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 part of the brainstem contains the main cardiovascular and respiratory centres?

  1. AMidbrain
  2. BMedulla oblongata
  3. CPons
  4. DCerebellum
2.Human anatomyHard

Which structure passes through the inguinal canal in males?

  1. AGreat saphenous vein
  2. BObturator nerve
  3. CFemoral artery
  4. DSpermatic cord
3.Human physiologyMedium

Which protein forms the mesh that stabilises a blood clot?

  1. AAlbumin
  2. BCollagen
  3. CFibrin
  4. DThrombin
4.Human physiologyHard

Which spirometry finding is typical of chronic obstructive pulmonary disease (COPD)?

  1. AFEV1/FVC ratio below 0.7
  2. BNormal FEV1 with low FVC
  3. CFEV1/FVC ratio above 0.9 with low FVC
  4. DIncreased peak expiratory flow
5.MicrobiologyMedium

Which virus family includes the viruses that cause cold sores and chickenpox?

  1. ACoronaviruses
  2. BPicornaviruses
  3. CHerpesviruses
  4. DRetroviruses
6.MicrobiologyHard

A patient taking high-dose immunosuppressive drugs after a transplant is due for vaccinations. Which vaccine type is generally avoided?

  1. AToxoid vaccines
  2. BInactivated whole-virus vaccines
  3. CLive attenuated vaccines
  4. DSubunit vaccines
7.PharmacologyMedium

A patient receiving chemotherapy has a very low neutrophil count and is going home. Which teaching is most important?

  1. AReport a temperature of 38 °C (100.4 °F) or higher straight away
  2. BEat plenty of raw shellfish to boost protein
  3. CTake paracetamol regularly so that fevers do not show
  4. DExercise in crowded gyms to build strength
8.PharmacologyHard

A man on long-term testosterone therapy has a rising haematocrit. Why is this important?

  1. ATestosterone raises red cell production, increasing clot risk
  2. BIt means the testosterone dose is too low
  3. CIt is a sign of liver failure caused by testosterone
  4. DA high haematocrit shows that the testosterone is not being absorbed
9.Arterial blood gas interpretationMedium

Practice values: Na⁺ 148 mmol/L, glucose 4.6 mmol/L and urea 2.9 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 148 mmol/Lglucose 4.6 mmol/Lurea 2.9 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard

A practice gas shows a metabolic alkalosis with HCO₃⁻ 36 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.

HCO₃⁻ 36 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVaccelerated junctional rhythmsinus tachycardiaatrial flutter, 2:1 conductionsupraventricular 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. 1accelerated junctional rhythm
  2. 2sinus tachycardia
  3. 3atrial flutter, 2:1 conduction
  4. 4supraventricular tachycardia (SVT)
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVidioventricular rhythmdual-chamber (AV) paced rhythmsecond-degree AV block type IIpaced rhythm, failure to capture

This patient has a pacemaker. Interpret the 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. 1idioventricular rhythm
  2. 2dual-chamber (AV) paced rhythm
  3. 3second-degree AV block type II
  4. 4paced rhythm, failure to capture
13.Evidence-based practice statisticsMedium
casescontrolsexposed8621abnot exposed224109cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 310 patients with a disease (cases) and 130 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 = 86b = 21c = 224d = 109
OR =
14.Evidence-based practice statisticsHard
CER13%EER?RRR 21%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 21% in relative terms (RRR = 21%). In a patient group the risk without treatment (CER) is 13% 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 = 21%CER = 13%
NNT =

3Solutions

1.Human anatomy

Answer: B · Medulla oblongata

Why this is the answer

The medulla oblongata contains the vital centres that regulate heart rate, blood pressure and breathing rhythm.

Answer

B · Medulla oblongata
2.Human anatomy

Answer: D · Spermatic cord

Why this is the answer

The inguinal canal carries the spermatic cord in males and the round ligament in females. The femoral vessels pass below the inguinal ligament instead.

Answer

D · Spermatic cord
3.Human physiology

Answer: C · Fibrin

Why this is the answer

Thrombin converts soluble fibrinogen into insoluble fibrin strands, which bind the platelet plug and trap red cells to form a stable clot.

Answer

C · Fibrin
4.Human physiology

Answer: A · FEV1/FVC ratio below 0.7

Why this is the answer

Narrowed airways slow expiration, so FEV1 falls more than FVC and the ratio drops below 0.7. In COPD this is not fully reversed by bronchodilators.

Answer

A · FEV1/FVC ratio below 0.7
5.Microbiology

Answer: C · Herpesviruses

Why this is the answer

Herpes simplex (cold sores) and varicella-zoster (chickenpox, shingles) are herpesviruses that remain latent in sensory ganglia and can reactivate.

Answer

C · Herpesviruses
6.Microbiology

Answer: C · Live attenuated vaccines

Why this is the answer

Live vaccines contain weakened organisms that can cause disease when the immune system is severely suppressed. Inactivated, toxoid and subunit vaccines are safe, although the response may be weaker.

Answer

C · Live attenuated vaccines
7.Pharmacology

Answer: A · Report a temperature of 38 °C (100.4 °F) or higher straight away

Why this is the answer

Neutropenic patients may have few signs of infection other than fever, and sepsis can develop quickly. A fever needs urgent assessment; good hand hygiene and avoiding sick contacts are also taught.

Answer

A · Report a temperature of 38 °C (100.4 °F) or higher straight away
8.Pharmacology

Answer: A · Testosterone raises red cell production, increasing clot risk

Why this is the answer

Testosterone increases erythropoiesis and can cause polycythaemia, which increases blood viscosity and thrombosis risk. Haematocrit is monitored, and the dose may be reduced or paused.

Answer

A · Testosterone raises red cell production, increasing clot risk
9.Arterial blood gas interpretation

Answer: Osm = 304 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 = 2·148 + 4.6 + 2.9 = 303.5 ≈ 304 mOsm/kg

Answer

Osm = 304 mOsm/kg
10.Arterial blood gas interpretation

Answer: PaCO₂ = 46.2 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

0.7·36 + 21 = 46.2 mmHg

Answer

PaCO₂ = 46.2 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 = 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: paced rhythm, failure to capture

Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.

Five steps

1. Rate: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: the pacing spikes are regular (every 0.92 s), but the R–R is not, because a QRS is missing after some spikes.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.16 s and wide after the captured spikes; 2 spikes are not followed by a QRS (no capture).

Interpretation

paced rhythm, failure to capture

Answer

paced rhythm, failure to capture

Failure to capture: the pacemaker fires but the heart does not respond. Textbooks treat it as urgent to report, because the patient may depend on the pacemaker.

13.Evidence-based practice statistics

Answer: OR = 1.99

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: 86/224 = 0.3839
controls: 21/109 = 0.1927

Odds ratio

OR = (86·109)/(21·224) = 1.993 ≈ 1.99

Answer

OR = 1.99

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

14.Evidence-based practice statistics

Answer: NNT = 37

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 = 0.13·0.21 = 0.0273

Number needed to treat

NNT = 1/0.0273 = 36.63 → 37

Answer

NNT = 37

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

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