KCET 2009 Biology Question Paper with Answer and Solution

(A) The incorrect statement is $A$.
In the Calvin cycle,$18$ molecules of $ATP$ are not synthesized during carbon fixation; rather,$ATP$ is consumed.
To produce one molecule of glucose ($6$ carbons),the Calvin cycle must turn $6$ times.
Each turn requires $3$ $ATP$ and $2$ $NADPH$ molecules.
Therefore,a total of $18$ $ATP$ and $12$ $NADPH$ are consumed,not synthesized,during the process.
Option $B$ is correct as $NADPH$ is used for the reduction of $1,3$-bisphosphoglycerate to glyceraldehyde-$3$-phosphate.
Option $C$ is correct as $RuBisCO$ is the enzyme responsible for the carboxylation of $RuBP$.
Option $D$ is correct as $3$-phosphoglycerate ($3$-$PGA$) is the first stable product of the Calvin cycle.

(D) is the correct answer.
In the experiment with $Hydrilla$,adding sodium bicarbonate $(NaHCO_3)$ increases the availability of carbon dioxide $(CO_2)$ in the water.
$CO_2$ is a key substrate for photosynthesis.
This enhanced availability of carbon dioxide improves the rate of photosynthesis,which,in turn,increases the amount of oxygen $(O_2)$ evolved as a byproduct.

(A) The correct answer is $A$.
Statement $A$ is false because the enzymes required for carbon fixation (the $Calvin$ cycle),such as $RuBisCO$,are located in the stroma of the chloroplasts,not in the grana.
The grana are primarily the sites for light-dependent reactions,where the thylakoid membranes contain the photosystems and electron transport chain components.
Statement $B$ is true: Non-cyclic photophosphorylation involves both $PS I$ and $PS II$ to produce $NADPH + H^+$.
Statement $C$ is true: The electron transport chain components are embedded within the thylakoid membranes.
Statement $D$ is true: Photosynthesis is a redox reaction where $H_2O$ is oxidized to $O_2$ and $CO_2$ is reduced to carbohydrates.

(D) The correct match is $A-r, B-s, C-p, D-q$.
$1$. Oxaloacetate $(A)$ is a $4-C$ dicarboxylic acid involved in the Krebs cycle.
$2$. Phosphoglyceraldehyde $(B)$ is a $3-C$ triose phosphate intermediate in glycolysis.
$3$. Isocitrate $(C)$ is a $6-C$ tricarboxylic acid formed during the Krebs cycle.
$4$. $\alpha$-Ketoglutarate $(D)$ is a $5-C$ dicarboxylic acid formed after the decarboxylation of isocitrate in the Krebs cycle.

(D) The correct answer is $D$.
Respiratory Quotient $(RQ)$ is defined as the ratio of the volume of $CO_2$ evolved to the volume of $O_2$ consumed during respiration.
$RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}}$.
The value of $RQ$ depends entirely on the nature of the respiratory substrate being oxidized.
For carbohydrates, the $RQ$ is $1.0$.
For fats, the $RQ$ is less than $1.0$ (approximately $0.7$).
For proteins, the $RQ$ is approximately $0.9$.

(C) $2, 4-D$ is a synthetic auxin and does not occur naturally in plants.
$GA$ (Gibberellins), $ABA$ (Abscisic acid), and $IAA$ (Indole-3-acetic acid) are all naturally occurring plant hormones.
Therefore, the correct option is $C$.

(C) Heparin.
$Heparin$ is a natural anticoagulant present in the blood that prevents the conversion of $prothrombin$ to $thrombin$ in an undamaged blood vessel.
This action prevents unnecessary blood clot formation by blocking the $thrombin$-mediated conversion of $fibrinogen$ to $fibrin$,thereby maintaining smooth blood flow in healthy,undamaged vessels.

(B) The correct answer is $A$ only.
$1$. Urea,uric acid,and ammonia are all nitrogenous waste products resulting from the metabolism of proteins and nucleic acids.
$2$. They differ significantly in their toxicity: ammonia is the most toxic,followed by urea,while uric acid is the least toxic.
$3$. Due to these differences in toxicity,they require varying amounts of water for excretion. Ammonia requires the most water,while uric acid requires the least.
$4$. Urea is primarily synthesized in the liver via the ornithine cycle,not in the kidneys.

(B) The correct answer is $B$.
Plasma proteins are large molecules that cannot pass through the glomerular filtration membrane during the process of ultrafiltration.
As a result,they are retained in the blood capillaries of the glomerulus.
In contrast,substances like urea,water,and glucose are small enough to pass through the filtration slits of the podocytes and the basement membrane,meaning their concentrations in the glomerular filtrate are initially similar to their concentrations in the blood plasma.

