KCET 2026 Biology Question Paper with Answer and Solution

(C) $1$. In an $Acrocentric$ chromosome, the centromere is situated close to one end, resulting in one extremely short arm and one very long arm.
$2$. $Metacentric$ chromosomes have the centromere in the middle.
$3$. $Submetacentric$ chromosomes have the centromere slightly away from the middle, resulting in unequal arms.
$4$. $Telocentric$ chromosomes have the centromere at the terminal end.

(B) Step $1$: The substrate binds to the active site of the enzyme $(d)$.
Step $2$: The substrate binds tightly to the enzyme, inducing a change in the active site of the enzyme $(b)$.
Step $3$: The active site of the enzyme, now in close proximity of the substrate, breaks the chemical bonds of the substrate and a new enzyme-product complex is formed $(a)$.
Step $4$: The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate $(c)$.
Therefore, the correct sequence is $d, b, a, c$.

(B) $1$. Meiosis is a reductional division where the chromosome number is halved.
$2$. In meiosis-$I$, the homologous chromosomes separate, resulting in two daughter cells, each containing half the number of chromosomes of the parent cell.
$3$. The parent cell is diploid $(2n = 8)$.
$4$. After meiosis-$I$, each daughter cell will have $n = 8 / 2 = 4$ chromosomes.
$5$. Therefore, the number of chromosomes in each daughter cell at the end of meiosis-$I$ is $4$.

(C) Cellulose is a linear polysaccharide polymer consisting of several hundred to many thousands of $D-glucose$ units linked by $\beta(1 \rightarrow 4)$ glycosidic bonds. It is the primary structural component of the plant cell wall.

(A) Step $1$: The Glomerular Filtration Rate $(GFR)$ is defined as the amount of filtrate formed by the kidneys per minute.
Step $2$: In a healthy adult human, the kidneys filter approximately $125 \text{ ml}$ of blood per minute.
Step $3$: Therefore, the normal $GFR$ is $125 \text{ ml/minute}$.

(A) Step $1$: Identify the function of plant hormones in pineapples.
Step $2$: Ethylene is a gaseous plant hormone known to promote fruit ripening and is specifically used in agriculture to induce flowering and synchronize fruit set in pineapples.
Step $3$: Therefore, the correct hormone is Ethylene.

(C) Step $1$: The trachea, primary, secondary, and tertiary bronchi, and initial bronchioles are supported by incomplete '$C$' shaped cartilaginous rings, not '$O$' shaped rings.
Step $2$: These rings prevent the collapse of the airways during inhalation.
Step $3$: Therefore, the statement claiming they are '$O$' shaped is incorrect.

(C) Step $1$: The '$P$' wave in an $ECG$ represents the electrical excitation or depolarisation of the atria, which leads to the contraction of both atria.
Step $2$: The '$QRS$' complex represents the depolarisation of the ventricles, which initiates the ventricular contraction.
Step $3$: The '$T$' wave represents the return of the ventricles from an excited to a normal state, i.e., repolarisation of the ventricles.
Step $4$: Therefore, Statement $I$ is incorrect because '$P$' wave represents atrial depolarisation, and Statement $II$ is correct because '$T$' wave represents ventricular repolarisation.

(C) Step $1$: Aldosterone is a mineralocorticoid hormone secreted by the adrenal cortex.
Step $2$: Its primary function is to regulate electrolyte balance and blood pressure.
Step $3$: It acts on the distal parts of the nephron, specifically the Distal Convoluted Tubule $(DCT)$ and the collecting duct.
Step $4$: It promotes the reabsorption of $Na^+$ and water into the blood, while facilitating the excretion of $K^+$ and phosphate ions.
Step $5$: Therefore, the correct site is the Distal Convoluted Tubule $(DCT)$.

(A) $1$. Fibrous joints are immovable joints found in the skull, e.g., $A$ = Cranial sutures.
$2$. Cartilaginous joints allow limited movement and are found between adjacent vertebrae, e.g., $B$ = Cartilaginous joint.
$3$. Synovial joints are freely movable. The joint between atlas and axis is a Pivot joint, e.g., $C$ = Pivot joint.
$4$. $A$ Hinge joint is a type of synovial joint, e.g., $D$ = Knee joint.
Therefore, the correct sequence is $A$ - Cranial sutures, $B$ - Cartilaginous joint, $C$ - Pivot joint, $D$ - Knee joint.

