AIIMS 2016 Biology Question Paper with Answer and Solution

(A) In mitosis,one cell divides to form two daughter cells.
If $n$ is the number of mitotic divisions,the number of cells produced is given by the formula $2^n$.
We need to find $n$ such that $2^n = 128$.
Since $128 = 2^7$,it follows that $n = 7$.
Therefore,$7$ mitotic divisions are required to produce $128$ cells from a single cell.

(A) The classification of Pteridophytes into four classes is as follows:
$1$. Psilotopsida: Example is $Psilotum$ $(V)$.
$2$. Lycopsida: Examples are $Selaginella$ $(III)$ and $Lycopodium$ $(IV)$.
$3$. Sphenopsida: Example is $Equisetum$ $(II)$.
$4$. Pteropsida: Examples are $Dryopteris$,$Pteris$,and $Adiantum$ $(I)$.
Therefore,the correct matching is $A-V; B-III, IV; C-II; D-I$. Since option $A$ is the closest representation of this classification,it is the correct choice.

(A) Phaeophyceae (brown algae) possess chlorophyll $a, c$,carotenoids,and xanthophylls (specifically fucoxanthin).
These pigments are responsible for the characteristic colour of these algae.
The variation in colour from olive green to various shades of brown depends upon the amount of xanthophyll pigments (fucoxanthin) present in the cells.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) The Assertion is correct because the typhlosole is an internal fold of the intestine that increases the surface area for absorption.
However,the Reason is incorrect because the typhlosole is a characteristic feature of the earthworm $(Pheretima)$,not the cockroach.
Therefore,the correct option is $D$.

(B) The correct matches are as follows:
- $A$. Coleorhiza: It is the protective sheath covering the radicle in monocot seeds $(IV)$.
- $B$. Apogamy: It is the development of a sporophyte directly from a gametophyte without the fusion of gametes $(I)$.
- $C$. Indusium: It is the thin,membranous protective covering of a sorus in ferns $(V)$.
- $D$. Caudex: It refers to an unbranched,columnar stem,often seen in palms or cycads,bearing a crown of leaves $(III)$.
Thus,the correct sequence is $A-IV, B-I, C-V, D-III$.

(A) leaf is considered simple when its lamina is entire or when the incisions do not reach the midrib or petiole. Even in simple leaves,the lamina can have various types of incisions that may reach up to half,more than half,or near the base or midrib. These incisions are classified based on the venation pattern (pinnate or palmate) as pinnatifid,palmatifid,pinnatipartite,palmatipartite,pinnatisect,or palmatisect. Therefore,both the Assertion and the Reason are correct,and the Reason explains why simple leaves can exhibit various forms of incisions.

(C) The increase in the girth or diameter of a plant stem is known as secondary growth.
This growth is facilitated by lateral meristems,which include the vascular cambium and the cork cambium (phellogen).
While phellogen is a type of lateral meristem,the term 'lateral meristem' is the broader,more accurate category that encompasses all tissues responsible for secondary growth in thickness.
Therefore,lateral meristem is the primary tissue responsible for the increase in girth.

(B) Based on the anatomy of the cockroach head:
$A$ represents the Ocellus (fenestra),which is a simple eye.
$B$ represents the Compound eye.
$C$ represents the Mandible,which is a hard,biting jaw.
$D$ represents the Maxilla,which helps in food manipulation.
$E$ represents the Labrum,which acts as the upper lip.
$F$ represents the Labium,which acts as the lower lip.
Therefore,the correct sequence is $A-$ Ocellus,$B-$ Compound eye,$C-$ Mandible,$D-$ Maxilla,$E-$ Labrum,$F-$ Labium.

