NEET 2019 Biology Question Paper with Answer and Solution

(C) The correct answer is $C$.
$1$. Viroids consist of free $RNA$ and lack a protein coat,which makes statement $A$ correct.
$2$. Viruses are obligate parasites because they require a host cell to replicate,making statement $B$ correct.
$3$. In viruses,the genetic material ($DNA$ or $RNA$) is the infective constituent,not the protein coat (capsid). The protein coat only protects the genetic material. Therefore,statement $C$ is incorrect.
$4$. Prions are infectious agents composed entirely of abnormally folded proteins,making statement $D$ correct.

(D) The glucose transporters $(GLUT)$ are a group of membrane proteins that facilitate the transport of glucose across the plasma membrane.
$GLUT-IV$ is the primary insulin-regulated glucose transporter found in adipose tissue and striated muscle (skeletal and cardiac muscle).
When insulin levels rise,$GLUT-IV$ is translocated from intracellular vesicles to the plasma membrane,thereby increasing the rate of glucose uptake into these cells.
In contrast,$GLUT-I$,$GLUT-II$,and $GLUT-III$ are generally considered insulin-independent transporters.

(B) The gastro-intestinal tract lining is protected from the corrosive action of digestive enzymes and acidic gastric juice by a thick layer of mucus.
This mucus is secreted by the $Goblet$ $Cells$,which are specialized glandular epithelial cells found in the mucosal epithelium of the gastro-intestinal tract.
$Chief$ $Cells$ (or $Peptic$ $Cells$) secrete proenzymes like $pepsinogen$.
$Oxyntic$ $Cells$ (or $Parietal$ $Cells$) secrete $HCl$ and $Intrinsic$ $Factor$.
Therefore,the correct option is $B$.

(B) $Pila$ is a mollusk that possesses a rasping organ for feeding called a radula $(iii)$.
$(b)$ $Bombyx$ (silkworm) is an arthropod that excretes through Malpighian tubules $(iv)$.
$(c)$ $Pleurobrachia$ belongs to the phylum Ctenophora and is characterized by the presence of eight external rows of ciliated comb plates $(ii)$.
$(d)$ $Taenia$ (tapeworm) is a platyhelminth that uses flame cells for excretion and osmoregulation $(i)$.
Therefore,the correct matching is $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(B) In a submetacentric chromosome,the centromere is situated slightly away from the middle,resulting in one shorter arm and one longer arm.
By convention,the shorter arm is designated as the $p$-arm (from the French 'petit',meaning small).
The longer arm is designated as the $q$-arm (the letter following $p$ in the alphabet).
Therefore,the shorter and longer arms are referred to as the $p$-arm and $q$-arm respectively.

(D) The site of perception of light or dark duration (photoperiod) is the leaves.
Although the flowering occurs at the shoot apex,the hormonal signal (florigen) is synthesized in the leaves upon exposure to the appropriate photoperiod and then translocated to the shoot apex to induce flowering.
Therefore,the leaves are considered the primary site of perception for photoperiodic stimuli.

(B) The $Hypothalamus$ is a small but vital part of the brain located at the base of the $Diencephalon$.
It acts as the body's thermostat,playing a crucial role in thermoregulation by maintaining the body's internal temperature at a constant level.
It also controls other essential functions such as hunger,thirst,and the release of hormones from the pituitary gland.

(C) In eukaryotic cells,$DNA$ is primarily found in the nucleus.
Additionally,semi-autonomous organelles like mitochondria and chloroplasts contain their own circular $DNA$.
Lysosomes are membrane-bound vesicles containing digestive enzymes and do not possess $DNA$.
Vacuoles are storage sacs surrounded by a membrane called the tonoplast and also do not contain $DNA$.
Therefore,the pair consisting of lysosomes and vacuoles is the correct answer as neither contains $DNA$.

(D) Xylem is a complex permanent tissue responsible for the conduction of water and dissolved minerals from the roots to the aerial parts of the plant.
In addition to water and mineral salts,xylem also transports some organic nitrogen (in the form of amino acids and related compounds) and plant hormones (such as cytokinins) from the roots to the shoot system.
Therefore,the most comprehensive description of xylem translocation is water,mineral salts,some organic nitrogen,and hormones.

(B) The formation of concentrated urine is primarily facilitated by the counter-current mechanism.
This mechanism involves the maintenance of a high osmolarity gradient in the inner medullary interstitium of the kidneys.
This gradient is created by the $NaCl$ and urea,which allows water to be reabsorbed from the collecting duct into the interstitium,thereby concentrating the urine.
Antidiuretic hormone $(ADH)$ further enhances this process by increasing the water permeability of the collecting duct.

(D) Lysosomes are membrane-bound vesicular structures formed by the process of packaging in the Golgi apparatus,not the endoplasmic reticulum.
Therefore,the statement that lysosomes are formed by the process of packaging in the endoplasmic reticulum is incorrect.
Lysosomes contain numerous hydrolytic enzymes (hydrolases - lipases,proteases,carbohydrases) which are optimally active at acidic $pH$.

(A) The concept of "Omnis cellula-e cellula" means that all cells arise from pre-existing cells.
This concept was first proposed by Rudolf Virchow in $1855$.
He modified the hypothesis of cell theory proposed by Schleiden and Schwann to give it a final shape.

(C) Hemodialysis is a process used to remove waste products like urea from the blood when kidneys fail. However,it is not a perfect substitute for natural kidney function.
$1$. The natural kidney produces the hormone erythropoietin,which stimulates the bone marrow to produce $RBC$s. Artificial kidneys cannot produce this hormone,leading to reduced $RBC$ production (anemia).
$2$. The natural kidney is responsible for the activation of Vitamin $D$,which is essential for the absorption of calcium ions from the gastrointestinal tract. In the absence of functional kidneys,this activation is impaired,leading to reduced calcium absorption.
$3$. Hemodialysis is specifically designed to remove nitrogenous wastes and excess potassium ions from the blood; therefore,options $(a)$ and $(b)$ are incorrect as these are the primary functions the machine successfully performs.
Thus,$(c)$ and $(d)$ are the correct consequences of long-term kidney failure that are not fully corrected by hemodialysis.

