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ICAR AICE-JRF/SRF Avian and Poultry Production Management: practice questions

ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
20written to this exam in this subject
11syllabus units

About ICAR AICE-JRF/SRF

ICAR 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.

Avian and Poultry Production Management in ICAR AICE-JRF/SRF

Deeper reasoning: sex-linked inheritance and autosexing, broiler breeder feed restriction and uniformity, hatchability troubleshooting from breakout patterns, photoperiod physiology and photorefractoriness, shell-quality physiology, and economics of integration.

Commercial broiler and layer production, brooding, lighting, hatchery operation, egg quality and biosecurity for chicken, duck, quail and turkey.

Our 20 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 5 medium, 15 hard, across 11 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 ICAR AICE-JRF/SRF and are not shown on any other exam's page.

Question 1 · hard

A gold (s/s) cockerel is mated to silver (S/W) hens. Silver is dominant and Z-linked. What colours are the day-old chicks?

  1. Sons silver, daughters gold
  2. Sons gold, daughters silver
  3. All chicks silver
  4. Half of each sex silver
Answer and working

A. Sons silver, daughters gold

Sons get Z-S from the dam and Z-s from the sire, so they are S/s and silver. Daughters get their only Z (s) from the sire and W from the dam, so they are gold. Reversing the result is the usual error when mammalian XY logic is applied.

Question 2 · hard

Oviposition in the hen is triggered mainly by:

  1. Arginine vasotocin and prostaglandins
  2. Oxytocin and relaxin
  3. Prolactin, progesterone and corticosterone
  4. FSH and LH alone
Answer and working

A. Arginine vasotocin and prostaglandins

Arginine vasotocin, the avian neurohypophyseal hormone, and prostaglandins from the ruptured follicle and shell gland drive uterine contraction and oviposition. Oxytocin and relaxin are the mammalian answer. LH drives ovulation, not oviposition, and prolactin is linked to broodiness.

Question 3 · medium

Photostimulating a pullet flock that is well below target body weight is most likely to cause:

  1. Small eggs, more prolapses and poor persistency of lay
  2. Larger eggs, higher peak production and better persistency
  3. Delayed sexual maturity by several weeks
  4. Increased broodiness
Answer and working

A. Small eggs, more prolapses and poor persistency of lay

Underweight pullets brought into lay produce small eggs, suffer more prolapse and cannibalism and do not sustain peak. Photostimulation should follow target age and body weight together.

Question 4 · hard

Why do integrators often report a 'corrected' FCR adjusted to a standard body weight when comparing broiler flocks?

  1. Heavier birds convert feed more efficiently, which exaggerates their advantage
  2. FCR rises naturally as birds get heavier, so flocks sold at different weights are not directly comparable
  3. Correction removes the effect of mortality on FCR
  4. Correction converts feed weight to dry matter
Answer and working

B. FCR rises naturally as birds get heavier, so flocks sold at different weights are not directly comparable

Maintenance takes a growing share of feed as birds grow, so FCR increases with body weight. Adjusting to a standard weight lets flocks sold at different weights be compared fairly. It does not correct for mortality or moisture.

Question 5 · hard

A hen averages 90% hen-day production, 62 g egg weight and 115 g feed intake per day. What is her feed conversion per kg of egg mass?

  1. 1.85 kg feed per kg egg
  2. 0.49 kg feed per kg egg
  3. 2.06 kg feed per kg egg
  4. 1.67 kg feed per kg egg
Answer and working

C. 2.06 kg feed per kg egg

Egg mass = 62 × 0.9 = 55.8 g/day. FCR = 115 ÷ 55.8 = 2.06. Using egg weight without the laying rate gives 1.85; 0.49 is the inverse; multiplying feed by the rate gives 1.67.

Question 6 · hard

Why are hatching eggs cooled soon after lay to below 'physiological zero'?