(A) Statement $(i)$ is incorrect because the special membranous structure formed by the extension of the plasma membrane in prokaryotes is called a mesosome,not a polysome. $A$ polysome is a chain of ribosomes attached to a single $mRNA$ molecule.
Statement $(ii)$ is incorrect because the smooth endoplasmic reticulum $(SER)$ is the major site for lipid synthesis,whereas the rough endoplasmic reticulum $(RER)$ is the site for protein and glycoprotein synthesis.
Statement $(iii)$ is correct because $RuBisCO$ (Ribulose bisphosphate carboxylase-oxygenase) is the most abundant protein in the biosphere.
Statement $(iv)$ is correct because the endomembrane system includes only the endoplasmic reticulum,Golgi complex,lysosomes,and vacuoles. Mitochondria,chloroplasts,and peroxisomes are not part of this system.
Therefore,statements $(iii)$ and $(iv)$ are correct.

(A) The centrosome is an organelle that serves as the main microtubule organizing center $(MTOC)$ of the animal cell.
It typically contains two cylindrical structures known as centrioles,which are oriented at right angles to each other.
These centrioles are composed of microtubules arranged in a $9 + 0$ pattern (nine triplets of microtubules).
Since the centrosome is defined by the presence of these centrioles,the Reason correctly explains the Assertion.

(D) Inorganic catalysts work efficiently at high temperature and high pressure.
Inorganic catalysts are substances that increase the rate of a chemical reaction without being consumed in the process.
Unlike enzymes (biological catalysts),which are sensitive to heat and denature at high temperatures,inorganic catalysts are stable and often require high thermal energy and pressure to function optimally.
An example of this is the use of iron as a catalyst in the Haber process for the production of ammonia,which operates at high temperature and high pressure.

(C) In the structure of a nucleotide like adenylic acid,the phosphate group is attached to the $5'$-carbon of the pentose sugar. This linkage is formed by a phosphoester bond,which is a type of ester bond.

(C) The Assertion is correct: $A$ prosthetic group is an organic compound or metal ion that is tightly bound to the apoenzyme (protein portion) and remains associated with it throughout the catalytic cycle.
The Reason is incorrect: $A$ complete,catalytically active enzyme consisting of the protein part (apoenzyme) and the non-protein part (co-factor/prosthetic group) is called a holoenzyme,not an apoenzyme. An apoenzyme is only the protein portion of the enzyme.

(A) glycosidic bond is a type of covalent bond that joins a carbohydrate molecule to another group,which may or may not be another carbohydrate.
These bonds are formed by a dehydration reaction (also known as a condensation reaction),where a water molecule is removed during the linkage of two monosaccharides.
In polysaccharides,individual monosaccharides are linked by these glycosidic bonds to form long chains.
Since the formation of the bond involves the process of dehydration,the Reason correctly explains the Assertion.

(D) Interkinesis is the stage between two meiotic divisions (meiosis $I$ and meiosis $II$),not between two mitotic divisions. Therefore,the Assertion is incorrect.
Interkinesis is indeed a short-lived stage where no $DNA$ replication occurs. Thus,the Reason is correct.
Since the Assertion is incorrect and the Reason is correct,the correct option is $D$.

(B) Diplotene is the longest and most active subphase of prophase $I$ of meiosis.
The beginning of diplotene is recognized by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers.
These $X$-shaped structures are called chiasmata.
Additionally,in the oocytes of some vertebrates,the diplotene stage can remain suspended for months or even years,a state known as dictyotene.
Since both statements are factually correct but the duration of the stage does not explain why chiasmata are present,the correct option is $B$.

(C) Active transport is a process that requires energy in the form of $ATP$ to move molecules against their concentration gradient.
Since the transport process slows down when energy production is inhibited,it indicates that the process is energy-dependent.
Therefore,the molecules are transported by active transport.

(C) Plants primarily absorb nitrogen as nitrates $(NO_3^-)$ or nitrites $(NO_2^-)$.
Phosphorus is absorbed mainly as phosphate ions ($H_2PO_4^-$ or $HPO_4^{2-}$).
Sulphur is absorbed by plants in the form of sulphate ions $(SO_4^{2-})$,not as sulphuric acid $(H_2SO_4)$.
Iron is absorbed in the form of ferric ions $(Fe^{3+})$.
Therefore,the pair Sulphur $\to H_2SO_4$ is mismatched.

