NEET 2015 Biology Question Paper with Answer and Solution

(B) Seed dormancy is primarily induced by the plant hormone Abscisic acid $(ABA)$.
$ABA$ acts as a growth inhibitor that prevents germination in seeds,buds,and underground storage organs,allowing them to survive unfavorable environmental conditions.
In contrast,Ethylene is often involved in breaking dormancy in certain seeds and buds.

(D) : $Mucor$ is a member of $Zygomycetes$ (the conjugation fungi) in which motile cells,e.g.,zoospores or planogametes,are absent.
Asexual reproduction takes place by the formation of non-motile mitospores called sporangiospores.
Sexual reproduction takes place by the formation of non-motile zygospores.
Therefore,the statement that $Mucor$ has biflagellate zoospores is incorrect.

(D) : Fimbriae are small bristle-like solid structures arising from the bacterial cell surface.
There are $300-400$ fimbriae per cell.
Their diameter is $3-10 \ nm$ while their length is $0.5-1.5 \ \mu m$.
Fimbriae are involved in attaching bacteria to solid surfaces (e.g.,rocks in a water body) or host tissues (e.g.,the urinary tract in $Neisseria \ gonorrhoeae$).
Some fimbriae cause the agglutination of $RBCs$.
They also help in the mutual clinging of bacteria.

(C) : Viroids are infectious $RNA$ particles which were discovered by $T$.$O$. Diener $(1971)$.
These are devoid of protein coat and cause diseases in plants only,e.g.,potato spindle tuber,chrysanthemum stunt etc.
Therefore,the statement that viroids were discovered by $D$.$J$. Ivanowsky is incorrect,as $D$.$J$. Ivanowsky discovered the causal organism of tobacco mosaic disease.

(C) The correct answer is $C$. Chrysophytes include diatoms and golden algae (desmids).
In diatoms,the cell walls form two thin overlapping shells,which fit together as in a soap box.
The walls are embedded with silica and thus the walls are indestructible.
The two shells are known as the epitheca (upper) and the hypotheca (lower).

(B) : Kingdom Monera consists of prokaryotic organisms,which are characterized by the absence of a nuclear envelope around the nucleus and the absence of membrane-bound cell organelles. Therefore,the statement that a nuclear membrane is present in Monera is incorrect.

(A) : Morels are Ascomycetes with edible ascocarps that have a fleshy,sponge-like conical cap or pileus and a stalk-like stipe,e.g.,$Morchella$ $esculenta$.
Truffles are also edible members of Ascomycetes with tuber-like subterranean ascocarps that are often dug out with the help of trained dogs and pigs,e.g.,$Tuber$ $aestivum$.
Therefore,the statement that morels and truffles are poisonous is incorrect,as they are considered delicacies.

(A) : Mycoplasma (Kingdom Monera) are the simplest and smallest free-living prokaryotes that are devoid of a cell wall. The plasma membrane forms the outer boundary of the cell in Mycoplasma.
$B$: Nostoc is a cyanobacterium (Kingdom Monera),in which the cell wall is composed of peptidoglycans.
$C$: Aspergillus is a fungus (Kingdom Fungi),in which the cell wall is mainly made of chitin.
$D$: Funaria is a bryophyte (Kingdom Plantae),in which the cell wall is cellulosic in nature.
Therefore,the correct answer is $A$.

(C) : $Anabaena$ is a prokaryotic organism.
It is a cyanobacterium (blue-green alga) which belongs to Kingdom $Monera$.
Like all other prokaryotes,it lacks a true nucleus and membrane-bound cell organelles.

(D) $Alternaria$ is a member of $Deuteromycetes$, which are commonly known as 'fungi imperfecti'.
In $Deuteromycetes$, the sexual reproduction stage is either absent or not yet discovered, which is why they are classified as imperfect fungi.
$Mucor$ belongs to $Phycomycetes$ (Zygomycetes).
$Agaricus$ is a saprophytic fungus, not parasitic.
$Phytophthora$ belongs to $Phycomycetes$ (Oomycetes), not $Basidiomycetes$.

(B) The correct answer is $B$.
In Phaeophyceae (brown algae),food is stored as complex carbohydrates,usually in the form of laminarin or mannitol.
In Rhodophyceae (red algae),food is stored as floridean starch,which is very similar to amylopectin and glycogen in structure.
Therefore,the statement that mannitol is stored food in Rhodophyceae is incorrect.

(B) : In gymnosperms (like $Pinus$),the male and female gametophytes do not have an independent free-living existence.
They remain within the sporangia retained on the sporophytes,$i.e.$,the female gametophyte (within the megasporangium) and the male gametophyte (within the microsporangium).

(A) Statement $A$ is correct: Lichens and mosses are pioneer species in primary succession on bare rocks.
Statement $B$ is incorrect: $Selaginella$ is a heterosporous pteridophyte,not homosporous.
Statement $C$ is incorrect: Coralloid roots in $Cycas$ contain nitrogen-fixing cyanobacteria (like $Nostoc$ or $Anabaena$),not $VAM$ (Vesicular Arbuscular Mycorrhiza),which is a fungal association.
Statement $D$ is correct: In bryophytes,the dominant phase is the gametophyte,while in pteridophytes,the dominant phase is the sporophyte.
Statement $E$ is correct: In gymnosperms,the male and female gametophytes do not have an independent free-living existence; they remain within the sporangia retained on the sporophytes.
Therefore,statements $A, D,$ and $E$ are correct.

(A) $Ectocarpus$ produces biflagellate gametes.
$Anabaena$ is a cyanobacterium and does not reproduce sexually.
$Spirogyra$ produces non-flagellated male gametes during conjugation,where the entire cell content functions as a gamete.
$Polysiphonia$ also produces non-flagellated spermatia.

