KCET 2022 Biology C4 Question Paper With Answer Key And Solutions PDF

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Shivam Yadav

Updated on - Nov 17, 2025

KCET 2022 Biology C-4 Question Paper with Answer Key pdf is available for download. The exam was conducted by Karnataka Examination Authority (KEA) on June 16, 2022. In terms of difficulty level, KCET Biology was of Easy to Moderate level. The question paper comprised a total of 60 questions.

KCET 2022 Biology (C-4) Question Paper with Answer Key

KCET Biology (C-4) Question Paper 2022 with Answer Key download iconDownload Check Solution

KCET 2022 Biology C4 Question Paper With Answer Key And Solutions PDF


Question 1:

Amniocentesis is a process to:

  • (1) . Determine the sex of the foetus
  • (2) . Determine any disease of heart
  • (3) . Know about the disease of brain
  • (4) . To grow cell on culture medium
Correct Answer: (1). Determine the sex of the foetus
View Solution

Step 1: Amniocentesis is a prenatal diagnostic procedure performed between 15–20 weeks of pregnancy.

Step 2: A small amount of amniotic fluid containing fetal cells is extracted using a needle under ultrasound guidance.

Step 3: The fetal cells are cultured and analyzed via karyotyping to detect chromosomal abnormalities, genetic disorders, and sex of the foetus (XX or XY).

Thus, it is used to determine the sex of the foetus. Quick Tip: Amniocentesis is banned in India for sex determination under PCPNDT Act to prevent female foeticide, but allowed for medical diagnosis.


Question 2:

The first human like being is:

  • (1) . Homo menthus
  • (2) . Homo erectus
  • (3) . Homo habilis
  • (4) . Homo sapiens
Correct Answer: (3). Homo habilis
View Solution

Step 1: The genus Homo evolved from Australopithecus around 2.4 million years ago.

Step 2: Homo habilis (2.4–1.4 mya) is the earliest species of Homo with brain size ~600–700 cm³ and evidence of stone tool use (Oldowan tools).

Step 3: Known as "handy man," it is considered the first human-like being. Homo erectus appeared later (~1.8 mya). Quick Tip: Evolutionary order: Australopithecus → Homo habilis → Homo erectus → Homo sapiens


Question 3:

XO type of sex determination and XY type of sex determination are the examples of:

  • (1) . Female Homogamety
  • (2) . Male Heterogamety
  • (3) . Female Heterogamety
  • (4) . Male Homogamety
Correct Answer: (3). Female Heterogamety
View Solution

Step 1: In XO-type (e.g., grasshoppers), females are XO (heterogametic) and males are XX (homogametic).

Step 2: In XY-type, males are XY (heterogametic).

Step 3: The question focuses on XO type, which is a classic example of female heterogamety. Quick Tip: - XY → Male heterogamety (humans)
- XO → Female heterogamety (insects)
- ZW → Female heterogamety (birds)


Question 4:

Example of Non-Mendelian disorder:

  • (1) . Cystic fibrosis
  • (2) . Haemophilia
  • (3) . Down’s syndrome
  • (4) . Thalassemia
Correct Answer: (3). Down’s syndrome
View Solution

Step 1: Mendelian disorders involve single-gene mutations.

Step 2: Down’s syndrome is caused by trisomy 21 (nondisjunction), a chromosomal disorder.

Step 3: Cystic fibrosis, haemophilia, thalassemia are Mendelian disorders.

Hence, Down’s syndrome is non-Mendelian. Quick Tip: Non-Mendelian: chromosomal, polygenic, mitochondrial.


Question 5:

Gynecomastia is a symptom of:

  • (1) . Cri-du-chat syndrome
  • (2) . Down’s syndrome
  • (3) . Klinefelter’s syndrome
  • (4) . Turner’s syndrome
Correct Answer: (3). Klinefelter’s syndrome
View Solution

Step 1: Klinefelter’s syndrome (47, XXY) affects males.

Step 2: Extra X chromosome causes reduced testosterone and estrogen excess → gynecomastia (breast development in males).

Step 3: Turner’s syndrome affects females; others do not cause gynecomastia. Quick Tip: Klinefelter: XXY → sterile male, gynecomastia, tall stature.


Question 6:

The affected male in the pedigree chart is symbolized by:

  • (1) . blacklozenge
  • (2) . square
  • (3) . bigcirc
  • (4) . blacksquare
Correct Answer: (4). blacksquare
View Solution

Step 1: Pedigree symbols:

- Males → square

- Females → circle

- Affected → filled

Step 2: Affected male = filled square (■). Quick Tip: Carrier: half-filled; Proband: arrow.


