Biomolecules practice · Mixed difficulty

Option B (Isoelectric point) is correct because the isoelectric point (pI) is the pH at which a molecule carries no net electrical charge.

Option C (Hydrogenation) is correct because hydrogenation adds hydrogen to the unsaturated bonds in triolein to convert it to tristearin, a saturated fat.

Correct Answer + Core Reason: Free anomeric carbon of $\beta \mathrm{ - D - }$ glucose. The reducing property of lactose is due to the presence of a free anomeric carbon in the glucose moiety.

Option A (Phosphoric acid) is correct because DNA consists of deoxyribose sugar, a nitrogenous base, and a phosphate group.

Option A (Increase in blood clotting time) is correct because Vitamin K is essential for blood clotting, so a deficiency leads to increased blood clotting time.

The correct answer is D ((A) is incorrect but (R) is correct) because maltose has an $\alpha$-glycosidic linkage, not a $\beta$-glycosidic linkage, while the reason correctly describes the linkage between the glucose units.

Option D (All of the above) is correct because fructose, being a ketohexose, contains five hydroxyl groups, three secondary alcoholic groups, and one ketone group with two primary alcoholic groups.

Option B (Ascorbic acid) is correct because bleeding gums are a symptom of scurvy, which is caused by a deficiency of Ascorbic acid (Vitamin C).

Option A is correct because serum proteins in blood act as buffers, resisting pH changes upon addition of small amounts of acid or base.

Option C (Testosterone) is correct because Testosterone is a steroid hormone, not a vitamin.

Option C (7.7 to 8.5) is the correct pOH range for the isoelectric point of the amphoteric ion of an amino acid, especially for amino acids with basic side chains.

The correct answer is C (6 membered glucose ring and 5 membered fructose ring) because cane sugar (sucrose) consists of a 6-membered glucose ring and a 5-membered fructose ring.

The correct answer is A (Fats and oils) because the iodine value is a measure of the degree of unsaturation in fats and oils.

Option C (Rearrangement) is correct because in alkaline medium, sugars undergo Lobry de Bruyn-van Ekenstein transformation, which is a rearrangement.

Option B (D represents configuration and (+) represents the dextrorotatory nature of molecule) is correct because 'D' refers to the configuration of the chiral carbon furthest from the aldehyde or ketone group, and '(+)' indicates that the molecule rotates plane-polarized light clockwise (dextrorotatory).

Option A ((A),(C) and (D) only) is correct because Valine, Lysine, and Threonine are all essential amino acids, meaning they cannot be synthesized by the human body and must be obtained through diet.

Option A (water) is correct because water constitutes approximately 70-90% of a cell's mass, making it the most abundant molecule.

Option D (Maltose) is correct because Molisch's test is positive for all carbohydrates, Benedict's test is positive for reducing sugars (like maltose), and Seliwanoff's test is positive for ketoses (like fructose).

The correct answer is A (Iron) because hemoglobin, the red coloring matter of blood, contains iron in a porphyrin ring system, which is crucial for oxygen transport.

Option D (All are correct) is correct because carbohydrates, phosphoric acid, and iron pigments can all act as cofactors or prosthetic groups in conjugated proteins.

Option D (Glycerides of unsaturated fatty acids) is correct because vegetable oils are esters of glycerol with unsaturated fatty acids.

Option C (+29.6°) is correct because the equilibrium constant and the specific rotation of the mixture can be used to calculate the specific rotation of beta-D-glucose.

Option C (Addition of antioxidant to oil minimizes rancidity) is true because antioxidants prevent the oxidation of oils, which causes rancidity.

Option A (The catalytic action of an enzyme is not specific) is correct because enzymes are highly specific catalysts, meaning they act on specific substrates.

Option A (8) is correct because with two amino acids, valine and proline, there are 2³ = 8 possible tripeptides.

Option B (Thymine) is correct because thymine is a nitrogenous base found in DNA, not RNA. RNA contains uracil instead.

Option A (Xerophythol) is correct because Xerophythol is another name for Vitamin A.

Option D (Sucrose) is correct because it lacks a free aldehyde or ketone group due to the glycosidic linkage involving both anomeric carbons.

Option C (glycogen) is correct because glycogen is the storage form of glucose in animals, hence its name "animal starch."

Option D (Sucrose) is correct because sucrose is a non-reducing sugar due to the glycosidic linkage involving both anomeric carbons, preventing it from reducing Tollens' reagent.