KCET 2021 Biology Question Paper with Answer and Solution

(A) is the correct sequence: Synapsis $\rightarrow$ Crossing over $\rightarrow$ Chiasmata $\rightarrow$ Terminalisation.
Prophase $I$ of meiosis is divided into $5$ sub-stages: leptotene,zygotene,pachytene,diplotene,and diakinesis.
$1$. Synapsis: Pairing of homologous chromosomes occurs during the zygotene stage.
$2$. Crossing over: Exchange of genetic material between non-sister chromatids of homologous chromosomes occurs during the pachytene stage.
$3$. Chiasmata: The $X$-shaped structures formed at the sites of crossing over become visible during the diplotene stage.
$4$. Terminalisation: The shifting of chiasmata towards the ends of the chromosomes occurs during the diakinesis stage.

(D) $PEP$ carboxylase $(PEPcase)$ is an enzyme that catalyzes the carboxylation of phosphoenolpyruvate $(PEP)$ in $C_4$ plants.
It is primarily involved in the initial carbon fixation step in $C_4$ plants and is not found in the mesophyll cells of $C_3$ plants.

(A) Mitochondrial matrix is the site for the Krebs' cycle $(i)$.
$(B)$ Cytoplasm is the site for Glycolysis $(iii)$.
$(C)$ $F_0$ and $F_1$ particles ($ATP$ synthase) are responsible for $ATP$ synthesis $(iv)$.
$(D)$ Inner mitochondrial membrane is the site for the Electron Transport Chain $(ETC)$ $(ii)$.
Therefore,the correct matching is $(A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)$.

(D) The correct answer is $D$ ($I$ and $V$).
Statement $I$ is incorrect because Kinetin is a derivative of adenine,which is a purine,not a pyrimidine.
Statement $II$ is correct; decapitation (removal of shoot tips) is used in tea plantations to promote lateral branching.
Statement $III$ is correct; ethylene is a gaseous plant hormone.
Statement $IV$ is correct; Gibberellic acid $(GA)$ is used to speed up the malting process in the brewing industry.
Statement $V$ is incorrect because Abscisic acid $(ABA)$ acts as a growth inhibitor,not a growth promoter.

(A) $P$ is the contraction of only left atria is incorrect because the $P$-wave represents the electrical excitation (depolarisation) of both atria,which leads to the contraction of both atria simultaneously.
$ECG$ or Electrocardiogram records the electrical activity of the heart to diagnose various cardiac conditions.
$1$. The $P$-wave represents atrial depolarisation,which leads to the contraction of both atria.
$2$. The $QRS$ complex represents ventricular depolarisation,which initiates ventricular contraction (systole).
$3$. The $T$-wave represents ventricular repolarisation,which marks the end of ventricular systole as the ventricles begin to relax.

(C) Cardiac output is defined as the volume of blood pumped by the heart per minute.
It is calculated by multiplying the stroke volume $(SV)$,which is the volume of blood ejected by the ventricle during each contraction,by the heart rate $(HR)$,which is the number of heart beats per minute.
The formula is: $\text{Cardiac Output} (CO) = SV \times HR$.
Given: Stroke volume $(SV)$ = $55 \ mL$ and Heart rate $(HR)$ = $70 \ \text{beats/min}$.
Therefore,$\text{Cardiac Output} = 55 \times 70 = 3850 \ mL$.

(A) The correct matching is as follows:
$A$. Catecholamines (e.g.,adrenaline and noradrenaline) are secreted by the adrenal medulla and are known as stress hormones,which help the body respond to emergency situations. Thus,$A-(iv)$.
$B$. $MSH$ (Melanocyte Stimulating Hormone) acts on melanocytes to regulate skin pigmentation. Thus,$B-(iii)$.
$C$. Thymosins are secreted by the thymus gland and play a major role in the differentiation of $T$-lymphocytes,which provide cell-mediated immunity. Thus,$C-(ii)$.
$D$. Melatonin is secreted by the pineal gland and plays a very important role in the regulation of a $24$-hour (diurnal) rhythm of our body,such as the sleep-wake cycle and body temperature. Thus,$D-(i)$.
Therefore,the correct sequence is $A-(iv), B-(iii), C-(ii), D-(i)$.

