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 nerve is compressed in carpal tunnel syndrome?

  1. AMusculocutaneous nerve
  2. BUlnar nerve
  3. CRadial nerve
  4. DMedian nerve
2.Human anatomyHard

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

  1. AInability to flex the knee
  2. BFoot drop
  3. CLoss of sensation on the sole of the foot
  4. DWeak hip abduction
3.Human physiologyMedium

What is the normal pH range of arterial blood?

  1. A6.80 to 7.00
  2. B7.50 to 7.65
  3. C7.35 to 7.45
  4. D7.00 to 7.20
4.Human physiologyHard

Which change shifts the oxygen–haemoglobin dissociation curve to the right?

  1. AHigher temperature
  2. BLower 2,3-BPG
  3. CHigher pH
  4. DLower CO₂
5.MicrobiologyMedium

Tetanus and diphtheria vaccines are toxoids. What does that mean?

  1. AThey contain an inactivated toxin that still triggers an immune response
  2. BThey contain live bacteria that cannot make toxin
  3. CThey contain the bacterial DNA only
  4. DThey contain ready-made antibodies against the toxin, giving passive immunity
6.MicrobiologyHard

In 1854 John Snow traced a cholera outbreak in London to the Broad Street water pump. Why is this study important?

  1. AIt led directly to the first cholera vaccine
  2. BIt proved that cholera is caused by a virus
  3. CIt showed that cholera spreads through contaminated water, founding modern epidemiology
  4. DIt showed that cholera spreads through "bad air" (miasma)
7.PharmacologyMedium

1000 mL of IV fluid is to be given over 8 hours with a pump. What rate should be set?

  1. A125 mL/h
  2. B250 mL/h
  3. C80 mL/h
  4. D100 mL/h
8.PharmacologyHard

Before starting rituximab, patients are tested for hepatitis B. Why?

  1. ARituximab can reactivate hepatitis B and cause liver failure
  2. BRituximab is made from the hepatitis B virus
  3. CRituximab is a vaccine against hepatitis B
  4. DHepatitis B infection makes rituximab completely ineffective
9.Arterial blood gas interpretationMedium
pH7.357.457.36PaCO₂mmHg354563HCO₃⁻mmol/L222634uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.36, PaCO₂ 63 mmHg (8.4 kPa) and HCO₃⁻ 34 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.36PaCO₂ 63 mmHg (8.4 kPa)HCO₃⁻ 34 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
10.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 28% oxygen shows PaO₂ 52 mmHg (6.9 kPa) and PaCO₂ 37 mmHg (4.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 28%PaO₂ 52 mmHg (6.9 kPa)PaCO₂ 37 mmHg (4.9 kPa)
A–a gradient = mmHg
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mV

This lead II strip shows sinus rhythm with one pause. Measure the pause, the R–R interval across it, in seconds.

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
pause = s
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVraterhythmP wavesPR intervalQRS duration

Interpret this lead II strip with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
  1. 1rate
  2. 2rhythm
  3. 3P waves
  4. 4PR interval
  5. 5QRS duration
13.Evidence-based practice statisticsMedium
Se79%Sp91%LR+ = ?

Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 79% and a specificity of 91%. Find the positive likelihood ratio LR+. Round to two decimal places.

Se = 79%Sp = 91%
LR+ =
14.Evidence-based practice statisticsHard
p (%)PPV (%)0.5%PPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 81% and a specificity of 93%. It is used in a population where the prevalence is 0.5%. Find the positive predictive value PPV as a percentage. Round to one decimal place.

Se = 81%Sp = 93%p = 0.5%
PPV = %

3Solutions

1.Human anatomy

Answer: D · Median nerve

Why this is the answer

The median nerve passes through the carpal tunnel under the flexor retinaculum; compression causes tingling in the thumb, index and middle fingers.

