NEET 2022 Biology Question Paper with Answer and Solution

(D) Option $A$ is incorrect because bacteria can be autotrophic (photosynthetic or chemosynthetic) or heterotrophic.
Option $B$ is incorrect because slime moulds are protists,not members of Kingdom $Monera$.
Option $C$ is incorrect because $Mycoplasma$ lack a cell wall.
Option $D$ is correct because $Cyanobacteria$ (blue-green algae) are photosynthetic autotrophs and are classified under Kingdom $Monera$ as prokaryotes.

(C) In an electrical synapse,the membranes of pre- and post-synaptic neurons are in very close proximity,allowing electrical current to flow directly from one neuron into the other. This makes impulse transmission across an electrical synapse very fast,often faster than in a chemical synapse.
In a chemical synapse,the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called the synaptic cleft. Transmission of impulses across chemical synapses is always slower than that across electrical synapses because it involves the release,diffusion,and binding of neurotransmitters.
Therefore,the statement that impulse transmission across a chemical synapse is always faster than that across an electrical synapse is incorrect.

(C) In the oocytes of some vertebrates,the $Diplotene$ stage of $Meiosis-I$ can remain suspended for months or even years.
This prolonged stage is known as the $Dictyotene$ stage.
During this phase,the chromosomes decondense and become transcriptionally active to synthesize materials required for the growth of the oocyte.
Therefore,the correct option is $C$.

(A) In the Calvin cycle,the reduction phase involves the conversion of $3$-phosphoglycerate ($3$-$PGA$) to glyceraldehyde $3$-phosphate (triose phosphate).
For the fixation of one molecule of $CO_2$,the reduction phase requires $1$ molecule of $ATP$ (for phosphorylation) and $1$ molecule of $NADPH$ (for reduction).
However,the overall stoichiometry for the production of one molecule of triose phosphate from $3$ molecules of $CO_2$ requires $6$ $ATP$ and $6$ $NADPH$ in the reduction phase.
Therefore,for the fixation of a single $CO_2$ molecule,the requirement is $1$ $ATP$ and $1$ $NADPH$ specifically for the reduction step.

(D) The $Pericycle$ is a layer of cells located between the $Endodermis$ and the vascular tissue in plant roots.
During the formation of lateral roots,cells of the $Pericycle$ undergo division to form the root primordium.
Similarly,during secondary growth in dicot roots,the $Pericycle$ cells located opposite to the protoxylem divide to form a portion of the vascular cambium ring.
Therefore,the $Pericycle$ is responsible for the initiation of both lateral roots and the vascular cambium.

(A) The correct matching is as follows:
$1$. Adenine is a nitrogenous base belonging to the category of $Purines$.
$2$. Anthocyanin is a secondary metabolite that acts as a $Pigment$.
$3$. Chitin is a structural $Polysaccharide$ found in the cell walls of fungi and the exoskeleton of arthropods.
$4$. Codeine is a secondary metabolite classified as an $Alkaloid$.
Therefore,the correct sequence is $(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)$.

(B) The correct matching is as follows:
$1$. $Chlamydomonas$ is a unicellular green alga,so $(a) - (iii)$.
$2$. $Cycas$ is a gymnosperm,so $(b) - (iv)$.
$3$. $Selaginella$ is a pteridophyte,so $(c) - (ii)$.
$4$. $Sphagnum$ is a moss (bryophyte),so $(d) - (i)$.
Therefore,the correct sequence is $(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)$.

(B) The provided floral diagram shows the following characteristics:
$1$. The flower is actinomorphic (radially symmetrical),indicated by the symbol $\oplus$.
$2$. It has four sepals in two whorls (valvate aestivation).
$3$. It has four petals in a cross-like arrangement (cruciform,valvate aestivation).
$4$. It has six stamens arranged in two whorls (tetradynamous condition: two outer short and four inner long).
$5$. The ovary is superior and bicarpellary,syncarpous with a false septum (replum).
These features are characteristic of the family $Brassicaceae$ (also known as $Cruciferae$).

(B) In the $TCA$ cycle (Krebs cycle),decarboxylation refers to the removal of a carbon atom in the form of $CO_2$.
$1$. The first decarboxylation occurs when isocitrate $(6C)$ is converted to $\alpha$-ketoglutarate $(5C)$ by the enzyme isocitrate dehydrogenase.
$2$. The second decarboxylation occurs when $\alpha$-ketoglutarate $(5C)$ is converted to succinyl-$CoA$ $(4C)$ by the $\alpha$-ketoglutarate dehydrogenase complex.
Therefore,decarboxylation occurs twice during a single $TCA$ cycle.

(B) The correct matches are as follows:
$a$. Porins are proteins that form huge pores in the outer membranes of mitochondria,plastids,and some bacteria. Thus,$a - iv$.
$b$. Leg-haemoglobin is a pigment found in the root nodules of leguminous plants,which gives them a pink colour. Thus,$b - i$.
$c$. $H^{+}$ accumulation occurs in the lumen of the thylakoid during the light reaction of photosynthesis. Thus,$c - ii$.
$d$. Respiration is considered an amphibolic pathway because it involves both catabolic (breakdown) and anabolic (synthesis) processes. Thus,$d - iii$.
Therefore,the correct sequence is $(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)$.

(B) Plant growth regulators are chemical substances that control growth and development in plants.
Cytokinins are a class of plant growth substances that promote cell division,or cytokinesis,in plant roots and shoots.
Chemically,cytokinins are adenine derivatives (specifically $N^6$-substituted adenine derivatives).
Examples include zeatin,which was isolated from corn kernels and coconut milk.
Therefore,the correct answer is $B$.

(D) The fruit wall of nuts,such as walnuts or almonds,is extremely hard and provides protection to the seed.
This hardness is due to the presence of $Sclerenchyma$ tissue,specifically a type of $Sclerenchyma$ cell known as $Sclereids$ (or stone cells).
$Sclereids$ are dead cells with highly thickened,lignified cell walls that provide mechanical support and rigidity to plant parts.
While $Sclereids$ are a type of $Sclerenchyma$,in the context of specific anatomical structures like fruit walls (endocarp) and seed coats,$Sclereids$ is the most precise answer.