(A) $Ustilago$ is an example of the class $Basidiomycetes$ (club fungi).
$Alternaria$,$Colletotrichum$,and $Trichoderma$ are all examples of the class $Deuteromycetes$ (imperfect fungi).
Therefore,$Ustilago$ is the odd one out.

(C) The correct answer is $C$.
In $1898$, the Dutch microbiologist $M. W. Beijerinck$ demonstrated that the extract of infected tobacco plants could cause infection in healthy plants.
He called this fluid extract "Contagium vivum fluidum", which translates to "infectious living fluid".

(A) $Laminaria$.
$Phaeophyceae$ (brown algae) typically possess a plant body differentiated into a root-like structure called a holdfast,a stem-like stalk called a stipe,and leaf-like photosynthetic organs called fronds.
$Laminaria$ is a well-known genus of brown algae belonging to the class $Phaeophyceae$.

(D) The correct answer is $D$.
Members of class $Aves$ are homeothermic or warm-blooded animals.
Their forelimbs are modified into wings to assist in flight,while their hindlimbs are adapted for walking,swimming,or perching.
The heart in birds is completely four-chambered,consisting of two atria and two ventricles.
They are oviparous (egg-laying) and exhibit direct development,meaning there is no larval stage.
Therefore,all the given statements $(I, II, III, IV)$ are correct.

(B) In epigynous flowers,the margin of the thalamus grows upward enclosing the ovary completely and getting fused with it,while other parts of the flower arise above the ovary.
Hence,the ovary is said to be inferior and other floral parts are superior.
Examples include flowers of guava,cucumber,and the ray florets of sunflower.

(A) The correct answer is $A$.
$A$ bacterial flagellum is composed of three distinct parts:
$1$. The basal body: This is the structure that anchors the flagellum into the cell wall and plasma membrane.
$2$. The hook: This is a curved structure that connects the basal body to the filament.
$3$. The filament: This is the longest portion that extends from the cell surface into the surrounding medium and is responsible for bacterial motility.

(D) $A-(iii), B-(iv), C-(ii), D-(i)$
The correct matching is as follows:
$A$. Trypsin is an enzyme that digests proteins.
$B$. $GLUT$-$4$ is a protein that enables glucose transport into cells.
$C$. Collagen is a structural protein that forms the intercellular ground substance in connective tissues.
$D$. Antibody is a protein produced by our body to fight against infectious agents and foreign pathogens.

(A) The given diagram represents the structure of a monocotyledonous seed (maize grain).
$A$ points to the coleoptile,which is the protective sheath covering the plumule.
$B$ points to the scutellum,which is a large,shield-shaped cotyledon.
$C$ points to the pericarp,which is the outer wall of the fruit fused with the seed coat.
$D$ points to the coleorrhiza,which is the protective sheath covering the radicle.
Therefore,the correct identification is $A$-Coleoptile,$B$-Scutellum,$C$-Pericarp,$D$-Coleorrhiza.

(D) During double fertilisation in angiosperms,one male gamete fuses with the egg cell to form a zygote (syngamy).
Another male gamete fuses with the two polar nuclei located in the central cell to form the primary endosperm nucleus $(PEN)$.
This process is known as triple fusion.
Therefore,the central cell is the cell of the female gametophyte that is involved in the formation of the $PEN$.

(C) Pollination by water (hydrophily) is quite rare in flowering plants and is limited to about $30$ genera, mostly monocotyledons.
In many aquatic plants such as $Vallisneria$ and $Hydrilla$, pollination occurs through water.
However, in many other aquatic plants like $Water \text{ } hyacinth$ $(Eichhornia)$ and $Water \text{ } lily$, the flowers emerge above the level of water and are pollinated by insects or wind, similar to most land plants.
Therefore, $Water \text{ } hyacinth$ does not show pollination by water.

