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 muscle is the main extensor of the elbow?
ACoracobrachialis
BBiceps brachii
CBrachialis
DTriceps brachii
2.Human anatomyHard
What is the correct order of structures that urine passes through after leaving the collecting ducts?
AMajor calyx, minor calyx, renal pelvis, ureter
BRenal pelvis, minor calyx, major calyx, ureter
CMinor calyx, major calyx, renal pelvis, ureter
DMinor calyx, renal pelvis, major calyx, ureter
3.Human physiologyMedium
Which hormonal event triggers ovulation?
AA rise in progesterone from the corpus luteum
BA fall in oestrogen
CA surge of luteinising hormone (LH)
DA surge of prolactin
4.Human physiologyHard
Which spirometry finding is typical of chronic obstructive pulmonary disease (COPD)?
AIncreased peak expiratory flow
BNormal FEV1 with low FVC
CFEV1/FVC ratio below 0.7
DFEV1/FVC ratio above 0.9 with low FVC
5.MicrobiologyMedium
In the chain of infection, what is the "reservoir"?
AThe place where the pathogen normally lives and multiplies
BThe person who is most at risk of infection
CThe way the pathogen enters a new host
DThe route by which the pathogen leaves the infected host's body
6.MicrobiologyHard
How do high-risk HPV types contribute to cervical cancer?
AThe virus destroys CD4 T cells in the cervix
BViral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
CThe virus kills cells so quickly that they are replaced by cancer cells
DThe virus produces an endotoxin that damages DNA
7.PharmacologyMedium
Which adverse effect of doxorubicin requires monitoring of heart function?
APericarditis caused by infection
BCumulative heart muscle damage leading to heart failure
CSevere hypertension during the infusion itself
DRapid rise in cholesterol
8.PharmacologyHard
Five days after starting clindamycin, a patient has frequent watery diarrhoea, abdominal cramps and a fever. What should the nurse suspect and do?
AA normal side effect; give loperamide and continue as usual
BFood poisoning; encourage a high-fibre diet and plenty of milk products
CC. difficile infection; send stool, use contact precautions, avoid antidiarrhoeals
DAn allergy to clindamycin; give an antihistamine and continue the antibiotic
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 8 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 8 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 38 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 38 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = s
12.ECG rhythm strip interpretationHard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? 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
1ventricular fibrillation
2asystole
3pulseless electrical activity (PEA)
4sinus tachycardia
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 68% and a specificity of 91%. Find the positive likelihood ratio LR+. Round to two decimal places.
Se = 68%Sp = 91%
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 93% and specificity 86%. The prevalence of the disease is 22%. How many true-positive results TP are expected?
N = 10 000p = 22%Se = 93%Sp = 86%
TP =
3Solutions
1.Human anatomy
Answer: D · Triceps brachii
Why this is the answer
Triceps brachii inserts on the olecranon of the ulna and extends the elbow. Biceps and brachialis are elbow flexors.
Answer
D · Triceps brachii
2.Human anatomy
Answer: C · Minor calyx, major calyx, renal pelvis, ureter
Why this is the answer
Collecting ducts open at the renal papillae into minor calyces, which join to form major calyces, which empty into the renal pelvis and then the ureter.
Answer
C · Minor calyx, major calyx, renal pelvis, ureter
3.Human physiology
Answer: C · A surge of luteinising hormone (LH)
Why this is the answer
Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.
Answer
C · A surge of luteinising hormone (LH)
4.Human physiology
Answer: C · FEV1/FVC ratio below 0.7
Why this is the answer
Narrowed airways slow expiration, so FEV1 falls more than FVC and the ratio drops below 0.7. In COPD this is not fully reversed by bronchodilators.
Answer
C · FEV1/FVC ratio below 0.7
5.Microbiology
Answer: A · 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
A · 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 · Cumulative heart muscle damage leading to heart failure
Why this is the answer
Anthracyclines such as doxorubicin can damage heart muscle, and the risk increases with cumulative dose. Heart function is assessed before and during treatment; the drug can also turn urine red.
Answer
B · Cumulative heart muscle damage leading to heart failure
8.Pharmacology
Answer: C · C. difficile infection; send stool, use contact precautions, avoid antidiarrhoeals
Why this is the answer
Broad-spectrum antibiotics, especially clindamycin, allow C. difficile to overgrow and release toxins. Stool is tested, contact precautions with soap-and-water handwashing are used, and antidiarrhoeals are avoided because they keep toxin in the gut.
Answer
C · C. difficile infection; send stool, use contact precautions, avoid antidiarrhoeals
9.Arterial blood gas interpretation
Answer: PaCO₂ = 20 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 = 1.5·8 + 8 = 20 mmHg
range: 18–22 mmHg
Answer
PaCO₂ = 20 mmHg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 47.6 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
0.7·38 + 21 = 47.6 mmHg
Answer
PaCO₂ = 47.6 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QT = 0.36 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
9 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 9 × 0.04 s = 0.36 s
Answer
QT = 0.36 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.92 s).
12.ECG rhythm strip interpretation
Answer: pulseless electrical activity (PEA)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 13 small boxes, 1500 ÷ 13 ≈ 115 per minute.
2. Rhythm: regular, every R–R is 0.52 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
sinus tachycardia
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
pulseless electrical activity (PEA)
Answer
pulseless electrical activity (PEA)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: LR+ = 7.56
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.
Positive likelihood ratio
LR+ = 0.68/(1 − 0.91) = 7.556 ≈ 7.56
Answer
LR+ = 7.56
14.Evidence-based practice statistics
Answer: TP = 2046
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.