KCET 2020 Biology Question Paper with Answer and Solution

(D) The correct answer is $D$.
Crossing over is the process of exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process occurs during the pachytene stage of prophase-$I$ of meiosis.
It is an enzyme-mediated process,and the enzyme involved in this process is known as recombinase.

(A) The correct answer is $A$.
In $C_3$ plants,the Calvin cycle ($C_3$ cycle) occurs exclusively in the mesophyll cells.
However,in $C_4$ plants,the process is compartmentalized: the initial $CO_2$ fixation occurs in the mesophyll cells to form a $4$-carbon compound,which is then transported to the bundle sheath cells.
Therefore,the $C_3$ cycle (Calvin cycle) takes place specifically in the bundle sheath cells of $C_4$ plants.

(A) The correct option is $A$.
Statement $I$ is correct: RuBisCO has a high affinity for $CO_2$,but it can also bind to $O_2$ when $CO_2$ concentration is low,leading to photorespiration.
Statement $II$ is incorrect: In $C_4$ plants,the Calvin cycle occurs in the chloroplasts of bundle sheath cells,but the statement implies a general rule or is often contrasted with $C_3$ plants where it occurs in mesophyll cells. However,statement $III$ is factually incorrect as per $NCERT$.
Statement $III$ is incorrect: Yeast poisons itself to death when the concentration of alcohol reaches about $13\%$,not $7\%$.
Statement $IV$ is correct: During aerobic respiration,oxygen acts as the final hydrogen acceptor at the end of the electron transport chain to form water.

(C) The correct answer is $C$.
In the citric acid cycle (Krebs cycle),the organic acids have different numbers of carbon atoms.
$1$. Succinic acid is a $4C$ organic acid.
$2$. $\alpha$-Ketoglutaric acid is a $5C$ organic acid.
$3$. Citric acid is a $6C$ organic acid.
Therefore,the sequence $4C, 5C, 6C$ corresponds to Succinic acid,$\alpha$-Ketoglutaric acid,and Citric acid respectively.

(A) The percentage distribution of different types of leucocytes (WBCs) in the blood of a healthy adult is as follows:
$1$. Neutrophils: $60-65\%$
$2$. Lymphocytes: $20-25\%$
$3$. Monocytes: $6-8\%$
$4$. Eosinophils: $2-3\%$
$5$. Basophils: $0.5-1\%$
Matching these with the given columns:
$A$. Neutrophils $\rightarrow (ii) 60-65\%$
$B$. Lymphocytes $\rightarrow (v) 20-25\%$
$C$. Monocytes $\rightarrow (i) 6-8\%$
$D$. Basophils $\rightarrow (iii) 0.5-1\%$
$E$. Eosinophils $\rightarrow (iv) 2-3\%$
Therefore, the correct matching is $(A)-(ii); (B)-(v); (C)-(i); (D)-(iii); (E)-(iv)$.

(D) The correct answer is $D$.
The human brain is divided into three main parts: forebrain,midbrain,and hindbrain.
The midbrain is situated between the thalamus/hypothalamus of the forebrain and the pons of the hindbrain.
The dorsal portion of the midbrain consists mainly of four round swellings (lobes) called corpora quadrigemina.

(C) estrogens.
Estrogens produce wide-ranging actions such as stimulation of growth and activities of female secondary sex organs,development of growing ovarian follicles,appearance of female secondary sex characters (e.g.,high pitch of voice,etc.),and mammary gland development.
Estrogens also regulate female sexual behavior.

(A) The correct matching is as follows:
$A$. Phycomycetes: $iii$. Albugo (Parasitic fungi on mustard).
$B$. Ascomycetes: $i$. Penicillium (Source of penicillin).
$C$. Basidiomycetes: $iv$. Puccinia (Rust fungus).
$D$. Deuteromycetes: $ii$. Alternaria (Imperfect fungi).
Therefore,the correct sequence is $(A)-(iii); (B)-(i); (C)-(iv); (D)-(ii)$.

(B) The division $Vertebrata$ is classified based on the presence or absence of jaws into two groups:
$1$. $Agnatha$: These are jawless vertebrates. Thus,$p$ corresponds to $Agnatha$.
$2$. $Gnathostomata$: These are jawed vertebrates. Thus,$q$ corresponds to $Gnathostomata$.
$Gnathostomata$ is further divided into two superclasses based on the presence of fins or limbs:
$1$. $Pisces$: These are aquatic vertebrates that bear fins for locomotion. Thus,$r$ corresponds to $Pisces$.
$2$. $Tetrapoda$: These are terrestrial vertebrates that bear limbs for locomotion. Thus,$s$ corresponds to $Tetrapoda$.
Therefore,the correct matching is $p$-Agnatha,$q$-Gnathostomata,$r$-Pisces,$s$-Tetrapoda.

