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 part of the brain is chiefly responsible for coordination and balance?
AHypothalamus
BMedulla oblongata
CThalamus
DCerebellum
2.Human anatomyHard
In portal hypertension, oesophageal varices form at which portosystemic anastomosis?
ABetween the superior rectal and middle rectal veins
BBetween the paraumbilical veins and the epigastric veins
CBetween the splenic vein and the left renal vein
DBetween the left gastric vein and the oesophageal tributaries of the azygos vein
3.Human physiologyMedium
What is the normal pH range of arterial blood?
A6.80 to 7.00
B7.35 to 7.45
C7.50 to 7.65
D7.00 to 7.20
4.Human physiologyHard
What is the main reason the ventricles do not contract at the same moment as the atria?
AThe Purkinje fibres conduct slowly
BThe ventricles have a longer refractory period only
CConduction is delayed at the atrioventricular node
DThe SA node fires twice per beat
5.MicrobiologyMedium
Which is a function of the complement system?
AProducing antibodies against newly encountered antigens
BMaking new red blood cells
CPresenting antigens to T cells
DForming pores in the membranes of some bacteria
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe bacteria multiply very fast inside the gut and invade the intestinal wall
BThe bacteria invade the bloodstream immediately
CThe bacteria made a heat-stable toxin in the food before it was eaten
DThe immune system reacts to the bacteria within minutes
7.PharmacologyMedium
How can the absorption of oral iron be improved?
ATake it at the same time as an antacid to protect the stomach
BTake it with a cup of strong tea
CTake it with a glass of milk
DTake it with vitamin C, such as orange juice, on an empty stomach
8.PharmacologyHard
A nurse calculates a child's dose and finds it is ten times higher than the usual range in the drug reference. What should the nurse do?
AWithhold it, recheck the calculation and clarify with the prescriber
BGive half the dose as a compromise
CGive one tenth of the dose without telling anyone
DGive the dose, because the prescriber is responsible for the order
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 27 mmHg (3.6 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
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
Interpret this lead II strip. Which interpretation can you defend from this single 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
1third-degree (complete) AV block
22:1 AV block (type undetermined)
3first-degree AV block
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 280 patients with a disease (cases) and 720 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 = 81b = 101c = 199d = 619
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 27% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.43. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.
control 27%RR = 1.43
NNH =
3Solutions
1.Human anatomy
Answer: D · Cerebellum
Why this is the answer
The cerebellum fine-tunes movement, posture and balance. Damage causes ataxia and an unsteady gait.
Answer
D · Cerebellum
2.Human anatomy
Answer: D · Between the left gastric vein and the oesophageal tributaries of the azygos vein
Why this is the answer
At the lower oesophagus, portal blood in the left gastric vein meets systemic blood in the oesophageal veins draining to the azygos system. High portal pressure dilates these veins.
Answer
D · Between the left gastric vein and the oesophageal tributaries of the azygos vein
3.Human physiology
Answer: B · 7.35 to 7.45
Why this is the answer
Arterial pH is normally 7.35–7.45. Below 7.35 is acidaemia and above 7.45 alkalaemia.
Answer
B · 7.35 to 7.45
4.Human physiology
Answer: C · Conduction is delayed at the atrioventricular node
Why this is the answer
The AV node conducts slowly (about 0.1 s delay), so atrial contraction completes ventricular filling before ventricular depolarisation begins.
Answer
C · Conduction is delayed at the atrioventricular node
5.Microbiology
Answer: D · Forming pores in the membranes of some bacteria
Why this is the answer
Complement proteins coat microbes for phagocytosis (opsonisation), attract immune cells and form the membrane attack complex that punches holes in some bacteria.
Answer
D · Forming pores in the membranes of some bacteria
6.Microbiology
Answer: C · The bacteria made a heat-stable toxin in the food before it was eaten
Why this is the answer
S. aureus produces enterotoxins in food left at warm temperatures. The toxin is already there when the food is eaten and survives reheating, so symptoms start quickly.
Answer
C · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: D · Take it with vitamin C, such as orange juice, on an empty stomach
Why this is the answer
Vitamin C keeps iron in its more absorbable ferrous form. Milk, antacids, tea and some antibiotics reduce iron absorption. If the stomach becomes upset, it may be taken with food, accepting lower absorption.
Answer
D · Take it with vitamin C, such as orange juice, on an empty stomach
8.Pharmacology
Answer: A · Withhold it, recheck the calculation and clarify with the prescriber
Why this is the answer
Doses outside the safe range must be questioned before administration. The nurse holds the dose, rechecks the weight and calculation and contacts the prescriber or pharmacist.
Answer
A · Withhold it, recheck the calculation and clarify with the prescriber
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 21.4 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
|27 − 40| = 13 mmHg → 1.3 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.3 = 21.4 mmol/L
Answer
HCO₃⁻ = 21.4 mmol/L
10.Arterial blood gas interpretation
Answer: AGc = 19.25 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.
Anion gap
AG = 137 − (103 + 21) = 13
Correction for albumin
13 + 0.25·(40 − 15) = 19.25 mmol/L
Answer
AGc = 19.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QTc = 0.396 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 = 9.5 × 0.04 = 0.38 s
RR = 23 × 0.04 = 0.92 s
Bazett’s formula
QTc = 0.38/√0.92 = 0.38/0.9592 = 0.396 s
Answer
QTc = 0.396 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (396 ms)
12.ECG rhythm strip interpretation
Answer: 2:1 AV block (type undetermined)
2:1 AV block: with two P waves for every QRS, no two conducted P waves follow each other, so you cannot see whether the PR would lengthen (type I) or stay constant (type II) before a non-conducted P wave. From one strip it is called a 2:1 AV block; a longer strip or a change in the conduction ratio decides the type. A narrow QRS makes type I more likely and a wide QRS type II, but neither proves it.
Five steps
1. Rate: R–R = 36 small boxes, 1500 ÷ 36 ≈ 42 per minute.
2. Rhythm: regular, every R–R is 1.44 s.
3. P waves: upright and regular (P–P 0.72 s, 83 per minute), two for every QRS.
4. PR interval: constant at 0.18 s in the conducted beats.
5. QRS: 0.06 s, narrow (below 0.12 s).
Why the type stays open
Every second P wave is not conducted, so no two conducted beats follow each other and a lengthening PR could never show.
2:1 AV block (type undetermined)
Answer
2:1 AV block (type undetermined)
A narrow QRS makes type I more likely and a wide QRS type II, but only a longer strip or a change in the conduction ratio decides.
13.Evidence-based practice statistics
Answer: OR = 2.49
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: 81/199 = 0.407
controls: 101/619 = 0.1632
Odds ratio
OR = (81·619)/(101·199) = 2.495 ≈ 2.49
Answer
OR = 2.49
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NNH = 9
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.