AIPMT 2010 Biology Question Paper with Answer and Solution

(A) The correct answer is $A$. The nucleic acid of a virus is surrounded by a protective protein coat called the capsid. The capsid is composed of small protein subunits known as capsomeres. In some viruses,this capsid is further covered by an outer layer called an envelope,which typically consists of a combination of lipids,proteins,and carbohydrates.

(A) : Protista includes all unicellular and colonial eukaryotes. The protistan cells are typically eukaryotic,possessing membrane-bound organelles like mitochondria,chloroplasts,Golgi bodies,endoplasmic reticulum,and a nucleus.
Protista is commonly known as the kingdom of unicellular eukaryotes.
Kingdom Fungi contains achlorophyllous,spore-producing,heterotrophic,multicellular,or multinucleate eukaryotic organisms (unicellular yeasts are also included in Fungi because their sexual reproduction is similar to that of some fungi).
Monerans are basically unicellular prokaryotes.
Archaea (ancient bacteria) are also a type of monerans which live in extreme environments like high temperature,high salt content,and acidic $pH$.

(A) The correct answer is $A$.
$1$. Hyperthermophilic organisms that thrive in extremely acidic environments (such as $pH \ 2$) and high temperatures are primarily classified under the domain $Archaea$,specifically the group known as thermoacidophiles.
$2$. $Archaebacteria$ are known for their unique cell wall structure that allows them to survive in extreme conditions like hot sulphur springs.
$3$. Certain $Eubacteria$ also exhibit remarkable adaptability and can survive in diverse,hostile environments including hot springs and acidic conditions,although they are distinct from $Archaea$ in their biochemical composition.

(B) The correct answer is $B$.
Many free-living bacteria and blue-green algae are capable of fixing atmospheric nitrogen.
$Rhodospirillum$ is a free-living,photosynthetic,anaerobic,nitrogen-fixing,non-sulphur bacterium.
It is capable of synthesizing its organic food in the presence of light and in the absence of $O_2$ through a process known as bacterial photosynthesis.
$Beijernickia$ and $Azotobacter$ are free-living but aerobic nitrogen-fixing bacteria.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.

(B) $Streptococcus$ is a bacterium included under Kingdom $Monera$.
$Monerans$ possess a prokaryotic cell organization in which membrane-bound organelles like mitochondria,endoplasmic reticulum,Golgi bodies,etc.,are absent.
All the other three,$i.e.$,$Saccharomyces$ (a fungus),$Chlamydomonas$ (an alga),and $Plasmodium$ (a protozoan protist),are eukaryotes containing true membrane-bound organelles.

(C) Based on the structure of a bacteriophage:
$A$ represents the Head,which contains the viral genetic material.
$B$ represents the Sheath (or tail tube),which is the contractile part of the tail.
$C$ represents the Collar,which connects the head to the tail.
$D$ represents the Tail fibres,which are used for attachment to the host cell surface.
Therefore,the correct identification is $A$-Head,$B$-Sheath,$C$-Collar,$D$-Tail fibres,which corresponds to option $C$.

(D) Statement $(i)$ is correct: Methanogens are archaebacteria found in marshy areas that produce methane $(CH_4)$.
Statement $(ii)$ is correct: Nostoc is a filamentous cyanobacterium (blue-green alga) that possesses heterocysts for atmospheric nitrogen fixation.
Statement $(iii)$ is incorrect: Chemosynthetic autotrophic bacteria oxidize inorganic substances (like nitrates,nitrites,or ammonia) to produce $ATP$ for energy,not to synthesize cellulose from glucose.
Statement $(iv)$ is correct: Mycoplasma are the smallest living cells,lack a cell wall,and are facultative anaerobes,meaning they can survive without oxygen.
Therefore,statements $(i), (ii),$ and $(iv)$ are correct.

(C) : Black stem rust is caused by $Puccinia graminis tritici$. The genus $Puccinia$ includes $700$ species, which cause rust diseases of many economic plants such as wheat, barley, oats, etc.
It is called a rust because of the reddish-brown color of the spores that are found chiefly upon the surface of the host leaves and stems.
$P. graminis$ is heteroecious, $i.e.$, requiring two hosts, wheat and barberry, for the completion of its normal life cycle.
According to the nature of the spores, the life cycle of $P. graminis$ is divided into five stages.
It is during the teleuto stage that the teliospores produce dark brown to black pustules on the surface of stems and leaves of the wheat, which results in 'black stem rust of wheat'.

(A) The correct answer is $A$.
Most eukaryotic algae possess a well-defined cell wall primarily composed of cellulose and other carbohydrates such as galactans and mannans.
These chemical components vary significantly among different algal groups,often including substances like xylan,alginic acid,silica,agar,pectin,and carrageenin.
In contrast,the cell wall of blue-green algae (cyanobacteria,which are prokaryotes) is composed of peptidoglycans (muropeptides),which are distinct from the cellulose-based walls of eukaryotic algae.

(D) $Sphagnum$ is a bryophyte in which the dominant phase or plant body is an independent and free-living gametophyte.
The sporophyte is parasitic over the gametophyte.
In $Pinus$ (a gymnosperm),mustard,and castor (angiosperms),the main plant body is sporophytic.
The gametophyte is highly reduced and is completely dependent on the sporophyte.

(C) : Monoecious plants are those that bear both male and female reproductive structures on the same individual plant.
In the case of $Pinus$, both male and female cones (strobili) are produced on the same tree, making it a monoecious plant.
In contrast, $Marchantia$ is dioecious (male and female thalli are separate), and both $Cycas$ and $Date \text{ } palm$ are dioecious, meaning male and female reproductive structures are found on separate plants.

(C) Based on the provided figures:
$A$ represents $Selaginella$,which is a pteridophyte.
$B$ represents $Equisetum$,which is a pteridophyte.
$C$ represents $Salvinia$,which is a heterosporous pteridophyte.
$D$ represents $Ginkgo$,which is a gymnosperm.
Therefore,the correct identification is given in option $C$.

