KCET 2015 Biology Question Paper with Answer and Solution

(A) is the correct answer.
Greenhouse crops such as tomatoes and bell pepper produce higher yields because they are grown in a $CO_2$ enriched atmosphere.
These plants are $C_3$ plants,which respond to higher concentrations of $CO_2$ by increasing their rate of photosynthesis,as $CO_2$ is often a limiting factor for them in normal atmospheric conditions.

(D) The energy of respiration is converted into heat.
In mammals,there are two types of adipose tissue: white and brown. 'Brown fat' cells contain a great many mitochondria,and brown adipose tissue is well supplied by blood capillaries. Brown adipose tissue produces a considerable amount of heat because lipids are respired there with little or no $ATP$ formation; most energy is released as heat. This heat is transported in the blood to the body organs.

(C) In the anaerobic respiration pathway in yeast,pyruvate is first converted into acetaldehyde and $CO_2$ by the enzyme pyruvate decarboxylase.
Then,acetaldehyde is reduced to ethanol by the enzyme alcohol dehydrogenase,using $NADH$ as a reducing agent.
Looking at the provided pathway diagram:
$1$. Pyruvate is converted to $C$ $(CO_2)$ and $B$ (Acetaldehyde).
$2$. Acetaldehyde $(B)$ is then converted to $A$ (Ethanol).
Therefore,the correct labels are $A$-Ethanol,$B$-Acetaldehyde,and $C-CO_2$.

(A) is the correct statement because oxygen acts as the final electron acceptor in the electron transport chain,combining with hydrogen ions to form water,which is essential for the removal of hydrogen.
$B$ is incorrect because pyruvate is formed in the cytosol during glycolysis.
$C$ is incorrect because fermentation involves the incomplete oxidation of glucose under anaerobic conditions.
$D$ is incorrect because during the conversion of succinyl-CoA to succinic acid in the Krebs cycle,a molecule of $GTP$ (guanosine triphosphate) is synthesised,not $ATP$.

(C) The correct sequence is as follows:
$1$. Gibberellin: It is known to hasten the maturity period in juvenile conifers.
$2$. Auxin: It plays a crucial role in controlling xylem differentiation.
$3$. Abscisic acid $(ABA)$: It increases the tolerance of plants to various kinds of stress,which is why it is also known as the 'stress hormone'.

(D) The correct answer is $D$.
In an $ECG$,the $P$-wave represents the electrical excitation (or depolarisation) of the atria.
The $QRS$ complex represents the depolarisation of the ventricles,which initiates the ventricular contraction (systole).
The $T$-wave represents the return of the ventricles from an excited state to a normal state,which is known as ventricular repolarisation.

(B) The correct answer is $B$. $A$ fall in glomerular blood flow,glomerular blood pressure,or $GFR$ can activate the $JG$ (juxta glomerular) cells to release renin.
Renin converts angiotensinogen in the blood to angiotensin $I$ and further to angiotensin $II$.
Angiotensin $II$,being a powerful vasoconstrictor,increases the glomerular blood pressure and thereby $GFR$.
Angiotensin $II$ also activates the adrenal cortex to release aldosterone.
Aldosterone causes reabsorption of $Na^+$ and water from the distal parts of the tubule.
This leads to an increase in blood pressure and $GFR$.
This complex mechanism is known as the Renin-Angiotensin mechanism.

(A) Based on the structure of the myosin monomer (meromyosin):
$A$ represents the Actin binding sites located on the head.
$B$ represents the Head (globular head).
$C$ represents the $ATP$ binding site located on the head.
$D$ represents the Cross arm (short arm connecting the head to the tail).
Therefore,the correct labeling is: $A$ = Actin binding site,$B$ = Head,$C$ = $ATP$ binding site,$D$ = Cross arm.

(B) is the incorrect pair.
$Gluconeogenesis$ is the process of synthesizing glucose from non-carbohydrate sources like amino acids and glycerol.
$Insulin$ is a hypoglycemic hormone that promotes $glycogenesis$ (conversion of glucose to glycogen) and glucose uptake by cells,thereby lowering blood glucose levels.
$Gluconeogenesis$ is actually promoted by hormones like $glucagon$ and $cortisol$.

