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 superficial vein in the front of the elbow is most often used for venepuncture?
AGreat saphenous vein
BMedian cubital vein
CBrachial vein
DAxillary vein
2.Human anatomyHard
Early appendicitis is often felt as vague pain around the umbilicus. Why?
AThe phrenic nerve carries the pain to the umbilicus
BThe appendix lies directly behind the umbilicus
CInflammation spreads first to the abdominal wall at the navel region
DIts visceral nerves enter the cord at T10, like the umbilical skin
3.Human physiologyMedium
Which ABO blood group has both anti-A and anti-B antibodies in the plasma?
AGroup O
BGroup A
CGroup AB
DGroup B
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt speeds up the SA node until tachycardia causes fainting
BIt hyperpolarises cardiac cells, so they can never fire
CIt blocks the binding of calcium to troponin
DIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.MicrobiologyMedium
How does an mRNA vaccine produce immunity?
AThe mRNA contains weakened live virus
BThe mRNA is inserted permanently into human DNA
CThe mRNA kills the virus directly in the blood
DBody cells use the mRNA to make a viral protein that the immune system learns to recognise
6.MicrobiologyHard
In Griffith's 1928 experiment, mice died after injection of heat-killed smooth (S) pneumococci mixed with live rough (R) pneumococci. What did this show?
AViruses carry genes between bacteria
BRough pneumococci are always more virulent than smooth ones
CA "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
DHeat does not kill pneumococci
7.PharmacologyMedium
A child weighing 20 kg is prescribed a drug at 15 mg/kg per dose. What is the dose?
A150 mg
B35 mg
C3000 mg
D300 mg
8.PharmacologyHard
During an IV glyceryl trinitrate (nitroglycerin) infusion for chest pain, the patient's blood pressure drops to 82/50 mmHg and he feels faint. What is the nurse's first action?
AStop or reduce the infusion as per protocol, lay the patient flat and notify the prescriber
BGive a sublingual nitroglycerin tablet and increase the rate to relieve the remaining chest pain
CSit the patient fully upright and give paracetamol for the headache
DIncrease the infusion rate to improve blood flow to the heart
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 55 mmHg (7.3 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic alkalosis with HCO₃⁻ 35 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 35 mmol/L
PaCO₂ = mmHg
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
Identify the rhythm on this 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
1torsades de pointes
2ventricular fibrillation
3ventricular tachycardia
4atrial flutter, 2:1 conduction
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 110 patients with a disease (cases) and 580 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 = 94b = 342c = 16d = 238
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 75% and a specificity of 91%. It is used in a population where the prevalence is 0.5%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 75%Sp = 91%p = 0.5%
PPV = %
3Solutions
1.Human anatomy
Answer: B · Median cubital vein
Why this is the answer
The median cubital vein crosses the cubital fossa between the cephalic and basilic veins. It is superficial, large and fairly fixed, so it is easy to puncture.
Answer
B · Median cubital vein
2.Human anatomy
Answer: D · Its visceral nerves enter the cord at T10, like the umbilical skin
Why this is the answer
Visceral pain from midgut structures is referred to the T10 dermatome around the umbilicus. When the parietal peritoneum becomes inflamed, pain localises to the right iliac fossa.
Answer
D · Its visceral nerves enter the cord at T10, like the umbilical skin
3.Human physiology
Answer: A · Group O
Why this is the answer
Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.
Answer
A · Group O
4.Human physiology
Answer: D · 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
D · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: D · Body cells use the mRNA to make a viral protein that the immune system learns to recognise
Why this is the answer
The mRNA is translated by cells into an antigen, such as a coronavirus spike protein. The immune system responds to it. The mRNA is broken down and does not enter the nucleus.
Answer
D · Body cells use the mRNA to make a viral protein that the immune system learns to recognise
6.Microbiology
Answer: C · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
Why this is the answer
Something from the dead S strain turned live R bacteria into capsule-forming, virulent S bacteria. Avery, MacLeod and McCarty later showed that this transforming principle was DNA.
Answer
C · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
7.Pharmacology
Answer: D · 300 mg
Why this is the answer
20 kg × 15 mg/kg = 300 mg. The result is also checked against the safe dose range and the adult maximum.
Answer
D · 300 mg
8.Pharmacology
Answer: A · Stop or reduce the infusion as per protocol, lay the patient flat and notify the prescriber
Why this is the answer
Hypotension is a major adverse effect of nitrates due to venous and arterial dilation. The infusion is stopped or reduced, the patient is positioned flat to improve venous return, and the prescriber is informed.
Answer
A · Stop or reduce the infusion as per protocol, lay the patient flat and notify the prescriber
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 29.25 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.
Change in PaCO₂
|55 − 40| = 15 mmHg → 1.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·1.5 = 29.25 mmol/L
Answer
HCO₃⁻ = 29.25 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 45.5 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.
Expected PaCO₂
0.7·35 + 21 = 45.5 mmHg
Answer
PaCO₂ = 45.5 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.49 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 = 11.5 × 0.04 = 0.46 s
RR = 22 × 0.04 = 0.88 s
Bazett’s formula
QTc = 0.46/√0.88 = 0.46/0.9381 = 0.49 s
Answer
QTc = 0.49 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (490 ms)
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: OR = 4.09
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: 94/16 = 5.875
controls: 342/238 = 1.437
Odds ratio
OR = (94·238)/(342·16) = 4.088 ≈ 4.09
Answer
OR = 4.09
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 4 %
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 = 0.75·0.005 = 0.00375
(1 − Sp)·(1 − p) = 0.09·0.995 = 0.08955
Positive predictive value
PPV = 0.00375/(0.00375 + 0.08955) = 0.04019 ≈ 4%
Answer
PPV = 4%
The lower the prevalence, the lower the PPV of the same test.