(D) The correct answer is $(D)$ Tepal.
The provided floral diagram represents the family Liliaceae,specifically the genus $Allium$ ($Allium$ $cepa$).
In this family,the calyx and corolla are not distinct,and the floral unit is collectively known as the perianth.
Each individual member of the perianth is referred to as a tepal.
Therefore,'$T$' points to a tepal.

(D) - presence of bulliform cells in the grass leaves.
In grasses,certain adaxial epidermal cells along the veins modify themselves into large,empty,colourless cells. These are called bulliform cells.
When the bulliform cells in the leaves have absorbed water and are turgid,the leaf surface is exposed.
When they are flaccid due to water stress,they make the leaves curl inwards to minimise water loss.

(A) The correct answer is $(A)$ Radial spoke.
In the structure of cilia and flagella,the axoneme consists of a $9+2$ arrangement of microtubules.
The peripheral microtubule doublets are connected to the central sheath,which surrounds the two central microtubules,by structures known as radial spokes.
There are $9$ radial spokes in total,one extending from each peripheral doublet to the central sheath.

(D) The correct answer is $D$ (oxygen).
Oxygen is the most abundant element by weight in the Earth's crust,accounting for approximately $46.6\%$ of its total mass.
Similarly,in the human body,oxygen is the most abundant element,making up about $65\%$ of the total body mass.

(C) In situ conservation refers to the conservation of species within their natural habitats.
This approach involves the protection and management of important components of biological diversity through a network of protected areas,such as National Parks,Wildlife Sanctuaries,and Biosphere Reserves.
Botanical gardens,on the other hand,are examples of ex situ conservation,where species are maintained outside their natural habitats.

(D) is the mismatch because synergids are haploid,not diploid.
$1$. The female gametophyte (embryo sac) consists of cells that are haploid $(n)$,including the egg cell,synergids,and antipodal cells.
$2$. The Zygote is formed by the fusion of a male gamete $(n)$ and an egg cell $(n)$,making it diploid $(2n)$.
$3$. The Primary Endosperm Nucleus $(PEN)$ is formed by the fusion of two polar nuclei $(n+n)$ and one male gamete $(n)$,resulting in a triploid $(3n)$ structure.

(A) The correct answer is $A-51$.
In flowering plants,the gametes are haploid $(n)$. Given that the chromosome number of gametes in apple is $n = 17$.
The Primary Endosperm Nucleus $(PEN)$ is formed by the fusion of two polar nuclei $(n + n)$ and one male gamete $(n)$,making it triploid $(3n)$.
Therefore,the chromosome number in the $PEN$ is $3 \times 17 = 51$.

(D) The exine of pollen grains is highly resistant to enzyme action.
Pollen grains possess a prominent two-layered wall. The hard outer layer,known as the exine,is composed of sporopollenin,which is one of the most resistant organic materials known to science.
It can withstand high temperatures,strong acids,and strong alkalis.
To date,no enzyme has been discovered that can degrade sporopollenin. This unique chemical property allows pollen grains to be well-preserved as fossils over long periods.

(D) $RNA$ interference $(RNAi)$ is a method of cellular defense in eukaryotes.
In this process,a double-stranded $RNA$ $(dsRNA)$ molecule is introduced,which is complementary to the target $mRNA$.
This $dsRNA$ binds to the specific $mRNA$ and silences it by preventing its translation into protein.
Therefore,the correct answer is $D$.

(D) $PCR$.
Polymerase Chain Reaction $(PCR)$ is a molecular biology technique that allows for the amplification and detection of very small amounts of $DNA$.
Because $PCR$ can amplify specific sequences of pathogen $DNA$ or $RNA$ (via reverse transcription),it enables the detection of disease-causing bacteria or viruses long before the symptoms of the disease appear in the patient.
This capability makes $PCR$ a powerful tool for early diagnosis and timely treatment.
Currently,$PCR$ is widely used to detect gene mutations in patients suspected of having cancer and to identify $HIV$ in patients suspected of having $AIDS$.