(B) The arrangement of sepals or petals in a floral bud with respect to the other members of the same whorl is called aestivation.
- Imbricate: In this type,the margins of sepals or petals overlap one another but not in any particular direction,as seen in $Cassia$ and $Gulmohur$.
- Valvate: When sepals or petals in a whorl just touch one another at the margin,without overlapping,as in $Calotropis$.
- Vexillary: In pea and bean flowers,there are five petals; the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel). This is called vexillary or papilionaceous aestivation.
- Twisted: If one margin of the appendage overlaps that of the next one and so on,as in $China$ $rose$,$lady's$ $finger$,and $cotton$.
Therefore,the correct matching is: $a-ii, b-i, c-iv, d-iii$.

(B) In plant tissue culture,the process where mature,differentiated parenchyma cells regain the capacity to divide mitotically is known as $Dedifferentiation$.
This process leads to the formation of a mass of undifferentiated cells called $Callus$.
$Differentiation$ is the process where cells undergo structural changes to mature into specialized cells.
$Redifferentiation$ is the process where dedifferentiated cells (like those in the callus) undergo further changes to become specialized again.
Therefore,the correct phenomenon is $Dedifferentiation$.

(B) During the $pachytene$ stage of $prophase-I$ in meiosis,the phenomenon of crossing over occurs.
Crossing over is the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process is mediated and catalyzed by an enzyme complex known as $recombinase$.
$Recombinase$ facilitates the breaking and rejoining of $DNA$ strands,ensuring genetic recombination.

(A) The $TCA$ cycle (Tricarboxylic Acid cycle),also known as the Krebs cycle,involves a series of enzymatic reactions in the mitochondrial matrix.
In this cycle,isocitrate $(6-C)$ undergoes oxidative decarboxylation to form $\alpha$-ketoglutaric acid $(5-C)$.
This is the only step in the cycle where a $5-C$ compound is produced.
Therefore,the correct answer is $\alpha$-ketoglutaric acid.

(D) In facilitated diffusion,carrier proteins are involved in the transport of molecules across the membrane.
When a molecule moves across the membrane independent of other molecules,it is called $Uniport$.
When two molecules move across the membrane in the same direction,it is called $Symport$.
When two molecules move across the membrane in opposite directions,it is called $Antiport$.
Therefore,the correct term for the movement of two molecules in the same direction is $Symport$.

(C) The ascent of xylem sap is primarily driven by the transpiration pull,which is generated by the evaporation of water from the leaves.
This process relies on the physical properties of water,specifically cohesion (attraction between water molecules) and adhesion (attraction of water molecules to the polar surfaces of xylem vessel elements).
Together with surface tension,these forces create a continuous water column that is pulled upward through the xylem as water evaporates from the stomata.
While root pressure contributes to the movement of water in some plants,it is not the primary mechanism for the ascent of sap in tall trees.

(B) rhizome is a modified underground stem that grows horizontally (prostrate) beneath the soil surface.
It is characterized by being thick,fleshy,and branched.
It possesses distinct nodes and internodes,where nodes often bear scale leaves and axillary buds.
The apical bud of a rhizome typically remains underground to facilitate continuous horizontal growth,not above the ground.
Therefore,the statement that the apical bud remains above the ground is incorrect.

(B) The enzyme $Nitrogenase$,which is responsible for biological nitrogen fixation,is highly sensitive to molecular oxygen $(O_2)$.
In leguminous plants,the root nodules contain a pink-colored pigment called $Leghemoglobin$.
$Leghemoglobin$ acts as an oxygen scavenger,binding to $O_2$ and maintaining a low oxygen concentration within the nodule.
This low oxygen environment is essential to protect the $Nitrogenase$ enzyme from oxidative damage,allowing it to function efficiently in the process of nitrogen fixation.

(C) Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called $Plasticity$.
For example,heterophylly in cotton,coriander,and larkspur is a classic example of plasticity where leaves of the juvenile plant are different in shape from those in mature plants.
$Redifferentiation$ is the process where dedifferentiated cells lose the ability to divide and mature to perform specific functions.
$Differentiation$ is the process by which meristematic cells undergo structural changes to become mature cells.
$Development$ is a term that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence.

(B) $Azotobacter$ and $Beijerinckia$ are well-known examples of free-living,aerobic,nitrogen-fixing bacteria found in the soil.
$Rhizobium$ and $Frankia$ are symbiotic nitrogen fixers.
$Anabaena$ is a cyanobacterium (often symbiotic or free-living,but not strictly aerobic in the same context as $Azotobacter$ for nitrogen fixation).
$Rhodospirillum$ is an anaerobic nitrogen fixer.
$Pseudomonas$ and $Thiobacillus$ are involved in denitrification,not nitrogen fixation.

(C) Primary proteins are referred to as polypeptides because they consist of a linear chain of amino acids linked together by peptide bonds.
In a polypeptide chain,the carboxyl group $(-COOH)$ of one amino acid reacts with the amino group $(-NH_2)$ of the next amino acid to form a peptide bond $(-CO-NH-)$.
Since the primary structure is defined by this specific sequence of amino acids held together by these covalent peptide bonds,the term 'polypeptide' is used to describe this linear arrangement.

(B) The diagram shows the green alga $Chara$.
$Chara$ belongs to the class Chlorophyceae.
Key characteristics of Chlorophyceae include:
$1$. They are commonly called green algae.
$2$. The plant body is usually grass green due to the dominance of pigments chlorophyll $a$ and $b$.
$3$. Food is stored in the form of starch.
$4$. $Chara$ is a monoecious plant,meaning it bears both male (antheridium) and female (oogonium) sex organs on the same plant body.
Evaluating the statements:
$(a)$ It is a member of Chlorophyceae: Correct.
$(b)$ Food is stored in the form of starch: Correct.
$(c)$ It is a monoecious plant showing oogonium and antheridium: Correct.
$(d)$ Food is stored in the form of laminarin or mannitol: Incorrect (this is characteristic of Phaeophyceae).
$(e)$ It shows dominance of pigments chlorophyll $a, c$ and Fucoxanthin: Incorrect (this is characteristic of Phaeophyceae).
Therefore,statements $(a), (b),$ and $(c)$ are correct.

