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 spinal nerve roots form the brachial plexus?

  1. AC3 to C7
  2. BT1 to T5
  3. CC1 to C4
  4. DC5 to T1
2.Human anatomyHard

A typical posterolateral disc herniation between L4 and L5 most often compresses which nerve root?

  1. AL3
  2. BL4
  3. CL5
  4. DS1
3.Human physiologyMedium

Since action potentials are all-or-none, how does a neuron signal a stronger stimulus?

  1. ABy making each action potential larger
  2. BBy changing the neurotransmitter it releases
  3. CBy making each action potential longer
  4. DBy firing action potentials more often
4.Human physiologyHard

Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?

  1. ACO occupies haemoglobin binding sites, so oxygen content falls
  2. BCO lowers the barometric pressure in the alveoli
  3. CCO blocks diffusion across the alveolar wall
  4. DCO stops oxygen from dissolving in plasma
5.MicrobiologyMedium

Which white blood cells are the main phagocytes in an acute bacterial infection and the main component of pus?

  1. ABasophils
  2. BNeutrophils
  3. CEosinophils
  4. DB lymphocytes
6.MicrobiologyHard

Prions differ from all other infectious agents because they:

  1. AContain no nucleic acid and consist of misfolded protein
  2. BAre killed by normal cooking and disinfection
  3. CContain only RNA, with no protein coat surrounding the genome
  4. DCan only infect bacteria
7.PharmacologyMedium

How does glyceryl trinitrate (GTN) relieve angina?

  1. AIt blocks beta-1 receptors and slows the heart
  2. BIt dissolves the clot in the coronary artery
  3. CIt increases the force of contraction so the heart pumps blood more efficiently
  4. DIt releases nitric oxide, mainly dilating veins and reducing cardiac workload
8.PharmacologyHard

A patient with severe chronic hyponatraemia receives hypertonic saline. Why is the rate of correction strictly limited?

  1. ARaising sodium too fast causes severe hypokalaemia and cardiac arrest
  2. BFast correction causes immediate kidney failure
  3. CRaising sodium too fast can cause osmotic demyelination in the brain
  4. DHypertonic saline is toxic to the liver at high rates
9.Arterial blood gas interpretationMedium

A practice gas shows a metabolic acidosis with HCO₃⁻ 12 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.

HCO₃⁻ 12 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard

A practice gas shows a metabolic alkalosis with HCO₃⁻ 39 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.

HCO₃⁻ 39 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVsupraventricular tachycardia (SVT)accelerated junctional rhythmventricular tachycardiaatrial fibrillation

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. 2accelerated junctional rhythm
  3. 3ventricular tachycardia
  4. 4atrial fibrillation
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVtorsades de pointesventricular tachycardiaatrial fibrillationatrial flutter, 2:1 conduction

Identify the rhythm on this lead II strip. 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. 1torsades de pointes
  2. 2ventricular tachycardia
  3. 3atrial fibrillation
  4. 4atrial flutter, 2:1 conduction
13.Evidence-based practice statisticsMedium
casesperson-yearsexposed49900not exposed421900

Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 900 person-years with 49 new cases, and unexposed people 1900 person-years with 42 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.

exposed: 49/900 PYnot exposed: 42/1900 PY
IRR =
14.Evidence-based practice statisticsHard
testdisease presentdisease absenttotalpositive?negativetotal10 000

Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 92% and specificity 79%. The prevalence of the disease is 8%. How many positive results n₊ are expected in total?

N = 10 000p = 8%Se = 92%Sp = 79%
n₊ =

3Solutions

1.Human anatomy

Answer: D · C5 to T1

Why this is the answer

The brachial plexus is formed by the ventral rami of C5, C6, C7, C8 and T1. It gives rise to the main nerves of the arm.

Answer

D · C5 to T1
2.Human anatomy

Answer: C · L5

Why this is the answer

A posterolateral L4–L5 herniation usually catches the traversing L5 root, causing weakness of big toe extension and numbness on the top of the foot.

Answer

C · L5
3.Human physiology

Answer: D · By firing action potentials more often

Why this is the answer

Stimulus strength is coded by the frequency of action potentials and by how many neurons are recruited, not by their amplitude.

Answer

D · By firing action potentials more often
4.Human physiology

Answer: A · CO occupies haemoglobin binding sites, so oxygen content falls

Why this is the answer

CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.

Answer

A · CO occupies haemoglobin binding sites, so oxygen content falls
5.Microbiology

Answer: B · Neutrophils

Why this is the answer

Neutrophils arrive quickly at sites of infection and engulf bacteria. Dead neutrophils, bacteria and tissue debris form pus.

Answer

B · Neutrophils
6.Microbiology

Answer: A · Contain no nucleic acid and consist of misfolded protein

Why this is the answer

Prions are misfolded proteins that induce normal proteins to misfold. They contain no DNA or RNA and are very resistant to standard sterilisation.

Answer

A · Contain no nucleic acid and consist of misfolded protein
7.Pharmacology

Answer: D · It releases nitric oxide, mainly dilating veins and reducing cardiac workload

Why this is the answer

GTN is converted to nitric oxide, which relaxes vascular smooth muscle through cGMP. Venodilation lowers preload, so the heart works less and needs less oxygen.

Answer

D · It releases nitric oxide, mainly dilating veins and reducing cardiac workload
8.Pharmacology

Answer: C · Raising sodium too fast can cause osmotic demyelination in the brain

Why this is the answer

In chronic hyponatraemia the brain adapts to low osmolality. Correcting sodium too rapidly shrinks brain cells and can cause osmotic demyelination (central pontine myelinolysis), so sodium is checked often.

Answer

C · Raising sodium too fast can cause osmotic demyelination in the brain
9.Arterial blood gas interpretation

Answer: PaCO₂ = 26 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 = 1.5·12 + 8 = 26 mmHg
range: 24–28 mmHg

Answer

PaCO₂ = 26 mmHg
10.Arterial blood gas interpretation

Answer: PaCO₂ = 48.3 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

0.7·39 + 21 = 48.3 mmHg

Answer

PaCO₂ = 48.3 mmHg

The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.

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: 0.06 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: ventricular tachycardia

Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.

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 identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).

Interpretation

ventricular tachycardia

Answer

ventricular tachycardia

In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.

13.Evidence-based practice statistics

Answer: IRR = 2.46

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.

Incidence rates

exposed: 49/900 = 54.44 per 1000 person-years
not exposed: 42/1900 = 22.11 per 1000 person-years

Rate ratio

IRR = 54.44/22.11 = 2.463 ≈ 2.46

Answer

IRR = 2.46
14.Evidence-based practice statistics

Answer: n₊ = 2668

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

10 000·0.08 = 800
10 000 − 800 = 9200

All positive results

TP = 800·0.92 = 736
FP = 9200·0.21 = 1932
n₊ = 736 + 1932 = 2668

Answer

n₊ = 2668

Of the 2668 positives only 736 have the disease (PPV 27.6%).

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