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 structure produces most of the cerebrospinal fluid?
AArachnoid granulations
BPineal gland
CChoroid plexus
DPia mater
2.Human anatomyHard
A typical posterolateral disc herniation between L4 and L5 most often compresses which nerve root?
AL3
BL5
CS1
DL4
3.Human physiologyMedium
Which protein forms the mesh that stabilises a blood clot?
AThrombin
BAlbumin
CCollagen
DFibrin
4.Human physiologyHard
Which mechanism mainly creates the concentration gradient in the medulla of the kidney?
AGlomerular filtration
BSecretion of urea into the proximal tubule
CADH acting on the descending limb
DThe countercurrent multiplier of the loop of Henle
5.MicrobiologyMedium
Which organism causes tuberculosis, and what makes it hard to stain with the Gram method?
ALegionella pneumophila, which lives inside amoebae
BMycoplasma pneumoniae, which has no cell wall to take up the stain
CStreptococcus pneumoniae, which has a capsule
DMycobacterium tuberculosis, which has a waxy mycolic acid wall
6.MicrobiologyHard
The minimum inhibitory concentration (MIC) of an antibiotic is:
AThe lowest concentration that prevents visible bacterial growth in vitro
BThe plasma level at which side effects begin
CThe highest dose a patient can tolerate
DThe lowest concentration that kills 99.9% of the bacterial inoculum in vitro
7.PharmacologyMedium
Warfarin works by:
ADirectly inhibiting thrombin
BBlocking the vitamin K-dependent synthesis of clotting factors
CActivating antithrombin
DBlocking platelet aggregation through irreversible inhibition of COX-1
8.PharmacologyHard
A ventilated ICU patient receives vecuronium, a neuromuscular blocker. Why must sedation and pain relief also be given?
ANeuromuscular blockers paralyse muscles but do not cause sedation or analgesia
BSedatives are needed to prevent vecuronium from causing seizures and agitation
CVecuronium causes severe pain at the injection site
DVecuronium stops working unless a sedative is also given
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 50 mmHg (6.7 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⁻.
A practice gas shows a respiratory acidosis: pH 7.34, PaCO₂ 62 mmHg (8.3 kPa), HCO₃⁻ 32 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.34PaCO₂ 62 mmHg (8.3 kPa)HCO₃⁻ 32 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = 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). A case-control study compares 290 patients with a disease (cases) and 570 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 53b = 205c = 237d = 365
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 70% and a specificity of 85%. It is used in a population where the prevalence is 10%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 70%Sp = 85%p = 10%
PPV = %
3Solutions
1.Human anatomy
Answer: C · Choroid plexus
Why this is the answer
The choroid plexus in the ventricles secretes most CSF. CSF is reabsorbed into venous blood through the arachnoid granulations.
Answer
C · Choroid plexus
2.Human anatomy
Answer: B · L5
Why this is the answer
A posterolateral L4–L5 herniation usually catches the traversing L5 root, causing weakness of big toe extension and numbness on the top of the foot.
Answer
B · L5
3.Human physiology
Answer: D · Fibrin
Why this is the answer
Thrombin converts soluble fibrinogen into insoluble fibrin strands, which bind the platelet plug and trap red cells to form a stable clot.
Answer
D · Fibrin
4.Human physiology
Answer: D · The countercurrent multiplier of the loop of Henle
Why this is the answer
The ascending limb actively transports NaCl into the interstitium without letting water follow, and the countercurrent flow multiplies this into a steep gradient that ADH then uses to concentrate urine.
Answer
D · The countercurrent multiplier of the loop of Henle
5.Microbiology
Answer: D · Mycobacterium tuberculosis, which has a waxy mycolic acid wall
Why this is the answer
M. tuberculosis has a cell wall rich in mycolic acids, so it resists Gram staining and is identified with acid-fast stains.
Answer
D · Mycobacterium tuberculosis, which has a waxy mycolic acid wall
6.Microbiology
Answer: A · 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
A · The lowest concentration that prevents visible bacterial growth in vitro
7.Pharmacology
Answer: B · Blocking the vitamin K-dependent synthesis of clotting factors
Why this is the answer
Warfarin inhibits vitamin K epoxide reductase, reducing synthesis of factors II, VII, IX and X. It is monitored with the INR and reversed with vitamin K.
Answer
B · Blocking the vitamin K-dependent synthesis of clotting factors
8.Pharmacology
Answer: A · Neuromuscular blockers paralyse muscles but do not cause sedation or analgesia
Why this is the answer
Neuromuscular blockers cause paralysis but leave awareness and pain sensation intact. Without sedation and analgesia, the patient could be awake and in pain while unable to move.
Answer
A · Neuromuscular blockers paralyse muscles but do not cause sedation or analgesia
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 25 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
|50 − 40| = 10 mmHg → 1 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·1 = 25 mmol/L
Answer
HCO₃⁻ = 25 mmol/L
10.Arterial blood gas interpretation
Answer: chronic
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 HCO₃⁻
acute: 24 + 1·2.2 = 26.2 mmol/L
chronic: 24 + 3.5·2.2 = 31.7 mmol/L
Compare
measured: 32 mmol/L → chronic
Answer
chronic
11.ECG rhythm strip interpretation
Answer: QTc = 0.48 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 11.5 × 0.04 = 0.46 s
RR = 23 × 0.04 = 0.92 s
Bazett’s formula
QTc = 0.46/√0.92 = 0.46/0.9592 = 0.48 s
Answer
QTc = 0.48 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (480 ms)
12.ECG rhythm strip interpretation
Answer: rhythm
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: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.62 to 0.94 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rhythm → atrial fibrillation
Answer
rhythm
13.Evidence-based practice statistics
Answer: OR = 0.4
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.
Odds of exposure
cases: 53/237 = 0.2236
controls: 205/365 = 0.5616
Odds ratio
OR = (53·365)/(205·237) = 0.3982 ≈ 0.4
Answer
OR = 0.4
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: PPV = 34.1 %
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 = 0.7·0.1 = 0.07
(1 − Sp)·(1 − p) = 0.15·0.9 = 0.135
Positive predictive value
PPV = 0.07/(0.07 + 0.135) = 0.3415 ≈ 34.1%
Answer
PPV = 34.1%
The lower the prevalence, the lower the PPV of the same test.