Biomolecules

Lecithin (phosphatidylcholine) is a phospholipid. It is an important component of cell membranes and is found abundantly in living tissues.

GLUT-4 – enables glucose transport into cells Insulin – hormone Trypsin – enzyme Collagen – intercellular ground substance Correct match: A-IV, B-I, C-II, D-III.

Lipase – hydrolyses ester bonds in lipids Nuclease – hydrolyses phosphodiester bonds in nucleic acids Protease – hydrolyses peptide bonds in proteins Amylase – hydrolyses glycosidic bonds in carbohydrates (starch) Correct match: A-II, B-IV, C-I, D-III.

The catalytic cycle of an enzyme: 1. The substrate binds to the active site of the enzyme (E). 2. The enzyme-substrate complex is formed (A). 3. The active site, in close proximity to the substrate, breaks the chemical bonds of the substrate (D). 4. The enzyme releases the products of the reaction (C). 5. The free enzyme is ready to bind with another molecule of substrate (B). So the sequence is E, A, D, C, B.

The competitive inhibitor competes with the substrate for the substrate binding site of the enzyme. Consequently, the substrate cannot bind and as a result, the enzyme action declines, e.g., inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure. Such competitive inhibitors are often used in the control of bacterial pathogens.

Both statements I and II are correct. Due to its close structural similarity with the substrate, a competitive inhibitor competes with the substrate for the substrate binding site of the enzyme. Consequently, the substrate cannot bind and as a result, the enzyme action declines, e.g., inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

Statement I is incorrect as the first amino acid is called as N-terminal amino acid and the last amino acid is called the C-terminal amino acid.

In a phosphoglyceride, fatty acids (saturated or unsaturated) are esterified to glycerol, and a phosphate group is also attached to the glycerol.

A basic amino acid has an extra amino group in its side chain.
D is lysine, $H_2N-CH(COOH)-(CH_2)_4-NH_2$, which has a second $-NH_2$ group at the end of its side chain, so it is basic.
A is glutamic acid (acidic, extra –COOH), B is serine (neutral, –OH group) and C is not an amino acid at all (it has no –COOH group).

A nucleotide is formed of a nitrogenous base + sugar + phosphate. Adenylic acid (a phosphate ester of adenosine) is a nucleotide. Uridine and Guanosine are nucleosides (base + sugar, no phosphate), and Guanine is a nitrogenous base alone.

Ligases catalyse the joining of two molecules by forming bonds such as C–O, C–S, C–N, or P–O, usually coupled with the breakdown of ATP. They do not catalyse the formation of C–C bonds.

Glutamic acid and aspartic acid are amino acids, not fatty acids. Lecithin is a phospholipid (a complex/derived lipid), not a fatty acid itself. Arachidonic acid and palmitic acid are true fatty acids. So the compounds that are not fatty acids are A, D and E.

As substrate concentration increases, the velocity of an enzyme-catalysed reaction rises initially and then plateaus at $V_{max}$, since all enzyme molecules become saturated with substrate and no free enzyme molecules remain to bind additional substrate. This gives a hyperbolic (rectangular hyperbola) curve.

Lyases catalyse the removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds (or, in the reverse reaction, adding groups to double bonds).

Primary structure of protein is its polypeptide chain (linear sequence of amino acids) (A-III); Secondary structure is characterised by alpha helix and beta sheet (B-IV); Tertiary structure is stabilised by disulphide bonds among other interactions (C-II); Quaternary structure is exemplified by human haemoglobin, made of multiple polypeptide subunits (D-I).

Glycine is the simplest amino acid, not a lipid — so the statement that glycine is an example of lipids is incorrect.

Aquaporin is a polypeptide/protein (a-iii); Asparagine is an amide (b-i); Abscisic acid is related to carotenoids (c-iv); Chitin is a polysaccharide (d-ii).

Protein is characterised by peptide bonds (A-IV); Unsaturated fatty acid is characterised by C=C double bonds (B-I); Nucleic acid is characterised by phosphodiester bonds (C-II); Polysaccharide is characterised by glycosidic bonds (D-III).

Inulin is a polymer of fructose (a fructan), used as a storage polysaccharide in some plants.

Lecithin is a phospholipid, a primary metabolite essential for basic cellular structure/function, not a secondary metabolite. (Curcumin, morphine, and anthocyanin are all secondary metabolites.)