(A) In the process of megasporogenesis,the megaspore mother cell $(MMC)$ undergoes meiosis to form four haploid megaspores.
Out of these four megaspores,three megaspores present towards the micropylar end degenerate,while only one megaspore present towards the chalazal end remains functional.
This functional megaspore develops into the female gametophyte (embryo sac).
The mature female gametophyte is typically $7$-celled and $8$-nucleated.

(B) The correct answer is $B$.
$Statement \ 1$ is correct because the microsporangium is surrounded by four wall layers: epidermis, endothecium, middle layers, and the innermost layer, which is the tapetum.
$Statement \ 2$ is correct because the cells of the tapetum possess dense cytoplasm and generally have more than one nucleus (multinucleate condition). Their primary function is to provide nourishment to the developing pollen grains.

(B) typical angiosperm anther is bilobed,with each lobe having two theca (dithecous).
In a transverse section,a typical microsporangium appears near-circular in outline.
It is generally surrounded by four wall layers: epidermis,endothecium,middle layers,and tapetum.
There are four microsporangia located at the corners,two in each lobe.
Eventually,these microsporangia develop into pollen sacs and extend longitudinally all through the length of an anther and are packed with pollen grains.

(C) The process of pollen-pistil interaction follows a specific sequence:
$1$. $II.$ Deposition of pollen grains on the stigma.
$2$. $IV.$ Development of the pollen tube through the style.
$3$. $V.$ Entry of the pollen tube into the ovule through the micropyle.
$4$. $III.$ Entry of the pollen tube into the embryo sac (synergids).
$5$. $I.$ Release of male gametes into the embryo sac for fertilization.
Therefore,the correct sequence is $II \to IV \to V \to III \to I$.

(C) The correct answer is $C$.
Gene silencing is a biological process in which the expression of a specific gene is suppressed or turned off.
This can be achieved using $RNA$ interference $(RNAi)$,which involves short interfering $RNA$ molecules that bind to complementary $mRNA$ and trigger its degradation.
Similarly,antisense $RNA$ technology involves introducing an $RNA$ molecule that is complementary to the target $mRNA$,thereby preventing its translation into protein.
Therefore,both $RNAi$ and antisense $RNA$ are effective methods for gene silencing.

(B) $PCR$.
Polymerase Chain Reaction $(PCR)$ is a molecular biology technique that allows for the amplification of specific $DNA$ sequences.
It enables the detection of pathogens like bacteria or viruses even when they are present in very low concentrations in a sample by creating millions of copies of their nucleic acid.

(A) $PCR$ stands for Polymerase Chain Reaction.
It is a molecular diagnostic technique used to amplify specific segments of $DNA$ in vitro.
In this process,a short region of a $DNA$ molecule is copied many times using the $DNA$ polymerase enzyme,allowing for the detection of pathogens like the $COVID-19$ virus even in small quantities.

(C) $\alpha-1$ antitrypsin is a protein that inhibits the enzyme elastase,which breaks down lung tissue.
Deficiency of this protein leads to emphysema,a chronic lung disease.
Therefore,$\alpha-1$ antitrypsin is used to treat emphysema by replacing the missing protein.

(D) The correct answer is $D$.
Population size is not a static parameter; it keeps fluctuating due to various ecological factors.
$1$. Food availability: Limited resources restrict population growth,while abundance promotes it.
$2$. Predation pressure: High predation rates reduce the population size of prey species.
$3$. Adverse weather: Extreme conditions like droughts,floods,or severe winters can significantly decrease population density.
Therefore,all the mentioned factors are responsible for the dynamic nature of population size.

(B) . Calotropis is a plant that produces highly poisonous cardiac glycosides,which protect it from herbivores. Thus,$A-iv$.
$B$. Pisaster is a starfish that acts as a predator in rocky intertidal communities of the American Pacific Coast,feeding on invertebrates. Thus,$B-i$.
$C$. The Monarch butterfly is highly distasteful to its predator (birds) because of a special chemical acquired during its caterpillar stage. Thus,$C-ii$.
$D$. Frogs are often cryptically coloured (camouflaged) to escape detection by predators. Thus,$D-iii$.
Therefore,the correct matching is $A-iv, B-i, C-ii, D-iii$.