(B) hypothalamus.
The hypothalamus is the part of the brain that acts as the body's thermostat and regulates essential homeostatic functions,including body temperature,hunger (appetite),and thirst.
$A$ tumor in the hypothalamus disrupts these regulatory centers,leading to the clinical presentation of abnormally low body temperature (hypothermia),loss of appetite,and extreme thirst (polydipsia).

(C) Both the statements $A$ and $B$ are correct and $B$ is the reason for $A$.
After hepatectomy,which involves the removal of a part or the whole liver,there is a rapid drop in blood sugar levels $(Statement A)$ because the liver's glycogen,which is the primary source of blood sugar $(Statement B)$,is no longer available to maintain blood glucose homeostasis.

(C) Prostaglandin.
Prostaglandins are not polypeptides; they are lipid compounds derived from fatty acids.
Insulin,antidiuretic hormone $(ADH)$,and oxytocin are all peptide hormones,which are composed of amino acids.

(D) endospores.
Endospores are highly resistant,dormant structures formed by certain bacteria under unfavorable environmental conditions.
These structures are capable of withstanding extreme heat,radiation,desiccation (dryness),and toxic chemicals.
They protect the bacterial $DNA$ and essential cellular components until favorable conditions return,allowing the bacterium to germinate and resume normal activity.

(B) thallus-like plant body,lack of vascular tissues and autotrophic nutrition.
Bryophytes resemble algae in their thallus-like plant body,lack of vascular tissues,and autotrophic nutrition.
Both groups have simple,non-differentiated structures and rely on photosynthesis for nutrition.
Neither group possesses the specialized vascular tissues ($xylem$ and $phloem$) found in higher plants for transporting water and nutrients.

(D) Echinodermata - Asteroidea - Starfish.
Starfish belongs to the phylum Echinodermata and the class Asteroidea,making this a correct match.
Spiders belong to the class Arachnida,not Insecta,within the phylum Arthropoda.
Unio is a bivalve mollusk belonging to the class Bivalvia,not Cephalopoda.
Planaria belongs to the class Turbellaria,not Trematoda,within the phylum Platyhelminthes.

(C) The correct answer is $C$.
Animals that give birth to young ones directly are known as viviparous animals.
$1$. Dolphin,kangaroo,bat,and cat are all mammals that exhibit viviparity.
$2$. Kiwi,ostrich,and penguin are birds,which are oviparous (egg-laying).
$3$. Platypus is a monotreme,which is an egg-laying mammal.
Therefore,the group containing only viviparous animals is Dolphin,kangaroo,bat,and cat.

(D) tortoise.
$Tortoises$ possess an exoskeletal shell made of bony plates covered by keratin and an endoskeleton comprising a backbone and internal bones.
This dual skeletal system provides protection and structural support.
This distinguishes $tortoises$ from $frogs$ (which have an endoskeleton only),$jellyfish$ (which lack both a true exoskeleton and endoskeleton),and $freshwater$ $mussels$ (which have an exoskeleton only).

(D) The correct matches are as follows:
$1$. Sunflower $(A)$ belongs to the family Compositae $(q)$.
$2$. Tulsi $(B)$ belongs to the family Labiatae $(r)$.
$3$. Coffee $(C)$ belongs to the family Rubiaceae $(s)$.
$4$. Vasaka $(D)$ belongs to the family Acanthaceae $(p)$.
Therefore,the correct matching is $A-q, B-r, C-s, D-p$.

(B) The correct answer is $B$.
Fleshy fruits with a stony endocarp are known as $drupes$.
In these fruits, the endocarp is hard and stony, which protects the seed inside.
This stony endocarp is surrounded by a fleshy mesocarp and a thin exocarp.
Common examples of $drupes$ include mango $(Mangifera \text{ } indica)$ and coconut $(Cocos \text{ } nucifera)$.

(A) The correct answer is $A$.
In the diagram of the maize grain:
$A$ represents the endosperm,which is the food-storing tissue.
$B$ represents the coleoptile,which is a protective sheath covering the plumule.
$C$ represents the scutellum,which is a large,shield-shaped cotyledon in monocots that absorbs nutrients from the endosperm.
$D$ represents the aleurone layer,which is the protein-rich outer layer of the endosperm.

(C) The correct answer is $A, C$ and $D$ only.
$1$. Ectoderm gives rise to the epidermis of the skin.
$2$. Mesoderm gives rise to muscles and the notochord.
$3$. Dermis is derived from the mesoderm,not the endoderm.
$4$. Enamel of teeth is derived from the ectoderm,not the endoderm.