(D) The correct answer is $D$. In $Fasciola$ (a member of Platyhelminthes),the body plan is known as the 'blind sac' body plan.
This means the digestive system has only a single opening that serves both as the mouth for ingestion (intake of food) and as the anus for egestion (removal of undigested waste).
In contrast,organisms like $Octopus$,$Asterias$,and $Ascidia$ possess a complete digestive tract with two separate openings (mouth and anus).

(C) $Spongilla$ is a common,widely distributed fresh water sponge belonging to the phylum $Porifera$.
$Spongilla$ possesses special flagellated cells known as choanocytes (collar cells) which line the spongocoel and the canals.
These cells are characteristic of sponges and are not found in the leech (annelid),dolphin (mammal),or penguin (bird).

(A) $Triploblastic$ is a condition describing an animal having a body composed of three embryonic germ layers: the ectoderm,mesoderm,and endoderm.
Most multicellular animals belonging to phylum $Platyhelminthes$ to phylum $Chordata$ are triploblastic.
Flatworms (phylum $Platyhelminthes$) are the first group of animals to exhibit triploblastic organization.
$Ctenophores$,sponges (phylum $Porifera$),and corals (phylum $Cnidaria$) are diploblastic,meaning they possess only two germ layers: the ectoderm and endoderm.

(A) $Acoelomates$ are animals that lack a body cavity or coelom. Examples include $Porifera$,$Cnidaria$,$Ctenophora$,and $Platyhelminthes$.
In $pseudocoelomates$,the body cavity is a $pseudocoelom$ or false coelom,which is not lined by mesoderm. Examples include $Aschelminthes$.
In $coelomates$,the body cavity is a true coelom,which is lined by mesoderm on both sides. Most phyla from $Annelida$ to $Chordata$ are $coelomates$.
$Molluscs$ and $Insects$ (Arthropoda) are $coelomates$,while $Flatworms$ are $acoelomates$.
Therefore,the statement that $Roundworms$ $(Aschelminthes)$ are $pseudocoelomates$ is correct.

(B) The correct option is $B$.
Nephridia are the primary excretory organs of the earthworm $(Pheretima)$.
Earthworms possess three types of nephridial structures based on their location: septal,integumentary,and pharyngeal nephridia.
These structures vary in their position and physiological function.
Septal and pharyngeal nephridia are enteronephric,meaning they discharge nitrogenous wastes into the alimentary canal.
Integumentary nephridia are exonephric,as they discharge nitrogenous wastes directly to the exterior of the body.

(D) : Animals belonging to Phylum $Chordata$ are fundamentally characterized by the presence of a notochord,a dorsal hollow nerve cord,and paired pharyngeal gill slits. Crocodile,penguin,whale,and dogfish are all chordates. All of them possess gill slits or have had them during embryonic development. Thus,paired gill slits are present in these animals at some stage of their life cycle.

(B) The correct answer is $B$. In basal placentation,the placenta develops at the base of the ovary,and a single ovule is attached to it. This type of placentation is commonly observed in families like $Asteraceae$ (formerly $Compositae$) and $Poaceae$.

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

(B) : If the gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level,it is called perigynous. The ovary here is said to be half-inferior,e.g.,plum,rose,and peach.

(A) : China rose of Family $Malvaceae$ possesses numerous stamens.
The filaments of the stamens are united into a single bundle or group, forming a staminal tube around the style.
Such a condition of stamens is technically termed as monadelphous.

(A) : $Opuntia$ is a xerophytic plant that lives in dry habitats.
It possesses fleshy organs where water and mucilage are stored.
The stem is modified into a flat,green,photosynthetic structure known as a phylloclade,which helps in reducing water loss and performing photosynthesis.

(D) The correct answer is $D$.
In plants like cotton,China rose,and lady's finger,the margins of sepals or petals overlap with the margin of the next one in a regular pattern.
This specific mode of arrangement of sepals or petals in a floral bud with respect to the other members of the same whorl is known as aestivation.
In the twisted type of aestivation,one margin of each petal overlaps the next one,creating a spiral or twisted appearance.
Image $D$ correctly represents this twisted aestivation.

(C) The correct floral formula for soybean (a member of the family Fabaceae) is: % ♀ K_{$(5)$} C_{$1$+$2$+$(2)$} A_{$(9)$+$1$} G_{$1$}.
Explanation:
$1$. %: Zygomorphic flower.
$2$. ♀: Bisexual flower.
$3$. K_{$(5)$}: Calyx with $5$ sepals,gamosepalous (fused).
$4$. C_{$1$+$2$+$(2)$}: Corolla with $5$ petals,polypetalous,showing vexillary aestivation ($1$ standard,$2$ wings,$2$ fused keels).
$5$. A_{$(9)$+$1$}: Androecium with $10$ stamens,diadelphous ($9$ fused,$1$ free).
$6$. G_{$1$}: Gynoecium with a single monocarpellary ovary,superior (indicated by the underline).

(B) Statement $(i)$ is correct: In vexillary aestivation (found in Fabaceae),the largest posterior petal is the standard (vexillum),two lateral petals are wings (alae),and two anterior petals form the keel (carina).
Statement $(ii)$ is correct: The floral formula for Liliaceae is $\oplus \text{ } \text{O}\mkern-11mu{^\nearrow} \text{ } P_{(3+3)} A_{3+3} \underline{G}_{(3)}$.
Statement $(iii)$ is incorrect: In pea flowers (Fabaceae),the stamens are diadelphous ($9+1$ arrangement),not monadelphous.
Statement $(iv)$ is incorrect: The floral formula for Solanaceae is $\oplus \text{ } \text{O}\mkern-11mu{^\nearrow} \text{ } K_{(5)} C_{(5)} A_{5} \underline{G}_{(2)}$. The formula provided in the image is incorrect.
Therefore,statements $(i)$ and $(ii)$ are correct. The correct option is $(b)$.

(C) : In old trees,the greater part of secondary xylem is dark brown due to the deposition of organic compounds like tannins,resins,oils,gums,aromatic substances,and essential oils in the central or innermost layers of the stem.
These substances make it hard,durable,and resistant to the attacks of microorganisms and insects.
This region comprises dead elements with highly lignified walls and is called heartwood.
The heartwood does not conduct water,but it gives mechanical support to the stem.
The peripheral region of the secondary xylem is lighter in color and is known as the sapwood.
It is involved in the conduction of water and minerals from root to leaf.

