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 bone of the skull contains the foramen magnum?
ATemporal bone
BParietal bone
CSphenoid bone
DOccipital bone
2.Human anatomyHard
A lesion of the left optic tract causes which visual field defect?
ARight homonymous hemianopia
BBlindness in the left eye only
CLeft homonymous hemianopia
DBitemporal hemianopia
3.Human physiologyMedium
According to Poiseuille's law, if the radius of an arteriole halves, its resistance to flow:
AIncreases about 16-fold
BIncreases 4-fold
CHalves
DDoubles
4.Human physiologyHard
After a head injury, a patient passes large volumes of very dilute urine and plasma sodium rises. What is the most likely cause?
ALack of ADH release (cranial diabetes insipidus)
BExcess aldosterone
CExcess ADH release (SIADH)
DUncontrolled diabetes mellitus
5.MicrobiologyMedium
In the chain of infection, what is the "reservoir"?
AThe route by which the pathogen leaves the infected host's body
BThe way the pathogen enters a new host
CThe person who is most at risk of infection
DThe place where the pathogen normally lives and multiplies
6.MicrobiologyHard
How do high-risk HPV types contribute to cervical cancer?
AThe virus kills cells so quickly that they are replaced by cancer cells
BViral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
CThe virus produces an endotoxin that damages DNA
DThe virus destroys CD4 T cells in the cervix
7.PharmacologyMedium
After naloxone reverses an opioid overdose, why must the patient be watched closely?
ANaloxone makes the opioid stay longer in the body
BNaloxone may wear off before the opioid does, so breathing can slow again
CNaloxone often causes severe low blood glucose a few hours after the dose
DNaloxone itself causes respiratory depression at high doses
8.PharmacologyHard
A teenager took a large overdose of diphenhydramine. He is agitated, hot, flushed and confused, with dry skin, dilated pupils and a full bladder. What is this pattern?
ACholinergic toxidrome
BOpioid toxidrome
CSerotonin syndrome
DAnticholinergic toxidrome
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 149 mmol/L, glucose 21.5 mmol/L and urea 18 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 149 mmol/Lglucose 21.5 mmol/Lurea 18 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 77 mmHg (10.3 kPa) and PaCO₂ 58 mmHg (7.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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
1sinus bradycardia
2junctional rhythm
3third-degree (complete) AV block
4asystole
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the positive likelihood ratio LR+. Round to two decimal places.
a = 69b = 115c = 41d = 245
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 27% in relative terms (RRR = 27%). In a patient group the risk without treatment (CER) is 37% 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 = 27%CER = 37%
NNT =
3Solutions
1.Human anatomy
Answer: D · Occipital bone
Why this is the answer
The foramen magnum is a large opening in the occipital bone through which the medulla oblongata continues as the spinal cord.
Answer
D · Occipital bone
2.Human anatomy
Answer: A · Right homonymous hemianopia
Why this is the answer
The left optic tract carries fibres from the left half of each retina, which see the right visual field. Its lesion causes loss of the right half of the field in both eyes.
Answer
A · Right homonymous hemianopia
3.Human physiology
Answer: A · Increases about 16-fold
Why this is the answer
Resistance is proportional to 1/r⁴, so halving the radius raises resistance by 2⁴ = 16. This is why small changes in arteriole diameter control blood flow so strongly.
Answer
A · Increases about 16-fold
4.Human physiology
Answer: A · Lack of ADH release (cranial diabetes insipidus)
Why this is the answer
Without ADH the collecting ducts stay impermeable to water, so dilute urine is produced and the patient loses free water, raising plasma sodium.
Answer
A · Lack of ADH release (cranial diabetes insipidus)
5.Microbiology
Answer: D · The place where the pathogen normally lives and multiplies
Why this is the answer
The chain of infection links the agent, reservoir, portal of exit, mode of transmission, portal of entry and susceptible host. Breaking any link stops transmission.
Answer
D · The place where the pathogen normally lives and multiplies
6.Microbiology
Answer: B · Viral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
Why this is the answer
E6 promotes the breakdown of p53 and E7 binds Rb, removing controls on cell division and DNA repair. Infected cells can then accumulate mutations and become cancerous over years.
Answer
B · Viral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
7.Pharmacology
Answer: B · Naloxone may wear off before the opioid does, so breathing can slow again
Why this is the answer
Naloxone acts for a shorter time than many opioids, especially long-acting and slow-release ones. Repeated doses or an infusion may be needed.
Answer
B · Naloxone may wear off before the opioid does, so breathing can slow again
8.Pharmacology
Answer: D · Anticholinergic toxidrome
Why this is the answer
Anticholinergic toxicity causes hyperthermia, dry flushed skin, mydriasis, urinary retention, tachycardia and delirium. Dry skin helps distinguish it from serotonin syndrome and sympathomimetic poisoning, which cause sweating.
Answer
D · Anticholinergic toxidrome
9.Arterial blood gas interpretation
Answer: Osm = 338 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 = 2·149 + 21.5 + 18 = 337.5 ≈ 338 mOsm/kg
Answer
Osm = 338 mOsm/kg
10.Arterial blood gas interpretation
Answer: A–a gradient = 21.62 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.
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 = 26 × 0.04 = 1.04 s
Bazett’s formula
QTc = 0.46/√1.04 = 0.46/1.02 = 0.451 s
Answer
QTc = 0.451 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (451 ms)
12.ECG rhythm strip interpretation
Answer: asystole
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: no countable ventricular rate.
2. Rhythm: none; an almost flat line.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: no QRS complexes.
Interpretation
asystole
Answer
asystole
A flat line is always checked: look at the patient, the leads and the gain, and view a second lead, because a loose electrode or fine VF can look like asystole.
13.Evidence-based practice statistics
Answer: LR+ = 1.96
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.
Sensitivity and specificity
Se = 69/110 = 0.6273
Sp = 245/360 = 0.6806
Positive likelihood ratio
LR+ = 0.6273/(1 − 0.6806) = 1.964 ≈ 1.96
Answer
LR+ = 1.96
14.Evidence-based practice statistics
Answer: NNT = 11
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 = 0.37·0.27 = 0.0999
Number needed to treat
NNT = 1/0.0999 = 10.01 → 11
Answer
NNT = 11
The same relative effect gives a smaller NNT when the baseline risk is higher.