(C) The correct statement is $III$.
Explanation of incorrect statements:
$I$. $Cuscuta$ is a parasitic plant that lacks chlorophyll and cannot perform photosynthesis; it is not a chlorophyllous endoparasite.
$II$. The human liverfluke ($Fasciola$ $hepatica$) requires two intermediate hosts (a snail and a fish) to complete its life cycle.
$IV$. In brood parasitism,the parasitic bird's eggs have evolved to resemble the host's eggs in size and color to avoid detection by the host,not the other way around.
$III$. Endoparasites exhibit extreme specialization,which makes their life cycles more complex as they must adapt to the host's internal environment and often require multiple hosts.

(A) The species-area relationship was proposed by Alexander von Humboldt.
The relationship is represented by the equation $S = CA^Z$.
Taking the logarithm on both sides,we get:
$\log S = \log(CA^Z)$
$\log S = \log C + \log(A^Z)$
$\log S = \log C + Z \log A$
Therefore,the correct equation is $\log S = \log C + Z \log A$.

(C) The correct answer is $C$ (Only $IV$).
Statement $I$ is correct because evolutionary processes like speciation require significant time to occur.
Statement $II$ is correct as tropical latitudes remain relatively undisturbed by seasonal variations,making them more constant and predictable.
Statement $III$ is correct because higher solar energy availability in the tropics leads to higher primary productivity.
Statement $IV$ is incorrect. Unlike tropical regions,temperate regions have been subjected to frequent glaciations in the past,which disrupted the environment and hindered species diversification.

(C) is the correct answer.
Advanced $ex \, situ$ conservation involves the preservation of genetic resources of species away from their natural habitat or area of origin.
This strategy is designed to conserve a larger amount of genetic material within a smaller physical space.
The various methods through which $ex \, situ$ conservation is achieved include:
$1$. Cryopreservation (storage of gametes or embryos at extremely low temperatures).
$2$. Plant tissue culture (propagation of plants from cells or tissues).
$3$. Seed banks (long-term storage of seeds).
$4$. $In \, vitro$ fertilisation (fertilisation outside the living organism).
Therefore, all the listed statements $(I, II, III, IV)$ are correct.

(C) The arguments for biodiversity conservation are categorized into three types:
$1$. Narrowly utilitarian: Humans derive countless direct economic benefits from nature such as food (cereals,pulses,fruits),firewood,fiber,construction material,industrial products (tannins,lubricants,dyes,resins,perfumes) and products of medicinal importance. Thus,$(A)$ matches with $(iii)$.
$2$. Broadly utilitarian: Biodiversity plays a major role in many ecosystem services that nature provides. The fast-dwindling Amazon forest is estimated to produce,through photosynthesis,$20$ percent of the total oxygen in the earth's atmosphere. Other services include pollination and aesthetic pleasures. Thus,$(B)$ matches with $(i)$.
$3$. Ethical: This argument relates to what we owe to millions of plant,animal and microbe species with whom we share this planet. Every species has an intrinsic value,even if it may not be of current or any economic value to us. We have a moral duty to care for their well-being and pass on our biological legacy in good order to future generations. Thus,$(C)$ matches with $(ii)$.
Therefore,the correct matching is $(A)-(iii), (B)-(i), (C)-(ii)$.

(B) The correct answer is $(B)$.
$A$ represents the acrosome,which contains enzymes that help in the penetration of the sperm into the ovum.
$B$ represents the middle piece,which contains numerous mitochondria arranged in a spiral. These mitochondria provide the energy $(ATP)$ required for the movement of the sperm.
$C$ represents the tail,which facilitates the movement of the sperm through the female reproductive tract.

