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 adrenal gland secretes adrenaline?

  1. AZona reticularis
  2. BAdrenal medulla
  3. CZona fasciculata
  4. DZona glomerulosa
2.Human anatomyHard

A footballer takes a blow to the outside of the planted knee. Which combination of injuries is classically called the "unhappy triad"?

  1. APosterior cruciate ligament, lateral collateral ligament and lateral meniscus
  2. BMedial collateral ligament, lateral collateral ligament and patella
  3. CAnterior cruciate ligament, medial collateral ligament and medial meniscus
  4. DAnterior cruciate ligament, posterior cruciate ligament and patellar ligament
3.Human physiologyMedium

Where does the final activation step of vitamin D (1-alpha hydroxylation) take place?

  1. AKidney
  2. BLiver
  3. CSkin
  4. DSmall intestine
4.Human physiologyHard

At the start of a fever, a patient shivers and feels cold even though body temperature is rising. Why?

  1. AThe thermoreceptors in the skin are destroyed
  2. BSweating has already cooled the body too much
  3. CThe body has lost its ability to produce heat
  4. DPyrogens raise the hypothalamic set point above the current body temperature
5.MicrobiologyMedium

Cholera causes profuse "rice-water" diarrhoea. What is the most important treatment?

  1. AAntiviral drugs
  2. BReplacing fluid and salts by oral or intravenous rehydration
  3. CStopping all oral fluid intake so that the bowel can rest and recover
  4. DAntifungal drugs
6.MicrobiologyHard

On MacConkey agar, pink colonies grow from a urine sample. What does this tell you?

  1. AThe organism is probably a Gram-negative rod that ferments lactose
  2. BThe organism is a Gram-positive coccus
  3. CThe organism is a fungus
  4. DThe organism does not ferment lactose, so it is probably Salmonella
7.PharmacologyMedium

Why should NSAIDs be used with caution in some people with asthma?

  1. AThey increase the clearance of salbutamol
  2. BThey block the anti-inflammatory effect of inhaled corticosteroids
  3. CThey can trigger bronchospasm in aspirin-sensitive asthma
  4. DThey cause mucus plugs by drying the airways
8.PharmacologyHard

Carboprost is a prostaglandin used for postpartum haemorrhage. In which woman should it be avoided?

  1. AA woman who had a vaginal birth
  2. BA woman who is breastfeeding
  3. CA woman with iron deficiency anaemia
  4. DA woman with asthma
9.Arterial blood gas interpretationMedium
pH7.357.457.43PaCO₂mmHg354528HCO₃⁻mmol/L222618uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.43, PaCO₂ 28 mmHg (3.7 kPa) and HCO₃⁻ 18 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.43PaCO₂ 28 mmHg (3.7 kPa)HCO₃⁻ 18 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
10.Arterial blood gas interpretationHard
pH7.357.457.19PaCO₂mmHg354538HCO₃⁻mmol/L222614appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.19, PaCO₂ 38 mmHg (5.1 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.19PaCO₂ 38 mmHg (5.1 kPa)HCO₃⁻ 14 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory 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 I (Wenckebach)atrial fibrillationfirst-degree AV blocksecond-degree AV block type II

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. 2atrial fibrillation
  3. 3first-degree AV block
  4. 4second-degree AV block type II
13.Evidence-based practice statisticsMedium
testdisease presentdisease absenttotalpositive174369543abnegative4615711617cdtotal22019402160

Practice exercise (made-up values, not clinical guidance). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the positive likelihood ratio LR+. Round to two decimal places.

a = 174b = 369c = 46d = 1571
LR+ =
14.Evidence-based practice statisticsHard
pre-test30%post-test?p′LR+

Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 30% for a disease. A test with sensitivity 62% and specificity 81% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.

pre-test 30%Se = 62%Sp = 81%positive
p′ = %

3Solutions

1.Human anatomy

Answer: B · 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

B · Adrenal medulla
2.Human anatomy

Answer: C · Anterior cruciate ligament, medial collateral ligament and medial meniscus

Why this is the answer

A valgus force with rotation tears the MCL, the ACL and the medial meniscus, which is firmly attached to the MCL.

