RCVS Certificate in Advanced Veterinary Practice (CertAVP) · UK · RCVS, with modules through accredited universities
Original practice questions written to the published syllabus · not past papersRCVS Certificate in Advanced Veterinary Practice (CertAVP) is a professional certification run by RCVS, with modules through accredited universities, for candidates in UK. Modular postgraduate certificate: 1 A module, 1 B module and 4 C modules. Designated certificates carry a synoptic oral exam.
Source: rcvs.org.uk.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Poisoning by metals, plants, mycotoxins, pesticides, non-protein nitrogen, venoms and pollutants, with residues in food and the antidotes used.
Our 17 RCVS Certificate in Advanced Veterinary Practice questions in this subject: 0 easy, 6 medium, 11 hard, across 13 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to RCVS Certificate in Advanced Veterinary Practice and are not shown on any other exam's page.
A 60 kg sheep eats 1.5 kg of feed a day containing 5 ppm selenium. What is its daily selenium intake per kg body weight?
B. 0.125 mg/kg
5 ppm = 5 mg/kg feed. 5 × 1.5 = 7.5 mg per day; 7.5 ÷ 60 = 0.125 mg/kg body weight. 7.5 is the total daily intake, not per kg body weight, the usual slip; 0.083 ignores the 1.5 kg eaten (5 ÷ 60) and 0.056 divides by the feed instead of multiplying (5 ÷ (1.5 × 60)).
A 450 kg cow drinks 50 L a day of water containing 8 mg/L fluoride. What is her fluoride intake from water, per kg body weight?
D. About 0.89 mg/kg per day
Intake = 8 mg/L × 50 L = 400 mg per day; 400 ÷ 450 = 0.89 mg/kg per day. 400 is the total, not per kg; 1.1 inverts the division (450 ÷ 400) and 0.018 ignores the 50 L drunk (8 ÷ 450).
A dairy cow drinks 60 L a day of well water containing 100 mg/L nitrate-nitrogen. How much nitrate (NO3) does she drink per day?
A. About 26.6 g
60 L × 100 mg/L = 6,000 mg = 6 g NO3-N; × 4.43 = 26.6 g NO3. 6 g is the nitrate-N, not nitrate, the usual reporting-basis slip; 1.35 g divides by 4.43 instead of multiplying, and 266 g converts 6,000 mg as 60 g.
Which specific antidote can be used for a dog with life-threatening arrhythmias after eating oleander leaves?
B. Digoxin-specific antibody fragments
Oleander glycosides resemble digoxin, and digoxin-specific Fab fragments cross-react with them. Pralidoxime and fomepizole target OPs and ethylene glycol respectively.
A 100 kg pig eats 2 kg a day of feed containing 300 ppb aflatoxin. What is its daily intake per kg body weight?
C. 6 micrograms/kg
300 ppb = 300 micrograms/kg feed. 300 × 2 = 600 micrograms per day; 600 ÷ 100 = 6 micrograms/kg. 600 is the total intake; 6 mg/kg confuses ppb with ppm; 3 micrograms/kg ignores the 2 kg eaten (300 ÷ 100).
A 10 L sheep spray at 0.05% active ingredient is to be made from a 50% concentrate. How much concentrate is needed?
A. 10 mL
C1V1 = C2V2: 50% × V1 = 0.05% × 10,000 mL, so V1 = 500 ÷ 50 = 10 mL. 100 mL is a tenfold error that would give 0.5%, a toxic overdose; 1 L enters 50% as 0.5 but 0.05% as 0.05, and 1 mL converts 10 L as 1,000 mL.
How is the amount of antivenom for an envenomated dog decided?
A. By venom load and clinical response, per the product label, not body weight
A small dog receives the same venom as a large one, so antivenom is dosed to neutralise venom and repeated as signs require.
Why is chloramphenicol prohibited in food-producing animals in many countries?
D. No ADI can be set, partly because of dose-independent aplastic anaemia in people
Idiosyncratic aplastic anaemia in humans is not dose-related, so no ADI or MRL can be established.
A 500 kg cow eats 12 kg of feed containing 50 ppm of a contaminant. What is her daily dose?
A. 1.2 mg/kg body weight
50 ppm = 50 mg/kg feed; 50 × 12 = 600 mg per day; 600 ÷ 500 = 1.2 mg/kg. 600 is the total intake, not per kg body weight; 0.1 ignores the 12 kg eaten (50 ÷ 500) and 4.2 divides 50 by 12 instead of multiplying.
A 10 kg dog eats five pieces of sugar-free gum, each containing 1 g of xylitol. What dose has it received?
B. 500 mg/kg
Total = 5 × 1 g = 5 g = 5,000 mg; 5,000 ÷ 10 = 500 mg/kg (0.5 g/kg). This is above the commonly cited threshold for hypoglycaemia (about 0.1 g/kg) and at the level cited for liver failure (about 0.5 g/kg). 50 mg/kg is a tenfold slip; 100 mg/kg counts only one piece (1 g ÷ 10 kg).
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
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