NEET UG — Full Mock Test

Solution: Pressure $=\frac{F}{A}$ has dimensions $[ML^{-1}T^{-2}]$. Energy density $=\frac{E}{V}=\frac{ML^2T^{-2}}{L^3}=[ML^{-1}T^{-2}]$, which matches pressure.

Solution: Maximum error $=2(1)+1(2)+1(3)=2+2+3=7\%$.

Solution: $H=\frac{u^2}{2g}=\frac{20^2}{2\times10}=\frac{400}{20}=20\,\text{m}$.

Solution: $R=\frac{u^2\sin 2\theta}{g}$ is maximum when $\sin 2\theta=1$, i.e. $2\theta=90^\circ$, so $\theta=45^\circ$.

Solution: $a=\frac{F}{m}=\frac{20}{4}=5\,\text{m/s}^2$.

Solution: $F=\frac{\Delta p}{\Delta t}=\frac{m\Delta v}{\Delta t}=\frac{2\times3}{0.6}=10\,\text{N}$.

Solution: $W=mgh=5\times10\times4=200\,\text{J}$.

Solution: $KE=\frac{1}{2}mv^2\propto v^2$. Doubling $v$ makes KE $2^2=4$ times the original.

Solution: All mass of a ring lies at distance $R$ from the axis, so $I=MR^2$.

Solution: The solid sphere has smaller moment of inertia ($\frac{2}{5}MR^2 < \frac{2}{3}MR^2$), giving larger linear acceleration, so it reaches the bottom first.

Solution: $g_h=\frac{g}{(1+h/R)^2}$. With $h=R$: $g_h=\frac{g}{(2)^2}=\frac{g}{4}$.

Solution: $v_e=\sqrt{\frac{2GM}{R}}$ depends on the planet's mass and radius and $G$, but not on the mass of the escaping body.

Solution: $Y=\frac{\text{stress}}{\text{strain}}$; strain is dimensionless, so $Y$ has the dimensions of stress (pressure), $[ML^{-1}T^{-2}]$.

Solution: Bernoulli's equation $P+\frac{1}{2}\rho v^2+\rho gh=\text{const}$ shows that where speed is higher, pressure is lower (at the same height).

Solution: $h=\frac{2T\cos\theta}{r\rho g}$, so the capillary rise is inversely proportional to the tube radius $r$.

Solution: Isothermal means constant temperature; for an ideal gas internal energy depends only on temperature, so $\Delta U=0$.

Solution: $\eta=1-\frac{T_C}{T_H}=1-\frac{300}{400}=1-0.75=0.25=25\%$.

Solution: $v_{rms}=\sqrt{\frac{3RT}{M}}\propto\sqrt{T}$, so rms speed varies as the square root of absolute temperature.

Solution: $T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{1}{\pi^2}}=2\pi\cdot\frac{1}{\pi}=2\,\text{s}$.

Solution: In SHM, $a=-\omega^2 x$; the magnitude is proportional to displacement and the acceleration is always directed towards the mean position.

Solution: $v=f\lambda=500\times0.4=200\,\text{m/s}$.

Solution: $F=\frac{kq_1q_2}{r^2}=\frac{9\times10^9\times2\times10^{-6}\times3\times10^{-6}}{(0.3)^2}=\frac{54\times10^{-3}}{0.09}=0.6\,\text{N}$.

Solution: By Gauss's law, the net charge enclosed inside the cavity is zero, so the electric field is zero everywhere inside a charged hollow conductor.

Solution: $C=\frac{K\varepsilon_0 A}{d}$. Inserting a dielectric ($K>1$) increases capacitance; larger $d$ or smaller $A$ would decrease it.

Solution: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1$, so $R=1\,\Omega$.

Solution: $v_d=\frac{eE\tau}{m}$, so drift velocity is directly proportional to the applied electric field $E$.

Solution: $F=qvB\sin\theta$. When velocity is parallel to $\vec{B}$, $\theta=0$, so $\sin\theta=0$ and the force is zero.

Solution: For a circular coil of $N$ turns, $B=\frac{\mu_0 N I}{2r}$ at its centre.

Solution: Diamagnetic materials have small negative susceptibility ($\chi<0$) and are weakly repelled by magnetic fields.

Solution: Cutting perpendicular to the length halves the length (and pole strength stays the same), so each piece has moment $\frac{M}{2}$.