(B) $1$. $A$ neuron is classified based on the number of processes (axons and dendrites) extending from the cell body.
$2$. In a unipolar neuron, there is only one process arising from the cell body, which acts as an axon.
$3$. These are typically found in the embryonic stage of development.
$4$. Therefore, they consist of a cell body and one axon.

(A) Step $1$: Gastrin is a peptide hormone primarily secreted by the $G$-cells located in the pyloric antrum of the stomach.
Step $2$: Upon release into the bloodstream, gastrin acts on the parietal cells (oxyntic cells) to stimulate the secretion of $HCl$ (hydrochloric acid).
Step $3$: Gastrin also stimulates the chief cells (peptic cells) to secrete pepsinogen.
Step $4$: Therefore, gastrin stimulates the secretion of both hydrochloric acid and pepsinogen.

(D) $1$. Bisexual: $\text{male}$
$2$. Actinomorphic: $\oplus$
$3$. Pentamerous: Parts in multiples of $5$
$4$. Gamosepalous: $K_{(5)}$
$5$. Gamopetalous: $C_{(5)}$
$6$. Epipetalous stamens: $A_5$ attached to $C_{(5)}$, represented as $C_{(5)}A_5$
$7$. Bicarpellary, syncarpous, superior ovary: $\underline{G}_{(2)}$
Combining these, the floral formula is $\oplus \, \text{male} \, K_{(5)} \, C_{(5)} \, A_5 \, \underline{G}_{(2)}$.

(C) Step $1$: In mitosis, the splitting of the centromere marks the beginning of $Anaphase$.
Step $2$: During $Anaphase$, the sister chromatids (now called daughter chromosomes) separate and move towards the opposite poles of the cell.
Step $3$: Since both observations describe these specific events, the stage is $Anaphase$.

(B) According to the rules of binomial nomenclature:
$1$. Biological names are generally in Latin and written in italics.
$2$. The first word represents the genus and the second word represents the specific epithet.
$3$. When handwritten, both words are separately underlined.
$4$. The genus starts with a capital letter, and the specific epithet starts with a small letter.
Therefore, statement $(B)$ is incorrect because it reverses the order of the genus and the specific epithet.

(D) $1$. The organism is eukaryotic and multicellular.
$2$. It is heterotrophic, which rules out Plantae.
$3$. It possesses a chitinous cell wall, which is a characteristic feature of the kingdom $Fungi$.
$4$. Therefore, organism '$X$' belongs to kingdom $Fungi$.

(A) Step $1$: In $Funaria$ (a moss), the gametophyte is the dominant, independent, and photosynthetic phase.
Step $2$: The sporophyte is dependent on the gametophyte for nutrition and is differentiated into foot, seta, and capsule.
Step $3$: Statement $(a)$ is correct as the gametophyte is dominant.
Step $4$: Statement $(b)$ is correct as the sporophyte is differentiated into foot, seta, and capsule.
Step $5$: Statement $(c)$ is incorrect because $gemmae$ are found in liverworts like $Marchantia$, not in $Funaria$.
Step $6$: Statement $(d)$ is incorrect because the sporophyte in $Funaria$ is dependent on the gametophyte, not independent.
Conclusion: Statements $(a)$ and $(b)$ are correct.

(C) Step $1$: Identify the excretory organs for each animal.
Step $2$: Leech $(Annelida)$ uses $Nephridia$ (iii) for excretion.
Step $3$: Locust $(Arthropoda)$ uses $Malpighian$ $tubules$ (iv) for excretion.
Step $4$: Liver fluke $(Platyhelminthes)$ uses $Flame$ $cells$ $(i)$ for excretion.
Step $5$: Balanoglossus $(Hemichordata)$ uses $Proboscis$ $gland$ (ii) for excretion.
Step $6$: Matching the pairs: $a-iii, b-iv, c-i, d-ii$.