(D) The correct answer is $(d)$.
Meristems are classified based on their position in the plant body into apical,intercalary,and lateral meristems.
Lateral meristems are responsible for secondary growth and are located on the lateral sides of the stem and root. Examples include vascular cambium (which consists of intrafascicular and interfascicular cambium) and cork cambium (phellogen).
Intercalary meristems are located between permanent tissues and are responsible for the elongation of internodes in grasses and other monocots. Therefore,intercalary meristem is not a lateral meristem.

(D) : The chief water conducting elements of xylem in gymnosperms are tracheids. These are elongated cells with tapering ends and are dead at maturity due to the deposition of lignin. These cells exhibit scalariform,annular,reticulate,or bordered pitted thickenings. Tracheids are the primary water-conducting xylem elements in both gymnosperms and pteridophytes. Generally,vessels are absent in gymnosperms,with rare exceptions such as $Gnetum$,$Welwitschia$,and $Ephedra$.

(D) In higher plants,the transport of food material (photosynthates) occurs through the phloem tissue. The primary conducting cells of the phloem are the sieve elements (sieve tube elements in angiosperms and sieve cells in gymnosperms). These cells are specialized for the translocation of organic nutrients from source to sink. Therefore,the correct option is $D$.

(B) : Simple squamous epithelium is composed of large flat cells whose edges fit closely together like the tiles in a floor,hence it is also called pavement epithelium.
The nuclei of the cells are flattened and often lie at the centre of the cells and cause bulgings of the cell surface.
The epithelium lines the blood vessels,lymph vessels,heart,terminal bronchioles,alveoli of the lungs,walls of the Bowman's capsules,and descending limbs of the loop of Henle.
In the blood vessels and heart,it is specifically called endothelium.

(B) The correct answer is $(B)$.
According to the fluid mosaic model,the plasma membrane is composed of a phospholipid bilayer in which protein molecules are embedded.
Lipids $(20-79\%)$,proteins $(20-70\%)$,carbohydrates $(1-5\%)$,and water $(20\%)$ are the main components of the plasma membrane.
Lipid molecules,primarily phospholipids,are amphipathic (having both hydrophobic and hydrophilic ends) and form a bilayer structure.
Protein molecules are distributed throughout this bilayer,either partially or totally embedded (intrinsic/integral proteins) or attached to the surface (extrinsic/peripheral proteins).

(C) : Cytoplasm is a granular,crystallo-colloidal complex that forms the living protoplasm of a cell,excluding its nucleus. It consists of proteins,nucleic acids,fats,carbohydrates,vitamins,minerals,waste metabolites,and all the cell organelles. It serves as the main site for various cellular activities such as respiration,nutrition,and storage.

(A) The correct answer is $A$.
Both mitochondria and chloroplasts are semi-autonomous organelles.
They contain their own circular $DNA$,which is responsible for producing their own $mRNA$,$tRNA$,and $rRNA$.
These organelles also possess their own $70S$ ribosomes,which allow them to synthesize some of their own proteins independently of the nucleus.

(A) : Plasmodesmata are fine cytoplasmic strands that connect the protoplasts of adjacent plant cells by passing through their cell walls.
Plasmodesmata are cylindrical in shape (about $20-40 \ nm$ in diameter) and are lined by the plasma membrane of the two adjacent cells.
They permit the passage between cells of substances including ions,sugars,amino acids,and macromolecules.

(D) The cytoskeleton is an elaborate network of filamentous proteinaceous structures consisting of microtubules,microfilaments,and intermediate filaments present in the cytoplasm. It plays a crucial role in various functions such as mechanical support,motility,and the maintenance of the shape of the cell. Therefore,the correct option is $(d)$.

(A) Based on the provided diagram:
$A$ points to the Rough Endoplasmic Reticulum $(RER)$,which is studded with ribosomes.
$B$ points to the Smooth Endoplasmic Reticulum $(SER)$,which lacks ribosomes.
$C$ points to the Nucleus,which is the large central organelle.
$D$ points to the Cytoplasm,the fluid surrounding the organelles.
Therefore,the correct matching is:
$A - (v)$ (Rough endoplasmic reticulum)
$B - (iv)$ (Smooth endoplasmic reticulum)
$C - (viii)$ (Nucleus)
$D - (iii)$ (Cytoplasm)
Thus,the correct option is $(A)$.

(D) During the early prophase of mitosis,the nucleus and the cell become spherical. The viscosity and refractivity of the cytoplasm increase. $DNA$ molecules condense to form shortened chromosomes. The endoplasmic reticulum $(ER)$ and the nucleolus begin to disappear.

(C) In figure $A$,the chromosomes have separated and are moving towards opposite poles,which is characteristic of the late anaphase stage of mitosis.
In figure $B$,the chromosomes are condensing and the nuclear envelope is beginning to disappear,which is characteristic of the prophase stage of mitosis.
Therefore,figure $A$ represents late anaphase and figure $B$ represents prophase. The correct option is $(c)$.

(D) The correct identification of the parts in the stomatal apparatus diagram is as follows:
$A$ - Epidermal cell: The outer cells surrounding the stomatal complex.
$B$ - Subsidiary cell: Specialized cells that surround the guard cells.
$C$ - Stomatal aperture: The central pore through which gas exchange occurs.
$D$ - Guard cell: The specialized kidney-shaped cells that regulate the opening and closing of the stomatal aperture.
Therefore,option $D$ is the correct answer.

(A) $Molybdenum$ is a micronutrient required in very small amounts by plants. It is essential for nodulation in legumes. It is a key component of the $Nitrate$ $reductase$ enzyme, which plays a crucial role in nitrogen fixation.

