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ASRB NET and ARS / SMS / STO Applied Animal Nutrition I: Ruminants: practice questions

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 papers
10worked here, free
20written to this exam in this subject
10syllabus units

About ASRB NET and ARS / SMS / STO

ASRB NET and ARS / SMS / STO (Agricultural Scientists Recruitment Board) is a government recruitment exam run by Agricultural Scientists Recruitment Board, for candidates in India. NET: 150 MCQs, 150 marks, 2 hours, -1/3 per wrong answer. ARS: preliminary 150, then descriptive mains 300, then interview 45.

Official site: pw.live.

We have not been able to confirm these details against the exam body's own specification, so fees, dates and pass marks are left out on purpose. Check the current format, eligibility, fees and dates on the official site.

Applied Animal Nutrition I: Ruminants in ASRB NET and ARS / SMS / STO

The rumen and how to feed cattle, buffalo, sheep and goats through each stage of life, including ration formulation by the Pearson square and least-cost methods.

Our 20 ASRB NET and ARS / SMS / STO questions in this subject: 2 easy, 10 medium, 8 hard, across 10 units.

10 practice questions, worked

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 ASRB NET and ARS / SMS / STO and are not shown on any other exam's page.

Question 1 · medium

The classical rumen ammonia-nitrogen concentration below which microbial protein synthesis becomes limited in vitro is about:

  1. 0.5 mg NH3-N per 100 ml
  2. 50 mg NH3-N per 100 ml
  3. 5 mg NH3-N per 100 ml
  4. 150 mg NH3-N per 100 ml
Answer and working

C. 5 mg NH3-N per 100 ml

Satter and Slyter's classical work gave about 5 mg NH3-N per 100 ml as the level for maximal microbial growth in vitro; higher levels may be needed for maximal fibre digestion on poor forage. Values ten times higher indicate excess nitrogen.

Question 2 · medium

What is the metabolic body weight of a 450 kg cow?

  1. 337.5 kg
  2. 59.9 kg
  3. 97.7 kg
  4. 112.5 kg
Answer and working

C. 97.7 kg

450^0.75 = 97.7 kg. 337.5 kg is 450 × 0.75 (exponent misread as a multiplier), 59.9 kg is 450^0.67, and 112.5 kg is 450/4.

Question 3 · hard

A concentrate mixture must contain 18 percent CP and must include 2 percent mineral mixture and 1 percent salt (no protein). The rest is maize (9 percent CP) and oilseed cake (42 percent CP). What percentage of the final mixture is maize?

  1. 72.7%
  2. 70.5%
  3. 28.1%
  4. 68.9%
Answer and working

D. 68.9%

The 97 percent variable portion must supply all protein, so its target is 18 ÷ 0.97 = 18.56 percent. Parts maize = 42 − 18.56 = 23.44; parts cake = 18.56 − 9 = 9.56; maize = 23.44/33 × 97 = 68.9 percent. 72.7 percent ignores the fixed items; 70.5 percent scales without raising the target; 28.1 percent is the cake.

Question 4 · medium

On a dry matter basis, a ration of wheat straw (4 percent CP) and concentrate (20 percent CP) must contain 10 percent CP in total DM. What proportion of the ration DM should be concentrate?

  1. 62.5%
  2. 37.5%
  3. 50.0%
  4. 40.0%
Answer and working

B. 37.5%

Parts straw = 20 − 10 = 10; parts concentrate = 10 − 4 = 6; total 16. Concentrate = 6/16 = 37.5 percent. 62.5 percent is the straw share; 50 percent assumes equal parts; 40 percent is a rounding guess.

Question 5 · hard

A grain mix of maize (9 percent CP) and wheat bran (15 percent CP) in a 1:1 ratio is to be combined with an oilseed cake (42 percent CP) to give 20 percent CP. What percentage of the final mixture is the cake?

  1. 73.3%
  2. 20.0%
  3. 26.7%
  4. 33.3%
Answer and working

C. 26.7%

Grain mix CP = (9 + 15)/2 = 12 percent. Parts grain = 42 − 20 = 22; parts cake = 20 − 12 = 8; total 30. Cake = 8/30 = 26.7 percent. 73.3 percent is the grain share; 33.3 percent comes from using maize alone (9 percent) as the grain value; 20 percent confuses the target CP with the cake share.