(B) The cell membrane is primarily composed of a lipid bilayer,which has a hydrophobic interior.
Substances that are hydrophobic or lipid-soluble can easily pass through the lipid bilayer via simple diffusion.
However,substances with a hydrophilic (water-loving) moiety are polar or charged and cannot easily pass through the non-polar,hydrophobic core of the lipid bilayer.
Therefore,hydrophilic substances require specialized transport proteins (like channels or carriers) to cross the cell membrane.

(B) According to the chemiosmotic hypothesis,$ATP$ synthesis is linked to the development of a proton gradient across the thylakoid membrane.
$(a)$ The splitting of water molecules occurs on the inner side of the thylakoid membrane,which releases protons $(H^+)$ into the lumen.
$(b)$ Protons accumulate within the lumen of the thylakoids due to water splitting and the pumping of protons from the stroma.
$(c)$ The primary electron acceptor transfers electrons to an electron transport system,which facilitates the movement of protons.
$(d)$ The $NADP$ reductase enzyme is located on the stroma side of the membrane. It uses electrons from $Fd$ and protons from the stroma to reduce $NADP^+$ to $NADPH + H^+$.
$(e)$ Protons decrease in number in the stroma because they are consumed by $NADP$ reductase and pumped into the lumen,making this statement incorrect.
Therefore,statements $(a), (b),$ and $(d)$ are correct.

(C) Prophase $I$ of meiosis is divided into five sub-stages: Leptotene,Zygotene,Pachytene,Diplotene,and Diakinesis.
$1$. $(b)$ Chromosomes become gradually visible under the microscope occurs during Leptotene.
$2$. $(a)$ Synapsis of homologous chromosomes occurs during Zygotene.
$3$. $(c)$ Crossing over between non-sister chromatids occurs during Pachytene.
$4$. $(e)$ Dissolution of the synaptonemal complex occurs during Diplotene.
$5$. $(d)$ Terminalisation of chiasmata occurs during Diakinesis.
Therefore,the correct sequence is $(b), (a), (c), (e), (d)$.

(B) Lysosomes contain hydrolytic enzymes (acid hydrolases) that are specifically adapted to function in an acidic environment,typically around $pH$ $5.0$.
These enzymes require an acidic medium to maintain their structural integrity and catalytic activity.
If the $pH$ inside the lysosome is increased to an alkaline level,the environment becomes unsuitable for these enzymes.
As a result,the hydrolytic enzymes will lose their functional conformation and become inactive.

(A) The enzyme Maltase acts on Maltose to break it down into two molecules of Glucose.
Option $A$ states that Maltose breaks down into Glucose + Galactose,which is incorrect.
Sucrose is broken down by Sucrase into Glucose and Fructose.
Lactose is broken down by Lactase into Glucose and Galactose.
Dipeptides are broken down by Dipeptidases into Amino acids.
Therefore,the reaction in option $A$ is incorrect.

(D) Mad cow disease (Bovine Spongiform Encephalopathy) in cattle and $Creutzfeldt-Jakob$ disease $(CJD)$ in humans are caused by prions.
Prions are abnormally folded proteins that can induce abnormal folding in normal proteins.
They are infectious agents that lack nucleic acids ($DNA$ or $RNA$) and are smaller than viruses.
Therefore,the correct option is $D$.

(B) The sliding filament theory states that muscle contraction occurs when thin actin filaments slide over thick myosin filaments.
During this process,the $A$-band remains constant in length because the length of the myosin filaments does not change.
The $I$-band shortens as the actin filaments are pulled toward the center of the sarcomere.
The $H$-zone also shortens or disappears.
Consequently,the overall length of the sarcomere decreases,leading to muscle contraction.

(C) Statement $I$ is correct. Amino acids are amphoteric because they contain both an acidic carboxyl group $(-COOH)$ and a basic amino group $(-NH_2)$. These groups can ionize depending on the $pH$ of the solution,leading to different structural forms.
Statement $II$ is incorrect. $A$ Zwitterion is a dipolar ion that has both positive and negative charges,resulting in a net charge of zero. This state occurs at a specific $pH$ known as the isoelectric point $(pI)$. At highly acidic $pH$,the amino acid exists primarily in a cationic form $(-NH_3^+)$,and at highly basic $pH$,it exists primarily in an anionic form $(-COO^-)$. Therefore,it does not exist as a Zwitterion at all acidic or basic $pH$ levels.

(C) The epithelium present in the bronchioles and Fallopian tubes is the ciliated epithelium.
Ciliated epithelium consists of cells that have hair-like projections called cilia on their free surface.
These cilia move particles or mucus in a specific direction over the epithelial surface.
In the Fallopian tubes,the cilia help in the movement of the ovum towards the uterus.
In the bronchioles,the cilia help in moving mucus and trapped dust particles out of the respiratory tract.

(A) Gout is a metabolic disorder of the skeletal system.
It is caused by the accumulation of uric acid crystals in the joints.
This accumulation leads to inflammation,swelling,and severe pain in the affected joints.
Therefore,the correct option is $A$.

(C) Vital Capacity $(VC)$ is defined as the maximum volume of air a person can breathe in after a forced expiration. This includes Expiratory Reserve Volume $(ERV)$,Tidal Volume $(TV)$,and Inspiratory Reserve Volume $(IRV)$.
Mathematically,$VC = ERV + TV + IRV$.
Statement $(a)$ is correct as it lists the components of $VC$.
Statement $(c)$ is correct as it defines $VC$ as the maximum volume of air a person can breathe in after forced expiration.
Statement $(e)$ is also correct because the maximum volume of air a person can breathe out after a forced inspiration is equivalent to the $VC$ (the total exchangeable air).
Therefore,statements $(a), (c),$ and $(e)$ are correct.

(B) The $Hepatic$ $portal$ $system$ is a specialized vascular connection that exists between the digestive tract and the liver.
It carries deoxygenated blood,which is rich in nutrients absorbed from the gastrointestinal tract,directly to the liver for processing,detoxification,and storage before it enters the systemic circulation via the hepatic vein.
This system is essential for metabolic regulation and filtering of substances absorbed from the gut.