(A) The correct answer is $A$ (Declining).
The age pyramid shown in the image represents a declining population.
In this type of pyramid,the proportion of pre-reproductive individuals (at the base) is smaller than the proportion of reproductive and post-reproductive individuals.
This indicates that the birth rate is low and the population is shrinking over time.

(C) The correct option is $C$.
In the Verhulst-Pearl logistic growth equation,$\frac{ dN }{ dt }= rN \left[\frac{ K - N }{ K }\right]$,the variables are defined as follows:
$N$ represents the population density at time $t$.
$r$ represents the intrinsic rate of natural increase.
$K$ represents the carrying capacity of the environment.
Therefore,the letter '$r$' denotes the intrinsic rate of natural increase.

(D) The correct answer is $D$ (Frog).
In the given food chain,the plant acts as the primary producer.
The insect acts as the primary consumer (herbivore) that feeds on the plant.
The link '$M$' represents the secondary consumer,which feeds on the insect.
$A$ frog is a common secondary consumer that feeds on insects and is subsequently preyed upon by snakes (tertiary consumers).
Therefore,the frog fits perfectly as the link '$M$' in this food chain.

(A) The process of Oogenesis begins with the multiplication of Oogonia $(p)$ in the fetal ovary.
These Oogonia undergo differentiation to form Primary oocytes $(q)$.
The Primary oocyte completes the first meiotic division to form a large Secondary oocyte $(r)$ and a small first polar body.
The Secondary oocyte then undergoes the second meiotic division to form a large Ovum $(s)$ and a second polar body.
Therefore,the correct sequence is $p$ - Oogonia,$q$ - Primary oocyte,$r$ - Secondary oocyte,$s$ - Ovum.

(A) The correct matching is as follows:
$(A)$ First month: The first sign of growing foetus is usually noticed by listening to the heart sound carefully through the stethoscope. Thus,$(A-iii)$.
$(B)$ Second month: By the end of the second month of pregnancy,the foetus develops limbs and digits. Thus,$(B-iv)$.
$(C)$ Fifth month: The first movements of the foetus and appearance of hair on the head are usually observed during the fifth month. Thus,$(C-ii)$.
$(D)$ Sixth month: By the end of $24$ weeks (second trimester),the body is covered with fine hair,eye-lids separate,and eyelashes are formed. Thus,$(D-i)$.
Therefore,the correct sequence is $A-iii, B-iv, C-ii, D-i$.

(A) The correct answer is $A$.
After ovulation,the ovum is captured by the fimbriae of the infundibulum.
From the infundibulum,the ovum moves towards the ampulla,which is the site of fertilization.
From the ampulla,it proceeds towards the isthmus.
If the fallopian tube is blocked at the ampullary region,the ovum cannot pass from the infundibulum to the isthmus.

(A) The correct answer is $A$.
Administration of progestogens or progestogen-estrogen combinations within $72$ hours of coitus has been found to be very effective as emergency contraceptives.
These are used to avoid possible pregnancy resulting from rape or casual unprotected intercourse.

(B) $GIFT$ ($Gamete$ $Intrafallopian$ $Transfer$) is a technique used for females who cannot produce or release viable ova but possess a healthy environment for fertilization and embryonic development.
In this procedure,ova are collected from a donor female and transferred into the fallopian tube of the recipient.
This allows fertilization to occur naturally within the body ($in$ $vivo$) of the recipient.
Therefore,option $B$ is the correct statement.

(A) The inheritance of $ABO$ blood groups is controlled by the gene $I$. The gene $I$ has three alleles: $I^A, I^B,$ and $i$.
If the man is heterozygous $(I^Ai)$ and the woman is heterozygous $(I^Bi)$,the cross is $I^Ai \times I^Bi$.
The possible genotypes of the offspring are $I^AI^B$ (blood group $AB$),$I^Ai$ (blood group $A$),$I^Bi$ (blood group $B$),and $ii$ (blood group $O$).
Therefore,the maximum possible blood groups among their children are $A, B, AB,$ and $O$.

(A) In a dihybrid cross,the phenotypic ratio in the $F_2$ generation is $9:3:3:1$.
The parental combinations are Round-Yellow $(9/16)$ and Wrinkled-Green $(1/16)$.
The non-parental (recombinant) combinations are Round-Green $(3/16)$ and Wrinkled-Yellow $(3/16)$.
Total proportion of non-parental characters = $3/16 + 3/16 = 6/16 = 3/8$.
Given that the total number of $F_2$ progenies is $64$.
Number of non-parental progenies = $(3/8) \times 64 = 3 \times 8 = 24$.

