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 supplies the diaphragm?

  1. AVagus nerve
  2. BAccessory nerve
  3. CPhrenic nerve
  4. DIntercostal nerves
2.Human anatomyHard

Why is the proximal pole of the scaphoid at risk of avascular necrosis after a waist fracture?

  1. AIts only artery runs through the lunate
  2. BIts blood supply enters mainly from the distal end
  3. CIt is drained by a single vein that clots easily
  4. DIt has no blood supply at all
3.Human physiologyMedium

What is anatomical dead space?

  1. AThe air that stays in the lungs after full exhalation (residual volume)
  2. BAlveoli that have collapsed completely
  3. CThe volume of the conducting airways where no gas exchange happens
  4. DThe space between the two pleural layers
4.Human physiologyHard

A patient hyperventilating during a panic attack has tingling fingers. What would the arterial blood gas most likely show?

  1. AHigh pH and high bicarbonate
  2. BLow pH and low bicarbonate
  3. CLow pH and high PaCO₂
  4. DHigh pH and low PaCO₂
5.MicrobiologyMedium

MRSA is resistant to methicillin because it has:

  1. ANo cell wall at all
  2. BAn altered penicillin-binding protein (PBP2a)
  3. CA polysaccharide capsule that blocks the entry of all beta-lactams
  4. DAn efflux pump for every antibiotic
6.MicrobiologyHard

A healthcare worker's blood test is positive for anti-HBs (hepatitis B surface antibody) and negative for HBsAg and anti-HBc. What does this most likely mean?

  1. APast natural infection that has cleared
  2. BCurrent acute hepatitis B infection
  3. CImmunity from hepatitis B vaccination
  4. DChronic hepatitis B infection
7.PharmacologyMedium

During a rapid infusion of vancomycin, a patient develops flushing and a red rash on the face, neck and upper body. What is the most likely cause and action?

  1. AA sign that the infection is spreading; give a second dose
  2. BA normal effect; increase the infusion rate to finish sooner
  3. CAcute kidney failure; stop all IV fluids and antibiotics at once
  4. DAn infusion reaction; stop or slow the infusion and inform the prescriber
8.PharmacologyHard

Which strategy is most effective at preventing errors with IV insulin infusions?

  1. AWriting "U" for units on the order
  2. BPreparing insulin infusions in advance and leaving them unlabelled
  3. CUsing insulin syringes to measure large IV volumes
  4. DAn independent double check and a smart pump with a drug library
9.Arterial blood gas interpretationMedium
pH7.357.457.11PaCO₂mmHg354578HCO₃⁻mmol/L222624uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.11, PaCO₂ 78 mmHg (10.4 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.11PaCO₂ 78 mmHg (10.4 kPa)HCO₃⁻ 24 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 91 mmHg (12.1 kPa) and PaCO₂ 58 mmHg (7.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 28%PaO₂ 91 mmHg (12.1 kPa)PaCO₂ 58 mmHg (7.7 kPa)
A–a gradient = mmHg
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 Fridericia’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 IIthird-degree (complete) AV blockfirst-degree AV blocksecond-degree AV block type I (Wenckebach)

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. 2third-degree (complete) AV block
  3. 3first-degree AV block
  4. 4second-degree AV block type I (Wenckebach)
13.Evidence-based practice statisticsMedium
casesperson-yearsexposed265200not exposed191900

Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 5200 person-years with 26 new cases, and unexposed people 1900 person-years with 19 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.

exposed: 26/5200 PYnot exposed: 19/1900 PY
IRR =
14.Evidence-based practice statisticsHard
pre-test60%post-test?p′LR−

Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 60% for a disease. A test with sensitivity 95% and specificity 72% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.

pre-test 60%Se = 95%Sp = 72%negative
p′ = %

3Solutions

1.Human anatomy

Answer: C · Phrenic nerve

Why this is the answer

The phrenic nerve (roots C3–C5) is the motor supply to the diaphragm. High cervical cord injury above C3 therefore stops spontaneous breathing.

