CBSE Class 11 Biology Question Paper Set 14 with Solutions

Answer: (a) Monocot stem. In a monocot stem the vascular bundles are scattered, conjoint and closed (no cambium between xylem and phloem).

Teacher's Note:
1. Conjoint means xylem and phloem lie on the same radius, and closed means there is no cambium.
2. Dicot stems have conjoint but open bundles, while roots have radial bundles.

Answer: (d) Ureotelic mode of excretion. Both excrete nitrogenous waste mainly as urea; frogs have external fertilization, nucleated RBCs and a three-chambered heart.

Teacher's Note:
1. Check each option separately for both animals instead of guessing from one.
2. Tadpoles are ammonotelic, but adult frogs become ureotelic.

Answer: (c) Suppression of growth of axillary buds by the presence of the apical bud. The apical bud produces auxin, which inhibits the lateral buds.

Teacher's Note:
1. Auxin made in the shoot tip is responsible for this effect.
2. This is why removing the tip (as in hedge trimming) makes the plant bushy.

Answer: (a). Bryophytes live on land, but the male gametes (antherozoids) are flagellated and need water to swim to the egg, just as amphibians need water to breed.

Teacher's Note:
1. The word amphibian refers to a dual life, on land and in water.
2. Options (b) and (c) are true facts about bryophytes but do not explain this name.

Answer: (b) Parathyroid. Parathyroid hormone (PTH) raises blood Ca²⁺ by promoting bone resorption and increasing reabsorption and absorption of calcium.

Teacher's Note:
1. PTH is the hypercalcemic hormone.
2. Do not confuse it with calcitonin of the thyroid, which lowers blood calcium.

Answer: (c) indica. It is a specific epithet (written in lower case, as in Mangifera indica); Solanum, Felis and Homo are genera.

Teacher's Note:
1. In a binomial name the first word (capital letter) is the genus and the second (small letter) is the species epithet.
2. A specific epithet never stands alone as a name.

Answer: (b) Cyanobacteria. They are blue-green algae (Nostoc, Anabaena) with gelatinous sheaths, nitrogen fixation in heterocysts, and they cause water blooms.

Teacher's Note:
1. Nitrogen fixation together with a gelatinous sheath is a strong clue for cyanobacteria.
2. Dinoflagellates also form blooms (red tides) but they are not nitrogen fixers.

Answer: (c) multicellular with no tissue organisation. Sycon is a sponge (Porifera), which has a cellular level of organisation.

Teacher's Note:
1. Porifera means pore-bearers and sponges have a canal system with choanocytes.
2. Gastrovascular cavity belongs to Cnidaria, not sponges.

Answer: (d) root cap, meristem, cell enlargement, cell maturation. The root cap protects the tip, then cells divide, then elongate, then differentiate.

Teacher's Note:
1. Think of the order as protect, divide, enlarge and mature.
2. Root hairs appear in the zone of maturation.

Answer: (c) DCT and afferent arteriole. The JG cells of the afferent arteriole and the macula densa of the DCT lie at the point where they touch.

Teacher's Note:
1. The JG apparatus secretes renin when blood pressure or GFR falls.
2. Remember that it is the distal tubule, not the proximal tubule, that touches the glomerular arterioles.

Answer: (b) proton gradient between the thylakoid lumen and the stroma. Protons accumulate in the lumen, and their flow back through CF₀-CF₁ (ATP synthase) drives ATP synthesis.

Teacher's Note:
1. Chemiosmosis links ATP synthesis to the movement of H⁺ ions across a membrane.
2. The lumen becomes acidic because of water splitting and proton pumping by plastoquinone.

Answer: (a). Monadelphous means filaments united into a single bundle (staminal tube), which is exactly what the reason states for China-rose (Hibiscus).

Teacher's Note:
1. Monadelphous, diadelphous and polyadelphous refer to one, two and many bundles of stamens.
2. Pea is diadelphous and lemon is polyadelphous.

Answer: (b). Platyhelminthes are dorso-ventrally flattened flatworms and Aschelminthes have round (cylindrical) bodies, but one fact does not explain the other.

Teacher's Note:
1. Two true statements are not automatically linked as cause and reason.
2. Ask whether R answers why A is true; here it only describes a different phylum.

Answer: (a). In the ascending limb, electrolytes are transported out but water cannot follow, so the filtrate becomes dilute (hypotonic).

