Medical Sciences: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 part of the adrenal gland secretes adrenaline?

  1. AZona fasciculata
  2. BZona reticularis
  3. CAdrenal medulla
  4. DZona glomerulosa
2.Human anatomyHard

Injury to the common fibular (peroneal) nerve at the neck of the fibula typically causes:

  1. AWeak hip abduction
  2. BLoss of sensation on the sole of the foot
  3. CFoot drop
  4. DInability to flex the knee
3.Human physiologyMedium

During which phase of the cardiac cycle are all four heart valves closed while ventricular pressure rises?

  1. ARapid ejection
  2. BAtrial systole
  3. CRapid filling
  4. DIsovolumetric contraction
4.Human physiologyHard

Why is severe hyperkalaemia dangerous for the heart?

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

Which bacterium is the most common cause of stomach ulcers?

  1. AHelicobacter pylori
  2. BEscherichia coli
  3. CClostridioides difficile
  4. DStaphylococcus aureus
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. AListeria monocytogenes
  2. BEscherichia coli
  3. CStreptococcus pneumoniae
  4. DNeisseria meningitidis
7.PharmacologyMedium

What makes an overdose of tricyclic antidepressants such as amitriptyline especially dangerous?

  1. ACardiac arrhythmias and seizures
  2. BKidney stones
  3. CSudden high blood glucose
  4. DSevere liver failure within hours
8.PharmacologyHard

Why is vincristine labelled "for intravenous use only – fatal if given by other routes"?

  1. AAccidental intrathecal injection causes fatal ascending neurotoxicity
  2. BSubcutaneous injection makes the drug inactive
  3. CIntramuscular injection causes severe and prolonged hypoglycaemia
  4. DOral vincristine causes immediate anaphylaxis
9.Arterial blood gas interpretationMedium
pH7.357.457.62PaCO₂mmHg354542HCO₃⁻mmol/L222642uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.62, PaCO₂ 42 mmHg (5.6 kPa) and HCO₃⁻ 42 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.62PaCO₂ 42 mmHg (5.6 kPa)HCO₃⁻ 42 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
10.Arterial blood gas interpretationHard
pH7.357.457.33PaCO₂mmHg354533HCO₃⁻mmol/L222617added normal-gap acidosispure high-gap acidosisadded metabolic alkalosis

A practice gas shows pH 7.33, PaCO₂ 33 mmHg (4.4 kPa) and HCO₃⁻ 17 mmol/L, with Na⁺ 138 and Cl⁻ 87 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.

pH 7.33PaCO₂ 33 mmHg (4.4 kPa)HCO₃⁻ 17 mmol/LNa⁺ 138 mmol/LCl⁻ 87 mmol/L
  1. 1added normal-gap acidosis
  2. 2pure high-gap acidosis
  3. 3added metabolic alkalosis
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mV

Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVsecond-degree AV block type IIjunctional rhythmthird-degree (complete) AV blockfirst-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 II
  2. 2junctional rhythm
  3. 3third-degree (complete) AV block
  4. 4first-degree AV block
13.Evidence-based practice statisticsMedium
casescontrolsexposed119136abnot exposed5194cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 170 patients with a disease (cases) and 230 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 = 119b = 136c = 51d = 94
OR =
14.Evidence-based practice statisticsHard
pre-test5%post-test?p′LR−

Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 93% and specificity 95% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.

pre-test 5%Se = 93%Sp = 95%negative
p′ = %

3Solutions

1.Human anatomy

Answer: C · Adrenal medulla

Why this is the answer

The adrenal medulla releases adrenaline and noradrenaline in response to sympathetic stimulation. The three zones of the cortex make steroid hormones.

Answer

C · Adrenal medulla
2.Human anatomy

Answer: C · Foot drop

Why this is the answer

The common fibular nerve supplies the dorsiflexors and evertors. Damage at the fibular neck causes foot drop and a high-stepping gait. The sole is supplied by the tibial nerve.

Answer

C · Foot drop
3.Human physiology

Answer: D · Isovolumetric contraction

Why this is the answer

After the AV valves close, the ventricles contract against closed semilunar valves. Pressure rises with no volume change until the aortic and pulmonary valves open.

Answer

D · Isovolumetric contraction
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: A · 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

A · Helicobacter pylori
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: A · Cardiac arrhythmias and seizures

Why this is the answer

Tricyclics block cardiac sodium channels (widening the QRS complex) and have anticholinergic effects. In overdose they cause arrhythmias, hypotension, seizures and coma.

Answer

A · Cardiac arrhythmias and seizures
8.Pharmacology

Answer: A · Accidental intrathecal injection causes fatal ascending neurotoxicity

Why this is the answer

Vincristine given intrathecally causes ascending paralysis and death. To prevent this, it is usually given in a small IV bag rather than a syringe and kept separate from intrathecal drugs.

Answer

A · Accidental intrathecal injection causes fatal ascending neurotoxicity
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 7.62: alkalaemia (above 7.45)
PaCO₂ 42 mmHg: normal (35–45)
HCO₃⁻ 42 mmol/L: high (22–26)

Compensation

Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 48.4–52.4 mmHg, so the measured 42 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.

uncompensated

Answer

uncompensated
10.Arterial blood gas interpretation

Answer: added metabolic alkalosis

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.

Anion gap

AG = 138 − (87 + 17) = 34

Delta ratio

(34 − 12)/(24 − 17) = 3.143
added metabolic alkalosis

Answer

added metabolic alkalosis
11.ECG rhythm strip interpretation

Answer: QTc = 0.375 s

QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.

Measure

QT = 9 × 0.04 = 0.36 s
RR = 23 × 0.04 = 0.92 s

Bazett’s formula

QTc = 0.36/√0.92 = 0.36/0.9592 = 0.375 s

Answer

QTc = 0.375 s

Within the usual limits (about 0.45 s for men, 0.46 s for women). (375 ms)

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 = 26 small boxes, 1500 ÷ 26 ≈ 58 per minute.
2. Rhythm: R–R regular (1.04 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: 0.10 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: OR = 1.61

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: 119/51 = 2.333
controls: 136/94 = 1.447

Odds ratio

OR = (119·94)/(136·51) = 1.613 ≈ 1.61

Answer

OR = 1.61

OR above 1: exposure is more common among the cases.

14.Evidence-based practice statistics

Answer: p′ = 0.4 %

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.

Likelihood ratio

LR− = (1 − 0.93)/0.95 = 0.07368

Pre-test odds

0.05/(1 − 0.05) = 0.05263

Post-test odds

0.05263·0.07368 = 0.003878

Post-test probability

p′ = 0.003878/(1 + 0.003878) = 0.003863 ≈ 0.4%

Answer

p′ = 0.4%

ExamProblems · examproblems.com/courses/medical-sciences