Solution: Lenz's law states that the induced current opposes the change in magnetic flux causing it, a consequence of conservation of energy.

Solution: For a sinusoidal AC, $I_{rms}=\frac{I_0}{\sqrt{2}}\approx0.707\,I_0$.

Solution: In a pure inductor, the current lags the applied voltage by $\frac{\pi}{2}$ (i.e. $90^\circ$).

Solution: EM waves are transverse: the electric and magnetic fields oscillate perpendicular to each other and to the direction of propagation.

Solution: In the EM spectrum, gamma rays have the highest frequency and therefore the shortest wavelength.

Solution: When the object is at the focus of a concave mirror, the reflected rays become parallel, so the image is formed at infinity.

Solution: $\beta=\frac{\lambda D}{d}$, so the fringe width is directly proportional to the wavelength $\lambda$.

Solution: $P=\frac{1}{f(\text{in m})}=\frac{1}{0.25}=+4\,\text{D}$ (positive for a converging lens).

Solution: Intensity determines the number of photons per second, hence the photoelectron count. Maximum KE and stopping potential depend on frequency, not intensity.

Solution: $\lambda=\frac{h}{p}$, so the de Broglie wavelength is inversely proportional to momentum $p$.

Solution: In Bohr's model, $r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}\propto n^2$.

Solution: Number of half-lives $=\frac{30}{10}=3$. Fraction remaining $=\left(\frac{1}{2}\right)^3=\frac{1}{8}$.

Solution: Nuclear forces are charge-independent, meaning they are nearly equal for p-p, p-n and n-n pairs. The reason directly explains the assertion.

Solution: At $0\,\text{K}$ the valence band is full and the conduction band empty with no thermal excitation, so an intrinsic semiconductor behaves as a perfect insulator.

Solution: Phosphorus is a pentavalent (group 15) element; it donates a free electron, producing an n-type semiconductor. Boron, aluminium and gallium are trivalent (p-type) dopants.

Solution: Moles of $O_2 = 5.6/22.4 = 0.25$ mol. Each $O_2$ molecule has 2 O atoms, so moles of O atoms $= 0.25 \times 2 = 0.50$ mol.

Solution: Radial nodes $= n - l - 1$. For $3s$: $n = 3$, $l = 0$, so radial nodes $= 3 - 0 - 1 = 2$.

Solution: For $n=2$, $l$ can be 0 or 1 only ($l = 0$ to $n-1$). $l=2$ is not allowed for $n=2$, so this set is invalid.

Solution: $E_n = -13.6/n^2$ eV. For $n=3$: $E_3 = -13.6/9 = -1.51$ eV.

Solution: Spontaneous when $\Delta G < 0$, i.e. $\Delta H < T\Delta S$. $T > \Delta H/\Delta S = 30000/100 = 300$ K.

Solution: Temperature is intensive (independent of amount of substance). Volume, enthalpy, and internal energy are extensive properties.

Solution: For an ideal gas at constant temperature, $\Delta U = 0$ (U depends only on T). By the first law $q = -w$; heat absorbed equals work done by the gas. $\Delta H = 0$ as well.

Solution: $K_p = K_c(RT)^{\Delta n}$ where $\Delta n = 2 - 4 = -2$. So $K_p = K_c(RT)^{-2}$.

Solution: HCl is a strong acid, fully dissociated. $[H^+] = 10^{-3}$ M. $\text{pH} = -\log(10^{-3}) = 3$.

Solution: For AgCl, $K_{sp} = s^2$. So $s = \sqrt{1.6 \times 10^{-10}} = 1.26 \times 10^{-5}$ M.

Solution: Let oxidation number of Cr be $x$. $2x + 7(-2) = -2$, so $2x = 12$, $x = +6$.

Solution: Depression in freezing point depends on the number of particles (van't Hoff factor i). $CaCl_2$ gives $i=3$ (highest particle count among these), so it shows the greatest depression and lowest freezing point.

Solution: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = +1.10$ V.

Solution: $Al^{3+} + 3e^- \rightarrow Al$ requires 3 F per mole. For 0.5 mol: $0.5 \times 3 = 1.5$ F.

Solution: Reducing to one-fourth requires two half-lives ($1 \rightarrow 1/2 \rightarrow 1/4$). Time $= 2 \times 20 = 40$ min.

Solution: K is in period 4, while Na, Mg, Al are in period 3. Atomic radius increases down a group and decreases across a period, so K has the largest atomic radius.