Teacher's Note:
1. The descending limb is the one permeable to water, so the filtrate becomes concentrated there.
2. The ascending limb dilutes the filtrate while the descending limb concentrates it.

Answer: (a). Scientific names are Latinised, and Latin is used because, being a dead language, its words keep a fixed meaning, so names remain stable worldwide.

Teacher's Note:
1. Stability of names is the main purpose of the binomial system.
2. Scientific names are written in italics, or underlined separately when handwritten.

Answer: 1. No, such plants could not carry out photosynthesis. Chlorophyll a forms the reaction centre of both photosystems (P700 and P680), where light energy is converted into chemical energy; chlorophyll b can only absorb light and pass the energy on to chlorophyll a.
2. Chlorophyll b, xanthophylls and carotenoids are accessory pigments. They absorb light of other wavelengths, pass the energy to chlorophyll a (widening the range of light used), and protect chlorophyll a from photo-oxidation.
OR
1. C₄ plants have Kranz anatomy: the vascular bundles are surrounded by a layer of large bundle sheath cells with thick walls, many chloroplasts and no intercellular spaces. The mesophyll cells are not differentiated into palisade and spongy tissue.
2. Advantage: CO₂ fixed in the mesophyll as oxaloacetate/malate is released inside the bundle sheath cells, where its concentration is high. This stops RuBisCO from acting as an oxygenase, so photorespiration is almost absent and productivity is higher.

Teacher's Note:
1. The reaction centre needs chlorophyll a, while all other pigments are only light harvesters.
2. For the OR part, remember that the name Kranz means wreath, describing the ring of bundle sheath cells.

Answer: (a) Competitive inhibition.
(b) Malonate is structurally similar to succinate, so it competes with the substrate for the active site of the enzyme. When it occupies the active site, succinate cannot bind and the enzyme activity falls. The inhibition can be reduced by raising the substrate concentration.

Teacher's Note:
1. A competitive inhibitor looks like the substrate and fights for the same active site.
2. Do not say that the enzyme is destroyed, because the effect is reversible.

Answer: 1. Blood group O is the universal donor.
2. Its red blood cells have neither antigen A nor antigen B, so the recipient's anti-A or anti-B antibodies cannot agglutinate them, whatever the recipient's group.

Teacher's Note:
1. Antigens are on the RBCs and antibodies are in the plasma.
2. AB is the universal recipient because it has no antibodies in the plasma.

Answer: 1. The two-kingdom system (Plantae and Animalia) placed bacteria, fungi and unicellular organisms like Euglena under plants or animals without proper basis, and did not separate prokaryotes from eukaryotes or unicellular from multicellular organisms.
2. The five-kingdom system (Monera, Protista, Fungi, Plantae, Animalia) uses cell structure, body organisation, mode of nutrition, reproduction and phylogenetic relationships. It separates prokaryotes from eukaryotes, and fungi (heterotrophs with chitinous cell walls) from plants, so it is more natural.

Teacher's Note:
1. Mention at least two criteria used by Whittaker, such as cell structure and mode of nutrition.
2. A good example is that fungi are not plants because they do not photosynthesise.

Answer: (a) The pericycle lies just inside the endodermis, outside the vascular bundles (it is the outermost layer of the stele).
(b) The protoxylem lies towards the periphery and the metaxylem towards the centre. This arrangement is called exarch. (The xylem and phloem form separate patches on alternate radii, i.e. radial.)
(c) The Casparian strips are deposits of waterproof suberin on the radial and tangential walls of endodermal cells. They block the apoplastic movement of water and solutes, so substances must pass through the cell membrane, which gives the plant control over what enters the xylem.

Teacher's Note:
1. Exarch means the oldest xylem is outside, which is the opposite of the endarch arrangement in stems.
2. Always connect the Casparian strip to endodermis and suberin.

Answer: 1. A: trichloroacetic acid (Cl₃CCOOH).
2. B: low molecular weight, about 18 to 800 daltons (below about 1000 daltons).
3. C, D, E, F: proteins, nucleic acids, polysaccharides and lipids (in any order).

Teacher's Note:
1. The acid-soluble pool holds small molecules like amino acids, sugars and nucleotides.
2. Lipids are in the insoluble part because membranes break into vesicles during grinding.

