RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Biology · 10 questions · 24 marks
Grind up any living tissue and analyse it chemically, and you find the same short list of molecule classes doing the work everywhere — carbohydrates, proteins, lipids and nucleic acids. This chapter takes that chemical inventory apart, showing how small building blocks such as amino acids and nucleotides polymerise into the macromolecules that build cells, and it closes with the enzymes that make every one of these reactions run fast enough to sustain life.
A dipeptide is formed when two amino acids join by a bond formed between the:
Answer
Amino group of one and the carboxyl group of the other, releasing water is correct — this dehydration (condensation) reaction forms the peptide bond, a covalent -CO-NH- linkage, and one water molecule is removed for every bond formed.
The blue-black colour produced when iodine solution is added to a food sample confirms the presence of:
Answer
Starch is correct — iodine molecules slot into the helical coils of amylose in starch to give the characteristic blue-black colour, a test that does not work with cellulose, which is unbranched and non-helical, or with simple sugars.
The pentose sugar found in RNA, distinguishing it from DNA, is:
Answer
Ribose is correct — RNA nucleotides carry ribose, which has a hydroxyl group at the 2' carbon, while DNA nucleotides carry deoxyribose, lacking that oxygen, which is the origin of the name deoxyribonucleic acid.
The maximum velocity (Vmax) of an enzyme-catalysed reaction is reached when:
Answer
All available active sites are saturated with substrate is correct — beyond this point adding more substrate cannot increase the rate further because every enzyme molecule is already working at capacity, so the rate-versus-substrate curve plateaus.
Assertion (A): Haemoglobin loses its oxygen-carrying function if it is heated strongly. Reason (R): Heating disrupts the weak bonds that hold a protein's tertiary and quaternary structure, so the molecule unfolds even though its primary sequence of amino acids is unchanged.
Answer
Both A and R are true and R is the correct explanation of A — denaturation destroys the three-dimensional shape on which biological activity depends, without breaking the covalent peptide bonds of the primary sequence, so the protein loses function while remaining chemically the 'same' chain.
Distinguish between a saturated and an unsaturated fatty acid, and state how this affects their physical state at room temperature.
Answer
A saturated fatty acid has a hydrocarbon chain in which every carbon-carbon bond is a single bond, so the chain is straight and packs tightly, making these fats solid at room temperature, as in most animal fats. An unsaturated fatty acid has one or more carbon-carbon double bonds, which introduce kinks in the chain that prevent tight packing, so these fats remain liquid at room temperature, as in most plant oils.
What are cofactors? Name and briefly describe the three types with one example each.
Answer
Cofactors are non-protein chemical components that many enzymes require in order to catalyse a reaction, since the protein part alone is not always sufficient. Inorganic cofactors called metal activators are simple ions such as zinc, which are needed for the activity of certain enzymes like carboxypeptidase. Coenzymes are organic molecules, often derived from vitamins, such as NAD or FAD, that bind loosely and transiently to the active site during the reaction and are then released, carrying electrons or chemical groups from one reaction to another. Prosthetic groups are organic cofactors that remain tightly and permanently bound to the enzyme protein throughout its functional life, such as the haem group bound within catalase.
Describe the four levels of protein structure with a suitable example at each level.
Answer
The primary structure of a protein is simply the linear sequence in which amino acids are joined one after another by peptide bonds along the polypeptide chain; it is this exact sequence, determined by the gene, that ultimately dictates every higher level of structure, and a single substitution in this sequence, as in sickle-cell haemoglobin, can change the protein's function drastically. The secondary structure arises when stretches of the chain fold locally into a regular, repeating pattern held together by hydrogen bonds between the backbone atoms; the two commonest patterns are the coiled alpha helix, seen in keratin, and the folded, sheet-like beta pleated sheet, seen in silk fibroin. The tertiary structure is the overall three-dimensional folding of the entire polypeptide into a compact globular or fibrous shape, produced by interactions among the R groups — hydrogen bonds, ionic bonds, disulphide bridges and hydrophobic interactions — and it is this precise folded shape that creates a functional active site, so tertiary structure is essential for biological activity, as seen in the compact folded shape of myoglobin. The quaternary structure appears only in proteins built from more than one polypeptide subunit, describing how the separate chains associate and arrange themselves relative to each other; haemoglobin illustrates this level well, being built of two alpha and two beta globin chains held together to form the complete, functional four-subunit molecule that carries oxygen in blood.
Explain the mechanism of enzyme action and describe how temperature, pH and substrate concentration each influence the rate of an enzyme-catalysed reaction.
Answer
An enzyme is a protein catalyst that speeds up a specific biochemical reaction by lowering its activation energy, without itself being used up in the process. The reaction begins when the substrate binds to a specific three-dimensional region of the enzyme called the active site, forming a temporary enzyme-substrate complex; the substrate is then converted into product, which is released, freeing the enzyme to bind another substrate molecule and repeat the cycle. Temperature affects the rate because enzymes are proteins whose activity depends on a precise folded shape: as temperature rises from a low value, molecular motion and collision frequency increase and the rate rises, reaching a peak at the enzyme's optimum temperature (commonly near 37 degrees Celsius for human enzymes); beyond this point the weak bonds holding the tertiary structure begin to break, the enzyme denatures, and activity falls sharply. pH has a similar bell-shaped effect, because the ionisation state of the amino acid side chains in and around the active site depends on the surrounding hydrogen ion concentration; each enzyme therefore has an optimum pH (pepsin works best around pH 2 in the stomach, while trypsin works best around pH 8 in the small intestine) and activity declines on either side of it as the active site's shape and charge distribution are disturbed. Substrate concentration affects the rate in a saturation-type curve: at low substrate concentration, increasing the substrate increases the rate almost proportionally because more active sites become occupied, but as concentration keeps rising an increasing fraction of enzyme molecules are already engaged, so the rate increase slows and eventually the curve flattens at Vmax once every active site is continuously occupied, at which point adding more substrate cannot raise the rate any further.
A student sets up four test tubes of the same starch solution and adds an equal volume of the same amylase extract to each, then places the tubes at 10°C, 37°C, 60°C and 37°C but with the pH adjusted strongly acidic. After twenty minutes she tests each tube with iodine solution. Tubes at 10°C and at pH-adjusted 37°C still turn blue-black; the tube at 37°C (neutral pH) shows no colour change; the tube at 60°C also still turns blue-black. (a) What does a blue-black colour with iodine indicate about the starch in that tube? (b) Explain why the 37°C neutral-pH tube shows no colour change. (c) Give the biochemical reason the 60°C tube still tests positive for starch. (d) Give the biochemical reason the acidic 37°C tube still tests positive for starch.
Answer
(a) A blue-black colour means starch is still present largely intact in that tube — the amylase has failed to break it down into smaller sugars. (b) At 37°C and neutral pH the amylase is working at or near its optimum temperature and pH, so it folds correctly and its active site functions efficiently, hydrolysing the starch into maltose and glucose units before the iodine test is applied; with no starch left, no blue-black colour appears. (c) At 60°C the enzyme has been denatured — the high temperature has broken the weak bonds maintaining its tertiary structure, distorting the active site so it can no longer bind starch effectively, so the starch remains undigested and gives a positive test. (d) At the strongly acidic pH the amylase is far outside its optimum range; the abnormal hydrogen ion concentration alters the ionisation of the active site's amino acid residues and disrupts its shape, inactivating the enzyme, so again the starch is left undigested and turns blue-black with iodine.
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