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 structure produces most of the cerebrospinal fluid?
APineal gland
BPia mater
CArachnoid granulations
DChoroid plexus
2.Human anatomyHard
Why does an inhaled foreign body more often lodge in the right main bronchus than the left?
AThe right bronchus is narrower and more horizontal
BThe left bronchus is closed by the epiglottis
CThe right bronchus is wider, shorter and more vertical
DThe right lung has stronger airflow due to the heart's position
3.Human physiologyMedium
During contraction of a skeletal muscle sarcomere, which band stays the same length?
AA band
BI band
CThe whole sarcomere
DH zone
4.Human physiologyHard
Which mechanism mainly creates the concentration gradient in the medulla of the kidney?
AADH acting on the descending limb
BGlomerular filtration
CSecretion of urea into the proximal tubule
DThe countercurrent multiplier of the loop of Henle
5.MicrobiologyMedium
Toxoplasma gondii is a risk in pregnancy. Which advice reduces exposure?
AAvoid contact with dogs
BAvoid all fruit and vegetables
CAvoid swimming pools and all other public bathing areas
DAvoid cleaning cat litter and avoid undercooked meat
6.MicrobiologyHard
Why are penicillins ineffective against Mycoplasma pneumoniae?
AIt lives only inside red blood cells
BIt has a thick waxy mycolic acid layer that blocks entry
CIt produces large amounts of beta-lactamase
DIt has no cell wall for penicillin to act on
7.PharmacologyMedium
A patient wants to stop paroxetine suddenly. What should the nurse explain?
BStopping suddenly causes the drug to work better for a while
CIt can be stopped at any time, as antidepressants cause no withdrawal effects
DIt can cause dizziness, "electric shock" feelings and irritability
8.PharmacologyHard
Why are patients on long-acting nitrates advised to have a nitrate-free period of several hours each day?
ATo prevent tolerance to the nitrate
BTo let the liver recover from toxicity
CTo reduce the risk of bleeding
DTo avoid nitrate accumulation in fat tissue
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 10 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 10 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 73 mmHg (9.7 kPa) and PaCO₂ 29 mmHg (3.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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
1first-degree AV block
2normal sinus rhythm
3junctional rhythm
4sinus tachycardia
12.ECG rhythm strip interpretationHard
This lead II strip shows second-degree AV block type I (Wenckebach). Measure the PR interval of the last conducted beat before a non-conducted P wave (the longest PR), in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
longest PR = s
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 280 patients received a new treatment and 420 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.
treatment: 31/280control: 118/420
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 76% and a specificity of 89%. It is used in a population where the prevalence is 3%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 76%Sp = 89%p = 3%
PPV = %
3Solutions
1.Human anatomy
Answer: D · Choroid plexus
Why this is the answer
The choroid plexus in the ventricles secretes most CSF. CSF is reabsorbed into venous blood through the arachnoid granulations.
Answer
D · Choroid plexus
2.Human anatomy
Answer: C · The right bronchus is wider, shorter and more vertical
Why this is the answer
The right main bronchus is wider, shorter and continues more directly in line with the trachea, so aspirated objects tend to fall into it.
Answer
C · The right bronchus is wider, shorter and more vertical
3.Human physiology
Answer: A · A band
Why this is the answer
In the sliding filament model, filaments slide past each other without shortening. The A band (myosin length) stays the same while the I band and H zone shorten.
Answer
A · A band
4.Human physiology
Answer: D · The countercurrent multiplier of the loop of Henle
Why this is the answer
The ascending limb actively transports NaCl into the interstitium without letting water follow, and the countercurrent flow multiplies this into a steep gradient that ADH then uses to concentrate urine.
Answer
D · The countercurrent multiplier of the loop of Henle
5.Microbiology
Answer: D · Avoid cleaning cat litter and avoid undercooked meat
Why this is the answer
Toxoplasma oocysts are shed in cat faeces and cysts are found in undercooked meat. Infection during pregnancy can damage the fetal brain and eyes.
Answer
D · Avoid cleaning cat litter and avoid undercooked meat
6.Microbiology
Answer: D · It has no cell wall for penicillin to act on
Why this is the answer
Mycoplasma lacks a cell wall, so beta-lactams have no target. Macrolides or tetracyclines, which act on the ribosome, are used instead.
Answer
D · It has no cell wall for penicillin to act on
7.Pharmacology
Answer: D · It can cause dizziness, "electric shock" feelings and irritability
Why this is the answer
Antidepressant discontinuation syndrome is common with short half-life drugs such as paroxetine and venlafaxine. The dose is tapered gradually under supervision.
Answer
D · It can cause dizziness, "electric shock" feelings and irritability
8.Pharmacology
Answer: A · To prevent tolerance to the nitrate
Why this is the answer
Tolerance to nitrates develops quickly with constant levels. An asymmetric dosing schedule or removing patches overnight keeps them effective.
Answer
A · To prevent tolerance to the nitrate
9.Arterial blood gas interpretation
Answer: PaCO₂ = 23 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 = 1.5·10 + 8 = 23 mmHg
range: 21–25 mmHg
Answer
PaCO₂ = 23 mmHg
10.Arterial blood gas interpretation
Answer: A–a gradient = 104.7 mmHg
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.
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: R–R = 23 small boxes, 1500 ÷ 23 ≈ 65 per minute.
2. Rhythm: regular, every R–R is 0.92 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.24 s, constant, longer than 0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
first-degree AV block
Answer
first-degree AV block
12.ECG rhythm strip interpretation
Answer:longest PR = 0.38 s
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.
March out the P waves
P–P = 0.96 s, regular; one P wave in every 4 is not conducted
PR in each group
PR₁ = 0.18 s, PR₂ = 0.30 s, PR₃ = 0.38 s
Answer
longest PR = 0.38 s
The biggest step in PR comes between the first and second beats of a group; later steps are smaller, which is why the R–R intervals shorten before the pause.
13.Evidence-based practice statistics
Answer: NNT = 6
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.
Event rates
EER = 31/280 = 0.1107
CER = 118/420 = 0.281
Number needed to treat
ARR = 0.281 − 0.1107 = 0.1702
NNT = 1/0.1702 = 5.874 → 6
Answer
NNT = 6
Treating 6 patients prevents about one event.
14.Evidence-based practice statistics
Answer: PPV = 17.6 %
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.
True and false positives (per person)
Se·p = 0.76·0.03 = 0.0228
(1 − Sp)·(1 − p) = 0.11·0.97 = 0.1067
Positive predictive value
PPV = 0.0228/(0.0228 + 0.1067) = 0.1761 ≈ 17.6%
Answer
PPV = 17.6%
The lower the prevalence, the lower the PPV of the same test.