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 sympathetic nervous system leaves the central nervous system from which spinal segments?
ACervical only
BThoracic and upper lumbar (T1–L2)
CSacral only (S2–S4)
DCranial and sacral
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?
ALoss of pain and temperature on both sides only
BLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
CRight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
DLoss of all sensation and movement on the left only
3.Human physiologyMedium
During which phase of the cardiac cycle are all four heart valves closed while ventricular pressure rises?
ARapid ejection
BIsovolumetric contraction
CAtrial systole
DRapid filling
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
BIt blocks the binding of calcium to troponin
CIt speeds up the SA node until tachycardia causes fainting
DIt hyperpolarises cardiac cells, so they can never fire
5.MicrobiologyMedium
Which is a function of the complement system?
AForming pores in the membranes of some bacteria
BPresenting antigens to T cells
CMaking new red blood cells
DProducing antibodies against newly encountered antigens
6.MicrobiologyHard
Which process can produce a completely new influenza A subtype that causes a pandemic?
AAntigenic shift through reassortment of gene segments
BTransformation with free DNA
CAntigenic drift through gradual accumulation of point mutations
DLysogenic conversion by a bacteriophage
7.PharmacologyMedium
Why is directly observed therapy (DOT) often used for tuberculosis?
ATB drugs must always be given intravenously in hospital
BThe drugs work only if a nurse is present
CPatients with TB cannot swallow tablets
DTreatment lasts months, and missed doses cause resistance
8.PharmacologyHard
Methotrexate for rheumatoid arthritis is taken once a week. A patient mistakenly takes the same dose every day. What is the main danger?
ABone marrow suppression and mucositis from methotrexate toxicity
BSevere hypertension from fluid retention
CHypoglycaemia because methotrexate strongly stimulates insulin release
DThe arthritis gets worse because of drug tolerance
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 137 mmol/L, glucose 12.5 mmol/L and urea 23.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 137 mmol/Lglucose 12.5 mmol/Lurea 23.6 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 146, Cl⁻ 106, HCO₃⁻ 21 mmol/L and albumin 17 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = s
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. The QRS complexes are regular: find the ventricular 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
ventricular rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 400 patients with a disease (cases) and 650 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 = 282b = 319c = 118d = 331
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 97% and specificity 75%. The prevalence of the disease is 2%. How many false-positive results FP are expected?
N = 10 000p = 2%Se = 97%Sp = 75%
FP =
3Solutions
1.Human anatomy
Answer: B · Thoracic and upper lumbar (T1–L2)
Why this is the answer
Sympathetic preganglionic neurons lie in the lateral horn from T1 to L2. The parasympathetic system has a craniosacral outflow.
Answer
B · Thoracic and upper lumbar (T1–L2)
2.Human anatomy
Answer: B · 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
B · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: B · Isovolumetric contraction
Why this is the answer
After the AV valves close, the ventricles contract against closed semilunar valves. Pressure rises with no volume change until the aortic and pulmonary valves open.
Answer
B · Isovolumetric contraction
4.Human physiology
Answer: A · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
Why this is the answer
High extracellular potassium makes the resting potential less negative. Many fast sodium channels then become inactivated, slowing conduction and risking fatal arrhythmias.
Answer
A · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: A · Forming pores in the membranes of some bacteria
Why this is the answer
Complement proteins coat microbes for phagocytosis (opsonisation), attract immune cells and form the membrane attack complex that punches holes in some bacteria.
Answer
A · Forming pores in the membranes of some bacteria
6.Microbiology
Answer: A · Antigenic shift through reassortment of gene segments
Why this is the answer
When two different influenza A viruses infect the same cell, their segments can mix, giving new haemagglutinin or neuraminidase combinations. Few people are immune, so pandemics can follow. Drift causes the smaller yearly changes.
Answer
A · Antigenic shift through reassortment of gene segments
7.Pharmacology
Answer: D · Treatment lasts months, and missed doses cause resistance
Why this is the answer
Standard TB treatment usually lasts six months or more with several drugs. Poor adherence leads to relapse and multidrug-resistant TB, so supervised dosing improves completion.
Answer
D · Treatment lasts months, and missed doses cause resistance
8.Pharmacology
Answer: A · Bone marrow suppression and mucositis from methotrexate toxicity
Why this is the answer
Methotrexate inhibits dihydrofolate reductase. Daily dosing causes toxicity in dividing tissues such as bone marrow and gut lining, which can be fatal. Folinic acid is used as a rescue treatment.
Answer
A · Bone marrow suppression and mucositis from methotrexate toxicity
9.Arterial blood gas interpretation
Answer: Osm = 310 mOsm/kg
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.
Calculated osmolality
Osm = 2·137 + 12.5 + 23.6 = 310.1 ≈ 310 mOsm/kg
Answer
Osm = 310 mOsm/kg
10.Arterial blood gas interpretation
Answer: AGc = 24.75 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.
Anion gap
AG = 146 − (106 + 21) = 19
Correction for albumin
19 + 0.25·(40 − 17) = 24.75 mmol/L
Answer
AGc = 24.75 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QT = 0.34 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
8.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 8.5 × 0.04 s = 0.34 s
Answer
QT = 0.34 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.80 s).
12.ECG rhythm strip interpretation
Answer: ventricular rate = 44 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 = 15 small boxes = 0.60 s → 1500 ÷ 15 = 100 ≈ 100 per minute
Ventricular rate
R–R = 34 small boxes = 1.36 s → 1500 ÷ 34 = 44.12 ≈ 44 per minute
Answer
ventricular rate = 44 bpm
Answer
ventricular rate = 44 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
13.Evidence-based practice statistics
Answer: OR = 2.48
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: 282/118 = 2.39
controls: 319/331 = 0.9637
Odds ratio
OR = (282·331)/(319·118) = 2.48
Answer
OR = 2.48
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: FP = 2450
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
10 000·0.02 = 200
10 000 − 200 = 9800
False positives
FP = 9800·0.25 = 2450
Answer
FP = 2450
Of the 2644 positives only 194 have the disease (PPV 7.3%).