(C) $1$. Polyadelphous condition refers to the arrangement where stamens are united into more than two bundles.
$2$. In $Mustard$ $(Brassicaceae)$, the condition is tetradynamous.
$3$. In $Chinarose$ $(Malvaceae)$, the condition is monadelphous (stamens united into one bundle).
$4$. In $Citrus$ $(Rutaceae)$, the stamens are united into more than two bundles, which is known as polyadelphous.
$5$. In $Pea$ $(Fabaceae)$, the condition is diadelphous (stamens united into two bundles).
$6$. Therefore, the correct answer is $Citrus$.

(B) $1$. In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells.
$2$. These cells are known as $Bulliform$ $cells$.
$3$. When these cells absorb water and become turgid, they make the leaf surface exposed.
$4$. When they lose water due to stress, they become flaccid, which causes the leaves to curl inwards to minimise water loss.

(B) Step $1$: In a frog, the oesophagus is a short tube that opens into the stomach.
Step $2$: The stomach continues as the intestine, rectum, and finally opens outside by the cloaca.
Step $3$: Option $B$ is incorrect because the oesophagus is described as a 'long' tube, whereas it is actually a 'short' tube in frogs.

(D) Step $1$: The plasma membrane is a selectively permeable barrier that regulates the transport of substances into and out of the cell.
Step $2$: Endocytosis is a process by which the plasma membrane engulfs external materials, which is a known function.
Step $3$: The plasma membrane facilitates the formation of intercellular junctions (like tight junctions or desmosomes) to connect adjacent cells.
Step $4$: Secretion (exocytosis) is a process mediated by the plasma membrane to release substances from the cell.
Step $5$: $DNA$ synthesis occurs within the nucleus (or cytoplasm in prokaryotes), not the plasma membrane. Therefore, $DNA$ synthesis is not a function of the plasma membrane.

(B) The wall of a microsporangium in an angiosperm consists of four layers:
$1$. The outermost layer is the $Epidermis$.
$2$. The second layer is the $Endothecium$.
$3$. The middle layers consist of $1-3$ layers of cells.
$4$. The innermost layer is the $Tapetum$.
Therefore, the correct sequence from outside to inside is $Epidermis \rightarrow Endothecium \rightarrow Middle \ layer \rightarrow Tapetum$.

(C) $1$. Cleistogamous flowers do not open at all, so anthers and stigma are not exposed. Thus, statement $(a)$ is incorrect.
$2$. Since they are closed, self-pollination occurs automatically, ensuring seed set even in the absence of pollinators. Thus, statement $(b)$ is correct.
$3$. They do not require pollinators because pollination occurs within the closed flower. Thus, statement $(c)$ is incorrect.
$4$. Due to self-pollination, there is no genetic recombination, so variations are not seen in the next generation. Thus, statement $(d)$ is correct.
$5$. Therefore, statements $(b)$ and $(d)$ are correct.

(B) The skeletal structure shown in the image represents a cannabinoid molecule, specifically Cannabinol or a related cannabinoid structure. Cannabinoids are a group of chemicals that interact with cannabinoid receptors in the brain. They are primarily known for their effects on the cardiovascular system of the body. Therefore, the correct option is $B$.

(B) The mammary glands contain alveoli which secrete milk.
Milk from the alveoli passes into the mammary tubules.
The tubules from each lobe join to form a mammary duct.
Several mammary ducts join to form a wider mammary ampulla, which is connected to the lactiferous duct through which milk is sucked out.
Thus, the correct sequence is: Alveoli $\rightarrow$ mammary tubules $\rightarrow$ mammary duct $\rightarrow$ ampulla $\rightarrow$ lactiferous ducts.

(D) Step $1$: The $Acrosome$ $(a)$ is a cap-like structure on the sperm head filled with enzymes that help in fertilization $(iii)$.
Step $2$: The $Head$ $(b)$ contains the elongated haploid nucleus $(iv)$.
Step $3$: The $Middle \ piece$ $(c)$ contains numerous mitochondria which produce energy for sperm movement $(i)$.
Step $4$: The $Tail$ $(d)$ facilitates sperm motility $(ii)$.
Step $5$: Matching these gives $a-iii, b-iv, c-i, d-ii$.

