KCET 2022 Biology D3 Question Paper With Answer Key And Solutions PDF

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Shivam Yadav

Updated on - Nov 17, 2025

KCET 2022 Biology D-3 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 (D-3) Question Paper with Answer Key

KCET Biology (D-3) Question Paper 2022 with Answer Key download iconDownload Check Solution

KCET 2022 Biology D3 Question Paper With Answer Key And Solutions PDF


Question 1:

Example for Non-Mendelian disorder:

  • (A) Thalassemia
  • (B) Cystic fibrosis
  • (C) Haemophilia
  • (D) Down's syndrome
Correct Answer: (D) Down's syndrome
View Solution

Step 1: Mendelian disorders → single gene mutations (autosomal/X-linked).
Step 2:
- Thalassemia, Cystic fibrosis, Haemophilia → gene mutations → Mendelian.
- Down's syndrome → trisomy 21 → chromosomal aneuploidy → non-Mendelian.
Hence, (D) is correct. Quick Tip: - Mendelian: Sickle cell, CF, Haemophilia - Non-Mendelian: Down’s, Turner’s, Klinefelter’s (chromosomal)


Question 2:

Gynecomastia is a symptom of

  • (A) Turner's syndrome
  • (B) Cri-du-chat syndrome
  • (C) Down's syndrome
  • (D) Kline Felter's syndrome
Correct Answer: (D) Kline Felter's syndrome
View Solution

Step 1: Klinefelter’s syndrome → XXY male.
Step 2: Symptoms:
- Tall, sterile, gynecomastia (breast development due to estrogen excess).
Step 3: Turner’s (XO) → female; others → no gynecomastia.
Hence, (D) is correct. Quick Tip: - Klinefelter’s (XXY): Male + gynecomastia - Turner’s (XO): Female, webbed neck


Question 3:

The affected male in the pedigree chart is symbolized by:

  • (A) blacksquare
  • (B) blacklozenge
  • (C) square
  • (D) bigcirc
Correct Answer: (A) blacksquare
View Solution

Step 1: Standard pedigree symbols:
- Male → square
- Female → circle
- Affected → shaded/filled
Step 2: Affected male → shaded square (■).
Hence, (A) is correct. Quick Tip: - Pedigree: – Male normal: □ – Male affected: ■ – Female normal: ○ – Female affected: ●


Question 4:

The given diagram represents:

  • (A) Nucleosome
  • (B) Mesosome
  • (C) Chromosome
  • (D) Ribosome
Correct Answer: (A) Nucleosome
View Solution

Step 1: Diagram shows DNA wrapped around histone core.
Step 2: This is "beads on a string" → nucleosome (basic chromatin unit).
Step 3:
- Mesosome: Bacterial
- Chromosome: Condensed
- Ribosome: Protein synthesis
Hence, nucleosome. Quick Tip: - Nucleosome: DNA + histone octamer (2 each H2A, H2B, H3, H4) - 11 nm fiber


Question 5:

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

  • (A) Progestogen
  • (B) FSH
  • (C) hCG
  • (D) Relaxin
Correct Answer: (B) FSH
View Solution

Step 1: Placental hormones:
- hCG, hPL, progesterone, estrogen, relaxin.
Step 2: FSH → anterior pituitary → stimulates follicles.
Hence, not secreted by placenta. Quick Tip: - Placenta: hCG, Progesterone, Relaxin - Pituitary: FSH, LH


Question 6:

Which of the following is correctly matched?

  • (A) Bulbil – Agave
  • (B) Spores – Sponge
  • (C) Conidia – Hydra
  • (D) Gemmules – Amoeba
Correct Answer: (A) Bulbil – Agave
View Solution

Step 1:
- Bulbil: Vegetative bud → Agave (correct).
- Gemmules: Sponges
- Conidia: Fungi (Penicillium)
- Spores: Ferns, fungi
Hence, only (A) is correct. Quick Tip: - Bulbil: Agave, Onion - Gemmule: Sponge - Conidia: Aspergillus


Question 7:

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

  • (A) RCH
  • (B) RTI
  • (C) MTP
  • (D) ART
Correct Answer: (D) ART
View Solution

Step 1:
- ART → Assisted Reproductive Technology (IVF, IUI, ZIFT, GIFT).
- RCH: Program
- MTP: Abortion
- RTI: Infection
Step 2: For infertility → ART.
Hence, (D). Quick Tip: - Infertility: ART (IVF, ICSI) - Contraception: MTP


Question 8:

Amniocentesis is a process to:

  • (A) To grow cell on culture medium
  • (B) Determine the sex of the foetus
  • (C) Determine any disease of heart
  • (D) Know about the disease of brain
Correct Answer: (B) Determine the sex of the foetus
View Solution

Step 1: Amniocentesis → amniotic fluid sampling (16–18 weeks).
Step 2: Fetal cells → karyotyping → sex determination, genetic disorders.
Step 3: Heart/brain → ultrasound; culture → secondary.
Hence, primary → sex & genetics. Quick Tip: - Amniocentesis: Fetal karyotype - Banned for sex selection in India


Question 9:

The first human like being is

  • (A) Homo sapiens
  • (B) Homo menthus
  • (C) Homo erectus
  • (D) Homo habilis
Correct Answer: (D) Homo habilis
View Solution

Step 1: Homo habilis (2.4–1.4 mya) → first in Homo genus.
Step 2: "Handy man" → used stone tools.
Step 3:
- H. erectus: Fire, upright
- H. sapiens: Modern
Hence, first human-like → H. habilis. Quick Tip: - Evolution: Australopithecus → H. habilis → H. erectus → H. sapiens


Question 10:

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

  • (A) Male Homogamety
  • (B) Female Homogamety
  • (C) Male Heterogamety
  • (D) Female Heterogamety
Correct Answer: (C) Male Heterogamety
View Solution

Step 1:
- XY: Male = XY (heterogametic)
- XO: Male = XO (heterogametic)
Step 2: In both, male produces two gamete types (X/Y or X/O).
Step 3: Female → XX (homogametic).
Hence, male heterogamety. Quick Tip: - Male heterogamety: XY, XO - Female heterogamety: ZW (birds)


Question 11:

One of the side-effects of the use of anabolic steroids in females

  • (A) Cirrhosis of liver
  • (B) Masculinisation
  • (C) Loss of memory
  • (D) Hallucination
Correct Answer: (B) Masculinisation
View Solution

Step 1: Anabolic steroids → synthetic testosterone.
Step 2: In females → androgenic effects:
- Deep voice
- Facial hair
- Enlarged clitoris → masculinisation.
Step 3: Liver damage possible, but masculinisation is primary in females.
Hence, (B). Quick Tip: - Females + steroids: Virilization (masculinisation) - Males: Gynecomastia, infertility


Question 12:

Which one of the following is an opiate narcotics?

  • (A) Amphetamines
  • (B) LSD
  • (C) Barbiturates
  • (D) Morphine
Correct Answer: (D) Morphine
View Solution

Step 1: Opiates → from opium poppy.
Step 2:
- Morphine, Heroin, Codeine → opiates.
- Amphetamines: Stimulants
- LSD: Hallucinogen
- Barbiturates: Sedatives
Hence, morphine. Quick Tip: - Opiates: Morphine, Heroin - Cannabinoids: THC - Cocaine: Stimulant


Question 13:

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

  • (A) Bacterium producing a large amount of CO₂
  • (B) Fungus that releases a lot of gases during metabolic activities
  • (C) Machine
  • (D) Bacterium that produces methane gas
Correct Answer: (A) Bacterium producing a large amount of CO₂
View Solution

Step 1: Swiss cheese → Propionibacterium shermanii.
Step 2: Ferments lactate → propionic acid + CO₂.
Step 3: CO₂ bubbles → large holes (eyes).
Step 4: Not methane, fungus, or machine.
Hence, (A). Quick Tip: - Swiss cheese eyes: CO₂ from Propionibacterium - Roquefort: Penicillium (blue veins)


Question 14:

Which vitamin is increased by ‘LAB’ in curd?

  • (A) Vitamin B₁₂
  • (B) Vitamin E
  • (C) Vitamin C
  • (D) Vitamin B
Correct Answer: (A) Vitamin B₁₂
View Solution

Step 1: LAB → Lactobacillus in curd.
Step 2: Synthesizes Vitamin B₁₂ (cobalamin) during fermentation.
Step 3:
- Vitamin C → destroyed by heat
- Vitamin E → not microbial
- Vitamin B → vague
Hence, B₁₂. Quick Tip: - Curd: ↑ B₁₂, B₂, folate - Yogurt: Probiotic + vitamins


Question 15:

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

  • (A) Amylase
  • (B) Lipase
  • (C) Renin
  • (D) Protease
Correct Answer: (B) Lipase
View Solution