Question 7:

The given diagram represents:


  • (1) . Mesosome
  • (2) . Chromosome
  • (3) . Ribosome
  • (4) . Nucleosome
Correct Answer: (4). Nucleosome
View Solution

Step 1: Nucleosome = DNA wrapped around histone octamer.

Step 2: H1 histone links nucleosomes.

Step 3: Diagram shows DNA + H1 → nucleosome. Quick Tip: Nucleosome → 11 nm → 30 nm fiber → chromosome


Question 8:

Which of the following hormones is not secreted by human placenta?

  • (1) . FSH
  • (2) . hCG
  • (3) . Relaxin
  • (4) . Progestogen
Correct Answer: (1). FSH
View Solution

Step 1: Placenta secretes: hCG, hPL, progestogen, relaxin.

Step 2: FSH is secreted by anterior pituitary.

Thus, FSH is not secreted by placenta. Quick Tip: hCG → maintains corpus luteum → progesterone.


Question 9:

Which of the following is correctly matched?

  • (1) . Spores - Sponge
  • (2) . Conidia - Hydra
  • (3) . Gemmules - Amoeba
  • (4) . Bulbil - Agave
Correct Answer: (4). Bulbil - Agave
View Solution

Step 1: Bulbil = vegetative propagule in Agave.

Step 2: Gemmules → sponges; Conidia → fungi; Spores → not in sponges.

Thus, Bulbil - Agave is correct. Quick Tip: Vegetative propagation: bulbil (Agave), runner (grass).


Question 10:

The technique advised by a doctor to overcome the problem of infertility:

  • (1) . RTI
  • (2) . MTP
  • (3) . ART
  • (4) . RCH
Correct Answer: (3). ART
View Solution

Step 1: ART (Assisted Reproductive Technology): IVF, IUI, ZIFT, GIFT — treats infertility.

Step 2: MTP → abortion; RTI → infection; RCH → health program.

Only ART treats infertility. Quick Tip: First test-tube baby: Louise Brown (1978) via IVF.


Question 11:

Which aspect forms the basis of DNA finger-printing?

  • (1) . The amount of DNA found in samples of blood, saliva and skin.
  • (2) . The ratio of purines and pyrimidines present in DNA.
  • (3) . The Sequence of DNA present in the ridges and grooves of finger-prints
  • (4) . The Satellite DNA showing high degree of repetition in DNA segments
Correct Answer: (4). The Satellite DNA showing high degree of repetition in DNA segments
View Solution

Step 1: DNA fingerprinting relies on variable number tandem repeats (VNTRs) in satellite DNA.

Step 2: These are highly repetitive, non-coding minisatellite regions with variable repeat counts among individuals.

Step 3: Restriction enzymes cut DNA, and gel electrophoresis separates fragments by size, creating unique banding patterns.

Thus, high degree of repetition in satellite DNA forms the basis. Quick Tip: DNA fingerprinting is used in paternity testing, forensics, and identification.


Question 12:

Identify the most infectious and fatal type of malarial parasite:

  • (1) . Plasmodium ovale
  • (2) . Plasmodium vivax
  • (3) . Plasmodium malariae
  • (4) . Plasmodium falciparum
Correct Answer: (4). Plasmodium falciparum
View Solution

Step 1: Plasmodium falciparum causes malignant tertian malaria.

Step 2: It leads to cerebral malaria, high parasitemia, and cytoadherence of infected RBCs to blood vessels.

Step 3: It has the highest mortality rate among malarial parasites.

P. vivax causes benign tertian; P. malariae causes quartan; P. ovale is mild. Quick Tip: Falciparum → cerebral malaria → most fatal.


Question 13:

The type of antibodies produced during the allergic reaction:

  • (1) . Ig M
  • (2) . Ig A
  • (3) . Ig E
  • (4) . Ig G
Correct Answer: (3). Ig E
View Solution

Step 1: Allergic reactions are Type I hypersensitivity mediated by IgE antibodies.

Step 2: Allergens bind to IgE on mast cells and basophils.

Step 3: This triggers degranulation → release of histamine, serotonin → allergic symptoms.

IgM, IgG, IgA are not involved in immediate hypersensitivity. Quick Tip: Antihistamines block histamine receptors in allergies.


Question 14:

One of the side-effect of the use of anabolic steroids in females:

  • (1) . Masculinisation
  • (2) . Loss of memory
  • (3) . Hallucination
  • (4) . Cirrhosis of liver
Correct Answer: (1). Masculinisation
View Solution

Step 1: Anabolic steroids are synthetic testosterone derivatives.

Step 2: In females, excess androgens cause virilization/masculinisation: deepening of voice, facial hair, menstrual irregularities.