(A) Seminal plasma is the fluid component of semen, which is rich in fructose, calcium, and certain enzymes. It is secreted by the accessory glands of the male reproductive system, which include the seminal vesicles, the prostate gland, and the bulbourethral glands. The epididymis is the site where sperm mature and are stored, but it does not contribute to the production of seminal plasma.

(D) $1$. Sexually Transmitted Infections (STIs) like $AIDS$, Hepatitis $B$, and Genital herpes are caused by viruses.
$2$. Viral infections are generally not completely curable because the virus integrates into the host genome or remains latent in the body.
$3$. While treatments exist to manage symptoms and reduce viral load, these specific infections cannot be eradicated from the body completely.
$4$. Therefore, $AIDS$ and Genital herpes are considered not completely curable.

(C) Step $1$: 'Saheli' is a new oral contraceptive for females developed by the Central Drug Research Institute $(CDRI)$ in Lucknow, India.
Step $2$: It is a non-steroidal preparation, meaning it does not contain hormones.
Step $3$: It is taken as a 'once-a-week' pill.
Step $4$: It is known for having very few side effects and high contraceptive value.
Step $5$: Therefore, the statement that it has 'low contraceptive value' is incorrect.

(B) $1$. The problem states that the female is normal, but the male partner has a low sperm count (oligospermia) and is unable to inseminate the female.
$2$. $IUI$ (Intra Uterine Insemination) is a technique used when the male partner is unable to inseminate the female or has a very low sperm count.
$3$. In this procedure, semen collected either from the husband or a healthy donor is artificially introduced into the uterus of the female.
$4$. Therefore, $IUI$ is the most appropriate $ART$ for this case.

(B) Step $1$: Perform the cross between the parents with genotypes $I^A I^B$ and $I^A I^O$.
Step $2$: The possible combinations are:
- $I^A I^A$ (Phenotype: Blood group $A$)
- $I^A I^O$ (Phenotype: Blood group $A$)
- $I^A I^B$ (Phenotype: Blood group $AB$)
- $I^B I^O$ (Phenotype: Blood group $B$)
Step $3$: Count the distinct genotypes: $I^A I^A$, $I^A I^O$, $I^A I^B$, and $I^B I^O$ (Total $4$ genotypes).
Step $4$: Count the distinct phenotypes: Blood group $A$, Blood group $AB$, and Blood group $B$ (Total $3$ phenotypes).
Conclusion: There are $4$ genotypes and $3$ phenotypes possible.

(B) Step $1$: Point mutation is a genetic mutation where a single nucleotide base is changed, inserted, or deleted from a $DNA$ or $RNA$ sequence.
Step $2$: Sickle cell anaemia is caused by a single base pair substitution in the gene for $ \beta $-globin chain of haemoglobin, where $ GAG $ is replaced by $ GUG $ at the sixth codon position.
Step $3$: This leads to the substitution of glutamic acid with valine, making it a classic example of point mutation.

(C) Step $1$: Statement $I$ is true. $A$ test cross is defined as the cross between an individual with a dominant phenotype (but unknown genotype) and a homozygous recessive parent to determine the genotype of the dominant individual.
Step $2$: Statement $II$ is false. In a test cross, if the unknown organism is homozygous dominant $(AA)$, all offspring will show the dominant trait $(Aa)$. If the unknown organism is heterozygous $(Aa)$, the offspring ratio will be $1:1$ (dominant to recessive). The ratio does not indicate the organism is homozygous dominant; rather, a $1:1$ ratio indicates it is heterozygous, and a $100\%$ dominant ratio indicates it is homozygous dominant.
Step $3$: Therefore, Statement $I$ is true and Statement $II$ is false.

(D) Step $1$: Sickle cell anaemia is a qualitative defect where a single amino acid substitution ($Glu$ to $Val$ at the $6^{th}$ position of the $\beta$-globin chain) occurs, leading to abnormal haemoglobin structure.
Step $2$: Thalassemia is a quantitative defect where there is reduced or absent synthesis of one of the globin chains ($\alpha$ or $\beta$) of the haemoglobin molecule.
Step $3$: Comparing the options, option $(D)$ correctly identifies that Thalassemia is caused by reduced synthesis of globin molecules.