(A) $Mycoplasmas$ are the smallest living cells known and they completely lack a cell wall.
They are pleomorphic,meaning they can change their shape.
Many $Mycoplasmas$ are pathogenic in animals and plants.
They are facultative anaerobes,meaning they can survive and grow in the absence of oxygen.

(C) The flow chart describes the following characteristics:
$1$. Kingdom: $Animalia$
$2$. Level of organisation: Tissue grade
$3$. Symmetry: Bilateral
$4$. Coelom: Acoelomate
Analysis of the options:
- $Hemichordata$: Organ-system level,bilateral symmetry,coelomate.
- $Aschelminthes$: Organ-system level,bilateral symmetry,pseudocoelomate.
- $Platyhelminthes$: Organ-system level,bilateral symmetry,acoelomate.
- $Ctenophora$: Tissue grade,biradial symmetry,acoelomate.
Note: While $Platyhelminthes$ are acoelomate and bilateral,they typically exhibit organ-system level of organisation. However,in many simplified classification schemes,$Platyhelminthes$ is the only phylum that fits the 'acoelomate' and 'bilateral' criteria among the choices provided. Therefore,$Platyhelminthes$ is the most appropriate answer.

(A) The correct answer is $A$ (capsule).
$Glycocalyx$ varies in composition and thickness among different bacteria.
If the $glycocalyx$ is loose and thin,it is known as the $slime$ $layer$.
If the $glycocalyx$ is thick and tough,it is known as the $capsule$.

(A) The correct statement is $A$.
Prokaryotes are unicellular organisms that lack membrane-bound cell organelles such as mitochondria,chloroplasts,and the endoplasmic reticulum.
Option $B$ is incorrect because prokaryotic cells lack a true nucleus.
Option $C$ is incorrect because cells arise from pre-existing cells (Omnis cellula-e cellula).
Option $D$ is incorrect because animal cells lack a cell wall.

(B) The structure shown in the image is $Uracil$.
$Uracil$ is a nitrogenous base,specifically a pyrimidine derivative,which is found in $RNA$ but not in $DNA$ (where it is replaced by $Thymine$).
The chemical structure consists of a six-membered ring containing two nitrogen atoms and two carbonyl groups,which matches the provided image.
Therefore,the correct option is $B$.

(D) The correct answer is $D$ (apomixis).
Apomixis is a form of asexual reproduction that mimics sexual reproduction.
In this process,seeds are produced without the fusion of male and female gametes (fertilisation).
Therefore,seeds formed through apomixis are genetically identical to the parent plant.

(C) $Phoenix$ $dactylifera$.
$A$ recent record of a $2000$ year old viable seed is that of the date palm,$Phoenix$ $dactylifera$,which was discovered during the archaeological excavation at King Herod's palace near the Dead Sea.

(B) The correct answer is $B$.
Continued self-pollination leads to the accumulation of recessive deleterious alleles in the population.
This process reduces the genetic diversity and results in a decrease in the biological fitness, vitality, and productivity of the offspring, a phenomenon known as $inbreeding \text{ } depression$.

(C) fragmentation.
Fragmentation is the process of breaking down detritus into smaller particles by the action of detritivores,such as earthworms,termites,etc.

(B) The correct answer is $(B)$ $0.03 \ J$.
According to the $10 \%$ law proposed by Lindeman $(1942)$,only $10 \%$ of the energy available at one trophic level is transferred to the next trophic level.
In the given food chain: Plant $\rightarrow$ Mice $\rightarrow$ Snake $\rightarrow$ Peacock.
$1$. Energy at producer level (Plant) = $30 \ J$.
$2$. Energy available to primary consumer (Mice) = $10 \%$ of $30 \ J = 3 \ J$.
$3$. Energy available to secondary consumer (Snake) = $10 \%$ of $3 \ J = 0.3 \ J$.
$4$. Energy available to tertiary consumer (Peacock) = $10 \%$ of $0.3 \ J = 0.03 \ J$.
Therefore,the peacock receives $0.03 \ J$ of energy.

