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 of these organs is retroperitoneal?
AStomach
BTransverse colon
CKidney
DSpleen
2.Human anatomyHard
Why can an inflamed gallbladder that irritates the diaphragm cause pain at the tip of the right shoulder?
ABile travels up the lymphatics to the shoulder
BThe phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
CThe vagus nerve supplies the skin of the shoulder
DThe gallbladder is attached to the right clavicle
3.Human physiologyMedium
In which form is most carbon dioxide carried in the blood?
AAs bicarbonate ions in plasma
BDissolved as CO₂ gas
CBound to the haem iron
DAs carbonic acid in platelets
4.Human physiologyHard
Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?
ACO stops oxygen from dissolving in plasma
BCO occupies haemoglobin binding sites, so oxygen content falls
CCO blocks diffusion across the alveolar wall
DCO lowers the barometric pressure in the alveoli
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
BPseudomonas aeruginosa
CClostridium tetani
DHelicobacter pylori
6.MicrobiologyHard
What did Pasteur's swan-neck flask experiment demonstrate?
AMicroorganisms need light to grow
BMicroorganisms come from other microorganisms, not from spontaneous generation
CAir itself produces microorganisms
DBoiling cannot kill microorganisms
7.PharmacologyMedium
Which drug is the antidote for an opioid overdose?
AAcetylcysteine
BAtropine
CFlumazenil
DNaloxone
8.PharmacologyHard
A postoperative patient using morphine PCA is very drowsy, hard to rouse and has a respiratory rate of 7 per minute. What should the nurse do first?
ALet the patient sleep and recheck in one hour
BEncourage the patient to press the button only when awake and in pain
CStop the PCA, try to rouse the patient and give naloxone as per protocol
DReduce the PCA dose and leave the pump running
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 80 mmHg (10.7 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
At sea level, a practice gas taken on 28% oxygen shows PaO₂ 71 mmHg (9.5 kPa) and PaCO₂ 53 mmHg (7.1 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
This lead II strip shows sinus rhythm with one pause. Measure the pause, the R–R interval across it, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
pause = 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). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 65 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 93% and a specificity of 89%. It is used in a population where the prevalence is 15%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 93%Sp = 89%p = 15%
PPV = %
3Solutions
1.Human anatomy
Answer: C · Kidney
Why this is the answer
The kidneys lie behind the parietal peritoneum. The stomach, spleen and transverse colon are intraperitoneal and have a mesentery or ligaments.
Answer
C · Kidney
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: A · As bicarbonate ions in plasma
Why this is the answer
About 70% of CO₂ is carried as bicarbonate. Carbonic anhydrase in red cells converts CO₂ and water to carbonic acid, which dissociates into H⁺ and HCO₃⁻.
Answer
A · As bicarbonate ions in plasma
4.Human physiology
Answer: B · CO occupies haemoglobin binding sites, so oxygen content falls
Why this is the answer
CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.
Answer
B · CO occupies haemoglobin binding sites, so oxygen content falls
5.Microbiology
Answer: B · 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
B · Pseudomonas aeruginosa
6.Microbiology
Answer: B · Microorganisms come from other microorganisms, not from spontaneous generation
Why this is the answer
Boiled broth in a flask with a curved neck stayed sterile because dust and microbes were trapped in the bend. When the neck was broken, microbes entered and grew, disproving spontaneous generation.
Answer
B · Microorganisms come from other microorganisms, not from spontaneous generation
7.Pharmacology
Answer: D · Naloxone
Why this is the answer
Naloxone reverses opioid-induced respiratory depression. Flumazenil reverses benzodiazepines, acetylcysteine treats paracetamol overdose and atropine treats bradycardia or organophosphate poisoning.
Answer
D · Naloxone
8.Pharmacology
Answer: C · Stop the PCA, try to rouse the patient and give naloxone as per protocol
Why this is the answer
Excessive sedation and a low respiratory rate signal opioid-induced respiratory depression. The opioid is stopped, the patient is stimulated and supported with oxygen, naloxone is given as per protocol and help is called.
Answer
C · Stop the PCA, try to rouse the patient and give naloxone as per protocol
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 28 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
|80 − 40| = 40 mmHg → 4 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·4 = 28 mmol/L
Answer
HCO₃⁻ = 28 mmol/L
10.Arterial blood gas interpretation
Answer: A–a gradient = 62.39 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.
Narrow-complex tachycardia and pauses: supraventricular tachycardia (SVT) is a regular tachycardia at about 150–250/min with narrow QRS complexes; the P waves are usually not visible because they are hidden in the QRS complex or the T wave. Sinus tachycardia in an adult at rest is usually below about 150/min and has a visible P wave before every QRS; atrial flutter with 2:1 conduction often runs at about 150/min and shows sawtooth waves. In sinus arrest the sinus node stops firing for a moment: a pause with no P wave and no QRS whose length is not a whole number of P–P cycles (a pause of exactly two or three cycles points to sinus exit block instead).
Measure
44 small boxes from R to R across the pause
Convert
pause = 44 × 0.04 s = 1.76 s
Compare with P–P
1.76 ÷ 0.72 ≈ 2.4 P–P cycles, not a whole number
Answer
pause = 1.76 s
No P wave appears in the pause, and the pause is not a whole number of P–P cycles: sinus arrest. A pause of exactly two or three cycles would point to sinus exit block.
12.ECG rhythm strip interpretation
Answer: PR interval
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 = 17 small boxes, 1500 ÷ 17 ≈ 88 per minute.
2. Rhythm: regular, every R–R is 0.68 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.30 s, constant, longer than 0.20 s.
5. QRS: 0.10 s, narrow (below 0.12 s).
First abnormal step
PR interval → first-degree AV block
Answer
PR interval
13.Evidence-based practice statistics
Answer: IR = 25.2 per 1000 person-years
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.
Person-years
PY = 480·2.5 + 150·2 + 430·2.5 = 2575
Incidence rate
IR = 65/2575·1000 = 25.2 per 1000 person-years
Answer
IR = 25.2 per 1000 person-years
14.Evidence-based practice statistics
Answer: PPV = 59.9 %
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.
True and false positives (per person)
Se·p = 0.93·0.15 = 0.1395
(1 − Sp)·(1 − p) = 0.11·0.85 = 0.0935
Positive predictive value
PPV = 0.1395/(0.1395 + 0.0935) = 0.5987 ≈ 59.9%
Answer
PPV = 59.9%
The lower the prevalence, the lower the PPV of the same test.