KCET 2023 Biology Question Paper with Answer and Solution

(C) $Musca$ $domestica$ belongs to the order $Diptera$.
$Mangifera$ $indica$ (Mango) belongs to the order $Sapindales$.
$Triticum$ $aestivum$ (Wheat) belongs to the order $Poales$.
$Homo$ $sapiens$ (Human) belongs to the order $Primata$.
Therefore,the correctly matched pair is $Musca$ $domestica$ : $Diptera$.

(D) The correct answer is $D$.
During the $zygotene$ stage of $prophase-I$ in $meiosis$,homologous chromosomes pair up in a process known as $synapsis$.
This pair of synapsed homologous chromosomes is referred to as a $bivalent$ or a $tetrad$ (because it consists of four chromatids).

(B) The correct option is $(B)$.
In the human heart's conducting system:
$p$ represents the Sino-Atrial Node $(SAN)$,which is located in the right atrium.
$q$ represents the Atrio-Ventricular Node $(AVN)$,located near the base of the interatrial septum.
$r$ represents the Bundle of His,which originates from the $AVN$ and passes through the interventricular septum.
$s$ represents the Interventricular septum,the wall separating the two ventricles.

(A) Cardiac output is the volume of blood pumped by each ventricle per minute.
Formula: $\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}$.
Given: $\text{Stroke Volume} = 70 \ mL$,Average $\text{Heart Rate} = 72 \ \text{beats/minute}$.
Cardiac output per minute = $70 \ mL \times 72 = 5040 \ mL/minute = 5.04 \ L/minute$.
To find the cardiac output in one hour $(60 \ \text{minutes})$:
$\text{Cardiac Output per hour} = 5.04 \ L/minute \times 60 \ \text{minutes} = 302.4 \ L/hour$.
Therefore,the correct option is $A$.

(B) The correct answer is $(B)$.
$ANF$ (Atrial Natriuretic Factor) is released by the atrial wall of the heart in response to increased blood flow.
It causes vasodilation (dilation of blood vessels),which leads to a decrease in blood pressure.
Therefore,the $ANF$ mechanism acts as a check or a negative feedback control on the Renin-Angiotensin mechanism,which is responsible for increasing blood pressure.

(C) The correct matching is as follows:
$1$. Bryophyta: $r$. Sphagnum (a moss).
$2$. Gymnosperm: $p$. Pinus (a conifer).
$3$. Algae: $s$. Ectocarpus (a brown alga).
$4$. Pteridophyta: $q$. Adiantum (a fern).
Therefore,the correct sequence is $1-r, 2-p, 3-s, 4-q$.

(B) The correct option is $B$.
In $Amoeba$,the contractile vacuole plays a vital role in maintaining water balance within the cell,a process known as osmoregulation.
It also helps in the removal of metabolic waste products,thus performing the function of excretion.
Therefore,the primary functions of the contractile vacuole are excretion and osmoregulation.

(C) The correct answer is $C$.
Flame cells (also known as protonephridia) are specialized excretory structures found in the members of the phylum $Platyhelminthes$ (flatworms).
These cells are primarily responsible for osmoregulation (maintaining water and salt balance) and the excretion of nitrogenous waste products from the body.

(B) The potato family is known as the Solanaceae family.
Its characteristic floral formula is $\oplus, \text{K}_{(5)}, \text{C}_{(5)}, \text{A}_5, \underline{\text{G}}_{(2)}$.
- $\oplus$: Actinomorphic (radially symmetrical).
- $\text{K}_{(5)}$: Calyx with $5$ sepals,gamosepalous (fused).
- $\text{C}_{(5)}$: Corolla with $5$ petals,gamopetalous (fused).
- $\text{A}_5$: Androecium with $5$ stamens,epipetalous (often attached to petals,though not explicitly shown in this simplified formula).
- $\underline{\text{G}}_{(2)}$: Gynoecium with $2$ carpels,syncarpous (fused),and superior ovary.

(C) The correct answer is $C$.
When the vascular cambium is present between the xylem and phloem,the vascular bundle is known as an open vascular bundle.
This type of vascular bundle is characteristic of dicotyledonous stems,as it allows for secondary growth.

(D) The correct matching is as follows:
$1$. Collagen: $s$. Intercellular ground substance (provides structural support).
$2$. Trypsin: $r$. Enzyme (a digestive enzyme that breaks down proteins).
$3$. Insulin: $q$. Hormone (regulates blood glucose levels).
$4$. Antibody: $p$. Fights infectious agents (part of the immune system).
Therefore,the correct sequence is $1-s, 2-r, 3-q, 4-p$.