(B) The correct answer is $B$ (Myometrium).
During parturition,the hormone oxytocin acts on the uterine muscles.
The $Myometrium$ is the thick middle layer of the uterus composed of smooth muscle tissue.
Under the influence of oxytocin,the $Myometrium$ undergoes strong rhythmic contractions,which facilitate the expulsion of the fetus from the uterus.

(D) Alveoli $ \rightarrow $ Mammary tubules $ \rightarrow $ Mammary duct $ \rightarrow $ Mammary ampulla $ \rightarrow $ Lactiferous duct.
$A$ functional mammary gland is a characteristic feature of all female mammals. The mammary glands are paired structures (breasts) that contain glandular tissue and a variable amount of fat. The glandular tissue of each breast is divided into $15-20$ mammary lobes containing clusters of cells called alveoli. The cells of the alveoli secrete milk,which is stored in the cavities (lumens) of the alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla,which is connected to the lactiferous duct through which milk is sucked out.

(A) is the incorrect statement.
Contraceptives are not regular requirements for the maintenance of reproductive health.
They are used specifically to prevent unwanted pregnancies and to space births,rather than to maintain general reproductive health.
Options $B$,$C$,and $D$ are correct statements regarding the role and potential side effects of contraceptives.

(C) The correct answer is $C$.
Intra-cytoplasmic sperm injection $(ICSI)$ is a specialized procedure in which a single sperm is directly injected into the ovum to form an embryo in the laboratory.
This technique is primarily used to treat cases of male infertility.

(D) The correct matching is as follows:
$(A)$ Autosomal trisomy: $(iv)$ Down's Syndrome (Trisomy of chromosome $21$).
$(B)$ Allosomal trisomy: $(iii)$ Klinefelter's Syndrome $(47, XXY)$.
$(C)$ Allosomal Monosomy: $(i)$ Turner's Syndrome $(45, XO)$.
$(D)$ Cystic fibrosis: $(ii)$ Mendelian disorder (Autosomal recessive trait).
Therefore,the correct sequence is $(A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)$.

(C) The correct matching is as follows:
$(A)$ Aneuploidy: It refers to the loss or gain of one or more chromosomes,resulting in an abnormal number of chromosomes $(ii)$.
$(B)$ Monoploidy: It refers to the presence of a single set of chromosomes $(n)$ in an organism $(iv)$.
$(C)$ Polyploidy: It refers to an increase in the whole set of chromosomes,such as $3n, 4n$,etc. $(i)$.
$(D)$ Diploidy: It refers to the presence of two sets of chromosomes $(2n)$ in an organism $(iii)$.
Therefore,the correct sequence is $A-ii, B-iv, C-i, D-iii$.

(B) The correct answer is $B$.
Parents are heterozygous for blood groups $A$ and $B$,meaning their genotypes are $I^A i$ and $I^B i$.
When these parents cross,the possible combinations of alleles in the offspring are:
$1. I^A I^B$ (Blood group $AB$)
$2. I^A i$ (Blood group $A$)
$3. I^B i$ (Blood group $B$)
$4. ii$ (Blood group $O$)
Therefore,the children can have any of the four blood groups: $A, B, AB,$ or $O$.

(B) The genotypes of the parents are $I^A I^B$ (husband) and $I^A I^O$ (wife).
To find the possible genotypes of the offspring,we perform a Punnett square cross:
- $I^A$ from father $\times$ $I^A$ from mother $\rightarrow$ $I^A I^A$ (Blood type $A$)
- $I^A$ from father $\times$ $I^O$ from mother $\rightarrow$ $I^A I^O$ (Blood type $A$)
- $I^B$ from father $\times$ $I^A$ from mother $\rightarrow$ $I^A I^B$ (Blood type $AB$)
- $I^B$ from father $\times$ $I^O$ from mother $\rightarrow$ $I^B I^O$ (Blood type $B$)
Possible genotypes: $I^A I^A, I^A I^O, I^A I^B, I^B I^O$ (Total $4$ genotypes).
Possible phenotypes: Blood type $A$,Blood type $AB$,Blood type $B$ (Total $3$ phenotypes).
Therefore,the correct answer is $4$ genotypes and $3$ phenotypes.