(B) The correct answer is $B$.
Macronutrients are essential elements present in plant tissues in large amounts,typically $1-10 \ mg$ per gram of dry weight.
These include carbon,hydrogen,oxygen,nitrogen,phosphorus,sulfur,potassium,calcium,and magnesium.
Micronutrients,or trace elements,are required in very small amounts,typically equal to or less than $0.1 \ mg$ per gram of dry matter.
These include iron,manganese,copper,molybdenum,zinc,boron,chlorine,and nickel.
Since magnesium is a macronutrient,it is not a micronutrient.

(D) The enzyme nitrogenase is highly sensitive to molecular oxygen and requires anaerobic conditions to function effectively. Leghaemoglobin acts as an oxygen scavenger to protect the nitrogenase enzyme from oxygen damage,creating an anaerobic environment within the root nodules. Therefore,the statement that nitrogenase is insensitive to oxygen is incorrect.

(B) The nitrogen cycle involves several key processes:
$1$. $A$ represents Denitrification,where nitrates in the soil are converted back into atmospheric nitrogen $(N_2)$ by bacteria like $Pseudomonas$.
$2$. $B$ represents Ammonification,where organic nitrogen from dead plants and animals is converted into ammonia $(NH_3)$ by decomposers.
$3$. $C$ represents Animals,which consume organic nitrogen from plants.
$4$. $D$ represents Plants,which absorb nitrogenous compounds from the soil.
Therefore,option $B$ is the correct sequence.

(D) The correct answer is $D$. Calvin,Benson,and their colleagues in California,$U.S.A.$,conducted experiments using radioactive $^{14}C$ in carbon dioxide to feed $Chlorella$ and $Scenedesmus$ (which are algae).
Radioactive carbon,$^{14}C$,has a half-life of $5568$ years,allowing the path of $CO_2$ fixation to be traced.
Algal suspension,while performing photosynthesis,was supplied with $^{14}CO_2$.
The algae were killed at specific intervals in near-boiling methanol,which immediately stopped photosynthetic activity due to the denaturation of enzymes.
After evaporating the alcohol and crushing the algae,the resulting paste was analyzed using two-dimensional paper chromatography.
Calvin and his co-workers discovered that after three seconds,radioactivity appeared in phosphoglyceric acid $(PGA)$.
Thus,$PGA$ was identified as the first stable product of photosynthesis in algae.

(D) $C_4$ plants are more efficient in photosynthesis because they have a mechanism to minimize photorespiration.
In $C_3$ plants,the enzyme $RuBisCO$ acts as an oxygenase in the presence of high $O_2$ concentrations,leading to photorespiration,which is a wasteful process that consumes energy and releases $CO_2$.
$C_4$ plants possess a specialized anatomy called $Kranz$ anatomy,which separates the initial $CO_2$ fixation from the Calvin cycle.
This spatial separation ensures that $RuBisCO$ always operates in a high $CO_2$ environment,effectively suppressing photorespiration.
Therefore,the lower rate of photorespiration in $C_4$ plants makes them more photosynthetically efficient compared to $C_3$ plants.

(A) Statement $(i)$ is incorrect because the $Z$ scheme involves both $PSII$ and $PSI$ working in series.
Statement $(ii)$ is correct because cyclic photophosphorylation involves only $PSI$ where electrons circulate within the photosystem.
Statement $(iii)$ is incorrect because cyclic photophosphorylation produces only $ATP$,not $NADPH_2$.
Statement $(iv)$ is correct because the stroma lamellae lack both $PSII$ and the enzyme $NADP$ reductase,which is why they only perform cyclic photophosphorylation.
Therefore,statements $(ii)$ and $(iv)$ are correct.

(C) The given pathway represents the $C_4$ cycle (Hatch-Slack pathway).
$A$ represents the primary fixation of $CO_2$ in the mesophyll cells,where $PEP$ (Phosphoenolpyruvate) combines with $CO_2$ to form a $C_4$ acid (Oxaloacetate).
$B$ represents the decarboxylation step in the bundle sheath cells,where the $C_4$ acid releases $CO_2$ to enter the Calvin cycle and converts into a $C_3$ acid.
$C$ represents the regeneration of $PEP$ from the $C_3$ acid in the mesophyll cells to continue the cycle.
Therefore,the correct sequence is Fixation $\rightarrow$ Decarboxylation $\rightarrow$ Regeneration.

(C) The correct answer is $C$.
Kranz anatomy is a specialized leaf structure found in $C_4$ plants.
In this anatomy,the mesophyll cells are undifferentiated and arranged in concentric layers around the vascular bundles.
The vascular bundles are surrounded by large bundle sheath cells,which are arranged in a wreath-like manner (the word 'Kranz' means wreath in German).
Examples of $C_4$ plants that exhibit Kranz anatomy include sugarcane,maize,and sorghum.

(B) : Fermentation is an energy-releasing metabolic process in which organic substrates,such as carbohydrates,are oxidized without the involvement of an external electron acceptor.
In this process,an endogenous electron acceptor (usually an organic compound) is used.
This is in contrast to aerobic respiration,where electrons are transferred to an exogenous electron acceptor,such as oxygen,via an electron transport chain.

(A) The scutellum is the tissue in a grass,wheat,or maize seed that lies between the embryo and the endosperm. It is a modified cotyledon,specialized for the digestion and absorption of the endosperm nutrients for the developing embryo.

(B) : Apomixis is a form of asexual reproduction that mimics sexual reproduction. In $Citrus$,apomictic embryos are formed through adventive embryony.
Adventive embryony is a type of apomixis where embryos develop directly from the diploid maternal sporophytic tissues of the ovule,such as the nucellus or integuments,without fertilization.
Since these cells are diploid and part of the parent plant,the resulting embryos are genetically identical to the parent plant.

(B) The correct answer is $B$.
Geitonogamy is the transfer of pollen grains from the anther to the stigma of another flower on the same plant.
Although geitonogamy is functionally cross-pollination involving a pollinating agent,genetically it is similar to autogamy since the pollen grains come from the same plant.
$A$: Xenogamy is the transfer of pollen grains from the anther to the stigma of a different plant.
$C$: Karyogamy is the fusion of two nuclei.
$D$: Autogamy is the transfer of pollen grains from the anther to the stigma of the same flower.