(A) The digestive system of a cockroach consists of the following structures:
$1$. $a$. Crop: It is a thin-walled sac used for the storage of food.
$2$. $b$. Proventriculus (or Gizzard): It has thick muscular walls and chitinous teeth,which help in grinding the food particles.
$3$. $c$. Hepatic caecae: These are $6-8$ blind tubules present at the junction of the foregut and midgut,which secrete digestive juices.
$4$. $d$. Malpighian tubules: These are $100-150$ yellow-colored thin filamentous structures present at the junction of the midgut and hindgut,which help in the removal of nitrogenous waste from the haemolymph.
Therefore,the correct matching is $(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)$.

(B) The pneumotaxic centre is a specialized region located in the $Pons$ of the human brain.
Its primary function is to moderate the functions of the respiratory rhythm centre.
It alters the respiratory rate by reducing the duration of inspiration,which subsequently increases the respiratory rate.

(B) $1$. Skeletal muscle fibres are multinucleated (syncytial) and arranged in parallel bundles,not uninucleated.
$2$. Cardiac muscle cells are connected by intercalated discs,which function as communication junctions (gap junctions) allowing the cells to contract as a single unit.
$3$. The walls of blood vessels are primarily composed of smooth muscle tissue,not columnar epithelium.
$4$. Smooth muscles are uninucleated and involuntary,not multinucleated.

(A) $FSH$ (Follicle Stimulating Hormone) is a peptide hormone. Peptide hormones are water-soluble and cannot pass through the lipid bilayer of the plasma membrane of target cells.
Therefore,they interact with specific membrane-bound receptors present on the surface of the target cell,which is stated in Assertion $(A)$.
Upon binding to the receptor,these hormones trigger the formation of second messengers like cyclic $AMP$ $(cAMP)$ or $Ca^{2+}$ inside the cell,which then regulate the cellular metabolism and physiological responses. This mechanism is correctly described in Reason $(R)$.
Since the inability of the hormone to enter the cell necessitates the use of a second messenger system to transmit the signal,Reason $(R)$ is the correct explanation for Assertion $(A)$.

(C) The heart structure varies among vertebrates:
$1$. $Scoliodon$ (Dogfish) is a cartilaginous fish,which has a two-chambered heart (one atrium and one ventricle).
$2$. $Hippocampus$ (Seahorse) is a bony fish,which also has a two-chambered heart.
$3$. $Chelone$ (Green sea turtle) is a reptile. Most reptiles,including turtles,possess a three-chambered heart consisting of two atria and one partially divided ventricle.
$4$. $Pteropus$ (Flying fox/Bat) is a mammal,which has a four-chambered heart.
Therefore,the correct answer is $Chelone$.

(A) During pregnancy,the metabolic rate of the mother increases significantly to support the developing fetus.
To meet these increased metabolic demands,the thyroid gland becomes more active,leading to an increased level of thyroxine in the maternal blood.
Therefore,Assertion $(A)$ is correct.
Reason $(R)$ is also correct because pregnancy involves various physiological and metabolic adaptations in the mother's body.
Since the increase in thyroxine is a direct consequence of the increased metabolic requirements during pregnancy,$(R)$ is the correct explanation of $(A)$.

(C) Cyclostomes are a group of jawless vertebrates.
Statement $(a)$ is correct: They lack scales and paired fins.
Statement $(b)$ is incorrect: They have a circular mouth,but they are jawless (agnathans).
Statement $(c)$ is correct: They bear $6-15$ pairs of gill slits for respiration.
Statement $(d)$ is incorrect: Cyclostomes are marine but migrate to fresh water for spawning (anadromous migration),after which they die.
Therefore,statements $(b)$ and $(d)$ are incorrect.

(D) Enamel is the hardest substance in the human body. It covers the crown of the tooth and provides a protective layer against mechanical wear and chemical erosion. It is essential for maintaining the structural integrity and shape of the teeth,allowing them to withstand the forces of mastication.

(A) Let us analyze each statement:
$(a)$ Bones provide a structural framework,supporting and protecting softer tissues and organs. This is a correct statement.
$(b)$ Limb bones are responsible for weight-bearing and locomotion. This is a correct statement.
$(c)$ Bone marrow,not ligaments,is the site of production of blood cells. This is an incorrect statement.
$(d)$ Adipose tissue is a type of loose connective tissue specialized to store fats. This is a correct statement.
$(e)$ Tendons attach skeletal muscles to bones,whereas ligaments attach one bone to another. This is an incorrect statement.
Therefore,statements $(a), (b),$ and $(d)$ are correct. The correct option is $A$.

(A) During the $Zygotene$ stage of $Meiosis-I$,homologous chromosomes pair up,a process known as $Synapsis$.
This pairing is facilitated by the formation of a $Synaptonemal$ $complex$.
The paired homologous chromosomes are called $Bivalents$ or $Tetrads$.
Therefore,the formation of $Bivalents$ or $Tetrads$ is a characteristic feature of the $Zygotene$ stage,associated with the $Synaptonemal$ $complex$.

(C) The taxonomical aid 'key' is used for identification of plants and animals based on the similarities and dissimilarities. Statement $(a)$ is true.
Key is analytical in nature because it involves the analysis of characters to identify an organism. Statement $(b)$ is true.
Keys are based on the contrasting characters generally in a pair called couplet. Statement $(c)$ is true.
Separate taxonomic keys are required for each taxonomic category such as family,genus,and species for identification purposes. Therefore,the same key cannot be used for all categories. Statement $(d)$ is false.
Each statement in the key is called a lead. Statement $(e)$ is true.
Thus,statements $(a), (b), (c),$ and $(e)$ are correct.

(D) The correct matching is as follows:
$1$. Multipolar neurons have one axon and two or more dendrites and are found in the cerebral cortex $(a - ii)$.
$2$. Bipolar neurons have one axon and one dendrite and are found in the retina of the eye $(b - iii)$.
$3$. Myelinated nerve fibres are found in spinal and cranial nerves $(c - iv)$.
$4$. Unmyelinated nerve fibres are commonly found in the autonomous and the somatic neural systems $(d - i)$.
Therefore,the correct sequence is $(a - ii, b - iii, c - iv, d - i)$.

(D) In cockroaches,the primary excretory organs are the $Malpighian$ $tubules$.
Other structures involved in the excretion or storage of nitrogenous wastes include the $Fat$ $body$,$Urecose$ $glands$,and $Nephrocytes$.
$Hepatic$ $caeca$ (also known as $Hepatopancreatic$ $caeca$) are blind tubules present at the junction of the foregut and midgut,which are primarily involved in the secretion of digestive juices,not excretion.
Therefore,$Hepatic$ $caeca$ is the correct answer.