(A) In the process of photophosphorylation,the cytochrome $b_6f$ complex acts as a proton pump.
It facilitates the movement of protons from the stroma into the thylakoid lumen,creating a proton gradient.
This proton gradient is essential for the functioning of $ATP$ synthase,which utilizes the energy of the proton motive force to synthesize $ATP$ from $ADP$ and inorganic phosphate.

(A) Statement $(i)$ is correct: Glycolysis involves ten enzymatic steps converting one molecule of glucose $(6C)$ into two molecules of pyruvate $(3C)$ with a net gain of $2\, ATP$.
Statement $(ii)$ is incorrect: Glucose undergoes partial oxidation to form two molecules of pyruvic acid,not one.
Statement $(iii)$ is incorrect: Glucose is phosphorylated to glucose-$6$-phosphate by the enzyme hexokinase,not phosphofructokinase.
Statement $(iv)$ is correct: The $EMP$ pathway is named after Embden,Meyerhof,and Parnas.
Statement $(v)$ is correct: $ATP$ is consumed during the phosphorylation of glucose to glucose-$6$-phosphate and fructose-$6$-phosphate to fructose-$1,6$-bisphosphate.
Therefore,statements $(i), (iv),$ and $(v)$ are correct.

(B) Statement $(i)$ is correct: $A$ single maize root apical meristem can produce more than $17,500$ new cells per hour.
Statement $(ii)$ is correct: The growth of a pollen tube is measured in terms of length.
Statement $(iii)$ is incorrect: The growth of a leaf is measured in terms of surface area,not volume.
Statement $(iv)$ is correct: Cells in a watermelon may increase in size by up to $3,50,000$ times.
Therefore,statements $(i), (ii),$ and $(iv)$ are correct,while $(iii)$ is incorrect.

(A) The physiological calorific value of carbohydrates is $4.0 \, kcal/g$,proteins is $4.0 \, kcal/g$,and fats is $9.0 \, kcal/g$. Lignin is a dietary fiber that cannot be digested by humans,so it provides $0 \, kcal$.
$1$. Raw sugar (carbohydrate): $5 \, g \times 4.0 \, kcal/g = 20.0 \, kcal$.
$2$. Albumin (protein): $4 \, g \times 4.0 \, kcal/g = 16.0 \, kcal$.
$3$. Total fat (pure ghee + vegetable ghee): $10 \, g + 2 \, g = 12 \, g$. Energy from fat: $12 \, g \times 9.0 \, kcal/g = 108.0 \, kcal$.
$4$. Lignin: $5 \, g \times 0 \, kcal/g = 0 \, kcal$.
Total energy = $20.0 + 16.0 + 108.0 + 0 = 144 \, kcal$.

(C) The Assertion is correct because $ptyalin$ (salivary amylase) acts on starch and converts it into maltose.
The Reason is incorrect because $sucrase$ (invertase) hydrolyses sucrose into glucose and fructose,not lactose. Lactose is hydrolysed by the enzyme $lactase$ into glucose and galactose.

(D) At the tissue level,cellular respiration consumes oxygen and produces carbon dioxide.
Consequently,the partial pressure (tension) of $O_2$ is low,and the partial pressure (tension) of $CO_2$ is high in the tissues compared to the blood.
According to the Bohr effect,high $CO_2$ concentration and low $O_2$ tension shift the oxygen-haemoglobin dissociation curve to the right,facilitating the dissociation of oxygen from oxyhaemoglobin.
Therefore,oxyhaemoglobin releases oxygen to the cells.