(C) Gregor Mendel studied $7$ pairs of contrasting traits in pea plants.
Among the given options,the recessive characters are: white flower color,constricted pod shape,yellow pod color,terminal flower position,dwarf plant height,wrinkled seed shape,and green seed color.
Therefore,$C$ (White flower) is the correct answer as it is a recessive trait,whereas inflated pod,green pod color,and axillary flower position are dominant traits.

(C) The correct answer is $C$.
In a $tRNA$ molecule,the amino acid binding site is located at the $3'$ end.
This end contains a conserved sequence $CCA$ where the specific amino acid is attached via an ester bond to the $3'-OH$ group of the terminal adenosine residue.

(C) The correct answer is $C$.
According to the semi-conservative model of $DNA$ replication demonstrated by Meselson and Stahl,the parental $DNA$ strands (labeled with $^{15}N$) separate,and each acts as a template for the synthesis of a new strand using $^{14}N$ from the medium.
In the first generation,each $DNA$ molecule consists of one heavy strand $(^{15}N)$ and one light strand $(^{14}N)$.
Since the two strands have different densities (one heavy and one light),the resulting $DNA$ molecule is a hybrid.
Because the hybrid $DNA$ contains one $^{14}N$ strand,it does not resemble the original parent $DNA$ (which was entirely $^{15}N$).

$(C)$. Griffith performed the transformation experiment, which led to the discovery of the "Transforming principle" $(A-iii)$.
$B$. Jacob and Monod proposed the "Lac operon" model to explain gene regulation in bacteria $(B-i)$.
$C$. Meselson and Stahl provided experimental evidence that "$DNA$ replicates semiconservatively" $(C-iv)$.
$D$. Hershey and Chase used bacteriophages to prove that "$DNA$ is the genetic material" $(D-ii)$.
Therefore, the correct match is $A-iii, B-i, C-iv, D-ii$.

(A) The experiment to demonstrate the semi-conservative replication of $DNA$ in eukaryotes was performed by Taylor and his colleagues in $1958$.
They used radioactive thymidine to detect the distribution of newly synthesized $DNA$ in the chromosomes.
This experiment was performed on the root meristem cells of the plant $Vicia$ $faba$ (fava bean).
By using autoradiography, they observed that the radioactive label was distributed in both chromatids of the chromosome, confirming the semi-conservative mode of $DNA$ replication.

(B) During transcription,the enzyme $RNA$ polymerase synthesizes mRNA using the $DNA$ template strand.
According to the principle of complementarity,the nitrogenous bases in the mRNA are complementary to the template strand of $DNA$.
In $DNA$,Adenine $(A)$ pairs with Thymine $(T)$,and Cytosine $(C)$ pairs with Guanine $(G)$.
However,in $RNA$,Uracil $(U)$ replaces Thymine $(T)$.
Therefore,the base pairing rule for transcription is:
$A$ $(DNA)$ $\rightarrow$ $U$ (mRNA)
$T$ $(DNA)$ $\rightarrow$ $A$ (mRNA)
$C$ $(DNA)$ $\rightarrow$ $G$ (mRNA)
$G$ $(DNA)$ $\rightarrow$ $C$ (mRNA)
Given $DNA$ template strand: $3'-ATGCTTCCGAAT-5'$
Complementary mRNA sequence: $5'-UACGAAGGCCUU-3'$
Thus,the correct option is $B$.

(D) The correct answer is $D$.
Techniques such as $CT$ (Computed Tomography),$MRI$ (Magnetic Resonance Imaging),and Radiology ($X$-rays) are widely used for the detection of cancers in internal organs.
$CT$ uses $X$-rays to generate a three-dimensional image of the internal organs.
$MRI$ uses strong magnetic fields and non-ionizing radiations to accurately detect pathological and physiological changes in the living tissue.
Therefore,all the listed techniques are useful for cancer detection.