Step 1: Oily stains → lipids.
Step 2: Lipase → hydrolyzes fats → glycerol + fatty acids.
Step 3:
- Amylase: Starch
- Renin: Milk clotting
- Protease: Proteins
Hence, lipase. Quick Tip: - Detergent enzymes: – Lipase → Oil – Protease → Blood – Amylase → Food


Question 16:

DNA replicates semicontinuously was first shown in:

  • (A) Plants
  • (B) Higher animals
  • (C) Escherichia coli
  • (D) Human cell
Correct Answer: (C) Escherichia coli
View Solution

Step 1: Meselson & Stahl (1958) → E. coli.
Step 2: \(^{15}\)N → \(^{14}\)N → hybrid DNA.
Step 3: Okazaki fragments → lagging strand → semicontinuous.
Hence, E. coli. Quick Tip: - Semiconservative: Meselson-Stahl - Semicontinuous: Okazaki (lagging strand)


Question 17:

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

  • (A) Salmonella typhimurium
  • (B) Streptococcus pneumoniae
  • (C) Escherichia coli
  • (D) Bacillus thuringiensis
Correct Answer: (B) Streptococcus pneumoniae
View Solution

Step 1: Griffith → S (smooth) and R (rough) strains.
Step 2: Heat-killed S + live R → virulent mice.
Step 3: Transforming principle → DNA.
Hence, S. pneumoniae. Quick Tip: - Griffith (1928): Transformation - Avery (1944): DNA is genetic material


Question 18:

The number of codons effective in coding twenty amino acids:

  • (A) 64
  • (B) 20
  • (C) 61
  • (D) 32
Correct Answer: (C) 61
View Solution

Step 1: Total codons = \(4^3 = 64\).
Step 2:
- 3 stop codons: UAA, UAG, UGA
- 61 sense codons → 20 amino acids (degenerate).
Hence, 61. Quick Tip: - 64 total - 61 code aa - 3 stop - AUG: Start + Met


Question 19:

Which aspect forms the basis of DNA finger-printing?

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

Step 1: DNA fingerprinting → VNTRs (Variable Number Tandem Repeats).
Step 2: Satellite DNA → highly repetitive, variable repeats.
Step 3: Unique pattern per individual.
Hence, (A). Quick Tip: - VNTRs/Minisatellites → DNA fingerprinting - Alec Jeffreys (1984)


Question 20:

Identify the most infectious and fatal type of malarial parasite:

  • (A) Plasmodium falciparum
  • (B) Plasmodium ovale
  • (C) Plasmodium vivax
  • (D) Plasmodium malariae
Correct Answer: (A) Plasmodium falciparum
View Solution

Step 1: P. falciparum → cerebral malaria, high parasitemia.
Step 2: Causes blackwater fever, organ failure.
Step 3: Most fatal (WHO).
Hence, (A). Quick Tip: - P. falciparum: Malignant tertian, fatal - P. vivax: Benign tertian, relapsing


Question 21:

The type of antibodies produced during the allergic reaction

  • (A) Ig G
  • (B) Ig M
  • (C) Ig A
  • (D) Ig E
Correct Answer: (D) Ig E
View Solution

Step 1: Allergic reaction → Type I hypersensitivity.
Step 2: Mediated by IgE bound to mast cells/basophils.
Step 3: Allergen → cross-links IgE → histamine release → allergy.
Step 4:
- IgG, IgM: Immunity
- IgA: Mucosal
Hence, IgE. Quick Tip: - Allergy: IgE → Mast cell degranulation - Anaphylaxis: Severe IgE reaction


Question 22:

ADA deficiency can be cured by

  • (A) Kidney Transplantation
  • (B) Heart Transplantation
  • (C) Bone-marrow Transplantation
  • (D) Liver Transplantation
Correct Answer: (C) Bone-marrow Transplantation
View Solution

Step 1: ADA deficiency → SCID (no T/B cells).
Step 2: Bone marrow → source of hematopoietic stem cells → produce healthy lymphocytes.
Step 3: First gene therapy → ADA gene into lymphocytes.
Hence, bone-marrow transplantation. Quick Tip: - ADA deficiency: SCID - Cure: Bone marrow transplant / Gene therapy


Question 23:

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

Step 1: Population change = (Births + Immigration) – (Deaths + Emigration)
Step 2:
- Births = 25
- Immigration = 2
- Deaths = 24
- 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 24:

The term “Molecular Scissors” refers to

  • (A) Restriction enzyme
  • (B) Taq Polymerase
  • (C) Polymerase-I
  • (D) Polymerase-II
Correct Answer: (A) Restriction enzyme
View Solution

