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. AOculomotor nerve (III)
  2. BFacial nerve (VII)
  3. CHypoglossal nerve (XII)
  4. DTrigeminal nerve (V)
2.Human anatomyHard

A pituitary tumour grows upwards and causes loss of both outer (temporal) halves of the visual field. Where is the visual pathway compressed?

  1. AOptic chiasm
  2. BLeft optic tract
  3. CRight optic nerve
  4. DOccipital cortex
3.Human physiologyMedium

Which cells in the stomach secrete hydrochloric acid?

  1. AChief cells
  2. BG cells
  3. CGoblet cells
  4. DParietal cells
4.Human physiologyHard

A young woman has muscle weakness that gets worse with repeated use and improves with an acetylcholinesterase inhibitor. What is the most likely mechanism?

  1. AA defect in the dystrophin protein of muscle
  2. BAntibodies against acetylcholine receptors at the neuromuscular junction
  3. CLoss of motor neurons in the spinal cord
  4. DDemyelination of central nervous system tracts
5.MicrobiologyMedium

Toxoplasma gondii is a risk in pregnancy. Which advice reduces exposure?

  1. AAvoid cleaning cat litter and avoid undercooked meat
  2. BAvoid contact with dogs
  3. CAvoid all fruit and vegetables
  4. DAvoid swimming pools and all other public bathing areas
6.MicrobiologyHard

Joseph Lister reduced infection after surgery in the 1860s. What was his main innovation?

  1. AGiving sulfonamide antibiotics after every operation
  2. BUsing steam autoclaves to sterilise all surgical instruments before each operation
  3. CUsing carbolic acid (phenol) as an antiseptic on wounds and instruments
  4. DVaccinating surgeons against staphylococci
7.PharmacologyMedium

Benzathine benzylpenicillin (penicillin G benzathine) is ordered. Which route is correct?

  1. AIntravenous push
  2. BSubcutaneous in the abdomen
  3. CDeep intramuscular only
  4. DIntravenous infusion over 30 minutes
8.PharmacologyHard

A patient who has taken lorazepam daily for many months stops it suddenly. Which serious complication may follow?

  1. ASevere bradycardia
  2. BWithdrawal seizures
  3. CPinpoint pupils and coma
  4. DProfound respiratory depression
9.Arterial blood gas interpretationMedium

Practice values: Na⁺ 152 mmol/L, glucose 30.7 mmol/L and urea 22.7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 152 mmol/Lglucose 30.7 mmol/Lurea 22.7 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
pH7.357.457.31PaCO₂mmHg354576HCO₃⁻mmol/L222637acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.31, PaCO₂ 76 mmHg (10.1 kPa), HCO₃⁻ 37 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.

pH 7.31PaCO₂ 76 mmHg (10.1 kPa)HCO₃⁻ 37 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVsinus tachycardiaaccelerated junctional rhythmventricular tachycardiasupraventricular tachycardia (SVT)

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. 1sinus tachycardia
  2. 2accelerated junctional rhythm
  3. 3ventricular tachycardia
  4. 4supraventricular tachycardia (SVT)
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVsecond-degree AV block type I (Wenckebach)third-degree (complete) AV blocksecond-degree AV block type IIfirst-degree 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 I (Wenckebach)
  2. 2third-degree (complete) AV block
  3. 3second-degree AV block type II
  4. 4first-degree AV block
13.Evidence-based practice statisticsMedium
peopleyears1600323701.534601.5new cases87

Practice exercise (made-up values, not clinical guidance). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 87 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.

600 × 3 years370 × 1.5 years460 × 1.5 years87 new cases
IR = per 1000 person-years
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 95% and specificity 99%. The prevalence of the disease is 8%. How many positive results n₊ are expected in total?

N = 10 000p = 8%Se = 95%Sp = 99%
n₊ =

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: A · Optic chiasm

Why this is the answer

Bitemporal hemianopia results from compression of the crossing nasal fibres at the optic chiasm, which lies directly above the pituitary gland.

Answer

A · Optic chiasm
3.Human physiology

Answer: D · Parietal cells

Why this is the answer

Parietal cells secrete HCl and intrinsic factor. Chief cells secrete pepsinogen and G cells secrete gastrin.

Answer

D · Parietal cells
4.Human physiology

Answer: B · Antibodies against acetylcholine receptors at the neuromuscular junction

Why this is the answer

In myasthenia gravis, autoantibodies reduce the number of functional acetylcholine receptors. Blocking acetylcholinesterase leaves more ACh to bind the remaining receptors.

Answer

B · Antibodies against acetylcholine receptors at the neuromuscular junction
5.Microbiology

Answer: A · Avoid cleaning cat litter and avoid undercooked meat

Why this is the answer

Toxoplasma oocysts are shed in cat faeces and cysts are found in undercooked meat. Infection during pregnancy can damage the fetal brain and eyes.

Answer

A · Avoid cleaning cat litter and avoid undercooked meat
6.Microbiology

Answer: C · Using carbolic acid (phenol) as an antiseptic on wounds and instruments

Why this is the answer

Lister applied carbolic acid to wounds, dressings and instruments, which greatly reduced post-operative infection. This started antiseptic surgery.

Answer

C · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
7.Pharmacology

Answer: C · Deep intramuscular only

Why this is the answer

Benzathine penicillin is a depot suspension given deep IM. Accidental intravenous injection can cause cardiorespiratory arrest and death.

Answer

C · Deep intramuscular only
8.Pharmacology

Answer: B · Withdrawal seizures

Why this is the answer

Abrupt withdrawal after long-term benzodiazepine use causes anxiety, insomnia, tremor and sometimes seizures. Long-term users should have the dose tapered gradually.

Answer

B · Withdrawal seizures
9.Arterial blood gas interpretation

Answer: Osm = 357 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 = 2·152 + 30.7 + 22.7 = 357.4 ≈ 357 mOsm/kg

Answer

Osm = 357 mOsm/kg
10.Arterial blood gas interpretation

Answer: chronic

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 HCO₃⁻

acute: 24 + 1·3.6 = 27.6 mmol/L
chronic: 24 + 3.5·3.6 = 36.6 mmol/L

Compare

measured: 37 mmol/L → chronic

Answer

chronic
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 = 8 small boxes, 1500 ÷ 8 ≈ 188 per minute.
2. Rhythm: regular, every R–R is 0.32 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: 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 = 35 small boxes, 1500 ÷ 35 ≈ 43 per minute.
2. Rhythm: R–R regular (1.40 s) and P–P regular (0.84 s), but the two are independent.
3. P waves: upright and regular at 71 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: 0.08 s, narrow (below 0.12 s).

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: IR = 28.6 per 1000 person-years

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.

Person-years

PY = 600·3 + 370·1.5 + 460·1.5 = 3045

Incidence rate

IR = 87/3045·1000 = 28.6 per 1000 person-years

Answer

IR = 28.6 per 1000 person-years
14.Evidence-based practice statistics

Answer: n₊ = 852

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 = 800·0.95 = 760
FP = 9200·0.01 = 92
n₊ = 760 + 92 = 852

Answer

n₊ = 852

Of the 852 positives only 760 have the disease (PPV 89.2%).

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