Step 3: Liver damage (cirrhosis) is possible but not specific to females; memory loss and hallucination are not typical. Quick Tip: Anabolic steroids → doping in sports → banned by WADA.


Question 15:

Which one of the following is a opiate narcotics?

  • (1) . LSD
  • (2) . Barbiturates
  • (3) . Morphine
  • (4) . Amphatamines
Correct Answer: (3). Morphine
View Solution

Step 1: Opiate narcotics are derived from opium poppy (Papaver somniferum).

Step 2: Morphine is a natural opiate alkaloid with strong analgesic and sedative effects.

Step 3: LSD (hallucinogen), barbiturates (sedative-hypnotics), amphetamines (stimulants) are not opiates. Quick Tip: Morphine → painkiller → highly addictive → leads to heroin.


Question 16:

The large holes in ‘Swiss-Cheese’ are made by a:

  • (1) . Fungus that releases a lot of gases during metabolic activities
  • (2) . Machine
  • (3) . Bacterium that produces methane gas
  • (4) . Bacterium producing a large amount of CO2
Correct Answer: (4). Bacterium producing a large amount of CO2
View Solution

Step 1: Swiss cheese is made using Propionibacterium shermanii.

Step 2: This bacterium ferments lactic acid → propionic acid + CO2.

Step 3: Trapped CO2 forms large holes (eyes) during ripening.

No methane or fungus is involved. Quick Tip: Roquefort cheese → Penicillium roqueforti (fungus).


Question 17:

Which vitamin is increases by ‘LAB’ in curd?

  • (1) . Vitamin E
  • (2) . Vitamin C
  • (3) . Vitamin B
  • (4) . Vitamin B12
Correct Answer: (4). Vitamin B12
View Solution

Step 1: Lactic Acid Bacteria (LAB) like Lactobacillus ferment milk into curd.

Step 2: During fermentation, they synthesize Vitamin B12 (cobalamin).

Step 3: Curd is a rich source of B12, especially for vegetarians.

Vitamins E, C, B (general) are not significantly increased. Quick Tip: Curd → probiotic → improves digestion and B12 absorption.


Question 18:

Enzyme which is useful to remove the oily stains in laundry?

  • (1) . Lipase
  • (2) . Renin
  • (3) . Protease
  • (4) . Amylase
Correct Answer: (1). Lipase
View Solution

Step 1: Oily stains are lipids (fats).

Step 2: Lipase enzyme hydrolyzes triglycerides into glycerol and fatty acids.

Step 3: It is added to biological detergents to remove oil/grease stains.

Protease (proteins), amylase (starch), renin (not used in detergents). Quick Tip: Biological detergents work best at 30–40°C.


Question 19:

DNA replicates semiconservatively was first shown in:

  • (1) . Higher animals
  • (2) . Esherichia coil
  • (3) . Human cell
  • (4) . Plants
Correct Answer: (2). Esherichia coil
View Solution

Step 1: Meselson and Stahl (1958) proved semiconservative replication.

Step 2: They used E. coli grown in ¹⁵NH₄Cl (heavy nitrogen).

Step 3: After one replication in ¹⁴N, DNA showed hybrid density → semiconservative model confirmed. Quick Tip: Semiconservative: each new DNA has one old + one new strand.


Question 20:

A series of experiments were conducted by Frederick Griffith in 1928, on transforming principle with:

  • (1) . Streptococcus pneumoniae
  • (2) . Escherichia coli
  • (3) . Bacillus thuringiensis
  • (4) . Salmonella typhimurium
Correct Answer: (1). Streptococcus pneumoniae
View Solution

Step 1: Griffith (1928) used S. pneumoniae (pneumococcus).

Step 2: Smooth (S) strain (virulent) + heat-killed S + Rough (R) strain (non-virulent) → live S strain in mice.

Step 3: Proved transforming principle (later identified as DNA by Avery et al.). Quick Tip: Griffith’s experiment → foundation of molecular genetics.


Question 21:

The number of codons effective in coding twenty amino acids:

  • (1) . 20
  • (2) . 61
  • (3) . 32
  • (4) . 64
Correct Answer: (2). 61
View Solution

Step 1: Total codons = 4³ = 64 (U, C, A, G).

Step 2: 3 are stop codons (UAA, UAG, UGA) → no amino acid.

Step 3: 1 start codon (AUG) codes for methionine but is also counted in 6(4).

Thus, 61 codons code for 20 amino acids (degenerate code). Quick Tip: Genetic code: triplet, degenerate, universal, non-overlapping.


Question 22:

An antibiotic resistance gene in a vector usually helps in the selection of:

  • (1) . Non-competent cells
  • (2) . Competent cells
  • (3) . Transformed cells
  • (4) . Non-recombinant cells
Correct Answer: (3). Transformed cells
View Solution

Step 1: Vector (e.g., pBR322) has antibiotic resistance genes (amp\^R, tet\^R).