(A) . Leaf shape
Mendel studied $7$ pairs of contrasting characters in pea plants. Leaf shape was not one of them. The characters studied by Mendel are summarized below:

$S. No.$	Character	Dominant Trait	Recessive Trait
$1.$	Plant height	Tall	Dwarf
$2.$	Flower colour	Purple	White
$3.$	Flower position	Axial	Terminal
$4.$	Pod shape	Inflated	Constricted
$5.$	Pod colour	Green	Yellow
$6.$	Seed shape	Round	Wrinkled
$7.$	Seed colour	Yellow	Green

(B) Pleiotropy.
When a single gene influences multiple phenotypic traits that are often unrelated,it is known as pleiotropy.
For instance,in pea plants,the gene that controls seed coat color also influences the color of flowers and the presence of pigmentation on the leaves.
Therefore,a single gene mutation results in multiple phenotypic expressions.

(D) is the correct answer.
In the $Lac$ $Operon$,the $i$ gene is a constitutive gene.
Constitutive genes are expressed continuously at a constant rate in a cell,regardless of the environmental conditions or the presence of substrates like $Lactose$.
The $i$ gene produces the repressor protein,which regulates the expression of the structural genes $(z, y, a)$ by binding to the operator region.
Since it is constitutive,it does not require an inducer to be transcribed,ensuring that the repressor protein is always available to keep the operon switched off in the absence of $Lactose$.

(B) The correct answer is $B$.
In a typical nucleosome,the $DNA$ helix is wrapped around a histone octamer.
This structure consists of approximately $200$ base pairs $(bp)$ of $DNA$.

(B) The process of transcription involves the synthesis of $mRNA$ from a $DNA$ template strand.
During this process,the base pairing rules are: $A$ (Adenine) pairs with $U$ (Uracil),$T$ (Thymine) pairs with $A$ (Adenine),$G$ (Guanine) pairs with $C$ (Cytosine),and $C$ (Cytosine) pairs with $G$ (Guanine).
Given $DNA$ template: $3'-ATGCATGCATGC-5'$.
Applying the base pairing rules:
$A \rightarrow U$
$T \rightarrow A$
$G \rightarrow C$
$C \rightarrow G$
Thus,the complementary $mRNA$ sequence is $5'-UACGUACGUACG-3'$.
Therefore,the correct option is $(B)$.

(D) The correct answer is $D$.
In eukaryotic cells,the primary transcript (pre-mRNA) contains both coding sequences called exons and non-coding sequences called introns.
During the process of $RNA$ processing or post-transcriptional modification,the non-coding introns are removed,and the exons are joined together in a specific order.
This process is known as splicing,which results in the formation of mature,functional mRNA.

(D) $tRNA$.
Transfer $RNA$ $(tRNA)$ acts as an adapter molecule.
It has an anticodon loop that base pairs with the codon on $mRNA$ and a $3'$ acceptor end that carries a specific amino acid to the ribosome for protein synthesis.

(D) is the correct answer.
$Brachiosaurus$ was one of the largest and tallest land dinosaurs, reaching heights of up to $12-16 \, m$ and weighing significantly more than $Tyrannosaurus \, rex$.
$Tyrannosaurus \, rex$ was a large carnivorous dinosaur, but in terms of overall size and mass among land dinosaurs, $Brachiosaurus$ (a sauropod) is considered much larger.
$Tyrannosaurus \, rex$ stood about $6 \, m$ $(20 \, feet)$ tall and possessed massive, dagger-like teeth.
Dinosaurs disappeared from the earth approximately $65 \, \text{million years ago (mya)}$.

(A) The correct answer is $A$ (disruptive).
Natural selection can lead to three main outcomes:
$1$. Stabilising selection: In this,more individuals acquire the mean character value,reducing variation.
$2$. Directional selection: In this,more individuals acquire values other than the mean character value,shifting the population towards one extreme.
$3$. Disruptive selection: In this,more individuals acquire peripheral character values at both ends of the distribution curve,while the intermediate (mean) individuals are selected against. Since the population shows only extreme tall and extreme dwarf plants with no intermediate forms,it is a clear case of disruptive selection.

(C) Malignant tumours are a mass of proliferating cells called neoplastic or tumour cells.
These cells grow very rapidly,invading and damaging the surrounding normal tissues.
Unlike benign tumours,malignant tumours do not remain confined to their original site and can spread to other parts of the body through a process called metastasis.

