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
Where does the spinal cord normally end in an adult?
AAt the coccyx
BAt about the level of T10
CAt about the level of vertebrae L1–L2
DAt the level of the sacrum
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?
ARight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
BLoss of pain and temperature on both sides only
CLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
DLoss of all sensation and movement on the left only
3.Human physiologyMedium
How is the ejection fraction calculated?
ACardiac output ÷ heart rate
BStroke volume ÷ end-diastolic volume
CEnd-diastolic volume ÷ end-systolic volume
DEnd-systolic volume ÷ stroke volume
4.Human physiologyHard
A pulmonary embolus blocks blood flow to a well-ventilated region of lung. How does the ventilation–perfusion (V/Q) ratio of that region change?
AIt becomes low, like in pneumonia
BIt becomes zero, acting as a shunt
CIt becomes very high, acting as dead space
DIt stays normal because ventilation falls too
5.MicrobiologyMedium
In the lysogenic cycle of a bacteriophage, what happens to the phage DNA?
AIt is destroyed by the bacterium
BIt is integrated into the bacterial chromosome as a prophage
CIt leaves the cell through the pilus
DIt is immediately used to make new phages that burst the cell
6.MicrobiologyHard
Several people vomit within a few hours of eating fried rice that was cooked the day before and kept at room temperature. Which organism is the classic cause?
AListeria monocytogenes
BClostridium botulinum
CBacillus cereus
DSalmonella Typhi
7.PharmacologyMedium
How should a hyoscine (scopolamine) patch for motion sickness be used?
ASwallow it if no water is available
BCut it in half for children without advice
CApply it behind the ear several hours before travel
DApply it to the chest only after the vomiting has started
8.PharmacologyHard
A patient is switched from oral to intravenous morphine. Why is the IV dose lower than the oral dose?
AThe oral route causes tolerance faster than the IV route
BIV morphine is a stronger chemical form of the drug
COral morphine has a large first-pass effect, so less reaches the blood
DIV morphine is excreted much more slowly by the kidney than oral morphine
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.54, PaCO₂ 51 mmHg (6.8 kPa) and HCO₃⁻ 42 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.54PaCO₂ 51 mmHg (6.8 kPa)HCO₃⁻ 42 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 138, Cl⁻ 100 and HCO₃⁻ 13 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 138 mmol/LCl⁻ 100 mmol/LHCO₃⁻ 13 mmol/L
delta ratio =
11.ECG rhythm strip interpretationMedium
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
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
1first-degree AV block
2sinus bradycardia
3second-degree AV block type I (Wenckebach)
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 60 patients with a disease (cases) and 660 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 = 22b = 106c = 38d = 554
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 47% in relative terms (RRR = 47%). In a patient group the risk without treatment (CER) is 18% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 47%CER = 18%
NNT =
3Solutions
1.Human anatomy
Answer: C · At about the level of vertebrae L1–L2
Why this is the answer
In adults the cord ends at the conus medullaris near L1–L2. Below that lies the cauda equina, which is why lumbar punctures are done at L3–L4 or L4–L5.
Answer
C · At about the level of vertebrae L1–L2
2.Human anatomy
Answer: C · 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
C · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: B · Stroke volume ÷ end-diastolic volume
Why this is the answer
Ejection fraction is the proportion of blood in the full ventricle that is ejected per beat. A normal value is about 55–70%.
Answer
B · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: C · It becomes very high, acting as dead space
Why this is the answer
Ventilation without perfusion gives a very high V/Q ratio (dead space). A shunt, such as in pneumonia, is the opposite: perfusion without ventilation.
Answer
C · It becomes very high, acting as dead space
5.Microbiology
Answer: B · It is integrated into the bacterial chromosome as a prophage
Why this is the answer
In lysogeny the phage genome integrates into the host chromosome and is copied with it. Later it may switch to the lytic cycle, making new phages and lysing the cell.
Answer
B · It is integrated into the bacterial chromosome as a prophage
6.Microbiology
Answer: C · Bacillus cereus
Why this is the answer
Bacillus cereus spores survive boiling. If cooked rice is left warm, the bacteria grow and make a heat-stable emetic toxin, so reheating does not make the rice safe.
Answer
C · Bacillus cereus
7.Pharmacology
Answer: C · Apply it behind the ear several hours before travel
Why this is the answer
The patch is applied behind the ear several hours before travel, and the hands are washed afterwards. Touching the eye with drug on the fingers can dilate the pupil and blur vision.
Answer
C · Apply it behind the ear several hours before travel
8.Pharmacology
Answer: C · Oral morphine has a large first-pass effect, so less reaches the blood
Why this is the answer
Much of an oral morphine dose is metabolised by the liver before reaching the systemic circulation. Equianalgesic tables are used, so the IV dose is a fraction of the oral dose.
Answer
C · Oral morphine has a large first-pass effect, so less reaches the blood
9.Arterial blood gas interpretation
Answer: partially compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.54: alkalaemia (above 7.45)
PaCO₂ 51 mmHg: high (35–45)
HCO₃⁻ 42 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: delta ratio = 1.182
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.
Anion gap
AG = 138 − (100 + 13) = 25
Delta ratio
(25 − 12)/(24 − 13) = 13/11 = 1.182
Answer
delta ratio = 1.182
Between 1 and 2: a pure high anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: pause = 2.2 s
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
55 small boxes from R to R across the pause
Convert
pause = 55 × 0.04 s = 2.2 s
Compare with P–P
2.20 ÷ 0.96 ≈ 2.3 P–P cycles, not a whole number
Answer
pause = 2.2 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: second-degree AV block type II
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: the P waves are regular; R–R is regular except for a long pause where a QRS is missing.
3. P waves: upright and regular (P–P 0.84 s, 71 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: constant at 0.16 s in the conducted beats.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
second-degree AV block type II
Answer
second-degree AV block type II
Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.
13.Evidence-based practice statistics
Answer: OR = 3.03
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: 22/38 = 0.5789
controls: 106/554 = 0.1913
Odds ratio
OR = (22·554)/(106·38) = 3.026 ≈ 3.03
Answer
OR = 3.03
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NNT = 12
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.
Absolute risk reduction
ARR = 0.18·0.47 = 0.0846
Number needed to treat
NNT = 1/0.0846 = 11.82 → 12
Answer
NNT = 12
The same relative effect gives a smaller NNT when the baseline risk is higher.