Answer: (a) Metaphase: chromosomes are aligned at the equatorial plate (metaphase plate), attached by their kinetochores to spindle fibres.
(b) Anaphase: the centromere splits and the sister chromatids separate and move to opposite poles.
(c) Zygotene of prophase I of meiosis I: homologous chromosomes pair up (synapsis) with the help of the synaptonemal complex.

Teacher's Note:
1. Write the stage and also what happens, because the question is about events.
2. Synapsis happens only in meiosis I and never in mitosis.

Answer: 1. The endomembrane system is the group of membranous organelles whose functions are coordinated, with materials passing between them directly or by vesicles. It includes the endoplasmic reticulum, Golgi apparatus, lysosomes and vacuoles. Mitochondria, chloroplasts and peroxisomes are not part of it.
2. Rough ER: ribosomes on its surface make proteins that are sent into the lumen for secretion or membrane use. Smooth ER synthesises lipids and steroids.
3. Golgi apparatus: modifies, packages and dispatches proteins and lipids (for example by glycosylation) and forms lysosomes. (Lysosomes: hydrolytic enzymes digest worn-out organelles and foreign material.)

Teacher's Note:
1. Name all the members of the system first, then give two functions clearly.
2. Mitochondria are a common wrong inclusion because they have a membrane, but their functions are not coordinated with the others.

Answer: (a) 1. The forelimbs are modified into wings. 2. The bones are long and hollow (pneumatic) with air cavities, and the body is streamlined and covered with feathers. (Air sacs connected to the lungs also help.)
(b)
1. Aves have feathers; mammals have hair.
2. Aves have a toothless beak; mammals have teeth of different types (heterodont).
3. Aves lay eggs (oviparous); most mammals give birth to young ones (viviparous) and have mammary glands.
4. Aves lack an external ear (pinna) and have no diaphragm; mammals have external ears and a muscular diaphragm.

Teacher's Note:
1. Give two adaptations in (a) and exactly four paired differences in (b).
2. Write differences as comparisons, not as separate descriptions.

Answer: Floral formula: Br, ⊕, ⚥, K₍₄₎ C₍₄₎ A₄ G̲₍₂₎ (the stamens are joined to the petals by a line drawn between C and A, the underline of G shows a superior ovary).
Floral diagram:
1. The mother axis is shown as a dot at the top and the bract as a small structure at the bottom, opposite to it.
2. Outer whorl: 4 sepals joined together, with their edges touching without overlapping (valvate).
3. Second whorl: 4 petals joined together, each overlapping the next in a twisted way, and the 4 stamens attached to the petals.
4. Centre: a gynoecium of 2 fused carpels (syncarpous) with 2 chambers, shown as a circle divided by a septum.

Teacher's Note:
1. In the formula, brackets mean fusion and a line under G means the ovary is superior, i.e. the flower is hypogynous.
2. Show the order from outside to inside: calyx, corolla, androecium and gynoecium.

Answer: (a) Abscisic acid (ABA) inhibits seed germination (it induces dormancy). Other function: it causes stomatal closure during water stress.
(b) Gibberellins cause bolting (rapid internode elongation) and flowering in rosette plants. Other function: they increase the length of grape stalks and of sugarcane stems (so sugar yield rises) and break seed dormancy.
(c) Ethylene ripens fruits. Other function: it promotes senescence and abscission of leaves and flowers (and breaks seed dormancy, promotes root hair growth).

Teacher's Note:
1. Match each hormone with one clearly different function to earn full marks.
2. ABA is called the stress hormone, and ethylene is the only gaseous plant hormone.

Answer: (a) The condition is erythroblastosis foetalis, due to Rh incompatibility. During the delivery of the first Rh-positive child, some foetal Rh-positive blood entered the Rh-negative mother, who made anti-Rh antibodies. In the second pregnancy with an Rh-positive foetus these antibodies cross the placenta and destroy the foetal RBCs, causing anaemia and jaundice.
(b) Soon after the birth of the first Rh-positive child (within 72 hours), the mother should have been given anti-Rh antibodies (anti-D) so that her immune system does not become sensitised.

Teacher's Note:
1. Rh-negative mother and Rh-positive foetus is the situation that matters, and the first child is usually not affected.
2. The preventive injection must be given to the mother, not to the baby.

