AIPMT 2004 Biology Question Paper with Answer and Solution

(C) Chloroplasts are double-membrane-bound organelles found in plant cells.
Inside the chloroplast,there is a system of flattened,sac-like structures called thylakoids.
The pigments,including chlorophyll,are embedded within the thylakoid membranes.
These pigments are essential for capturing light energy during the process of photosynthesis.
Therefore,the correct location of chlorophyll is the thylakoids.

(C) In the resting state of a neural membrane,the axonal membrane is significantly more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$.
Due to the concentration gradient,$K^+$ ions tend to diffuse out of the cell,while $Na^+$ ions are prevented from entering.
However,if we consider the permeability and the electrochemical gradient,the resting membrane potential is primarily maintained by the leakage of $K^+$ out of the cell.
If the question asks which ion is allowed to enter the cell during specific conditions or if we look at the permeability of the membrane,$Na^+$ is generally restricted,but the question implies the movement across the membrane.
Given the options provided and the standard physiological context,$Na^+$ is the ion that would enter the cell if the membrane permeability were to change (as in an action potential),but in the resting state,the membrane is selectively permeable to $K^+$.
However,based on standard textbook questions of this type,the correct answer is $Na^+$ entering the cell when the membrane becomes permeable during depolarization.

(A) The $Vagus$ nerve (Cranial nerve $X$) is a key component of the parasympathetic nervous system.
It innervates the heart,lungs,and the digestive tract (including the stomach and intestines) and regulates functions like heart rate,gastrointestinal motility,and glandular secretions (such as pancreatic secretion).
Tongue movements are primarily controlled by the $Hypoglossal$ nerve (Cranial nerve $XII$).
Therefore,injury to the $Vagus$ nerve will not affect tongue movements.

(A) Hormones are classified based on their chemical nature.
$1$. Epinephrine (also known as Adrenaline) is derived from the amino acid Tyrosine.
$2$. Progesterone and Estrogen are steroid hormones derived from cholesterol.
$3$. Prostaglandins are lipid derivatives (fatty acids).
Therefore,Epinephrine is the correct answer.

(A) $1$. $Luteinizing$ $hormone$ $(LH)$ is essential for ovulation in females. Its deficiency leads to the failure of ovulation.
$2$. $Diabetes$ $insipidus$ is caused by the deficiency of $Vasopressin$ $(ADH)$,not $Insulin$.
$3$. $Tetany$ is caused by the deficiency of $Parathyroid$ $hormone$ $(PTH)$,not $Thyroxine$.
$4$. $Diabetes$ $mellitus$ is caused by the deficiency of $Insulin$,not $Parathyroid$ $hormone$.
Therefore,the correct match is $Luteinizing$ $hormone$ and failure of ovulation.

(D) The term 'constant of oil' refers to the specific physical and chemical properties of oils,such as the iodine value,which measures the degree of unsaturation. Erucic acid $(C_{22}H_{42}O_2)$ is a monounsaturated omega-$9$ fatty acid found in high concentrations in rapeseed and mustard oils. It is often used as a marker or constant in the analysis of these specific oils to determine their purity and composition.

(A) The elemental analysis of plant tissues reveals that carbon $(C)$,hydrogen $(H)$,and oxygen $(O)$ are the most abundant elements found in plants. Among these,carbon is the primary structural component of all organic molecules,including carbohydrates,proteins,lipids,and nucleic acids,which constitute the bulk of the plant's dry weight. Therefore,carbon is present in the highest amount compared to other elements like nitrogen,manganese,or iron.

(B) Crossing over is a biological process that occurs during the $pachytene$ stage of $prophase-I$ of $meiosis$.
It involves the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process leads to genetic recombination,which is essential for creating genetic diversity in offspring.
Therefore,the correct answer is the non-sister chromatids of a bivalent.

(B) The cell cycle consists of two main phases: Interphase and $M$-phase (Mitosis).
Interphase is further divided into $G_1$,$S$,and $G_2$ phases.
$1$. $G_1$ phase (Gap $1$): The cell grows and prepares for $DNA$ replication.
$2$. $S$ phase (Synthesis): $DNA$ replication occurs,doubling the amount of $DNA$ per cell.
$3$. $G_2$ phase (Gap $2$): The cell prepares for mitosis by synthesizing proteins and organelles.
$4$. $M$-phase: The actual division of the nucleus and cytoplasm occurs.
Therefore,the statement that $DNA$ replication occurs in the $S$ phase is correct.

