EBVS European College diplomas (e.g. DipECVS, DipECVIM-CA) · Europe · European Board of Veterinary Specialisation and the individual colleges
Original practice questions written to the published syllabus · not past papersEBVS European College diplomas (e.g. DipECVS, DipECVIM-CA) is a specialist certification exam run by European Board of Veterinary Specialisation and the individual colleges, for candidates in Europe. College-specific: residency and exam.
Source: ecvd.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
The laboratory and field methods that find parasites in faeces, blood and skin, the immune response to parasites, parasitic zoonoses and integrated control. Advanced and postgraduate level.
Our 15 EBVS European College diplomas questions in this subject: 0 easy, 1 medium, 14 hard, across 8 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 EBVS European College diplomas and are not shown on any other exam's page.
A flotation device reads two chambers of 1 mL each from a suspension of 5 g of faeces made up to 50 mL. What is its analytical sensitivity (EPG represented by one egg)?
A. 5 EPG
Factor = 50 mL ÷ 2 mL read ÷ 5 g = 5 EPG per egg. 10 EPG would apply if only one chamber were read; 50 EPG is the usual McMaster factor; 4.5 EPG uses the 45 mL of added fluid instead of the 50 mL total.
A practice offers flock monitoring by composite faecal egg count (equal weights from ten lambs pooled). Which statement about composite counts is correct?
C. They estimate the group mean well but cannot show which individuals are high shedders
Pooling equal weights gives an estimate of the mean at lower cost, but information on distribution between animals is lost. It does not destroy eggs, and the pooled result is a mean, not ten individual counts.
A coprophagic dog's flotation shows large numbers of Eimeria oocysts and some strongyle-type eggs. The dog is healthy. The most likely explanation is:
B. Spurious parasitism
Eimeria does not infect dogs (canine coccidia are Cystoisospora); oocysts and strongyle eggs of herbivores pass through coprophagic dogs unchanged. Pseudoparasites are objects that only resemble parasite stages, whereas these are genuine parasite stages of another host. The spurious parasitism here comes from eating herbivore faeces.
A flock has itchy sheep but surface scrapings are negative. Which additional test can detect Psoroptes ovis infestation at flock level, even early in infestation?
C. An ELISA for serum antibodies against Psoroptes ovis
A serological ELISA detects antibodies to P. ovis and is used as a flock test in some countries, including before obvious lesions. Mite eggs are not counted in faeces, and Knott's test detects microfilariae.
A test has sensitivity 90% and specificity 95%. It is used where 10% of animals are infected. What is the positive predictive value?
C. 67%
Per 1,000 animals: 100 infected give 90 true positives; 900 uninfected give 45 false positives. PPV = 90 ÷ 135 = 67%. 90% is the sensitivity, 95% the specificity, and 99% is close to the negative predictive value (855 ÷ 865).
A serosurvey finds 25% of animals positive with a test of sensitivity 80% and specificity 95%. Using the Rogan–Gladen estimator, the true prevalence is about:
D. 26.7%
True prevalence = (AP + Sp − 1) ÷ (Se + Sp − 1) = (0.25 + 0.95 − 1) ÷ (0.80 + 0.95 − 1) = 0.20 ÷ 0.75 = 26.7%. 25% is the uncorrected apparent prevalence; 20% multiplies AP by Se; 31.3% divides AP by Se only, ignoring false positives.
Two independent tests with sensitivities 80% and 70% are used in parallel (an animal is positive if either test is positive). The combined sensitivity is:
C. 94%
In parallel, an infected animal is missed only if both tests miss it: 0.2 × 0.3 = 0.06, so sensitivity = 94%. 56% (0.8 × 0.7) is the combined sensitivity of serial testing; 75% is a simple average; 150% adds them.
Which factor is NOT a recognised reason why vaccines against parasites have been slow to develop?
A. Parasites have too few antigens for the immune system to recognise
Parasites have many antigens; the problem is finding protective ones. Multiple life stages, immune evasion and competition from cheap drugs are the real obstacles.
Ten treated lambs have post-treatment counts (EPG) of 0, 0, 50, 0, 100, 0, 0, 50, 0, 0; the untreated control mean is 400 EPG. Using arithmetic means, what is the FECR?
A. 95%
Treated mean = 200 ÷ 10 = 20 EPG. FECR = 100 × (1 − 20 ÷ 400) = 95%. 5% is the fraction remaining; 83% averages only the three positive lambs (200 ÷ 3 = 66.7 EPG); 50% uses the group total of 200 instead of the mean.
Demodex mites are found in deep scrapings from an adult dog that has developed generalised demodicosis for the first time at 8 years of age. What is the most important next step besides treating the mites?
B. Look for underlying disease or immunosuppressive treatment
Demodex canis is a normal follicular inhabitant; generalised disease starting in adulthood suggests failing immunity from endocrine disease, neoplasia or immunosuppressive drugs. It is not zoonotic or contagious between adult dogs.
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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