(A) $Lub$ and $Dub$ are two heart sounds produced during each cardiac cycle.
$Lub$ is the first heart sound,which is produced due to the closure of the atrioventricular valves (tricuspid and bicuspid valves) at the beginning of ventricular systole.
It is a low-pitched sound of longer duration (approximately $0.15 \,sec$).
Since the assertion states that $Lub$ is produced during the cardiac cycle and the reason correctly identifies its cause as the closure of the atrioventricular valves,both statements are correct and the reason explains the assertion.

(B) In mammals,urea is produced in the liver through the ornithine cycle (urea cycle) as a byproduct of amino acid metabolism.
Since the liver is the site of urea synthesis,the blood leaving the liver via the hepatic vein contains the highest concentration of urea compared to any other blood vessel in the body.
The renal vein carries blood away from the kidneys,which have already filtered the urea out of the blood,so it contains the lowest concentration of urea.
The dorsal aorta carries oxygenated blood to various parts of the body,and the hepatic portal vein carries blood from the digestive tract to the liver,neither of which has as high a concentration of urea as the hepatic vein.

(C) The ornithine cycle (also known as the urea cycle) is a series of biochemical reactions that occur in the liver to convert toxic ammonia into urea.
In the final step of this cycle,the enzyme arginase catalyzes the hydrolysis of arginine.
This reaction produces ornithine and urea.
Ornithine is then recycled back into the mitochondria to continue the cycle,while urea is excreted from the body.

(A) The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes like sodium. As the filtrate moves down,water is reabsorbed into the interstitium,which increases the concentration of solutes,making the filtrate hypertonic.
The ascending limb of the loop of Henle is impermeable to water but permeable to electrolytes like sodium. As the filtrate moves up,electrolytes are actively or passively transported out of the tubule into the interstitium,which decreases the concentration of solutes,making the filtrate hypotonic.
Therefore,both the Assertion and the Reason are correct,and the Reason explains why the tonicity changes in the respective limbs.

(A) In birds,the two clavicles fuse with a single inter-clavicle to form a $V$-shaped bone known as the 'wishbone' or 'furcula' (also called the 'merrythought' bone). This structure is part of the pectoral girdle,which is also known as the shoulder girdle.

(C) The assertion is correct because muscle contraction involves the formation and breaking of cross-bridges between myosin heads and actin filaments,known as the sliding filament theory.
The reason is incorrect because muscle contraction is initiated by a signal sent by the $Central \ Nervous \ System$ $(CNS)$,not the peripheral nervous system,via a motor neuron.
Therefore,the assertion is correct,but the reason is incorrect.

(D) Nerve conduction,or the transmission of a nerve impulse along an axon,primarily relies on the movement of $Na^+$ and $K^+$ ions across the neuronal membrane.
During the resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,leading to a rapid influx of $Na^+$ into the cell,which causes depolarization.
Following this,the membrane becomes permeable to $K^+$ ions,leading to an efflux of $K^+$ out of the cell,which causes repolarization.
Thus,$Na^+$ and $K^+$ are the essential ions for the generation and conduction of nerve impulses.

(A) In a reflex arc,the pathway of a nerve impulse is as follows:
$1$. $A$ represents the sense organ (receptor) which detects the stimulus.
$2$. $B$ represents the sensory nerve (afferent neuron) which carries the impulse towards the central nervous system.
$3$. $C$ represents the dorsal horn of the spinal cord grey matter,where sensory neurons enter.
$4$. $D$ represents the interneuron (relay neuron) which connects sensory and motor neurons.
$5$. $E$ represents the ventral horn of the spinal cord grey matter,where motor neurons exit.
$6$. $F$ represents the motor nerve (efferent neuron) which carries the impulse away from the central nervous system.
$7$. $G$ represents the effector (muscle or gland) which performs the response.
Therefore,the correct sequence is $A =$ sense organ,$B =$ sensory nerve,$C =$ dorsal horn,$D =$ interneuron,$E =$ ventral horn,$F =$ motor nerve,$G =$ effector.