Step 1: Molecular scissors → cut DNA at specific sites.
Step 2: Restriction enzymes (e.g., EcoRI) → recognize palindromes → cleave.
Step 3: Used in cloning.
Step 4: Polymerases → synthesize, not cut.
Hence, restriction enzyme. Quick Tip: - Restriction enzyme → "Molecular scissors" - DNA ligase → "Molecular glue"


Question 25:

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

  • (A) Completion of replication
  • (B) Palindromic sequence
  • (C) Initiator codon at 5' end
  • (D) Deletion mutation
Correct Answer: (B) Palindromic sequence
View Solution

Step 1: Sequence: \(5'\)-GAATTC-\(3'\) \(3'\)-CTTAAG-\(5'\)
Step 2: Reads same on complementary strand in reverse → palindrome.
Step 3: Recognized by EcoRI.
Hence, palindromic sequence. Quick Tip: - Palindrome: 5'-GAATTC-3' = 3'-CTTAAG-5' - EcoRI cuts between G and A


Question 26:

Gel electrophoresis is used for

  • (A) Separation of DNA fragments according to their size.
  • (B) Cutting of DNA into fragments.
  • (C) Construction of recombinant DNA by joining with cloning vectors.
  • (D) Isolation of DNA molecule.
Correct Answer: (A) Separation of DNA fragments according to their size.
View Solution

Step 1: Agarose gel → DNA migrates to anode (negative charge).
Step 2: Smaller fragments move faster.
Step 3:
- Cutting → restriction enzyme
- Joining → ligase
- Isolation → extraction
Hence, separation by size. Quick Tip: - Gel electrophoresis: – Small DNA → far – Large DNA → near well


Question 27:

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

  • (A) Non-recombinant cells
  • (B) Non-competent cells
  • (C) Competent cells
  • (D) Transformed cells
Correct Answer: (D) Transformed cells
View Solution

Step 1: Vector → ampᵣ gene.
Step 2: Only transformed cells grow on ampicillin.
Step 3: Non-recombinants also grow → use insertional inactivation.
Hence, first selects transformants. Quick Tip: - Selectable marker: ampᵣ → transformants - Blue-white: X-gal → recombinants (white)


Question 28:

Silencing of specific mRNA in RNAi is by

  • (A) ssDNA
  • (B) dsDNA
  • (C) SSRN
  • (D) dsRNA
Correct Answer: (D) dsRNA
View Solution

Step 1: RNAi → dsRNA introduced.
Step 2: Dicer → cuts → siRNA.
Step 3: RISC → degrades target mRNA.
Step 4: SSRN → not a term.
Hence, dsRNA. Quick Tip: - RNAi: dsRNA → Dicer → siRNA → RISC → mRNA cleavage


Question 29:

Cry-IAC effectively controls,

  • (A) Root nematode
  • (B) Ring worm
  • (C) Cotton bollworms
  • (D) Corn borer
Correct Answer: (C) Cotton bollworms
View Solution

Step 1: Cry-IAC → Bt toxin.
Step 2: Targets lepidopteran larvae (moths/butterflies).
Step 3: Cotton bollworm (Helicoverpa) → lepidopteran.
Step 4:
- Cry-IIAB: Nematodes
- Corn borer: Different Cry gene
Hence, cotton bollworms. Quick Tip: - Cry-I: Lepidoptera (bollworm) - Cry-II: Diptera + Lepidoptera - Cry-III: Coleoptera


Question 30:

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

  • (A) Less than 3.5 micrometers in diameter
  • (B) Less than 7.5 micrometers in diameter
  • (C) Less than 2.5 micrometers in diameter
  • (D) Less than 4.5 micrometers in diameter
Correct Answer: (C) Less than 2.5 micrometers in diameter
View Solution

Step 1: PM₂.₅ → particles < 2.5 μm in diameter.
Step 2: Small size → penetrate deep into lungs → alveoli → enter bloodstream.
Step 3: Cause cardiovascular disease, lung cancer, COPD.
Step 4: PM₁₀ (>2.5 μm) → trapped in upper airways.
Hence, < 2.5 μm → greatest risk. Quick Tip: - PM₂.₅: < 2.5 μm → alveoli + blood - PM₁₀: < 10 μm → nose/throat - WHO: PM₂.₅ < 10 μg/m³ (annual mean)


Question 31:

Maintenance of constant internal environment is called as

  • (A) Thermoregulation
  • (B) Osmoregulation
  • (C) Metastasis
  • (D) Homeostasis
Correct Answer: (D) Homeostasis
View Solution