(A) During the $M$ phase,specifically at the end of telophase,the nuclear envelope reforms around the daughter chromosomes. This process is initiated as the chromosomes begin to decondense (lose their condensed state) and become chromatin again. The nuclear lamina and nuclear pore complexes assemble to enclose the genetic material,effectively reforming the nuclear envelope.

(A) To count the number of chromosomes in a cell,the best stage is $Metaphase$.
During $Metaphase$,the chromosomes align at the equatorial plate (metaphase plate) of the cell.
At this stage,the chromosomes are most condensed,clearly visible,and distinct,making it the ideal phase for morphological study and counting of chromosomes.

(A) Chlorophyll is the primary photosynthetic pigment responsible for capturing light energy.
In the chloroplast,chlorophyll molecules are embedded within the thylakoid membranes.
These thylakoid membranes are stacked to form structures known as $Grana$.
Therefore,chlorophyll is specifically located in the $Grana$ of the chloroplast,where the light-dependent reactions of photosynthesis occur.

(A) When plants grow in low-intensity light (shade conditions),they often undergo morphological and physiological adaptations to maximize light capture. One such adaptation is the development of broader leaves or an increase in the size of photosynthetic units to increase the surface area for light absorption. This allows the plant to maintain an efficient rate of photosynthesis despite the limited light availability.

(C) In $C_3$ plants,the primary carbon fixation step occurs during the Calvin cycle.
During this process,the enzyme $RuBisCO$ catalyzes the carboxylation of $RuBP$ $(Ribulose-1,5-bisphosphate)$ with $CO_2$.
This reaction results in the formation of two molecules of $3$-phosphoglyceric acid ($3$-$PGA$),which is a $3$-carbon compound.
$3$-$PGA$ is the first stable product of the Calvin cycle in $C_3$ plants.

(B) The duodenum secretes two primary hormones,$Secretin$ and $Cholecystokinin$ $(CCK)$,which play a crucial role in the digestive process.
$Secretin$ stimulates the pancreas to release bicarbonate-rich pancreatic juice and also stimulates the Brunner's glands (located in the submucosa of the duodenum) to secrete mucus and enzymes.
$Cholecystokinin$ $(CCK)$ stimulates the gallbladder to release bile and the pancreas to release digestive enzymes.
Therefore,the correct option is $B$.

(D) The respiratory rhythm center in the medulla region of the brain is primarily responsible for the regulation of respiration. This center is highly sensitive to $CO_2$ and hydrogen ions. An increase in the concentration of $CO_2$ in the blood stimulates the chemosensitive area,which in turn signals the respiratory rhythm center to make necessary adjustments. This results in an increase in the rate and depth of breathing to eliminate excess $CO_2$ from the body. Therefore,breathing becomes faster and deeper.

(D) Carboxyhemoglobin is formed when carbon monoxide $(CO)$ binds to hemoglobin in the blood.
Carbon monoxide has an affinity for hemoglobin that is approximately $200-250$ times greater than that of oxygen $(O_2)$.
When $CO$ is inhaled,it binds to hemoglobin to form a stable compound called carboxyhemoglobin,which prevents the blood from carrying oxygen effectively to the body tissues.
Therefore,high levels of carboxyhemoglobin in the blood indicate exposure to high concentrations of carbon monoxide.

(C) The heart's rhythmic contraction is initiated by a specialized patch of cardiac muscle fibers called the $SA$ node (Sinoatrial node).
It is located in the right upper corner of the right atrium.
Because it generates the action potential that sets the pace for the heart's contraction,it is known as the natural pacemaker of the heart.

(A) To separate blood into plasma and blood cells,the blood must be kept in a liquid state (prevented from clotting).
$1$. Heparin and sodium oxalate are anticoagulants that prevent blood clotting by removing or chelating calcium ions,which are essential for the coagulation cascade.
$2$. $A$ sterile test tube is standard for collection.
$3$. Calcium carbonate is a source of calcium ions. Adding calcium to blood promotes the coagulation process,which would cause the blood to clot,making it impossible to separate plasma and cells effectively. Therefore,a test tube containing calcium carbonate should not be used.

