Medical Sciences: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Arterial blood gas interpretation

Normal ranges
pH 7.35–7.45, PaCO₂ 35–45 mmHg (4.7–6.0 kPa), HCO₃⁻ 22–26 mmol/L
ROME
Respiratory: PaCO₂ opposite to pH; Metabolic: HCO₃⁻ equal to pH
Winter’s formula
PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2
Metabolic alkalosis
PaCO₂ ≈ 0.7·HCO₃⁻ + 21
Anion gap
AG = Na⁺ − (Cl⁻ + HCO₃⁻) ≈ 8–12
Albumin-corrected gap
AG + 2.5·(4.0 − alb g/dL) = AG + 0.25·(40 − alb g/L)
Delta ratio
(AG − 12)/(24 − HCO₃⁻)
P/F ratio
PaO₂/FiO₂
Alveolar gas equation
PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8
Serum osmolality
2·Na⁺ + glucose + urea (2·Na⁺ + glucose/18 + BUN/2.8)

Method

  1. Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
  2. Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
  3. 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.
  4. In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
  5. 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

  1. 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.
  2. Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
  3. 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.
  4. 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.
  5. 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

  1. Draw the 2 × 2 table and label a, b, c and d before you calculate.
  2. Read sensitivity and specificity down the columns, predictive values along the rows.
  3. For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
  4. For a likelihood ratio, convert the probability to odds, multiply, and convert back.
  5. 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 cranial nerve controls most of the muscles of facial expression?

  1. ATrigeminal nerve (V)
  2. BFacial nerve (VII)
  3. CHypoglossal nerve (XII)
  4. DOculomotor nerve (III)
2.Human anatomyHard

Which muscle is the main flexor of the hip joint?

  1. AAdductor magnus
  2. BGluteus maximus
  3. CIliopsoas
  4. DBiceps femoris
3.Human physiologyMedium

What does the QRS complex on an ECG represent?

  1. AAtrial repolarisation
  2. BVentricular depolarisation
  3. CAtrial depolarisation
  4. DVentricular repolarisation
4.Human physiologyHard

Why is severe hyperkalaemia dangerous for the heart?

  1. AIt blocks the binding of calcium to troponin
  2. BIt hyperpolarises cardiac cells, so they can never fire
  3. CIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
  4. DIt speeds up the SA node until tachycardia causes fainting
5.MicrobiologyMedium

Why are pregnant women advised to avoid unpasteurised soft cheeses?

  1. AThey contain too much lactose for the fetus
  2. BThey may contain tapeworm eggs
  3. CThey often contain the hepatitis C virus, which crosses the placenta
  4. DThey may contain Listeria, which can harm the fetus
6.MicrobiologyHard

A teenager has fever, neck stiffness and a non-blanching purple rash. CSF shows Gram-negative diplococci. Which organism is most likely?

  1. AStreptococcus pneumoniae
  2. BListeria monocytogenes
  3. CEscherichia coli
  4. DNeisseria meningitidis
7.PharmacologyMedium

A patient starting niacin (nicotinic acid) complains of flushing and warmth after doses. Which advice may reduce this?

  1. ATake the niacin with a hot drink so that it dissolves and acts faster
  2. BTake aspirin 30 minutes before the niacin, if the prescriber agrees
  3. CDouble the dose to build tolerance faster
  4. DTake it on an empty stomach in a hot bath
8.PharmacologyHard

A patient who takes clonidine stopped it suddenly two days ago. He is now agitated, has a headache and his blood pressure is very high. What is the most likely explanation?

  1. ANormal variation unrelated to the medication
  2. BClonidine toxicity due to slow excretion by the kidneys
  3. CRebound hypertension from abrupt clonidine withdrawal
  4. DAn allergic reaction to clonidine
9.Arterial blood gas interpretationMedium

A practice gas shows a metabolic acidosis with HCO₃⁻ 18 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.

HCO₃⁻ 18 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 21% oxygen shows PaO₂ 94 mmHg (12.5 kPa) and PaCO₂ 29 mmHg (3.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 21%PaO₂ 94 mmHg (12.5 kPa)PaCO₂ 29 mmHg (3.9 kPa)
A–a gradient = mmHg
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVsupraventricular tachycardia (SVT)atrial fibrillationaccelerated junctional rhythmsinus tachycardia

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
  1. 1supraventricular tachycardia (SVT)
  2. 2atrial fibrillation
  3. 3accelerated junctional rhythm
  4. 4sinus tachycardia
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVtranscutaneous pacingimmediate defibrillationCPR (non-shockable rhythm)synchronised cardioversion

A patient without a palpable pulse has this organised rhythm on the monitor (pulseless electrical activity). Which treatment do adult resuscitation guidelines name for 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
  1. 1transcutaneous pacing
  2. 2immediate defibrillation
  3. 3CPR (non-shockable rhythm)
  4. 4synchronised cardioversion
13.Evidence-based practice statisticsMedium
casescontrolsexposed98154abnot exposed212166cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 310 patients with a disease (cases) and 320 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.

a = 98b = 154c = 212d = 166
OR =
14.Evidence-based practice statisticsHard
control11%treatment?