Step 2: Only transformed cells (took up recombinant plasmid) grow on antibiotic medium.

Step 3: Non-transformed cells die → selection of transformed cells. Quick Tip: Insertional inactivation: recombinant plasmid loses tet^R → white colonies (blue-white screening).


Question 23:

Silencing of specific mRNA in RNAi is by:

  • (1) . dsDNA
  • (2) . SSRNA
  • (3) . dsRNA
  • (4) . ssDNA
Correct Answer: (3). dsRNA
View Solution

Step 1: RNA interference (RNAi) is a gene silencing mechanism.

Step 2: dsRNA (double-stranded) is introduced → processed by Dicer into siRNA.

Step 3: siRNA + RISC → degrades complementary mRNA → gene silenced. Quick Tip: RNAi used in pest-resistant plants (Bt cotton) and nematode control (flavr savr tomato).


Question 24:

Cry-IAC effectively controls:

  • (1) . Ring worm
  • (2) . Cotton bollworms
  • (3) . Corn borer
  • (4) . Root nematode
Correct Answer: (2). Cotton bollworms
View Solution

Step 1: Cry-IAC and Cry-IAB genes from Bacillus thuringiensis (Bt).

Step 2: Produce crystal proteins toxic to lepidopteran larvae (bollworm, armyworm).

Step 3: Cry-IAC specifically controls cotton bollworms (Helicoverpa armigera).

Cry-IIAB → corn borer; Cry gene for nematodes not common. Quick Tip: Bt cotton → first GM crop in India (2002).


Question 25:

ADA deficiency can be cured by:

  • (1) . Heart Transplantation
  • (2) . Bone-marrow Transplantation
  • (3) . Liver Transplantation
  • (4) . Kidney Transplantation
Correct Answer: (2). Bone-marrow Transplantation
View Solution

Step 1: ADA (Adenosine Deaminase) deficiency → SCID (Severe Combined Immunodeficiency).

Step 2: ADA enzyme is required in T-lymphocytes (from bone marrow).

Step 3: Bone marrow transplantation provides healthy stem cells → produce functional ADA.

First gene therapy (1990) also used for ADA deficiency. Quick Tip: Gene therapy: retroviral vector → insert normal ADA gene.


Question 26:

Average natality rate in our village is 25, average mortality is 24, immigration 2 and emigration 3 and the net increase in population is :

  • (A) 27
  • (B) 0
  • (C) 5
  • (D) 10
Correct Answer: (B) 0
View Solution

Step 1: Population change = (Births + Immigration) – (Deaths + Emigration)
Step 2:
- Births (natality) = 25
- Deaths (mortality) = 24
- Immigration = 2
- Emigration = 3
Step 3: \[ \Delta P = (25 + 2) - (24 + 3) = 27 - 27 = 0 \]
Hence, net increase = 0. Quick Tip: - Population growth: \(\Delta P = (B + I) - (D + E)\) - Zero growth: \(B + I = D + E\)


Question 27:

The term "Molecular Scissors" refers to:

  • (1) . Taq polymerase
  • (2) . Polymerase-I
  • (3) . Polymerase-II
  • (4) . Restriction enzyme
Correct Answer: (4). Restriction enzyme
View Solution

Step 1: Restriction enzymes (e.g., EcoRI, HindIII) cut DNA at specific recognition sites.

Step 2: They act as molecular scissors in genetic engineering.

Step 3: Used to cut vector and insert DNA for cloning. Quick Tip: EcoRI → 5'-GAATTC-3' → sticky ends.


Question 28:

What does the sample of given base sequence represent? 5' - GAATTC - 3' 3' - CTTAAG - 5'

  • (1) . Palindromic sequence
  • (2) . Initiator codon at 5' end
  • (3) . Deletion mutation
  • (4) . Completion of replication
Correct Answer: (1). Palindromic sequence
View Solution

Step 1: Sequence: 5'-GAATTC-3' and 3'-CTTAAG-5' → reads same forward and backward on complementary strand.

Step 2: This is a palindromic sequence.

Step 3: Recognized by EcoRI restriction enzyme. Quick Tip: Palindrome: reads same in 5'→3' on both strands.


Question 29:

Gel electrophoresis is used for:

  • (1) . Cutting of DNA into fragments
  • (2) . Construction of recombinant DNA by joining with cloning vectors
  • (3) . Isolation of DNA molecule
  • (4) . Separation of DNA fragments according to their size
Correct Answer: (4). Separation of DNA fragments according to their size
View Solution

Step 1: Gel electrophoresis uses agarose gel and electric field.