(D) The correct answer is $D$.
According to the $NCERT$ textbook diagram representing the global biodiversity of vertebrates:
$A$ represents Fishes (the largest portion).
$B$ represents Mammals.
$C$ represents Birds.
$D$ represents Reptiles.
$E$ represents Amphibians.
This distribution reflects the relative species diversity among these vertebrate groups.

$(B)$ . The pair "Endemism - Species confined to one region and also found in other regions" is incorrectly matched.
Endemic species are those that are restricted to a particular geographical area or region and are not found naturally anywhere else. They are not found in other regions. Therefore, the statement provided in option $B$ is incorrect.

(C) The correct answer is $C$.
$Ex$ $situ$ conservation refers to the conservation of selected rare plants or animals in locations outside their natural habitats.
$Ex$ $situ$ conservation methods include offsite collections such as botanical gardens, zoological parks, seed banks, tissue culture, and cryopreservation.
Biosphere reserves are examples of $in$ $situ$ conservation, where the entire ecosystem is protected within its natural environment.

(C) The correct option is $C$.
$FSH$ (Follicle Stimulating Hormone) acts on the Sertoli cells of the testes.
It stimulates the secretion of some factors which help in the process of spermiogenesis,which is the transformation of spermatids into spermatozoa.

(B) The correct answer is $B$.
During the menstrual cycle,the uterine wall undergoes significant structural and functional changes.
The innermost layer of the uterus,known as the $endometrium$,is the site of these cyclical changes.
It undergoes proliferation,secretory activity,and eventually sloughing off (menstruation) if fertilization does not occur.

(D) The correct answer is $D$.
In human development,the heart is formed after one month of pregnancy.
By the end of the second month of pregnancy,the foetus develops limbs and digits.
Therefore,the limbs and digits are developed after $8$ weeks.

(A) hormone releasing $IUDs$.
Progestasert is a hormone-releasing $IUD$ that releases progesterone.
It makes the uterus unsuitable for implantation and the cervix hostile to the sperms by increasing the viscosity of cervical mucus,which prevents sperm entry.

(A) Assisted Reproductive Technology $(ART)$ refers to a set of medical procedures used to treat infertility.
Common $ART$ techniques include:
$1$. In-vitro fertilisation and embryo transfer $(IVF-ET)$.
$2$. Gamete Intra Fallopian Transfer $(GIFT)$.
$3$. Intra Cytoplasmic Sperm Injection $(ICSI)$.
$4$. Artificial insemination $(AI)$.
$5$. Zygote Intra Fallopian Transfer $(ZIFT)$.
Option $A$ mentions 'Zygote Extra Fallopian Transfer',which is not a recognized medical procedure. The correct term is $ZIFT$ (Zygote Intra Fallopian Transfer). Therefore,'Zygote Extra Fallopian Transfer' is not an $ART$ technique.

(B) The correct answer is $B$.
Down's syndrome is an autosomal aneuploidy caused by the presence of an extra copy of chromosome-$21$.
In this case,the child has $3$ copies of chromosome-$21$,where $2$ copies share the same alleles as one of the mother's chromosomes.
If non-disjunction occurs during meiosis-$I$,the homologous chromosomes fail to separate,resulting in a gamete containing both maternal homologs.
Since these $2$ chromosomes are identical to the mother's alleles (due to the presence of homologous chromosomes that did not segregate),it indicates that the non-disjunction event occurred during maternal meiosis-$I$.

(D) In pedigree analysis,a single horizontal line between two individuals represents mating,while double horizontal lines represent consanguineous mating or mating between close relatives. Therefore,the correct option is $D$.

(A) is the correct answer because it is not a case of aneuploidy.
$1$. Phenylketonuria $(PKU)$ is an inborn error of metabolism.
$2$. It is an autosomal recessive disorder caused by a mutation in the gene encoding the enzyme phenylalanine hydroxylase.
$3$. This enzyme is required to convert the amino acid phenylalanine into tyrosine in the liver.
$4$. Because the defective gene affects multiple phenotypic traits (such as mental retardation and reduction in hair and skin pigmentation),it is a classic example of pleiotropy.

