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 rotator cuff consists of supraspinatus, infraspinatus, teres minor and which other muscle?
ASubscapularis
BTeres major
CLatissimus dorsi
DDeltoid
2.Human anatomyHard
An indirect inguinal hernia enters the inguinal canal at which point?
AThrough the posterior wall of the canal, medial to the inferior epigastric vessels
BThrough the superficial ring only, without entering the canal
CBelow the inguinal ligament, through the femoral canal
DThrough the deep inguinal ring, lateral to the inferior epigastric vessels
3.Human physiologyMedium
What is anatomical dead space?
AThe air that stays in the lungs after full exhalation (residual volume)
BThe space between the two pleural layers
CThe volume of the conducting airways where no gas exchange happens
DAlveoli that have collapsed completely
4.Human physiologyHard
Why does breathing 100% oxygen correct hypoxaemia from a true right-to-left shunt only poorly?
AShunted blood is already fully saturated
BPure oxygen damages haemoglobin
CThe extra oxygen is all exhaled as CO₂
DThe shunted blood never comes into contact with alveolar gas
5.MicrobiologyMedium
Which organism is a common cause of infection in burns and in people with cystic fibrosis, and can produce a green pigment?
AStreptococcus pyogenes
BHelicobacter pylori
CClostridium tetani
DPseudomonas aeruginosa
6.MicrobiologyHard
A urinary catheter has become colonised by bacteria in a biofilm. Why is the catheter usually replaced instead of just giving antibiotics?
ABacteria in biofilms are protected and much harder for antibiotics to kill
BAntibiotics damage the catheter material
CBiofilms contain only viruses, so antibiotics do not work
DBiofilm bacteria cannot cause disease
7.PharmacologyMedium
When should a patient take a once-daily proton pump inhibitor such as omeprazole?
AStraight after a large evening meal
BOnly when heartburn is severe
CAbout 30–60 minutes before the first meal of the day
DWith a glass of milk at bedtime, when acid output is highest
8.PharmacologyHard
Before starting long-term bisphosphonate treatment, a patient is advised to have a dental check-up. Why?
ABisphosphonates dissolve tooth enamel within weeks
BThey are linked to osteonecrosis of the jaw after dental surgery
CBisphosphonates cause severe gum overgrowth
DBisphosphonates stain the teeth black if taken for many years
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 26 mmHg (3.5 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⁻ 42 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 42 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 flutter, 2:1 conduction
2sinus bradycardia
3sinus tachycardia
4atrial fibrillation
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 980 patients received a new treatment and 420 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 69/980control: 59/420
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 88% and specificity 87%. The prevalence of the disease is 14%. How many false-positive results FP are expected?
N = 10 000p = 14%Se = 88%Sp = 87%
FP =
3Solutions
1.Human anatomy
Answer: A · Subscapularis
Why this is the answer
SITS: supraspinatus, infraspinatus, teres minor and subscapularis. Teres major is close by but is not part of the cuff.
Answer
A · Subscapularis
2.Human anatomy
Answer: D · Through the deep inguinal ring, lateral to the inferior epigastric vessels
Why this is the answer
An indirect hernia passes through the deep ring, lateral to the inferior epigastric vessels, and runs along the canal. A direct hernia pushes through the posterior wall medial to them.
Answer
D · Through the deep inguinal ring, lateral to the inferior epigastric vessels
3.Human physiology
Answer: C · The volume of the conducting airways where no gas exchange happens
Why this is the answer
Anatomical dead space (about 150 ml) is air in the nose, trachea and bronchi. It is moved with every breath but never reaches the alveoli.
Answer
C · The volume of the conducting airways where no gas exchange happens
4.Human physiology
Answer: D · The shunted blood never comes into contact with alveolar gas
Why this is the answer
Shunted blood bypasses ventilated alveoli and mixes deoxygenated blood into arterial blood. Extra inspired oxygen cannot reach it, unlike in V/Q mismatch or hypoventilation.
Answer
D · The shunted blood never comes into contact with alveolar gas
5.Microbiology
Answer: D · Pseudomonas aeruginosa
Why this is the answer
Pseudomonas aeruginosa is an opportunistic pathogen of burns, ventilated patients and cystic fibrosis lungs. It is naturally resistant to many antibiotics.
Answer
D · Pseudomonas aeruginosa
6.Microbiology
Answer: A · Bacteria in biofilms are protected and much harder for antibiotics to kill
Why this is the answer
In a biofilm, bacteria live in a protective matrix, grow slowly and can tolerate high antibiotic levels. Removing the device removes the source of infection.
Answer
A · Bacteria in biofilms are protected and much harder for antibiotics to kill
7.Pharmacology
Answer: C · About 30–60 minutes before the first meal of the day
Why this is the answer
Proton pump inhibitors block pumps that are active, so they work best when taken before breakfast. Food then activates the pumps when the drug level is high.
Answer
C · About 30–60 minutes before the first meal of the day
8.Pharmacology
Answer: B · They are linked to osteonecrosis of the jaw after dental surgery
Why this is the answer
Osteonecrosis of the jaw is a rare but serious adverse effect, mostly after invasive dental work. Dental problems are treated before starting, and good oral hygiene is maintained.
Answer
B · They are linked to osteonecrosis of the jaw after dental surgery
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 17 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
|26 − 40| = 14 mmHg → 1.4 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.4 = 17 mmol/L
Answer
HCO₃⁻ = 17 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 50.4 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
0.7·42 + 21 = 50.4 mmHg
Answer
PaCO₂ = 50.4 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
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: 0.06 s, narrow (below 0.12 s).
Interpretation
sinus tachycardia
Answer
sinus tachycardia
12.ECG rhythm strip interpretation
Answer: atrial rate = 83 bpm
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.
Atrial rate
P–P = 18 small boxes = 0.72 s → 1500 ÷ 18 = 83.33 ≈ 83 per minute
Ventricular rate
R–R = 44 small boxes = 1.76 s → 1500 ÷ 44 = 34.09 ≈ 34 per minute
Answer
atrial rate = 83 bpm
Answer
atrial rate = 83 bpm
Third-degree block with a wide-complex ventricular escape (usually 20–40 per minute).
13.Evidence-based practice statistics
Answer: RR = 0.5
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.
Event rates
EER = 69/980 = 0.07041
CER = 59/420 = 0.1405
Relative risk
RR = 0.07041/0.1405 = 0.5012 ≈ 0.5
Answer
RR = 0.5
14.Evidence-based practice statistics
Answer: FP = 1118
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
10 000·0.14 = 1400
10 000 − 1400 = 8600
False positives
FP = 8600·0.13 = 1118
Answer
FP = 1118
Of the 2350 positives only 1232 have the disease (PPV 52.4%).