(D) The correct answer is $D$.
In the life cycle of $Plasmodium$,the sexual stage involves the formation of gametocytes in the human host.
These gametocytes are ingested by the female $Anopheles$ mosquito during a blood meal.
Fertilization and the development of the zygote occur within the stomach (midgut) of the mosquito,leading to the formation of an ookinete,which then develops into oocysts.

(A) The correct answer is $A$ (Dopamine).
Cocaine is a coca alkaloid obtained from the coca plant, $Erythroxylum$ $coca$, which is native to South America.
It interferes with the transport of the neurotransmitter dopamine by blocking its reuptake into the presynaptic neuron, leading to an accumulation of dopamine in the synaptic cleft and causing a sense of euphoria.

(D) The correct answer is $D$.
Passive immunity is a type of immunity that occurs when ready-made antibodies are introduced directly into the body to provide immediate protection against foreign pathogens or toxins.
In the case of a snake bite,the patient is injected with an antivenom,which contains preformed antibodies specifically designed to neutralize the snake venom.
Since the body's own immune system is not actively involved in producing these antibodies,this process is classified as passive immunization.

(A) The correct answer is $A$.
Humans cannot digest cellulose because they lack the specific bacteria known as methanogens in their digestive tract.
These methanogens produce the enzyme cellulase,which is essential for the fermentation and breakdown of cellulose.
Methanogens are anaerobic bacteria that produce large amounts of methane,along with $CO_2$ and $H_2$,during the digestion process.
These bacteria are naturally present in the rumen of ruminant animals (like cattle),where they facilitate the breakdown of cellulose,thereby playing a crucial role in the nutrition of these animals.

(C) is the correct answer. Secondary treatment is highly significant because it reduces the $BOD$ level of sewage.
Primary treatment involves the physical removal of large and small particles through filtration and sedimentation.
The effluent from primary treatment is then passed into large aeration tanks for secondary treatment (biological treatment).
In these tanks,aerobic microbes consume the major part of the organic matter present in the effluent.
This significant reduction in organic matter leads to a decrease in the Biochemical Oxygen Demand $(BOD)$ of the sewage,making it safer to discharge into water bodies.

(D) $PCR$ stands for Polymerase Chain Reaction.
It is a molecular biology technique used to amplify a single copy or a few copies of a segment of $DNA$ across several orders of magnitude,generating thousands to millions of copies of a particular $DNA$ sequence.
In cases where the collected $DNA$ sample is insufficient for analysis like $DNA$ fingerprinting,$PCR$ is used to amplify the available $DNA$ to a sufficient quantity.

(D) The correct answer is $D$.
In agarose gel electrophoresis,$DNA$ fragments are separated based on their size.
Since $DNA$ molecules are negatively charged,they move towards the anode $(+)$ when an electric field is applied.
The agarose gel acts as a molecular sieve,allowing smaller fragments to move faster and further through the gel matrix compared to larger fragments.
In the provided diagram,the bands closer to the wells are larger because they move slowly,while the bands further away from the wells are smaller because they move faster.
Therefore,$M$ represents the largest $DNA$ bands (closest to the wells),and $N$ (labeled as $S$ in the diagram) represents the smallest $DNA$ bands (farthest from the wells).

(A) The correct answer is $A$.
Biolistics,also known as the gene gun method,is specifically designed for the transformation of plant cells.
In this technique,microparticles of gold or tungsten coated with foreign $DNA$ are bombarded into the target cells at high velocity.
These particles penetrate the rigid cell walls and membranes of plant cells to deliver the genetic material directly into the cytoplasm or nucleus without causing significant damage to the cell.

(A) The Polymerase Chain Reaction $(PCR)$ consists of three main steps: Denaturation,Annealing,and Extension.
$1$. Denaturation: The double-stranded $DNA$ is heated to high temperatures $(94-98^{\circ}C)$ to separate the strands. This is a physical process,not enzyme-dependent.
$2$. Annealing: Primers bind to the single-stranded $DNA$ template at lower temperatures $(50-65^{\circ}C)$. This is also not enzyme-dependent.
$3$. Extension: The enzyme $Taq$ $DNA$ polymerase adds nucleotides to the primer,extending the new $DNA$ strand. This step is strictly dependent on the activity of the $DNA$ polymerase enzyme.
Therefore,the correct answer is Extension only.