KCET 2014 Biology Question Paper with Answer and Solution

(B) The correct answer is $B$.
During the $S$-phase (Synthesis phase) of the cell cycle,$DNA$ replication occurs within the nucleus.
Simultaneously,the centrosome,which is responsible for organizing microtubules,also duplicates in the cytoplasm.
This ensures that each daughter cell receives a pair of centrosomes during cell division.

(B) The correct answer is $B$. Both statements $A$ and $B$ are correct.
Photorespiration is a wasteful process in $C_3$ plants.
In $C_3$ plants,some $O_2$ binds to $RuBisCO$ instead of $CO_2$,which decreases the efficiency of $CO_2$ fixation.
In this pathway,$RuBP$ binds with $O_2$ to form one molecule of phosphoglycerate $(3C)$ and one molecule of phosphoglycolate $(2C)$.
During the photorespiratory pathway,there is no synthesis of sugars,$ATP$,or $NADPH$. Instead,it leads to the release of $CO_2$ and the consumption of $ATP$.

(A) The correct matching is as follows:
$1$. Phosphoenol pyruvate $(PEP)$ is a $3$-carbon compound $(t)$.
$2$. Ribulose biphosphate $(RuBP)$ is a $5$-carbon compound $(s)$.
$3$. Oxaloacetic acid $(OAA)$ is a $4$-carbon compound $(r)$.
$4$. Acetyl coenzyme $A$ is a $2$-carbon compound $(q)$.
Therefore,the correct sequence is $A-t, B-s, C-r, D-q$.

(C) $A-q, B-t, C-p, D-r, E-s$
$1$. Residual Volume $(RV)$ is the volume of air remaining in the lungs even after a forcible expiration, which is approximately $1100 \text{ ml} - 1200 \text{ ml}$.
$2$. Inspiratory Reserve Volume $(IRV)$ is the additional volume of air, a person can inspire by a forcible inspiration, which is $2500 \text{ ml} - 3000 \text{ ml}$.
$3$. Vital Capacity $(VC)$ is the maximum volume of air a person can breathe in after a forced expiration, which is $4000 \text{ ml} - 4600 \text{ ml}$.
$4$. Expiratory Reserve Volume $(ERV)$ is the additional volume of air, a person can expire by a forcible expiration, which is $1000 \text{ ml} - 1100 \text{ ml}$.
$5$. Inspiratory Capacity $(IC)$ is the total volume of air a person can inspire after a normal expiration, which is $3000 \text{ ml} - 3500 \text{ ml}$.

(D) The correct answer is $D$.
The knee joint is an example of a hinge joint.
$A$ hinge joint is a type of synovial joint that allows movement primarily in one plane,similar to the hinge of a door.
It is characterized by the presence of a fluid-filled synovial cavity between the articulating surfaces of the two bones,which facilitates smooth movement.

(B) The correct answer is $B$ (Cortisol).
$1$. Cortisol is a glucocorticoid secreted by the adrenal cortex.
$2$. In addition to its primary role in carbohydrate metabolism,it plays a crucial role in maintaining the cardiovascular system and kidney functions.
$3$. Cortisol is well-known for producing anti-inflammatory reactions and suppressing the immune response,which is why it is often used clinically to treat inflammatory conditions and autoimmune diseases.
$4$. It also stimulates the production of red blood cells $(RBCs)$.

(C) $Marchantia$ is considered a heterothallic plant because it is dioecious.
In $Marchantia$,the male and female reproductive structures (antheridiophore and archegoniophore) are borne on separate individual thalli.
Therefore,the plant body is unisexual,which is referred to as dioecious or heterothallic.

(A) The correct matching is as follows:
$1$. $A$. Glycogen: It is the stored food material in fungi,such as $Agaricus$ $(t)$.
$2$. $B$. Pyrenoids: These are proteinaceous storage bodies found in the chloroplasts of green algae,such as $Spirogyra$ $(s)$.
$3$. $C$. Laminarin and mannitol: These are the characteristic stored food products of brown algae,such as $Sargassum$ $(p)$.
$4$. $D$. Floridean starch: This is the stored food material in red algae,which is very similar to amylopectin and glycogen in structure,such as $Polysiphonia$ $(r)$.
Therefore,the correct sequence is $A-t, B-s, C-p, D-r$.