Solution: Expected order is Be > B (Be has stable filled $2s^2$). Across period 2: $B < Be < C < N$. B has lower IE than Be due to easy removal of the $2p$ electron.

Solution: Chlorine has the most negative electron gain enthalpy. F is anomalously low due to its small size and high electron-electron repulsion in the compact 2p subshell.

Solution: $SF_4$ has 4 bond pairs and 1 lone pair on S, giving $sp^3d$ hybridization and a see-saw shape.

Solution: $O_2$ has 16 electrons. Bond order $= (N_b - N_a)/2 = (10 - 6)/2 = 2$. $O_2$ is also paramagnetic due to two unpaired electrons in $\pi^*$ orbitals.

Solution: $NH_3$ is pyramidal with a lone pair, giving a net dipole moment. $CO_2$ (linear), $BF_3$ (trigonal planar), and $CCl_4$ (tetrahedral) are symmetrical and have zero dipole moment.

Solution: $H-C\equiv C-H$ has 2 C-H sigma bonds, 1 C-C sigma bond (total 3 sigma) and 2 pi bonds in the triple bond.

Solution: Reducing character of group 15 hydrides increases down the group as the E-H bond weakens. $BiH_3$ is the strongest reducing agent.

Solution: $H_3PO_3$ has two P-OH bonds (dibasic) and one P-H bond, which makes it a reducing agent. $H_3PO_4$ is tribasic and not a reducing agent.

Solution: As the number of terminal oxygen atoms increases, the negative charge on the conjugate base is delocalised over more oxygen atoms, stabilising it and increasing acid strength.

Solution: $Sc^{3+}$ has a $3d^0$ configuration; with no d electrons, no d-d transitions are possible, so it is colourless. The others have partially filled d orbitals and are coloured.

Solution: Due to lanthanoid contraction, the expected increase in radius from Zr to Hf is offset, so Zr and Hf have nearly the same atomic and ionic radii and very similar chemistry.

Solution: Six $NH_3$ ligands (hexaammine), Co with oxidation state +3 balancing three $Cl^-$. The cationic complex uses the metal name: hexaamminecobalt(III) chloride.

Solution: Fe is +2 ($d^6$). With the strong field ligand $CN^-$, electrons pair up in the lower $t_{2g}$ set ($t_{2g}^6 e_g^0$), giving zero unpaired electrons (diamagnetic, low spin).

Solution: In the spectrochemical series, $CN^-$ is among the strongest field ligands, causing the largest crystal field splitting. The order is $F^- < H_2O < NH_3 < CN^-$.

Solution: Carbocation stability increases with the number of alkyl groups due to +I effect and hyperconjugation: tertiary > secondary > primary > methyl.

Solution: Per Markovnikov's rule, H adds to the carbon with more H atoms, and Br to the more substituted carbon, forming the more stable secondary carbocation. The major product is 2-bromopropane.

Solution: Benzene has 6 delocalised pi electrons ($4n+2$ with $n=1$), is cyclic, planar, and fully conjugated, satisfying Huckel's rule. Cyclobutadiene ($4\pi$) is antiaromatic; cyclohexane is non-aromatic.

Solution: Friedel-Crafts acylation of benzene with acetyl chloride ($CH_3COCl$) in the presence of anhydrous $AlCl_3$ gives acetophenone ($C_6H_5COCH_3$).

Solution: $S_N1$ rate depends on carbocation stability. Tert-butyl chloride forms the most stable tertiary carbocation, so it reacts fastest by the $S_N1$ mechanism.

Solution: In chlorobenzene, the lone pair on Cl delocalises into the ring, giving the C-Cl bond partial double-bond character. This shorter, stronger bond resists nucleophilic substitution.

Solution: The phenoxide ion delocalises its negative charge over the aromatic ring through resonance, making it more stable than the ethoxide ion. Hence phenol is more acidic than ethanol.

Solution: Tollens' reagent oxidises aldehydes (giving a silver mirror) but not ketones. Acetone is a ketone, so it gives a negative Tollens' test.

Solution: The Cannizzaro reaction is given by aldehydes lacking an alpha-hydrogen. Benzaldehyde has no alpha-H and undergoes self oxidation-reduction with concentrated NaOH.

Solution: In aqueous medium the order (balancing +I effect, solvation and steric factors) is $(CH_3)_2NH > CH_3NH_2 > NH_3$. Dimethylamine is the strongest base of these three in water.