(A) The abundance of formed elements in human blood is as follows:
$1$. Erythrocytes $(RBCs)$: $5.0-5.5$ million/$mm^3$ (Most abundant).
$2$. Platelets: $150,000-350,000$ per $mm^3$.
$3$. Neutrophils: $60-65\%$ of total $WBCs$.
$4$. Monocytes: $6-8\%$ of total $WBCs$.
$5$. Eosinophils: $2-3\%$ of total $WBCs$.
Comparing the absolute numbers, the order of abundance is: Erythrocytes > Platelets > Neutrophils > Monocytes > Eosinophils.
Thus, the correct sequence is $(c), (a), (b), (e), (d)$.

(A) Statement $I$ is correct: $DNA$ polymerases are enzymes that synthesize $DNA$ by adding nucleotides only to the $3^{\prime}$ end of a growing strand. Therefore,polymerization always occurs in the $5^{\prime} \rightarrow 3^{\prime}$ direction.
Statement $II$ is correct: Because $DNA$ strands are antiparallel and $DNA$ polymerase can only synthesize in the $5^{\prime} \rightarrow 3^{\prime}$ direction,one strand (the leading strand) is synthesized continuously toward the replication fork,while the other strand (the lagging strand) is synthesized discontinuously in short segments called Okazaki fragments away from the replication fork.
Conclusion: Both statements are correct.

(C) Hydrarch succession takes place in wetter areas and the successional series progress from hydric to the mesic conditions.
In hydrarch succession,the pioneer species are the small phytoplanktons (such as blue-green algae,green algae,and diatoms).
These organisms are the first to colonize the bare water body,eventually leading to the formation of more complex communities.

(D) Statement $I$ is incorrect because both Sickle cell anaemia and Haemophilia are recessive disorders. Specifically,Sickle cell anaemia is an autosomal recessive trait,while Haemophilia is an $X$-linked recessive trait.
Statement $II$ is correct because both conditions affect the blood. Sickle cell anaemia involves the formation of abnormal haemoglobin leading to distorted red blood cells,and Haemophilia is a clotting factor deficiency disorder.
Therefore,Statement $I$ is incorrect and Statement $II$ is correct.

(A) Restriction enzymes,specifically those like $EcoRI$,cut the $DNA$ strands at specific points known as recognition sequences or palindromic sites.
When these enzymes cut the $DNA$ a little away from the centre of the palindromic site,they leave single-stranded portions at the ends.
These overhanging stretches are called sticky ends because they can form hydrogen bonds with their complementary cut counterparts.
Thus,Assertion $(A)$ is correct as it describes the formation of sticky ends.
Reason $(R)$ is also correct because the formation of these sticky ends is specifically due to the staggered cuts made away from the centre of the palindromic sequence.
Therefore,$(R)$ is the correct explanation of $(A)$.

(D) According to the data provided in the $NCERT$ textbook regarding the biodiversity of the Amazon rainforest,the estimated number of species is as follows:
$1$. Invertebrates: $1,25,000$
$2$. Plants: $40,000$
$3$. Fishes: $3,000$
$4$. Birds: $1,300$
$5$. Mammals: $427$
Therefore,the descending order is: Invertebrates $(b) >$ Plants $(a) >$ Fishes $(c) >$ Birds $(e) >$ Mammals $(d)$.
Thus,the correct sequence is $(b) > (a) > (c) > (e) > (d)$.

(A) The process of ecological succession begins in a bare area where no life previously existed.
$1$. The first group of organisms that colonize such a barren area are known as pioneer species.
$2$. Examples include lichens on bare rocks or phytoplankton in a new pond.
$3$. These species modify the environment,making it suitable for subsequent species to establish.
$4$. Therefore,the correct answer is Pioneer species.

(A) The typical mature embryo sac of angiosperms is $7$-celled and $8$-nucleate.
It consists of the following components:
$1$. An egg apparatus at the micropylar end,which includes one egg cell and two synergids.
$2$. Three antipodal cells at the chalazal end.
$3$. $A$ large central cell containing two polar nuclei,which eventually fuse to form a diploid secondary nucleus.
Therefore,the correct composition is one egg cell,two synergids,three antipodal cells,and two polar nuclei.

(B) Ecological succession is the gradual and predictable change in the species composition of a given area.
Regardless of whether the succession begins in a dry area (xerarch) or a wet area (hydrarch),it eventually leads to a stable,mature community known as the climax community.
In both cases,the climax community is characterized by moderate moisture conditions,which is referred to as a $Mesic$ environment.
Therefore,all successions proceed to a $Mesic$ climax community.

(D) The separation of $DNA$ fragments based on their size is achieved through a technique called $Gel$ $electrophoresis$.
In this process,$DNA$ fragments are separated by forcing them to move through a gel matrix (usually agarose) under an electric field.
Since $DNA$ molecules are negatively charged,they move towards the anode (positive electrode).
The smaller fragments move faster and travel further through the pores of the gel compared to larger fragments,allowing for effective separation.

(A) The World Summit on Sustainable Development was held in $2002$ in Johannesburg,South Africa.
During this summit,$190$ countries pledged their commitment to achieve a significant reduction in the current rate of biodiversity loss at global,regional,and local levels by the year $2010$.

(A) In artificial hybridization,the goal is to ensure that only the desired pollen grains reach the stigma.
$(a)$ If the plant produces unisexual flowers,the female flower buds do not need to be bagged because they do not contain anthers,thus preventing self-pollination naturally.
$(b)$ Unisexual flowers do not have stamens (anthers),so emasculation (removal of anthers) is not required.
$(c)$ This statement is incorrect; bagging is done immediately after emasculation to prevent unwanted pollination,not after cross-pollination.
$(d)$ This is the standard procedure for bisexual flowers to prevent self-pollination.
$(e)$ This statement is incorrect; protogyny (stigma maturing before anthers) can be used for cross-pollination by selecting appropriate parents.
Therefore,statements $(a), (b),$ and $(d)$ are correct,but since the options provided require selecting the best fit,we analyze the standard procedure: $(a)$ and $(b)$ are correct. Statement $(d)$ is the standard definition of bagging. Thus,$(a), (b),$ and $(d)$ would be the ideal set,but based on the provided options,the most accurate selection is $(a), (b)$ and $(d)$ which is not listed,however,$(a), (b)$ and $(c)$ is often cited in specific contexts where bagging is re-applied. Re-evaluating the standard $NCERT$ text: $(a)$ and $(b)$ are definitely correct. $(d)$ is the definition of bagging. Given the options,$(a), (b)$ and $(d)$ is the correct logic,but since it is not an option,we select the closest logical grouping.

