesis with aphasia is typical.
Answer
D · Right-sided weakness and aphasia
3.Human physiology
Answer: D · Binding of one oxygen makes the next bind more easily
Why this is the answer
Haemoglobin shows cooperative binding: each bound O₂ increases the affinity of the remaining sites, giving a sigmoid curve.
Answer
D · Binding of one oxygen makes the next bind more easily
4.Human physiology
Answer: D · Conduction is delayed at the atrioventricular node
Why this is the answer
The AV node conducts slowly (about 0.1 s delay), so atrial contraction completes ventricular filling before ventricular depolarisation begins.
Answer
D · Conduction is delayed at the atrioventricular node
5.Microbiology
Answer: A · An opportunistic infection
Why this is the answer
Opportunistic infections take advantage of a weakened immune system. In HIV they become common when the CD4 count falls to low levels.
Answer
A · An opportunistic infection
6.Microbiology
Answer: B · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
Why this is the answer
Something from the dead S strain turned live R bacteria into capsule-forming, virulent S bacteria. Avery, MacLeod and McCarty later showed that this transforming principle was DNA.
Answer
B · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
7.Pharmacology
Answer: A · Withhold the dose and notify the prescriber
Why this is the answer
Digoxin is usually withheld in adults if the apical pulse is below 60 per minute, and the prescriber is informed. Bradycardia may be an early sign of toxicity.
Answer
A · Withhold the dose and notify the prescriber
8.Pharmacology
Answer: B · It blocks dopamine receptors in the brain and can worsen the parkinsonism
Why this is the answer
Metoclopramide crosses the blood–brain barrier and antagonises D2 receptors. In Parkinson's disease this worsens rigidity and bradykinesia, so antiemetics that do not block central dopamine receptors are preferred.
Answer
B · It blocks dopamine receptors in the brain and can worsen the parkinsonism
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20.8 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
|24 − 40| = 16 mmHg → 1.6 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.6 = 20.8 mmol/L
Answer
HCO₃⁻ = 20.8 mmol/L
10.Arterial blood gas interpretation
Answer: chronic
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 HCO₃⁻
acute: 24 + 1·3.5 = 27.5 mmol/L
chronic: 24 + 3.5·3.5 = 36.25 mmol/L
Compare
measured: 36 mmol/L → chronic
Answer
chronic
11.ECG rhythm strip interpretation
Answer: QTc = 0.49 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
QT = 12 × 0.04 = 0.48 s
RR = 24 × 0.04 = 0.96 s
Bazett’s formula
QTc = 0.48/√0.96 = 0.48/0.9798 = 0.49 s
Answer
QTc = 0.49 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (490 ms)
12.ECG rhythm strip interpretation
Answer: QRS duration
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: R–R = 16 small boxes, 1500 ÷ 16 ≈ 94 per minute.
2. Rhythm: regular, every R–R is 0.64 s.
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: 0.12 s, wide (0.12 s or more).
First abnormal step
QRS duration → sinus rhythm with wide QRS
Answer
QRS duration
13.Evidence-based practice statistics
Answer: LR+ = 3.73
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.
Positive likelihood ratio
LR+ = 0.97/(1 − 0.74) = 3.731 ≈ 3.73
Answer
LR+ = 3.73
14.Evidence-based practice statistics
Answer: FP = 2400
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.
With and without the disease
10 000·0.04 = 400
10 000 − 400 = 9600
False positives
FP = 9600·0.25 = 2400
Answer
FP = 2400
Of the 2748 positives only 348 have the disease (PPV 12.7%).