(C) Step $1$: $James \ Watson$ and $Francis \ Crick$ proposed the double-helix model of $DNA$ in $1953$.
Step $2$: $Maurice \ Wilkins$ and $Rosalind \ Franklin$ provided the $X$-ray diffraction data that was crucial for determining the structure.
Step $3$: $Erwin \ Chargaff$ provided the base-pairing rules ($A=T$ and $G=C$), which were essential for the model.
Step $4$: $Meselson$ and $Stahl$ performed the experiment in $1958$ to prove the semi-conservative mode of $DNA$ replication, which occurred after the model was proposed.
Step $5$: Therefore, $Meselson$ and $Stahl$ had no contribution to the development of the double-helix model.

(C) According to Chargaff's rule, the amount of adenine $(A)$ is equal to the amount of thymine $(T)$, and the amount of cytosine $(C)$ is equal to the amount of guanine $(G)$.
Given: $C = 28\%$.
Since $C = G$, then $G = 28\%$.
The total percentage of all bases is $A + T + C + G = 100\%$.
Since $A = T$, we can write $2A + 2C = 100\%$.
Substituting the value of $C$: $2A + 2(28\%) = 100\%$.
$2A + 56\% = 100\%$.
$2A = 100\% - 56\% = 44\%$.
$A = 22\%$.
Therefore, the percentage of adenine is $22\%$.

(A) The correct sequence of steps in $DNA$ fingerprinting is:
$1$. Isolation of $DNA$ $(d)$
$2$. Digestion of $DNA$ by restriction endonucleases $(a)$
$3$. Separation of $DNA$ fragments by electrophoresis $(e)$
$4$. Blotting of $DNA$ fragments to nitrocellulose membrane $(c)$
$5$. Autoradiography $(b)$
Thus, the correct sequence is $(d)$, $(a)$, $(e)$, $(c)$, $(b)$.

(C) Step $1$: The Big Bang Theory suggests that the Universe originated from a singular, extremely hot, and dense point.
Step $2$: As the Universe expanded, it cooled down, allowing for the formation of subatomic particles and eventually simple atoms like Hydrogen and Helium.
Step $3$: Therefore, Hydrogen and Helium gases formed after the expansion began, not before the explosion.
Step $4$: Thus, the statement that 'Hydrogen and Helium gases formed before the explosion' is incorrect.

(D) Hugo De Vries proposed the mutation theory of evolution. According to him, evolution is a discontinuous process caused by large, sudden, and heritable changes in the genetic material, which he termed as 'saltation' or 'single-step large mutation'. Unlike Darwin, who believed in the accumulation of small variations, De Vries emphasized that these large mutations are the primary cause of speciation.

(B) The chronological order of human evolution is as follows:
$1$. $Dryopithecus$ (approx. $15$ million years ago)
$2$. $Australopithecus$ (approx. $2$ million years ago)
$3$. $Homo$ $habilis$ (first human-like being)
$4$. $Homo$ $erectus$ (evolved from $Homo$ $habilis$)
$5$. $Neanderthal$ $man$ (lived near East and Central Asia)
Therefore, the correct sequence is $e \rightarrow c \rightarrow a \rightarrow b \rightarrow d$.

(B) Step $1$: The exine of pollen grains is composed of a highly resistant organic material called $Sporopollenin$.
Step $2$: $Sporopollenin$ is one of the most resistant organic materials known. It can withstand high temperatures, strong acids, and alkali.
Step $3$: No enzyme is known that degrades $Sporopollenin$. Due to this, pollen grains are well preserved as fossils.

(B) $1$. Pneumonia is a respiratory disease caused by bacteria such as $Streptococcus \ pneumoniae$ and $Haemophilus \ influenzae$.
$2$. $Wuchereria \ malayi$ causes filariasis.
$3$. $Salmonella \ typhi$ causes typhoid fever.
$4$. $Trichophyton$ is a fungus responsible for ringworm infections.
$5$. Therefore, $Haemophilus \ influenzae$ is the correct causative organism for pneumonia among the given options.