(D) The axonal membrane of a neuron in a resting state is significantly more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$. Therefore,the Assertion $(A)$ is incorrect.
In a resting state,the neuron is polarized and does not conduct any nerve impulse. Therefore,the Reason $(R)$ is correct.

(B) The thymus gland is known as the 'Throne of immunity' because it plays a vital role in the development of the immune system.
It is the primary site for the maturation and differentiation of $T-$lymphocytes (or $T-$cells).
These $T-$cells are essential for cell-mediated immunity and also help in regulating the production of antibodies by $B-$cells.

(A) Flora is a taxonomic aid that provides the actual account of habitat and distribution of plants in a given area.
It serves as an index to the plant species found in a particular area,which facilitates the correct identification of those plants.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) Statement $(iii)$ is incorrect. The duration of the reproductive phase varies significantly among different organisms depending on their species,environmental conditions,and life history strategies. It is not uniform across all living beings.

(D) Pollen grains are generally spherical and possess a prominent two-layered wall.
The hard outer layer,called the exine,is made up of sporopollenin,which is one of the most resistant organic materials known.
It can withstand high temperatures,strong acids,and strong alkalis,and no enzyme is known that degrades sporopollenin.

(A) Double fertilization is a unique and characteristic feature of angiosperms.
It involves two distinct fusion events:
$1$. Syngamy: One male gamete fuses with the egg cell to form a diploid zygote $(2n)$.
$2$. Triple Fusion: The second male gamete fuses with the two polar nuclei (or the diploid secondary nucleus) to form the primary endosperm nucleus $(PEN)$,which is triploid $(3n)$.
Since the process is defined by these two specific fusion events,the Reason correctly explains the Assertion.

(C) During copulation (coitus),semen is released by the penis into the vagina (insemination).
The human male ejaculates about $200 - 300$ million sperms during a single coitus.
For normal fertility,it is essential that at least $60\%$ of all sperms must have normal shape and size.
Furthermore,for the sperms to reach the ovum and fertilize it,at least $40\%$ of them must show vigorous (energetic) motility.

(C) Based on the anatomical structure of the human foetus within the uterus:
$A$ represents the Umbilical cord,which contains blood vessels that transport nutrients and oxygen to the foetus.
$B$ represents the Placental villi (chorionic villi),which are finger-like projections that increase the surface area for exchange between maternal and foetal blood.
$C$ represents the Yolk sac,which is an extra-embryonic membrane.
$D$ represents the Plug of mucus in the cervix,which acts as a protective barrier during pregnancy.
Therefore,the correct labeling is $A \to$ Umbilical cord with its vessels,$B \to$ Placental villi,$C \to$ Yolk sac,$D \to$ Plug of mucus in cervix.

(D) Testicular lobules are the structural compartments found within the testes,each containing $1-3$ highly coiled seminiferous tubules.
These lobules are the sites of spermatogenesis (production of sperm),not fertilization.
Fertilization is the process of fusion of male and female gametes,which typically occurs in the ampullary region of the fallopian tube in the female reproductive system.
Therefore,the Assertion is correct,but the Reason is incorrect.

(D) $ICSI$ (Intracytoplasmic sperm injection) - Formation of embryo by directly injecting sperm into the ovum $(III)$.
$IUI$ (Intrauterine insemination) - Artificial introduction of semen into the vagina or uterus $(I)$.
$IUT$ (Intrauterine transfer) - Transfer of embryo with more than $8$ blastomeres into the uterus $(V)$.
$GIFT$ (Gamete intrafallopian transfer) - Transfer of ovum collected from a donor into the fallopian tube where fertilization occurs $(II)$.
$ZIFT$ (Zygote intrafallopian transfer) - Transfer of the zygote or early embryo (with up to $8$ blastomeres) into a fallopian tube $(IV)$.
Therefore, the correct match is $A-III, B-I, C-V, D-II, E-IV$.

