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 cranial nerve supplies the trapezius and sternocleidomastoid muscles?
AVagus nerve (X)
BFacial nerve (VII)
CHypoglossal nerve (XII)
DAccessory nerve (XI)
2.Human anatomyHard
Which muscle "unlocks" the fully extended knee by rotating the femur laterally on the tibia at the start of flexion?
ABiceps femoris
BPopliteus
CSartorius
DGastrocnemius
3.Human physiologyMedium
What is the main effect of vagus nerve stimulation on the heart?
AIt increases the heart rate
BIt increases ventricular contractility strongly
CIt slows the heart rate
DIt constricts the coronary arteries
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
ANo change because the valve is not part of the ventricle
BThinning of the wall with dilation
CConcentric hypertrophy with a thicker wall
DAtrophy of the muscle due to less work
5.MicrobiologyMedium
Which antibody class is the first to rise in a new infection?
AIgE
BIgM
CIgG
DIgA
6.MicrobiologyHard
A patient taking high-dose immunosuppressive drugs after a transplant is due for vaccinations. Which vaccine type is generally avoided?
AInactivated whole-virus vaccines
BToxoid vaccines
CSubunit vaccines
DLive attenuated vaccines
7.PharmacologyMedium
Why should a patient rinse the mouth after using an inhaled corticosteroid?
ATo stop the drug being absorbed too quickly into the blood
BTo make the drug work faster in the lungs
CTo prevent a fall in blood glucose
DTo reduce the risk of oral thrush (candidiasis) and hoarseness
8.PharmacologyHard
A non-competitive (irreversible) antagonist changes a dose–response curve by:
AShifting the curve to the right with the same maximum
BMaking the curve steeper without other changes
CShifting the curve to the left
DLowering the maximum response
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 139 mmol/L, glucose 28.7 mmol/L and urea 4.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 139 mmol/Lglucose 28.7 mmol/Lurea 4.6 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.23, PaCO₂ 84 mmHg (11.2 kPa), HCO₃⁻ 34 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 7.23PaCO₂ 84 mmHg (11.2 kPa)HCO₃⁻ 34 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
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). A case-control study compares 250 patients with a disease (cases) and 110 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 = 62b = 19c = 188d = 91
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 72% and a specificity of 99%. It is used in a population where the prevalence is 3%. Find the negative predictive value NPV as a percentage. Round to one decimal place.
Se = 72%Sp = 99%p = 3%
NPV = %
3Solutions
1.Human anatomy
Answer: D · Accessory nerve (XI)
Why this is the answer
The accessory nerve (XI) supplies sternocleidomastoid and trapezius. Damage weakens head turning and shoulder shrugging.
Answer
D · Accessory nerve (XI)
2.Human anatomy
Answer: B · Popliteus
Why this is the answer
At full extension the knee locks by medial rotation of the femur. Popliteus laterally rotates the femur on the fixed tibia to unlock it so flexion can begin.
Answer
B · Popliteus
3.Human physiology
Answer: C · It slows the heart rate
Why this is the answer
The vagus releases acetylcholine on muscarinic receptors in the SA and AV nodes, slowing pacemaker firing and AV conduction.
Answer
C · It slows the heart rate
4.Human physiology
Answer: C · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.
Answer
C · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: B · IgM
Why this is the answer
IgM appears first in a primary response; IgG follows and provides long-term protection. A positive IgM therefore suggests recent infection.
Answer
B · IgM
6.Microbiology
Answer: D · Live attenuated vaccines
Why this is the answer
Live vaccines contain weakened organisms that can cause disease when the immune system is severely suppressed. Inactivated, toxoid and subunit vaccines are safe, although the response may be weaker.
Answer
D · Live attenuated vaccines
7.Pharmacology
Answer: D · To reduce the risk of oral thrush (candidiasis) and hoarseness
Why this is the answer
Steroid left in the mouth and throat suppresses local immunity and can cause candidiasis and a hoarse voice. Rinsing and spitting, and using a spacer, reduce this risk.
Answer
D · To reduce the risk of oral thrush (candidiasis) and hoarseness
8.Pharmacology
Answer: D · Lowering the maximum response
Why this is the answer
A competitive antagonist shifts the curve to the right but higher agonist doses reach the same maximum. A non-competitive antagonist reduces the maximum response.
Answer
D · Lowering the maximum response
9.Arterial blood gas interpretation
Answer: Osm = 311 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 = 2·139 + 28.7 + 4.6 = 311.3 ≈ 311 mOsm/kg
Answer
Osm = 311 mOsm/kg
10.Arterial blood gas interpretation
Answer: acute on 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 + 1·4.4 = 28.4 mmol/L
chronic: 24 + 3.5·4.4 = 39.4 mmol/L
Compare
measured: 34 mmol/L → acute on chronic
Answer
acute on chronic
11.ECG rhythm strip interpretation
Answer: QTc = 0.439 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 = 8.5 × 0.04 = 0.34 s
RR = 15 × 0.04 = 0.60 s
Bazett’s formula
QTc = 0.34/√0.60 = 0.34/0.7746 = 0.439 s
Answer
QTc = 0.439 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (439 ms)
12.ECG rhythm strip interpretation
Answer: atrial rate = 100 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
atrial rate = 100 bpm
Answer
atrial rate = 100 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
13.Evidence-based practice statistics
Answer: OR = 1.58
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: 62/188 = 0.3298
controls: 19/91 = 0.2088
Odds ratio
OR = (62·91)/(19·188) = 1.58
Answer
OR = 1.58
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NPV = 99.1 %
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.