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
CChoroid plexus
DArachnoid granulations
2.Human anatomyHard
In which part of the small intestine are vitamin B12 and bile salts mainly absorbed?
ADuodenum
BCaecum
CJejunum
DTerminal ileum
3.Human physiologyMedium
What causes the first heart sound (S1)?
AClosure of the aortic and pulmonary valves
BClosure of the mitral and tricuspid valves
COpening of the aortic valve
DBlood filling the atria
4.Human physiologyHard
Which combination of signs is expected in a patient in early hypovolaemic shock after major bleeding?
ATachycardia, cool pale skin and reduced urine output
BBradycardia, cool pale skin and high urine output
CBradycardia, warm flushed skin and high urine output
DTachycardia, warm flushed skin and high urine output
5.MicrobiologyMedium
Which statement is one of Koch's postulates?
AThe organism must be found in healthy people as well
BThe disease must respond to antibiotics
CThe organism must be visible with the naked eye
DThe organism must cause the same disease when introduced into a healthy susceptible host
6.MicrobiologyHard
Why must a negative-sense RNA virus such as influenza carry its own RNA-dependent RNA polymerase into the host cell?
AHost ribosomes cannot read any viral genes
BThe polymerase is needed to integrate the genome into host DNA
CThe polymerase protects the virus from antibodies
DHost cells cannot make mRNA from an RNA template
7.PharmacologyMedium
Why do patients taking clozapine need regular blood counts?
AIt can cause agranulocytosis
BIt changes the blood group
CIt often causes iron-deficiency anaemia
DIt raises the platelet count dangerously
8.PharmacologyHard
A drug is ordered at 5 micrograms/kg/min for an 80 kg patient. The infusion contains 1,600 micrograms per mL. What rate in mL/h is needed?
A25 mL/h
B150 mL/h
C4 mL/h
D15 mL/h
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.16, PaCO₂ 70 mmHg (9.3 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.
pH 7.16PaCO₂ 70 mmHg (9.3 kPa)HCO₃⁻ 24 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 70 mmHg (9.3 kPa) and PaCO₂ 32 mmHg (4.3 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 Fridericia’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
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). In a cohort study, exposed people contributed 3900 person-years with 111 new cases, and unexposed people 1900 person-years with 36 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 111/3900 PYnot exposed: 36/1900 PY
IRR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 67% and specificity 83% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 5%Se = 67%Sp = 83%positive
p′ = %
3Solutions
1.Human anatomy
Answer: C · 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
C · Choroid plexus
2.Human anatomy
Answer: D · Terminal ileum
Why this is the answer
The terminal ileum absorbs vitamin B12 (bound to intrinsic factor) and reabsorbs bile salts. The duodenum mainly absorbs iron.
Answer
D · Terminal ileum
3.Human physiology
Answer: B · Closure of the mitral and tricuspid valves
Why this is the answer
S1 ("lub") is caused by closure of the atrioventricular valves at the start of systole. S2 ("dub") is closure of the semilunar valves.
Answer
B · Closure of the mitral and tricuspid valves
4.Human physiology
Answer: A · Tachycardia, cool pale skin and reduced urine output
Why this is the answer
Baroreceptor-driven sympathetic activation raises heart rate and constricts skin vessels. Reduced renal perfusion, ADH and aldosterone lower urine output.
Answer
A · Tachycardia, cool pale skin and reduced urine output
5.Microbiology
Answer: D · The organism must cause the same disease when introduced into a healthy susceptible host
Why this is the answer
Koch's postulates say the microbe must be found in every case, grown in pure culture, cause the disease in a healthy host and be recovered again from that host. Some pathogens cannot be cultured, so the postulates have limits.
Answer
D · The organism must cause the same disease when introduced into a healthy susceptible host
6.Microbiology
Answer: D · Host cells cannot make mRNA from an RNA template
Why this is the answer
A negative-sense genome must first be copied into positive-sense mRNA. Human cells have no enzyme that copies RNA from RNA, so the virus brings its own polymerase.
Answer
D · Host cells cannot make mRNA from an RNA template
7.Pharmacology
Answer: A · It can cause agranulocytosis
Why this is the answer
Clozapine can cause severe neutropenia (agranulocytosis). Regular white cell and neutrophil counts allow treatment to be stopped before life-threatening infection occurs.
Answer
A · It can cause agranulocytosis
8.Pharmacology
Answer: D · 15 mL/h
Why this is the answer
5 × 80 = 400 micrograms/min, × 60 = 24,000 micrograms/h, ÷ 1,600 micrograms/mL = 15 mL/h. Weight-based infusions are double-checked because errors are easy to make.
Answer
D · 15 mL/h
9.Arterial blood gas interpretation
Answer: uncompensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.16: acidaemia (below 7.35)
PaCO₂ 70 mmHg: high (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)
Compensation
uncompensated
Answer
uncompensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 61.12 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 = 9.5 × 0.04 = 0.38 s
RR = 25 × 0.04 = 1.00 s
Fridericia’s formula
QTc = 0.38/∛1.00 = 0.38/1 = 0.38 s
Answer
QTc = 0.38 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (380 ms)
12.ECG rhythm strip interpretation
Answer: atrial rate = 79 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 = 19 small boxes = 0.76 s → 1500 ÷ 19 = 78.95 ≈ 79 per minute
Ventricular rate
R–R = 34 small boxes = 1.36 s → 1500 ÷ 34 = 44.12 ≈ 44 per minute
Answer
atrial rate = 79 bpm
Answer
atrial rate = 79 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
13.Evidence-based practice statistics
Answer: IRR = 1.5
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.
Incidence rates
exposed: 111/3900 = 28.46 per 1000 person-years
not exposed: 36/1900 = 18.95 per 1000 person-years
Rate ratio
IRR = 28.46/18.95 = 1.502 ≈ 1.5
Answer
IRR = 1.5
14.Evidence-based practice statistics
Answer: p′ = 17.2 %
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.