Answer

C · Phrenic nerve
2.Human anatomy

Answer: B · Its blood supply enters mainly from the distal end

Why this is the answer

Branches of the radial artery enter the scaphoid mainly distally and flow backwards. A fracture through the waist can cut off the proximal pole, causing avascular necrosis.

Answer

B · Its blood supply enters mainly from the distal end
3.Human physiology

Answer: C · The volume of the conducting airways where no gas exchange happens

Why this is the answer

Anatomical dead space (about 150 ml) is air in the nose, trachea and bronchi. It is moved with every breath but never reaches the alveoli.

Answer

C · The volume of the conducting airways where no gas exchange happens
4.Human physiology

Answer: D · High pH and low PaCO₂

Why this is the answer

Hyperventilation blows off CO₂ and causes acute respiratory alkalosis. Alkalosis increases calcium binding to albumin, lowering ionised calcium and causing tingling.

Answer

D · High pH and low PaCO₂
5.Microbiology

Answer: B · An altered penicillin-binding protein (PBP2a)

Why this is the answer

The mecA gene codes for PBP2a, which has low affinity for beta-lactams, so the bacterium can keep building its cell wall despite methicillin.

Answer

B · An altered penicillin-binding protein (PBP2a)
6.Microbiology

Answer: C · Immunity from hepatitis B vaccination

Why this is the answer

The vaccine contains HBsAg, so a vaccinated person makes anti-HBs but no anti-HBc. Past natural infection gives both anti-HBs and anti-HBc, and ongoing infection shows HBsAg.

Answer

C · Immunity from hepatitis B vaccination
7.Pharmacology

Answer: D · An infusion reaction; stop or slow the infusion and inform the prescriber

Why this is the answer

Rapid vancomycin infusion releases histamine, causing flushing and itching of the upper body (formerly called "red man syndrome"). It is managed by stopping or slowing the infusion, and sometimes with an antihistamine.

Answer

D · An infusion reaction; stop or slow the infusion and inform the prescriber
8.Pharmacology

Answer: D · An independent double check and a smart pump with a drug library

Why this is the answer

Independent double checks, standard concentrations and smart pumps with dose limits reduce insulin errors. "U" for units is avoided because it can be misread as a zero.

Answer

D · An independent double check and a smart pump with a drug library
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 7.11: acidaemia (below 7.35)
PaCO₂ 78 mmHg: high (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)

Compensation

uncompensated

Answer

uncompensated
10.Arterial blood gas interpretation

Answer: A–a gradient = 36.14 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 − 58/0.8 = 199.6 − 72.5 = 127.1 mmHg

Gradient

A–a = 127.1 − 91 = 36.14 mmHg

Answer

A–a gradient = 36.14 mmHg
11.ECG rhythm strip interpretation

Answer: QTc = 0.46 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 = 11.5 × 0.04 = 0.46 s
RR = 25 × 0.04 = 1.00 s

Fridericia’s formula

QTc = 0.46/∛1.00 = 0.46/1 = 0.46 s

Answer

QTc = 0.46 s

Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (460 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 = 29 small boxes, 1500 ÷ 29 ≈ 52 per minute.
2. Rhythm: R–R regular (1.16 s) and P–P regular (0.64 s), but the two are independent.
3. P waves: upright and regular at 94 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: IRR = 0.5

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: 26/5200 = 5 per 1000 person-years
not exposed: 19/1900 = 10 per 1000 person-years

Rate ratio

IRR = 5/10 = 0.5

Answer

IRR = 0.5
14.Evidence-based practice statistics

Answer: p′ = 9.4 %

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 − 0.95)/0.72 = 0.06944

Pre-test odds

0.6/(1 − 0.6) = 1.5

Post-test odds

1.5·0.06944 = 0.1042

Post-test probability

p′ = 0.1042/(1 + 0.1042) = 0.09434 ≈ 9.4%

Answer

p′ = 9.4%

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