Step 1: Homeostasis → maintaining stable internal conditions (pH, temperature, ions, glucose).
Step 2:
- Thermoregulation: Temperature
- Osmoregulation: Water/salt balance
- Metastasis: Cancer spread
Step 3: Homeostasis encompasses all.
Hence, homeostasis. Quick Tip: - Homeostasis → "Same standing" (Walter Cannon) - Negative feedback: Most common


Question 32:

The animals which are active during day time :

  • (A) Vesporal
  • (B) Cresporal
  • (C) Diurnal
  • (D) Auroral
Correct Answer: (C) Diurnal
View Solution

Step 1:
- Diurnal: Active during day (e.g., humans, eagles).
- Nocturnal: Night (owls)
- Crepuscular: Dawn/dusk (rabbits)
Step 2: Vesporal, Cresporal, Auroral → not standard terms.
Hence, diurnal. Quick Tip: - Diurnal → Day - Nocturnal → Night - Crepuscular → Twilight


Question 33:

Which of the following statement is incorrect related to biomes ?

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

Step 1: Deserts: High temperature + very low rainfall (<25 cm/year).
Step 2: (D) says more rainfall + low temperature → incorrect.
Step 3:
- (A): Grasslands → moderate rain, warm
- (B): Tundra → cold, low precipitation
- (C): Temperature + precipitation → biome distribution
Hence, (D) is incorrect. Quick Tip: - Desert: Hot, arid - Tundra: Cold, arid - Rainforest: Hot, wet


Question 34:

The amount of Photosynthetically active radiation captured by plants is

  • (A) 60 – 70 percent
  • (B) 12 – 20 percent
  • (C) 20 – 30 percent
  • (D) 2 – 10 percent
Correct Answer: (D) 2 – 10 percent
View Solution

Step 1: PAR (400–700 nm) → ~50% of solar radiation.
Step 2: Plants capture 2–10% of PAR for GPP.
Step 3: Most energy lost: reflection, heat, transmission.
Step 4: NPP → ~1–2% of solar energy.
Hence, 2–10%. Quick Tip: - PAR: 50% solar - GPP: 2–10% of PAR - NPP: 0.2–2% solar


Question 35:

The given graph represents

  • (A) Population growth
  • (B) Species area relationship
  • (C) Enzyme activity
  • (D) Growth of organisms
Correct Answer: (B) Species area relationship
View Solution

Step 1: Equation: \( S = cA^Z \)
- \( S \): Species richness
- \( A \): Area
- \( c, Z \): Constants
Step 2: Species-area relationship (Alexander von Humboldt).
Step 3: Log form: \(\log S = \log c + Z \log A\) → straight line.
Hence, species-area relationship. Quick Tip: - \( S = cA^Z \) → Species-area curve - Z: 0.1–0.2 (mainland), 0.6–1.0 (islands)


Question 36:

Cuscuta is an example of

  • (A) Ectoparasitism
  • (B) Predation
  • (C) Brood Parasitism
  • (D) Endoparasitism
Correct Answer: (A) Ectoparasitism
View Solution

Step 1: Cuscuta (dodder) → yellow, leafless vine.
Step 2: Attaches externally to host stem → haustoria penetrate → absorb nutrients.
Step 3:
- Ectoparasite: Lives outside host
- Endoparasite: Inside
- Brood parasitism: Cuckoo
Hence, ectoparasitism. Quick Tip: - Cuscuta: Stem holoparasite → ectoparasite - Rafflesia: Root endoparasite


Question 37:

Cell theory was formulated by

  • (A) Schleiden and Robert Brown
  • (B) Schwann and Robert Brown
  • (C) Schleiden and Schwann
  • (D) Robert Hook and Robert Brown
Correct Answer: (C) Schleiden and Schwann
View Solution

Step 1: Cell Theory (1838–1839):
- M.J. Schleiden → all plants made of cells.
- Theodor Schwann → all animals made of cells.
Step 2: Postulates:
1. All living organisms composed of cells.
2. Cell is basic unit of life.
Step 3: Virchow (1855): "Omnis cellula-e-cellula".
Hence, Schleiden and Schwann. Quick Tip: - Cell Theory: – Schleiden (1838): Plants – Schwann (1839): Animals – Virchow: Cell division


Question 38:

The type of Polysaccharide present in a cotton fibre

  • (A) Cellulose
  • (B) Glycogen
  • (C) Starch
  • (D) Inulin
Correct Answer: (A) Cellulose
View Solution