(B) The correct answer is $B$. Metastasis is the most feared property of malignant tumors.
$1$. Malignant tumors consist of a mass of proliferating cells known as neoplastic or tumor cells.
$2$. These cells grow rapidly,invading and damaging the surrounding normal tissues.
$3$. As these cells actively divide,they can detach from the primary tumor and spread to distant organs of the body through the blood and lymph systems.
$4$. This process of spreading and developing secondary tumors at distant sites is called metastasis,which makes cancer difficult to treat and life-threatening.

(C) The correct answer is $C$.
$Pneumonia$ is a respiratory infection that affects the alveoli of the lungs.
It is primarily caused by pathogenic bacteria such as $Streptococcus$ $pneumoniae$ and $Haemophilus$ $influenzae$.

(B) $Erythroxylum$ $coca$ is the correct answer.
$Erythroxylum$ $coca$ is a shrub native to South America.
The leaves of this plant are used to extract a stimulant drug known as cocaine or coca alkaloid,which interferes with the transport of the neurotransmitter dopamine.

(D) passive immunisation.
When a person is exposed to deadly microbes (like those causing tetanus or snake bites),there is no time to wait for the body to produce its own antibodies. In such cases,preformed antibodies or antitoxins are directly injected into the patient's body to provide an immediate immune response. This type of immunity is known as passive immunisation.

(A) The correct answer is $A$.
Large holes in Swiss cheese are formed due to the production of a large amount of $CO_2$ by the bacterium $Propionibacterium$ $shermanii$ during the fermentation process.

(A) The correct matching is: $(A)$-$(iii)$; $(B)$-$(i)$; $(C)$-$(iv)$; $(D)$-$(ii)$.
$1$. Cyclosporin-$A$ is produced by the fungus $Trichoderma$ $polysporum$. It is used as an immuno-suppressive agent in organ transplant patients.
$2$. Streptokinase is produced by the bacterium $Streptococcus$ and is used as a 'clot buster' for removing clots from the blood vessels of patients who have undergone myocardial infarction.
$3$. Statins are produced by the yeast $Monascus$ $purpureus$. They act as blood cholesterol-lowering agents by competitively inhibiting the enzyme responsible for cholesterol synthesis.
$4$. Penicillin was the first antibiotic discovered,obtained from the fungus $Penicillium$ $notatum$.

(C) - Swiss cheese.
$Swiss$ cheese is produced by the fermentation of lactic acid by the bacterium $Propionibacterium$ $shermanii$,which produces propionic acid,acetic acid,and carbon dioxide.
The acids provide flavor to the cheese,and the carbon dioxide,which becomes trapped in the curd,produces the characteristic holes in the cheese.
In contrast,$Penicillin$ is obtained from the fungus $Penicillium$ $notatum$,$Statins$ are produced by the yeast $Monascus$ $purpureus$,and $Cyclosporin-A$ is produced by the fungus $Trichoderma$ $polysporum$.

(D) The correct answer is $(D)$ both $(a)$ and $(b)$.
In India,the technology of biogas production was developed mainly due to the efforts of the Indian Agricultural Research Institute $(IARI)$ and the Khadi and Village Industries Commission $(KVIC)$.
$IARI$ conducted significant research on the microbial processes involved in biogas production.
$KVIC$ played a crucial role in the $1960s$ by developing and promoting practical models for biogas plants suitable for rural areas.

(C) The given sequence $5'-GAATTC-3'$ and $3'-CTTAAG-5'$ is a palindromic sequence.
In molecular biology,a palindromic $DNA$ sequence is a sequence of nucleotides that reads the same in the $5'$ to $3'$ direction on one strand as it does in the $5'$ to $3'$ direction on the complementary strand.
This specific sequence is the recognition site for the restriction enzyme $EcoRI$.

(C) The correct answer is $C$.
The $rop$ gene,which codes for the proteins involved in the replication of the plasmid $pBR322$ in $E. coli$,is located at the restriction site of $Pvu II$.
The repressor of primer $(rop)$ gene codes for the $rop$ protein,which regulates the copy number of the plasmid and modulates its replication process.