Step 2: DNA (negatively charged) moves toward anode.

Step 3: Smaller fragments move faster → separation by size. Quick Tip: Ethidium bromide + UV → orange bands.


Question 30:

The amount of Photosynthetically active radiation captured by plants is:

  • (1) . 12 - 20 percent
  • (2) . 20 - 30 percent
  • (3) . 2 - 10 percent
  • (4) . 60 - 70 percent
Correct Answer: (3). 2 - 10 percent
View Solution

Step 1: PAR (Photosynthetically Active Radiation) = 400–700 nm.

Step 2: Plants capture only 2–10% of incident solar radiation for photosynthesis.

Step 3: Most energy lost as heat, reflection, or transmission. Quick Tip: GPP = Gross Primary Productivity; NPP = GPP – Respiration.


Question 31:

The given graph represents:


  • (1) . Species area relationship
  • (2) . Enzyme activity
  • (3) . Growth of organisms
  • (4) . Population growth
Correct Answer: (1). Species area relationship
View Solution

Step 1: The equation S = cA\^z is the species-area relationship (Alexander von Humboldt).

Step 2: S = species richness, A = area, c and z are constants.

Step 3: Logarithmic form: log S = log c + z log A → straight line on log-log plot.

Thus, the graph represents species-area relationship. Quick Tip: z ≈ 0.1–0.2 (mainlands), 0.2–0.3 (islands).


Question 32:

Cuscuta is an example of:

  • (1) . Predation
  • (2) . Broad Parasitism
  • (3) . Endoparasitism
  • (4) . Ectoparasitism
Correct Answer: (4). Ectoparasitism
View Solution

Step 1: Cuscuta (dodder) is a holoparasitic plant.

Step 2: It grows externally on host, penetrates haustoria into vascular tissue.

Step 3: Lives outside host → ectoparasite (not inside like endoparasite). Quick Tip: Cuscuta → yellow/orange twining stems → no chlorophyll.


Question 33:

Particulates of \phantom{abcde} size pose greatest risk to human health.

  • (1) . Less than 7.5 micrometers in diameter
  • (2) . Less than 2.5 micrometers in diameter
  • (3) . Less than 4.5 micrometers in diameter
  • (4) . Less than 3.5 micrometers in diameter
Correct Answer: (2). Less than 2.5 micrometers in diameter
View Solution

Step 1: PM2.5 (≤ 2.5 μm) are fine particulates.

Step 2: They penetrate deep into lungs and enter bloodstream.

Step 3: Cause respiratory and cardiovascular diseases → greatest health risk. Quick Tip: PM10 (>(2).5 μm) → upper respiratory tract; PM(2).5 → alveoli.


Question 34:

Maintenance of constant internal environment is called as:

  • (1) . Osmoregulation
  • (2) . Metastasis
  • (3) . Homeostasis
  • (4) . Thermoregulation
Correct Answer: (3). Homeostasis
View Solution

Step 1: Homeostasis = maintenance of stable internal conditions (pH, temperature, ions).

Step 2: Involves feedback mechanisms (negative/positive).

Step 3: Osmoregulation and thermoregulation are parts of homeostasis. Quick Tip: Claude Bernard: "constancy of internal milieu".


Question 35:

The animals which are active during day time:

  • (1) . Cresporal
  • (2) . Diurnal
  • (3) . Auroral
  • (4) . Vesporal
Correct Answer: (2). Diurnal
View Solution

Step 1: Diurnal = active during daytime (e.g., humans, birds).

Step 2: Nocturnal = night; Crepuscular = dawn/dusk.

Step 3: Auroral and vesporal are not standard terms. Quick Tip: Diurnal → opposite of nocturnal.


Question 36:

Which of the following statement is incorrect related to biomes?

  • (1) . Low temperature and less rainfall is a characteristics of Tundra biomes.
  • (2) . Variation in temperature and mean precipitation accounts for the major biomes.
  • (3) . More rainfall and low temperature is the characteristics of deserts.
  • (4) . High temperature and minimum rainfall help to form grasslands.
Correct Answer: (3). More rainfall and low temperature is the characteristics of deserts. Peak
View Solution

Step 1: Deserts → high temperature, very low rainfall (<25 cm/year).

Step 2: Option 3 says more rainfall and low temperature → incorrect.

Step 3:
- Tundra: cold, low precipitation
- Biomes depend on temp + rainfall
- Grasslands: moderate rain, warm Quick Tip: Desert: <25 cm rain; Tundra: permafrost.