Answer

C · Anterior cruciate ligament, medial collateral ligament and medial meniscus
3.Human physiology

Answer: A · Kidney

Why this is the answer

Vitamin D is made in the skin or eaten, hydroxylated in the liver, and then activated to calcitriol in the kidney. Kidney failure therefore causes vitamin D deficiency.

Answer

A · Kidney
4.Human physiology

Answer: D · Pyrogens raise the hypothalamic set point above the current body temperature

Why this is the answer

Pyrogens trigger prostaglandin E₂ in the hypothalamus, raising the set point. Until body temperature reaches the new set point, heat-gain responses such as shivering occur.

Answer

D · Pyrogens raise the hypothalamic set point above the current body temperature
5.Microbiology

Answer: B · Replacing fluid and salts by oral or intravenous rehydration

Why this is the answer

Cholera toxin makes gut cells secrete large amounts of water and salt. Rapid rehydration saves most lives; antibiotics only shorten the illness.

Answer

B · Replacing fluid and salts by oral or intravenous rehydration
6.Microbiology

Answer: A · The organism is probably a Gram-negative rod that ferments lactose

Why this is the answer

MacConkey agar contains bile salts and crystal violet that inhibit Gram-positive bacteria (selective) and lactose with a pH indicator (differential). Lactose fermenters such as E. coli produce acid and turn pink.

Answer

A · The organism is probably a Gram-negative rod that ferments lactose
7.Pharmacology

Answer: C · They can trigger bronchospasm in aspirin-sensitive asthma

Why this is the answer

In aspirin-exacerbated respiratory disease, COX inhibition increases leukotriene production, causing bronchospasm. These patients should avoid aspirin and other non-selective NSAIDs.

Answer

C · They can trigger bronchospasm in aspirin-sensitive asthma
8.Pharmacology

Answer: D · A woman with asthma

Why this is the answer

Carboprost is a PGF2-alpha analogue that can cause bronchospasm. Oxytocin is first-line, and carboprost is avoided in asthma.

Answer

D · A woman with asthma
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 7.43: within 7.35–7.45
PaCO₂ 28 mmHg: low (35–45)
HCO₃⁻ 18 mmol/L: low (22–26)

Compensation

fully compensated

Answer

fully compensated
10.Arterial blood gas interpretation

Answer: added respiratory acidosis

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

1.5·14 + 8 = 29 → 27–31 mmHg

Compare

measured: 38 mmHg → added respiratory acidosis

Answer

added respiratory acidosis
11.ECG rhythm strip interpretation

Answer: QTc = 0.459 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 = 10 × 0.04 = 0.40 s
RR = 19 × 0.04 = 0.76 s

Bazett’s formula

QTc = 0.40/√0.76 = 0.40/0.8718 = 0.459 s

Answer

QTc = 0.459 s

Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (459 ms)

12.ECG rhythm strip interpretation

Answer: second-degree AV block type I (Wenckebach)

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: 4 QRS complexes in 6 s, × 10 = 40 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 1.00 s, 60 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.20 → 0.32 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).

Interpretation

second-degree AV block type I (Wenckebach)

Answer

second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics

Answer: LR+ = 4.16

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.

Sensitivity and specificity

Se = 174/220 = 0.7909
Sp = 1571/1940 = 0.8098

Positive likelihood ratio

LR+ = 0.7909/(1 − 0.8098) = 4.158 ≈ 4.16

Answer

LR+ = 4.16
14.Evidence-based practice statistics

Answer: p′ = 58.3 %

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+ = 0.62/(1 − 0.81) = 3.263

Pre-test odds

0.3/(1 − 0.3) = 0.4286

Post-test odds

0.4286·3.263 = 1.398

Post-test probability

p′ = 1.398/(1 + 1.398) = 0.5831 ≈ 58.3%

Answer

p′ = 58.3%

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