(A) The correct answer is $A$ (round worm).
The diagram illustrates a pseudocoelom,which is a body cavity that is not completely lined by mesoderm. Instead,the mesoderm is present as scattered pouches between the ectoderm and endoderm.
$1$. Round worms (Phylum Aschelminthes) are characterized by the presence of a pseudocoelom.
$2$. Earthworms (Phylum Annelida) and cockroaches (Phylum Arthropoda) are eucoelomate (true coelomates).
$3$. Tapeworms (Phylum Platyhelminthes) are acoelomate (lack a body cavity).

(C) Option $C$ is correct. In some prokaryotes like cyanobacteria,there are other membranous extensions into the cytoplasm called chromatophores which contain pigments.
Explanation of other options:
- Option $A$: The core of a cilium or flagellum is called the axoneme,not the basal body.
- Option $B$: Elaioplasts store oils and fats,while amyloplasts store starch. Aleuroplasts store proteins.
- Option $D$: Acrocentric chromosomes have two arms,one very short and one very long,with the centromere situated close to one end.

(D) $RuBisCO$ (Ribulose bisphosphate carboxylase-oxygenase) is the most abundant protein in the plant world, as it is essential for the process of photosynthesis.
$Collagen$ is the most abundant protein in the animal world, providing structural support to tissues.
Therefore, $RuBisCO$ of plants and $Collagen$ of animals are the most abundant proteins in the living world.

(C) The correct option is $C$.
$A$ represents $GnRH$ (Gonadotropin-releasing hormone),which is secreted by the hypothalamus.
$GnRH$ stimulates the anterior pituitary gland to release gonadotropins,which are $FSH$ (Follicle-stimulating hormone) and $LH$ (Luteinizing hormone).
$B$ represents $FSH$,which acts on the ovaries to stimulate the growth of ovarian follicles.
$C$ represents $LH$,which acts on the testes to stimulate Leydig cells to secrete androgens (testosterone).

(B) In males,one primary spermatocyte undergoes meiosis to produce $4$ functional,viable sperm cells.
In females,one primary oocyte undergoes meiosis,but due to unequal cytoplasmic division,it produces only $1$ functional,viable ovum and $2$ or $3$ non-functional polar bodies that eventually degenerate.
Therefore,the ratio of gametes produced from one male primary sex cell to one female primary sex cell is $4:1$.

(A) Based on the standard diagram of the $L$.$S$. of a grass embryo (monocot embryo) as provided in the $NCERT$ textbook:
- $A$ represents the Epiblast (the small,rudimentary second cotyledon).
- $B$ represents the Radicle (the embryonic root).
- $C$ represents the Coleoptile (the protective sheath covering the plumule).
- $D$ represents the Scutellum (the large,shield-shaped cotyledon).
- The structure covering the radicle is the Coleorrhiza.
- The top portion represents the Shoot apex (plumule).
Comparing these labels with the given options,option $A$ provides the correct identification for the parts labeled in the standard diagram of a grass embryo.

(C) The correct answer is $C$.
In a mature pollen grain,the outer protective layer is known as the exine,which is composed of a highly resistant organic material called sporopollenin.
However,at certain specific locations,the exine is absent or very thin,and these regions lack sporopollenin.
These specific regions are termed as germ pores.
During the process of pollen germination,the pollen tube emerges from the pollen grain through these germ pores.

(B) In castor and maize plants,both male and female flowers are present on the same plant (monoecious condition).
Since the flowers are unisexual,self-pollination within the same flower (autogamy) is prevented.
However,pollen grains can be transferred from the anther of one flower to the stigma of another flower on the same plant,which is known as geitonogamy.
Therefore,autogamy is prevented but geitonogamy is not.

(B) The correct answer is $B$.
Antipodal cells are formed after the mitotic division of the functional megaspore within the embryo sac.
Since the megaspore is haploid $(n)$,the antipodal cells formed from it are also haploid $(n)$,not diploid $(2n)$.
Therefore,the pair 'Antipodals - $2n$' is incorrect.

(A) Somatic hybridisation is a technique in plant tissue culture where protoplasts from two different plant species are fused to form a hybrid protoplast.
To achieve this,the cell walls and the middle lamella must be removed or digested using enzymes like cellulase and pectinase.
Once the cell walls are removed,the resulting protoplasts are fused to create a somatic hybrid,which can then be regenerated into a new plant.
Therefore,the correct process involves the digestion of cell walls and middle lamella before fusion.

