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 artery is most commonly used to feel the pulse at the wrist?
ARadial artery
BUlnar artery
CAxillary artery
DBrachial artery
2.Human anatomyHard
A stab wound cuts the left half of the spinal cord at T10. Which pattern of loss is expected below the lesion?
ALoss of pain and temperature on both sides only
BLoss of all sensation and movement on the left only
CLeft-sided weakness and loss of proprioception, right-sided loss of pain and temperature
DRight-sided weakness and loss of proprioception, left-sided loss of pain and temperature
3.Human physiologyMedium
Why do myelinated axons conduct action potentials faster than unmyelinated ones?
AThe impulse jumps between nodes of Ranvier
BMyelin contains more sodium channels
CMyelin makes the axon thinner
DMyelin releases neurotransmitter along the axon
4.Human physiologyHard
Constriction of the efferent arteriole of a glomerulus, with the afferent arteriole unchanged, tends to:
ALower glomerular capillary pressure and GFR
BHave no effect on GFR at all
CRaise glomerular capillary pressure and GFR
DStop filtration completely
5.MicrobiologyMedium
Why are pregnant women advised to avoid unpasteurised soft cheeses?
AThey may contain tapeworm eggs
BThey often contain the hepatitis C virus, which crosses the placenta
CThey contain too much lactose for the fetus
DThey may contain Listeria, which can harm the fetus
6.MicrobiologyHard
Which hepatitis viruses are mainly spread through contaminated food and water?
AHepatitis B and D
BHepatitis B and C
CHepatitis A and E
DHepatitis C and E
7.PharmacologyMedium
An IV site is cool, swollen and pale, and the infusion is running slowly. What should the nurse do?
AIncrease the infusion rate to push through and clear the blockage
BStop the infusion, remove the cannula and elevate the limb
CFlush the cannula forcefully with saline
DApply a tight bandage over the site
8.PharmacologyHard
A patient with angina is taught to use sublingual glyceryl trinitrate (nitroglycerin). Which statement shows a need for further teaching?
A"I will replace my tablets before the expiry date."
B"I will sit down before I take the tablet, because it can make me feel faint or dizzy."
C"I may get a headache after taking it."
D"If the pain is no better after one tablet, I will wait an hour before calling for help."
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.44, PaCO₂ 52 mmHg (6.9 kPa) and HCO₃⁻ 34 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.44PaCO₂ 52 mmHg (6.9 kPa)HCO₃⁻ 34 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 142, Cl⁻ 112, HCO₃⁻ 15 mmol/L and albumin 28 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Interpret this lead II rhythm strip. Which rhythm is it? 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 arrhythmia
2atrial flutter, 4:1 conduction
3atrial fibrillation
4atrial flutter, 2:1 conduction
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 randomised trial, 780 patients received a new treatment and 790 received the control treatment. The table shows how many in each group had a side effect. Find the number needed to harm NNH: how many patients must be treated for one extra patient to have the side effect? Round up to a whole number of patients.
treatment: 316/780control: 213/790
NNH =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 78% and a specificity of 83%. 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 = 78%Sp = 83%p = 0.5%
PPV = %
3Solutions
1.Human anatomy
Answer: A · Radial artery
Why this is the answer
The radial artery lies superficially on the thumb side of the wrist, against the radius, which makes the pulse easy to feel.
Answer
A · Radial artery
2.Human anatomy
Answer: C · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
Why this is the answer
In Brown-Séquard syndrome the corticospinal tract and dorsal columns are lost on the same side, while the spinothalamic tract, which crosses near entry, causes contralateral pain and temperature loss.
Answer
C · Left-sided weakness and loss of proprioception, right-sided loss of pain and temperature
3.Human physiology
Answer: A · The impulse jumps between nodes of Ranvier
Why this is the answer
Myelin insulates the axon, so action potentials are only generated at the nodes of Ranvier. This saltatory conduction is much faster.
Answer
A · The impulse jumps between nodes of Ranvier
4.Human physiology
Answer: C · Raise glomerular capillary pressure and GFR
Why this is the answer
Efferent constriction holds blood in the glomerulus, raising its pressure and GFR. Angiotensin II acts this way, which is why ACE inhibitors can lower GFR.
Answer
C · Raise glomerular capillary pressure and GFR
5.Microbiology
Answer: D · They may contain Listeria, which can harm the fetus
Why this is the answer
Listeria monocytogenes can grow in chilled foods and cross the placenta. It can cause miscarriage, stillbirth or severe neonatal infection.
Answer
D · They may contain Listeria, which can harm the fetus
6.Microbiology
Answer: C · Hepatitis A and E
Why this is the answer
Hepatitis A and E spread by the faecal–oral route. B, C and D spread through blood and body fluids and can cause chronic infection.
Answer
C · Hepatitis A and E
7.Pharmacology
Answer: B · Stop the infusion, remove the cannula and elevate the limb
Why this is the answer
These signs suggest infiltration. The infusion is stopped and the cannula removed, the limb is elevated, and a new cannula is placed elsewhere; vesicant drugs need the extravasation protocol.
Answer
B · Stop the infusion, remove the cannula and elevate the limb
8.Pharmacology
Answer: D · "If the pain is no better after one tablet, I will wait an hour before calling for help."
Why this is the answer
Unrelieved chest pain may be a heart attack, so help must not be delayed. In US teaching, emergency services are called if pain is not better 5 minutes after the first dose; UK advice is to repeat the dose after 5 minutes and call 999 if pain persists 5 minutes after the second. Sitting down prevents falls from hypotension, and headache is a common effect.
Answer
D · "If the pain is no better after one tablet, I will wait an hour before calling for help."
9.Arterial blood gas interpretation
Answer: fully compensated
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 7.44: within 7.35–7.45
PaCO₂ 52 mmHg: high (35–45)
HCO₃⁻ 34 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 42.8–46.8 mmHg, so the measured 52 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: AGc = 18 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 = 142 − (112 + 15) = 15
Correction for albumin
15 + 0.25·(40 − 28) = 18 mmol/L
Answer
AGc = 18 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: atrial fibrillation
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Five steps
1. Rate: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.46 to 1.00 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: atrial rate = 100 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 = 15 small boxes = 0.60 s → 1500 ÷ 15 = 100 ≈ 100 per minute
Ventricular rate
R–R = 34 small boxes = 1.36 s → 1500 ÷ 34 = 44.12 ≈ 44 per minute
Answer
atrial rate = 100 bpm
Answer
atrial rate = 100 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
13.Evidence-based practice statistics
Answer: NNH = 8
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.
Event rates
EER = 316/780 = 0.4051
CER = 213/790 = 0.2696
Number needed to harm
ARI = 0.4051 − 0.2696 = 0.1355
NNH = 1/0.1355 = 7.38 → 8
Answer
NNH = 8
About one in every 8 treated patients has the side effect because of the treatment.
14.Evidence-based practice statistics
Answer: PPV = 2.3 %
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 = 0.78·0.005 = 0.0039
(1 − Sp)·(1 − p) = 0.17·0.995 = 0.1692
Positive predictive value
PPV = 0.0039/(0.0039 + 0.1692) = 0.02254 ≈ 2.3%
Answer
PPV = 2.3%
The lower the prevalence, the lower the PPV of the same test.