Solution: Only primary amines react with chloroform and alcoholic KOH to give foul-smelling isocyanides (carbylamines). It is a confirmatory test for primary amines.

Solution: Aniline reacts with nitrous acid (from $NaNO_2/HCl$) at low temperature ($0-5^\circ C$) to form benzene diazonium chloride ($C_6H_5N_2^+Cl^-$).

Solution: Glucose has a free aldehyde (hemiacetal) group and reduces Tollens'/Fehling's reagent. Sucrose is non-reducing (anomeric carbons linked); starch and cellulose are polysaccharides.

Solution: Amino acids are joined by peptide (amide) bonds formed between the carboxyl group of one amino acid and the amino group of another, with loss of water.

Solution: Vitamin C (ascorbic acid) is water-soluble; its deficiency causes scurvy. Vitamins A, D, and K are fat-soluble.

Solution: In a binomial name the first word denotes the genus (Mangifera) and the second word denotes the specific epithet (indica). NCERT calls the second component the specific epithet.

Solution: As per NCERT, Order is the assemblage of families which exhibit a few similar characters. The order Polymoniales includes families Convolvulaceae and Solanaceae.

Solution: Logistic growth with limited resources gives a sigmoid (S-shaped) Verhulst-Pearl logistic curve. Exponential growth with unlimited resources gives a J-shaped curve.

Solution: Whittaker (1969) used cell structure, body organisation, mode of nutrition, reproduction and phylogenetic relationships. Cell wall presence alone was not a criterion; it varies within kingdoms.

Solution: Euglenoids (e.g., Euglena) lack a cell wall (having a protein-rich pellicle), are photosynthetic in sunlight but heterotrophic in darkness — hence a bridge between plants and animals.

Solution: NCERT states Mycoplasma completely lack a cell wall, are the smallest living cells known, can survive without oxygen and many are pathogenic.

Solution: In bryophytes the main plant body is the haploid gametophyte; the sporophyte is dependent on it. In pteridophytes, gymnosperms and angiosperms the diploid sporophyte is dominant.

Solution: In gymnosperms ovules are not enclosed within an ovary wall and remain exposed before and after fertilisation; the resulting seeds are 'naked', i.e., not covered by a fruit.

Solution: NCERT attributes the red colour of red algae (Rhodophyceae) to the predominance of the red pigment r-phycoerythrin. Fucoxanthin is characteristic of brown algae.

Solution: In imbricate aestivation the margins overlap but not in any particular direction, as in Cassia and Gulmohur. Twisted aestivation has a regular overlapping direction.

Solution: Fabaceae (papilionaceous) has a diadelphous androecium, stamens (9)+1 — nine stamens fused into one bundle and the tenth stamen remaining free.

Solution: Coconut is a fibrous drupe; its edible white kernel and coconut water are the endosperm. The fibrous husk is the mesocarp and the hard shell is the endocarp.

Solution: Endodermal cells of the root bear Casparian strips — band-like thickenings of suberin on their radial and tangential walls — which regulate water entry into the vascular tissue.

Solution: In dicot stems vascular bundles are conjoint, collateral and open (cambium present, allowing secondary growth), arranged in a ring. Monocot stems have scattered, closed bundles lacking cambium.

Solution: Phellogen cuts off phellem (cork) outwards and phelloderm (secondary cortex) inwards. Phellogen, phellem and phelloderm together constitute the periderm.

Solution: Lysosomes are membrane-bound vesicles formed by the Golgi apparatus, rich in hydrolytic enzymes (lipases, proteases, carbohydrases) that are optimally active at acidic pH.

Solution: The fluid mosaic model holds that the quasi-fluid nature of the lipid bilayer enables lateral movement of proteins; membrane fluidity supports functions like endocytosis and cell growth.

Solution: A 70S ribosome consists of a larger 50S and a smaller 30S subunit. The eukaryotic 80S ribosome is made of 60S and 40S subunits.

Solution: DNA replication occurs in the S (synthesis) phase of interphase; DNA content per cell doubles, while chromosome number stays the same.

Solution: During zygotene the homologous chromosomes start pairing (synapsis) forming the synaptonemal complex. Crossing over occurs later, during pachytene.

Solution: In mitotic anaphase the centromeres split and sister chromatids move to opposite poles. In anaphase I of meiosis, homologous chromosomes (not sister chromatids) separate.