(B) : Pollination by wind is called anemophily,and plants that utilize this method are called anemophilous plants.
Anemophilous flowers are typically small and inconspicuous.
They produce a large number of pollen grains that are small,dry,and lightweight,allowing them to be carried over long distances (up to $1300 \ km$) by the wind.
The stigma is often feathery or brush-like to effectively trap airborne pollen.
These flowers do not produce nectar or scent,as they do not need to attract biotic pollinators.
Grasses are common examples of anemophilous plants.

(C) The correct identification of the structures is as follows:
$A$ represents the offset of water hyacinth $(Eichhornia)$,which is a vegetative propagule.
$B$ represents the antheridiophore of $Marchantia$,a male reproductive structure in liverworts.
$C$ represents the antipodal cells found at the chalazal end of the mature embryo sac in angiosperms.
$D$ represents the oogonium,which is the female sex organ in the green alga $Chara$.
Therefore,option $C$ is the correct choice.

(B) $Pistia$ (water lettuce) is an aquatic plant that reproduces vegetatively through structures known as offsets.
An offset is a short,thick runner that grows horizontally at the water surface.
It produces a rosette of leaves at the apex and adventitious roots at the base,eventually developing into a new independent plant.

(C) : Sertoli cells (named after the Italian histologist $Enrico$ $Sertoli$) are located within the walls of the seminiferous tubules of the testes.
They are large,pale cells that provide structural support and nutrition to the developing germ cells.
Specifically,they anchor and nourish the spermatids,which become partially embedded within them during the process of spermiogenesis.

(B) The correct answer is $B$.
$Vasa$ $efferentia$ are a series of fine ciliated ductules that arise from the $rete$ $testis$.
These ductules transport spermatozoa from the $rete$ $testis$ to the $epididymis$ (which then leads to the $vas$ $deferens$).
Therefore,they serve as the connecting pathway between the $rete$ $testis$ and the $epididymis$.
Among the given options,$B$ is the most appropriate description of the pathway segment involving the $rete$ $testis$ and the subsequent duct system.

(A) The correct answer is $A$.
Seminal plasma is the fluid component of semen,which is ejaculated from the penis during sexual climax.
It is primarily secreted by the seminal vesicles,prostate gland,and bulbourethral (Cowper's) glands.
Seminal plasma is rich in fructose,calcium,and various enzymes.
Fructose serves as the primary energy source for sperm,while calcium and enzymes help in sperm motility and activation.
Additionally,it provides a medium for sperm transport,lubricates the female reproductive tract,and neutralizes the acidic environment of the vagina to protect the sperm.

(B) : In human beings,after one month of pregnancy,the embryo's heart is formed.
By the end of the second month of pregnancy,the foetus develops limbs and digits.
By the end of $12$ weeks (first trimester),most of the major organ systems are formed.
The first movements of the foetus and appearance of hair on the head are usually observed during the fifth month.
By the end of $24$ weeks (second trimester),the body is covered with fine hair,eyelids separate,and eyelashes are formed.
By the end of nine months of pregnancy,the foetus is fully developed and is ready for delivery.

(B) The correct answer is $B$. Oogenesis begins with the division of oogonia (gamete mother cells),which give rise to primary oocytes. These primary oocytes enter prophase $I$ of meiosis and are temporarily arrested at this stage. As the follicle matures,the primary oocyte completes its first meiotic division to form a haploid secondary oocyte and a first polar body. The secondary oocyte then begins the second meiotic division but arrests at metaphase $II$. This second meiotic division is only completed after the sperm penetrates the secondary oocyte during fertilization. This process results in the formation of a second polar body and a mature haploid ovum (ootid).

(B) The correct answer is $B$.
$Acrosome$ is a cap-like structure located at the anterior end of the sperm head.
It contains various hydrolytic enzymes,collectively known as sperm lysins.
During the process of fertilisation,these enzymes are released to dissolve the protective layers of the ovum (such as the $zona$ $pellucida$),which facilitates the entry of the sperm into the egg.
Failure of the acrosome reaction prevents the sperm from penetrating the egg,which is a common cause of male infertility.

(A) : $A$ morula is an embryo at an early stage of embryonic development,consisting of cells (called blastomeres) in a solid ball contained within the zona pellucida.
During cleavage,the zygote undergoes rapid mitotic divisions without any significant growth in size.
Because there is no net growth,the total volume of cytoplasm remains approximately the same as that of the original zygote,while the number of cells increases.
However,since each cell division involves $DNA$ replication,the total amount of $DNA$ in the morula is significantly higher than in the uncleaved zygote.
Therefore,the morula has almost equal quantity of cytoplasm as an uncleaved zygote but much more $DNA$.

(B) The correct answer is $(b)$.
Each Fallopian tube is about $10-12 \ cm$ long and extends from the periphery of each ovary to the uterus.
The part closer to the ovary is the funnel-shaped infundibulum.
The edges of the infundibulum possess finger-like projections called fimbriae,which help in the collection of the ovum after ovulation.
The infundibulum leads to a wider part of the oviduct called the ampulla.
The last part of the oviduct,the isthmus,has a narrow lumen and it joins the uterus.

(B) The correct answer is $B$.
Parturition is induced by a complex neuroendocrine mechanism.
The signals for parturition originate from the fully developed foetus and the placenta,which induce mild uterine contractions known as the foetal ejection reflex.
This reflex triggers the release of oxytocin from the maternal pituitary gland.
Oxytocin acts on the uterine muscles and causes stronger uterine contractions,which in turn stimulates further secretion of oxytocin.
The stimulatory reflex between the uterine contraction and oxytocin secretion continues,resulting in progressively stronger contractions.
This ultimately leads to the expulsion of the baby out of the uterus through the birth canal.

(D) The correct answer is $(d)$.
Implantation in the endometrial uterine wall takes place at the blastocyst stage of embryonic development.
Before implantation,the blastomeres of the early blastocyst arrange themselves into an outer layer called the trophoblast and an inner group of cells attached to the trophoblast called the inner cell mass.
It is the trophoblast layer through which the blastocyst attaches to the endometrium,and the inner cell mass differentiates into the embryo.