(C) In most flowering plants,the nucellus is completely consumed during the development of the embryo.
However,in some seeds such as beet and black pepper,a part of the nucellus remains persistent.
This residual,persistent nucellus is known as the perisperm.

(B) The chromosomal theory of inheritance was proposed by $Sutton$ and $Boveri$ in $1902$.
This theory states that chromosomes are the vehicles of heredity and that the segregation of a pair of Mendelian factors (genes) during meiosis is parallel to the segregation of a pair of homologous chromosomes.
$Thomas$ $Morgan$ is known for his experimental verification of this theory using $Drosophila$ $melanogaster$ (fruit fly).
$Gregor$ $Mendel$ is the father of genetics who proposed the laws of inheritance.
$Robert$ $Brown$ is known for the discovery of the cell nucleus.

$(A)$. Sacred groves are tracts of forest which are held in high esteem and protected by local communities, such as the Khasi and Jaintia Hills in Meghalaya $(a-iv)$.
$b$. Zoological parks are facilities where animals are kept within enclosures, displayed to the public, and in which they may also be bred. This is a method of $Ex-situ$ conservation $(b-iii)$.
$c$. Nile perch is an example of an alien species introduced into Lake Victoria, which caused the extinction of more than $200$ species of cichlid fish $(c-i)$.
$d$. The Amazon rainforest is often referred to as the 'lungs of the planet' because it is estimated to contribute about $20\%$ of the total oxygen in the earth's atmosphere through photosynthesis $(d-ii)$.
Therefore, the correct matching is $(a-iv, b-iii, c-i, d-ii)$.

(A) The correct matches are as follows:
$1$. Gene gun $(a)$: It is a biolistic method used for the direct transfer of foreign $DNA$ into host cells $(ii)$.
$2$. Gene therapy $(b)$: It involves the replacement of a faulty or defective gene with a normal, healthy gene to treat a genetic disorder $(i)$.
$3$. Gene cloning $(c)$: It is the process of creating multiple identical copies of a specific $DNA$ fragment or molecule $(iv)$.
$4$. Genome $(d)$: It refers to the entire set of genetic material $(DNA)$ present in the cells of an organism $(iii)$.
Therefore, the correct matching is $(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$.

(C) The correct matches are based on the standard values provided in the $NCERT$ textbook for the Molecular Basis of Inheritance:
$1$. Bacteriophage $\phi \times 174$ has $5386$ nucleotides $(a-ii)$.
$2$. Bacteriophage lambda has $48502$ base pairs $(b-i)$.
$3$. Escherichia coli $(E. coli)$ has $4.6 \times 10^6$ base pairs $(c-iv)$.
$4$. The haploid content of human $DNA$ is $3.3 \times 10^9$ base pairs $(d-iii)$.
Therefore,the correct matching is $(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)$.

(C) Apomixis is a form of asexual reproduction that mimics sexual reproduction,where seeds are produced without fertilization.
In hybrid crops,the main problem is that the desirable traits segregate in the next generation,meaning farmers must buy new hybrid seeds every year.
If an apomictic gene is introduced into these hybrids,the seeds produced will be clones of the parent plant.
Consequently,there will be no segregation of characters,and farmers can harvest and reuse these seeds for subsequent plantings without losing the hybrid vigor or desired traits.

(C) Tropical forests are characterized by high biodiversity and a stable climate that supports plant growth throughout the year.
Because of the favorable climatic conditions,many plant species in tropical regions produce fruits at different times of the year.
This continuous availability of food resources (fruits) supports a large population of frugivorous (fruit-eating) birds,as they do not face food scarcity during any season.
Therefore,the primary reason for the abundance of frugivorous birds in tropical forests is the availability of fruits throughout the year.

(D) The symptoms described,such as a small round head,furrowed tongue,partially open mouth,broad palm with a characteristic palm crease,and retarded physical,psychomotor,and mental development,are characteristic of $Down$ syndrome.
$Down$ syndrome is a genetic disorder caused by the presence of an extra copy of chromosome $21$,which is known as Trisomy $21$.
This results in a total of $47$ chromosomes instead of the normal $46$.

(B) To isolate genetic material $(DNA)$ from plant cells,the cell wall must be broken down. Plant cell walls are primarily composed of cellulose,so the enzyme $(a)$ Cellulase is required.
Fungal cell walls are composed of chitin,so the enzyme $(b)$ Chitinase is required to break them down and release the genetic material.
Therefore,the correct pair is $(a)$ Cellulase and $(b)$ Chitinase.

(A) The correct matches based on ecological data provided in the $NCERT$ textbook are:
- $a$. Carbon dissolved in oceans: $71 \%$ of the global carbon is found dissolved in oceans $(ii)$.
- $b$. Annual fixation of carbon through photosynthesis: Approximately $4 \times 10^{13} \ kg$ of carbon is fixed annually by photosynthesis $(iii)$.
- $c$. $PAR$ (Photosynthetically Active Radiation) captured by plants: Plants capture only $2$ to $10 \%$ of the incident $PAR$ $(iv)$.
- $d$. Productivity of oceans: The annual net primary productivity of the oceans is approximately $55$ billion tons $(i)$.
Therefore,the correct sequence is $(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)$.

(C) In pea plants,green pod colour $(G)$ is dominant over yellow $(g)$,and inflated pod shape $(I)$ is dominant over constricted $(i)$.
The parental cross is between green-inflated $(GGII)$ and yellow-constricted $(ggii)$.
The $F_1$ generation is $GgIi$ (green-inflated).
In a dihybrid cross,the $F_2$ phenotypic ratio is $9:3:3:1$.
The phenotypes are:
$9$ (Green,Inflated) : $3$ (Green,Constricted) : $3$ (Yellow,Inflated) : $1$ (Yellow,Constricted).
The total number of parts is $9+3+3+1 = 16$.
The proportion of yellow and inflated pods is $3/16$.
Percentage = $(3/16) \times 100 = 18.75 \%$.