Question 37:

Different types of respiratory organs like gills, book gills, book lungs and trachea are present in:

  • (1) . Annelids
  • (2) . Sponges
  • (3) . Molluscs
  • (4) . Arthropods
Correct Answer: (4). Arthropods
View Solution

Step 1: Arthropods show diversity in respiration:

- Aquatic: gills (crustaceans)

- Terrestrial: book lungs (scorpions), trachea (insects)

- Some: book gills (horseshoe crabs)

Step 2: Molluscs → ctenidia; Annelids → skin. Quick Tip: Trachea → air tubes in insects.


Question 38:

Which of the following plants is used to extract Colchicine?

  • (1) . Tulip
  • (2) . Colchicum
  • (3) . Aloe
  • (4) . Asparagus
Correct Answer: (2). Colchicum
View Solution

Step 1: Colchicine is a mitotic poison from Colchicum autumnale (autumn crocus).

Step 2: Inhibits spindle formation → used in polyploidy breeding.

Step 3: Not from tulip, aloe, or asparagus. Quick Tip: Colchicine → doubles chromosomes → polyploid plants.


Question 39:

Rows of S-shaped setae in the body of earthworm are present in all the segments, except:

  • (1) . The first segment
  • (2) . The last segment
  • (3) . The first and last segment
  • (4) . The first, last and clitellum
Correct Answer: (4). The first, last and clitellum
View Solution

Step 1: Earthworm (Pheretima) has S-shaped setae for locomotion.

Step 2: Present in all segments except:

- 1st segment (peristomium)

- Last segment (anus)

- Clitellum (14–16 segments, glandular, no setae)

Step 3: Thus, first, last, and clitellum lack setae. Quick Tip: Clitellum → cocoon formation in reproduction.


Question 40:

Cell theory was formulated by:

  • (1) . Schwann and Robert Brown
  • (2) . Schleiden and Schwann
  • (3) . Robert Hook and Robert Brown
  • (4) . Schleiden and Robert Brown
Correct Answer: (2). Schleiden and Schwann
View Solution

Step 1: Matthias Schleiden (1838): all plants made of cells.

Step 2: Theodor Schwann (1839): all animals made of cells.

Step 3: Together formulated cell theory (1839).

Robert Brown → nucleus (1831); Hooke → cell (1665). Quick Tip: Cell theory: (1). All organisms cellular, (2). Cell is basic unit.


Question 41:

The type of Polysaccharide present in a cotton fibre:

  • (1) . Glycogen
  • (2) . Starch
  • (3) . Insulin
  • (4) . Cellulose
Correct Answer: (4). Cellulose
View Solution

Step 1: Cotton fibre is made of cell wall material.

Step 2: Plant cell walls are composed of cellulose (β-1,4-glucose polymer).

Step 3: Cotton is >90% cellulose → used in textiles.

Glycogen (animal), starch (plant storage), insulin (protein hormone). Quick Tip: Cellulose → structural polysaccharide; humans cannot digest.


Question 42:

Enzyme involved in crossing over:

  • (1) . Ligase
  • (2) . Recombinase
  • (3) . Polymerase
  • (4) . Endonuclease
Correct Answer: (2). Recombinase
View Solution

Step 1: Crossing over occurs in pachytene of meiosis.

Step 2: Recombinase (e.g., Spo11, DMC1) mediates DNA strand exchange.

Step 3: Forms chiasmata → genetic recombination.

Endonuclease initiates break; ligase seals. Quick Tip: Synaptonemal complex → holds homologues during crossing over.


Question 43:

Kranz anatomy can be seen in:

  • (1) . Maize
  • (2) . Tomato
  • (3) . Potato
  • (4) . Pea
Correct Answer: (1). Maize
View Solution

Step 1: Kranz anatomy = wreath-like; seen in C4 plants.

Step 2: Bundle sheath cells with chloroplasts surround vascular bundle.

Step 3: Maize is C4 → has Kranz anatomy.

Tomato, potato, pea → C3 plants. Quick Tip: C4 plants: high light, hot climate; Hatch-Slack pathway.


Question 44:

Respiratory quotient of glucose is:

  • (1) . 1.0
  • (2) . 0
  • (3) . 0.7
  • (4) . 0.9
Correct Answer: (1). 1.0
View Solution

Step 1: RQ = CO₂ produced / O₂ consumed

Step 2: Glucose: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Step 3: RQ = 6 / 6 = (1).0

Fats → 0.7; proteins → 0.9. Quick Tip: RQ > 1 → anaerobic; RQ = 0 → no respiration.


Question 45:

A person suddenly starts coughing while swallowing food. This coughing would have been due to improper movement of:

  • (1) . Tongue
  • (2) . Epiglottis
  • (3) . Diaphragm
  • (4) . Neck
Correct Answer: (2). Epiglottis
View Solution

Step 1: Epiglottis = cartilage flap covering trachea during swallowing.