(C) In the $ABO$ blood grouping system,the presence of antigens on the surface of red blood cells determines the blood group.
$1$. Individuals with blood group $A$ have antigen $A$ and antibody $b$ in their plasma.
$2$. Individuals with blood group $B$ have antigen $B$ and antibody $a$ in their plasma.
$3$. Individuals with blood group $AB$ have both antigens $A$ and $B$ on their red blood cells but do not have any antibodies ($a$ or $b$) in their plasma.
$4$. Individuals with blood group $O$ do not have any antigens on their red blood cells but have both antibodies $a$ and $b$ in their plasma.
Therefore,the blood group with both antigens present and no antibodies is $AB$.

(D) The aquatic fern $Azolla$ maintains a symbiotic relationship with the cyanobacterium $Anabaena$ $azollae$.
This cyanobacterium lives in the leaf cavities of the fern.
It fixes atmospheric nitrogen,which is then utilized by the fern for its growth.
Therefore,$Anabaena$ is the correct answer.

(B) The correct answer is $B$.
$1$. Laterite soils are rich in iron and aluminum oxides,making the statement in $A$ correct.
$2$. Terra rossa is a reddish clay soil produced by the weathering of limestone; it is not specifically associated with roses,making the statement in $B$ incorrect.
$3$. Chernozems are black-colored soils containing a high percentage of humus and are known as some of the most fertile soils in the world,making the statement in $C$ correct.
$4$. Black soils (Regur soils) are typically rich in calcium carbonate,magnesium,potash,and lime,making the statement in $D$ correct.

(A) $Nitrogenase$ is a critical enzyme complex responsible for the biological nitrogen fixation process,where atmospheric $N_2$ is converted into ammonia $(NH_3)$.
This process is primarily carried out by diazotrophic bacteria,such as $Rhizobium$,which live in symbiotic association with the root nodules of leguminous plants.
If $Nitrogenase$ enzymes are destroyed,the biological reduction of atmospheric nitrogen cannot take place.
Therefore,leguminous plants will be unable to fix atmospheric nitrogen into a usable form.

(A) The enzyme $Nitrogenase$ is essential for the process of biological nitrogen fixation,where atmospheric $N_2$ is converted into $NH_3$ (ammonia).
Legume plants form a symbiotic relationship with $Rhizobium$ bacteria,which contain the $Nitrogenase$ enzyme to fix atmospheric nitrogen.
If $Nitrogenase$ is inactivated,the plant will be unable to perform biological nitrogen fixation.
Therefore,nitrogen fixation in legume plants will not occur.

(B) The peristome is a specialized structure consisting of a fringe of teeth-like projections located at the mouth of the capsule in mosses.
These teeth help in the gradual dispersal of spores.
Peristome teeth are classified into two main types based on their structure:
$1$. Nematodontous: These are solid structures composed of bundles of dead cells. Examples include $Polytrichum$,$Pogonatum$,and $Tetraphis$.
$2$. Orthodontous: These are composed of thin,membranous,transversely barred teeth,formed by the thickened portions of the cell walls of adjacent cells.
Since both the definition of the peristome and its classification are correct,the Assertion and Reason are both true,but the Reason does not explain why the peristome is a fringe of teeth-like projections.

(A) During the $M-$phase of the cell cycle,specifically during telophase,the nuclear envelope begins to reform around the daughter chromosomes.
Before this reformation,the chromosomes,which were highly condensed during metaphase and anaphase,must undergo decondensation to return to their chromatin state.
Simultaneously,the nuclear lamina,which is a dense fibrillar network inside the nucleus,must reassemble to provide structural support to the newly forming nuclear envelope.
Therefore,the decondensation of chromosomes and the reassembly of the nuclear lamina are the critical events that precede the reformation of the nuclear envelope.

(A) In a $Hemitropous$ ovule,the body of the ovule is placed transversely at a right angle $(90^{\circ})$ to the funicle.
In this type,the nucellus and the embryo sac lie at a right angle to the funicle.
$Orthotropous$ ovules are straight,$Anatropous$ ovules are inverted $(180^{\circ})$,and $Campylotropous$ ovules have a curved body where the nucellus is bent.