(D) The movement of sugars in the phloem is described as $bi-directional$.
This is because the source (where sugars are produced,usually leaves) and the sink (where sugars are consumed or stored,such as roots,fruits,or developing buds) can change depending on the season or the plant's developmental stage.
For example,during spring,sugar stored in roots (acting as a source) moves upward to developing buds (acting as a sink),whereas during summer,sugar produced in leaves (acting as a source) moves downward to roots (acting as a sink).
Therefore,phloem transport is not restricted to a single direction.

(B) Muscular dystrophy is a group of genetic diseases that cause progressive weakness and loss of muscle mass. It is an inherited disorder caused by mutations in genes responsible for muscle structure and function.
Tetany is caused by low calcium levels in body fluids.
Myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction.
Botulism is a rare but serious illness caused by a toxin that attacks the body's nerves.

(A) The features provided are:
$1$. Organ system level of organisation: Found in Platyhelminthes to Chordata.
$2$. Bilateral symmetry: Found in Platyhelminthes to Chordata (except adult Echinoderms).
$3$. True coelomates with segmentation of body: This is a specific characteristic of the phyla Annelida,Arthropoda,and Chordata.
- Annelids show metameric segmentation.
- Arthropods show segmentation of the body into head,thorax,and abdomen.
- Chordates show segmentation in the form of metameric muscles and skeletal structures.
Molluscs are coelomates but do not exhibit true body segmentation. Therefore,the correct group is Annelida,Arthropoda,and Chordata.

(D) The correct answer is $D$.
$1$. Morels and truffles are edible fungi belonging to the class Ascomycetes,which are considered delicacies.
$2$. $Claviceps$ is a genus of fungi that produces various alkaloids and is the source of $LSD$ (Lysergic acid diethylamide).
$3$. In Ascomycetes,conidia are asexual spores produced exogenously (outside) on conidiophores,while ascospores are sexual spores produced endogenously (inside) within asci.
$4$. Yeasts are unicellular fungi and do not possess filamentous bodies or hyphae. Filamentous bodies with threadlike hyphae are characteristic of multicellular fungi like molds (e.g.,$Rhizopus$ or $Penicillium$).

(C) In parietal placentation,the ovules develop on the inner wall of the ovary or on the peripheral part.
In this type of placentation,the ovary is one-chambered (unilocular) but it may become two-chambered due to the formation of a false septum (replum).

(B) The Respiratory Quotient $(RQ)$ is defined as the ratio of the volume of $CO_2$ evolved to the volume of $O_2$ consumed during respiration.
For fatty acids like tripalmitin $(C_{51}H_{98}O_6)$, the chemical equation for aerobic respiration is:
$2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O + \text{Energy}$
Calculating the $RQ$:
$RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}} = \frac{102}{145} \approx 0.7$
Therefore, the $RQ$ value for tripalmitin is $0.7$.

(D) The cell cycle is a series of events that take place in a cell leading to its division and duplication of its $DNA$ to produce two daughter cells.
It is divided into two main phases: Interphase and $M$-phase (Mitosis).
Interphase is further divided into three sub-phases: $G_{1}$ phase (Gap $1$),$S$ phase (Synthesis),and $G_{2}$ phase (Gap $2$).
The sequence starts with $G_{1}$ phase,where the cell grows and prepares for $DNA$ replication.
This is followed by the $S$ phase,where $DNA$ replication occurs.
Next is the $G_{2}$ phase,where the cell prepares for mitosis.
Finally,the $M$ phase (Mitosis) occurs,where the cell divides.
Therefore,the correct sequence is $G_{1} \rightarrow S \rightarrow G_{2} \rightarrow M$.

(D) Thiobacillus is a genus of bacteria known for its role in the nitrogen cycle,specifically in the process of denitrification.
Denitrification is the process by which nitrate $(NO_3^-)$ is reduced to nitrogen gas $(N_2)$ or nitrous oxide $(N_2O)$.
This process is carried out by denitrifying bacteria such as Pseudomonas and Thiobacillus in soil,which helps in returning nitrogen to the atmosphere.

(C) The phenomenon of retention of the female gametophyte with the developing young embryo on the parent sporophyte is a significant evolutionary precursor to the seed habit.
In $Pteridophytes$,specifically in heterosporous species like $Selaginella$ and $Salvinia$,the female gametophytes are not fully independent but are retained on the parent sporophyte for variable periods.
This development represents an important step in evolution towards the seed habit found in $Gymnosperms$ and $Angiosperms$.

(D) In humans,there are $12$ pairs of ribs.
$1$. The first $7$ pairs are called 'true ribs' or vertebrosternal ribs because they are attached dorsally to the thoracic vertebrae and ventrally to the sternum with the help of hyaline cartilage.
$2$. The $8^{\text{th}}$,$9^{\text{th}}$,and $10^{\text{th}}$ pairs of ribs do not articulate directly with the sternum but join the $7^{\text{th}}$ rib with the help of hyaline cartilage; these are called vertebrochondral or 'false ribs'.
$3$. The last $2$ pairs ($11^{\text{th}}$ and $12^{\text{th}}$) are not connected ventrally and are therefore called 'floating ribs' or vertebral ribs.
Thus,option $D$ is the correct statement.

(C) Concanavalin $A$ is a secondary metabolite classified as a lectin.
Secondary metabolites are chemical compounds produced by plants,fungi,or microbes that are not essential for basic survival but provide ecological advantages.
Lectins are proteins that bind to specific carbohydrates.
Therefore,the correct option is $C$.

(B) $Pinus$ exhibits an obligate symbiotic association with fungi known as mycorrhizae.
This association is essential for the plant because the fungal hyphae help in the absorption of water and mineral nutrients from the soil,which the $Pinus$ roots are unable to do efficiently on their own.
Without this association,the $Pinus$ seedling cannot obtain sufficient nutrients to germinate and establish itself in the environment.
Therefore,the correct option is $B$.