Step 2: Food enters larynx → triggers cough reflex.

Step 3: Improper closure of epiglottis → food in windpipe → coughing. Quick Tip: Choking → Heimlich maneuver.


Question 46:

Binomial nomenclature is introduced by:

  • (1) . John Ray
  • (2) . Carolus Linnaeus
  • (3) . Lamarck
  • (4) . Bentham and Hooker
Correct Answer: (2). Carolus Linnaeus
View Solution

Step 1: Carolus Linnaeus (1753) in Species Plantarum.

Step 2: Binomial nomenclature: Genus + species (e.g., Homo sapiens).

Step 3: Standardized naming → universal system. Quick Tip: ICBN (plants), ICZN (animals).


Question 47:

Bovine spongiform encephalopathy is caused by:

  • (1) . Fungi
  • (2) . Viroids
  • (3) . Prions
  • (4) . Virus
Correct Answer: (3). Prions
View Solution

Step 1: BSE (Mad Cow Disease) → neurodegenerative.

Step 2: Caused by prions (PrP\^Sc) → misfolded proteins.

Step 3: No nucleic acid → infects brain → spongy appearance. Quick Tip: Prions → Kuru, CJD, Scrapie.


Question 48:

Phycoerythrin and Floridean starch is found in:

  • (1) . Red algae
  • (2) . Blue – green algae
  • (3) . Green algae
  • (4) . Brown algae
Correct Answer: (1). Red algae
View Solution

Step 1: Red algae (Rhodophyta) → red pigment phycoerythrin.

Step 2: Store food as floridean starch (similar to glycogen).

Step 3: Found in deep water → absorb blue light. Quick Tip: Agar from Gelidium, Gracilaria.


Question 49:

Menstrual cycle exhibited by:

  • (1) . Tiger
  • (2) . Cow
  • (3) . Rat
  • (4) . Apes
Correct Answer: (4). Apes
View Solution

Step 1: Menstrual cycle = shedding of uterine lining (menstruation).

Step 2: Seen in primates (humans, apes, monkeys).

Step 3: Others have estrous cycle (no bleeding). Quick Tip: Human cycle: 28 days; apes similar.


Question 50:

An example of dioecious plant:

  • (1) . Papaya
  • (2) . Cucurbita
  • (3) . Coconut
  • (4) . Mango
Correct Answer: (1). Papaya
View Solution

Step 1: Dioecious = male and female flowers on separate plants.

Step 2: Papaya → male and female trees.

Step 3: Cucurbita, coconut, mango → monoecious. Quick Tip: Date palm, Marchantia → dioecious.


Question 51:

Stalk of the Stamen is:

  • (1) . Peduncle
  • (2) . Filament
  • (3) . Pedicel
  • (4) . Petiole
Correct Answer: (2). Filament
View Solution

Step 1: Stamen = male reproductive organ of flower.

Step 2: Consists of anther (pollen sac) and filament (stalk).

Step 3: Filament supports anther.

Peduncle (inflorescence stalk), pedicel (flower stalk), petiole (leaf stalk). Quick Tip: Androecium = all stamens.


Question 52:

The ovule of angiosperm is technically known as:

  • (1) . Megaspore
  • (2) . Megasporangium
  • (3) . Megasporophyl
  • (4) . Megaspore mother cell
Correct Answer: (2). Megasporangium
View Solution

Step 1: Ovule = integumented megasporangium.

Step 2: Contains nucellus (megasporangium tissue).

Step 3: Produces megaspore via meiosis from megaspore mother cell.

Megasporophyll = carpel. Quick Tip: Ovule → seed after fertilization.


Question 53:

Typical mature embryosac of angiosperm is:

  • (1) . 8 nucleated 1 celled structure
  • (2) . 8 nucleated 8 celled structure
  • (3) . 8 nucleated 7 celled structure
  • (4) . 7 nucleated 8 celled structure
Correct Answer: (3). 8 nucleated 7 celled structure
View Solution

Step 1: Polygonum type (most common).

Step 2: 7 cells: egg, 2 synergids, 3 antipodals, 1 central cell (2 polar nuclei).

Step 3: 8 nuclei: central cell has 2 nuclei.

Thus, 8-nucleated, 7-celled. Quick Tip: Embryosac → female gametophyte.


Question 54:

One of the 2000 years old viable seed, discovered during the archeological excavation at King Herold’s near dead sea:

  • (1) . Lupin
  • (2) . Sunflower
  • (3) . Phoenix dactylifera
  • (4) . Maize
Correct Answer: (3). Phoenix dactylifera
View Solution

Step 1: Date palm (Phoenix dactylifera) seed from Masada (Israel).