(D) In a typical $28$-day menstrual cycle,the cycle is divided into phases.
$1$. The menstrual phase occurs from day $1$ to $5$.
$2$. The follicular (proliferative) phase occurs from day $6$ to $13$.
$3$. Ovulation occurs on approximately the $14$th day,which marks the end of the proliferative phase.
$4$. The luteal (secretory) phase occurs from day $15$ to $28$.
Therefore,ovulation occurs at the end of the proliferative phase.

(B) Mitochondrial $DNA$ is inherited exclusively through the cytoplasm of the egg cell (maternal inheritance).
Since the parent with the mitochondrial mutation is used as the male,the sperm contributes only nuclear $DNA$ and virtually no cytoplasm to the zygote.
Therefore,the mitochondrial mutation is not transmitted to the $F_1$ generation,and consequently,it will not be present in the $F_2$ generation.
Thus,the mutation is not observed in any offspring.

(C) point mutation is a change in a single base pair of $DNA$.
When a purine is replaced by another purine (e.g.,Adenine to Guanine) or a pyrimidine is replaced by another pyrimidine,it is called a transition.
When a purine is replaced by a pyrimidine or vice versa,it is called a transversion.
However,since the question asks for the general category of mutation involving a single base change,'Point mutation' is the most appropriate classification for a substitution mutation.

(C) Independent assortment is a principle described by Mendel,which states that genes for different traits segregate independently during gamete formation.
However,when two genes are located close together on the same chromosome,they tend to be inherited together,a phenomenon known as linkage.
In $Drosophila$,$T.H. Morgan$ observed that genes $A$ and $B$ did not show independent assortment because they were physically linked on the same chromosome.
Therefore,the lack of independent assortment is due to linkage.

(C) In humans,males have only one $X$-chromosome ($XY$ genotype). Therefore,any recessive gene located on the $X$-chromosome is always expressed in males because there is no corresponding allele on the $Y$-chromosome to mask its effect. This condition is known as 'hemizygous'. In females,due to the presence of two $X$-chromosomes ($XX$ genotype),a recessive gene is expressed only if it is present on both $X$-chromosomes.

(A) The genotype of the male is $AaBbX^hY$.
Since genes $A$ and $B$ are autosomal and heterozygous,the possible gametes for these alleles are $AB, Ab, aB, ab$ with a probability of $1/4$ each.
The male is hemizygous for the $hemophilia$ gene $(X^hY)$,meaning he carries the $X^h$ chromosome and the $Y$ chromosome.
The probability of a sperm receiving the $X^h$ chromosome is $1/2$,and the probability of receiving the $Y$ chromosome is $1/2$.
For a sperm to be $abh$,it must contain the $a$ allele,the $b$ allele,and the $h$ allele (which is carried on the $X^h$ chromosome).
The probability of getting $ab$ from the autosomal genes is $1/4$.
The probability of getting the $X^h$ chromosome is $1/2$.
Therefore,the total probability is $1/4 \times 1/2 = 1/8$.

(A) The cross is between $RRTt \times RRTt$.
$1$. For the fruit color gene $(RR \times RR)$,all offspring will be $RR$ (red-fruited).
$2$. For the height gene $(Tt \times Tt)$,the offspring genotypes will be $1 TT : 2 Tt : 1 tt$.
$3$. Phenotypically,$3/4$ $(75\%)$ will be tall ($TT$ and $Tt$) and $1/4$ $(25\%)$ will be dwarf $(tt)$.
$4$. Combining these,all offspring will have red fruit,and $75\%$ will be tall.
Therefore,$75\%$ of the offspring will be tall and red-fruited.

(B) Colorblindness is an $X$-linked recessive disorder.
Let $X^C$ represent the allele for normal vision and $X^c$ represent the allele for colorblindness.
The woman's father was colorblind $(X^cY)$,so she must be a carrier $(X^CX^c)$.
The man is normal $(X^CY)$.
The cross is: $X^CX^c \times X^CY$.
The possible offspring are: $X^CX^C$ (normal daughter),$X^CX^Y$ (normal son),$X^CX^c$ (carrier daughter),and $X^cY$ (colorblind son).
Among the sons,$50\%$ are normal $(X^CY)$ and $50\%$ are colorblind $(X^cY)$.