(B) Thomas Hunt Morgan conducted experiments on Drosophila melanogaster to study linkage and recombination.
In cross $I$,the genes for yellow body $(y)$ and white eyes $(w)$ are tightly linked,resulting in a low frequency of recombination,which is $1.3\%$.
In cross $II$,the genes for white eyes $(w)$ and miniature wings $(m)$ are loosely linked,resulting in a higher frequency of recombination,which is $37.2\%$.
Therefore,the percentage of recombinants produced in cross $I$ and cross $II$ are $1.3\%$ and $37.2\%$ respectively.

(A) The term 'degenerate' in the context of the genetic code usually refers to the property where multiple codons code for the same amino acid. However,in many textbooks,the term 'degenerate' is incorrectly used to refer to 'stop codons' or 'nonsense codons'. Based on the provided options,the question asks for the set of stop codons.
Stop codons (also known as nonsense or termination codons) do not code for any amino acids and signal the end of protein synthesis.
The three stop codons are $UAA$ (ochre),$UAG$ (amber),and $UGA$ (opal).

(B) For a gene with two alleles,$A$ and $a$,if the frequency of $A$ is $p$ and the frequency of $a$ is $q$,then the frequencies of the three possible genotypes ($AA, Aa,$ and $aa$) are expressed by the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$.
Here,$p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.
Given that $p = 0.7$,we know that $p + q = 1$,so $q = 1 - 0.7 = 0.3$.
The frequency of the heterozygous genotype $Aa$ is represented by $2pq$.
Substituting the values: $2pq = 2 \times 0.7 \times 0.3 = 0.42$.

(B) The Assertion is correct because,according to the Oparin-Haldane hypothesis,the first organic molecules synthesized in the primitive atmosphere and oceans were simple monomers like amino acids and nucleotides,which polymerized to form proteins and nucleic acids.
The Reason is also correct because the early life forms (protobionts) originated and evolved in the primitive aquatic environment (the 'primordial soup').
However,the fact that life originated in water is not the direct scientific explanation for why proteins and nucleic acids were the specific organic compounds required for the origin of life. Therefore,the Reason does not explain the Assertion.

(B) $Yersinia pestis$ (formerly known as $Pasteurella pestis$) is the bacterium that causes Bubonic Plague.
This pathogen is primarily transmitted to humans through the bite of infected rat fleas,specifically the species $Xenopsylla cheopis$.
$Cimex$ refers to bedbugs,$Pediculus$ refers to lice,and $Aedes$ is a genus of mosquitoes known for transmitting diseases like Dengue and Chikungunya.
Therefore,the correct vector for the transmission of $Yersinia pestis$ is $Xenopsylla$.

(D) Tetanus is caused by the bacterium $Clostridium$ $tetani$.
$Pasteurella$ $pestis$ (also known as $Yersinia$ $pestis$) is the causative agent of plague,not tetanus.
Therefore,the pair $Tetanus - Pasteurella$ $pestis$ is incorrectly matched.

(B) The Assertion is correct. Cocaine is an alkaloid obtained from the coca plant, $Erythroxylum \text{ } coca$. It interferes with the transport of the neurotransmitter dopamine,resulting in a potent stimulating action on the central nervous system,which produces a sense of euphoria and increased energy.
The Reason is also a correct statement regarding immunology. Active immunity is indeed developed when a host is exposed to antigens,which can occur either through natural infection or through intentional exposure via immunization (vaccination).
However,the Reason does not explain the mechanism of action of cocaine. Therefore,both statements are correct,but the Reason is not the correct explanation of the Assertion.

(D) Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.
Cross-breeding is a breeding technique in which a superior male of one breed is mated with a superior female of another breed.
This method allows the desirable qualities of two different breeds to be combined into a single progeny.

(D) The correct matches are as follows:
$1$. $A$. Escherichia coli is widely used in biotechnology for the production of human insulin $(III)$.
$2$. $B$. Rhizobium melilotae is a nitrogen-fixing bacterium associated with the $Nif$ gene $(I)$.
$3$. $C$. Bacillus thuringiensis produces toxins used as a biological insecticide $(V)$.
$4$. $D$. Pseudomonas putida is known for its ability to degrade hydrocarbons in crude oil $(II)$.
Therefore,the correct combination is $A-III, B-I, C-V, D-II$.