(B) In many monocot leaves,such as grasses,certain adaxial epidermal cells along the veins modify themselves into large,empty,colourless cells called $bulliform$ cells.
When these cells absorb water and are turgid,the leaf surface is exposed.
However,during very dry weather,these $bulliform$ cells lose water due to stress and become flaccid.
This loss of turgidity causes the leaves to curl inwards to minimize the surface area exposed to the atmosphere,thereby reducing the rate of transpiration and preventing water loss.

(C) The correct matches are as follows:
$(a)$ Crypts of Lieberkuhn are found in the mucosa of the small intestine between the bases of the villi.
$(b)$ Glisson's Capsule is the thin connective tissue sheath that covers each hepatic lobule in the liver.
$(c)$ Islets of Langerhans are the endocrine part of the pancreas.
$(d)$ Brunner's Glands are submucosal glands found specifically in the duodenum of the small intestine.
Therefore, the correct sequence is: $(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)$.

(C) The correct matches are as follows:
$1$. Insulin: Deficiency or resistance leads to $Diabetes \text{ } mellitus$ $(a-v)$.
$2$. Thyroxin: Deficiency leads to $Goitre$ $(b-iv)$.
$3$. Corticoids: Deficiency of glucocorticoids/mineralocorticoids leads to $Addison's \text{ } disease$ $(c-i)$.
$4$. Growth Hormone: Hypersecretion in adults leads to $Acromegaly$ $(d-iii)$.
Therefore,the correct sequence is $(a-v, b-iv, c-i, d-iii)$.

(B) The process of glycolysis begins with the phosphorylation of glucose to form glucose-$6$-phosphate.
This reaction consumes one molecule of $ATP$ and is catalyzed by the enzyme $Hexokinase$ (or $Glucokinase$ in the liver).
This is the first irreversible step of the glycolytic pathway,which effectively traps glucose inside the cell.

(D) The cornea is the anterior,transparent part of the eye-ball.
It is composed of a dense matrix of collagen fibers,which provides structural integrity and transparency.
It is avascular,meaning it lacks blood vessels,which helps maintain its transparency.
It is considered the most sensitive portion of the eye due to the presence of a high density of nerve endings.
Therefore,statement $D$ is correct.

(B) Statement $(A)$ is true: Prosthetic groups are organic compounds or metal ions that are tightly bound to the apoenzyme (the protein part of the enzyme).
Statement $(B)$ is false: $A$ complete,catalytically active enzyme consisting of the protein part (apoenzyme) and its non-protein cofactor (such as a prosthetic group) is called a holoenzyme,not an apoenzyme. The apoenzyme refers specifically to the inactive protein portion of the enzyme.

(B) The respiratory disorder described is $Asthma$.
$Asthma$ is a difficulty in breathing causing wheezing due to inflammation of the bronchi and bronchioles.
It is often triggered by allergens and pollutants present in the air.
Option $A$ refers to nasal polyps.
Option $C$ refers to $Emphysema$.
Option $D$ refers to $Respiratory$ $Distress$ $Syndrome$.

(D) In gymnosperms,the phloem is composed of sieve cells and albuminous cells. Unlike angiosperms,gymnosperms do not possess sieve tubes and companion cells. Instead,they have sieve cells which are less specialized than sieve tubes,and albuminous cells which perform functions analogous to companion cells. Therefore,the correct answer is $D$.

(A) Pineapple plants have a long vegetative growth phase before flowering.
To induce flowering artificially and ensure year-round production,specific plant growth regulators are used.
$Auxins$ (like $NAA$ or $2,4-D$) and $Ethylene$ are widely used in commercial pineapple cultivation to promote synchronous flowering.
$Ethylene$ is particularly effective in inducing flowering in pineapple,while $Auxins$ help in fruit development and uniform ripening.
Therefore,the combination of $Auxin$ and $Ethylene$ is the correct choice for increasing yield.

(A) The $G_0$ phase,also known as the quiescent stage,is a state where cells exit the cell cycle.
In this phase,cells remain metabolically active but do not proliferate unless called upon to do so depending on the requirement of the organism.
Therefore,cells in the $G_0$ phase are considered to have exited the cell cycle.

(D) The correct matches are as follows:
$(a)$ Saprophyte: Organisms that obtain nutrients from dead and decaying organic matter,leading to the decomposition of dead organic materials $(ii)$.
$(b)$ Parasite: Organisms that live on or inside other living organisms (hosts) and derive nutrients from them,thus living on living plants or animals $(iii)$.
$(c)$ Lichens: These represent a symbiotic association between algae (phycobiont) and fungi (mycobiont) $(iv)$.
$(d)$ Mycorrhiza: This is a symbiotic association of fungi with the roots of higher plants $(i)$.
Therefore,the correct sequence is $(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)$.

(C) Cardiac output is defined as the volume of blood pumped by each ventricle per minute. It is calculated as the product of stroke volume and heart rate.
Stroke volume is the amount of blood pumped by the ventricle in one beat,which is the difference between the end-diastolic volume $(EDV)$ and the end-systolic volume $(ESV)$.
Given: $EDV = 100 \; mL$,$ESV = 50 \; mL$.
Stroke volume = $EDV - ESV = 100 \; mL - 50 \; mL = 50 \; mL$.
Cardiac output = $5 \; L = 5000 \; mL$.
Formula: $\text{Cardiac output} = \text{Stroke volume} \times \text{Heart rate}$.
$5000 \; mL = 50 \; mL \times \text{Heart rate}$.
Heart rate = $5000 / 50 = 100$ beats per minute.

(D) Ciliated epithelium consists of cells that bear cilia on their free surfaces. The function of these cilia is to move particles or mucus in a specific direction over the epithelium. In humans,ciliated epithelial cells are primarily found in the inner surface of hollow organs like the bronchioles and the Fallopian tubes. In the bronchioles,they help in moving mucus and trapped dust particles out of the respiratory tract,while in the Fallopian tubes,they assist in the movement of the ovum towards the uterus.

(D) Annual rings are formed due to the differential activity of the vascular cambium throughout the year,which is influenced by seasonal climatic variations.
In temperate regions,the cambium is very active during the spring season,producing a large number of xylem elements with wide vessels,known as spring wood or early wood,which is lighter in color.
In the autumn season,the cambium becomes less active and produces fewer xylem elements with narrow vessels,known as autumn wood or late wood,which is darker in color.
These two bands together constitute an annual ring.
Therefore,annual rings are highly prominent in trees of temperate regions where distinct seasonal variations occur.
Thus,the statement that 'Annual rings are not prominent in trees of temperate region' is incorrect.