(A) The male accessory glands include paired seminal vesicles,a prostate gland,and paired bulbourethral glands.
These glands produce secretions that constitute the seminal plasma.
Seminal plasma is rich in fructose,calcium,and certain enzymes,which provide nutrition and protection to the sperm.
The secretions of the bulbourethral glands also assist in the lubrication of the penis.

(B) The correct answer is $(b)$. Assisted reproductive technologies $(ART)$ include a number of special techniques that assist infertile couples to have children.
An important technique of $ART$ is the test-tube baby programme.
The baby produced by conceiving in a culture dish and nursing in the uterus is called a test-tube baby.
This method involves in vitro fertilization $(IVF)$,i.e.,fertilization of male and female gametes outside the body in conditions similar to those in the body,followed by embryo transfer $(ET)$.
Zygote or early embryo up to $8$ blastomeres is transferred into the Fallopian tube,a process known as Zygote Intra Fallopian Transfer $(ZIFT)$.
Embryos with more than $8$ blastomeres are transferred to the uterus to complete their further development.

(D) : Amniocentesis is the withdrawal of a sample of the fluid (amniotic fluid) surrounding a foetus in the uterus by piercing the amniotic sac through the abdominal wall,under direct ultrasound guidance.
As the amniotic fluid contains cells from the foetus,cell cultures enable chromosome patterns to be studied so that prenatal diagnosis of chromosomal abnormalities can be made.
Certain metabolic errors and other abnormalities,such as spina bifida,can also be diagnosed prenatally from analysis of the cells or of the fluid.
As this technique also helps in the detection of the sex of the unborn foetus,it has been legally banned in many regions to prevent female foeticide.

(C) $Cu$ ions released from copper-releasing $IUDs$ (e.g.,$CuT$,$Cu7$,$Multiload$ $375$) act by suppressing the motility and fertilizing capacity of sperms.
While $IUDs$ also increase the phagocytosis of sperms within the uterus,the primary effect of the copper ions specifically is the inhibition of sperm motility and their ability to fertilize the egg.

(A) The correct answer is $(A)$. In human pedigree analysis,standard symbols are used to represent family relationships and genetic traits:
$1$. $A$ square represents a male,and a circle represents a female.
$2$. $A$ horizontal line connecting a square and a circle represents mating.
$3$. $A$ double horizontal line between a square and a circle represents mating between relatives (consanguineous mating).
$4$. An open (unshaded) symbol represents an unaffected individual.
$5$. $A$ solid (shaded) symbol represents an affected individual.
Based on the provided options:
- Option $(A)$ correctly shows a double horizontal line,which represents mating between relatives.
- Option $(B)$ shows a circle,which represents a female,not a male.
- Option $(C)$ shows a square,which represents a male,not a female.
- Option $(D)$ shows a diamond,which represents an individual of unspecified sex,not an affected male.

(C) : The three alleles $I^A, I^B$ and $i$ of gene $I$ in the $ABO$ blood group system can produce six different genotypes and four different phenotypes as shown below:

Genotypes	Phenotypes
$I^AI^A, I^Ai$	Blood group $A$
$I^BI^B, I^Bi$	Blood group $B$
$I^AI^B$	Blood group $AB$
$ii$	Blood group $O$

Thus, there are four distinct phenotypes: $A, B, AB$ and $O$.

(D) : Linkage is the phenomenon of certain genes staying together during inheritance through generations without any change or separation due to their being present on the same chromosome.
Linked genes occur in the same chromosome.
The strength of the linkage between two genes is inversely proportional to the distance between the two.
$i.e.$,two linked genes show a higher frequency of crossing over (recombination) if the distance between them is higher and a lower frequency if the distance is small.

(A) : $A$ test cross is performed to determine the genotype of an organism showing the dominant phenotype. In a typical test cross,an organism with a dominant phenotype whose genotype is unknown is crossed with an individual that is homozygous recessive for the trait being investigated. The resulting progeny can be analyzed to determine whether the parent was homozygous dominant or heterozygous. This method is more precise than self-crossing for genotype determination.

(C) The correct answer is $(c)$.
According to Mendel's law of dominance,in heterozygous individuals,a character is represented by two contrasting factors called alleles or allelomorphs which occur in pairs.
Out of the two contrasting alleles,only one is able to express its effect in the individual,which is called the dominant factor or dominant allele.
The other allele,which does not show its effect in the heterozygous individual,is called the recessive factor or recessive allele.
Option $(c)$ cannot be explained by the law of dominance.
It is explained by the law of segregation (or law of purity of gametes),which states that alleles do not show any blending and both characters recover as such in the $F_2$ generation.

(C) : The three alleles $I^A$,$I^B$,and $i$ of gene $I$ in the $ABO$ blood group system can produce six different genotypes and four different phenotypes as shown below:

Genotypes	Phenotypes
$I^AI^A, I^Ai$	Blood group $A$
$I^BI^B, I^Bi$	Blood group $B$
$I^AI^B$	Blood group $AB$
$ii$	Blood group $O$

Thus,there are a total of four possible phenotypes: $A, B, AB,$ and $O$.

(C) The pedigree shows an affected father and an unaffected mother producing both affected and unaffected offspring. This pattern is consistent with an autosomal dominant trait.
Let the dominant allele be $A$ and the recessive allele be $a$.
The affected father must be heterozygous $(Aa)$ because he has an unaffected son $(aa)$.
The unaffected mother must be homozygous recessive $(aa)$.
The cross is $Aa \times aa$.
The offspring genotypes are $Aa$ (affected) and $aa$ (unaffected).
Since the mother is $aa$,she contributes an $a$ allele to all children. The father contributes either $A$ or $a$.
Therefore,the female parent is homozygous recessive $(aa)$,and the male parent is heterozygous $(Aa)$.
Option $(A)$ is incorrect because the female parent is homozygous recessive.
Option $(B)$ is incorrect because they can have a normal daughter $(aa)$.
Option $(C)$ is correct because if the trait were colour blindness ($X$-linked recessive),an affected father would pass the trait to all his daughters,but here he has an unaffected daughter (if we assume the circle represents a daughter,though the pedigree shows two affected daughters and one unaffected son and one affected son). Actually,looking at the pedigree,the father is affected and the mother is unaffected. They have two affected daughters,one unaffected son,and one affected son. This is consistent with an autosomal dominant trait. If it were $X$-linked recessive,an affected father $(X^aY)$ and normal mother $(X^AX^A)$ would produce normal daughters $(X^AX^a)$ and normal sons $(X^AY)$. Thus,it cannot be $X$-linked recessive (colour blindness).