(B) In the secondary treatment of sewage,activated sludge flocs consist of masses of bacteria associated with fungal filaments to form a mesh-like structure.
These flocs require a constant supply of oxygen for the aerobic decomposition of organic matter.
If the oxygen availability is reduced,the oxygen cannot diffuse into the interior of the flocs.
Consequently,the centre of the flocs becomes anoxic (lacking oxygen).
This leads to the death of the aerobic bacteria residing in the centre,which eventually causes the structural integrity of the flocs to fail,leading to their breakage.

(C) palindromic sequence in $DNA$ is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (e.g.,$5' \rightarrow 3'$).
Restriction enzymes recognize specific palindromic sequences to cut the $DNA$.
In option $(c)$,the sequence $5'-GAATTC-3'$ on one strand corresponds to $3'-CTTAAG-5'$ on the complementary strand.
When read in the $5' \rightarrow 3'$ direction,both strands yield the sequence $GAATTC$,which is a classic recognition site for the restriction enzyme $EcoRI$.

(A) transgenic food crop which may help in solving the problem of night blindness in developing countries is golden rice.
Golden rice is a genetically modified rice variety that has been engineered to produce elevated levels of beta-carotene.
Beta-carotene is a precursor of vitamin $A$,which helps in preventing night blindness.
This modification gives the rice its characteristic golden colour.

(A) Salinity is defined as the concentration of salts in water,measured in parts per thousand $(ppt)$.
In the sea,the salinity typically ranges between $30-35$ $ppt$.
In contrast,inland waters (freshwater) have a salinity of less than $5$ $ppt$.
Some hypersaline lagoons can have salinities exceeding $100$ $ppt$.

(D) The main source of energy for an ecosystem is the radiant energy or light energy derived from the sun.
$50\%$ of the total solar radiation that falls on Earth is Photosynthetically Active Radiation $(PAR)$.
The light energy is converted into chemical energy in the form of sugar by photosynthesis.
The chemical equation is: $6H_2O + 6CO_2 + \text{Light} \to C_6H_{12}O_6 + 6O_2$.
Plants utilize only $2-10\%$ of the incident $PAR$ for the process of photosynthesis.

(B) Net primary productivity $(NPP)$ is the rate of organic matter built up or stored by producers in their bodies per unit time and area. It is calculated as $NPP = GPP - R$,where $GPP$ is Gross Primary Productivity and $R$ is the energy lost due to respiration.
Secondary productivity is defined as the rate of formation of new organic matter by consumers (heterotrophs) per unit time and area.
Both statements are scientifically correct,but the Reason does not explain why $NPP$ is defined as $GPP$ minus respiration. Therefore,the correct option is $B$.

(D) The species-area relationship is given by the equation $\log S = \log C + Z \log A$.
From the graph,the slope $(Z)$ is $1.15$ (as indicated by the value $S=1.15$ in the provided image).
For an area of $100,000\, km^2$,the number of species $(S_1)$ is approximately $70$.
For an area of $1\, km^2$,the number of species $(S_2)$ is approximately $5$.
The number of species lost is $S_1 - S_2 = 70 - 5 = 65$.
The percentage of species lost is $(65 / 70) \times 100 \approx 92.8\%$,which is approximately $93\%$.
Therefore,$93\%$ of the bird species will be lost.

(B) The solubility of gases in liquids is inversely proportional to temperature.
As the temperature of a liquid increases,the kinetic energy of the gas molecules increases,causing them to escape from the liquid phase more easily.
Since lake $A$ has a higher temperature than lake $B$,the solubility of oxygen in lake $A$ will be lower.
Conversely,lake $B$,having a lower temperature,will have a higher rate of oxygen dissolution compared to lake $A$.