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 head of the femur articulates with which structure?
AGlenoid cavity
BGreater trochanter
CObturator foramen
DAcetabulum
2.Human anatomyHard
Why does a displaced fracture of the femoral neck carry a high risk of avascular necrosis of the femoral head?
AIt damages the sciatic nerve that controls blood flow
BIt tears the retinacular branches of the medial circumflex femoral artery
CIt cuts the main trunk of the femoral artery
DIt blocks the femoral vein and causes congestion
3.Human physiologyMedium
Which protein forms the mesh that stabilises a blood clot?
AFibrin
BAlbumin
CCollagen
DThrombin
4.Human physiologyHard
A patient with diabetic ketoacidosis breathes deeply and rapidly. What pattern is expected on the blood gas?
AHigh pH, low bicarbonate and low PaCO₂
BLow pH, high bicarbonate and high PaCO₂
CLow pH, low bicarbonate and low PaCO₂
DLow pH, normal bicarbonate and high PaCO₂
5.MicrobiologyMedium
How does an autoclave sterilise instruments?
AWith ultraviolet light only
BWith alcohol vapour at room temperature
CWith steam under pressure at about 121 °C
DWith dry heat at about 60 °C
6.MicrobiologyHard
Which mechanism of antibiotic resistance does a bacterium use when it pumps the drug out of the cell?
AEnzymatic inactivation
BTarget modification
CEfflux pumps
DReduced porin uptake only
7.PharmacologyMedium
A patient with diabetes is sweaty and shaky. Blood glucose is 3.0 mmol/L (54 mg/dL). The patient is awake and can swallow. What should the nurse do first?
AGive a large meal high in fat and protein to keep glucose up for longer
BGive 15–20 g of fast-acting carbohydrate, such as juice
CGive the next scheduled dose of insulin
DGive intramuscular glucagon at once
8.PharmacologyHard
A patient has SIADH with mild hyponatraemia and no neurological symptoms. What is usually the first-line treatment?
AA rapid bolus of hypertonic saline
BDesmopressin
CLarge volumes of IV glucose 5%
DFluid restriction
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 134 mmol/L, glucose 20.4 mmol/L and urea 7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 134 mmol/Lglucose 20.4 mmol/Lurea 7 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.19, PaCO₂ 19 mmHg (2.5 kPa), HCO₃⁻ 7 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.19PaCO₂ 19 mmHg (2.5 kPa)HCO₃⁻ 7 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
11.ECG rhythm strip interpretationMedium
This lead II strip shows atrial flutter. Measure the flutter-wave interval and find the atrial rate in beats per minute.
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
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 250 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 = 102b = 145c = 148d = 105
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 20% in relative terms (RRR = 20%). In a patient group the risk without treatment (CER) is 28% 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 = 20%CER = 28%
NNT =
3Solutions
1.Human anatomy
Answer: D · Acetabulum
Why this is the answer
The acetabulum is the cup-shaped socket of the hip bone, formed by the ilium, ischium and pubis. The glenoid cavity is the shoulder's socket.
Answer
D · Acetabulum
2.Human anatomy
Answer: B · It tears the retinacular branches of the medial circumflex femoral artery
Why this is the answer
The femoral head is supplied mainly by retinacular branches of the medial circumflex femoral artery running along the neck. A displaced neck fracture tears them and the head can die.
Answer
B · It tears the retinacular branches of the medial circumflex femoral artery
3.Human physiology
Answer: A · 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: C · With steam under pressure at about 121 °C
Why this is the answer
Pressurised steam reaches about 121 °C and kills all microorganisms, including spores, typically in about 15 minutes or more depending on the load.
Answer
C · With steam under pressure at about 121 °C
6.Microbiology
Answer: C · Efflux pumps
Why this is the answer
Efflux pumps actively export antibiotics such as tetracyclines and fluoroquinolones, keeping the inside concentration too low to work.
Answer
C · Efflux pumps
7.Pharmacology
Answer: B · Give 15–20 g of fast-acting carbohydrate, such as juice
Why this is the answer
Conscious hypoglycaemia is treated with 15–20 g of fast-acting carbohydrate, with glucose rechecked after about 15 minutes. Fat slows absorption, and glucagon is kept for patients who cannot swallow.
Answer
B · Give 15–20 g of fast-acting carbohydrate, such as juice
8.Pharmacology
Answer: D · Fluid restriction
Why this is the answer
In SIADH, excess ADH makes the kidneys retain water. Restricting fluid intake raises sodium gradually; hypertonic saline is kept for severe symptomatic cases, and hypotonic fluids and desmopressin would worsen it.
Answer
D · Fluid restriction
9.Arterial blood gas interpretation
Answer: Osm = 295 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 = 2·134 + 20.4 + 7 = 295.4 ≈ 295 mOsm/kg
Answer
Osm = 295 mOsm/kg
10.Arterial blood gas interpretation
Answer: appropriate compensation
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 PaCO₂
1.5·7 + 8 = 18.5 → 16.5–20.5 mmHg
Compare
measured: 19 mmHg → appropriate compensation
Answer
appropriate compensation
11.ECG rhythm strip interpretation
Answer: atrial rate = 250 bpm
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.
Flutter waves
F–F = 6 small boxes = 0.24 s → 1500 ÷ 6 = 250 per minute
QRS complexes
R–R = 24 small boxes = 0.96 s → 1500 ÷ 24 = 63 per minute
conduction ratio = 250 ÷ 63 = 4:1
Answer
atrial rate = 250 bpm
12.ECG rhythm strip interpretation
Answer: atrial rate = 83 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 = 18 small boxes = 0.72 s → 1500 ÷ 18 = 83.33 ≈ 83 per minute
Ventricular rate
6 QRS complexes in 6 s × 10 = 60 per minute
Answer
atrial rate = 83 bpm
Answer
atrial rate = 83 bpm
Second-degree type II: 3 P waves for every 2 QRS complexes, with a constant PR.
13.Evidence-based practice statistics
Answer: OR = 0.5
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: 102/148 = 0.6892
controls: 145/105 = 1.381
Odds ratio
OR = (102·105)/(145·148) = 0.4991 ≈ 0.5
Answer
OR = 0.5
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: NNT = 18
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 = 0.28·0.2 = 0.056
Number needed to treat
NNT = 1/0.056 = 17.86 → 18
Answer
NNT = 18
The same relative effect gives a smaller NNT when the baseline risk is higher.