Terpenoides are exemplified by diterpenes (A-II); Lectins are exemplified by Concanavalin A (B-IV); Alkaloids are exemplified by codeine (C-I); Toxins are exemplified by ricin (D-III).

Statement I is correct. The source key also treats Statement II as correct, reasoning that zwitterion formation takes place at the isoelectric point, which can be at acidic or basic pH depending on the amino acid.

Many enzymes need metal ions as cofactors (activators) for their activity.
Tyrosinase is a copper-containing enzyme. It oxidises the amino acid tyrosine to DOPA and dopaquinone in the pathway of melanin pigment formation (and causes browning of cut fruits).
Carbonic anhydrase requires zinc ($Zn^{2+}$). Lactic dehydrogenase uses the coenzyme NAD, and tryptophanase uses pyridoxal phosphate.
So copper is necessarily associated with tyrosinase.

Reference: NCERT Class XI, page 105.

A competitive inhibitor increases the $K_m$ of the enzyme: a higher substrate concentration is needed to reach $\frac{1}{2}V_{max}$.
So the statement that it decreases $K_m$ is incorrect.

Malonate resembles succinate and competes with it for the active site of succinic dehydrogenase (competitive inhibition).
Competitive inhibition is reversed by adding excess substrate, so the statement that excess succinate does not reverse it is not correct.

A good storage material should stay in the cell for a long time without disturbing the working of the cell.
(2) Starch is chemically non-reactive: it has very few free reducing ends, so it does not take part in the reactions of the cell and can be stored safely for long periods.
(4) Starch is osmotically inactive: it is a large, insoluble polymer, so it does not change the osmotic concentration of the cell and does not draw water into it, even when stored in large amounts.
(1) is not a property of starch: because it is insoluble it is not easily translocated; carbohydrate is translocated mainly as sucrose.
(3) Being easily digested by animals is of no use to the plant as a storage property.
(5) Being synthesized during photosynthesis does not make it a good storage material.
So the properties that make starch useful as a storage material are (2) and (4).

The graph plots potential energy (y-axis) against the progress of the reaction (x-axis), so the label B on the y-axis is the potential energy.
The substrate must pass through a high-energy, unstable intermediate structure before forming the product; the peak of the curve, labelled A, is this transition state.
The difference in energy between the substrate and the transition state is the activation energy.
The taller curve is for the reaction without the enzyme, so the larger energy gap, C, is the activation energy without enzyme.
The enzyme lowers the energy barrier; the smaller gap under the lower curve, D, is the activation energy with enzyme.
Hence A – transition state, B – potential energy, C – activation energy without enzyme, D – activation energy with enzyme.

The figure shows a pentose sugar with –OH groups on both the 2' and 3' carbons, so it is ribose.
A sugar joined to a nitrogenous base without a phosphate group is a nucleoside; with a phosphate it would be a nucleotide.
Uracil is a base found with ribose, so the category is nucleoside and X is uracil.

Lecithin is a phospholipid: a glyceride in which phosphorus and a phosphorylated organic compound are present; it is found in the cell membrane. This description is correct.
Alanine is an $\alpha$-amino acid: the amino group and the carboxyl (acidic) group are both attached to the same carbon, the $\alpha$-carbon, not anywhere in the molecule.
Palmitic acid is a saturated fatty acid with 16 carbon atoms (including the carboxyl carbon), not an unsaturated 18-carbon acid.
Adenylic acid is a nucleotide: adenosine (adenine + ribose) with a phosphate group, not with glucose phosphate.
Hence only lecithin is correctly characterised.

Structure A has a glycerol backbone with two fatty acid chains ($R_1$, $R_2$) joined by ester bonds and a phosphate group linked to choline, $-CH_2-CH_2-N^+(CH_3)_3$. This is the phospholipid lecithin.
Lecithin is a component of the cell membrane, so this identification and function are correct.
A is not a triglyceride, because a triglyceride has three fatty acids and no phosphate group.
Structure B is a purine with two fused rings and an $-NH_2$ group: it is adenine. Adenine is a nitrogenous base, not a nucleotide (a nucleotide also has a sugar and a phosphate).
B is not uracil, which is a single-ring pyrimidine and is found in RNA, not in DNA.
Hence the correct option is A : lecithin – a component of cell membrane.