(B) - preventing the process of translation of mRNA.
$RNA$ interference $(RNAi)$ is a biological process in which $RNA$ molecules inhibit gene expression or translation by neutralizing targeted mRNA molecules.
In the case of $Meloidogyne \ incognita$ infecting tobacco plants,$dsRNA$ is introduced into the host cells.
This $dsRNA$ initiates the $RNAi$ pathway,which leads to the degradation of the specific mRNA of the nematode,thereby preventing its translation into essential proteins required for the parasite's survival.

(D) $ADA$ stands for adenosine deaminase.
This enzyme is crucial for the proper functioning of the immune system.
The deficiency of this enzyme leads to a condition known as Severe Combined Immunodeficiency $(SCID)$.
In patients with $SCID$, the body lacks functional $T$-lymphocytes, which are essential for mounting an immune response.
Consequently, the immune system is unable to function normally.

(B) The incorrect statement is $B$.
$1$. The pyramid of energy is always upright because energy flow in an ecosystem is unidirectional and follows the $10\%$ law, where energy is lost as heat at each successive trophic level.
$2$. It is impossible for a pyramid of energy to be inverted because energy cannot be created at higher trophic levels.
$3$. The pyramid of biomass in the sea is inverted because the standing crop of phytoplankton (producers) is smaller than that of the zooplankton and fish (consumers) at any given time.
$4$. Pyramids of number and biomass can be upright or inverted depending on the ecosystem structure.

(D) The correct answer is $D$.
Habitat destruction,defined as the elimination or alteration of the conditions necessary for animals and plants to survive,is the primary cause of species extinction and endangerment.
It not only impacts individual species but also the overall health of the global ecosystem.
Habitats are destroyed or fragmented by various human activities such as industrial development,deforestation,and pollution.
While over-hunting and poaching also have harmful effects,habitat loss remains the most significant threat to biodiversity.

(A) The correct answer is $A$. The statement that the richest reservoirs of animal and plant life with few or no threatened species are called "biodiversity hotspots" is incorrect.
Biodiversity hotspots are regions that are characterized by high levels of species richness and high degree of endemism, but they are also the most threatened regions due to habitat loss and fragmentation.
They are identified based on three main criteria:
$(i)$ Degree of endemism (presence of species found nowhere else).
$(ii)$ Degree of threat to the habitat due to degradation and fragmentation.
$(iii)$ High number of species or species diversity.

(C) The correct answer is $C$.
During pregnancy,the placenta acts as an endocrine tissue and produces several hormones,including human chorionic gonadotropin $(hCG)$,human placental lactogen $(hPL)$,estrogens,and progestogens.
$hCG$,$hPL$,and relaxin are hormones that are produced in women only during pregnancy.
Additionally,the levels of other hormones such as estrogens,progestogens,cortisol,prolactin,and thyroxine are increased several-fold in the maternal blood to support fetal growth,metabolic changes,and the maintenance of pregnancy.

(B) Population explosion is primarily caused by a rapid increase in the human population size.
This is driven by a significant decline in death rates,including a reduction in maternal mortality rate $(MMR)$ and infant mortality rate $(IMR)$,while birth rates remain relatively high.
Therefore,the correct option is $(B)$.

(C) The correct answer is $C$.
$IUDs$ (Intrauterine Devices) are effective contraceptive methods used by females.
Copper-releasing $IUDs$ (like $CuT$,$Cu7$,and $Multiload$ $375$) release copper ions $(Cu^{2+})$ into the uterus.
These ions suppress sperm motility and the fertilizing capacity of sperms.
Additionally,they increase the phagocytosis of sperms within the uterus,thereby preventing fertilization.

(D) Down's syndrome is an example of aneuploidy of autosome.
Down's syndrome is a genetic disorder caused by the presence of an extra copy of chromosome number $21$ (trisomy $21$).
Since chromosome $21$ is an autosome,this condition is classified as an autosomal aneuploidy.
It occurs due to the nondisjunction of chromosomes during meiosis,leading to a cell having $47$ chromosomes instead of the normal $46$.

(D) is the correct answer.
Sickle-cell anaemia is an autosomal recessive genetic disorder.
It is caused by a point mutation in the gene coding for the $\beta$-globin chain of haemoglobin.
Specifically,the $6^{\text{th}}$ amino acid of the $\beta$-globin chain,which is normally glutamic acid,is replaced by valine.
This substitution changes the structure of the haemoglobin molecule,causing red blood cells to become sickle-shaped under low oxygen conditions.