(B) Pathogenic bacteria often acquire resistance to antibiotics through the presence of $R$-plasmids (resistance plasmids).
These are small,circular,extrachromosomal $DNA$ molecules that exist independently of the bacterial chromosome.
Plasmids carry genes that encode enzymes capable of degrading or modifying antibiotics,thereby rendering them ineffective.
Since bacteria can transfer these plasmids to other bacteria through conjugation,antibiotic resistance can spread rapidly within a population.
Therefore,the correct option is $B$.

(B) The transgenic cow '$Rosie$',produced in $1997$,was the first transgenic cow. Its milk contained the human protein alpha-lactalbumin. This protein makes the milk nutritionally more balanced for human babies compared to natural cow milk,as it closely mimics the composition of human breast milk.

(D) Graft rejection is a process where the recipient's immune system recognizes the transplanted organ as foreign ('non-self').
This recognition is primarily mediated by $T$-lymphocytes,which are the key components of the cell-mediated immune response.
When $T$-cells identify the foreign antigens on the graft,they initiate an immune attack to destroy the transplanted tissue.
Therefore,the cell-mediated immune response is the primary mechanism responsible for graft rejection.

(C) In the original Hershey and Chase experiment,they used radioactive phosphorus $(^{32}P)$ to label $DNA$ and radioactive sulfur $(^{35}S)$ to label proteins.
Since $DNA$ enters the bacterial cell during infection by bacteriophages and proteins do not,they observed radioactive phosphorus inside the bacterial cells,proving $DNA$ is the genetic material.
If the composition were swapped,$DNA$ would contain sulfur and proteins would contain phosphorus.
Upon infection,the $DNA$ (now containing radioactive sulfur) would enter the bacterial cells.
Therefore,radioactive sulfur would be detected inside the bacterial cells,while radioactive phosphorus would remain outside with the protein coat.

(B) According to Gause's Competitive Exclusion Principle,two species competing for the same limiting resource cannot coexist indefinitely. However,species can avoid competition and coexist by adopting 'resource partitioning'. This involves choosing different foraging patterns,such as feeding at different times of the day or visiting different types of flowers,thereby reducing direct competition for the same resource.

(A) Intra Uterine Devices (IUDs) are contraceptive methods inserted into the uterus by doctors or expert nurses.
They are categorized as non-medicated IUDs (e.g.,Lippes loop),copper-releasing IUDs (e.g.,CuT,Cu7,Multiload $375$),and hormone-releasing IUDs (e.g.,Progestasert,$LNG$-$20$).
Progestogens are a class of hormones used in contraceptive pills or implants,but they are not classified as IUDs themselves.
Therefore,Progestogens is the correct answer.

(B) The correct matches are as follows:
$1$. Chlamydomonas produces motile asexual spores called Zoospores $(a-ii)$.
$2$. Penicillium reproduces through asexual spores known as Conidia $(b-i)$.
$3$. Hydra reproduces by the formation of external Buds $(c-iv)$.
$4$. Sponges (like Sycon) reproduce through internal buds called Gemmules $(d-iii)$.
Therefore,the correct matching is $a-ii, b-i, c-iv, d-iii$.

(B) Primary production is defined as the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.
It is expressed in terms of weight $(g/m^2)$ or energy $(kcal/m^2)$.

(C) Species that are restricted to a specific geographic area and are not found naturally anywhere else in the world are known as $Endemic$ species.
Western Ghats are a biodiversity hotspot known for a high degree of endemism,meaning many of its flora and fauna are unique to that region.
$Threatened$ species are those likely to become extinct in the near future.
$Keystone$ species are those that have a disproportionately large effect on their environment relative to their abundance.
$Vulnerable$ species are those that are likely to become endangered unless the circumstances threatening their survival and reproduction improve.

(B) Assertion $(A)$ is correct because $Spirulina$ is a cyanobacterium (blue-green alga) that can be grown on wastewater to reduce pollution by absorbing nutrients and producing biomass.
Reason $(R)$ is also correct because $Spirulina$ is indeed a rich source of proteins,vitamins,minerals,and other nutrients,which makes it a popular single-cell protein $(SCP)$ source.
However,the nutritional value of $Spirulina$ (Reason) is not the reason why it is used to reduce environmental pollution (Assertion). The ability to reduce pollution is due to its metabolic activity and nutrient uptake capacity,not its nutritional composition.
Therefore,both statements are correct,but $(R)$ is not the correct explanation of $(A)$.

(C) The mammary glands are paired structures that contain glandular tissue and variable amounts of fat. The glandular tissue of each breast is divided into $15-20$ mammary lobes containing clusters of cells called alveoli.
The path of milk secretion from the alveoli to the exterior is as follows:
$1$. Alveoli $(c)$ produce milk.
$2$. The cells of alveoli secrete milk into the mammary tubules $(e)$.
$3$. The tubules of each lobe join to form a mammary duct $(a)$.
$4$. Several mammary ducts join to form a wider mammary ampulla $(d)$,which is connected to the lactiferous duct $(b)$ through which milk is sucked out.
Therefore,the correct sequence from proximal to distal is: $(c) \rightarrow (e) \rightarrow (a) \rightarrow (d) \rightarrow (b)$.

(C) In pedigree analysis,standard symbols are used to represent family relationships and traits.
$1$. $A$ diamond shape represents an individual of unspecified sex.
$2$. $A$ filled (shaded) symbol (square or circle) represents an affected individual.
$3$. $A$ mating (marriage) is represented by a horizontal line connecting a square (male) and a circle (female).
$4$. Consanguineous mating (mating between close relatives) is represented by two horizontal lines connecting the square and the circle.
$5$. The symbol shown in option $C$ represents a normal mating,not consanguineous mating. Therefore,option $C$ is the incorrect match.

