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. AFacial nerve (VII)
  2. BOculomotor nerve (III)
  3. CTrigeminal nerve (V)
  4. DHypoglossal nerve (XII)
2.Human anatomyHard

A stroke affecting the left middle cerebral artery in a right-handed person most likely causes:

  1. ALeft-sided weakness and aphasia
  2. BIsolated loss of balance
  3. CRight-sided weakness and aphasia
  4. DRight-sided weakness and left visual neglect
3.Human physiologyMedium

Which hormonal event triggers ovulation?

  1. AA surge of prolactin
  2. BA surge of luteinising hormone (LH)
  3. CA fall in oestrogen
  4. DA rise in progesterone from the corpus luteum
4.Human physiologyHard

Why does ketoacidosis typically develop in untreated type 1 diabetes but rarely in type 2?

  1. AType 2 patients cannot make ketones at all
  2. BKetones are only made in children
  3. CEven low insulin levels suppress lipolysis and ketogenesis
  4. DType 1 patients have more glucagon receptors
5.MicrobiologyMedium

Which bacterium is the most common cause of stomach ulcers?

  1. AClostridioides difficile
  2. BEscherichia coli
  3. CStaphylococcus aureus
  4. DHelicobacter pylori
6.MicrobiologyHard

Which bacterial toxin causes the muscle spasms of tetanus?

  1. ABotulinum toxin, which blocks acetylcholine release at the neuromuscular junction
  2. BTetanospasmin, which blocks inhibitory neurotransmitter release
  3. CEndotoxin from the outer membrane
  4. DCholera toxin, which activates adenylate cyclase
7.PharmacologyMedium

Which patient should usually not receive a live vaccine?

  1. AA healthy adult nurse who works on a children's ward
  2. BA patient with mild seasonal allergies
  3. CA healthy 1-year-old child
  4. DA patient receiving chemotherapy for leukaemia
8.PharmacologyHard

Why can an ACE inhibitor cause acute kidney injury in a patient with bilateral renal artery stenosis?

  1. AACE inhibitors constrict the afferent arteriole
  2. BThe drug crystallises in the renal tubules
  3. CGFR depends on angiotensin II constricting the efferent arteriole
  4. DACE inhibitors are directly toxic to the proximal tubule cells in the kidney
9.Arterial blood gas interpretationMedium
pH7.357.457.28PaCO₂mmHg354553HCO₃⁻mmol/L222624uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.28, PaCO₂ 53 mmHg (7.1 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.28PaCO₂ 53 mmHg (7.1 kPa)HCO₃⁻ 24 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
10.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 35% oxygen shows PaO₂ 73 mmHg (9.7 kPa) and PaCO₂ 38 mmHg (5.1 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 35%PaO₂ 73 mmHg (9.7 kPa)PaCO₂ 38 mmHg (5.1 kPa)
A–a gradient = mmHg
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVjunctional rhythmnormal sinus rhythmfirst-degree AV blocksinus 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. 1junctional rhythm
  2. 2normal sinus rhythm
  3. 3first-degree AV block
  4. 4sinus tachycardia
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVsecond-degree AV block type IIjunctional rhythmfirst-degree AV blockthird-degree (complete) AV block

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
  1. 1second-degree AV block type II
  2. 2junctional rhythm
  3. 3first-degree AV block
  4. 4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Se85%Sp64%LR− = ?

Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 85% and a specificity of 64%. Find the negative likelihood ratio LR−. Round to two decimal places.

Se = 85%Sp = 64%
LR− =
14.Evidence-based practice statisticsHard
CER30%EER?RRR 40%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 40% in relative terms (RRR = 40%). In a patient group the risk without treatment (CER) is 30% 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 = 40%CER = 30%
NNT =

3Solutions

1.Human anatomy

Answer: A · 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

A · Facial nerve (VII)
2.Human anatomy

Answer: C · Right-sided weakness and aphasia

Why this is the answer

The left MCA supplies the left motor cortex (face and arm of the right side) and Broca's and Wernicke's areas, so right hemiparesis with aphasia is typical.