Answer: (a) Simple (endemic) goitre, an enlargement of the thyroid gland.
(b) Deficiency of iodine in the diet (common in hilly regions), which lowers thyroid hormone production; the pituitary then secretes more TSH, which enlarges the gland.
(c) In pregnancy it causes defective development and maturation of the baby, such as stunted growth (cretinism), mental retardation, low intelligence, deaf-mutism, abnormal skin, and also abortions and stillbirths.
(d) Exophthalmic goitre (Graves' disease) is a form of hyperthyroidism in which the thyroid is enlarged and the eyeballs protrude; it also raises the metabolic rate, heart rate and sweating.
(e) T₃ is triiodothyronine and T₄ is thyroxine (tetraiodothyronine).

Teacher's Note:
1. Link the cause (iodine lack) to the effect (more TSH and a swollen gland) for full marks.
2. Simple goitre is hypothyroid, while exophthalmic goitre is hyperthyroid.

Answer: (a) Leptotene, zygotene, pachytene, diplotene and diakinesis.
(b) Pachytene is marked by the appearance of recombination nodules, the site where crossing over between non-sister chromatids of homologous chromosomes takes place; the bivalent is now seen as a tetrad.
(c) The synaptonemal complex forms during zygotene and dissolves during diplotene.
(d) Reduction occurs at anaphase I (meiosis I), because the homologous chromosomes of each bivalent separate and go to opposite poles, so each daughter nucleus receives only one chromosome of each pair, whereas in meiosis II sister chromatids separate with no change in number.

Teacher's Note:
1. Learn the five sub-stages in order with the mnemonic of one key event for each.
2. Meiosis I is reductional because homologues separate, while meiosis II is like mitosis.

Answer: 1. Site: thylakoid membranes of the chloroplast (grana), where both photosystems PS II and PS I work in series.
2. Light excites PS II (P680); electrons are passed to the primary acceptor and then through plastoquinone, the cytochrome b₆f complex and plastocyanin to PS I. The electrons lost by PS II are replaced by the splitting of water: 2H₂O → 4H⁺ + O₂ + 4e⁻.
3. In PS I (P700) light again excites the electrons, which pass to a primary acceptor and then through ferredoxin to NADP reductase, which reduces NADP⁺ to NADPH + H⁺.
4. The movement of electrons down the chain and the proton gradient built up in the lumen drive ATP synthase to form ATP.
5. Why non-cyclic: the electron flow is one-way, from water to NADP⁺, and the electron does not return to the photosystem it left, unlike cyclic photophosphorylation. Both ATP and NADPH are formed and O₂ is released.

Teacher's Note:
1. Describe the zig-zag path of electrons as the Z-scheme, with PS II first and PS I second.
2. Remember that cyclic photophosphorylation makes only ATP, with no NADPH or oxygen.

Answer: 1. Temperature: activity rises with temperature up to an optimum; at high temperature the enzyme is denatured and loses activity, while low temperature makes it inactive but not destroyed.
2. pH: each enzyme has an optimum pH (for example pepsin about 2, trypsin about 8); extremes change the shape of the enzyme and reduce activity.
3. Substrate concentration: the rate of reaction increases with substrate concentration until all active sites are occupied; it then reaches Vmax.
4. Inhibitors: substances that resemble the substrate (competitive inhibitors) bind the active site and reduce the rate.
5. Activation energy: it is less when an enzyme is present. The enzyme forms an enzyme-substrate complex, which makes the transition state easier to reach, so the reaction needs less energy than the uncatalysed reaction.

Teacher's Note:
1. Give each factor one point with a clear effect, and mention the optimum for temperature and pH.
2. Enzymes only lower the activation energy and do not change the final equilibrium or the energy of reactants and products.

Answer: (a) The liver (with gall bladder storing bile) and the pancreas.
(b) In the stomach, hydrochloric acid and gastric juice from the stomach wall digest the food (mainly proteins) into chyme. In the intestine (duodenum), bile from the gall bladder emulsifies fats and pancreatic juice digests carbohydrates and proteins; final digestion occurs in the intestine.
(c) The inner wall of the intestine has numerous finger-like folds called villi and microvilli, which greatly increase the surface area for absorption of digested food. The undigested waste goes to the rectum and is passed out through the cloaca.

Teacher's Note:
1. Bile contains no enzymes and only helps in emulsifying fats.
2. Increased surface area is the key idea whenever absorption is asked.