(C) $DNA$ fingerprinting is a technique used to identify individuals by their respective $DNA$ profiles.
It involves the molecular analysis of $DNA$ samples to detect variations in the nucleotide sequences.
These variations,known as polymorphisms,are unique to each individual (except in identical twins).
Therefore,it is defined as the molecular analysis of different $DNA$ samples using specific techniques like $VNTR$ analysis.

(C) During the process of transcription,$DNA$ acts as a template for the synthesis of $m-RNA$.
According to the base-pairing rules,$Adenine$ $(A)$ pairs with $Uracil$ $(U)$ in $RNA$,and $Thymine$ $(T)$ pairs with $Adenine$ $(A)$.
Given $DNA$ sequence: $ATACG$.
- $A$ pairs with $U$
- $T$ pairs with $A$
- $A$ pairs with $U$
- $C$ pairs with $G$
- $G$ pairs with $C$
Therefore,the resulting $m-RNA$ sequence is $UA UGC$.

(A) According to Chargaff's rules for double-stranded $DNA$,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of guanine $(G)$ is equal to the amount of cytosine $(C)$.
Therefore,$A = T$ and $G = C$.
This implies that $A + G = T + C$.
When we take the ratio of purines $(A+G)$ to pyrimidines $(C+T)$,we get $\frac{A+G}{C+T} = 1$.
Since $A+G$ is always equal to $C+T$,this ratio remains constant for a given species.

(D) The telomeres are the terminal regions of eukaryotic chromosomes that protect the ends from degradation and fusion.
These regions consist of highly conserved,repetitive $DNA$ sequences.
In most eukaryotes,including humans,these telomeric sequences are characterized by a high content of $G$ (Guanine) nucleotides.
For example,the human telomeric repeat sequence is $5'-TTAGGG-3'$,which is rich in Guanine.
Therefore,telomeres are Guanine-rich.

(A) gene is a segment of $DNA$ that codes for a specific polypeptide or protein.
When a mutation occurs at the gene level,the sequence of nucleotides in the $DNA$ changes.
This altered $DNA$ sequence leads to the transcription of an altered $mRNA$ sequence.
Consequently,the translation process results in the synthesis of a protein with a different amino acid sequence,which alters the protein's structure and function.
Since proteins are responsible for expressing traits (phenotypes),a change in protein structure leads to a change in the organism's traits.

(D) In bacteria,$DNA$ replication is typically bidirectional. Starting from the origin of replication $(ori)$,the $DNA$ polymerase enzyme creates replication forks that move in both directions,resulting in $DNA$ synthesis occurring in both directions from the origin.

(D) The $t-RNA$ (transfer $RNA$) molecule is known to have a clover-leaf secondary structure.
In this structure, the molecule folds into loops, including the anticodon loop, the $D$-loop, and the $T\psi C$-loop.
This structure is essential for its function in protein synthesis, where it carries specific amino acids to the ribosome based on the codon sequence of $m-RNA$.

(D) The cyanobacterium $Anabaena$ $azollae$ is a nitrogen-fixing organism that lives in a symbiotic association with the water fern $Azolla$.
It resides within the leaf cavities of $Azolla$,providing fixed nitrogen to the plant in exchange for carbohydrates and a protected environment.
This association is widely used as a biofertilizer in rice cultivation.

(D) $Anthesis$ is defined as the period during which a flower is fully open and functional. It refers to the process of the opening of a flower bud into a mature flower,allowing for pollination to occur. Therefore,the correct option is $D$.

(C) Modern methods for dating fossils and determining evolutionary timelines have moved beyond simple radio-carbon dating. $Electron$ $Spin$ $Resonance$ $(ESR)$ is a powerful technique used to date geological and archaeological materials by measuring the accumulation of trapped electrons. Additionally,the analysis of ancient $DNA$ $(aDNA)$ extracted from fossils provides precise information about the evolutionary relationships and divergence times of various organisms. Therefore,$ESR$ and $DNA$ analysis are the most accurate modern methods used in evolutionary biology.

(D) Molecular evidence,specifically comparing $DNA$ sequences,provides the most accurate phylogenetic relationships. Studies involving $DNA$ extracted from sex chromosomes,autosomes,and mitochondria have shown that the genetic similarity between humans and chimpanzees is approximately $98-99\%$. This extensive genomic analysis confirms that humans are more closely related to chimpanzees than to any other hominid ape.