(A) The alimentary canal of a cockroach is divided into three regions: foregut,midgut,and hindgut.
$1$. The foregut includes the mouth,pharynx,oesophagus,crop (for food storage),and gizzard (for grinding food).
$2$. The midgut is the site of digestion and absorption.
$3$. The hindgut includes the ileum,colon,and rectum.
Therefore,the correct sequence starting from the mouth is: Pharynx $\rightarrow$ Oesophagus $\rightarrow$ Crop $\rightarrow$ Gizzard $\rightarrow$ Ileum $\rightarrow$ Colon $\rightarrow$ Rectum.

(B) Steroid hormones are lipid-soluble and can easily pass through the plasma membrane of target cells.
Once inside the cytoplasm,they bind to specific intracellular receptors (usually present in the cytoplasm or nucleus) to form a hormone-receptor complex.
This hormone-receptor complex then enters the nucleus and binds to specific sequences of $DNA$ (hormone response elements).
This binding regulates gene expression by promoting or inhibiting the transcription of specific genes into $mRNA$,which subsequently leads to the synthesis of specific proteins that alter cellular activities.

(A) Expiratory Capacity $(EC)$ is defined as the total volume of air a person can expire after a normal inspiration.
It is calculated as the sum of Tidal Volume $(TV)$ and Expiratory Reserve Volume $(ERV)$.
Formula: $EC = TV + ERV$.
Given: $TV = 500 \; mL$ and $ERV = 1000 \; mL$.
Calculation: $EC = 500 \; mL + 1000 \; mL = 1500 \; mL$.
The Residual Volume $(RV)$ is not required to calculate the Expiratory Capacity.
Therefore,the correct answer is $1500 \; mL$.

(B) The correct answer is $B$. The enzymes of the electron transport chain $(ETC)$ are embedded in the inner mitochondrial membrane,not the outer membrane. The outer membrane is permeable to small molecules like ions,$ATP$,and nutrients due to the presence of porins. The inner membrane forms infoldings called cristae,which increase the surface area for biochemical reactions. The mitochondrial matrix contains a single circular $DNA$ molecule,a few $RNA$ molecules,and $70S$ ribosomes.

(A) The Electrocardiogram $(ECG)$ is a graphical representation of the electrical activity of the heart during a cardiac cycle.
$1$. $(a) \; P-$wave represents the electrical excitation (or depolarisation) of the atria, which leads to the contraction of both the atria.
$2$. $(b) \; QRS$ complex represents the depolarisation of the ventricles, which initiates the ventricular contraction.
$3$. $(c) \; T-$wave represents the return of the ventricles from an excited to a normal state (repolarisation). The end of the $T-$wave marks the end of systole.
$4$. $(d) \; \text{Reduction in the size of } T-$wave is often associated with coronary ischemia, where the heart muscle does not receive enough oxygen.
Therefore, the correct matching is: $(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)$.

(B) According to the rules of binomial nomenclature,the scientific name consists of two parts: the genus name and the specific epithet.
$1$. The genus name starts with a capital letter,and the specific epithet starts with a small letter.
$2$. When the name of the author who first described the species is written after the specific epithet,it is written in abbreviated form in Roman script.
$3$. For Mango,the scientific name is Mangifera indica,and since Carolus Linnaeus first described it,it is written as Mangifera indica Linn.

(B) Secretory cells are specialized to synthesize and export proteins or other substances.
The $Golgi$ complex plays a crucial role in the packaging,modification,and sorting of proteins and lipids for secretion.
Therefore,cells that are highly active in secretion,such as pancreatic acinar cells or goblet cells,contain a very high number of $Golgi$ complexes to handle the extensive processing and packaging requirements.
While the endoplasmic reticulum is involved in synthesis,the $Golgi$ complex is the primary organelle responsible for the final packaging and dispatch of secretory products.

(C) The non-membranous nucleoplasmic structures present in the nucleus are known as nucleoli (singular: nucleolus).
These structures are not bounded by any membrane.
The nucleolus is the primary site for the active synthesis of ribosomal $RNA$ $(rRNA)$.
Ribosomal proteins are imported from the cytoplasm and assembled with $rRNA$ within the nucleolus to form ribosomal subunits.

(B) Organisms that possess only $70S$ ribosomes are prokaryotes (e.g.,bacteria).
Prokaryotic cells lack a well-defined nucleus and membrane-bound organelles.
The genetic material in prokaryotes consists of a single,double-stranded,circular $DNA$ molecule that is not associated with histone proteins (naked $DNA$).
Therefore,option $B$ is the correct description of the genetic material found in such organisms.

(B) In meiosis $I$,the homologous chromosomes separate,but the sister chromatids remain attached at the centromere.
Starting with a diploid cell $(2n)$ with $4C$ amount of $DNA$ (after $S$ phase),meiosis $I$ results in two daughter cells.
Each daughter cell receives one set of chromosomes,but each chromosome still consists of two sister chromatids.
Therefore,each daughter cell has $n$ chromosomes and $2C$ amount of $DNA$.
$A$ haploid gamete (formed after meiosis $II$) has $n$ chromosomes and $1C$ amount of $DNA$.
Thus,the daughter cells after meiosis $I$ have twice the amount of $DNA$ $(2C)$ compared to a haploid gamete $(1C)$.

(B) Lecithin is a type of phospholipid,specifically a phosphatidylcholine.
Phospholipids are complex lipids that contain a phosphate group in addition to fatty acids and an alcohol (usually glycerol).
They are essential components of biological membranes,forming the lipid bilayer.
Therefore,the main constituent or the category to which Lecithin belongs is a phospholipid.

(B) Nitrogenous bases are classified into two types: Purines and Pyrimidines.
Purines are double-ring structures,which include Adenine $(A)$ and Guanine $(G)$.
Pyrimidines are single-ring structures,which include Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$.
$DNA$ contains Adenine,Guanine,Cytosine,and Thymine.
$RNA$ contains Adenine,Guanine,Cytosine,and Uracil.
Therefore,the purines present in both $DNA$ and $RNA$ are Adenine and Guanine.