(C) . Loss of cell-mediated immunity.
Damage to the thymus in a child may lead to the loss of cell-mediated immunity. The thymus is the primary lymphoid organ responsible for the development and maturation of $T$-lymphocytes ($T$-cells). Since $T$-cells are essential for orchestrating cell-mediated immune responses,their absence or dysfunction due to thymic damage significantly impairs this branch of the immune system.

(A) $A-14$.
Pollen grains are haploid reproductive cells produced by meiosis,containing $n$ chromosomes.
Given that the diploid number of the plant is $2n = 28$,the haploid number is $n = 28 / 2 = 14$.
In tissue culture,when pollen grains are cultured to form callus,the cells of the callus retain the same genetic constitution as the pollen grains.
Therefore,the cells of the callus will be haploid and contain $14$ chromosomes.

(B) The correct answer is $B$ (shoot apex culture).
In plant tissue culture,the shoot apical meristem is generally free from viral infections even if the rest of the plant is infected.
This is because the virus multiplication rate is slower than the rate of cell division in the meristematic region.
By excising and culturing the shoot apex,scientists can regenerate healthy,virus-free clones of the parent plant.
This technique is widely used in agriculture to produce disease-free planting material for various crops.

(A) The correct answer is $A$.
The introduction of high-yielding varieties (HYVs) is considered the greatest threat to genetic diversity in agricultural crops.
This is because the widespread adoption of a few genetically uniform high-yielding varieties often leads to the replacement and loss of traditional,locally adapted,and diverse crop varieties (landraces).
This process,known as genetic erosion,significantly reduces the overall genetic diversity within the agricultural ecosystem.
While the extensive use of fertilizers,insecticides,and pesticides can negatively impact soil health and local ecosystems,the replacement of diverse germplasm with uniform HYVs has a more direct and irreversible impact on genetic diversity.

(C) Botanical garden.
In situ conservation refers to the conservation of species within their natural habitats.
National parks,sanctuaries,and biosphere reserves are examples of in situ conservation areas.
Botanical gardens are examples of ex situ conservation,where plants are conserved outside their natural habitats.

(A) In a primary spermatocyte,the diploid number of chromosomes is $2n = 16$.
During the first meiotic division $(Meiosis-I)$,the homologous chromosomes separate,resulting in the formation of two secondary spermatocytes.
Each secondary spermatocyte receives $n = 8$ chromosomes.
Since each chromosome consists of $2$ sister chromatids,the total number of chromatids in each secondary spermatocyte is $8 \times 2 = 16$.

(C) primary oocyte in the Graafian follicle.
Vitellogenesis is the process of yolk formation and deposition.
It occurs in the primary oocyte within the Graafian follicle during oogenesis in the ovary.
This stage involves the accumulation of yolk,which includes proteins,fats,phospholipids,carbohydrates,and glycogen,all of which are essential for the development of the embryo.

(D) The correct sequence of events in the human female reproductive cycle is: $A \rightarrow C \rightarrow E \rightarrow D \rightarrow B$.
$1$. $A$ (secretion of $FSH$): The cycle begins with the secretion of Follicle Stimulating Hormone $(FSH)$ from the anterior pituitary.
$2$. $C$ (growth of the follicle and oogenesis): $FSH$ stimulates the growth of ovarian follicles and the process of oogenesis.
$3$. $E$ (sudden increase in the level of $LH$): As the follicle matures,estrogen levels rise,leading to a surge in Luteinizing Hormone $(LH)$.
$4$. $D$ (ovulation): The $LH$ surge triggers the rupture of the Graafian follicle and the release of the ovum.
$5$. $B$ (growth of corpus luteum): After ovulation,the remnants of the Graafian follicle transform into the corpus luteum.

(A) The correct answer is $A$ $(0\%)$.
Given:
$1$. Both husband and wife have normal vision.
$2$. Both fathers were colour blind $(X^CY)$.
$3$. Both mothers were normal and did not carry the gene for colour blindness $(XX)$.
Analysis:
- The wife's father was colour blind $(X^CY)$,so she must have inherited the $X^C$ chromosome from him. Since she has normal vision,she must be a carrier $(X^CX)$.
- The husband's father was colour blind $(X^CY)$,but the husband inherited his $X$ chromosome from his mother (who was $XX$). Therefore,the husband is normal $(XY)$.
Cross:
Parents: $XY$ (Husband) $\times$ $X^CX$ (Wife)
Gametes: $X, Y$ and $X^C, X$
Offspring:
- $X^CX$ (Carrier Daughter)
- $XX$ (Normal Daughter)
- $X^CY$ (Colour blind Son)
- $XY$ (Normal Son)
Conclusion:
From the cross,the daughters are either carrier $(X^CX)$ or normal $(XX)$. None of the daughters are colour blind. Therefore,the probability of their daughters becoming colour blind is $0\%$.