Step 1: Cotton fiber → seed hair of Gossypium.
Step 2: >90% cellulose (β-1,4-glucan).
Step 3:
- Glycogen: Animal storage
- Starch: Plant storage
- Inulin: Fructan (dahlia)
Hence, cellulose. Quick Tip: - Cotton: Cellulose (95%) - Wool: Protein - Jute: Lignin + cellulose


Question 39:

Enzyme involved in crossing over

  • (A) Endonuclease
  • (B) Ligase
  • (C) Recombinase
  • (D) Polymerase
Correct Answer: (C) Recombinase
View Solution

Step 1: Crossing over → pachytene of meiosis.
Step 2: Recombinase (Spo11, RAD51) → forms Holliday junction.
Step 3:
- Endonuclease → cuts
- Ligase → joins
- Polymerase → fills
But recombinase initiates.
Hence, recombinase. Quick Tip: - Crossing over: – Synaptonemal complex → Recombinase → Chiasma


Question 40:

Kranz anatomy can be seen in

  • (A) Pea
  • (B) Maize
  • (C) Tomato
  • (D) Potato
Correct Answer: (B) Maize
View Solution

Step 1: Kranz anatomy → C4 plants.
Step 2:
- Bundle sheath with chloroplasts
- Mesophyll radial
Step 3: Maize → C4; others → C3.
Hence, maize. Quick Tip: - Kranz: C4 (maize, sugarcane) - C3: Pea, tomato, potato


Question 41:

Respiratory quotient of glucose is

  • (A) 0.9
  • (B) 1.0
  • (C) 0
  • (D) 0.7
Correct Answer: (B) 1.0
View Solution

Step 1: RQ = \(\frac{CO_2}{O_2}\)
Step 2: Glucose: \[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \]
Step 3: \[ RQ = \frac{6}{6} = 1.0 \]
Hence, 1.0. Quick Tip: - RQ: – Carbs: 1.0 – Fats: 0.7 – Proteins: 0.9 – Anaerobic: ∞


Question 42:

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

  • (A) Neck
  • (B) Tongue
  • (C) Epiglottis
  • (D) Diaphragm
Correct Answer: (C) Epiglottis
View Solution

Step 1: Epiglottis → flap over larynx.
Step 2: Closes during swallowing → food to esophagus.
Step 3: If fails → food in trachea → cough reflex.
Hence, epiglottis. Quick Tip: - Swallowing: – Epiglottis closes larynx – Peristalsis → esophagus


Question 43:

Binomial nomenclature is introduced by

  • (A) Bentham and Hooker
  • (B) John Ray
  • (C) Carolus Linnaeus
  • (D) Lamarck
Correct Answer: (C) Carolus Linnaeus
View Solution

Step 1: Binomial → Genus + species.
Step 2: Linnaeus → "Species Plantarum" (1753), "Systema Naturae" (1758).
Step 3:
- Bentham & Hooker → classification
- John Ray → early binomial
Hence, Linnaeus. Quick Tip: - Binomial: Linnaeus - ICBN/ICZN: Rules - Homo sapiens: Linnaeus


Question 44:

Bovine spongiform encephalopathy is caused by

  • (A) Virus
  • (B) Fungi
  • (C) Viroids
  • (D) Prions
Correct Answer: (D) Prions
View Solution

Step 1: BSE → Mad Cow Disease.
Step 2: Caused by prions → misfolded PrP.
Step 3: No nucleic acid → resistant.
Step 4: vCJD in humans.
Hence, prions. Quick Tip: - Prions: – BSE, CJD, Kuru, Scrapie – Protein only


Question 45:

Phycocrythrin and Floridean starch is found in

  • (A) Brown algae
  • (B) Red algae
  • (C) Blue – green algae
  • (D) Green algae
Correct Answer: (B) Red algae
View Solution

Step 1: Red algae (Rhodophyta):
- Phycoerythrin (red pigment)
- Floridean starch (storage)
Step 2:
- Brown: Fucoxanthin
- BGA: Phycocyanin
- Green: Chlorophyll a,b
Hence, red algae. Quick Tip: - Red algae: – Phycoerythrin – Floridean starch – Agar, carrageenan


Question 46:

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

  • (A) Arthropods
  • (B) Annelids
  • (C) Sponges
  • (D) Molluscs
Correct Answer: (A) Arthropods
View Solution

Step 1: Arthropods:
- Gills: Crabs
- Book gills: Horseshoe crab
- Book lungs: Scorpions
- Trachea: Insects
Step 2: Others lack diversity.
Hence, arthropods. Quick Tip: - Arthropod respiration: – Gills, book gills, book lungs, trachea


Question 47:

Which of the following plant is used to extract Colchicine?