Answer

C · Right-sided weakness and aphasia
3.Human physiology

Answer: B · A surge of luteinising hormone (LH)

Why this is the answer

Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.

Answer

B · A surge of luteinising hormone (LH)
4.Human physiology

Answer: C · Even low insulin levels suppress lipolysis and ketogenesis

Why this is the answer

Insulin strongly inhibits lipolysis and hepatic ketone production. In type 1 there is almost no insulin, so fatty acids flood the liver and ketones build up.

Answer

C · Even low insulin levels suppress lipolysis and ketogenesis
5.Microbiology

Answer: D · Helicobacter pylori

Why this is the answer

H. pylori colonises the stomach lining and produces urease to neutralise acid locally. It causes most peptic ulcers and is treated with antibiotics plus a PPI.

Answer

D · Helicobacter pylori
6.Microbiology

Answer: B · Tetanospasmin, which blocks inhibitory neurotransmitter release

Why this is the answer

Tetanospasmin from Clostridium tetani blocks the release of GABA and glycine from inhibitory interneurons, causing spastic paralysis. Botulinum toxin, by contrast, causes flaccid paralysis.

Answer

B · Tetanospasmin, which blocks inhibitory neurotransmitter release
7.Pharmacology

Answer: D · A patient receiving chemotherapy for leukaemia

Why this is the answer

Live attenuated vaccines can cause serious infection in immunosuppressed people and are generally avoided in pregnancy. Inactivated vaccines are safe but may give a weaker response.

Answer

D · A patient receiving chemotherapy for leukaemia
8.Pharmacology

Answer: C · GFR depends on angiotensin II constricting the efferent arteriole

Why this is the answer

When renal perfusion is low, angiotensin II constricts the efferent arteriole to maintain glomerular pressure. Blocking it lets the pressure and GFR fall, so kidney function can drop sharply.

Answer

C · GFR depends on angiotensin II constricting the efferent arteriole
9.Arterial blood gas interpretation

Answer: uncompensated

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 7.28: acidaemia (below 7.35)
PaCO₂ 53 mmHg: high (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)

Compensation

uncompensated

Answer

uncompensated
10.Arterial blood gas interpretation

Answer: A–a gradient = 129 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 = 0.35·713 − 38/0.8 = 249.5 − 47.5 = 202 mmHg

Gradient

A–a = 202 − 73 = 129 mmHg

Answer

A–a gradient = 129 mmHg
11.ECG rhythm strip interpretation

Answer: normal sinus rhythm

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 = 18 small boxes, 1500 ÷ 18 ≈ 83 per minute.
2. Rhythm: regular, every R–R is 0.72 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.12 s, constant, within 0.12–0.20 s.
5. QRS: 0.10 s, narrow (below 0.12 s).

Interpretation

normal sinus rhythm

Answer

normal sinus rhythm
12.ECG rhythm strip interpretation

Answer: third-degree (complete) AV block

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: R–R = 45 small boxes, 1500 ÷ 45 ≈ 33 per minute.
2. Rhythm: R–R regular (1.80 s) and P–P regular (0.68 s), but the two are independent.
3. P waves: upright and regular at 88 per minute, but not linked to the QRS complexes; some hide in a QRS or a T wave.
4. PR interval: varies from beat to beat; there is no fixed relation.
5. QRS: 0.14 s, wide (0.12 s or more).

Interpretation

third-degree (complete) AV block

Answer

third-degree (complete) AV block

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: LR− = 0.23

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 − 0.85)/0.64 = 0.2344 ≈ 0.23

Answer

LR− = 0.23
14.Evidence-based practice statistics

Answer: NNT = 9

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 = 0.3·0.4 = 0.12

Number needed to treat

NNT = 1/0.12 = 8.333 → 9

Answer

NNT = 9

The same relative effect gives a smaller NNT when the baseline risk is higher.

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