(A) The wrong statement is: $A$. Alleles $b$ and $c$ also produce sugar.
Explanation:
$1$. The $ABO$ blood grouping in humans is controlled by the gene $I$,which has three alleles: $I^{A}$,$I^{B}$,and $i$.
$2$. The alleles $I^{A}$ and $I^{B}$ produce specific enzymes (glycosyltransferases) that add sugar polymers to the surface of red blood cells,forming antigens $A$ and $B$,respectively.
$3$. The allele $i$ does not produce any sugar or antigen.
$4$. When $I^{A}$ and $I^{B}$ are present together,both express themselves equally,a phenomenon known as co-dominance.
$5$. $I^{A}$ and $I^{B}$ are dominant over $i$,meaning if $I^{B}$ and $i$ are present,the phenotype will be blood group $B$ (due to $I^{B}$ expression).

(B) is the correct answer: $XO$ type of sex determines male sex in grasshopper.
In $Drosophila$,the $XX$ condition (homozygous) results in a female,while $XY$ results in a male.
In birds,the $ZZ$ condition (homozygous) results in a male,while $ZW$ results in a female.
In humans,Klinefelter's syndrome is characterized by an $XXY$ genotype ($47$ chromosomes),not $XO$.

(B) The correct answer is $B$.
$DNA$ polymorphism refers to the occurrence of an inheritable mutation in a population at a frequency greater than $0.01$ $(1\%)$.
These variations are the basis of genetic mapping and $DNA$ fingerprinting.

(D) $AUG$ is a specific codon that serves two primary roles in protein synthesis.
First,it acts as the initiation codon,signaling the start of translation for the ribosome.
Second,it codes for the amino acid methionine $(Met)$.
Therefore,the correct option is $D$.

(A) $UAA$ is one of the three stop codons (nonsense codons) which do not code for any amino acid.
If the codon for the $25^{\text{th}}$ amino acid is mutated to $UAA$,the translation process will be terminated at that specific position because no $tRNA$ molecule can recognize the stop codon to bring the next amino acid.
Consequently,the ribosome will release the growing peptide chain after the $24^{\text{th}}$ amino acid.
Therefore,a polypeptide chain consisting of $24$ amino acids will be formed.

(B) . $VNTR$ (Variable Number Tandem Repeats) are used in $DNA$ fingerprinting $(A-q)$.
$B$. Introns and exons are processed during the splicing of $hnRNA$ $(B-s)$.
$C$. Dystrophin is known as the largest gene in humans $(C-p)$.
$D$. Satellite $DNA$ is separated from the main genomic $DNA$ as a peak during density gradient centrifugation,often referred to as bulk $DNA$ $(D-r)$.
Therefore,the correct matching is $A-q, B-s, C-p, D-r$.

(B) The four nitrogenous bases present in $DNA$ are adenine,guanine,cytosine,and thymine.
Thymine is chemically known as $5$-methyl uracil.
Unlike $RNA$,which contains uracil,$DNA$ contains thymine as its specific nitrogenous base.

(A) Total length of $DNA$ = $3.2 \text{ Kbp} = 3200 \text{ bp}$.
According to Chargaff's rule,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of guanine $(G)$ is equal to cytosine $(C)$.
Given,$A = 820$,therefore $T = 820$.
Total $A + T$ content = $820 + 820 = 1640$.
Total $G + C$ content = Total base pairs - Total $A + T$ content = $3200 - 1640 = 1560$.
Since $G = C$,the number of cytosine bases = $\frac{1560}{2} = 780$.

(D) In eukaryotes,$RNA$ polymerase-$I$ is responsible for the transcription of $rRNA$ genes,specifically $28S$,$18S$,and $5.8S$ $rRNA$.
$5S$ $rRNA$ is transcribed by $RNA$ polymerase-$III$.
Therefore,the correct option is $(D)$.

