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 two regions of the adult spine normally show a lordosis (curve concave towards the back)?
AThoracic and sacral
BCervical and lumbar
CThoracic and lumbar
DCervical and thoracic
2.Human anatomyHard
Why can an inflamed gallbladder that irritates the diaphragm cause pain at the tip of the right shoulder?
AThe gallbladder is attached to the right clavicle
BBile travels up the lymphatics to the shoulder
CThe vagus nerve supplies the skin of the shoulder
DThe phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
3.Human physiologyMedium
Which receptor is stimulated when the patellar tendon is tapped in the knee-jerk reflex?
ANociceptor
BGolgi tendon organ
CMuscle spindle
DPacinian corpuscle
4.Human physiologyHard
In pernicious anaemia, autoimmune destruction of gastric parietal cells leads to anaemia because:
AIntrinsic factor is lost, so vitamin B12 is not absorbed
BGastrin is lost, so the bone marrow is no longer stimulated
CPepsin is lost, so haemoglobin cannot be made
DAcid is lost, so iron is never absorbed in any form
5.MicrobiologyMedium
Which bacterium causes whooping cough (pertussis)?
AHaemophilus influenzae
BCorynebacterium diphtheriae
CBordetella pertussis
DMycoplasma pneumoniae
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
In which patient is bupropion contraindicated?
AA patient with high blood cholesterol and a family history of it
BA patient with a seizure disorder or an eating disorder
CA patient who wants to stop smoking
DA patient with seasonal allergies
8.PharmacologyHard
A patient with asthma starts timolol eye drops for glaucoma and soon becomes wheezy. What is the most likely explanation?
ATimolol activates histamine receptors in the lungs
BThe eye drops contain an allergen that causes hay fever and asthma
CThe wheeze is unrelated to the eye drops
DTimolol is absorbed and blocks beta-2 receptors in the airways
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 29 mmHg (3.9 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 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⁻ 33 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 33 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
1sinus bradycardia
2junctional rhythm
3normal sinus rhythm
4sinus tachycardia
12.ECG rhythm strip interpretationHard
Identify the rhythm on this lead II strip. 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
2torsades de pointes
3atrial flutter, 4:1 conduction
4ventricular fibrillation
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 83% and a specificity of 86%. Find the positive likelihood ratio LR+. Round to two decimal places.
Se = 83%Sp = 86%
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 84% and specificity 93%. The prevalence of the disease is 27%. How many positive results n₊ are expected in total?
N = 10 000p = 27%Se = 84%Sp = 93%
n₊ =
3Solutions
1.Human anatomy
Answer: B · Cervical and lumbar
Why this is the answer
The cervical and lumbar regions are lordotic (secondary curves). The thoracic and sacral regions are kyphotic (primary curves).
Answer
B · Cervical and lumbar
2.Human anatomy
Answer: D · 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
D · The phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
3.Human physiology
Answer: C · Muscle spindle
Why this is the answer
Tapping the tendon stretches quadriceps muscle spindles. Their afferents synapse directly on motor neurons in a monosynaptic reflex that contracts the muscle.
Answer
C · Muscle spindle
4.Human physiology
Answer: A · Intrinsic factor is lost, so vitamin B12 is not absorbed
Why this is the answer
Parietal cells secrete intrinsic factor, needed for B12 absorption in the terminal ileum. B12 deficiency impairs DNA synthesis and causes megaloblastic anaemia.
Answer
A · Intrinsic factor is lost, so vitamin B12 is not absorbed
5.Microbiology
Answer: C · Bordetella pertussis
Why this is the answer
Bordetella pertussis causes long coughing fits with a "whoop". It is most dangerous for young babies, which is why vaccination in pregnancy is offered in many countries.
Answer
C · Bordetella pertussis
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: B · A patient with a seizure disorder or an eating disorder
Why this is the answer
Bupropion increases the risk of seizures, especially with bulimia or anorexia, where electrolyte disturbances are common. It is used for depression and as a smoking cessation aid.
Answer
B · A patient with a seizure disorder or an eating disorder
8.Pharmacology
Answer: D · Timolol is absorbed and blocks beta-2 receptors in the airways
Why this is the answer
Ophthalmic timolol is a non-selective beta blocker and can reach the circulation in amounts that cause bronchospasm or bradycardia. It is generally avoided in asthma.
Answer
D · Timolol is absorbed and blocks beta-2 receptors in the airways
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 18.5 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
|29 − 40| = 11 mmHg → 1.1 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.1 = 18.5 mmol/L
Answer
HCO₃⁻ = 18.5 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 44.1 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
0.7·33 + 21 = 44.1 mmHg
Answer
PaCO₂ = 44.1 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: sinus bradycardia
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 = 32 small boxes, 1500 ÷ 32 ≈ 47 per minute.
2. Rhythm: regular, every R–R is 1.28 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus bradycardia
Answer
sinus bradycardia
12.ECG rhythm strip interpretation
Answer: ventricular fibrillation
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: no countable ventricular rate.
2. Rhythm: chaotic, no repeating pattern.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: no QRS complexes.
Interpretation
ventricular fibrillation
Answer
ventricular fibrillation
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: LR+ = 5.93
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+ = 0.83/(1 − 0.86) = 5.929 ≈ 5.93
Answer
LR+ = 5.93
14.Evidence-based practice statistics
Answer: n₊ = 2779
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
10 000·0.27 = 2700
10 000 − 2700 = 7300
All positive results
TP = 2700·0.84 = 2268
FP = 7300·0.07 = 511
n₊ = 2268 + 511 = 2779
Answer
n₊ = 2779
Of the 2779 positives only 2268 have the disease (PPV 81.6%).