(A) The correct answer is $A$.
In the embryo sac, the $Egg$ $cell$ is haploid $(n)$ before fertilization and becomes a diploid $(2n)$ zygote after fertilization.
The $Central$ $cell$ contains two polar nuclei (each $n$) and is effectively diploid $(2n)$ before fertilization; after fusion with a male gamete $(n)$, it becomes a triploid $(3n)$ primary endosperm nucleus $(PEN)$.
$Synergids$ and $Antipodals$ are haploid cells that degenerate after fertilization without undergoing a change in ploidy.

(B) The correct answer is $B$.
In angiosperms,the process of double fertilization involves two events: syngamy and triple fusion.
The primary endosperm nucleus $(PEN)$ is formed by the fusion of two haploid polar nuclei (which form the diploid secondary nucleus) and one haploid male gamete.
Since this process involves the fusion of three haploid nuclei,it is termed as triple fusion.
This results in the formation of a triploid $(3n)$ primary endosperm nucleus.

(D) Each stamen has a bilobed,dithecous anther,which means it contains $4$ microsporangia.
Total number of microsporangia in $10$ stamens = $10 \times 4 = 40$.
Each microsporangium contains $5$ pollen mother cells (PMCs).
Total number of PMCs in the flower = $40 \times 5 = 200$.
Each pollen mother cell undergoes meiosis to produce $4$ pollen grains.
Total number of pollen grains = $200 \times 4 = 800$.

(B) Detritivores are organisms that obtain nutrients by consuming detritus,which consists of decaying organic matter,including plant and animal remains and fecal matter.
Examples of detritivores include earthworms,termites,woodlice,and millipedes.
They play a crucial role in the ecosystem by breaking down complex organic materials into simpler substances,thereby facilitating the process of decomposition and nutrient cycling.
Therefore,an animal feeding on decaying organic matter is called a detritivore.

(D) The incorrect statement is $D$.
Green plants capture only about $1\%$ of the solar energy that falls on their leaves.
This energy is used in the process of photosynthesis to convert light energy into chemical energy,which is then stored as food.
The other statements $(A, B, C)$ are correct: energy is lost as heat at each trophic level,energy transfer follows the $10\%$ law,and energy flow in an ecosystem is always unidirectional.

(D) The correct matches are as follows:
$1$. Standing state refers to the amount of inorganic nutrients (like nitrogen,phosphorus,etc.) present in the soil at a given time $(1-q)$.
$2$. Pioneer species are the first organisms to colonize a previously bare area during ecological succession $(2-r)$.
$3$. Detritivores are organisms that break down detritus into smaller particles (e.g.,earthworms) $(3-s)$.
$4$. Standing crop refers to the total mass of living material (biomass) present in an ecosystem at a given time $(4-p)$.
Therefore,the correct combination is $1-q, 2-r, 3-s, 4-p$.

(B) The correct answer is $B$.
According to David Tilman's long-term ecosystem experiments,plots with more species showed less year-to-year variation in total biomass.
This indicates that higher biodiversity contributes to greater ecosystem stability and productivity over time.

(D) The correct option is $D$.
In the human menstrual cycle:
$1$. The Menstruation phase is characterized by the shedding of the endometrial lining,known as menses.
$2$. The Follicular phase involves the maturation of follicles and culminates in the $LH$ surge,which triggers ovulation.
$3$. The Luteal phase follows ovulation,where the ruptured follicle transforms into the corpus luteum,which secretes progesterone to maintain the endometrium,leading to its regeneration and preparation for potential implantation.

(D) The correct answer is $D$.
The clitoris is a tiny,finger-like structure that lies at the upper junction of the two labia minora,situated above the urethral opening.
It is a highly sensitive organ homologous to the penis in males.

(A) The male sex accessory ducts consist of the $Rete$ $testis$,$vasa$ $efferentia$,$epididymis$,and $vas$ $deferens$.
These ducts are responsible for the transport and storage of sperms from the $testis$ to the outside through the $urethra$.
$Seminal$ $vesicle$ is an accessory gland,not a duct.

(A) Medical Termination of Pregnancy $(MTP)$ is considered relatively safe during the first trimester of pregnancy.
This is because the fetus is not yet well-developed,and the surgical or medical procedure carries significantly lower risks of complications for the mother compared to later stages of pregnancy.

