ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR
Original practice questions written to the published syllabus · not past papersICAR AICE-JRF/SRF (PhD) is a PhD, fellowship and lectureship exam run by NTA for ICAR, for candidates in India. Computer-based, 2 hours, 120 questions: Part A 20 (general agriculture, animal husbandry, current events), Part B 50 (core group), Part C 50 (specialised subject). +4 / -1, maximum 480.
Source: icar.org.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Deeper chemistry and process questions: casein micelle and kappa-casein cleavage, whey protein denaturation, heat-stable enzymes and UHT defects, D and z values, lactose crystallisation in condensed milk, WPNI, yield equations (Van Slyke), mass balance standardisation and HACCP application.
Milk composition and quality testing, adulteration, heat processing, indigenous and fermented dairy products, dairy standards, packaging and quality control.
Our 26 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 13 medium, 13 hard, across 11 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 ICAR AICE-JRF/SRF and are not shown on any other exam's page.
Which statement about the freezing point of milk is correct?
D. It is nearly constant, around -0.52 to -0.54 degrees C, because it follows the osmotic pressure of blood
Milk is isotonic with blood, so its freezing point depression is nearly constant and set by lactose and soluble salts. Added water raises it towards zero, not lower. Fat and colloidal casein contribute almost nothing to freezing point depression.
Which fatty acid is characteristic of ruminant milk fat and underlies the Reichert-Meissl value?
C. Butyric acid (C4:0)
Butyric acid is found in useful amounts only in ruminant milk fat, and it is steam-volatile and water-soluble, which is what the RM value measures. Lauric acid is high in coconut oil; palmitic acid is abundant in milk fat but also in many other fats.
A UHT process holds milk for 4 s at 140 degrees C. The target spore has D140 = 0.8 s. How many decimal reductions are achieved?
A. 5
Log reductions = time/D = 4/0.8 = 5. 3.2 is 4 x 0.8; 0.2 is 0.8/4; 50 misplaces the decimal in D (4/0.08).
Why is sweetened condensed milk seeded with very fine lactose powder during cooling?
C. To form many tiny lactose crystals so the product does not feel sandy
Lactose is supersaturated in the cooled product; seeding creates many small crystals below the threshold of grittiness. It does not change the sucrose ratio.
Why does ghee develop a granular texture when cooled slowly?
A. High-melting glycerides crystallise slowly into large grains in liquid fat
Graininess is a fat crystallisation phenomenon; slow cooling favours large crystals of high-melting triglycerides. Ghee has almost no water, protein or lactose left.
During churning of cream into butter, which change takes place in the emulsion?
C. Oil-in-water emulsion of cream inverts to water-in-oil emulsion of butter
Agitation breaks membranes, partly crystalline globules clump, and the phase inverts; buttermilk is released. Butter is not a protein gel.
A cooperative wants to mechanise khoa production while keeping its traditional texture. Which equipment is most suited?
A. A scraped-surface heat exchanger or conical vat with rotating scrapers
Khoa needs open-pan style concentration with constant scraping to prevent burning; scraped-surface and scraper-vat systems reproduce this. Spray drying gives powder, not khoa.
What is the basis of the protocooperation between the two yoghurt bacteria?
C. The lactobacillus releases peptides and amino acids; the streptococcus supplies formic acid and CO2
Proteolytic L. bulgaricus supplies amino acids to S. thermophilus, which supplies formic acid and CO2 to the lactobacillus; together they acidify faster than either alone.
Using the Van Slyke formula, Y = (0.93F + C - 0.1) x 1.09 / (1 - W), estimate Cheddar yield per 100 kg milk with 3.6% fat, 2.5% casein and cheese moisture of 37%.
A. About 9.9 kg
0.93 x 3.6 = 3.348; + 2.5 - 0.1 = 5.748; x 1.09 = 6.265; / 0.63 = 9.94 kg. Dividing by 0.37 (moisture instead of 1 - W) gives 16.9 kg; 6.3 kg omits the moisture correction; 10.4 kg drops the 0.93 fat-recovery factor.
What does an ATP bioluminescence swab of a cleaned surface detect?
D. Organic residue from food soil and microbes, not specific organisms
ATP comes from any biological material, so the swab checks cleanliness rapidly but cannot identify or count specific bacteria.
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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