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

Where does fertilisation most commonly take place?

  1. ABody of the uterus
  2. BAmpulla of the uterine tube
  3. COvary
  4. DCervix
2.Human anatomyHard

A patient has a drooping eyelid, a dilated pupil and an eye that rests "down and out". Which nerve is most likely damaged?

  1. AOptic nerve (II)
  2. BAbducens nerve (VI)
  3. CTrochlear nerve (IV)
  4. DOculomotor nerve (III)
3.Human physiologyMedium

Which ABO blood group has both anti-A and anti-B antibodies in the plasma?

  1. AGroup O
  2. BGroup B
  3. CGroup AB
  4. DGroup A
4.Human physiologyHard

Which mechanism mainly creates the concentration gradient in the medulla of the kidney?

  1. AThe countercurrent multiplier of the loop of Henle
  2. BADH acting on the descending limb
  3. CGlomerular filtration
  4. DSecretion of urea into the proximal tubule
5.MicrobiologyMedium

Which bacterium causes whooping cough (pertussis)?

  1. ACorynebacterium diphtheriae
  2. BBordetella pertussis
  3. CHaemophilus influenzae
  4. DMycoplasma pneumoniae
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. BSubunit vaccines
  3. CLive attenuated vaccines
  4. DInactivated whole-virus vaccines
7.PharmacologyMedium

A patient is starting olanzapine. Which monitoring is most important over the following months?

  1. AWeight, blood glucose and blood lipids
  2. BINR and platelet count
  3. CHearing tests and vision tests every month
  4. DPeak expiratory flow
8.PharmacologyHard

A patient with epilepsy needs pain relief. Why is tramadol used with caution?

  1. AIt speeds up the metabolism of all antiepileptic drugs
  2. BIt causes a sharp rise in blood glucose
  3. CIt lowers the seizure threshold
  4. DIt always causes respiratory arrest in epilepsy
9.Arterial blood gas interpretationMedium
pH7.357.457.29PaCO₂mmHg354560HCO₃⁻mmol/L222628uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.29, PaCO₂ 60 mmHg (8.0 kPa) and HCO₃⁻ 28 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.29PaCO₂ 60 mmHg (8.0 kPa)HCO₃⁻ 28 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
10.Arterial blood gas interpretationHard

Practice values: Na⁺ 136, Cl⁻ 104 and HCO₃⁻ 13 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.

Na⁺ 136 mmol/LCl⁻ 104 mmol/LHCO₃⁻ 13 mmol/L
delta ratio =
11.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mVfirst-degree AV blocknormal sinus rhythmjunctional rhythmsinus bradycardia

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. 1first-degree AV block
  2. 2normal sinus rhythm
  3. 3junctional rhythm
  4. 4sinus bradycardia
12.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVfirst-degree AV blocksecond-degree AV block type IIsinus bradycardiathird-degree (complete) 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. 1first-degree AV block
  2. 2second-degree AV block type II
  3. 3sinus bradycardia
  4. 4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Se76%Sp61%LR− = ?

Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 76% and a specificity of 61%. Find the negative likelihood ratio LR−. Round to two decimal places.

Se = 76%Sp = 61%
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 84% and specificity 62% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.

pre-test 30%Se = 84%Sp = 62%positive
p′ = %

3Solutions

1.Human anatomy

Answer: B · Ampulla of the uterine tube

Why this is the answer

Fertilisation usually occurs in the ampulla of the uterine (fallopian) tube. The embryo then travels to the uterus to implant.

Answer

B · Ampulla of the uterine tube
2.Human anatomy

Answer: D · Oculomotor nerve (III)

Why this is the answer

The oculomotor nerve supplies most eye muscles, levator palpebrae and the pupil constrictor. Lesion leaves lateral rectus and superior oblique unopposed, so the eye goes down and out.

Answer

D · Oculomotor nerve (III)
3.Human physiology

Answer: A · Group O

Why this is the answer

Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.

Answer

A · Group O
4.Human physiology

Answer: A · The countercurrent multiplier of the loop of Henle

Why this is the answer

The ascending limb actively transports NaCl into the interstitium without letting water follow, and the countercurrent flow multiplies this into a steep gradient that ADH then uses to concentrate urine.

Answer

A · The countercurrent multiplier of the loop of Henle
5.Microbiology

Answer: B · Bordetella pertussis

Why this is the answer

Bordetella pertussis causes long coughing fits with a "whoop". It is most dangerous for young babies, which is why vaccination in pregnancy is offered in many countries.

Answer

B · Bordetella pertussis
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 · Weight, blood glucose and blood lipids

Why this is the answer

Olanzapine and several other atypical antipsychotics cause weight gain, insulin resistance and dyslipidaemia. Weight, waist size, glucose and lipids are checked regularly.

Answer

A · Weight, blood glucose and blood lipids
8.Pharmacology

Answer: C · It lowers the seizure threshold

Why this is the answer

Tramadol can cause seizures, especially at high doses or with other drugs that lower the seizure threshold. Its serotonergic action also raises the risk of serotonin syndrome.

Answer

C · It lowers the seizure threshold
9.Arterial blood gas interpretation

Answer: partially 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 7.29: acidaemia (below 7.35)
PaCO₂ 60 mmHg: high (35–45)
HCO₃⁻ 28 mmol/L: high (22–26)

Compensation

partially compensated

Answer

partially compensated
10.Arterial blood gas interpretation

Answer: delta ratio = 0.6364

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 = 136 − (104 + 13) = 19

Delta ratio

(19 − 12)/(24 − 13) = 7/11 = 0.6364

Answer

delta ratio = 0.6364

Below 1: an added normal anion gap acidosis.

11.ECG rhythm strip interpretation

Answer: sinus bradycardia

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.

Five steps

1. Rate: R–R = 33 small boxes, 1500 ÷ 33 ≈ 45 per minute.
2. Rhythm: regular, every R–R is 1.32 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).

Interpretation

sinus bradycardia

Answer

sinus bradycardia
12.ECG rhythm strip interpretation

Answer: second-degree AV block type II

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: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: the P waves are regular; R–R is regular except for a long pause where a QRS is missing.
3. P waves: upright and regular (P–P 0.96 s, 63 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: constant at 0.24 s in the conducted beats.
5. QRS: 0.12 s, wide (0.12 s or more).

Interpretation

second-degree AV block type II

Answer

second-degree AV block type II

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: LR− = 0.39

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.

Negative likelihood ratio

LR− = (1 − 0.76)/0.61 = 0.3934 ≈ 0.39

Answer

LR− = 0.39
14.Evidence-based practice statistics

Answer: p′ = 48.6 %

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+ = 0.84/(1 − 0.62) = 2.211

Pre-test odds

0.3/(1 − 0.3) = 0.4286

Post-test odds

0.4286·2.211 = 0.9474

Post-test probability

p′ = 0.9474/(1 + 0.9474) = 0.4865 ≈ 48.6%

Answer

p′ = 48.6%

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