Step 2: Germinated in 2005 → ~2000 years old.

Step 3: Oldest viable seed known. Quick Tip: Lupin seed → 2000-year record (before date palm).


Question 55:

The testis are situated outside the abdominal cavity in scortum as it helps to:

  • (1) . Regulates harmone secretion
  • (2) . Store sperm
  • (3) . Release sperm
  • (4) . Maintain the low temperature
Correct Answer: (4). Maintain the low temperature
View Solution

Step 1: Spermatogenesis requires 2–3°C below body temperature.

Step 2: Scrotum keeps testes outside abdomen.

Step 3: Pampiniform plexus regulates temperature.

Sperm stored in epididymis. Quick Tip: Cryptorchidism → undescended testes → infertility.


Question 56:

Identify the odd one from the following:

  • (1) . Isthamus
  • (2) . Fimbriae
  • (3) . Labia minora
  • (4) . Infundibulum
Correct Answer: (3). Labia minora
View Solution

Step 1: Isthmus, infundibulum, fimbriae → parts of fallopian tube.

Step 2: Labia minora → external genitalia (vulva).

Step 3: Odd one → labia minora. Quick Tip: Fallopian tube: infundibulum → ampulla → isthmus → uterus.


Question 57:

In which month of gestation, the first movements of foetus and appearance of hair on its head is observed?

  • (1) . 8th month
  • (2) . 1st month
  • (3) . 4th month
  • (4) . 5th month
Correct Answer: (4). 5th month
View Solution

Step 1: Quickening (first fetal movement felt by mother) → 5th month.

Step 2: Lanugo (fine hair on head) appears around 5th month.

Step 3: Heartbeat detectable by 1st month; limbs by 8th week. Quick Tip: Gestation: 9 months = 3 trimesters.


Question 58:

The most abundant type of WBC cells:

  • (1) . monocytes
  • (2) . Basophils
  • (3) . Neutrophils
  • (4) . Eosinophils
Correct Answer: (3). Neutrophils
View Solution

Step 1: Neutrophils → 60–65% of total WBCs.

Step 2: First responders in acute inflammation.

Step 3: Phagocytic; multilobed nucleus.

Basophils (0–1%), eosinophils (2–3%), monocytes (2–8%). Quick Tip: WBC count: 4000–11000/mm(3).


Question 59:

Filtration of blood during urine formation takes place in:

  • (1) . Glomerulus
  • (2) . DCT
  • (3) . PCT
  • (4) . Collecting duct
Correct Answer: (1). Glomerulus
View Solution

Step 1: Ultrafiltration occurs in glomerulus.

Step 2: High pressure → plasma filters into Bowman’s capsule.

Step 3: Forms glomerular filtrate.

Reabsorption in PCT/DCT; secretion in DCT. Quick Tip: GFR ≈ 125 ml/min.


Question 60:

Corpus callosum connects the:

  • (1) . Spinal cord with the brain
  • (2) . Two lobes of cerebellum
  • (3) . Two cerebral hemispheres
  • (4) . Cerebrum and cerebellum
Correct Answer: (3). Two cerebral hemispheres
View Solution

Step 1: Corpus callosum = bundle of white matter fibers.

Step 2: Connects left and right cerebral hemispheres.

Step 3: Allows communication between hemispheres. Quick Tip: Split-brain patients → corpus callosum severed.

KCET Questions

  • 1.
    When a change in the gene frequency of population occurs by chance, it is called

      • Genetic recombination
      • Genetic drift
      • Founder effect
      • Gene migration

    • 2.

      In a practical examination, the following pedigree chart was given as a spotter for identification. The students identify the given pedigree chart as

        • Sex-linked dominant
        • Sex-linked recessive
        • Autosomal dominant
        • Autosomal recessive

      • 3.

        Identify the types of aestivation in corolla labelled as 'a', 'b', 'c' and 'd'

          • a-Vexillary, b-Imbricate, c-Twisted, d-Valvate
          • a-Vexillary, b-Imbricate, c-Valvate, d-Twisted
          • a-Vexillary, b-Twisted, c-Imbricate, d-Valvate
          • a-Imbricate, b-Valvate, c-Vexillary, d-Twisted

        • 4.
          Which of the following enzymes increases the permeability of the bacterial cell to lactose?

            • Transacetylase
            • Amylase
            • $\beta$-galactosidase
            • Permease

          • 5.
            Darwin’s finches represent one of the best examples of ____

              • Chemical evolution
              • Genetic equilibrium
              • Seasonal migration
              • Adaptive radiation

            • 6.
              How may types of gametes will be formed by a parent with genotype ‘AaBbCc’?

                • 8
                • 12
                • 6
                • 4

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