Answer

D · Median nerve
2.Human anatomy

Answer: B · 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

B · Foot drop
3.Human physiology

Answer: C · 7.35 to 7.45

Why this is the answer

Arterial pH is normally 7.35–7.45. Below 7.35 is acidaemia and above 7.45 alkalaemia.

Answer

C · 7.35 to 7.45
4.Human physiology

Answer: A · Higher temperature

Why this is the answer

Higher temperature, CO₂, H⁺ (lower pH) and 2,3-BPG lower haemoglobin's affinity and shift the curve right, releasing more oxygen to active tissues.

Answer

A · Higher temperature
5.Microbiology

Answer: A · They contain an inactivated toxin that still triggers an immune response

Why this is the answer

A toxoid is a toxin treated (for example with formaldehyde) so it is harmless but still antigenic. The immune system makes antibodies that neutralise the real toxin.

Answer

A · They contain an inactivated toxin that still triggers an immune response
6.Microbiology

Answer: C · It showed that cholera spreads through contaminated water, founding modern epidemiology

Why this is the answer

Snow mapped cases around the pump and linked them to its water. Removing the pump handle is a classic example of using epidemiology to find and stop a source, before the bacterium was identified.

Answer

C · It showed that cholera spreads through contaminated water, founding modern epidemiology
7.Pharmacology

Answer: A · 125 mL/h

Why this is the answer

1000 mL ÷ 8 h = 125 mL/h. Infusion pumps are set in mL per hour.

Answer

A · 125 mL/h
8.Pharmacology

Answer: A · Rituximab can reactivate hepatitis B and cause liver failure

Why this is the answer

Rituximab destroys CD20-positive B cells, weakening control of latent hepatitis B. Reactivation can cause fulminant hepatitis, so carriers receive monitoring or antiviral prophylaxis.

Answer

A · Rituximab can reactivate hepatitis B and cause liver failure
9.Arterial blood gas interpretation

Answer: fully compensated

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 7.36: within 7.35–7.45
PaCO₂ 63 mmHg: high (35–45)
HCO₃⁻ 34 mmol/L: high (22–26)

Compensation

fully compensated

Answer

fully compensated
10.Arterial blood gas interpretation

Answer: A–a gradient = 101.4 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 = 0.28·713 − 37/0.8 = 199.6 − 46.25 = 153.4 mmHg

Gradient

A–a = 153.4 − 52 = 101.4 mmHg

Answer

A–a gradient = 101.4 mmHg
11.ECG rhythm strip interpretation

Answer: pause = 1.68 s

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).

Measure

42 small boxes from R to R across the pause

Convert

pause = 42 × 0.04 s = 1.68 s

Compare with P–P

1.68 ÷ 0.72 ≈ 2.3 P–P cycles, not a whole number

Answer

pause = 1.68 s

No P wave appears in the pause, and the pause is not a whole number of P–P cycles: sinus arrest. A pause of exactly two or three cycles would point to sinus exit block.

12.ECG rhythm strip interpretation

Answer: PR interval

Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.

Five steps

1. Rate: R–R = 20 small boxes, 1500 ÷ 20 ≈ 75 per minute.
2. Rhythm: regular, every R–R is 0.80 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.32 s, constant, longer than 0.20 s.
5. QRS: 0.06 s, narrow (below 0.12 s).

First abnormal step

PR interval → first-degree AV block

Answer

PR interval
13.Evidence-based practice statistics

Answer: LR+ = 8.78

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.

Positive likelihood ratio

LR+ = 0.79/(1 − 0.91) = 8.778 ≈ 8.78

Answer

LR+ = 8.78
14.Evidence-based practice statistics

Answer: PPV = 5.5 %

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.

True and false positives (per person)

Se·p = 0.81·0.005 = 0.00405
(1 − Sp)·(1 − p) = 0.07·0.995 = 0.06965

Positive predictive value

PPV = 0.00405/(0.00405 + 0.06965) = 0.05495 ≈ 5.5%

Answer

PPV = 5.5%

The lower the prevalence, the lower the PPV of the same test.

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