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?
AGreat saphenous vein
BMedian cubital vein
CAxillary vein
DBrachial vein
2.Human anatomyHard
A stab wound cuts the left half of the spinal cord at T10. Which pattern of loss is expected below the lesion?
ALoss of pain and temperature on both sides only
BRight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
CLoss of all sensation and movement on the left only
DLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiologyMedium
Which ABO blood group has both anti-A and anti-B antibodies in the plasma?
AGroup A
BGroup O
CGroup B
DGroup AB
4.Human physiologyHard
During exercise, blood flow to active skeletal muscle rises greatly. What is the main cause?
ASympathetic stimulation constricts the muscle arterioles via alpha receptors
BThe heart pumps blood only to the muscles
CThe parasympathetic system dilates all arteries
DLocal metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles
5.MicrobiologyMedium
Pneumocystis jirovecii pneumonia in a person with advanced HIV is an example of:
AAn opportunistic infection
BA vector-borne infection
CA zoonotic infection
DA food-borne infection
6.MicrobiologyHard
Primary syphilis typically presents with:
AA widespread rash on the palms and soles
BA painless ulcer (chancre) at the site of infection
CA painful cluster of small blisters on the genitals or the mouth
DDamage to the heart and brain
7.PharmacologyMedium
Sumatriptan is contraindicated in which patient?
AA patient with a recent ankle sprain
BA patient with iron deficiency anaemia
CA patient with coronary artery disease
DA patient with mild asthma
8.PharmacologyHard
A toddler swallowed many of his mother's iron tablets. He has vomiting and bloody diarrhoea. Which drug may be used to bind iron?
ADeferoxamine (desferrioxamine)
BAcetylcysteine
CNaloxone
DActivated charcoal given by mouth
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 142 mmol/L, glucose 14.7 mmol/L and urea 9.8 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 142 mmol/Lglucose 14.7 mmol/Lurea 9.8 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 40% oxygen shows PaO₂ 85 mmHg (11.3 kPa) and PaCO₂ 34 mmHg (4.5 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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
2atrial fibrillation
3junctional rhythm
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
This lead II strip shows a conduction problem between the atria and the ventricles. 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
1second-degree AV block type I (Wenckebach)
2second-degree AV block type II
3third-degree (complete) AV block
4first-degree AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 5200 person-years with 343 new cases, and unexposed people 2200 person-years with 48 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 343/5200 PYnot exposed: 48/2200 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 80% and a specificity of 93%. It is used in a population where the prevalence is 15%. Find the negative predictive value NPV as a percentage. Round to one decimal place.
Se = 80%Sp = 93%p = 15%
NPV = %
3Solutions
1.Human anatomy
Answer: B · 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
B · Median cubital vein
2.Human anatomy
Answer: D · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
Why this is the answer
In Brown-Séquard syndrome the corticospinal tract and dorsal columns are lost on the same side, while the spinothalamic tract, which crosses near entry, causes contralateral pain and temperature loss.
Answer
D · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: B · Group O
Why this is the answer
Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.
Answer
B · Group O
4.Human physiology
Answer: D · Local metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles
Why this is the answer
Active hyperaemia: metabolites produced by working muscle relax arteriolar smooth muscle locally, overriding sympathetic vasoconstriction in that tissue.
Answer
D · Local metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles
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 painless ulcer (chancre) at the site of infection
Why this is the answer
Treponema pallidum causes a painless chancre in the primary stage. The secondary stage brings rash and systemic symptoms, and tertiary syphilis can damage the heart and nervous system. Painful blisters suggest herpes.
Answer
B · A painless ulcer (chancre) at the site of infection
7.Pharmacology
Answer: C · A patient with coronary artery disease
Why this is the answer
Triptans can constrict coronary arteries and cause angina or myocardial infarction. They are avoided in coronary disease, uncontrolled hypertension and previous stroke.
Answer
C · A patient with coronary artery disease
8.Pharmacology
Answer: A · Deferoxamine (desferrioxamine)
Why this is the answer
Iron poisoning can cause GI bleeding, shock, metabolic acidosis and liver failure. Deferoxamine chelates iron and is used in serious poisoning; activated charcoal does not bind iron.
Answer
A · Deferoxamine (desferrioxamine)
9.Arterial blood gas interpretation
Answer: Osm = 309 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 = 2·142 + 14.7 + 9.8 = 308.5 ≈ 309 mOsm/kg
Answer
Osm = 309 mOsm/kg
10.Arterial blood gas interpretation
Answer: A–a gradient = 157.7 mmHg
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.
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 = 31 small boxes, 1500 ÷ 31 ≈ 48 per minute.
2. Rhythm: regular, every R–R is 1.24 s.
3. P waves: inverted, just before each QRS.
4. PR interval: 0.10 s, shorter than 0.12 s.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
junctional rhythm
Answer
junctional rhythm
12.ECG rhythm strip interpretation
Answer: second-degree AV block type I (Wenckebach)
AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.
Five steps
1. Rate: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 0.92 s, 65 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: lengthens 0.18 → 0.32 → 0.40 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type I (Wenckebach)
Answer
second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics
Answer: IRR = 3.02
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: 343/5200 = 65.96 per 1000 person-years
not exposed: 48/2200 = 21.82 per 1000 person-years
Rate ratio
IRR = 65.96/21.82 = 3.023 ≈ 3.02
Answer
IRR = 3.02
14.Evidence-based practice statistics
Answer: NPV = 96.3 %
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.