Practice exercise (made-up values, not clinical guidance). Without a drug, 11% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.83. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.

control 11%RR = 1.83
NNH =

3Solutions

1.Human anatomy

Answer: B · Facial nerve (VII)

Why this is the answer

The facial nerve (cranial nerve VII) supplies the muscles of facial expression. Damage causes facial palsy, as in Bell's palsy.

Answer

B · Facial nerve (VII)
2.Human anatomy

Answer: C · Iliopsoas

Why this is the answer

Iliopsoas (psoas major and iliacus) is the strongest hip flexor. Gluteus maximus extends the hip.

Answer

C · Iliopsoas
3.Human physiology

Answer: B · Ventricular depolarisation

Why this is the answer

The P wave is atrial depolarisation, the QRS complex ventricular depolarisation and the T wave ventricular repolarisation. Atrial repolarisation is hidden in the QRS.

Answer

B · Ventricular depolarisation
4.Human physiology

Answer: C · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias

Why this is the answer

High extracellular potassium makes the resting potential less negative. Many fast sodium channels then become inactivated, slowing conduction and risking fatal arrhythmias.

Answer

C · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
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: D · Neisseria meningitidis

Why this is the answer

Neisseria meningitidis is a Gram-negative diplococcus causing meningitis and septicaemia with a petechial or purpuric rash. It needs urgent antibiotics and contacts may need prophylaxis.

Answer

D · Neisseria meningitidis
7.Pharmacology

Answer: B · Take aspirin 30 minutes before the niacin, if the prescriber agrees

Why this is the answer

Niacin causes prostaglandin-mediated vasodilation and flushing. Aspirin beforehand, taking it with food and avoiding hot drinks and alcohol reduce this effect.

Answer

B · Take aspirin 30 minutes before the niacin, if the prescriber agrees
8.Pharmacology

Answer: C · Rebound hypertension from abrupt clonidine withdrawal

Why this is the answer

Clonidine is a central alpha-2 agonist. Stopping it suddenly lets sympathetic activity surge, causing rebound hypertension, so it is tapered gradually.

Answer

C · Rebound hypertension from abrupt clonidine withdrawal
9.Arterial blood gas interpretation

Answer: PaCO₂ = 35 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 = 1.5·18 + 8 = 35 mmHg
range: 33–37 mmHg

Answer

PaCO₂ = 35 mmHg
10.Arterial blood gas interpretation

Answer: A–a gradient = 19.48 mmHg

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.

Alveolar oxygen

PAO₂ = 0.21·713 − 29/0.8 = 149.7 − 36.25 = 113.5 mmHg

Gradient

A–a = 113.5 − 94 = 19.48 mmHg

Answer

A–a gradient = 19.48 mmHg
11.ECG rhythm strip interpretation

Answer: supraventricular tachycardia (SVT)

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. Narrow-complex tachycardia and pauses: supraventricular tachycardia (SVT) is a regular tachycardia at about 150–250/min with narrow QRS complexes; the P waves are usually not visible because they are hidden in the QRS complex or the T wave. Sinus tachycardia in an adult at rest is usually below about 150/min and has a visible P wave before every QRS; atrial flutter with 2:1 conduction often runs at about 150/min and shows sawtooth waves. In sinus arrest the sinus node stops firing for a moment: a pause with no P wave and no QRS whose length is not a whole number of P–P cycles (a pause of exactly two or three cycles points to sinus exit block instead).

Five steps

1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none visible; at this rate they are hidden in the QRS complexes or the T waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s), so the rhythm starts above the ventricles.

Interpretation

supraventricular tachycardia (SVT)

Answer

supraventricular tachycardia (SVT)
12.ECG rhythm strip interpretation

Answer: CPR (non-shockable rhythm)

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 = 30 small boxes, 1500 ÷ 30 ≈ 50 per minute.
2. Rhythm: regular, every R–R is 1.20 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
sinus bradycardia

What the patient shows

no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm

Answer

CPR (non-shockable rhythm)

Answer

CPR (non-shockable rhythm)

PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.

13.Evidence-based practice statistics

Answer: OR = 0.5

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.

Odds of exposure

cases: 98/212 = 0.4623
controls: 154/166 = 0.9277

Odds ratio

OR = (98·166)/(154·212) = 0.4983 ≈ 0.5

Answer

OR = 0.5

OR below 1: exposure is less common among the cases, which suggests a protective effect.

14.Evidence-based practice statistics

Answer: NNH = 11

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.

Risk with the drug

0.11·1.83 = 0.2013

Absolute risk increase

ARI = 0.2013 − 0.11 = 0.0913

Number needed to harm

NNH = 1/0.0913 = 10.95 → 11

Answer

NNH = 11

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