(b) is correct: beri-beri is a deficiency disease caused by the lack of thiamine (vitamin $B_1$), which is needed as a coenzyme in carbohydrate metabolism.
(d) is correct: it occurs mainly where the staple diet is polished rice, because thiamine is present in the husk and outer layers of the grain, which are removed in polishing.
(e) is correct: the symptoms include pain due to neuritis, paralysis, muscle wasting, progressive oedema, mental deterioration and finally heart failure.
(a) is not correct: beri-beri is not a disease specific to sub-Saharan Africa.
(c) is not correct: a disorder in infants and children due to protein deficiency describes kwashiorkor, not beri-beri.
So the group containing all three correct statements is b, d and e.

Vitamin $B_1$ (thiamine) deficiency causes beri-beri; correctly matched.
Vitamin $B_{12}$ (cyanocobalamin) deficiency causes pernicious anaemia; correctly matched.
Vitamin $B_6$ (pyridoxine) deficiency causes dermatitis, anaemia, convulsions and loss of appetite (anorexia) with nausea; this pairing is acceptable.
Vitamin $B_2$ (riboflavin) deficiency causes cheilosis (cracks at the corners of the mouth), glossitis and dermatitis. Pellagra (the 3 Ds: dermatitis, diarrhoea and dementia) is caused by deficiency of niacin (vitamin $B_3$ or nicotinic acid), not riboflavin.
So the pair that is not correctly matched is Vitamin $B_2$ – Pellagra.
Note: the answer key of the source gives Vitamin $B_6$ – loss of appetite; it is changed here because pellagra is definitely a niacin deficiency disease, not a riboflavin deficiency disease.

Vitamin C (ascorbic acid) deficiency causes scurvy (bleeding gums, poor wound healing); correctly matched.
Vitamin $B_{12}$ (cyanocobalamin) deficiency causes pernicious anaemia; correctly matched.
Beri-beri is caused by deficiency of vitamin $B_1$ (thiamine). Vitamin $B_6$ (pyridoxine) deficiency causes dermatitis, anaemia and convulsions. So 'Vitamin $B_6$ – Beri-beri' is not correctly matched.
As per the answer key, the pair not correctly matched is Vitamin $B_6$ – Beri-beri.
Note: 'Vitamin $B_2$ – Pellagra' is also not strictly correct, since pellagra is caused by deficiency of niacin (vitamin $B_3$), while riboflavin ($B_2$) deficiency causes cheilosis and glossitis. The answer key gives $B_6$ – Beri-beri, which is the clearly wrong pair.

Only a few elements make up almost the whole mass of a living organism.
Oxygen (about 65 percent), carbon (about 18.5 percent), hydrogen (about 9.5 percent) and nitrogen (about 3.3 percent) together make up about 96 percent of the body mass.
The next most abundant elements by mass are calcium (about 1.5 percent) and phosphorus (about 1 percent).
Calcium and phosphorus are present in large amounts in bones and teeth; phosphorus is also a part of nucleic acids, ATP and phospholipids.
These six elements together account for about 98 to 99 percent of the mass.
Sulphur, sodium and magnesium are each present in amounts of less than about 0.3 percent.
So, going by mass, the two remaining elements are calcium and phosphorus, as given in the answer key.

The Michaelis constant ($K_m$) is the substrate concentration at which the rate of an enzyme-catalysed reaction is half of its maximum velocity ($V_{max}/2$).
$K_m$ is a measure of the affinity of the enzyme for its substrate.
A low $K_m$ means that the enzyme reaches half its maximum rate at a low substrate concentration, i.e. it has a high affinity for the substrate and is more efficient.
A high $K_m$ means a low affinity.
So the catalytic efficiencies of two different enzymes can be compared by their $K_m$ values.
The pH optimum, the molecular size and the mere formation of product do not give such a comparison.

A reducing sugar has a free (or potentially free) aldehyde or ketone group, so it can reduce Tollens' reagent and Fehling's solution.
Galactose is a monosaccharide (an aldohexose). In solution its ring opens to give a free aldehyde group, so it is a reducing sugar.
Gluconic acid is formed by oxidation of the aldehyde group of glucose to –COOH, so it no longer has a free aldehyde group.
In $\beta$-methyl galactoside, the anomeric –OH is converted into a –OCH$_3$ group (a glycoside), so the ring cannot open; it is non-reducing.
In sucrose, the anomeric carbons of glucose and fructose are joined to each other, so it is non-reducing.
So galactose is the reducing sugar.