(B) $CNG$ (Compressed Natural Gas) is considered a better fuel than diesel for several reasons:
$1$. It cannot be adulterated like petrol or diesel,ensuring purity.
$2$. It burns more efficiently,leaving very little unburnt residue.
$3$. It is cheaper than diesel.
Therefore,statements $(a)$,$(c)$,and $(d)$ are correct.
Statement $(b)$ is incorrect because filling $CNG$ often takes more time due to pressure requirements.
Statement $(e)$ is incorrect because $CNG$ is highly inflammable,though it is safer due to its lighter-than-air nature which allows it to disperse quickly in case of a leak.

(D) The introduction of foreign $DNA$ into plant cells is typically achieved through methods like $Agrobacterium$-mediated transformation,the use of a gene gun (biolistics),and disarmed pathogen vectors.
Bacteriophages are viruses that infect bacteria. While they are used as vectors for cloning $DNA$ into bacterial cells,they are not used to introduce foreign $DNA$ directly into plant cells.
Therefore,the correct answer is $D$.

(C) In the process of spermatogenesis,one primary spermatocyte undergoes meiosis-$I$ to form two secondary spermatocytes.
Each secondary spermatocyte then undergoes meiosis-$II$ to form two spermatids.
Thus,one secondary spermatocyte produces two spermatozoa.
To form $400\,million$ spermatozoa,the number of secondary spermatocytes required is calculated as:
$\text{Number of secondary spermatocytes} = \frac{\text{Total spermatozoa}}{2} = \frac{400\,million}{2} = 200\,million$.

(B) Haemophilia is an $X$-linked recessive disorder.
Let $X^H$ be the normal allele and $X^h$ be the haemophilic allele.
The mother of the girl is haemophilic, so her genotype is $X^h X^h$. She must pass one $X^h$ chromosome to her daughter. Since the girl is normal, her genotype must be $X^H X^h$ (carrier).
The male has no ancestral history of haemophilia, so his genotype is $X^H Y$.
Cross: $X^H X^h$ (carrier female) $\times$ $X^H Y$ (normal male).
Offspring genotypes: $X^H X^H$ (normal daughter), $X^H X^h$ (carrier daughter), $X^H Y$ (normal son), $X^h Y$ (haemophilic son).
Possible phenotypes: Normal daughter, carrier daughter, normal son, and haemophilic son.
Thus, the possible phenotypes include normal daughter, normal son, carrier daughter, and haemophilic son. Options $(b)$ and $(c)$ correctly describe these possibilities.

(C) Intrauterine devices $(IUDs)$ function through various mechanisms:
$1$. Statement $(a)$ is incorrect because $IUDs$ actually increase the phagocytosis of sperm within the uterus.
$2$. Statement $(b)$ is correct; copper ions released from copper-releasing $IUDs$ (like $CuT$) suppress sperm motility and their fertilizing capacity.
$3$. Statement $(c)$ is incorrect because $IUDs$ make the cervix hostile to sperm,preventing their entry.
$4$. Statement $(d)$ is correct; copper ions suppress the fertilization capacity of sperm.
$5$. Statement $(e)$ is incorrect because $IUDs$ are inserted by doctors or expert nurses in the uterine cavity and do not require surgical intervention.
Therefore,statements $(b)$ and $(d)$ are correct.

(B) Statement $(a)$ is correct: Agarose is a natural polymer extracted from seaweeds like Gelidium and Gracilaria.
Statement $(b)$ is correct: In agarose-gel electrophoresis,$DNA$ fragments are separated based on their size (molecular weight) through the sieving effect of the gel matrix.
Statement $(c)$ is correct: $DNA$ molecules are negatively charged due to the phosphate backbone. Therefore,they move towards the anode (positively charged electrode) when an electric field is applied.
Statement $(d)$ is incorrect: $DNA$ does not move towards the cathode (negatively charged electrode).
Thus,statements $(a), (b),$ and $(c)$ are correct.

(C) The mutation theory of evolution was proposed by Hugo de Vries.
According to this theory,evolution is a discontinuous process caused by large,sudden,and heritable changes called mutations.
These mutations are random and directionless,unlike the small directional variations proposed by Darwin.
Speciation occurs due to a single step large mutation,which is also known as saltation.
Therefore,the correct statement is that a single step large mutation is a cause of speciation.

(A) The codon on $mRNA$ is $5' UAC 3'$.
According to the base-pairing rules,the anticodon on $tRNA$ is complementary to the codon on $mRNA$.
Complementary base pairing occurs in an antiparallel orientation ($5' \rightarrow 3'$ of codon pairs with $3' \rightarrow 5'$ of anticodon).
Given codon: $5' UAC 3'$.
The complementary sequence is $3' AUG 5'$.
By convention,anticodon sequences are written in the $5' \rightarrow 3'$ direction.
Reversing $3' AUG 5'$ gives $5' GUA 3'$.

(C) Inbreeding refers to the mating of more closely related individuals within the same breed for $4-6$ generations.
$1$. It increases homozygosity,which is essential for evolving pure lines in any animal.
$2$. It helps in the accumulation of superior genes and the elimination of less desirable genes by exposing harmful recessive genes that are then removed by selection.
$3$. Therefore,the statement that it decreases homozygosity is incorrect,as inbreeding actually increases homozygosity.

(D) The innate immunity consists of four types of barriers:
$1$. Physical barriers: Skin and mucus coating of the respiratory,gastrointestinal,and urogenital tracts prevent the entry of microorganisms. Thus,$c - ii$.
$2$. Physiological barriers: Acid in the stomach $(HCl)$,saliva in the mouth,and tears from eyes prevent microbial growth. Thus,$d - iv$.
$3$. Cellular barriers: Certain types of leukocytes (WBCs) like polymorpho-nuclear leukocytes ($PMNL$-neutrophils) and monocytes and natural killer cells in the blood as well as macrophages in tissues can phagocytose and destroy microbes. Thus,$a - iii$.
$4$. Cytokine barriers: Virus-infected cells secrete proteins called interferons which protect non-infected cells from further viral infection. Thus,$b - i$.
Therefore,the correct matching is $(a) - (iii), (b) - (i), (c) - (ii), (d) - (iv)$.

(B) According to Chargaff's rule,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of cytosine $(C)$ is equal to the amount of guanine $(G)$.
Given that $A = 30\%$,therefore $T$ must also be $30\%$.
Given that $C = 20\%$,therefore $G$ must also be $20\%$.
Thus,the percentage composition of $T$ is $30\%$ and $G$ is $20\%$.