(D) The presence of gills in the tadpole stage of a frog is an example of recapitulation or evolutionary evidence. According to the theory of evolution,organisms often show ancestral traits during their embryonic or larval development. Since frogs are amphibians that evolved from fish-like ancestors which possessed gills for respiration,the tadpole retains this ancestral feature. This indicates that frogs evolved from gill-bearing ancestors.

(B) According to the Oparin-Haldane hypothesis,the primitive atmosphere of the Earth was a reducing atmosphere.
It contained gases like methane $(CH_4)$,ammonia $(NH_3)$,hydrogen $(H_2)$,and water vapor $(H_2O)$.
Free oxygen $(O_2)$ was absent in the early atmosphere because it was highly reactive and would have immediately combined with other elements to form oxides.
Therefore,the correct option is $B$.

(C) $T$-lymphocytes are a type of white blood cell that plays a central role in cell-mediated immunity.
They originate in the bone marrow and mature in the thymus gland.
There are three main functional types of $T$-lymphocytes:
$1$. Helper $T$-cells $(T_h)$: These activate other immune cells.
$2$. Cytotoxic $T$-cells $(T_c)$: These directly kill infected or cancerous cells.
$3$. Suppressor $T$-cells $(T_s)$: These regulate the immune response and prevent autoimmunity by suppressing the activity of other immune cells.
Therefore,option $C$ is the correct statement.

(B) $1$. $Streptomyces$ is a genus of bacteria known for producing various antibiotics.
$2$. $Spirulina$ is a cyanobacterium used as a source of single-cell protein $(SCP)$.
$3$. $Rhizobium$ is a symbiotic bacterium that fixes atmospheric nitrogen, acting as a biofertilizer.
$4$. $Serratia$ is a genus of Gram-negative bacteria; it is not related to 'drug addicts'. The term 'drug addict' is unrelated to this biological classification. Therefore, the pair $Serratia - \text{Drug addict}$ is incorrectly matched.

(D) The correct answer is $D$.
$Glossina \text{ } palpalis$ (Tsetse fly) is the vector for African sleeping sickness (Trypanosomiasis).
$Culex \text{ } pipiens$ is a common vector for $Wuchereria \text{ } bancrofti$, which causes Filariasis.
$Aedes \text{ } aegypti$ is the primary vector for Yellow fever and Dengue.
$Anopheles \text{ } stephensi$ is a vector for Malaria, not Leishmaniasis. Leishmaniasis (Kala-azar) is transmitted by the sandfly, $Phlebotomus$.

(C) The wheat production revolution in India during the $1960$s was primarily driven by the introduction of semi-dwarf wheat varieties. These varieties were developed by Dr. Norman $E$. Borlaug and brought to India by Dr. $M$.$S$. Swaminathan. The key feature of these varieties was a mutation that resulted in a significant reduction in plant height (semi-dwarf trait),which prevented lodging (falling over) and allowed the plants to support heavy,grain-filled ears of wheat,thereby drastically increasing yield.

(A) The development of resistance in pests against chemical pesticides or biological control agents is primarily due to $Random \text{ } mutation$.
When a large population of pests is exposed to a control agent, most individuals die.
However, due to random mutations occurring in the population, some individuals may possess a genetic variation that confers resistance.
These resistant individuals survive and reproduce, passing the resistance gene to their offspring, leading to a population that is resistant to the agent.

(A) $ELISA$ (Enzyme-Linked Immunosorbent Assay) is a diagnostic technique based on the principle of antigen-antibody interaction.
In this technique,an enzyme is linked to an antibody or antigen to detect the presence of a specific pathogen (like a virus).
The most commonly used enzyme as a key reagent in $ELISA$ is alkaline phosphatase or horseradish peroxidase,which produces a color change upon reacting with its substrate,indicating the presence of the virus.

(D) Restriction endonucleases are enzymes that cut $DNA$ at specific recognition sequences. These enzymes are naturally produced by bacteria as a defense mechanism to protect themselves from bacteriophage infections by cleaving the foreign viral $DNA$. In biotechnology,they are essential tools for creating recombinant $DNA$ molecules.

