ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR
Original practice questions written to the published syllabus · not past papersBiomolecules, enzymes and the metabolic pathways that turn feed into energy, milk, meat and eggs, including the ruminant's special dependence on volatile fatty acids and gluconeogenesis.
In the exams: ICAR AICE-JRF core group 08; ICAR AIEEA-PG 12 (Veterinary Biochemistry); China basic paper 12
ICAR AICE-JRF/SRF (PhD), core group 08
Deeper regulation: hormonal control of metabolism, second messengers, ruminant gluconeogenesis from propionate, ketogenesis and fatty liver, analytical techniques and their principles.
How our questions for ICAR AICE-JRF/SRF spread across the units of Veterinary Biochemistry. Difficulty: 0 easy, 7 medium, 22 hard.
| Ruminant metabolism: volatile fatty acids, gluconeogenesis and ketogenesis | 6 | |
| Enzymes and enzyme kinetics | 4 | |
| Lipid metabolism | 4 | |
| Carbohydrate metabolism | 3 | |
| Cell signalling | 3 | |
| Nucleic acids | 2 | |
| Vitamins: biochemical roles | 2 | |
| Biochemical laboratory techniques | 2 | |
| Biomolecules and their interactions | 1 | |
| Protein and amino-acid metabolism | 1 | |
| Hormones | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Aspartic acid has pKa values of 2.09 (alpha-carboxyl), 3.86 (side-chain carboxyl) and 9.82 (amino). What is its isoelectric point?
C. 2.975
For an acidic amino acid the neutral (zwitterionic) form lies between the two carboxyl dissociations, so pI = (2.09 + 3.86)/2 = 2.975. Averaging the alpha-carboxyl and amino pKa gives 5.955; averaging side chain and amino gives 6.840; averaging all three gives 5.257.
Further reading: Nelson DL, Cox MM, Lehninger Principles of Biochemistry, 8th ed.
Complete aerobic oxidation of one glucose yields about 30 ATP rather than 32 when cytosolic NADH enters mitochondria by which route?
A. The glycerol-3-phosphate shuttle
The glycerol-3-phosphate shuttle passes electrons to FAD, giving 1.5 ATP per cytosolic NADH instead of 2.5, so the two glycolytic NADH give 2 fewer ATP. The malate-aspartate shuttle preserves NADH, giving 32. The carnitine shuttle moves acyl groups.
Further reading: Nelson DL, Cox MM, Lehninger Principles of Biochemistry, 8th ed.
Cyanide from wilted or regrowing sorghum kills ruminants mainly by inhibiting which enzyme?
B. Cytochrome c oxidase
Cyanide binds ferric iron in cytochrome a3 of complex IV, halting electron transport; venous blood stays bright red because tissues cannot use oxygen. Aconitase is blocked by fluorocitrate from fluoroacetate; ATP synthase by oligomycin.
Further reading: Nelson DL, Cox MM, Lehninger Principles of Biochemistry, 8th ed.
Why is niacin a dietary requirement for cats, even when tryptophan intake is high?
D. High picolinic carboxylase activity diverts tryptophan metabolites away from NAD
In most mammals part of tryptophan is converted to niacin, but in cats picolinic carboxylase is so active that almost none is. They absorb tryptophan normally and use NAD+ like other species.
Further reading: Kaneko JJ, Harvey JW, Bruss ML (eds), Clinical Biochemistry of Domestic Animals, 6th ed.
In high-yielding dairy cows in early lactation, the somatotropic axis is described as 'uncoupled'. What does this mean?
A. Growth hormone is high but hepatic growth hormone receptors and IGF-1 are low
Reduced hepatic GH receptor expression means high GH fails to stimulate IGF-1, so IGF-1 does not feed back to lower GH. High GH and low insulin favour lipolysis and direct glucose to the udder (homeorhesis).
Further reading: Kaneko JJ, Harvey JW, Bruss ML (eds), Clinical Biochemistry of Domestic Animals, 6th ed.
When cellular energy falls, AMP-activated protein kinase (AMPK) phosphorylates and inhibits which enzyme, reducing fatty acid synthesis?
C. Acetyl-CoA carboxylase
AMPK inactivates acetyl-CoA carboxylase; malonyl-CoA falls, releasing inhibition of CPT-I so fatty acid oxidation rises. It does not inhibit CPT-I itself.
Further reading: Nelson DL, Cox MM, Lehninger Principles of Biochemistry, 8th ed.
If a cow absorbed 30 mol of propionate in a day and all of it were converted to glucose, what is the theoretical maximum glucose produced (molar mass 180.16 g/mol)?
B. About 2.7 kg (15 mol)
Two three-carbon propionate molecules make one six-carbon glucose: 30/2 = 15 mol x 180.16 g/mol = 2702 g, about 2.7 kg. 30 mol assumes one glucose per propionate; 10 mol divides by 3; 7.5 mol divides by 4.
Further reading: Bergman EN, Physiological Reviews 70:567-590 (1990)
A cow produces 30 kg milk per day containing 4.8% lactose (molar mass 342.3 g/mol). How much glucose (180.16 g/mol) is needed for lactose synthesis alone?
C. About 1.52 kg
Lactose = 30 x 0.048 = 1.44 kg = 1440/342.3 = 4.21 mol. Each lactose needs two glucose (one becomes galactose), so 8.41 mol x 180.16 = 1516 g, about 1.52 kg. 0.76 kg counts one glucose per lactose; 1.44 kg equates the mass of lactose with glucose; 3.03 kg doubles again.
Further reading: Nelson DL, Cox MM, Lehninger Principles of Biochemistry, 8th ed.
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