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 is the largest nerve in the human body?
AFemoral nerve
BMedian nerve
CSciatic nerve
DVagus nerve
2.Human anatomyHard
Where does the thoracic duct usually empty its lymph into the blood?
ARight atrium directly
BJunction of the right subclavian and right internal jugular veins
CInferior vena cava at the level of the kidneys
DJunction of the left subclavian and left internal jugular veins
3.Human physiologyMedium
Which hormone makes the gallbladder contract after a fatty meal?
ACholecystokinin (CCK)
BInsulin
CGastrin
DSecretin
4.Human physiologyHard
In the renin–angiotensin–aldosterone system, where is angiotensin I mainly converted to angiotensin II?
AIn the kidneys' juxtaglomerular cells
BIn the adrenal cortex
CIn the lungs
DIn the liver
5.MicrobiologyMedium
Which cells does HIV mainly infect and destroy?
ANeutrophils
BRed blood cells
CPlatelets
DCD4 T helper cells
6.MicrobiologyHard
How do retroviruses such as HIV copy their genome into the host DNA?
AThe viral RNA is translated into DNA by ribosomes
BHost DNA polymerase copies the RNA without help
CThe viral DNA is copied directly by host RNA polymerase
DReverse transcriptase makes DNA from viral RNA, and integrase inserts it
7.PharmacologyMedium
What makes an overdose of tricyclic antidepressants such as amitriptyline especially dangerous?
ACardiac arrhythmias and seizures
BSevere liver failure within hours
CSudden high blood glucose
DKidney stones
8.PharmacologyHard
A patient receives high-dose IV aciclovir for herpes encephalitis. Which nursing action reduces the main risk to the kidneys?
AKeep the patient well hydrated and give the infusion slowly
BGive a loop diuretic before each dose
CGive the infusion as a rapid push
DRestrict oral and IV fluids to prevent brain swelling
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 129 mmol/L, glucose 20.3 mmol/L and urea 24.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 129 mmol/Lglucose 20.3 mmol/Lurea 24.6 mmol/L
Osm = mOsm/kg
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
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
1junctional rhythm
2sinus tachycardia
3first-degree AV block
4sinus bradycardia
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
1second-degree AV block type II
2second-degree AV block type I (Wenckebach)
3third-degree (complete) AV block
4first-degree AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 87% and a specificity of 63%. Find the negative likelihood ratio LR−. Round to two decimal places.
Se = 87%Sp = 63%
LR− =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 31% in relative terms (RRR = 31%). In a patient group the risk without treatment (CER) is 35% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 31%CER = 35%
NNT =
3Solutions
1.Human anatomy
Answer: C · Sciatic nerve
Why this is the answer
The sciatic nerve (L4–S3) leaves the pelvis through the greater sciatic foramen and supplies the hamstrings and all the muscles below the knee. Skin on the inner side of the leg is supplied by the saphenous nerve instead.
Answer
C · Sciatic nerve
2.Human anatomy
Answer: D · Junction of the left subclavian and left internal jugular veins
Why this is the answer
The thoracic duct drains lymph from both legs, the abdomen and the left side of the upper body into the left venous angle. The right lymphatic duct drains the rest into the right venous angle.
Answer
D · Junction of the left subclavian and left internal jugular veins
3.Human physiology
Answer: A · Cholecystokinin (CCK)
Why this is the answer
Fat and protein in the duodenum release CCK, which contracts the gallbladder and stimulates pancreatic enzyme secretion.
Answer
A · Cholecystokinin (CCK)
4.Human physiology
Answer: C · In the lungs
Why this is the answer
Renin (kidney) cleaves angiotensinogen (liver) to angiotensin I, which ACE, mainly on pulmonary capillary endothelium, converts to angiotensin II.
Answer
C · In the lungs
5.Microbiology
Answer: D · CD4 T helper cells
Why this is the answer
HIV binds CD4 and a co-receptor on T helper cells. Loss of these cells weakens both antibody and cell-mediated immunity, leading to AIDS.
Answer
D · CD4 T helper cells
6.Microbiology
Answer: D · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
Why this is the answer
Reverse transcriptase converts the RNA genome into DNA, which integrase inserts into the host genome as a provirus. Both enzymes are drug targets.
Answer
D · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
7.Pharmacology
Answer: A · Cardiac arrhythmias and seizures
Why this is the answer
Tricyclics block cardiac sodium channels (widening the QRS complex) and have anticholinergic effects. In overdose they cause arrhythmias, hypotension, seizures and coma.
Answer
A · Cardiac arrhythmias and seizures
8.Pharmacology
Answer: A · Keep the patient well hydrated and give the infusion slowly
Why this is the answer
Aciclovir can crystallise in the tubules and cause acute kidney injury. Good hydration, slow infusion and monitoring of kidney function reduce this risk.
Answer
A · Keep the patient well hydrated and give the infusion slowly
9.Arterial blood gas interpretation
Answer: Osm = 303 mOsm/kg
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.
Calculated osmolality
Osm = 2·129 + 20.3 + 24.6 = 302.9 ≈ 303 mOsm/kg
Answer
Osm = 303 mOsm/kg
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: first-degree AV block
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: R–R = 25 small boxes, 1500 ÷ 25 ≈ 60 per minute.
2. Rhythm: regular, every R–R is 1.00 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.30 s, constant, longer than 0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
first-degree AV block
Answer
first-degree AV block
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 4 is not followed by a QRS.
4. PR interval: lengthens 0.20 → 0.32 → 0.38 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: LR− = 0.21
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.
Negative likelihood ratio
LR− = (1 − 0.87)/0.63 = 0.2063 ≈ 0.21
Answer
LR− = 0.21
14.Evidence-based practice statistics
Answer: NNT = 10
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.
Absolute risk reduction
ARR = 0.35·0.31 = 0.1085
Number needed to treat
NNT = 1/0.1085 = 9.217 → 10
Answer
NNT = 10
The same relative effect gives a smaller NNT when the baseline risk is higher.