ASRB NET and ARS / SMS / STO (Agricultural Scientists Recruitment Board) · India · Agricultural Scientists Recruitment Board
Original practice questions written to the published syllabus · not past papersWhat feed is made of and how its value is measured: proximate analysis, energy partition, protein quality, minerals, vitamins, anti-nutritional factors and feed processing.
In the exams: ICAR AICE-JRF core group 05; ICAR AIEEA-PG 14 (Animal Nutrition); TNPSC Nutrition 20; RPSC (feeds of Rajasthan); EGEL Plus nutrition 13; ASRB NET (numerical problems, per coaching sources)
How our questions for ASRB NET and ARS / SMS / STO spread across the units of Principles of Animal Nutrition & Feed Technology. Difficulty: 8 easy, 12 medium, 16 hard.
| Energy partition: GE, DE, ME and NE | 13 | |
| Principles of feed formulation | 7 | |
| Measures of protein quality: BV, NPU, CP and DCP | 6 | |
| Proximate principles and Weende analysis | 5 | |
| Rumen-degradable and undegradable protein | 3 | |
| Water | 1 | |
| Unconventional feeds | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A concentrate analyses (as fed): moisture 10%, CP 12%, EE 3%, CF 8%, total ash 7%. Its nitrogen-free extract on an as-fed basis is:
B. 60%
NFE = 100 - (10 + 12 + 3 + 8 + 7) = 60%. Leaving out moisture gives 70%, a common slip; 66.7% is the value on DM basis, not as fed; forgetting to subtract CF gives 68%, the total carbohydrate.
Further reading: McDonald et al., Animal Nutrition, 7th ed.
A green fodder contains 20% dry matter and 9% crude protein on a DM basis. Its crude protein on an as-fed (fresh) basis is:
C. 1.8%
As-fed = DM-basis value x DM fraction = 9 x 0.20 = 1.8%. Dividing by the DM fraction (45.0%) converts the wrong way; 7.2% multiplies by the moisture fraction (0.80) instead of the DM fraction.
Further reading: McDonald et al., Animal Nutrition, 7th ed.
A straw has NDF 62%, ADF 38% and ADL 6% (DM basis, ash ignored). Its hemicellulose and cellulose contents are, respectively:
D. 24% and 32%
Hemicellulose = NDF - ADF = 62 - 38 = 24%. Cellulose = ADF - ADL = 38 - 6 = 32%. Taking all of ADF as cellulose (38%) forgets that ADF also contains lignin, and 56% (NDF - ADL) is hemicellulose plus cellulose together.
Further reading: Van Soest, Nutritional Ecology of the Ruminant, 2nd ed.
A bullock eats 10.0 kg DM a day and voids 3.8 kg faecal DM. The apparent DM digestibility coefficient is:
A. 62.0%
DC = (10.0 - 3.8) / 10.0 x 100 = 62.0%. 38.0% is the indigestible fraction; dividing by faecal output (163.2%) uses the wrong denominator; 263.2% is intake divided by faecal output (10.0/3.8).
Further reading: McDonald et al., Animal Nutrition, 7th ed.
A calf consumes 120 g N; faecal N is 36 g, of which 12 g is metabolic faecal N. The true digestibility of nitrogen is:
D. 80.0%
True DC = [120 - (36 - 12)] / 120 x 100 = 80.0%. 70.0% is apparent digestibility; subtracting MFN as an extra loss (60.0%) goes in the wrong direction; 90.0% subtracts only the 12 g MFN from intake.
Further reading: McDonald et al., Animal Nutrition, 7th ed.
In a nitrogen balance study a pig eats 20 g N a day, excreting 5 g in faeces and 6 g in urine. The apparent biological value of the protein is:
B. 60.0%
Absorbed = 20 - 5 = 15 g; retained = 15 - 6 = 9 g. BV = 9/15 x 100 = 60.0%. 45.0% is NPU (retained/intake); 75.0% is apparent N digestibility; 40.0% is the share of absorbed N lost in urine (6/15).
Further reading: McDonald et al., Animal Nutrition, 7th ed.
A rat eats 16 g N over a trial; faecal N is 4 g and urinary N 4 g. Net protein utilisation (apparent) is:
D. 50.0%
NPU = N retained / N intake = (16 - 4 - 4) / 16 x 100 = 50.0%. 66.7% is BV; 75.0% is digestibility; 33.3% is the share of absorbed N lost in urine (4/12).
Further reading: McDonald et al., Animal Nutrition, 7th ed.
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