(D) The correct answer is $D$.
Phylum $Porifera$ (sponges) is characterized by a cellular level of body organization.
Their body wall contains a meshwork of cells and internal cavities (spongocoel) lined with specialized flagellated cells known as $choanocytes$ or collar cells,which are responsible for filtering food particles from the water.
The development in this phylum is indirect,meaning it includes a free-swimming larval stage (such as $amphiblastula$ or $parenchymula$) before reaching the adult form,which aids in the dispersal of the species.

(D) : Metagenesis refers to the alternation of generation between asexual and sexual phases of an organism.
In this process,the organism exists in two different forms.
For example,in $Obelia$ (a coelenterate),the polyp form reproduces asexually to produce medusae,and the medusa form reproduces sexually to produce polyps.

(B) $Petromyzon$ (Lamprey) belongs to the Class $Cyclostomata$ of Phylum $Chordata$.
It is a jawless fish that migrates from the ocean to fresh water to lay eggs (anadromous migration).
The eggs hatch in about $3$ weeks into minute,transparent larvae called ammocoetes.
After metamorphosis,the young lampreys swim down to the sea,where they remain for $3$ or $4$ years before reaching maturity.
Once they reach maturity,they migrate back to fresh water streams or rivers to spawn and then die.

(A) : $Trichinella$ $spiralis$ is a minute nematode parasite that exhibits viviparity,meaning it produces live young (larvae) instead of eggs.
The adults of $T. spiralis$ reside in the human small intestine,where the females release large numbers of larvae directly.
These larvae penetrate the intestinal wall and migrate through the bloodstream to muscles,causing trichinosis or trichiniasis,which presents with symptoms such as diarrhea,nausea,vertigo,muscle pain,and fever.
Humans typically become infected after consuming undercooked meat containing the parasite's larval cysts.

(D) None of the options provided is correct without exceptions.
$1$. In $(a)$, $Cyclostomata$ do not have paired appendages.
$2$. In $(b)$, the skin of $Aves$ is dry and non-glandular (except for the oil gland at the base of the tail).
$3$. In $(c)$, while $Mammalia$ are characterized by mammary glands and hair, not all mammals have pinnae (e.g., monotremes) or two pairs of limbs (e.g., whales and dolphins lack hindlimbs).
$4$. In $(d)$, while $Chondrichthyes$ generally have gills without an operculum, there are exceptions like $Chimaera$ (ratfish), which possesses an operculum.

(A) : Duck-billed platypus is an egg-laying mammal (oviparous).
It is found in the rivers of eastern Australia and Tasmania.
It is a beaver-like monotreme about $50-60 \ cm$ long and is well-adapted to live in water.
Usually,two eggs are laid at a time.
The female curls around them for incubation and remains inactive for about two weeks.
Newly hatched young ones are very immature,naked,blind,and each is $2.5 \ cm$ long.
In contrast,whales,bats,and elephants are placental mammals and are viviparous.

(A) : The exoskeleton,composed of a chitinous cuticle,is the primary factor that has enabled insects to colonize land and diversify into almost all possible habitats. It provides essential protection,structural support,and significantly helps in preventing desiccation (water loss).

(D) The plants listed are China rose,mustard,brinjal,potato,guava,cucumber,onion,and tulip.
$1$. China rose: Superior ovary (Hypogynous flower).
$2$. Mustard: Superior ovary (Hypogynous flower).
$3$. Brinjal: Superior ovary (Hypogynous flower).
$4$. Potato: Superior ovary (Hypogynous flower).
$5$. Guava: Inferior ovary (Epigynous flower).
$6$. Cucumber: Inferior ovary (Epigynous flower).
$7$. Onion: Superior ovary (Hypogynous flower).
$8$. Tulip: Superior ovary (Hypogynous flower).
Thus,China rose,mustard,brinjal,potato,onion,and tulip have a superior ovary. The total count is $6$.

(D) : Axile placentation occurs in syncarpous pistils. The ovary is partitioned into two or more chambers. Placentae occur in the central region where the septa meet so that an axile column bearing ovules is formed,$e.g.$,shoe flower (pentalocular),lemon (multilocular),etc.

(D) : $Pistia$ (water lettuce) is a floating aquatic plant.
In aquatic plants,roots are generally poorly developed and do not participate in the absorption of water.
Water is absorbed directly through the general body surface of these plants.

(B) In perigynous flowers,the gynoecium is situated in the center and other floral parts are located on the rim of the thalamus almost at the same level.
In this condition,the ovary is said to be half-inferior.
Examples of perigynous flowers include $plum$,$rose$,and $peach$.
Therefore,the correct option is $(b)$.

(D) The flowers of the family $Fabaceae$ (subfamily $Papilionoideae$) exhibit a butterfly-shaped corolla,known as a papilionaceous corolla.
This arrangement consists of five petals: one posterior,large,outermost petal called the $standard$ or $vexillum$; two lateral petals called $wings$ or $alae$; and two anterior,fused petals called the $keel$ or $carina$.
The $keel$ petals enclose the stamens and the carpel.
This characteristic structure is found in plants like $Indigofera$,bean,gram,and pea.

(C) : In xerophytic plants, the leaves modify into sharp, pointed spines, $e.g.$, $Aloe$, $Solanum$ $surattense$, $Opuntia$, $Asparagus$ etc.
This modification is either for the protection of the plant, to reduce transpiration, or for both.