(D) For the industrial production of enzymes or other products on a large scale,microbes are grown in large vessels known as bioreactors.
Bioreactors provide optimal growth conditions such as temperature,$pH$,substrate,salts,vitamins,and oxygen to achieve the desired product.
$BOD$ incubators are used for laboratory-scale incubation,while sludge digesters are used in sewage treatment plants.
Therefore,the correct equipment for large-scale industrial production is the bioreactor.

(A) Golden rice is a genetically modified variety of rice $(Oryza \text{ } sativa)$ that has been engineered to produce beta-carotene, which is a precursor of Vitamin $A$.
This modification was achieved by introducing genes from the daffodil plant $(Narcissus \text{ } pseudonarcissus)$ and the bacterium $Erwinia \text{ } uredovora$ into the rice genome.
Therefore, it is enriched with Vitamin $A$ and contains genes derived from the daffodil plant.

(C) $In$ $situ$ conservation refers to the protection of species in their natural habitat.
Biosphere Reserves,Wildlife Sanctuaries,and Sacred Groves are all examples of $in$ $situ$ conservation as they protect organisms within their natural environment.
Botanical Gardens are examples of $ex$ $situ$ conservation,where plants are maintained outside their natural habitats for research,education,and conservation purposes.

(D) The genetic code is read in triplets (codons). $A$ change in the reading frame (frameshift mutation) occurs when the number of nucleotides inserted or deleted is not a multiple of $3$.
If $3$ nucleotides (or a multiple of $3$) are inserted or deleted,the reading frame remains unchanged,although one or more amino acids may be added or removed.
In the given $mRNA$ sequence $5'-AAC-AGC-GGU-GCU-AUU-3'$,deleting $GGU$ (which consists of $3$ nucleotides) from the $7^{th}, 8^{th}$,and $9^{th}$ positions removes exactly one codon.
This results in the loss of one amino acid,but the subsequent codons remain in the same reading frame.
Therefore,option $D$ is the correct condition.

(D) Nuclear waste is extremely radioactive and hazardous to living organisms.
It must be disposed of with extreme caution.
The most accepted and scientifically suitable method for the disposal of nuclear waste is to bury it within rocks deep below the Earth's surface.
This prevents the leakage of radioactive materials into the biosphere and ensures long-term containment.

(B) The correct matches are as follows:
$1$. $(a)$ Lactobacillus: Produces $(ii)$ Curd by fermenting milk.
$2$. $(b)$ Saccharomyces cerevisiae: Commonly known as Brewer's yeast,it is used to produce $(iv)$ Bread.
$3$. $(c)$ Aspergillus niger: $A$ fungus used for the commercial production of $(iii)$ Citric Acid.
$4$. $(d)$ Acetobacter aceti: $A$ bacterium used for the production of $(v)$ Acetic Acid.
Therefore,the correct matching is $(a)-(ii), (b)-(iv), (c)-(iii), (d)-(v)$. The correct option is $B$.

(C) The genetic map unit is known as the Centimorgan $(cM)$,named in honor of Thomas Hunt Morgan.
One Centimorgan $(1 \ cM)$ is defined as the distance between two gene loci on a chromosome such that there is a $1 \%$ frequency of recombination (crossing over) between them.
Therefore,a map distance of $1 \ cM$ corresponds to $1 \%$ recombination frequency.
This unit is essential for mapping the relative positions of genes on chromosomes.

(C) Intra-Uterine Devices (IUDs) are effective contraceptive methods.
They are categorized into three types:
$1$. Non-medicated IUDs: e.g.,Lippes Loop.
$2$. Copper-releasing IUDs: e.g.,Copper $T$ $(CuT)$,Copper $7$ $(Cu7)$,Multiload $375$.
$3$. Hormone-releasing IUDs: These release hormones that make the uterus unsuitable for implantation and the cervix hostile to the sperms. Examples include Progestasert and $LNG-20$.
Therefore,the correct option is $C$.

(A) Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
$Trichoderma$ species are free-living fungi that are very common in the root ecosystem.
They are effective biocontrol agents of several plant pathogens.
$Chlorella$ is an alga,$Anabaena$ is a cyanobacterium (often used as a biofertilizer),and $Lactobacillus$ is a bacterium used in the production of curd from milk.
Therefore,the correct option is $A$.

(A) Expressed Sequence Tags $(ESTs)$ are a strategy used in the Human Genome Project to identify all the genes that are expressed as $RNA$.
These are short sequences of $cDNA$ (complementary $DNA$) that are generated by sequencing either one or both ends of an expressed gene transcript.
Since they represent the parts of the genome that are actively transcribed into $RNA$,they serve as a useful tool for gene discovery and mapping.

(D) Colostrum is the yellowish fluid produced by the mother during the initial days of lactation.
It is rich in antibodies,specifically Immunoglobulin $A$ $(IgA)$.
These antibodies provide passive immunity to the newborn infant,protecting them from various infections during their early life.
Therefore,the correct option is $D$.

(D) Inbreeding refers to the mating of more closely related individuals within the same breed for $4-6$ generations.
$1$. It increases homozygosity,which is necessary if we want to evolve a pureline in any animal.
$2$. Inbreeding exposes harmful recessive genes that are eliminated by selection.
$3$. It also helps in the accumulation of superior genes and the elimination of less desirable genes.
$4$. However,continued inbreeding,especially close inbreeding,usually reduces fertility and productivity.
Option $D$ is technically incorrect because the phrasing in the original option was grammatically flawed and misleading regarding the process of selection; however,in the context of standard biology questions,all statements $A, B,$ and $C$ are standard facts. Upon review,the statement in option $D$ is often presented as a benefit,but the question asks for the incorrect statement. Actually,all statements $A, B, C,$ and $D$ are generally considered correct in the context of animal breeding theory. If forced to choose,option $D$ is the most poorly phrased,but scientifically,inbreeding does help in the accumulation of superior genes.

