Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
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.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
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
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.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
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.
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.
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
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
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 type of epithelium lines most of the trachea?
ASimple cuboidal epithelium
BTransitional epithelium
CPseudostratified ciliated columnar epithelium
DStratified squamous epithelium
2.Human anatomyHard
Which structure is found in the posterior mediastinum?
AOesophagus
BHeart
CThymus
DAscending aorta
3.Human physiologyMedium
Which hormone is released by the adrenal cortex in response to ACTH?
AAldosterone, as the main response
BAdrenaline
CCortisol
DThyroxine
4.Human physiologyHard
According to the length–tension relationship, why does a skeletal muscle fibre produce less force when overstretched?
AThe muscle runs out of ATP when stretched
BThe neuromuscular junction stops working
CActin and myosin overlap less, so fewer cross-bridges form
DCalcium is no longer released from the sarcoplasmic reticulum when stretched
5.MicrobiologyMedium
What does an ELISA test detect?
AThe DNA sequence of the whole bacterial genome
BBacterial growth on solid agar plates
CThe shape of bacteria under the microscope
DSpecific antigens or antibodies, using an enzyme that causes a colour change
6.MicrobiologyHard
Primary syphilis typically presents with:
ADamage to the heart and brain
BA painless ulcer (chancre) at the site of infection
CA widespread rash on the palms and soles
DA painful cluster of small blisters on the genitals or the mouth
7.PharmacologyMedium
During an oxytocin infusion, contractions become very frequent and the fetal heart rate shows late decelerations. What is the nurse's first action?
AIncrease the infusion to finish labour quickly
BGive a second oxytocin bolus
CPosition the woman flat on her back
DStop the oxytocin infusion
8.PharmacologyHard
A noradrenaline infusion is running through a peripheral cannula. The skin around the site becomes pale, cold and hard. What should the nurse do?
ALeave the infusion running and recheck in 4 hours
BStop the infusion at that site and follow the extravasation protocol
CIncrease the rate to improve blood flow to the area
DApply heat and massage the area, then continue the infusion
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 75 mmHg (10.0 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
Practice values: Na⁺ 135, Cl⁻ 89 and HCO₃⁻ 10 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 135 mmol/LCl⁻ 89 mmol/LHCO₃⁻ 10 mmol/L
delta ratio =
11.ECG rhythm strip interpretationMedium
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
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
1idioventricular rhythm
2paced rhythm, failure to capture
3dual-chamber (AV) paced rhythm
4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 210 patients with a disease (cases) and 500 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 = 48b = 185c = 162d = 315
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 84% and specificity 86% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 5%Se = 84%Sp = 86%positive
p′ = %
3Solutions
1.Human anatomy
Answer: C · Pseudostratified ciliated columnar epithelium
Why this is the answer
Respiratory epithelium is pseudostratified ciliated columnar with goblet cells. Mucus traps particles and cilia move it up towards the pharynx.
Answer
C · Pseudostratified ciliated columnar epithelium
2.Human anatomy
Answer: A · Oesophagus
Why this is the answer
The posterior mediastinum contains the oesophagus, descending aorta, azygos veins and thoracic duct. The heart and ascending aorta are in the middle mediastinum; the thymus is in the superior and anterior mediastinum.
Answer
A · Oesophagus
3.Human physiology
Answer: C · Cortisol
Why this is the answer
ACTH from the anterior pituitary stimulates cortisol release from the zona fasciculata. Adrenaline comes from the adrenal medulla.
Answer
C · Cortisol
4.Human physiology
Answer: C · Actin and myosin overlap less, so fewer cross-bridges form
Why this is the answer
Maximum force occurs at an optimal sarcomere length with best filament overlap. Overstretching reduces overlap; too short a length makes filaments collide.
Answer
C · Actin and myosin overlap less, so fewer cross-bridges form
5.Microbiology
Answer: D · Specific antigens or antibodies, using an enzyme that causes a colour change
Why this is the answer
In an enzyme-linked immunosorbent assay, antibodies linked to an enzyme bind the target. Adding the substrate produces a colour, showing whether the antigen or antibody is present.
Answer
D · Specific antigens or antibodies, using an enzyme that causes a colour change
6.Microbiology
Answer: B · A painless ulcer (chancre) at the site of infection
Why this is the answer
Treponema pallidum causes a painless chancre in the primary stage. The secondary stage brings rash and systemic symptoms, and tertiary syphilis can damage the heart and nervous system. Painful blisters suggest herpes.
Answer
B · A painless ulcer (chancre) at the site of infection
7.Pharmacology
Answer: D · Stop the oxytocin infusion
Why this is the answer
Excessive contractions reduce placental blood flow and cause fetal distress. The oxytocin is stopped first, then the woman is turned on her side, IV fluids and oxygen are given as per protocol and the provider is informed.
Answer
D · Stop the oxytocin infusion
8.Pharmacology
Answer: B · Stop the infusion at that site and follow the extravasation protocol
Why this is the answer
Extravasation of a vasopressor can cause tissue necrosis. The infusion is stopped and moved to another site, and phentolamine, an alpha blocker, may be injected into the area to reverse vasoconstriction.
Answer
B · Stop the infusion at that site and follow the extravasation protocol
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.5 mmol/L
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.
Change in PaCO₂
|75 − 40| = 35 mmHg → 3.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.5 = 27.5 mmol/L
Answer
HCO₃⁻ = 27.5 mmol/L
10.Arterial blood gas interpretation
Answer: delta ratio = 1.714
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 = 135 − (89 + 10) = 36
Delta ratio
(36 − 12)/(24 − 10) = 24/14 = 1.714
Answer
delta ratio = 1.714
Between 1 and 2: a pure high anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: QTc = 0.426 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 = 10.5 × 0.04 = 0.42 s
RR = 24 × 0.04 = 0.96 s
Fridericia’s formula
QTc = 0.42/∛0.96 = 0.42/0.9865 = 0.426 s
Answer
QTc = 0.426 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (426 ms)
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: 4 QRS complexes in 6 s, × 10 = 40 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 = 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.
Odds of exposure
cases: 48/162 = 0.2963
controls: 185/315 = 0.5873
Odds ratio
OR = (48·315)/(185·162) = 0.5045 ≈ 0.5
Answer
OR = 0.5
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: p′ = 24 %
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.