(A) In a woody dicot stem,the arrangement of tissues from the periphery (outer side) to the center (inner side) is as follows:
$1$. Phellem (cork): This is the outermost layer of the periderm.
$2$. Secondary cortex (phelloderm): This lies just inside the phellem and phellogen.
$3$. Secondary phloem: This is located outside the vascular cambium.
$4$. Wood (Secondary xylem): This is located inside the vascular cambium.
Therefore,the correct sequence from outer to inner side is: $(iv)$ Phellem $\rightarrow$ $(i)$ Secondary cortex $\rightarrow$ $(iii)$ Secondary phloem $\rightarrow$ $(ii)$ Wood (Secondary xylem).
Thus,the correct option is $(A)$.

(A) The correct answer is $A$.
In $monocot$ roots, the vascular bundles are radial, meaning $xylem$ and $phloem$ are arranged in separate patches along different radii.
$A$ key feature is that these vascular bundles are 'closed', which means there is no $cambium$ present between the $xylem$ and $phloem$.
Because $cambium$ is absent, $monocot$ roots do not undergo secondary growth.

(D) In plants,vascular bundles are classified as open or closed based on the presence or absence of cambium.
In dicotyledonous stems,vascular bundles contain cambium between the xylem and phloem,which allows for secondary growth; these are called open vascular bundles.
In monocotyledonous stems,the cambium is absent between the xylem and phloem,meaning they cannot undergo secondary growth.
Therefore,vascular bundles in monocotyledons are referred to as closed vascular bundles.

(D) The correct answer is $D$. In cockroaches, the body cells release nitrogenous waste into the haemolymph primarily as $potassium \text{ } urate$. The $Malpighian \text{ } tubules$ then extract these metabolic wastes, including $potassium \text{ } urate$, water, and carbon dioxide from the haemolymph. Within the $Malpighian \text{ } tubules$, these are converted into uric acid. $A$ significant portion of water and salts is reabsorbed by the cells of the $Malpighian \text{ } tubules$ and returned to the haemolymph, while the remaining uric acid is excreted through the alimentary canal.

(D) : Most cells in animal tissues (with the exception of a few terminally differentiated cells such as skeletal muscle cells and blood cells) are in communication with their adjoining cells via gap junctions.
At the place where a gap junction is present,the membranes of two adjacent cells are separated by a uniform narrow gap of about $2-4 \ nm$.
The gap is spanned by channel-forming proteins called connexins,which allow inorganic ions and other small water-soluble molecules to pass directly from the cytoplasm of one cell to the cytoplasm of another cell.

(A) The correct answer is $A$.
In the body of a cockroach,the exoskeleton consists of hardened plates called sclerites.
The dorsal plates are called terga,the ventral plates are called sterna,and the lateral plates are called pleura.
These sclerites are joined to each other by a thin,flexible,and tough cuticle known as the arthrodial membrane,which allows for movement between the segments.

(D) The correct matching is as follows:
$A$. Thylakoids: These are flattened membranous sacs found within the stroma of chloroplasts,involved in light-dependent reactions $(iii)$.
$B$. Cristae: These are the infoldings of the inner mitochondrial membrane that increase the surface area for cellular respiration $(iv)$.
$C$. Cisternae: These are the disc-shaped,flattened sacs that make up the Golgi apparatus $(i)$.
$D$. Chromatin: This is the condensed,thread-like structure of $DNA$ and proteins found in the nucleus $(ii)$.
Therefore,the correct sequence is $A-(iii), B-(iv), C-(i), D-(ii)$.

(C) : $A$ prokaryotic cell is characterized by the absence of an organized nucleus and membrane-bound cell organelles.
$DNA$ is naked,i.e.,without a nuclear envelope,and lies coiled in the cytoplasm,which is commonly called the nucleoid or genophore.
Mesosomes,plasma membrane,and $70S$ ribosomes are present in a prokaryotic cell.

(B) The correct answer is $B$.
Membrane-bound organelles include the endoplasmic reticulum,nuclei,lysosomes,Golgi apparatus,and mitochondria.
Ribosomes are non-membrane-bound ribonucleoprotein particles.
Chromosomes are structures composed of $DNA$ and proteins found within the nucleus and are not considered membrane-bound organelles.

(D) The correct answer is $D$.
$1$. Membrane-bound organelles include the endoplasmic reticulum,nucleus,lysosomes,Golgi apparatus,vacuoles,and mitochondria.
$2$. Ribosomes are non-membrane-bound organelles composed of ribonucleoproteins,found in both prokaryotic and eukaryotic cells.
$3$. Mesosomes are infoldings of the plasma membrane in prokaryotes,which are membrane-bound structures.

(D) : Ribosomes are small,spherical,non-membrane-bound organelles found in living cells that serve as the site for protein synthesis.
Ribosomes are composed of two subunits,one large and one small,which consist of ribosomal $RNA$ $(rRNA)$ and proteins.
Unlike the nucleus,mitochondria,and chloroplasts,which contain their own genetic material,ribosomes do not contain any $DNA$.

(B) : Recent developments have shown that the nuclear membrane is derived from the rough endoplasmic reticulum $(RER)$. During cell division,the nuclear membrane disintegrates. The nuclear envelope transmembrane proteins are absorbed into the $RER$. Once the division is completed,the $RER$ reassembles the nuclear envelope.

(D) The correct answer is $D$.
In chloroplasts,the inner membrane encloses a fluid-filled space called the stroma.
Within the stroma,there are organized,flattened,membranous sacs called thylakoids.
These thylakoids are stacked like a pile of coins to form structures known as grana (singular: granum).
Stroma lamellae are the membrane channels that connect the thylakoids of different grana.
Cristae are the infoldings of the inner mitochondrial membrane,not chloroplasts.