(B) The correct pathway of sperm transport in the male reproductive system is as follows:
$1$. Sperm are produced in the $Seminiferous \text{ } tubules$.
$2$. They move into the $Rete \text{ } testis$.
$3$. From there, they pass through the $Vasa \text{ } efferentia$.
$4$. They are then stored and matured in the $Epididymis$.
$5$. During ejaculation, they travel through the $Vas \text{ } deferens$.
$6$. They enter the $Ejaculatory \text{ } duct$ (formed by the duct of the seminal vesicle and $Vas \text{ } deferens$).
$7$. Finally, they pass through the $Urethra$ and exit the body via the $Urethral \text{ } meatus$.

(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.
Given $p = 0.4$,then $q = 1 - 0.4 = 0.6$.
The genotype frequencies are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p^2$ represents the frequency of homozygous dominant individuals $(AA)$,
$2pq$ represents the frequency of heterozygous individuals $(Aa)$,
and $q^2$ represents the frequency of homozygous recessive individuals $(aa)$.
Calculating the values:
$p^2 = (0.4)^2 = 0.16 (AA)$
$2pq = 2 \times 0.4 \times 0.6 = 0.48 (Aa)$
$q^2 = (0.6)^2 = 0.36 (aa)$
Therefore,the frequencies are $0.16(AA), 0.48(Aa),$ and $0.36(aa)$.

(B) Klinefelter's syndrome is a genetic disorder caused by the presence of an additional $X$ chromosome,resulting in a karyotype of $47, XXY$.
Individuals with this condition exhibit overall masculine development but also show feminine characteristics such as gynaecomastia (development of breast tissue).
These individuals are sterile due to the inability to produce functional sperm.
In contrast,Turner's syndrome $(45, XO)$ involves a missing $X$ chromosome in females,while Down's syndrome $(47, XX/XY + 21)$ is a trisomy of chromosome $21$.

(D) The greenhouse effect is a natural process that warms the Earth's surface.
When the Sun's energy reaches the Earth's atmosphere,some of it is reflected back to space and the rest is absorbed and re-radiated by greenhouse gases.
$CO_2$ (Carbon dioxide) and $CH_4$ (Methane) are the primary greenhouse gases contributing to global warming.
$CO_2$ accounts for about $60\%$ of the total greenhouse effect,while $CH_4$ accounts for about $20\%$.
Therefore,the correct pair is Carbon dioxide and Methane.

(A) The $Montreal$ protocol was signed in $1987$ to control the emission of ozone-depleting substances,specifically chlorofluorocarbons $(CFCs)$.
This international treaty was designed to protect the ozone layer by phasing out the production and consumption of numerous substances that are responsible for ozone depletion.
In contrast,the $Kyoto$ protocol was focused on reducing greenhouse gas emissions to combat global warming.

(D) The phenomenon where the female gamete develops into an embryo without fertilization is known as $Parthenogenesis$.
In the context of plants,this is often associated with $Apomixis$.
$Autogamy$ refers to self-pollination.
$Parthenocarpy$ refers to the development of fruit without fertilization.
$Syngamy$ is the fusion of male and female gametes (fertilization).

(C) Sexually transmitted diseases $(STDs)$ like Gonorrhoea,Syphilis,and Chlamydiasis are caused by bacteria and can be completely cured if detected early and treated with appropriate antibiotics.
However,viral infections such as Genital herpes,Human Immunodeficiency Virus $(HIV)$ infection,and Hepatitis-$B$ are not completely curable.
Genital herpes is caused by the Herpes Simplex Virus $(HSV)$ and remains in the body for life,with periodic outbreaks,making it incurable.

(D) The rejection of a transplanted organ,such as a kidney graft,is primarily mediated by the recipient's immune system recognizing the donor tissue as foreign.
This process is driven by $T$-lymphocytes,which are the key components of the cell-mediated immune response.
Specifically,cytotoxic $T$-cells identify the $MHC$ (Major Histocompatibility Complex) antigens on the surface of the transplanted cells as non-self and initiate an attack to destroy the graft.
Therefore,the cell-mediated immune response is responsible for graft rejection.

(C) The genetic code is considered nearly universal,meaning that the same codons specify the same amino acids in almost all organisms,from bacteria to humans.
Because of this universality,a human gene (such as the one for insulin) can be inserted into a bacterial cell (like $E. coli$) using recombinant $DNA$ technology.
The bacterial machinery recognizes the human codons and translates them into the correct human protein (insulin),as the genetic language is shared across species.

(C) Alfred Sturtevant,a student of $T$.$H$. Morgan,used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on the chromosome.
This technique is known as genetic mapping or linkage mapping.
$1$ map unit corresponds to $1\%$ recombination frequency.

(B) Restriction endonucleases are enzymes that recognize specific palindromic nucleotide sequences in $DNA$ and cut the sugar-phosphate backbone at specific sites on both strands of the $DNA$ molecule.
Option $A$ is correct because these enzymes act at specific internal positions.
Option $B$ is incorrect because restriction endonucleases cut both strands of the $DNA$ double helix,not just one.
Option $C$ is correct as they cleave the phosphodiester bonds in the sugar-phosphate backbone.
Option $D$ is correct as they identify specific palindromic sequences.

(A) The most significant cause of biodiversity loss and the extinction of species is $Habitat \text{ } loss \text{ } and \text{ } fragmentation$.
As human populations increase, the demand for land grows, leading to the destruction of natural habitats such as tropical rainforests.
When large habitats are broken into smaller fragments, species that require large territories or those that are migratory are severely affected, leading to population decline and eventual extinction.

(B) Hugo de Vries proposed the Mutation Theory of evolution.
According to him,evolution is a saltatory process,meaning it occurs through large,sudden,and discontinuous variations.
These mutations are random and directionless,unlike the small and directional variations proposed by Darwin in his theory of natural selection.

(D) In $Antirrhinum$ (Snapdragon),the inheritance of flower colour is an example of incomplete dominance.
$1$. The Principle of Dominance states that one allele masks the expression of another,which is not observed here as the phenotype is intermediate (pink).
$2$. Incomplete dominance results in a phenotypic ratio of $1:2:1$ ($1$ Red : $2$ Pink : $1$ White) in the $F_{2}$ generation,which matches the genotypic ratio.
$3$. The Law of Segregation (Mendel's First Law) states that alleles separate during gamete formation. This law is universal and applies to all sexually reproducing organisms,including $Antirrhinum$. Therefore,the statement that the Law of Segregation does not apply is incorrect.

