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 structure do sperm mature and get stored after leaving the testis?
ASeminal vesicle
BEpididymis
CProstate gland
DSeminiferous tubules
2.Human anatomyHard
An indirect inguinal hernia enters the inguinal canal at which point?
AThrough the deep inguinal ring, lateral to the inferior epigastric vessels
BThrough the superficial ring only, without entering the canal
CBelow the inguinal ligament, through the femoral canal
DThrough the posterior wall of the canal, medial to the inferior epigastric vessels
3.Human physiologyMedium
Which ABO blood group has both anti-A and anti-B antibodies in the plasma?
AGroup AB
BGroup O
CGroup A
DGroup B
4.Human physiologyHard
Which adaptation develops over days to weeks at high altitude?
ALower ventilation to save oxygen
BA fall in haematocrit
CIncreased renal bicarbonate retention
DHigher erythropoietin levels and more red blood cells
5.MicrobiologyMedium
Which antibody class is the first to rise in a new infection?
AIgE
BIgG
CIgM
DIgA
6.MicrobiologyHard
A patient taking high-dose immunosuppressive drugs after a transplant is due for vaccinations. Which vaccine type is generally avoided?
AToxoid vaccines
BLive attenuated vaccines
CInactivated whole-virus vaccines
DSubunit vaccines
7.PharmacologyMedium
1000 mL of IV fluid is to be given over 8 hours with a pump. What rate should be set?
A100 mL/h
B125 mL/h
C80 mL/h
D250 mL/h
8.PharmacologyHard
A woman using the combined oral contraceptive pill starts carbamazepine for epilepsy. What is the main concern?
ACarbamazepine toxicity because the pill strongly blocks its metabolism
BNo concern, because the drugs do not interact
CContraceptive failure because the hormones are metabolised faster
DSevere hypoglycaemia from the combination
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 14 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 14 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 44 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 44 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
1sinus tachycardia
2atrial flutter, 4:1 conduction
3atrial flutter, 3:1 conduction
4atrial flutter, 2:1 conduction
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
1first-degree AV block
2third-degree (complete) AV block
3second-degree AV block type I (Wenckebach)
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 730 patients received a new treatment and 870 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the absolute risk reduction ARR as a percentage. Round to one decimal place.
treatment: 23/730control: 70/870
ARR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 89% and a specificity of 99%. It is used in a population where the prevalence is 10%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 89%Sp = 99%p = 10%
PPV = %
3Solutions
1.Human anatomy
Answer: B · Epididymis
Why this is the answer
Sperm are made in the seminiferous tubules and then mature and are stored in the epididymis before passing into the ductus deferens.
Answer
B · Epididymis
2.Human anatomy
Answer: A · Through the deep inguinal ring, lateral to the inferior epigastric vessels
Why this is the answer
An indirect hernia passes through the deep ring, lateral to the inferior epigastric vessels, and runs along the canal. A direct hernia pushes through the posterior wall medial to them.
Answer
A · Through the deep inguinal ring, lateral to the inferior epigastric vessels
3.Human physiology
Answer: B · Group O
Why this is the answer
Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.
Answer
B · Group O
4.Human physiology
Answer: D · Higher erythropoietin levels and more red blood cells
Why this is the answer
Hypoxia stimulates EPO and red cell production, raising oxygen-carrying capacity. Hyperventilation causes alkalosis, which the kidneys compensate by excreting bicarbonate.
Answer
D · Higher erythropoietin levels and more red blood cells
5.Microbiology
Answer: C · IgM
Why this is the answer
IgM appears first in a primary response; IgG follows and provides long-term protection. A positive IgM therefore suggests recent infection.
Answer
C · IgM
6.Microbiology
Answer: B · Live attenuated vaccines
Why this is the answer
Live vaccines contain weakened organisms that can cause disease when the immune system is severely suppressed. Inactivated, toxoid and subunit vaccines are safe, although the response may be weaker.
Answer
B · Live attenuated vaccines
7.Pharmacology
Answer: B · 125 mL/h
Why this is the answer
1000 mL ÷ 8 h = 125 mL/h. Infusion pumps are set in mL per hour.
Answer
B · 125 mL/h
8.Pharmacology
Answer: C · Contraceptive failure because the hormones are metabolised faster
Why this is the answer
Carbamazepine induces CYP3A4, speeding up the breakdown of oestrogen and progestogen. Contraceptive levels fall and pregnancy can occur, so another method is advised.
Answer
C · Contraceptive failure because the hormones are metabolised faster
9.Arterial blood gas interpretation
Answer: PaCO₂ = 29 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 = 1.5·14 + 8 = 29 mmHg
range: 27–31 mmHg
Answer
PaCO₂ = 29 mmHg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 51.8 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
0.7·44 + 21 = 51.8 mmHg
Answer
PaCO₂ = 51.8 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: atrial flutter, 4:1 conduction
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 = 20 small boxes, 1500 ÷ 20 ≈ 75 per minute.
2. Rhythm: regular, every R–R is 0.80 s.
3. P waves: none; regular sawtooth flutter waves every 0.20 s (300 per minute), 4 for each QRS.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 4:1 conduction
Answer
atrial flutter, 4:1 conduction
12.ECG rhythm strip interpretation
Answer: second-degree AV block type II
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: the P waves are regular; R–R is regular except for a long pause where a QRS is missing.
3. P waves: upright and regular (P–P 0.80 s, 75 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: constant at 0.24 s in the conducted beats.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type II
Answer
second-degree AV block type II
Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.
13.Evidence-based practice statistics
Answer: ARR = 4.9 %
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 = 23/730 = 0.03151
CER = 70/870 = 0.08046
Absolute risk reduction
ARR = 0.08046 − 0.03151 = 0.04895 ≈ 4.9%
Answer
ARR = 4.9%
14.Evidence-based practice statistics
Answer: PPV = 90.8 %
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 = 0.89·0.1 = 0.089
(1 − Sp)·(1 − p) = 0.01·0.9 = 0.009
Positive predictive value
PPV = 0.089/(0.089 + 0.009) = 0.9082 ≈ 90.8%
Answer
PPV = 90.8%
The lower the prevalence, the lower the PPV of the same test.