(D) : Smooth endoplasmic reticulum $(SER)$ is a system of smooth membranes (i.e.,membranes not having ribosomes) within the cytoplasm of plant and animal cells.
It forms a link between the cell and nuclear membranes.
It is the site of important metabolic reactions,including phospholipid and fatty acid synthesis.
In animal cells,lipid-like steroidal hormones are also synthesized.

(D) The centromere is a specialized region of a chromosome that attaches to the spindle fibers during cell division.
Based on the position of the centromere,chromosomes are classified as follows:
$1$. Metacentric: Centromere is in the middle,forming two equal arms.
$2$. Sub-metacentric: Centromere is slightly away from the middle,resulting in one shorter and one longer arm.
$3$. Acrocentric: Centromere is situated close to one end,resulting in one extremely short arm and one very long arm.
$4$. Telocentric: Centromere is at the terminal end of the chromosome.
Therefore,the correct answer is acrocentric.

(B) The correct answer is $B$.
Inclusion bodies are non-membrane-bound structures present in the cytoplasm of prokaryotic cells that serve as storage sites for reserve materials.
Examples of inclusion bodies include $Glycogen$ $granules$, $Phosphate$ $granules$, and $Cyanophycean$ $granules$.
$Polysomes$ (or $polyribosomes$) are not inclusion bodies; they are structures formed by the aggregation of multiple ribosomes attached to a single $mRNA$ strand during protein synthesis.

(A) The correct answer is $A$.
Chitin is a structural polysaccharide that forms the exoskeleton of arthropods.
It is a complex carbohydrate (a nitrogen-containing polysaccharide) in which $N$-acetyl glucosamine monomers are linked together by $(1, 4)$ $\beta$-glycosidic linkages.
This chitinous exoskeleton provides structural strength,protection,and elasticity to the bodies of arthropods.

(B) phosphodiester bond is a chemical bond that joins the $3'$ carbon atom of one sugar molecule and the $5'$ carbon atom of another in a nucleic acid chain ($DNA$ or $RNA$).
- In a polypeptide,amino acids are linked by peptide bonds.
- In a diglyceride,fatty acids are linked to glycerol by ester bonds.
- In a polysaccharide,monosaccharides are linked by glycosidic bonds.
- Nucleic acids (polynucleotides) are formed by linking nucleotides through phosphodiester bonds.
Therefore,the correct option is $B$.

(B) The correct answer is $B$. Competitive inhibition is a reversible process where the inhibitor competes with the normal substrate for the active site of the enzyme.
In competitive inhibition,the inhibitor is structurally similar to the substrate,allowing it to bind to the active site. This binding prevents the substrate from binding,thereby reducing the enzyme's affinity for the substrate.
The $K_m$ value (Michaelis constant) represents the substrate concentration at which the reaction velocity is half of the maximum velocity $(V_{max})$. $A$ higher $K_m$ indicates lower affinity. Since a competitive inhibitor reduces the apparent affinity of the enzyme for the substrate,it increases the $K_m$ value,not decreases it. Therefore,statement $B$ is incorrect.
Statement $A$ is correct because the inhibitor does not affect the catalytic step $(k_{cat})$ once the enzyme-substrate complex is formed. Statement $C$ is correct as the binding is reversible. Statement $D$ is correct because the inhibitor is not a substrate and is not chemically modified by the enzyme.

(D) The correct sequence is $(ii), (i), (iii), (iv)$.
Prophase-$I$ of meiosis is divided into five sub-stages: Leptotene,Zygotene,Pachytene,Diplotene,and Diakinesis.
$(ii)$ Synapsis (pairing of homologous chromosomes) occurs during the Zygotene stage.
$(i)$ Crossing over (exchange of genetic material between non-sister chromatids) occurs during the Pachytene stage.
$(iii)$ Terminalisation of chiasmata (shifting of chiasmata towards the ends of chromosomes) occurs during the Diakinesis stage.
$(iv)$ Disappearance of the nucleolus also occurs during the Diakinesis stage,marking the end of Prophase-$I$.

(A) In a diploid somatic cell during the $G_1$ phase,the $DNA$ content is $2C$ and the chromosome number is $2n$.
In a haploid gamete,the $DNA$ content is $1C$ and the chromosome number is $n$.
The $S$ phase is characterized by the replication of $DNA$,which doubles the amount of $DNA$ per cell.
Therefore,in a somatic cell after the $S$ phase,the $DNA$ content becomes $4C$,while the chromosome number remains $2n$.
Comparing this to a gamete ($1C$ $DNA$ and $n$ chromosomes),the somatic cell has $4$ times the $DNA$ ($4C$ vs $1C$) and $2$ times the number of chromosomes ($2n$ vs $n$).

(D) The correct matching is as follows:
$A$. Synapsis aligns homologous chromosomes occurs during the Zygotene stage of Prophase $I$ of Meiosis $(ii)$.
$B$. Synthesis of $RNA$ and protein occurs during the $G_2$ phase of the cell cycle $(iii)$.
$C$. The action of the enzyme recombinase occurs during the Pachytene stage of Prophase $I$ of Meiosis $(v)$.
$D$. During Anaphase $I$,homologous chromosomes separate,but centromeres do not split,and chromatids move towards opposite poles $(iv)$.
Therefore,the correct sequence is $A-(ii), B-(iii), C-(v), D-(iv)$.

(B) $Root$ pressure is a positive pressure that develops in the $xylem$ sap of the roots of certain plants. It is a direct manifestation of active water absorption,where mineral ions are actively transported from the soil into the root hairs,creating a water potential gradient that draws water into the roots.

(A) : The seeds of monocotyledonous plants have only one cotyledon.
In the family $Poaceae$ (e.g.,wheat,maize,etc.),this cotyledon is called the scutellum,which is situated towards the lateral side of the embryonal axis.
It provides nourishment to the developing embryo.

