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
In which part of the nephron does most reabsorption of water and sodium take place?
ADistal convoluted tubule
BProximal convoluted tubule
CCollecting duct
DLoop of Henle
2.Human anatomyHard
During a lumbar puncture, which ligament is the last one pierced before the needle enters the epidural space?
ALigamentum flavum
BSupraspinous ligament
CInterspinous ligament
DPosterior longitudinal ligament
3.Human physiologyMedium
Which cells in the stomach secrete hydrochloric acid?
AGoblet cells
BG cells
CChief cells
DParietal cells
4.Human physiologyHard
At the start of a fever, a patient shivers and feels cold even though body temperature is rising. Why?
ASweating has already cooled the body too much
BPyrogens raise the hypothalamic set point above the current body temperature
CThe body has lost its ability to produce heat
DThe thermoreceptors in the skin are destroyed
5.MicrobiologyMedium
Which organism is a Gram-negative diplococcus that causes a sexually transmitted infection?
ANeisseria gonorrhoeae
BTrichomonas vaginalis
CChlamydia trachomatis
DTreponema pallidum
6.MicrobiologyHard
Joseph Lister reduced infection after surgery in the 1860s. What was his main innovation?
AGiving sulfonamide antibiotics after every operation
BUsing carbolic acid (phenol) as an antiseptic on wounds and instruments
CUsing steam autoclaves to sterilise all surgical instruments before each operation
DVaccinating surgeons against staphylococci
7.PharmacologyMedium
A patient is starting lisinopril. Which rare but serious adverse effect should the nurse teach the patient to report immediately?
ADarker skin after exposure to sunlight
BSwelling of the lips, tongue or face
CA harmless rash on the palms of the hands
DAn increase in appetite in the evening
8.PharmacologyHard
The volume of distribution of a drug is very large (for example 500 L in an adult). What does this suggest?
AThe drug stays almost entirely in the plasma compartment
BThe drug cannot be given intravenously
CThe drug is quickly excreted by the kidneys
DThe drug is extensively taken up into tissues
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.19, PaCO₂ 19 mmHg (2.5 kPa) and HCO₃⁻ 7 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.19PaCO₂ 19 mmHg (2.5 kPa)HCO₃⁻ 7 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 144, Cl⁻ 105, HCO₃⁻ 22 mmol/L and albumin 32 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
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
This lead II strip shows a conduction problem between the atria and the ventricles. 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
1third-degree (complete) AV block
2second-degree AV block type I (Wenckebach)
3second-degree AV block type II
4first-degree AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 760 patients received a new treatment and 200 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 69/760control: 26/200
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 70% and specificity 84%. The prevalence of the disease is 28%. How many positive results n₊ are expected in total?
N = 10 000p = 28%Se = 70%Sp = 84%
n₊ =
3Solutions
1.Human anatomy
Answer: B · Proximal convoluted tubule
Why this is the answer
The proximal convoluted tubule reabsorbs about two thirds of the filtered sodium and water, plus nearly all glucose and amino acids.
Answer
B · Proximal convoluted tubule
2.Human anatomy
Answer: A · Ligamentum flavum
Why this is the answer
The needle passes skin, supraspinous and interspinous ligaments, then ligamentum flavum into the epidural space, then dura and arachnoid into the subarachnoid space.
Answer
A · Ligamentum flavum
3.Human physiology
Answer: D · Parietal cells
Why this is the answer
Parietal cells secrete HCl and intrinsic factor. Chief cells secrete pepsinogen and G cells secrete gastrin.
Answer
D · Parietal cells
4.Human physiology
Answer: B · Pyrogens raise the hypothalamic set point above the current body temperature
Why this is the answer
Pyrogens trigger prostaglandin E₂ in the hypothalamus, raising the set point. Until body temperature reaches the new set point, heat-gain responses such as shivering occur.
Answer
B · Pyrogens raise the hypothalamic set point above the current body temperature
5.Microbiology
Answer: A · Neisseria gonorrhoeae
Why this is the answer
Neisseria gonorrhoeae appears as pairs of Gram-negative cocci, often inside neutrophils. Treponema is a spirochaete, Chlamydia is an intracellular bacterium and Trichomonas is a protozoan.
Answer
A · Neisseria gonorrhoeae
6.Microbiology
Answer: B · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
Why this is the answer
Lister applied carbolic acid to wounds, dressings and instruments, which greatly reduced post-operative infection. This started antiseptic surgery.
Answer
B · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
7.Pharmacology
Answer: B · Swelling of the lips, tongue or face
Why this is the answer
ACE inhibitors can cause angioedema, which can obstruct the airway and needs emergency care. It is linked to raised bradykinin levels and can occur even after long use.
Answer
B · Swelling of the lips, tongue or face
8.Pharmacology
Answer: D · The drug is extensively taken up into tissues
Why this is the answer
A Vd far above total body water means the plasma concentration is low because most of the drug sits in tissues (for example fat or muscle), as with digoxin or amiodarone.
Answer
D · The drug is extensively taken up into tissues
9.Arterial blood gas interpretation
Answer: partially 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 7.19: acidaemia (below 7.35)
PaCO₂ 19 mmHg: low (35–45)
HCO₃⁻ 7 mmol/L: low (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: AGc = 19 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 = 144 − (105 + 22) = 17
Correction for albumin
17 + 0.25·(40 − 32) = 19 mmol/L
Answer
AGc = 19 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QTc = 0.459 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 = 11 × 0.04 = 0.44 s
RR = 23 × 0.04 = 0.92 s
Bazett’s formula
QTc = 0.44/√0.92 = 0.44/0.9592 = 0.459 s
Answer
QTc = 0.459 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (459 ms)
12.ECG rhythm strip interpretation
Answer: second-degree AV block type I (Wenckebach)
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.
Five steps
1. Rate: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 0.88 s, 68 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.16 → 0.30 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type I (Wenckebach)
Answer
second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics
Answer: RR = 0.7
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 = 69/760 = 0.09079
CER = 26/200 = 0.13
Relative risk
RR = 0.09079/0.13 = 0.6984 ≈ 0.7
Answer
RR = 0.7
14.Evidence-based practice statistics
Answer: n₊ = 3112
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.
With and without the disease
10 000·0.28 = 2800
10 000 − 2800 = 7200
All positive results
TP = 2800·0.7 = 1960
FP = 7200·0.16 = 1152
n₊ = 1960 + 1152 = 3112
Answer
n₊ = 3112
Of the 3112 positives only 1960 have the disease (PPV 63%).