Solution: In C4 plants the primary CO2 acceptor in mesophyll cells is the 3-carbon molecule phosphoenolpyruvate (PEP), with PEP carboxylase as the enzyme. RuBP is the acceptor in the Calvin cycle.

Solution: Photolysis of water is linked to Photosystem II and occurs on the inner (lumen) side of the thylakoid membrane, releasing O2, protons and electrons.

Solution: C4 plants show Kranz anatomy: large bundle-sheath cells form rings around vascular bundles, with many chloroplasts, thick walls and no intercellular spaces.

Solution: NCERT states a net gain of 38 ATP molecules from the complete aerobic oxidation of one glucose molecule under the assumptions specified in the text.

Solution: The Krebs cycle occurs in the mitochondrial matrix, whereas glycolysis occurs in the cytoplasm and the electron transport chain lies on the inner mitochondrial membrane.

Solution: The 'Evil Quartet' — the four major causes of biodiversity loss — are habitat loss and fragmentation, over-exploitation, alien species invasions, and co-extinctions. Genetic recombination is not among them.

Solution: Abscisic acid (ABA) promotes senescence and abscission, induces seed dormancy and is called the stress hormone because it closes stomata and increases stress tolerance.

Solution: NCERT notes that gibberellins increase the length of the sugarcane stem, raising the yield by as much as 20 tonnes per acre.

Solution: Ethylene is a gaseous PGR that hastens fruit ripening, promotes senescence and abscission, breaks seed and bud dormancy, and induces flowering in pineapple.

Solution: The tapetum is the innermost anther wall layer; it nourishes developing pollen grains and its cells have dense cytoplasm, often with more than one nucleus.

Solution: In double fertilisation one male gamete fuses with the egg (syngamy) forming the diploid zygote, while the other fuses with the two polar nuclei (triple fusion) forming the triploid primary endosperm nucleus.

Solution: A mature embryo sac is 7-celled and 8-nucleate: three antipodals, two synergids, one egg cell and one central cell with two polar nuclei (8 nuclei but 7 cells).

Solution: A dihybrid cross gives an F2 phenotypic ratio of 9:3:3:1, the basis of Mendel's Law of Independent Assortment.

Solution: In Antirrhinum (snapdragon), a cross of red (RR) and white (rr) gives pink (Rr) in F1 — incomplete dominance, with an intermediate F1 phenotype.

Solution: A test cross crosses an individual of unknown genotype showing the dominant phenotype with the homozygous recessive parent, revealing whether it is homozygous or heterozygous.

Solution: Hershey and Chase (1952) used radiolabelled bacteriophages (32P for DNA, 35S for protein) to show that DNA, not protein, enters the host bacterium and is the genetic material.

Solution: Adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine through three hydrogen bonds, as described by the Watson–Crick double helix model.

Solution: The i gene (regulatory gene) codes for the repressor protein, which binds the operator and switches off transcription of the structural genes z, y and a in the absence of lactose.

Solution: NCERT states the large holes in Swiss cheese are due to a large amount of CO2 produced by a bacterium named Propionibacterium sharmanii.

Solution: Rhizobium is a symbiotic nitrogen-fixing bacterium that resides in the root nodules of leguminous plants and fixes atmospheric nitrogen into forms usable by the plant.

Solution: Cyclosporin A, an immunosuppressant for organ-transplant patients, is produced by Trichoderma polysporum. Statins, which lower blood cholesterol, come from Monascus purpureus.

Solution: In commensalism one species benefits and the other is unaffected, e.g., an orchid growing as an epiphyte on a mango branch, or barnacles living on a whale.

Solution: A single tree (producer) supporting a large number of insects (primary consumers) gives an inverted pyramid of numbers. The pyramid of energy is always upright.

Solution: In situ conservation protects species in their natural habitats by conserving whole ecosystems (biosphere reserves, national parks, sanctuaries, sacred groves). Ex situ uses botanical gardens, zoos and seed banks.

Solution: NCERT states that Coelenterata (Cnidaria) are characterised by the presence of cnidoblasts/cnidocytes on the tentacles and body, used for anchorage, defence and capturing prey. Comb plates are the diagnostic feature of Ctenophora.

Solution: NCERT specifically notes that bioluminescence is well marked in Ctenophora (sea walnuts / comb jellies). They are exclusively marine and use comb plates for locomotion.