(D) Coelacanth.
In $1938$,a fish caught in South Africa was identified as a Coelacanth,a species previously thought to be extinct.
These lobe-finned fish are considered the evolutionary precursors to the first amphibians that transitioned to living on both land and water.
Consequently,they are recognized as the ancestors of modern-day amphibians,such as frogs and salamanders.

(D) divergent evolution.
Divergent evolution is the process by which two or more species evolve from a common ancestral species due to different environmental adaptations.
This process highlights the presence of common ancestry among the resulting species.

(B) The correct answer is $B$. Sporozoites do not multiply in the blood.
After entering the human body through the bite of an infected female $Anopheles$ mosquito,the sporozoites initially reach the liver and multiply within the liver cells.
Subsequently,they attack the red blood cells $(RBCs)$,causing them to rupture.
The rupture of $RBCs$ releases a toxic substance called haemozoin,which is responsible for the chill and high fever recurring every $3$ to $4$ days.
Therefore,the statement that sporozoites multiply in the blood is incorrect.

(D) $Smack$ (heroin) is a depressant obtained from the latex of the poppy plant, $Papaver$ $somniferum$.
$Crack$ (cocaine) is a stimulant obtained from the leaves of the coca plant, $Erythroxylum$ $coca$.
Therefore, the correct answer is $D$.

(C) Natural killer $(NK)$ cells are a type of white blood cell that plays a major role in the host-rejection of both tumors and virally infected cells.
In the context of innate immunity,the human body has four types of barriers:
$1$. Physical barriers: e.g.,skin,mucus coating of the epithelium.
$2$. Physiological barriers: e.g.,acid in the stomach,saliva in the mouth,tears from eyes.
$3$. Cellular barriers: e.g.,polymorphonuclear leukocytes ($PMNL$-neutrophils),monocytes,macrophages,and natural killer lymphocytes.
$4$. Cytokine barriers: e.g.,interferons.
Therefore,natural killer lymphocytes are classified under cellular barriers.

(A) - Establishing the potential of penicillin as an effective antibiotic.
$Penicillium$ was the first antibiotic discovered by Alexander Fleming $(1928)$. He found that the fungus $Penicillium$ $notatum$ or its extract could inhibit the growth of the bacterium $Staphylococcus$ $aureus$.
However, its full potential as an effective antibiotic was established much later by Ernst Chain and Howard Florey.
This antibiotic was extensively used to treat American soldiers wounded in World War $II$.
Fleming, Chain, and Florey were awarded the Nobel Prize in $1945$ for this discovery.

(C) The correct answer is $C$.
Biogas is a renewable fuel produced when organic matter,such as food or animal waste,is broken down by microbes under anaerobic conditions.
Biogas primarily consists of methane $(50-70\%)$,carbon dioxide $(30-40\%)$,and traces of nitrogen,hydrogen sulphide,and hydrogen.

(B) $BOD$ stands for Biochemical Oxygen Demand.
It is defined as the amount of oxygen that would be consumed if all the organic matter in $1000 \ mL$ ($1$ litre) of water were oxidised by bacteria.
It is a measure of the organic matter present in the water sample; higher $BOD$ indicates higher pollution levels.

(D) filtration and sedimentation.
Primary sewage water treatment entails the physical removal of particles through filtration and sedimentation of solid waste present in the water.
The purpose of primary treatment is to settle down large and small particles by gravity,removing floatable debris and grit,which reduces the pollution load to ease the subsequent secondary treatment process.

(A) Plasmid.
Plasmids are extra-chromosomal,self-replicating,circular,$dsDNA$ molecules.
Plasmid vectors are typically used to clone small $DNA$ fragments,usually ranging from $100$ to $1000$ $bp$.

(B) Both $A$ and $B$ are correct statements.
$1$. Gel electrophoresis is a technique used to separate $DNA$ fragments based on their size (molecular weight) and charge.
$2$. Elution is the process of extracting the separated $DNA$ fragments from the agarose gel matrix.
$3$. Ethidium bromide $(EtBr)$ is a fluorescent dye used to stain $DNA$. When exposed to $UV$ radiation,the $DNA$ bands stained with $EtBr$ appear as bright orange fluorescent bands,allowing for their visualization and subsequent extraction.