(C) : $A$ test cross is performed to determine the unknown genotype of an organism showing a dominant phenotype.
In a typical test cross, the organism with the dominant phenotype is crossed with a homozygous recessive parent.
This method allows researchers to identify whether the dominant individual is homozygous or heterozygous by analyzing the phenotypic ratio of the offspring.
If the offspring show a $1:1$ ratio of dominant to recessive phenotypes, the parent was heterozygous.
If all offspring show the dominant phenotype, the parent was homozygous dominant.

(C) The given situation is an example of incomplete dominance,where the phenotype in the $F_1$ generation does not resemble either of the two parents.
In $Antirrhinum$ (snapdragon),the cross between a red-flowered plant $(RR)$ and a white-flowered plant $(rr)$ results in pink-flowered plants $(Rr)$ in the $F_1$ generation.
When two pink-flowered plants $(Rr)$ are hybridized,the cross is $Rr \times Rr$.
The resulting offspring genotypes are $1 RR$ (red) : $2 Rr$ (pink) : $1 rr$ (white).
Therefore,the genotype of the two plants used for hybridization is $Rr$.

(D) The central dogma of molecular biology,proposed by $F.H.C.$ Crick in $1958$,describes the unidirectional flow of genetic information from $DNA$ to $mRNA$ and then to protein.
However,$H.$ Temin and $D.$ Baltimore discovered that in certain retroviruses,such as $HIV$,genetic information can flow from $RNA$ to $DNA$.
This process is known as reverse transcription or Teminism.
It is catalyzed by the enzyme reverse transcriptase.
Therefore,$HIV$ does not follow the standard central dogma of molecular biology.

(C) : Palindromic nucleotide sequences in the $DNA$ molecule are groups of bases that form the same sequence when read in both forward and backward direction.
In the given question,only option $(c)$ represents a palindromic sequence,which is the recognition site for the restriction enzyme $EcoRI$.
It reads $GAATTC$ from $5' \rightarrow 3'$ on both strands,allowing it to be cut at the middle by the enzyme.

(B) The genetic code is non-ambiguous. $A$ non-ambiguous code means that there is no ambiguity regarding a particular codon. One codon specifies only one amino acid and not any other. There are $64$ codons in total. Out of these $64$,$3$ are stop codons (nonsense codons),which do not code for any amino acid,while the remaining $61$ code for one of the $20$ amino acids. Each codon is specific to a single amino acid,thus the code is not ambiguous.

(C) The correct statements are $(ii)$ and $(iv)$.
Explanation:
$(i)$ Incorrect: Allolactose (an isomer of lactose) acts as an inducer,not glucose or galactose. It binds to the repressor to inactivate it.
$(ii)$ Correct: In the absence of the inducer (lactose),the repressor protein binds to the operator region,preventing $RNA$ polymerase from transcribing the structural genes.
$(iii)$ Incorrect: The $z$-gene codes for $\beta$-galactosidase,while the $y$-gene codes for permease.
$(iv)$ Correct: The lac operon model was proposed by Francois Jacob and Jacques Monod in $1961$ for which they were awarded the Nobel Prize.

(C) The $3' \rightarrow 5'$ phosphodiester bond is formed between the phosphate group,which is attached to the $5'$ carbon of the sugar residue of one nucleotide,and the $3'$ hydroxyl group of the sugar residue of the adjacent nucleotide.
This linkage creates the sugar-phosphate backbone of the polynucleotide chain.
Therefore,it serves to join one nucleotide with another nucleotide.

(B) The correct answer is $B$. The $lac$ operon (an inducible operon) in $E. coli$ consists of a promoter,an operator,a regulator gene ($i$ gene),and three structural genes: $z, y,$ and $a$.
These structural genes code for the enzymes $\beta$-galactosidase,$\beta$-galactoside permease,and $\beta$-galactoside transacetylase,respectively.
The $i$ gene codes for a repressor protein that regulates the expression of these structural genes.
In the presence of an inducer (allolactose),the repressor is inactivated,allowing $RNA$ polymerase to transcribe the structural genes.

(B) : Unlike in prokaryotes where transcription and translation take place in the same compartment,in eukaryotes,the primary transcript is first processed in the nucleus and then transported outside of the nucleus.
Since the primary transcripts of eukaryotes contain both expressing genes (exons) and non-expressing genes (introns),they undergo splicing of introns,followed by capping and tailing at the $5'$-end and $3'$-end,respectively.

(B) is the correct statement.
$1$. The gene for producing insulin is present in every somatic cell of the body because all cells originate from the zygote via mitosis and contain the same genetic information. However,it is only expressed in the beta cells of the pancreas.
$2$. Centrioles (not centromeres) are found in animal cells and produce asters during cell division.
$3$. $A$ nucleosome consists of a core of eight histone proteins wrapped by $DNA$,not nucleotides.
$4$. $DNA$ is a polymer of nucleotides,not a structure consisting of a core of eight histones.

(C) Darwin's finches are a classic example of adaptive radiation.
Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats).
In the Galapagos Islands,Darwin observed that many varieties of finches evolved from a single ancestral species.
These finches adapted to different niches based on the availability of food resources such as insects,seeds,and cactus,leading to the diversification of their beak shapes and sizes.

(C) Wings of butterfly and birds are analogous organs,which are the result of convergent evolution.
$(B)$ Miller's experiment used $CH_4, H_2, NH_3$ and water vapour $(H_2O)$ to produce amino acids.
$(C)$ Vermiform appendix is a vestigial organ and serves as an anatomical evidence of evolution.
$(D)$ According to Darwin,evolution occurred due to small variations and survival of the fittest.
Comparing these with the options,option $(C)$ correctly fills the blanks for statements $(B)$ and $(C)$.