(A) $(A) :$ $Sargassum$ is a brown alga. In brown algae,asexual reproduction occurs by means of spores,and sexual reproduction varies from isogamy,anisogamy to oogamy. Binary fission is a common method of asexual reproduction in unicellular organisms like bacteria and amoeba,not in $Sargassum$.

(B) The filiform apparatus is a specialized mass of finger-like projections of the cell wall extending into the cytoplasm.
It is a characteristic feature of the synergids (helper cells) located in the micropylar region of the embryo sac.
Its primary function is to guide the pollen tube into the synergids within the ovule towards the egg cell.

(A) : In angiosperms,microsporogenesis,i.e.,the formation of microspores (or pollen grains),occurs by the meiotic divisions of diploid microspore mother cells (or pollen mother cells). Microsporogenesis takes place in the anther.
Megasporogenesis,i.e.,the formation of megaspores,occurs by the meiotic divisions of diploid megaspore mother cells. Megasporogenesis takes place in the ovule.
Both processes involve meiosis to reduce the chromosome number from diploid $(2n)$ to haploid $(n)$.

(D) The correct answer is $D$.
Coconut water is a classic example of free-nuclear endosperm.
During the development of the coconut,the primary endosperm nucleus undergoes successive nuclear divisions without cytokinesis,resulting in the formation of a large number of free nuclei.
This liquid,known as coconut water,is the free-nuclear endosperm.
Later,cellularization occurs in the peripheral region,forming the solid,multicellular endosperm known as the coconut kernel or coconut meal.

(B) : Parthenocarpic fruits are fruits that develop without the process of fertilization. These fruits are naturally seedless. For example, $\text{Banana}$ is a well-known parthenocarpic fruit.

(C) The protoplast of the male gametophyte divides mitotically to produce two unequal cells: a small generative cell and a large vegetative cell.
The generative cell divides later into two non-motile male gametes (or sperms).
Thus,the mature male gametophyte in angiosperms produces two sperms and a vegetative cell.
The vegetative cell later grows to produce the pollen tube.

(D) To sustain animal visits,flowers provide rewards to the animals. The most common floral rewards are nectar and pollen grains. Nectar is a sugary fluid that provides energy,while pollen grains are rich in nutrients and serve as a protein source for many pollinators. Therefore,the correct option is $(d)$.

(C) $Geitonogamy$ involves the transfer of pollen grains from the anther of one flower to the stigma of another flower on the same plant.
Since the pollen must travel from one flower to another,it requires a pollinating agent (like insects,wind,or water).
However,it is genetically similar to $autogamy$ because both flowers belong to the same plant and therefore share the same genotype.

(B) : Honey is made from nectar through a process of regurgitation and evaporation. Honeybees transform saccharides (carbohydrates) into honey by regurgitating it a number of times,until it is partially digested. The bees perform the regurgitation and digestion as a group. After the last regurgitation,the aqueous solution still has high water content; the process continues by evaporation of much of the water and enzymatic transformation. Honey is produced by bees as a food source,not by digesting pollen.

(C) The correct answer is $C$.
An ovule is an integumented megasporangium found in spermatophytes,which develops into a seed after fertilization.
The stalk of the ovule is called the funiculus or funicle.
The point of attachment of the body of the ovule to the funiculus is known as the hilum.
In a mature seed,the hilum appears as a scar,representing the point where the seed was attached to the funicle.

(D) The correct answer is $D$.
In flowering plants,the style is the part of the carpel that the pollen tube must traverse to reach the ovule.
Styles are categorized into two types: hollow and solid.
In hollow styles,the stylar canal is lined by glandular cells.
In contrast,a solid style contains a central core of specialized cells known as transmitting tissue.
This transmitting tissue is composed of thin-walled cells that provide nutrition and a pathway for the pollen tube to grow through the style towards the ovary.

(D) The correct answer is $D$. In human females,the secondary oocyte is released from the ovary during ovulation. The process of meiosis-$II$ remains arrested at the metaphase-$II$ stage. The completion of meiosis-$II$ occurs only when a sperm enters the secondary oocyte during the process of fertilisation. The entry of the sperm triggers the degradation of $MPF$ ($M$-phase promoting factor) and activates the $APC$ ($Anaphase$ promoting complex),allowing the cell to complete the second meiotic division,resulting in the formation of a mature ovum and a second polar body.

(B) The $Zona \ pellucida$ is a thick,transparent,and acellular glycoprotein layer that surrounds the plasma membrane of an oocyte.
It is secreted by both the oocyte and the surrounding follicular cells.
Other layers like the $Granulosa$ cells and $Theca \ interna$ are composed of living cells,while the $Stroma$ refers to the connective tissue framework of the ovary.

(C) The correct answer is $C$.
Ovulation in human females involves the release of the secondary oocyte from the Graafian follicle,typically occurring around the $14^{th}$ day of the menstrual cycle.
This process is triggered by a rapid surge in Luteinizing Hormone $(LH)$ and Follicle Stimulating Hormone $(FSH)$,known as the $LH$ surge.
Before ovulation,the levels of estradiol (estrogen) reach their peak to stimulate the $LH$ surge; therefore,a decrease in estradiol is not associated with the event of ovulation itself,as the surge is dependent on high estrogen levels.

(D) : Ectopic pregnancy is a complication of pregnancy in which implantation of the embryo takes place at a site other than the uterus.
Signs and symptoms include abdominal pain and vaginal bleeding.
Most ectopic pregnancies $(90\%)$ occur in the Fallopian tube,which are known as tubal pregnancies.