(B) $LNG-20$ is a hormone-releasing $IUD$ (Intra Uterine Device).
It releases levonorgestrel,a synthetic progestogen,which makes the uterus unsuitable for implantation and the cervix hostile to sperm.
In contrast,Multiload $375$ is a copper-releasing $IUD$,and Lippes loop is a non-medicated $IUD$.

(B) $ZIFT$ stands for Zygote Intra Fallopian Transfer.
In this assisted reproductive technology,the zygote or early embryos (up to $8$ blastomeres) are transferred into the fallopian tube of the female.

(A) $1$. Haemophilia is an $X$-linked recessive disorder. $A$ female carrier $(X^HX^h)$ has a $50\%$ chance of passing the defective $X^h$ chromosome to her sons,who will then express the disease.
$2$. Thalassemia is a quantitative problem because it involves the reduced synthesis of one of the globin chains of the haemoglobin molecule.
$3$. $A$ change in the whole set of chromosomes is called polyploidy,whereas aneuploidy refers to the gain or loss of one or more individual chromosomes.
$4$. Sickle cell anaemia is a qualitative problem because it results from a point mutation in the gene coding for the $\beta$-globin chain,leading to the substitution of glutamic acid with valine,which alters the structure and function of the haemoglobin molecule.

(D) The correct answer is $D$.
In the male heterogametic type of sex determination,the male individual produces two different types of gametes (e.g.,$X$ and $Y$ in humans or $X$ and $O$ in some insects),which is why they are called heterogametic. The female,in this case,is homogametic,meaning she produces only one type of gamete $(X)$.

(C) In a Mendelian cross between a homozygous dominant parent $(AA)$ and a homozygous recessive parent $(aa)$,the $F_1$ generation consists of all heterozygous individuals $(Aa)$ expressing the dominant trait.
When $F_1$ individuals are self-crossed $(Aa \times Aa)$,the $F_2$ generation shows a phenotypic ratio of $3:1$ (dominant to recessive).
Therefore,the dominant trait is expressed in both $F_1$ and $F_2$ generations,while the recessive trait is masked in $F_1$ and only reappears in the $F_2$ generation.

(B) Statement $(1)$ is correct because it describes the phenomenon of pleiotropy,where a single gene influences multiple phenotypic traits.
Statement $(2)$ is incorrect. The $AB$-blood group in humans is an example of co-dominance,but it is regulated by multiple alleles $(I^A, I^B, i)$,not just one dominant and one recessive allele. Specifically,$I^A$ and $I^B$ are co-dominant,and $i$ is recessive.

(D) Sturtevant.
Alfred Sturtevant,a student of Thomas Hunt Morgan,developed the first genetic map in $1913$ based on the frequency of recombination between gene pairs on the same chromosome.

(D) The correct answer is $D$.
In eukaryotes,monocistronic structural genes possess interrupted coding sequences,which is why they are referred to as 'split genes'.
The coding sequences or expressed sequences are defined as $Exons$.
$Exons$ are the sequences that appear in the mature or processed $RNA$.
These $Exons$ are interrupted by non-coding sequences known as $Introns$.

(C) $Lac-Operon$ model was elucidated by Francois Jacob and Jacques Monod.
$Lac-Operon$ is a cluster of genes through which $Escherichia coli$ catabolizes lactose.
It was first proposed by $F. Jacob$ and $J. Monod$, who were also awarded the Nobel Prize in Physiology or Medicine in $1965$.

(C) The correct answer is $C$.
Histones are positively charged proteins because they contain a very high proportion of basic amino acids,specifically $Lysine$ and $Arginine$.
These amino acids possess positively charged side chains at physiological $pH$.
Because of this positive charge,histones can effectively bind to the negatively charged phosphate backbone of $DNA$ to form nucleosomes.

(B) During the process of transcription,the enzyme $DNA$-dependent $RNA$ polymerase is responsible for synthesizing $RNA$.
This enzyme has a strict requirement to catalyze the polymerization of nucleotides only in the $5' \rightarrow 3'$ direction.
Since the two strands of $DNA$ are antiparallel,the strand that runs in the $3' \rightarrow 5'$ direction serves as the template.
This allows the newly synthesized $RNA$ strand to be formed in the $5' \rightarrow 3'$ direction,which is complementary and antiparallel to the template strand.

