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 valve lies between the left atrium and the left ventricle?
APulmonary valve
BAortic valve
CTricuspid valve
DMitral (bicuspid) valve
2.Human anatomyHard
Why can an inflamed gallbladder that irritates the diaphragm cause pain at the tip of the right shoulder?
AThe vagus nerve supplies the skin of the shoulder
BThe phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
CBile travels up the lymphatics to the shoulder
DThe gallbladder is attached to the right clavicle
3.Human physiologyMedium
Which receptors respond mainly to a fall in arterial oxygen levels?
ABaroreceptors in the carotid sinus
BStretch receptors in the lungs
CPeripheral chemoreceptors in the carotid and aortic bodies
DCentral chemoreceptors in the medulla
4.Human physiologyHard
Constriction of the efferent arteriole of a glomerulus, with the afferent arteriole unchanged, tends to:
AHave no effect on GFR at all
BLower glomerular capillary pressure and GFR
CStop filtration completely
DRaise glomerular capillary pressure and GFR
5.MicrobiologyMedium
Which cells does HIV mainly infect and destroy?
APlatelets
BRed blood cells
CCD4 T helper cells
DNeutrophils
6.MicrobiologyHard
The minimum inhibitory concentration (MIC) of an antibiotic is:
AThe lowest concentration that kills 99.9% of the bacterial inoculum in vitro
BThe plasma level at which side effects begin
CThe highest dose a patient can tolerate
DThe lowest concentration that prevents visible bacterial growth in vitro
7.PharmacologyMedium
Which statement about allopurinol is correct?
AIt increases uric acid production to flush out crystals
BIt lowers uric acid over time and does not relieve the pain of an acute attack
CIt works within an hour to relieve joint pain
DIt is taken only during an acute attack and stopped as soon as the pain settles
8.PharmacologyHard
A patient on methadone is also prescribed diazepam for anxiety. What is the main safety concern?
AMethadone causes diazepam to be excreted very quickly
BSevere hypertension from the combination
CAdditive sedation and respiratory depression
DDiazepam blocks the effect of methadone completely
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 146 mmol/L, glucose 25.5 mmol/L and urea 21.4 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 146 mmol/Lglucose 25.5 mmol/Lurea 21.4 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.10, PaCO₂ 33 mmHg (4.4 kPa), HCO₃⁻ 10 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.10PaCO₂ 33 mmHg (4.4 kPa)HCO₃⁻ 10 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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
2sinus tachycardia
3atrial flutter, 2:1 conduction
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
A patient without a palpable pulse has this organised rhythm on the monitor (pulseless electrical activity). Which treatment do adult resuscitation guidelines name for 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
1synchronised cardioversion
2immediate defibrillation
3transcutaneous pacing
4CPR (non-shockable rhythm)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 3700 person-years with 278 new cases, and unexposed people 4200 person-years with 126 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 278/3700 PYnot exposed: 126/4200 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 46% in relative terms (RRR = 46%). In a patient group the risk without treatment (CER) is 20% 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 = 46%CER = 20%
NNT =
3Solutions
1.Human anatomy
Answer: D · Mitral (bicuspid) valve
Why this is the answer
The mitral (bicuspid) valve separates the left atrium and left ventricle; the tricuspid valve does the same on the right side.
Answer
D · Mitral (bicuspid) valve
2.Human anatomy
Answer: B · The phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
Why this is the answer
Diaphragmatic irritation is carried by the phrenic nerve (C3–C5). The brain refers it to skin supplied by the same segments, via the supraclavicular nerves over the shoulder.
Answer
B · The phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
3.Human physiology
Answer: C · Peripheral chemoreceptors in the carotid and aortic bodies
Why this is the answer
The carotid and aortic bodies detect low PaO₂ (and also high CO₂ and H⁺). The central chemoreceptors respond to CO₂ but not directly to oxygen.
Answer
C · Peripheral chemoreceptors in the carotid and aortic bodies
4.Human physiology
Answer: D · Raise glomerular capillary pressure and GFR
Why this is the answer
Efferent constriction holds blood in the glomerulus, raising its pressure and GFR. Angiotensin II acts this way, which is why ACE inhibitors can lower GFR.
Answer
D · Raise glomerular capillary pressure and GFR
5.Microbiology
Answer: C · CD4 T helper cells
Why this is the answer
HIV binds CD4 and a co-receptor on T helper cells. Loss of these cells weakens both antibody and cell-mediated immunity, leading to AIDS.
Answer
C · CD4 T helper cells
6.Microbiology
Answer: D · The lowest concentration that prevents visible bacterial growth in vitro
Why this is the answer
The MIC is the lowest concentration that inhibits visible growth. The minimum bactericidal concentration (MBC) is the lowest that kills 99.9% of the bacteria.
Answer
D · The lowest concentration that prevents visible bacterial growth in vitro
7.Pharmacology
Answer: B · It lowers uric acid over time and does not relieve the pain of an acute attack
Why this is the answer
Allopurinol reduces urate production and is taken daily to prevent attacks. Acute attacks are treated with NSAIDs, colchicine or corticosteroids, and a new rash on allopurinol must be reported.
Answer
B · It lowers uric acid over time and does not relieve the pain of an acute attack
8.Pharmacology
Answer: C · Additive sedation and respiratory depression
Why this is the answer
Combining opioids with benzodiazepines greatly increases the risk of sedation, respiratory depression and death. If both are needed, the lowest doses are used and the patient is monitored closely.
Answer
C · Additive sedation and respiratory depression
9.Arterial blood gas interpretation
Answer: Osm = 339 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 = 2·146 + 25.5 + 21.4 = 338.9 ≈ 339 mOsm/kg
Answer
Osm = 339 mOsm/kg
10.Arterial blood gas interpretation
Answer: added respiratory 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.
Expected PaCO₂
1.5·10 + 8 = 23 → 21–25 mmHg
Compare
measured: 33 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: sinus tachycardia
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 = 14 small boxes, 1500 ÷ 14 ≈ 107 per minute.
2. Rhythm: regular, every R–R is 0.56 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus tachycardia
Answer
sinus tachycardia
12.ECG rhythm strip interpretation
Answer: CPR (non-shockable rhythm)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 13 small boxes, 1500 ÷ 13 ≈ 115 per minute.
2. Rhythm: regular, every R–R is 0.52 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
sinus tachycardia
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
CPR (non-shockable rhythm)
Answer
CPR (non-shockable rhythm)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: IRR = 2.5
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.
Incidence rates
exposed: 278/3700 = 75.14 per 1000 person-years
not exposed: 126/4200 = 30 per 1000 person-years
Rate ratio
IRR = 75.14/30 = 2.505 ≈ 2.5
Answer
IRR = 2.5
14.Evidence-based practice statistics
Answer: NNT = 11
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 = 0.2·0.46 = 0.092
Number needed to treat
NNT = 1/0.092 = 10.87 → 11
Answer
NNT = 11
The same relative effect gives a smaller NNT when the baseline risk is higher.