(B) $1$. Lymphoid organs are classified into primary and secondary lymphoid organs.
$2$. Primary lymphoid organs include the $Bone \ marrow$ and $Thymus$, where lymphocytes mature and differentiate.
$3$. Secondary lymphoid organs include the $Spleen$, $Lymph \ nodes$, $Tonsils$, $Peyer’s \ patches$, and $Appendix$. These are sites where lymphocytes interact with antigens and proliferate.
$4$. Option $B$ contains only secondary lymphoid organs.

(B) Step $1$: Biological response modifiers are substances that stimulate or restore the ability of the immune system to fight infection and disease.
Step $2$: $\alpha$-Interferons are cytokines produced by virus-infected cells that protect non-infected cells from further viral infection.
Step $3$: They are also used as biological response modifiers in cancer treatment because they activate the immune system to destroy tumor cells.
Step $4$: Therefore, the correct option is $\alpha$-Interferon.

(B) Step $1$: $LAB$ (Lactic Acid Bacteria) grow in milk and convert it into curd.
Step $2$: During this process, $LAB$ also produce acids that coagulate and partially digest the milk proteins.
Step $3$: $A$ small amount of $LAB$ added to fresh milk as an inoculum or starter contains millions of $LAB$, which at suitable temperatures multiply and convert milk to curd, which also improves its nutritional quality by increasing the content of Vitamin $B_{12}$.

(C) $BOD$ stands for Biochemical Oxygen Demand. It is defined as the amount of dissolved oxygen that would be consumed by microorganisms in $1 \text{ litre}$ of water if all the organic matter in it were oxidized by bacteria. Therefore, it is measured by the amount of oxygen consumed.

(D) Step $1$: Analyze Statement $I$. Baculoviruses are pathogens that attack insects and other arthropods, not nematodes. Therefore, Statement $I$ is false.
Step $2$: Analyze Statement $II$. Baculoviruses are excellent candidates for integrated pest management because they are species-specific and have no negative impact on plants, mammals, birds, fish, or even non-target insects. This makes them ideal for ecologically sensitive areas. Therefore, Statement $II$ is true.
Step $3$: Conclusion. Since Statement $I$ is false and Statement $II$ is true, the correct option is $D$.

(C) $1$. Genetic engineering is the technique that involves the manipulation of genetic material ($DNA$ or $RNA$) to alter the chemistry of genetic material.
$2$. This modified genetic material is then introduced into host organisms.
$3$. The introduction of this material changes the phenotype of the host organism.
$4$. Therefore, the correct option is $C$.

(C) Step $1$: In the plasmid $pBR322$, the $BamHI$ restriction site is located within the $tet^R$ (tetracycline resistance) gene.
Step $2$: If a desired $DNA$ fragment is inserted at the $BamHI$ site, the $tet^R$ gene is inactivated due to insertional inactivation.
Step $3$: Consequently, the recombinant $E. coli$ will lose its resistance to tetracycline and will not be able to grow in a medium containing tetracycline.
Step $4$: Therefore, Statement $I$ is true, and Statement $II$ is false.

(A) Step $1$: Visualization of $DNA$ in gel electrophoresis is done using $Ethidium \ bromide$ followed by exposure to $UV$ radiation.
Step $2$: Precipitation of $DNA$ is achieved by adding chilled ethanol to the solution.
Step $3$: Breaking of the bacterial cell wall requires the enzyme $Lysozyme$.
Step $4$: Breaking of the fungal cell wall requires the enzyme $Chitinase$.
Step $5$: Matching these gives: $a-iv, b-iii, c-ii, d-i$.

(C) $1$. Emphysema is a chronic respiratory disease caused by the deficiency of the protein $\alpha - 1 \text{ antitrypsin}$.
$2$. Transgenic animals, such as sheep, have been developed to produce this human protein in their milk.
$3$. This protein is extracted and used for the treatment of emphysema.
$4$. Therefore, the correct option is $\alpha - 1 \text{ antitrypsin}$.