(A) : Spermatogonia are diploid cells $(2n)$ which mature into primary spermatocytes $(2n)$ through growth. They then produce two haploid secondary spermatocytes $(n)$ by meiosis $I$. Each secondary spermatocyte $(n)$ completes meiosis $II$ and produces two spermatids $(n)$. Each spermatid $(n)$ develops into a spermatozoan or sperm $(n)$. Similarly,in females,oogonia are the diploid cells $(2n)$ from which,through meiosis,haploid polar bodies $(n)$ and a single ovum $(n)$ are produced.

(C) : The sperms in the female genital tract are made capable of fertilising the egg by secretions of the female genital tract.
These secretions remove coating substances deposited on the surface of the sperms,particularly those on the acrosome.
Thus,the receptor sites on the acrosome are exposed and the sperm becomes active to penetrate the egg.
This phenomenon of sperm activation in mammals is known as capacitation.

(C) : Hysterectomy is the surgical removal of the uterus. It may also involve the removal of the cervix,ovaries,Fallopian tubes,and other surrounding structures.

(D) $GIFT$ stands for Gamete Intra Fallopian Transfer.
This is an assisted reproductive technology used for females who cannot produce an ovum but can provide a suitable environment for fertilisation and further development.
In this technique,gametes (sperm and ovum) are collected from donors and transferred into the Fallopian tube of the female recipient.
Fertilisation occurs naturally inside the female body (in vivo fertilization).

(B) The correct answer is $(b)$ Encephalitis.
$1.$ Trichomoniasis is a common sexually transmitted infection caused by the parasite $Trichomonas$ $\text{vaginalis}$.
$2.$ Syphilis is a bacterial infection typically spread by sexual contact, caused by the bacterium $Treponema$ $\text{pallidum}$.
$3.$ Acquired Immuno Deficiency Syndrome $(AIDS)$ is a chronic, potentially life-threatening condition caused by the Human Immunodeficiency Virus $(HIV)$, which can be transmitted through sexual contact.
$4.$ Encephalitis is an inflammation of the brain, usually caused by a viral infection (such as herpes simplex virus, West Nile virus, or rabies) or an autoimmune reaction. It is not classified as a sexually transmitted disease.

(A) The correct answer is $A$.
Chikungunya virus is an arbovirus primarily transmitted to humans through the bite of infected $Aedes$ $aegypti$ or $Aedes$ $albopictus$ mosquitoes.
It is not sexually transmitted through semen.
In contrast,Ebola virus,Hepatitis $B$ virus,and Human immunodeficiency virus $(HIV)$ have been documented to be present in semen and can be transmitted through sexual contact.

(B) : The phenomenon of expression of both the alleles in a heterozygote is called codominance.
Both alleles do not show a dominance-recessive relationship and are able to express themselves independently when present together.
As a result,the heterozygous condition has a phenotype that is a combination of both homozygous genotypes.
For example,the alleles for blood group $A$ $(I^A)$ and blood group $B$ $(I^B)$ are codominant,so when they occur together in an individual,they produce blood group $AB$.

(D) : Mendel considered the following $7$ pairs of contrasting characters in his experiments on pea plants:

$1$. Seed shape	Round $(R)$ / Wrinkled $(r)$
$2$. Seed colour	Yellow $(Y)$ / Green $(y)$
$3$. Flower colour	Violet $(V)$ / White $(v)$
$4$. Pod shape	Inflated $(I)$ / Constricted $(i)$
$5$. Pod colour	Green $(G)$ / Yellow $(g)$
$6$. Flower position	Axial $(A)$ / Terminal $(a)$
$7$. Stem height	Tall $(T)$ / Dwarf $(t)$

Mendel did not use pod length as a character in his experiments. Therefore, the correct option is $D$.

(B) colour blind man has the genotype $X^cY$,and a normal woman has the genotype $XX$.
When they marry,the offspring are: $X^cX$ (carrier daughter) and $XY$ (normal son).
All daughters are carriers $(X^cX)$ and all sons are normal $(XY)$.
When a carrier daughter $(X^cX)$ marries a normal man $(XY)$,the possible genotypes for their children are: $XX$ (normal daughter),$X^cX$ (carrier daughter),$XY$ (normal son),and $X^cY$ (colour blind son).
The probability of having a colour blind son (the grandson of the original couple) is $1/4$ or $0.25$.

(B) The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as pleiotropy.
$A$ gene that exhibits multiple phenotypic effects due to its ability to control the expression of two or more characters is called a pleiotropic gene.
In human beings,pleiotropy is exhibited by conditions such as sickle cell anaemia and phenylketonuria.

(C) The correct answer is $(C)$ Autosomal dominant.
Analysis of the pedigree:
$1$. The trait appears in every generation (vertical transmission),which is a characteristic of dominant traits.
$2$. Affected individuals have at least one affected parent.
$3$. Both males and females are affected,and affected fathers can pass the trait to their sons,which rules out $X$-linked recessive inheritance.
$4$. Since the trait does not skip generations and appears in both sexes with equal frequency,it is identified as an autosomal dominant trait.

(C) The correct answer is $C$.
Linkage is the phenomenon where genes located on the same chromosome tend to be inherited together through generations without separation.
While Sutton and Boveri $(1902-1903)$ proposed the chromosomal theory of inheritance and Bateson and Punnett $(1906)$ observed the phenomenon in sweet peas, it was $T.H. Morgan$ who coined the term "linkage" and provided experimental proof using the fruit fly, $Drosophila$ $\text{melanogaster}$.

(A) : Translocation is a chromosomal abnormality caused by the rearrangement of parts between non-homologous chromosomes.
It results in the transfer of a segment of $DNA$ from one chromosome to another,which may cause a gene to move from one linkage group to another.