  • (A) Asparagus
  • (B) Tulip
  • (C) Colchicum
  • (D) Aloe
Correct Answer: (C) Colchicum
View Solution

Step 1: Colchicine → from Colchicum autumnale (autumn crocus).
Step 2: Inhibits spindle → polyploidy.
Hence, Colchicum. Quick Tip: - Colchicine: Colchicum autumnale - Use: Polyploidy, gout


Question 48:

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

  • (A) the first, last and clitellum
  • (B) the first segment
  • (C) the last segment
  • (D) the first and last segment
Correct Answer: (A) the first, last and clitellum
View Solution

Step 1: Setae → locomotion.
Step 2: Absent in:
- 1st segment
- Last segment
- Clitellum (14–16)
Step 3: Clitellum → cocoon formation.
Hence, first, last, clitellum. Quick Tip: - Earthworm setae: – Absent: 1st, last, 14–16 – S-shaped, 4 pairs/segment


Question 49:

The ovule of angiosperm is technically known as:

  • (A) Megaspore mother cell
  • (B) Megaspore
  • (C) Megasporangium
  • (D) Megasporephyll
Correct Answer: (C) Megasporangium
View Solution

Step 1: Ovule → integumented megasporangium.
Step 2:
- Megasporangium → produces megaspores.
- Megaspore mother cell → diploid cell inside.
- Megaspore → haploid product.
- Megasporophyll → carpel.
Step 3: Ovule = megasporangium + integuments.
Hence, megasporangium. Quick Tip: - Ovule = Megasporangium - Nucellus = Megasporangium tissue


Question 50:

Typical mature embryosac of angiosperm is

  • (A) 7 nucleated 8 celled structure
  • (B) 8 nucleated 1 celled structure
  • (C) 8 nucleated 8 celled structure
  • (D) 8 nucleated 7 celled structure
Correct Answer: (A) 7 nucleated 8 celled structure
View Solution

Step 1: Polygonum type (common):
- 3 antipodals
- 2 synergids
- 1 egg cell
- 1 central cell (2 polar nuclei → fuse → 1 nucleus)
Step 2: Total 8 cells, 7 nuclei.
Hence, 7 nucleated, 8 celled. Quick Tip: - Embryosac: – 8 cells – 7 nuclei (central cell: 2n → 1 nucleus)


Question 51:

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

  • (A) Maize
  • (B) Lupin
  • (C) Sunflower
  • (D) Phoenix dactylifera
Correct Answer: (D) Phoenix dactylifera
View Solution

Step 1: Date palm (Phoenix dactylifera) seed → ~2000 years old.
Step 2: Excavated at Masada, Israel → germinated in 2005.
Step 3: Named "Methuselah".
Hence, Phoenix dactylifera. Quick Tip: - Oldest viable seed: Date palm (2000 yrs) - Lupin: Arctic (10,000 yrs, unconfirmed)


Question 52:

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

  • (A) Maintain the low temperature
  • (B) Regulates hormone secretion
  • (C) Store sperm
  • (D) Release sperm
Correct Answer: (A) Maintain the low temperature
View Solution

Step 1: Spermatogenesis requires 2–3°C below body temperature (35°C).
Step 2: Scrotum → thermoregulation via pampiniform plexus, cremaster muscle.
Step 3: Hormone, storage, release → not primary reason.
Hence, low temperature. Quick Tip: - Scrotum: 35°C for sperm - Cryptorchidism: Undescended → infertility


Question 53:

Identify the odd one from the following:

  • (A) Infundibulum
  • (B) Isthmus
  • (C) Fimbriae
  • (D) Labia minora
Correct Answer: (D) Labia minora
View Solution

Step 1: Infundibulum, Fimbriae, Isthmus → parts of fallopian tube.
Step 2: Labia minora → external genitalia (vulva).
Hence, labia minora is odd. Quick Tip: - Fallopian tube: – Fimbriae → Infundibulum → Ampulla → Isthmus


Question 54:

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

  • (A) 5th month
  • (B) 8th month
  • (C) 1st month
  • (D) 4th month
Correct Answer: (A) 5th month
View Solution