Solution: NCERT states the first clinical gene therapy was given in 1990 to a girl with ADA (adenosine deaminase) deficiency, an enzyme crucial for the immune system; the deficiency causes Severe Combined Immunodeficiency (SCID).

Solution: NCERT describes pseudostratified epithelium where the nuclei lie at different levels giving a false (pseudo) impression of layering, yet every cell contacts the basement membrane. It lines the trachea and large bronchi.

Solution: Cardiac muscle of the heart is striated and involuntary; its cells are joined end to end by intercalated discs. Skeletal muscle is striated but voluntary; smooth (visceral) muscle is unstriated and involuntary.

Solution: NCERT states the cockroach has an open circulatory system; the colourless blood (lacking respiratory pigment) is termed haemolymph and bathes the visceral organs directly in sinuses.

Solution: NCERT states collagen is the most abundant protein in the animal world, whereas RuBisCO (Ribulose bisphosphate Carboxylase-Oxygenase) is the most abundant protein in the whole of the biosphere.

Solution: NCERT explains that enzymes increase reaction rate by lowering the activation energy barrier (the energy difference between substrate and the transition state), without altering the equilibrium of the reaction.

Solution: NCERT states amino acids are linked by peptide bonds, formed between the carboxyl (–COOH) group of one amino acid and the amino (–NH2) group of the next. Glycosidic bonds link monosaccharides; phosphodiester bonds link nucleotides.

Solution: NCERT states salivary amylase (ptyalin) hydrolyses about 30% of starch into maltose in the buccal cavity at pH 6.8. Pepsin acts on proteins in the stomach; trypsin in the small intestine.

Solution: NCERT states HCl secreted by parietal (oxyntic) cells converts inactive pepsinogen into active pepsin and provides the acidic pH (1.8) optimal for pepsin. Enterokinase activates trypsinogen, not pepsinogen.

Solution: NCERT describes four layers from outside (lumen-distal) to inside (lumen-proximal): serosa, muscularis, sub-mucosa and mucosa. The mucosa is the innermost layer lining the lumen.

Solution: Inspiratory Capacity (IC) = Tidal Volume + Inspiratory Reserve Volume, i.e., the total volume a person can inspire after a normal expiration. Vital Capacity includes expiratory reserve as well.

Solution: NCERT states about 70% of CO2 is transported as bicarbonate, around 20-25% bound to haemoglobin as carbamino-haemoglobin, and about 7% dissolved in plasma.

Solution: NCERT states carbonic anhydrase is present in very high concentration in RBCs and in minute amounts in plasma; it catalyses the reversible reaction between CO2 and water to form carbonic acid (and then bicarbonate).

Solution: NCERT states the Sino-atrial node (SAN), located in the right upper corner of the right atrium, has the highest rate of action potential generation (70-75 per minute) and is therefore called the pacemaker.

Solution: NCERT states group AB individuals have both antigens A and B on their RBCs but no anti-A or anti-B antibodies in plasma, so they can accept blood from all groups, making them universal recipients.

Solution: NCERT states thrombocytes (platelets) are cell fragments produced from megakaryocytes; they release factors involved in the coagulation or clotting of blood. A reduced platelet count can lead to clotting disorders.

Solution: NCERT classifies humans as ureotelic; ammonia produced by metabolism is converted into less toxic urea in the liver via the ornithine cycle and excreted by the kidneys. Ammonotelic animals excrete ammonia; uricotelic animals excrete uric acid.

Solution: NCERT states the PCT helps to maintain the ionic balance and acid-base balance of body fluids by selective secretion of H+, NH3 (ammonia) and K+ ions into the filtrate and by reabsorption of HCO3- from it.

Solution: NCERT states ADH (vasopressin) facilitates water reabsorption from the latter parts of the tubule, preventing diuresis. A fall in body fluid volume activates osmoreceptors that stimulate ADH release from the neurohypophysis.

Solution: NCERT's sliding filament theory states that contraction occurs as the thin (actin) filaments slide over the thick (myosin) filaments; the length of the A-band remains constant while the I-band and H-zone shorten.

Solution: NCERT states that in the resting state, a subunit of troponin masks the active binding sites for myosin on the actin filaments; tropomyosin runs along the actin and, with troponin, covers these sites until Ca2+ binds troponin.

Solution: NCERT states the human skull is composed of 8 cranial bones (plus 14 facial bones) and the axial skeleton comprises 80 bones (skull, vertebral column, sternum and ribs).