(A) : Genes are the units of inheritance and contain the information that is required to express a particular trait in an organism.
Alternating forms of a single gene which code for a pair of contrasting traits are known as alleles.
For example,two alleles determine the height of a pea plant (tall and dwarf).

(A) Multiple alleles refer to the existence of more than two alternative forms of a gene occupying the same locus on a homologous chromosome. Since an individual can only carry two alleles for a given trait (one on each homologous chromosome),multiple alleles can only be observed at the population level. Therefore,they are always present at the same locus of the chromosome.

(A) The man has blood group $A$,so his genotype can be $I^AI^A$ or $I^AI^O$.
The woman has blood group $B$,so her genotype can be $I^BI^B$ or $I^BI^O$.
If we consider the heterozygous condition for both parents $(I^AI^O \times I^BI^O)$,the possible combinations for the offspring are:
$1. I^AI^B$ (Blood group $AB$)
$2. I^AI^O$ (Blood group $A$)
$3. I^BI^O$ (Blood group $B$)
$4. I^OI^O$ (Blood group $O$)
Therefore,all four blood groups ($A, B, AB,$ and $O$) are possible in their offspring.

(B) Mendel selected $7$ pairs of true-breeding contrasting characters in pea plants $(Pisum \text{ } sativum)$ for his hybridization experiments. These characters are summarized in the table below:

$1$. Character: Seed shape (Dominant: Round,Recessive: Wrinkled)

$2$. Character: Seed colour (Dominant: Yellow,Recessive: Green)

$3$. Character: Flower colour (Dominant: Violet,Recessive: White)

$4$. Character: Pod shape (Dominant: Inflated,Recessive: Constricted)

$5$. Character: Pod colour (Dominant: Green,Recessive: Yellow)

$6$. Character: Flower position (Dominant: Axial,Recessive: Terminal)

$7$. Character: Stem height (Dominant: Tall,Recessive: Dwarf)

(D) : The abnormal baby has an extra $X$ chromosome,thus it must have been produced by the fusion of an abnormal $XX$ ovum with a normal $X$ sperm.
Abnormal $XX$ sperm is not possible because males have an $XY$ genotype; if they produce abnormal sperms,then $XY$ sperms and $O$ sperms would be produced.
If the fusion of multiple gametes had occurred (either two ova with one sperm or two sperms with one ovum),the human baby would have a triploid genotype,not the trisomy of sex chromosomes.

(B) : Chargaff's rules are applicable only for double-stranded $DNA$ molecules.
These are not applicable for single-stranded $DNA$ or $RNA$ molecules.
Chargaff's rules state that $DNA$ helices contain equal molar ratios of $A$ and $T$,and $G$ and $C$.
This is because in a double-stranded $DNA$ molecule,complementary base pairing occurs between $A$ and $T$,and $C$ and $G$ base pairs.
This specific complementary base pairing is not possible in the case of a single-stranded $RNA$ molecule.
Thus,Chargaff's rules are not applicable to $RNA$.

(A) The correct order from largest to smallest is: Genome > Chromosome > Gene > Nucleotide.
$1$. Genome: The entire set of genetic instructions found in an organism.
$2$. Chromosome: $A$ thread-like structure of nucleic acids and protein found in the nucleus,carrying genetic information in the form of $DNA$.
$3$. Gene: $A$ specific segment of $DNA$ located on a chromosome that codes for a functional product (protein or $RNA$).
$4$. Nucleotide: The basic structural unit of $DNA$,consisting of a pentose sugar,a phosphate group,and a nitrogenous base.
Therefore,option $A$ is correct.

(D) : Satellite $DNA$ is a part of repetitive $DNA$ which consists of long repetitive nucleotide sequences in tandem,forming a separate fraction during density ultracentrifugation.
$DNA$ fingerprinting involves identifying differences in specific regions of $DNA$ sequences known as repetitive $DNA$,where a small stretch of $DNA$ is repeated many times.
These repetitive $DNA$ sequences are separated from bulk genomic $DNA$ as distinct peaks during density gradient centrifugation.
The bulk $DNA$ forms a major peak,while the smaller peaks are referred to as satellite $DNA$.
Depending on base composition ($A:T$ rich or $G:C$ rich),segment length,and the number of repetitive units,satellite $DNA$ is classified into categories such as microsatellites and minisatellites.
These sequences typically do not code for any proteins,yet they constitute a large portion of the human genome.
They exhibit a high degree of polymorphism,which forms the basis of $DNA$ fingerprinting.
Since $DNA$ from every tissue (such as blood,hair follicles,skin,bone,saliva,sperm,etc.) of an individual shows the same degree of polymorphism,it serves as a highly useful identification tool in forensic applications.

(D) $lac$ operon regulation is described as negative and inducible.
$1$. Negative control: The $lac I$ gene produces a repressor protein that binds to the operator region,preventing $RNA$ polymerase from transcribing the structural genes. This keeps the operon in an 'off' state under normal conditions.
$2$. Inducible: The presence of an inducer (allolactose) binds to the repressor protein,causing a conformational change that prevents the repressor from binding to the operator. This allows transcription to proceed.
$3$. Therefore,because the default state is 'off' due to a repressor and it is turned 'on' by an inducer,it is classified as negative and inducible.