(C) In domesticated fowls,sex determination follows the $ZW-ZZ$ type mechanism,where the female is heterogametic $(ZW)$ and the male is homogametic $(ZZ)$. Therefore,the sex of the progeny depends on the type of egg (whether it carries $Z$ or $W$) rather than the sperm. Thus,statement $C$ is incorrect.
- Male fruit flies $(Drosophila)$ are heterogametic $(XY)$,so statement $A$ is correct.
- Male grasshoppers are $XO$ type,meaning $50 \%$ of sperms lack a sex chromosome,so statement $B$ is correct.
- Human males are $XY$,where the $Y$ chromosome is significantly shorter than the $X$ chromosome,so statement $D$ is correct.

(C) Polyblend is a fine powder of recycled modified plastic.
It was developed by Ahmed Khan of Bangalore.
When mixed with bitumen,it enhances the water-repellent properties of the bitumen and helps in the construction of roads with increased life span and resistance to water damage.
Therefore,it has proved to be a good material for the construction of roads.

(A) The secondary oocyte remains arrested at the metaphase-$II$ stage of meiosis-$II$ until a sperm makes contact with the zona pellucida of the ovum.
Upon the entry of the sperm into the cytoplasm of the secondary oocyte,the completion of meiosis-$II$ is triggered.
This process results in the formation of a large haploid ovum (ootid) and a small second polar body.
Therefore,the extrusion of the second polar body occurs after the entry of the sperm but before the fusion of the male and female pronuclei (fertilization).

(B) The Earth Summit,also known as the United Nations Conference on Environment and Development $(UNCED)$,was held in Rio de Janeiro in $1992$.
This historic summit was called for the conservation of biodiversity and the sustainable utilization of its benefits.
It aimed to address the urgent need for global cooperation in protecting the environment and ensuring sustainable development for future generations.

(B) In the process of recombinant $DNA$ technology,the isolation of genetic material $(DNA)$ involves several steps.
After the removal of other biomolecules like $RNA$,proteins,and lipids using specific enzymes,the purified $DNA$ remains in the aqueous phase.
To precipitate the $DNA$ out of this mixture,chilled ethanol is added.
This causes the $DNA$ to form a fine thread-like precipitate,which can be collected by spooling.

(A) $1$. Lactational amenorrhea is a natural method based on the fact that ovulation does not occur during the period of intense lactation following parturition. This is regulated by high levels of prolactin,which inhibits $GnRH$ secretion,thus preventing ovulation. This is a hormonal-based physiological state.
$2$. Pills (Oral contraceptives) contain either progestogens or a combination of progestogen and estrogen,which inhibit ovulation and implantation.
$3$. Emergency contraceptives (e.g.,morning-after pills) contain high doses of progestogens or progestogen-estrogen combinations to prevent or delay ovulation.
$4$. Barrier methods (e.g.,condoms,diaphragms) are physical devices that prevent the meeting of sperm and ovum; they do not involve hormones.
$5$. $CuT$ (Copper-$T$) is an Intrauterine Device $(IUD)$ that releases copper ions to suppress sperm motility and fertilizing capacity; it does not involve hormones.
Therefore,Lactational amenorrhea,Pills,and Emergency contraceptives all involve a role of hormones.

(B) Heroin,also known as diacetylmorphine,is a semi-synthetic opioid drug.
It is chemically synthesized from morphine,which is extracted from the latex of the poppy plant,$Papaver$ $somniferum$.
The process involves the acetylation of morphine,where two acetyl groups are added to the morphine molecule.
This chemical modification makes heroin more lipid-soluble than morphine,allowing it to cross the blood-brain barrier more rapidly.

(B) Stabilizing selection is a type of natural selection in which the population mean stabilizes on a particular non-extreme trait value.
In this case,the newborns with an average weight ($3$ to $3.3 \; kg$) have a high survival rate $(97 \%)$,while those with extreme weights (very low: $2$ to $2.5 \; kg$ or very high: $4.5$ to $5 \; kg$) have a very high mortality rate $(99 \%)$.
Since the selection process favors the intermediate phenotype and eliminates the extreme phenotypes,it is known as Stabilizing Selection.

(D) An ecological pyramid is a graphical representation of the relationship between different trophic levels in an ecosystem.
$1$. The pyramid of energy is always upright because energy flow is unidirectional and decreases at each successive trophic level.
$2$. The pyramid of numbers in a grassland is typically upright.
$3$. The pyramid of biomass in a forest is generally upright.
$4$. The pyramid of biomass in a sea is generally inverted because the biomass of phytoplankton (producers) is much lower than that of the zooplankton and small fish (consumers) that feed on them at any given time.

(D) In flowering plants, after fertilization, the following changes occur:
$1$. The $Ovary$ develops into a $Fruit$.
$2$. The $Zygote$ develops into an $Embryo$.
$3$. The $Central \text{ } cell$ (or $Primary \text{ } Endosperm \text{ } Cell$) develops into $Endosperm$.
$4$. The $Ovules$ develop into $Seeds$, not the $Embryo \text{ } sac$. The $Embryo \text{ } sac$ is the structure present inside the $Ovule$ before fertilization.
Therefore, the statement '$Ovules$ develop into $Embryo \text{ } sac$' is incorrect.

(B) In some seeds,such as black pepper and beet,remnants of the nucellus are persistent. This residual,persistent nucellus is known as the $Perisperm$.

(C) $Bacillus$ $thuringiensis$ $(Bt)$ produces insecticidal protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. The protein exists as an inactive protoxin but becomes an active toxin when ingested by the insect. The activation of the toxin is triggered by the alkaline $pH$ of the insect's gut, which solubilizes the crystals. Once activated, the toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis, eventually leading to the death of the insect.