(C) The genotype of the heterozygous parent is $Rr$.
When this plant is selfed $(Rr \times Rr)$,the resulting progeny genotypes are $RR, Rr, Rr, rr$.
The phenotypic ratio is $3$ round : $1$ wrinkled.
The parental phenotype is 'round' (since the parent is heterozygous for round seeds).
Out of the total $1600$ seeds,the number of round-seeded plants is $\frac{3}{4} \times 1600 = 1200$.
Therefore,$1200$ seedlings will exhibit the parental phenotype.

(D) $Statement A$ is correct because according to the Law of Segregation,alleles of a gene pair segregate from each other during gamete formation,such that each gamete carries only one allele for a given trait.
$Statement B$ is correct because during the anaphase stage of mitosis,the centromere splits and the sister chromatids separate,moving towards opposite poles of the cell.
Since $Statement B$ describes a process of cell division (mitosis) and $Statement A$ describes the principle of inheritance (meiosis/gametogenesis),$Statement B$ is not the reason for $Statement A$.

(B) The correct sequence is $C-A-E-B-D$.
The steps involved in $DNA$ fingerprinting are as follows:
$1$. Isolation of $DNA$ from the sample cell.
$2$. Amplification of $DNA$ (if required) using $PCR$.
$3$. Digestion of $DNA$ by restriction endonucleases $(RENS)$ $(C)$.
$4$. Separation of $DNA$ fragments by gel electrophoresis $(A)$.
$5$. Transferring (blotting) of separated $DNA$ fragments to synthetic membranes like nitrocellulose or nylon $(E)$.
$6$. Hybridisation using labelled variable number tandem repeat $(VNTR)$ probe $(B)$.
$7$. Detection of hybridised $DNA$ fragments by autoradiography $(D)$.

(C) is the correct answer.
In the 'Lac' operon model,lactose (or allolactose) acts as an inducer.
When lactose is present in high concentration in the growth medium,it binds to the repressor protein.
This binding changes the conformation of the repressor protein,preventing it from binding to the operator gene.
Consequently,the $RNA$ polymerase is free to bind to the promoter region,allowing the transcription of structural genes into polycistronic $mRNA$.

(A) The correct answer is $A$.
Oswald Avery, Colin MacLeod, and Maclyn McCarty worked to determine the biochemical nature of the 'transforming principle' in Griffith's experiment.
They purified biochemicals (proteins, $DNA$, $RNA$) from the heat-killed '$S$' cells to see which ones could transform live '$R$' cells into '$S$' cells.
They discovered that digestion with $proteases$ and $RNAases$ did not affect transformation, meaning neither protein nor $RNA$ was the genetic material.
However, digestion with $DNAase$ inhibited the transformation, proving that $DNA$ is the substance that causes the transformation of '$R$' strain into '$S$' strain.
Therefore, the experiment that conclusively proved $DNA$ is the genetic material involved the addition of $DNAase$ to the mixture, which prevented the transformation.

(A) $Statement B$ is correct and $Statement A$ is wrong.
In eukaryotes,the primary transcript (pre-$mRNA$) undergoes significant post-transcriptional modifications such as splicing,capping,and tailing before it is translated.
$Statement A$ is incorrect because the primary transcript is not translated directly.
$Statement B$ is correct because the $hnRNA$ (heterogeneous nuclear $RNA$) contains both coding sequences (exons) and non-coding sequences (introns),which must be removed via splicing.

(D) divergent evolution.
$Thorns$ of $Bougainvillea$ and $tendrils$ of $Cucurbita$ are modified branches and are $axillary$ in position.
This means $axillary$ branches in $Bougainvillea$ are modified into $thorns$ for protection from browsing animals, and in $Cucurbita$ into $tendrils$ for climbing.
These are called $homologous$ $organs$ and they are the result of $divergent$ $evolution$, i.e., they have a common ancestry but perform different functions.

(A) The correct answer is $A$.
Stanley $L$. Miller and Harold $C$. Urey conducted a landmark experiment in $1953$ to test the chemical origin of life theory.
They created a closed system simulating the conditions of the primitive Earth.
The mixture used in the spark-discharge apparatus consisted of methane $(CH_4)$,ammonia $(NH_3)$,hydrogen $(H_2)$,and water vapour $(H_2O)$ in a ratio of $2:2:1$ (for $CH_4:NH_3:H_2$).
This experiment successfully produced simple amino acids like glycine,alanine,and aspartic acid,demonstrating that organic molecules could be synthesized from inorganic precursors under prebiotic conditions.