(A) Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats).
Darwin's finches,Australian marsupials,and placental mammals are classic examples of adaptive radiation as they show divergent evolution from a common ancestor.
The long-necked giraffe is not an example of adaptive radiation; its evolution is typically explained by the theory of natural selection (Lamarckism or Darwinism) regarding the adaptation of a specific trait within a lineage rather than the diversification of a common ancestor into multiple new species.

(C) According to the Hardy-Weinberg principle,the equation is $p^2 + 2pq + q^2 = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.
Given that the frequency of recessive individuals $(q^2)$ is $0.16$.
Therefore,$q = \sqrt{0.16} = 0.4$.
Since $p + q = 1$,we have $p = 1 - 0.4 = 0.6$.
The frequency of heterozygous individuals is represented by $2pq$.
$2pq = 2 \times 0.6 \times 0.4 = 0.48$.
Thus,the frequency of heterozygous individuals is $0.48$.

(C) The correct matches are as follows:
$1$. $Haemophilus$ $influenzae$ causes Pneumonia $(r)$.
$2$. $Entamoeba$ $histolytica$ causes Amoebiasis $(s)$.
$3$. $Plasmodium$ $falciparum$ causes Malignant malaria $(p)$.
$4$. $Wuchereria$ $bancrofti$ causes Elephantiasis $(q)$.
Therefore,the correct sequence is $1-r, 2-s, 3-p, 4-q$.

(B) is the correct match.

Disease	Pathogen	Main organ
Ringworm	Fungus	Skin

$1$. Ringworm is a common fungal infection caused by genera such as $Microsporum$,$Trichophyton$,and $Epidermophyton$. It primarily affects the skin,nails,and scalp,causing dry,scaly lesions.
$2$. Dysentery is caused by $Entamoeba$ $histolytica$ (a protozoan) and affects the large intestine,not the liver.
$3$. Typhoid is caused by $Salmonella$ $typhi$ (a bacterium) and affects the small intestine,not the lungs.
$4$. Filariasis is caused by $Wuchereria$ $bancrofti$ (a filarial worm) and typically affects the lymphatic vessels of the lower limbs,not the small intestine.

(B) The correct answer is $B$.
$Pneumonia$ is a respiratory infection caused by bacteria such as $Streptococcus$ $pneumoniae$ and $Haemophilus$ $influenzae$.
The symptoms of pneumonia include difficulty in breathing,high fever,chills,cough,and headache.
In severe cases,the lips and fingernails may turn gray or bluish in color.

(C) The correct answer is $C$.
Roquefort cheese is ripened by growing a specific fungus,$Penicillium \ roqueforti$,on it.
This fungus provides the cheese with its characteristic flavor and texture.

(D) The correct statement is $D$. The first step in $PCR$ is denaturation,which involves heating the $DNA$ to approximately $94-98^{\circ}C$ to separate the two strands of the gene of interest.
Option $A$ is incorrect because $DNA$ from different organisms can be joined (recombinant $DNA$ technology).
Option $B$ is incorrect because genetic engineering is widely and successfully used in plants (e.g.,$Bt$ cotton).
Option $C$ is incorrect because there are ethical and safety concerns (biosafety) associated with $r-DNA$ technology.

(A) $DNA$ amplification.
$PCR$ is a very sensitive technique that allows rapid amplification of a specific segment of $DNA$.
$PCR$ makes billions of copies of a specific $DNA$ fragment or gene,which allows detection and identification of gene sequences using visual techniques based on size and charge.

(C) The correct option is $C$.
To clone a bacterial gene in animal cells,the following tools and techniques are required:
$I$. Endonuclease: Used to cut the $DNA$ at specific sites.
$II$. Ligase: Used to join the $DNA$ fragments.
$IV$. Microinjection: $A$ direct method for introducing foreign $DNA$ into animal cells.
$VIII$. Electrophoresis: Used for the separation and isolation of $DNA$ fragments.
Note: $A. tumefaciens$ is used for plants,Gene gun is used for plants,Lysozyme is used for bacterial cell wall degradation,and Cellulase is used for plant cell wall degradation.

(C) - Elution.
Competent cells are those that are capable of taking up foreign $DNA$ from their environment. Methods to make cells competent include chemical treatment (e.g.,$CaCl_2$),heat shock,micro-injection,biolistics (gene gun),and the use of disarmed pathogen vectors. Elution is a process used in chromatography or gel electrophoresis to extract a substance from a solid or a gel,and it is not a method for transforming host cells.