(D) In angiosperms,the process of double fertilization occurs.
After the pollen tube enters the synergid,it releases two male gametes into the cytoplasm of the synergid.
One male gamete moves towards the egg cell and fuses with its nucleus to form a diploid zygote $(n + n = 2n)$,a process known as syngamy.
The second male gamete moves towards the two polar nuclei located in the central cell and fuses with them to form a triploid primary endosperm nucleus $(PEN)$ $(n + n + n = 3n)$,a process known as triple fusion.
Thus,one male gamete fuses with the egg and the other fuses with the central cell nuclei.

(C) In the $Lac$ operon:
$1$. The $i$ gene codes for the repressor protein,which regulates the operon.
$2$. The $z$ gene codes for $\beta-\text{galactosidase}$,which hydrolyzes lactose into glucose and galactose.
$3$. The $y$ gene codes for permease,which increases the permeability of the cell to $\beta-\text{galactosides}$.
$4$. The $a$ gene codes for transacetylase,which transfers an acetyl group to $\beta-\text{galactosides}$.
Matching these:
$(a) - (iii)$
$(b) - (i)$
$(c) - (iv)$
$(d) - (ii)$
Therefore,the correct sequence is $(iii), (i), (iv), (ii)$.

(C) The brain capacities of the given hominids are as follows:
$1$. $(a)$ Homo habilis: Their brain capacity was between $650-800 \text{ cc}$. Thus,$(a) - (iii)$.
$2$. $(b)$ Homo neanderthalensis: They had a brain capacity of approximately $1400 \text{ cc}$. Thus,$(b) - (iv)$.
$3$. $(c)$ Homo erectus: Their brain capacity was about $900 \text{ cc}$. Thus,$(c) - (i)$.
$4$. $(d)$ Homo sapiens: Modern humans have a brain capacity of approximately $1350 \text{ cc}$. Thus,$(d) - (ii)$.
Therefore,the correct matching is $(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)$.

(D) Typhoid fever is a bacterial infection caused by the pathogen $Salmonella \ typhi$.
This bacterium typically enters the small intestine through contaminated food and water and migrates to other organs through the blood.
The standard diagnostic test used to confirm the presence of $Salmonella \ typhi$ antibodies in the patient's serum is the $Widal \ test$.
Therefore,the correct pair is $Salmonella \ typhi$ and $Widal \ test$.

(B) Statins are produced by the yeast $Monascus$ $purpureus$.
They act by competitively inhibiting the enzyme responsible for the synthesis of cholesterol in the body.
Therefore,they are used as blood cholesterol-lowering agents.
Cyclosporin $A$ is an immunosuppressive agent.
Streptokinase is used as a 'clot buster' for removing clots from the blood vessels of patients who have undergone myocardial infarction.
Lipases are used in detergent formulations to remove oily stains from laundry.

(B) Biocontrol agents are organisms that are used to control plant diseases and pests.
$1$. $Bacillus$ $thuringiensis$ $(Bt)$ is a bacterium used as a bio-insecticide to control butterfly caterpillars.
$2$. $Trichoderma$ species are free-living fungi that are very common in the root ecosystems and are effective biocontrol agents of several plant pathogens.
$3$. $Baculoviruses$ are pathogens that attack insects and other arthropods. The majority of $baculoviruses$ used as biological control agents are in the genus $Nucleopolyhedrovirus$.
Therefore,the correct group of biocontrol agents is $Trichoderma$,$Baculovirus$,and $Bacillus$ $thuringiensis$.

(A) The correct sequence of events in human reproduction is as follows:
$1$. Gametogenesis: Formation of gametes (sperm and ovum).
$2$. Gamete transfer: Insemination (transfer of sperm into the female reproductive tract).
$3$. Syngamy: Fertilization (fusion of male and female gametes to form a zygote).
$4$. Zygote: The single-celled diploid structure formed after fertilization.
$5$. Cell division (Cleavage): The zygote undergoes mitotic divisions to form a morula and blastocyst.
$6$. Cell differentiation: The cells of the embryo organize into specific layers and tissues.
$7$. Organogenesis: The development of organs from the differentiated tissues.

(C) Oxytocin is a hormone produced by the hypothalamus and released by the posterior pituitary gland.
It plays a crucial role in two major physiological processes:
$1$. Foetal ejection reflex: During parturition (childbirth),oxytocin acts on the uterine muscles to cause strong contractions,which helps in the expulsion of the baby.
$2$. Milk ejection reflex: During lactation,oxytocin acts on the smooth muscles of the mammary glands (myoepithelial cells) to cause the ejection of milk.
Therefore,the correct answer is $C$.

(D) During the luteal phase of the menstrual cycle,the corpus luteum secretes high levels of progesterone and estrogen.
These hormones exert a strong negative feedback effect on the anterior pituitary gland and the hypothalamus.
As a result,the secretion of gonadotropins,specifically $FSH$ (Follicle Stimulating Hormone) and $LH$ (Luteinizing Hormone),is significantly inhibited.
Since $FSH$ is required for the recruitment and development of new ovarian follicles,its low levels during the luteal phase prevent the development of any new follicles.

(D) When more than one adaptive radiation appeared to have occurred in an isolated geographical area (representing different habitats),one can call this convergent evolution.
In Australia,marsupials and placental mammals show many similar characteristics due to evolution in similar environmental conditions,despite being phylogenetically distinct.
This phenomenon,where unrelated species evolve similar traits as an adaptation to similar environments,is known as convergent evolution.

(C) The correct matching is as follows:
$(a) XX-XO$ method of sex determination is observed in insects like grasshoppers,where males are heterogametic $(XO)$ and females are homogametic $(XX)$. Thus,$(a) - (iii)$.
$(b) XX-XY$ method of sex determination is observed in humans and Drosophila,where females are homogametic $(XX)$ and males are heterogametic $(XY)$. Thus,$(b) - (iv)$.
$(c)$ Karyotype $45$ (specifically $44 + XO$) results in Turner's syndrome,which is a chromosomal disorder in females. Thus,$(c) - (i)$.
$(d) ZW-ZZ$ method of sex determination is observed in birds,where females are heterogametic $(ZW)$ and males are homogametic $(ZZ)$. Thus,$(d) - (ii)$.
Therefore,the correct sequence is $a-iii, b-iv, c-i, d-ii$.