(A) According to Chargaff's rule,for double-stranded $DNA$,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of guanine $(G)$ is equal to cytosine $(C)$,i.e.,$A = T$ and $G = C$.
Given that cytosine $(C)$ is $17\%$,therefore,guanine $(G)$ must also be $17\%$.
The total percentage of all four bases is $100\%$.
So,$A + T + G + C = 100\%$.
Substituting the known values: $A + T + 17\% + 17\% = 100\%$.
$A + T + 34\% = 100\%$.
$A + T = 100\% - 34\% = 66\%$.
Since $A = T$,the percentage of adenine $(A)$ is $66\% / 2 = 33\%$ and the percentage of thymine $(T)$ is $66\% / 2 = 33\%$.
Thus,the percentages are $G = 17\%, A = 33\%, T = 33\%$.

(D) $(D) :$ Analogous organs are those that perform similar functions but differ in their structural details and evolutionary origin.
Analogous structures are the result of convergent evolution.
The wings of an insect and the wings of a bird are considered analogous because,while they both serve the function of flight,their basic anatomical structures are entirely different (insect wings are outgrowths of the exoskeleton,while bird wings are modified forelimbs).
Therefore,they represent convergent evolution.

(D) : Natural selection is the most widely accepted theory concerning the principal causal mechanism of evolutionary change,proposed by $Charles \ Darwin$ and $Alfred \ Russel \ Wallace$.
It results from differential reproduction,where some members of a population produce abundant offspring,some only a few,and others none,compared to other phenotypes in the same population.
This determines the relative share of different genotypes that individuals possess and propagate in a population.
Industrial melanism is a classic example that supports evolution by natural selection.
It is an adaptation where moths living in industrial areas developed melanin pigments to match their bodies to the soot-covered tree trunks,thereby increasing their survival rate against predators.

(C) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation under specific conditions,which include random mating,a large population size,no migration (gene flow),no mutations,and no natural selection.
If individuals mate selectively (non-random mating),the allele frequencies will change over generations,and the population will not be in Hardy-Weinberg equilibrium.
Therefore,the correct option is $C$.

(B) The correct answer is $B$.
Genetic variation in sexually reproducing organisms primarily arises from the reshuffling of genes.
This occurs through three main processes:
$1$. Independent assortment of chromosomes during meiosis.
$2$. Crossing over,which leads to the exchange of genetic material between homologous chromosomes.
$3$. Random fertilization of gametes.
Collectively,these processes are referred to as recombination. While mutations and genetic drift also contribute to variation,recombination is the most frequent and consistent mechanism driving variation in a sexually reproducing population.

(B) : The brain capacity gradually increased from early human ancestors to modern man.
$Homo \text{ } habilis$ had a brain capacity of $650-800 \text{ } cc$, which increased to around $900 \text{ } cc$ in $Homo \text{ } erectus$.
The Neanderthal man $(Homo \text{ } neanderthalensis)$ had a brain capacity of $1400 \text{ } cc$, which evolved to around $1450 \text{ } cc$ in $Homo \text{ } sapiens$.

(B) : Serum globulins are proteins that include gamma globulins (antibodies) and a variety of enzymes and carrier/transport proteins.
Serum protein electrophoresis $(SPEP)$ is used to analyze the profile of these globulins by separating them based on size and charge.
There are four major groups identified: alpha-$1$ globulins,alpha-$2$ globulins,beta globulins,and gamma globulins.
Since the gamma fraction primarily consists of immunoglobulins (antibodies),a deficiency in antibodies is directly reflected by a low level of serum globulins.

(A) $IgA$ is the primary immunoglobulin found in human milk,particularly in colostrum. It provides passive immunity to the infant by protecting the mucosal surfaces of the gastrointestinal tract from pathogens. While $IgG$ is the most abundant immunoglobulin in human serum,$IgA$ is the predominant antibody in secretions,including breast milk.

(A) : Babesiosis is a malaria-like parasitic disease caused by infection with $Babesia$,a parasitic protozoan.
Babesiosis has long been recognized as a disease of cattle and other domestic animals,until human forms of babesiosis were discovered.
$Babesia$ parasites reproduce in the red blood cells of mammals and cause haemolytic anaemia,which is quite similar to malaria.
The parasite is transmitted by ticks.

(D) The correct answer is $(d)$.
Cell-mediated immune response $(CMIS)$ is primarily mediated by $T$-lymphocytes.
These cells are responsible for graft rejection because the immune system recognizes the proteins present on the surface of the transplanted tissue or organ as foreign (non-self).
Upon recognition,the $T$-cells initiate a specific immune attack against the transplanted organ,leading to its rejection.

(A) The correct matches are as follows:
$(A)$ Tuberculosis: The $BCG$ vaccine uses a live attenuated (harmless) strain of Mycobacterium bovis,which acts as a harmless bacterium $(iv)$.
$(B)$ Whooping cough (Pertussis): The vaccine typically uses killed Bordetella pertussis bacteria $(iii)$.
$(C)$ Diphtheria: The vaccine is a toxoid,which is an inactivated toxin $(ii)$ produced by the bacteria.
$(D)$ Polio: The Salk vaccine uses inactivated (killed) poliovirus,while the Sabin vaccine uses live attenuated (harmless) virus $(i)$.
Therefore,the correct sequence is $A-iv, B-iii, C-ii, D-i$.

(C) $Entamoeba \ histolytica$ (Gr.,$entos$: within + $amoeba$: change + $histos$: tissue + $lysis$: dissolve) is the causative organism of amoebic dysentery or amoebiasis in humans.
It is a microscopic endoparasite of humans.
It is commonly found in the upper part of the large intestine (colon) and can often lodge in the liver,lungs,brain,and testes.
In its life cycle,it occurs in three distinct forms: $(i)$ trophozoite or magna form,$(ii)$ precystic or minuta form,and $(iii)$ cystic form.
The trophozoite is the most active,motile,and feeding form,which is pathogenic to humans.
It lives in the mucous and submucous layers of the colon and feeds on these layers and erythrocytes.