Question 6 · easy

Least-cost ration formulation is solved mathematically by:

  1. linear programming
  2. the Pearson square
  3. the Weende scheme
  4. regression analysis
Answer and working

A. linear programming

Least-cost formulation minimises total cost subject to nutrient and inclusion constraints, a linear programming problem solved by the simplex method or equivalent. The Pearson square handles only one nutrient between two feeds; Weende is a feed analysis scheme.

Question 7 · medium

In a herd where mature cows weigh 600 kg, what is the target breeding weight for heifers at 55 percent of mature body weight?

  1. 270 kg
  2. 510 kg
  3. 330 kg
  4. 550 kg
Answer and working

C. 330 kg

0.55 × 600 = 330 kg. 270 kg is 45 percent, 510 kg is the 85 percent post-calving target, and 550 kg confuses 55 percent with 550 kg.

Question 8 · hard

A close-up ration contains (DM basis) 0.12 percent Na, 1.20 percent K, 0.65 percent Cl and 0.40 percent S. Using DCAD = (Na + K) − (Cl + S) in mEq/kg DM, with atomic weights Na 23, K 39.1, Cl 35.5, S 32.1 and S taken as divalent, what is the DCAD?

  1. 51 mEq/kg
  2. 0.27 mEq/kg
  3. 791 mEq/kg
  4. -73 mEq/kg
Answer and working

D. -73 mEq/kg

Na = 0.12 × 10000 ÷ 23 = 52; K = 1.20 × 10000 ÷ 39.1 = 307; Cl = 0.65 × 10000 ÷ 35.5 = 183; S = 0.40 × 10000 ÷ 32.1 × 2 = 249. DCAD = 52 + 307 − 183 − 249 = -73 mEq/kg. Forgetting the valence of S gives 51; using percentages directly gives 0.27; adding all four ions gives 791.

Question 9 · hard

A grass silage sample has total nitrogen of 2.5 g/100 g DM and ammonia-nitrogen of 0.35 g/100 g DM. What is ammonia-N as a percentage of total N, and how is the silage judged?

  1. 14 percent, moderate protein breakdown, between good and poor
  2. 0.9 percent, excellent fermentation and almost no proteolysis
  3. 71 percent, which is typical of a well-made, wilted silage
  4. 14 percent, but this shows that the silage is too dry to ferment
Answer and working

A. 14 percent, moderate protein breakdown, between good and poor

0.35 ÷ 2.5 × 100 = 14 percent. Below about 10 percent is good and above about 15 percent poor, so 14 percent indicates moderate proteolysis. 0.9 percent comes from multiplying instead of dividing (0.35 × 2.5 = 0.875); 71 percent is 2.5 ÷ 0.35 × 10; NH3-N does not indicate dryness.

Question 10 · hard

A lucerne hay has 35 percent ADF and 50 percent NDF. Using DDM = 88.9 − 0.779 × ADF, DMI = 120 ÷ NDF and RFV = DDM × DMI ÷ 1.29, what is its relative feed value?

  1. 115
  2. 148
  3. 133
  4. 62
Answer and working

A. 115

DDM = 88.9 − 27.3 = 61.6; DMI = 120 ÷ 50 = 2.4 percent of BW; RFV = 61.6 × 2.4 ÷ 1.29 = 115. 148 forgets to divide by 1.29; 133 swaps ADF and NDF; 62 stops at DDM.

A study plan for Applied Animal Nutrition I: Ruminants in ASRB NET and ARS / SMS / STO

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.

  1. First, the heavy units (60% of the 20 questions): Ration formulation by the Pearson square (5, 25%); Least-cost ration formulation (3, 15%); Evaluation of silage and hay (2, 10%); Feeding of heifers (2, 10%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Nutrient requirements and feeding standards (ICAR 2013, NRC and NASEM) (2, 10%); Rumen manipulation (2, 10%); Area-specific mineral mixtures (1, 5%); Clinical nutrition (1, 5%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Feeding of dry animals (1, 5%); Rumen ecosystem and function (1, 5%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Evaluation of silage and hay (2 hard of 2), Feeding of dry animals (1 hard of 1), Feeding of heifers (1 hard of 2).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Bank the easy marks: Least-cost ration formulation (1 easy of 3); Ration formulation by the Pearson square (1 easy of 5).

    Do these early in revision so they are fast on the day.

  6. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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