(B) The correct option is $(B) I^{A}I^{O}$ and $I^{O}I^{O}$.
For the offspring to have only $A$ or $O$ blood groups,one parent must have the genotype $I^{A}I^{O}$ (blood group $A$) and the other must have $I^{O}I^{O}$ (blood group $O$).
When these parents are crossed,the possible genotypes of the offspring are $I^{A}I^{O}$ (blood group $A$) and $I^{O}I^{O}$ (blood group $O$).
This confirms that the offspring will only possess blood groups $A$ or $O$.

(D) Let the allele for red flowers be $R$ and the allele for white flowers be $r$. Since red is dominant,the genotype of the red-flowered parent is $RR$ and the white-flowered parent is $rr$.
When these are crossed,the $F_1$ generation plants have the genotype $Rr$ (all red flowers).
When $F_1$ plants are selfed $(Rr \times Rr)$,the resulting $F_2$ generation genotypes are $RR$,$Rr$,$Rr$,and $rr$ in a $1:2:1$ ratio.
The phenotypic ratio is $3$ red-flowered plants to $1$ white-flowered plant.
Therefore,the proportion of plants producing white flowers is $1/4$.

(A) The correct answer is $A$.
$DNA$ profiling or $DNA$ fingerprinting requires nucleated cells to extract genomic $DNA$.
$Leucocytes$ (white blood cells) are nucleated cells found in blood,making them the ideal source for $DNA$ extraction.
$Platelets$ are cell fragments lacking a nucleus.
$Erythrocytes$ (mature red blood cells) in humans lack a nucleus and organelles.
$Serum$ is the liquid component of blood remaining after clotting,which does not contain cells or $DNA$.

(D) The correct answer is $D$.
During $DNA$ replication,the process begins with the breaking of hydrogen bonds between the nitrogenous bases of the two $DNA$ strands.
This unwinds the double helix and allows each strand to serve as a template for the synthesis of a new complementary strand.

(B) $RNA$ contains ribose sugar in its nucleotide structure.
$ATP$ (adenosine triphosphate) also contains ribose sugar,as it is a nucleoside triphosphate.
$RNA$ polymerase is an enzyme (protein) and does not itself contain ribose sugar,although it interacts with $RNA$ which does.
Thus,ribose sugar is present in both $RNA$ and $ATP$.

$(A)$ Statement $A$ is wrong because a monocistronic mRNA, typically found in eukaryotes, contains the genetic information for only a single polypeptide chain.
Polycistronic mRNA, which is common in prokaryotes, is the type that can produce multiple polypeptide chains from a single mRNA strand.
Statement $B$ is correct because the process of translation is terminated by the presence of specific stop codons (terminator codons like $UAA$, $UAG$, or $UGA$) on the mRNA molecule, which signal the release of the polypeptide chain from the ribosome.

(B) . Variations are inherited from parents to offspring through genes.
Option $(B)$ is incorrect because Darwinism does not address the mechanism of genetic inheritance. Charles Darwin was unaware of the principles of genetics; it was Gregor Mendel's later work that explained how traits are passed through genes. Darwinism focuses on the principles of natural selection,which include:
$(A)$ Organisms with advantageous traits are better suited to their environment and reproduce more successfully.
$(C)$ There is natural variation within each species.
$(D)$ Organisms tend to produce more offspring than the environment can support,leading to competition for limited resources.
These principles collectively drive evolutionary change through the survival of the fittest.

(A) rhinitis.
The symptoms of frequent nasal discharge,nasal congestion,reddening of eyes,and watery eyes are indicative of rhinitis.
Rhinitis is the inflammation of the mucous membrane of the nose,which can be triggered by viral infections or allergic reactions to allergens like pollen,dust,or dander.

(A) $Interferons$ are antiviral proteins produced by virus-infected cells to protect non-infected cells from further viral infection.
Among the given options,measles is a viral disease caused by the $Measles$ $virus$.
Tetanus and anthrax are bacterial diseases,and malaria is a protozoan disease.
Therefore,the production of $interferons$ is a characteristic response to a viral infection like measles.

(D) subjecting the $DNA$ to polymerase chain reaction.
To obtain a sufficient amount of $DNA$ from a small fragment,the polymerase chain reaction $(PCR)$ is the best method.
$PCR$ amplifies the extracted $DNA$,allowing for a larger quantity to be analyzed.
Gel electrophoresis is used to separate $DNA$ fragments based on their size.
Restriction endonucleases are enzymes that cut $DNA$ at specific recognition sequences.
Hybridizing with a $DNA$ probe is used to detect specific $DNA$ sequences.
None of these other techniques are used for the amplification of $DNA$.