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 type of epithelium lines most of the trachea?
ASimple cuboidal epithelium
BPseudostratified ciliated columnar epithelium
CTransitional epithelium
DStratified squamous epithelium
2.Human anatomyHard
Injury to the common fibular (peroneal) nerve at the neck of the fibula typically causes:
ALoss of sensation on the sole of the foot
BFoot drop
CWeak hip abduction
DInability to flex the knee
3.Human physiologyMedium
Which ion enters the cell to cause the depolarisation phase of a typical neuronal action potential?
APotassium (K⁺)
BCalcium (Ca²⁺)
CChloride (Cl⁻)
DSodium (Na⁺)
4.Human physiologyHard
Why does left ventricular failure cause pulmonary oedema?
AThe right ventricle pumps less blood into the lungs
BRaised left atrial pressure increases pulmonary capillary hydrostatic pressure
CPlasma oncotic pressure in the lungs rises sharply
DThe bronchial arteries become blocked
5.MicrobiologyMedium
Tetanus and diphtheria vaccines are toxoids. What does that mean?
AThey contain live bacteria that cannot make toxin
BThey contain an inactivated toxin that still triggers an immune response
CThey contain ready-made antibodies against the toxin, giving passive immunity
DThey contain the bacterial DNA only
6.MicrobiologyHard
Corynebacterium diphtheriae causes diphtheria only if it produces diphtheria toxin. When does it make this toxin?
AOnly after it has conjugated with E. coli and received a plasmid
BOnly when it is grown at high temperature
CWhenever it is exposed to antibiotics
DWhen it carries a bacteriophage with the toxin gene
7.PharmacologyMedium
How can the absorption of oral iron be improved?
ATake it with a glass of milk
BTake it with a cup of strong tea
CTake it at the same time as an antacid to protect the stomach
DTake it with vitamin C, such as orange juice, on an empty stomach
8.PharmacologyHard
A patient who uses insulin is taught sick-day rules. Which statement shows a need for further teaching?
A"If I am ill and not eating, I will stop my insulin completely."
B"I will seek help if I keep vomiting or have ketones."
C"I will keep drinking fluids, even if I cannot eat much."
D"I will check my blood glucose and ketones more often while I am ill."
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 8 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 8 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows pH 7.32, PaCO₂ 36 mmHg (4.8 kPa) and HCO₃⁻ 18 mmol/L, with Na⁺ 136 and Cl⁻ 98 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
Interpret this lead II rhythm strip. 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
1atrial fibrillation
2atrial flutter, 2:1 conduction
3sinus rhythm with PVCs
4sinus arrhythmia
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the lead II strip. 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
1second-degree AV block type II
2idioventricular rhythm
3paced rhythm, failure to capture
4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 4800 person-years with 557 new cases, and unexposed people 4500 person-years with 131 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 557/4800 PYnot exposed: 131/4500 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 85% and specificity 75%. The prevalence of the disease is 1%. How many false-positive results FP are expected?
N = 10 000p = 1%Se = 85%Sp = 75%
FP =
3Solutions
1.Human anatomy
Answer: B · Pseudostratified ciliated columnar epithelium
Why this is the answer
Respiratory epithelium is pseudostratified ciliated columnar with goblet cells. Mucus traps particles and cilia move it up towards the pharynx.
Answer
B · Pseudostratified ciliated columnar epithelium
2.Human anatomy
Answer: B · Foot drop
Why this is the answer
The common fibular nerve supplies the dorsiflexors and evertors. Damage at the fibular neck causes foot drop and a high-stepping gait. The sole is supplied by the tibial nerve.
Answer
B · Foot drop
3.Human physiology
Answer: D · Sodium (Na⁺)
Why this is the answer
Voltage-gated Na⁺ channels open and Na⁺ rushes in, depolarising the membrane. K⁺ outflow then repolarises it.
Answer
D · Sodium (Na⁺)
4.Human physiology
Answer: B · Raised left atrial pressure increases pulmonary capillary hydrostatic pressure
Why this is the answer
A failing left ventricle backs up pressure into the left atrium and pulmonary veins. High capillary hydrostatic pressure pushes fluid into the lung interstitium and alveoli.
Answer
B · Raised left atrial pressure increases pulmonary capillary hydrostatic pressure
5.Microbiology
Answer: B · They contain an inactivated toxin that still triggers an immune response
Why this is the answer
A toxoid is a toxin treated (for example with formaldehyde) so it is harmless but still antigenic. The immune system makes antibodies that neutralise the real toxin.
Answer
B · They contain an inactivated toxin that still triggers an immune response
6.Microbiology
Answer: D · When it carries a bacteriophage with the toxin gene
Why this is the answer
The tox gene is carried by a bacteriophage. When the phage is integrated as a prophage, the bacterium can produce toxin. Strains without the phage are non-toxigenic.
Answer
D · When it carries a bacteriophage with the toxin gene
7.Pharmacology
Answer: D · Take it with vitamin C, such as orange juice, on an empty stomach
Why this is the answer
Vitamin C keeps iron in its more absorbable ferrous form. Milk, antacids, tea and some antibiotics reduce iron absorption. If the stomach becomes upset, it may be taken with food, accepting lower absorption.
Answer
D · Take it with vitamin C, such as orange juice, on an empty stomach
8.Pharmacology
Answer: A · "If I am ill and not eating, I will stop my insulin completely."
Why this is the answer
Stress hormones during illness often raise glucose, so basal insulin is usually continued and doses may need adjusting. Stopping insulin in type 1 diabetes can lead to ketoacidosis.
Answer
A · "If I am ill and not eating, I will stop my insulin completely."
9.Arterial blood gas interpretation
Answer: PaCO₂ = 20 mmHg
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.
Winter’s formula
PaCO₂ = 1.5·8 + 8 = 20 mmHg
range: 18–22 mmHg
Answer
PaCO₂ = 20 mmHg
10.Arterial blood gas interpretation
Answer: pure high-gap acidosis
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 = 136 − (98 + 18) = 20
Delta ratio
(20 − 12)/(24 − 18) = 1.333
pure high-gap acidosis
Answer
pure high-gap acidosis
11.ECG rhythm strip interpretation
Answer: atrial fibrillation
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Five steps
1. Rate: 10 QRS complexes in 6 s, × 10 = 100 per minute.
2. Rhythm: irregularly irregular, R–R from 0.40 to 0.78 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: paced rhythm, failure to capture
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: 6 QRS complexes in 6 s, × 10 = 60 per minute.
2. Rhythm: the pacing spikes are regular (every 0.84 s), but the R–R is not, because a QRS is missing after some spikes.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.16 s and wide after the captured spikes; one spike is not followed by a QRS (no capture).
Interpretation
paced rhythm, failure to capture
Answer
paced rhythm, failure to capture
Failure to capture: the pacemaker fires but the heart does not respond. Textbooks treat it as urgent to report, because the patient may depend on the pacemaker.
13.Evidence-based practice statistics
Answer: IRR = 3.99
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.
Incidence rates
exposed: 557/4800 = 116 per 1000 person-years
not exposed: 131/4500 = 29.11 per 1000 person-years
Rate ratio
IRR = 116/29.11 = 3.986 ≈ 3.99
Answer
IRR = 3.99
14.Evidence-based practice statistics
Answer: FP = 2475
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
10 000·0.01 = 100
10 000 − 100 = 9900
False positives
FP = 9900·0.25 = 2475
Answer
FP = 2475
Of the 2560 positives only 85 have the disease (PPV 3.3%).