  1. It kills bacteria on the shell
  2. Embryo development effectively stops below it, holding the embryo at rest
  3. It hardens the shell and cuticle so eggs survive handling and transport
  4. It prevents any water loss from the egg through the shell pores during storage
Answer and working

B. Embryo development effectively stops below it, holding the embryo at rest

Physiological zero, usually quoted in the low-to-mid 20s °C, is the temperature below which development effectively halts. Storing eggs below it but not too cold holds the embryo until setting. Cooling does not sterilise shells or stop all water loss.

Question 7 · medium

Embryos from a breeder flock show 'clubbed down' and increased mid-incubation mortality. Which breeder diet deficiency is classically associated with this?

  1. Riboflavin
  2. Vitamin A
  3. Biotin
  4. Selenium
Answer and working

A. Riboflavin

Clubbed down, where down feathers fail to emerge from their sheaths and curl, is a classic sign of riboflavin (vitamin B2) deficiency in the embryo. Biotin deficiency is associated with skeletal deformities such as a parrot beak.

Question 8 · hard

An egg weighs 60 g and its thick albumen height is 7.5 mm. Using HU = 100 log10(H − 1.7 W^0.37 + 7.57), what is its Haugh unit value and US grade?

  1. 117.8, grade AA
  2. 86.6, grade A
  3. 86.6, grade AA
  4. 86.6, grade B
Answer and working

C. 86.6, grade AA

W^0.37 = 60^0.37 = 4.549; 1.7 × 4.549 = 7.733; 7.5 − 7.733 + 7.57 = 7.337; log10 = 0.8655; HU = 86.6. HU ≥ 72 is AA in the US standard. Omitting the weight term gives 117.8, an impossible value above 100 for a normal egg. Grade A covers 60–72 and grade B is below 60.

Question 9 · medium

Why does the thick albumen become thinner during storage of table eggs?

  1. Water moves from the yolk into the albumen by osmosis and dilutes the thick gel
  2. Bacteria digest the ovomucin within a few days
  3. Calcium leaches from the shell into the albumen
  4. CO2 loss through the shell raises albumen pH, and the ovomucin gel breaks down
Answer and working

D. CO2 loss through the shell raises albumen pH, and the ovomucin gel breaks down

CO2 escapes through the pores, the albumen pH rises, and the gel structure of the thick white thins, lowering Haugh units. Water moves from the albumen into the yolk, not the reverse, which is why the yolk flattens and its index falls. Warm storage speeds all of this.

Question 10 · hard

Why is the date of live infectious bursal disease (IBD) vaccination in broilers often calculated from maternal antibody titres in day-old chicks?

  1. Maternal antibody increases with age, so vaccination must be done at day-old before titres rise further
  2. Maternal antibody blocks vaccine given too early, so it is timed for when titres have fallen
  3. Maternal antibody has no effect on live vaccines
  4. Titres reveal the chicks' sex and body weight, which together determine the vaccine dose
Answer and working

B. Maternal antibody blocks vaccine given too early, so it is timed for when titres have fallen

Maternally derived antibody decays with a predictable half-life. Given too early, a live vaccine is neutralised; given too late, field virus may infect first. Estimating the decline from day-old titres sets the vaccination day.

A study plan for Avian and Poultry Production Management in ICAR AICE-JRF/SRF

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 (50% of the 20 questions): Avian physiology (3, 15%); Broiler management (3, 15%); Egg quality (2, 10%); Incubation (2, 10%).

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

  2. Then the middle band: Layer management (2, 10%); Lighting programmes (2, 10%); Poultry breeds and strains (2, 10%); Biosecurity (1, 5%).

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

  3. Then the lighter units: Candling and hatchability (1, 5%); Hatchery operations (1, 5%); Poultry economics and marketing (1, 5%).

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

  4. Leave time for the hard ones: Avian physiology (3 hard of 3), Incubation (2 hard of 2), Layer management (2 hard of 2).

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

  5. 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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