Step 1: Quickening → first fetal movement → 18–20 weeks (~5th month).
Step 2: Lanugo (fine hair) → appears on head → 5th month.
Step 3: 4th month → limbs form; 8th → fat deposition.
Hence, 5th month. Quick Tip: - 5th month: Quickening + Lanugo - 7–8th month: Lanugo sheds


Question 55:

The most abundant type of WBC cells

  • (A) Eosinophils
  • (B) Monocytes
  • (C) Basophils
  • (D) Neutrophils
Correct Answer: (D) Neutrophils
View Solution

Step 1: WBC differential:
- Neutrophils: 50–70%
- Lymphocytes: 20–40%
- Monocytes: 2–8%
- Eosinophils: 1–4%
- Basophils: 0.5–1%
Step 2: Neutrophils → first responders.
Hence, most abundant. Quick Tip: - WBC order: Neutrophils > Lymphocytes > Monocytes > Eosinophils > Basophils ("Never Let Monkeys Eat Bananas")


Question 56:

Filtration of blood during urine formation takes place in

  • (A) Collecting duct
  • (B) Glomerulus
  • (C) DCT
  • (D) PCT
Correct Answer: (B) Glomerulus
View Solution

Step 1: Ultrafiltration → glomerulus (Bowman’s capsule).
Step 2: High pressure → plasma filters → filtrate.
Step 3:
- PCT/DCT → reabsorption
- Collecting duct → concentration
Hence, glomerulus. Quick Tip: - Glomerulus: Filtration - PCT: 65% reabsorption - Loop of Henle: Concentration


Question 57:

Corpus Callosum connects the

  • (A) Cerebrum and cerebellum
  • (B) Spinal cord with the brain
  • (C) Two lobes of cerebellum
  • (D) Two cerebral hemispheres
Correct Answer: (D) Two cerebral hemispheres
View Solution

Step 1: Corpus callosum → white matter bundle.
Step 2: Connects left and right cerebral hemispheres.
Step 3: Allows interhemispheric communication.
Hence, two cerebral hemispheres. Quick Tip: - Corpus callosum: Left ↔ Right cerebrum - Split-brain: Severed → no communication


Question 58:

Menstrual cycle is exhibited by:

  • (A) Apes
  • (B) Tiger
  • (C) Cow
  • (D) Rat
Correct Answer: (A) Apes
View Solution

Step 1: Menstrual cycle → uterine bleeding if no pregnancy.
Step 2: Seen in higher primates: humans, apes, Old World monkeys.
Step 3: Others → estrous cycle (no bleeding).
Hence, apes.


Question 59:

An example of dioecious plant:

  • (A) Mango
  • (B) Papaya
  • (C) Cucurbita
  • (D) Coconut
Correct Answer: (B) Papaya
View Solution

Step 1: Dioecious → male and female flowers on separate plants.
Step 2: Papaya → male, female, hermaphrodite plants.
Step 3: Others → monoecious or bisexual.
Hence, papaya. Quick Tip: - Dioecious: Papaya, Date palm - Monoecious: Cucurbita, Coconut


Question 60:

Stalk of the Stamen is:

  • (A) Petiole
  • (B) Peduncle
  • (C) Filament
  • (D) Pedicel
Correct Answer: (C) Filament
View Solution

Step 1: Stamen = anther + filament.
Step 2: Filament → stalk supporting anther.
Step 3:
- Petiole: Leaf stalk
- Peduncle: Inflorescence stalk
- Pedicel: Flower stalk
Hence, filament. Quick Tip: - Stamen: Filament (stalk) + Anther - Pollen sac → inside anther

KCET Questions

  • 1.
    A colourblind man marries a carrier woman. The percentage of their colourblind progeny in the next generation will be ____

      • 75\%
      • 100\%
      • 25\%
      • 50\%

    • 2.
      How may types of gametes will be formed by a parent with genotype ‘AaBbCc’?

        • 8
        • 12
        • 6
        • 4

      • 3.

        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

        • 4.

          Identify which one of the given pair of options is correct with respect to Down's syndrome and Turner's syndrome

            • c
            • d
            • a
            • b

          • 5.
            In which of the following, HIV replicates and produces its progeny viruses?

              • Killer T-lymphocytes
              • Suppressor T-lymphocytes
              • Helper T-lymphocytes
              • Memory T-lymphocytes

            • 6.
              Choose the correct statement from the following:

                • Charles Darwin travelled around the world in a ship called HMS Beagle
                • There has been gradual evolution of life forms
                • According to Darwin, fitness refers to physical fitness only
                • Fossils are remains of hard parts of life forms found in rocks

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