(D) : The most apparent change during the evolutionary history of modern man $(Homo \ sapiens)$ is the significant increase in brain size.
The brain capacity gradually increased from early human ancestors to modern humans.
$Homo \ habilis$ had a brain capacity of $650-800 \ c.c.$,which increased to $900 \ c.c.$ in $Homo \ erectus$.
True humans,including modern man,displayed a consistent and gradual increase in cranial capacity.
$Neanderthal$ man had a brain capacity of approximately $1400 \ c.c.$,which evolved to an average of $1450 \ c.c.$ $(1300-1600 \ c.c.)$ in the living modern man $(Homo \ sapiens \ sapiens)$.

(B) : Ringworm (tinea) is a fungal infection of the skin,scalp,or nails. It is caused by dermatophyte fungi,specifically species of $Microsporum$,$Trichophyton$,and $Epidermophyton$. These fungi can also affect animals,which act as a source of infection for humans. The disease spreads through direct contact or via contaminated materials. Lesions often form partial or complete rings and cause intense itching. The condition is treated using antifungal agents administered orally or applied topically.

(D) The correct answer is $D$.
Widal test (developed by $G.F.I$ Widal) is a serological agglutination test used for the diagnosis of typhoid fever.
It detects the presence of antibodies against the $Salmonella$ $typhi$ bacterium in the patient's serum.
This test helps in confirming the infection caused by $Salmonella$ $typhi$.

(D) $AIDS$ (Acquired Immuno Deficiency Syndrome) is a syndrome caused by the retrovirus $HIV$ (Human Immunodeficiency Virus).
The virus specifically targets and destroys a subgroup of lymphocytes known as helper $T$-cells (or $CD4$ lymphocytes).
This destruction leads to a severe suppression of the body's immune response.
$HIV$ is transmitted through infected blood,semen,and vaginal fluids.
The major routes of transmission include unprotected sexual intercourse,sharing of contaminated needles (common in intravenous drug abuse),and the administration of contaminated blood or blood products.
It is important to note that $HIV$ is not transmitted through casual contact like eating together.
While a combination of antiretroviral drugs can delay the progression of the disease for many years,there is currently no cure that can fully eradicate $AIDS$.

(B) : Morphine is a potent opioid analgesic used mainly to relieve severe and persistent pain,particularly in terminally ill patients or those who have undergone surgery.
It also induces feelings of euphoria.
It is administered by mouth,injection,or in suppositories.
Common side effects are nausea,vomiting,constipation,and drowsiness.
With regular use,tolerance develops and dependence may occur.

(A) $MRI$ is the safest technique for the detection of cancers.
$1$. Radiography $(X-ray)$ and $CT$ scans use ionizing radiation, which can be harmful to tissues and potentially increase the risk of cancer.
$2$. Histopathological studies are invasive techniques that require biopsy, which involves surgical removal of tissue.
$3$. $MRI$ uses strong magnetic fields and non-ionizing radio waves to accurately detect pathological and physiological changes in living tissues without exposing the patient to harmful radiation.

(C) $Plasmodium$ is a protozoan parasite responsible for malaria in humans.
During the life cycle of $Plasmodium$,the parasite multiplies rapidly inside the $RBCs$ (erythrocytes).
When these $RBCs$ rupture,they release the parasite along with a toxic substance called $haemozoin$.
This release of $haemozoin$ into the bloodstream is responsible for the recurring chills and high fever associated with malaria.

(B) : Breeding of crops with higher levels of vitamins and minerals or higher protein and healthier fats is called biofortification. This is the most practical approach to improve public health.

(D) : Sewage water can be purified by passing it through sewage treatment plants with the action of heterotrophic microorganisms. There are three stages of this treatment: primary,secondary,and tertiary. Primary treatment removes floating and suspended solids from sewage through two processes of filtration and sedimentation. First,floating matter is removed through sequential filtration. The filtrate is kept in large open settling tanks where grit settles down. The sediment is called primary sludge,while the supernatant is called effluent. The primary sludge traps a lot of microbes and debris. In secondary treatment,the primary effluent is taken to aeration tanks. $A$ large number of aerobic heterotrophic microbes grow in the aeration tank. They form flocs. Flocs are masses of bacteria held together by slime and fungal filaments to form mesh-like structures. The microbes digest a lot of organic matter,converting it into microbial biomass and releasing a lot of minerals. As the $BOD$ of the waste matter is reduced to $10-15\%$ of raw sewage,it is passed into a settling tank. Thus,secondary treatment is more or less biological. The sediment of the settling tank is called activated sludge. The remaining is passed into a large tank called an anaerobic sludge digester. It is designed for continuous operation. The aerobic microbes present in the sludge get killed. Anaerobic microbes digest the organic mass as well as aerobic microbes of the sludge. They are of two types: non-methanogenic and methanogenic. Methanogenic bacteria produce a mixture of gases containing methane,$H_2S$,and $CO_2$.

(D) : The natural method of pest and pathogen control involving the use of viruses, bacteria, and other organisms (which are their natural predators) is called biocontrol or biological control.
For example, the free-living fungus $Trichoderma$ exerts biocontrol over several plant pathogens to control plant diseases.
$Baculoviruses$ (mostly of the genus $Nucleopolyhedrovirus$) are also used as biocontrol agents, but they are primarily used for the control of insects and arthropods.
$Bacillus \text{ } thuringiensis$ is a soil bacterium used as a biopesticide.
$Glomus$ species are the most common fungal partners of mycorrhiza residing in the roots of higher plants.

(B) $Azospirillum$ is a nitrogen-fixing bacterium that forms a loose association with the roots of various plants,including paddy ($Oryza$ $sativa$).
It is commonly used as a biofertilizer in paddy fields.
Inoculation of paddy fields with $Azospirillum$ helps in increasing the crop yield and reduces the requirement for chemical nitrogen fertilizers.