(C) The variation in soil temperature is most significant in deserts.
$1$. Deserts are characterized by extreme conditions,including high solar radiation and very low moisture content in the soil.
$2$. Due to the lack of vegetation cover and low water content,the soil in deserts heats up rapidly during the day and loses heat quickly at night,leading to a high diurnal range of temperature.
$3$. In contrast,forests,wetlands,and grasslands have more vegetation cover and moisture,which act as thermal buffers,stabilizing the soil temperature.

(D) foundation species is a species that plays a major role in creating or maintaining a habitat that supports other species. These species often have a large biomass and significantly influence the community structure by providing physical habitat or resources. Therefore,option $D$ is the correct description of a foundation species.

(D) To ensure the restoration of a degraded ecosystem, it is essential to reduce the factors that cause instability and promote resilience.
An ecosystem that is highly stable and highly resilient is better equipped to recover from disturbances.
Stability refers to the ability of an ecosystem to remain unchanged despite disturbances, while resilience refers to the capacity of an ecosystem to recover after a disturbance.
Therefore, to restore an ecosystem, we must aim for $High \text{ stability}$ and $High \text{ resilience}$ by stopping activities that degrade these properties.

(D) In any ecosystem,the total amount of organic matter produced by producers through photosynthesis is known as Gross Primary Productivity $(GPP)$.
Net Primary Productivity $(NPP)$ is the amount of biomass available to consumers after accounting for the energy lost by producers during respiration $(R)$.
The relationship is defined as: $NPP = GPP - R$.
Since $GPP$ represents the total energy captured before any losses,it is always higher than $NPP$ or any other form of productivity in an ecosystem.

(B) The most effective method for conserving biodiversity, including plant diversity, is $in-situ$ conservation.
$In-situ$ conservation involves protecting the entire ecosystem where the species naturally occur.
Biosphere reserves are large protected areas designed to conserve the ecosystem, species, and genetic diversity in their natural habitat.
In contrast, tissue culture, botanical gardens, and seed banks are examples of $ex-situ$ conservation, which are secondary methods used when natural habitats are threatened or destroyed.
Therefore, the establishment of biosphere reserves is the most effective approach for long-term conservation.

(D) The Red Panda $(Ailurus fulgens)$ is classified as $Endangered$ by the $IUCN$ Red List of Threatened Species.
This classification is due to a significant decline in their population caused by habitat loss,fragmentation,and poaching.

(D) Fluoride pollution,when present in drinking water at concentrations higher than $1.5 \ ppm$,primarily affects the skeletal system and teeth. This condition is known as fluorosis. In teeth,it causes dental fluorosis,which leads to mottling or discoloration of the enamel. Therefore,the correct option is $D$.

(D) The Bhopal gas tragedy,which occurred on the night of December $2-3, 1984$,was caused by the accidental release of methyl isocyanate $(MIC)$ gas from the Union Carbide pesticide plant.
$MIC$ is highly reactive with water. When water entered the storage tank,it triggered an exothermic reaction that produced a large amount of heat and pressure,causing the tank to rupture and release the toxic gas into the atmosphere.

(B) The correct pair is $B$ (Kyoto Protocol - Climate change).
$1$. The Basel Convention is related to the control of transboundary movements of hazardous wastes and their disposal.
$2$. The Kyoto Protocol is an international treaty that extends the $1992$ United Nations Framework Convention on Climate Change $(UNFCCC)$ that commits state parties to reduce greenhouse gas emissions,based on the scientific consensus that global warming is occurring and that human-made $CO_2$ emissions are driving it.
$3$. The Montreal Protocol is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion.
$4$. The Ramsar Convention is an international treaty for the conservation and sustainable use of wetlands.

(B) Lead $(Pb)$ is a toxic heavy metal that can accumulate in the human body.
According to medical and environmental health standards,a blood lead level of $30 \mu g / 100 \ ml$ is considered the warning concentration,above which it can cause significant health issues such as neurological damage,anemia,and developmental delays.
Therefore,the correct option is $B$.

(D) The Government of India has implemented the Ethanol Blended Petrol $(EBP)$ programme to reduce dependence on imported crude oil and to lower carbon emissions.
Initially,the government permitted a $5\%$ blending of ethanol (alcohol) with petrol.
Over time,this target has been increased to $10\%$ and further to $20\%$ (E20) to promote sustainable fuel usage.
Given the options provided,$5\%$ was the initial standard percentage permitted for blending.