(B) The correct answer is $B$.
Cancer cells lose the property of contact inhibition.
Contact inhibition is a mechanism that prevents normal cells from dividing when they come into contact with other cells.
Cancer cells ignore these signals and continue to divide uncontrollably,leading to the formation of a tumor.
Therefore,the statement that they exhibit contact inhibition is incorrect.

(B) The correct answer is $B$.
Malaria is caused by a protozoan parasite belonging to the genus $Plasmodium$.
The infective stage of the $Plasmodium$ parasite for human beings is the $sporozoite$.
These $sporozoites$ are stored in the salivary glands of the female $Anopheles$ mosquito and are injected into the human bloodstream during the mosquito's blood meal.

(A) $Ascaris$ is a common roundworm parasite that infects the human intestine.
Symptoms of ascariasis include internal bleeding,muscular pain,fever,anemia,and blockage of the intestinal passage due to the accumulation of a large number of worms.
Therefore,the correct option is $A$.

(D) is the correct answer.
Heroin,commonly known as smack,is chemically diacetylmorphine.
It is a white,odorless,bitter,crystalline compound.
It is extracted from the latex of the poppy plant ($Papaver$ $somniferum$) and is prepared by the acetylation of morphine.

(C) Statement $I$ is incorrect because Heroin (diacetylmorphine) is obtained by the acetylation of morphine,not the other way around.
Statement $II$ is correct because cannabinoids are a group of chemicals that interact with cannabinoid receptors in the brain and are well-known for their significant effects on the cardiovascular system of the body.

(D) fungus.
Cheese can be classified on the basis of its texture, hardness, and ripening process.
'Roquefort cheese' is a semi-soft cheese.
In this cheese, spores of the fungus $Penicillium roqueforti$ are added to the curd before the final stages of cheese production to provide its characteristic flavor and texture.

(B) The correct matches are as follows:
$1$. Citric acid is produced by the fungus $Aspergillus$ $niger$ $(A-r)$.
$2$. Cyclosporin $A$ is an immunosuppressive agent produced by the fungus $Trichoderma$ $polysporum$ $(B-s)$.
$3$. Statins are blood-cholesterol lowering agents produced by the yeast $Monascus$ $purpureus$ $(C-q)$.
$4$. Gobar gas (biogas) is produced by methanogens like $Methanobacterium$ $(D-p)$.
Therefore,the correct matching is $A-r, B-s, C-q, D-p$. The correct option is $B$.

(A) is the correct answer.
In a sewage treatment plant,during secondary treatment,a large number of aerobic heterotrophic microbes grow in the aeration tank.
These microbes include bacteria and some filamentous fungi and yeasts,which form $Flocs$.
$Flocs$ are mesh-like structures formed by the interweaving of fungal filaments and bacterial cells.

(A) palindromic $DNA$ sequence is a sequence of base pairs that reads the same when the orientation of reading is the same,i.e.,$5'$ to $3'$ on both strands.
Let's analyze the options:
$A$: $5'-GGATCC-3'$ and $3'-GGTACC-5'$. The complementary strand of $5'-GGATCC-3'$ is $3'-CCTAGG-5'$. Since $3'-GGTACC-5'$ does not match $3'-CCTAGG-5'$,this is not a palindrome.
$B$: $5'-GAATTC-3'$ and $3'-CTTAAG-5'$. This is a palindrome (EcoRI site).
$C$: $5'-GCGGCCGC-3'$ and $3'-CGCCGGCG-5'$. This is a palindrome (NotI site).
$D$: $5'-CCCGGG-3'$ and $3'-GGGCCC-5'$. This is a palindrome (SmaI site).
Therefore,option $A$ is the correct answer.

(C) The desirable characteristics for a cloning vector (plasmid) are:
$1$. Origin of replication $(ori)$: $A$ site at which replication can be initiated $(C)$.
$2$. Selectable markers: One or more identifiable marker genes $(D)$ to identify and eliminate non-transformants.
$3$. Cloning sites: One or more unique restriction sites $(E)$ for the insertion of foreign $DNA$.
Therefore,the correct characteristics are $C, D,$ and $E$. Option $A$ is incorrect because plasmids must replicate inside the host cell. Option $B$ is a feature of expression vectors,not necessarily cloning vectors.

(B) $EcoRI$ is a restriction endonuclease enzyme that cuts $DNA$ between $G$ and $A$ in the base sequence $5'-GAATTC-3'$.