Solution: NCERT states that on stimulation, the membrane becomes freely permeable to Na+; the rapid influx of Na+ reverses the polarity (inside becomes positive, outside negative), producing depolarisation and the action potential.

Solution: NCERT states the hypothalamus contains centres that control body temperature, urge for eating and drinking; it also secretes hormones (releasing/inhibiting). The medulla controls respiration and cardiovascular reflexes.

Solution: NCERT states rods contain the purplish-red pigment rhodopsin (visual purple), which contains a derivative of vitamin A, and are responsible for twilight (scotopic) vision. Cones are responsible for daylight and colour vision.

Solution: NCERT states the islets of Langerhans contain alpha-cells secreting glucagon (a hyperglycaemic hormone) and beta-cells secreting insulin (a hypoglycaemic hormone). Glucagon stimulates glycogenolysis and gluconeogenesis.

Solution: NCERT states iodine is essential for the normal rate of synthesis of thyroid hormones (T3 and T4); its dietary deficiency leads to goitre, an enlargement of the thyroid gland.

Solution: NCERT states the adrenal medulla secretes adrenaline (epinephrine) and noradrenaline (norepinephrine), collectively called catecholamines. They are rapidly secreted in response to stress and are termed emergency hormones or hormones of Fight or Flight, so R correctly explains A.

Solution: NCERT states the rupture of the Graafian follicle and release of the ovum (ovulation) is induced by a rapid surge of LH (the LH surge) during the middle of the cycle (around the 14th day).

Solution: NCERT states the remaining parts of the Graafian follicle transform into the corpus luteum, which secretes large amounts of progesterone essential for maintenance of the endometrium needed for implantation.

Solution: NCERT states the blastocyst has an outer trophoblast layer (which attaches to the endometrium) and an inner group of cells called the inner cell mass, which differentiates into the embryo proper.

Solution: NCERT states oral pills containing progestogens (or progestogen-oestrogen combinations) inhibit ovulation and implantation, and alter the quality of cervical mucus to retard/prevent entry of sperms. Saheli is a once-a-week non-steroidal pill.

Solution: NCERT defines ZIFT as Zygote Intra Fallopian Transfer, in which the zygote or early embryo (up to 8 blastomeres) is transferred into the fallopian tube. Transfer of an embryo with more than 8 blastomeres into the uterus is called IUT.

Solution: NCERT states that although amniocentesis can detect genetic disorders, it is statutorily banned for sex determination because it is being misused to abort the normal female foetus (female foeticide).

Solution: NCERT states Plasmodium falciparum causes the most serious, malignant malaria, which can prove fatal. The malarial parasite is transmitted by the bite of the infected female Anopheles mosquito.

Solution: NCERT describes pBR322 as a well-known plasmid cloning vector with an origin of replication (ori), selectable markers (ampicillin and tetracycline resistance genes) and recognition sites for restriction enzymes.

Solution: NCERT states HIV (a retrovirus) enters macrophages and helper T-lymphocytes (TH), where it replicates; the progressive decrease in the number of helper T-cells weakens immunity, leading to AIDS.

Solution: NCERT states that homologous organs have similar structure and origin but different functions (e.g., forelimbs of whale, bat, cheetah, human); they indicate divergent evolution. Analogous organs have different structure but similar function (convergent evolution).

Solution: NCERT states the Bt toxin gene (cry genes) has been cloned from the bacterium Bacillus thuringiensis and expressed in plants such as Bt cotton to provide resistance against insects. The protein crystals contain a toxic insecticidal protein.

Solution: NCERT describes industrial melanism: before industrialisation white-winged moths predominated; after industrialisation, dark-winged (melanic) moths were better camouflaged on soot-covered trees and were selected, illustrating natural selection.

Solution: NCERT states restriction endonucleases recognise a specific palindromic sequence in DNA and cut both strands a little away from the centre, producing sticky ends; this site-specific cutting earns them the name molecular scissors. DNA ligase joins fragments.

Solution: NCERT states PCR uses a thermostable DNA polymerase (Taq polymerase, isolated from the bacterium Thermus aquaticus) which remains active during the high temperature of denaturation, enabling repeated cycles of amplification.

Solution: NCERT states the transgenic cow Rosie produced human protein-enriched milk (2.4 grams per litre) containing the human alpha-lactalbumin, which is nutritionally more balanced for human babies than natural cow milk.