(B) Step $1$: Two $DNA$ sequences corresponding to $A$ and $B$ chains of human insulin are prepared.
Step $2$: These sequences are introduced into plasmids of $E. coli$ to produce the chains.
Step $3$: The $A$ and $B$ chains are produced separately in the bacterial cultures.
Step $4$: The chains are extracted and combined by creating disulphide bonds to form mature insulin.
The correct sequence is $(a) \rightarrow (c) \rightarrow (b) \rightarrow (d)$.

(C) Step $1$: Statement $I$ is true. $Adenosine \ deaminase$ $(ADA)$ is an enzyme essential for the proper functioning of the immune system.
Step $2$: Statement $II$ is false. While bone marrow transplantation is one method, $ADA$ deficiency can also be treated by enzyme replacement therapy, where functional $ADA$ is injected into the patient. Gene therapy is another potential curative approach.
Step $3$: Since Statement $I$ is true and Statement $II$ is false, the correct option is $C$.

(A) $1$. The exponential growth model assumes that resources are unlimited in the environment.
$2$. In this model, the population grows in a geometric fashion $(dN/dt = rN)$.
$3$. Because resources are unlimited, the population continues to grow indefinitely, and a stationary phase is never reached.
$4$. Option $(A)$ states that resources are limited, which is a characteristic of the logistic growth model, not the exponential growth model. Therefore, it is the incorrect statement.

(B) $1$. Age pyramids represent the distribution of various age groups in a population.
$2$. Pyramid $A$ has a bell-shaped structure where pre-reproductive and reproductive age groups are roughly equal, indicating a stable population.
$3$. Pyramid $B$ is urn-shaped, where the pre-reproductive age group is smaller than the reproductive age group, indicating a declining population.
$4$. Pyramid $C$ is triangular, where the pre-reproductive age group is the largest, indicating an expanding population.
$5$. Thus, the correct sequence is $A$ - stable, $B$ - declining, $C$ - expanding.

(B) The process of decomposition involves five main steps in a specific sequence:
$1$. Fragmentation: Detritivores break down detritus into smaller particles.
$2$. Leaching: Water-soluble inorganic nutrients go down into the soil horizon.
$3$. Catabolism: Bacterial and fungal enzymes degrade detritus into simpler inorganic substances.
$4$. Humification: Formation of dark-colored amorphous substance called humus.
$5$. Mineralization: Degradation of humus by microbes to release inorganic nutrients.
Therefore, the correct sequence is Fragmentation $\rightarrow$ Leaching $\rightarrow$ Catabolism $\rightarrow$ Humification $\rightarrow$ Mineralization.

(B) Step $1$: The total annual Net Primary Productivity $(NPP)$ of the biosphere is approximately $170$ billion tons (dry weight).
Step $2$: Out of this, the productivity on land is approximately $115$ billion tons.
Step $3$: The productivity in the oceans is approximately $55$ billion tons.
Step $4$: Therefore, the values for land and ocean respectively are $115$ and $55$ billion tons.

(B) $1$. Observe the pedigree: The trait appears in every generation (vertical transmission), which is characteristic of a dominant trait.
$2$. Check for sex-linkage: An affected mother has an unaffected son, and an affected father would pass the trait to all daughters if it were $X$-linked dominant. Here, an affected mother has an unaffected son, which is possible in both autosomal and $X$-linked dominant traits. However, the transmission from an affected mother to both sons and daughters, and the presence of the trait in both sexes, strongly suggests an autosomal dominant pattern.
$3$. Conclusion: Since the trait does not skip generations and appears in both males and females, it is autosomal dominant.

(D) Step $1$: Analyze Statement $I$. Tropical latitudes harbor significantly more species biodiversity compared to temperate regions due to favorable climatic conditions. Thus, Statement $I$ is false.
Step $2$: Analyze Statement $II$. Tropical environments are characterized by being less seasonal, relatively more constant, and predictable, which allows for niche specialization and higher biodiversity. Thus, Statement $II$ is true.
Step $3$: Conclusion. Since Statement $I$ is false and Statement $II$ is true, the correct option is $D$.