KCET 2022 Biology C1 Question Paper With Answer Key And Solutions PDF

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Shivam Yadav

Updated on - Nov 17, 2025

KCET 2022 Biology C-1 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-1) Question Paper with Answer Key

KCET Biology (C-1) Question Paper 2022 with Answer Key download iconDownload Check Solution

KCET 2022 Biology C1 Question Paper With Answer Key And Solutions PDF


Question 1:

Which aspect forms the basis of DNA finger-printing?

  • (A) The Satellite DNA showing high degree of repetition in DNA segments
  • (B) The ratio of purines and pyrimidines present in DNA
  • (C) The amount of DNA found in samples of blood, saliva and skin
  • (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 uses VNTRs (Variable Number Tandem Repeats) in satellite DNA.

Step 2: These are highly repetitive, non-coding regions with variable repeat lengths.

Step 3: Produce unique banding patterns after restriction digestion and gel electrophoresis. Quick Tip: Alec Jeffreys (1984) → father of DNA fingerprinting.


Question 2:

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 causes malignant tertian malaria.

Step 2: High parasitemia, cytoadherence → cerebral malaria.

Step 3: Highest mortality rate. Quick Tip: Falciparum → ring forms, banana gametocytes.


Question 3:

The type of antibodies produced during the allergic reaction:

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

Step 1: Type I hypersensitivity → mediated by IgE.

Step 2: Binds to mast cells → histamine release on allergen exposure.

Step 3: Causes asthma, hay fever, anaphylaxis. Quick Tip: Desensitization → allergen immunotherapy.


Question 4:

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

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

Step 1: Anabolic steroids → synthetic androgens.

Step 2: In females → virilization: hirsutism, voice deepening, clitoral enlargement.

Step 3: Liver damage possible but not gender-specific. Quick Tip: Nandrolone, stanozolol → common anabolic steroids.


Question 5:

Which one of the following is an opiate narcotics?

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

Step 1: Opiates → derived from opium (Papaver somniferum).

Step 2: Morphine → principal alkaloid → analgesic, sedative.

Step 3: LSD (hallucinogen), amphetamines (stimulant), barbiturates (sedative). Quick Tip: Codeine, heroin → semi-synthetic opiates.


Question 6:

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

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

Step 1: Propionibacterium shermanii used in Swiss cheese.

Step 2: Ferments lactate → propionic acid + CO₂.

Step 3: CO₂ trapped → forms eyes (holes). Quick Tip: Emmental cheese → large eyes.


Question 7:

Which vitamin is increased by ‘LAB’ in curd?

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

Step 1: Lactic Acid Bacteria (LAB) synthesize B₁₂ during milk fermentation.

Step 2: Curd → rich source of cobalamin.

Step 3: Important for vegetarians. Quick Tip: Yogurt → probiotic + B₁₂.


Question 8:

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

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

Step 1: Oily stains → triglycerides.

Step 2: Lipase hydrolyzes fats → glycerol + fatty acids.

Step 3: Used in biological detergents. Quick Tip: Tide, Ariel → contain lipase, protease.


Question 9:

DNA replicates semiconservatively was first shown in:

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

Step 1: Meselson-Stahl experiment (1958).

Step 2: Used E. coli with ¹⁵N (heavy) → switched to ¹⁴N.

Step 3: Density gradient → hybrid band → semiconservative. Quick Tip: Conservative, dispersive → ruled out.


Question 10:

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

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

Step 1: Griffith’s transformation experiment.

Step 2: S strain (virulent) + heat-killed S + R strain → virulent mice.

Step 3: Transforming principle = DNA (later proved). Quick Tip: Avery, MacLeod, McCarty (1944) → DNA is genetic material.


Question 11:

The number of codons effective in coding twenty amino acids:

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

Step 1: Total codons = 4³ = 64.

Step 2: 3 stop codons (UAA, UAG, UGA).

Step 3: 61 codons code for 20 amino acids (degenerate code). Quick Tip: AUG → start + methionine.


Question 12:

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

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

Step 1: Antibiotic resistance gene (e.g., amp\^R) is a selectable marker in vectors.

Step 2: Only transformed cells (that uptake plasmid) survive on antibiotic medium.

Step 3: Non-transformed cells die → selection of transformed cells. Quick Tip: Blue-white screening → insertional inactivation.


Question 13:

Silencing of specific mRNA in RNAi is by

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

Step 1: RNA interference (RNAi) → gene silencing.

Step 2: dsRNA is processed by Dicer → siRNA.

Step 3: siRNA + RISC → degrades target mRNA. Quick Tip: RNAi → natural defense against viruses.


Question 14:

Cry-IAC effectively controls

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

Step 1: Cry-IAC → Bt toxin from Bacillus thuringiensis.

Step 2: Toxic to lepidopteran larvae (bollworm).

Step 3: Used in Bt cotton. Quick Tip: Cry-IAc, Cry-IAb → bollworm; Cry-IIAb → corn borer.


Question 15:

ADA deficiency can be cured by

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

Step 1: ADA deficiency → SCID (no T-cells).

Step 2: Bone marrow contains stem cells → produce ADA enzyme.

Step 3: Bone marrow transplant restores immunity. Quick Tip: First gene therapy (1990) → ADA deficiency.


Question 16:

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

Step 1: Net change = (Births + Immigration) – (Deaths + Emigration)

Step 2: = (25 + 2) – (24 + 3) = 27 – 27 = 0

Step 3: Natural increase = 1; Net migration = –1 → 0 Quick Tip: ΔP = (B – D) + (I – E)


Question 17:

The term “Molecular Scissors” refers to

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

Step 1: Restriction enzymes cut DNA at specific sites.

Step 2: Called molecular scissors in genetic engineering.

Step 3: e.g., EcoRI, HindIII. Quick Tip: Sticky ends → cohesive ends.


Question 18:

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 reads same forward and backward on complementary strands.

Step 2: 5'-GAATTC-3' = 3'-CTTAAG-5' reversed → palindrome.

Step 3: Recognition site for EcoRI. Quick Tip: Palindrome → symmetry in DNA.


Question 19:

Gel electrophoresis is used for

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

Step 1: Agarose gel + electric field.

Step 2: DNA (–ve charge) moves to anode.

Step 3: Smaller fragments move faster → size-based separation. Quick Tip: Ethidium bromide → UV visualization.


Question 20:

The amount of Photosynthetically active radiation captured by plants is

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

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

Step 2: Plants capture only 2–10% of incident solar energy.

Step 3: Most lost as heat, reflection, or unused wavelengths. Quick Tip: Only ~1% of solar energy reaches Earth; plants use fraction of PAR.


Question 21:

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 → species-area relationship (Humboldt).

Step 2: S = species richness, A = area.

Step 3: Log form: log S = log c + z log A → straight line. Quick Tip: z = 0.1–0.2 (mainland), 0.2–0.3 (islands).


Question 22:

Cuscuta is an example of

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

Step 1: Cuscuta (dodder) → parasitic plant.

Step 2: Grows externally on host → haustoria penetrate phloem.

Step 3: Ectoparasite (outside host body). Quick Tip: Cuscuta → yellow twining stem, no chlorophyll.


Question 23:

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

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

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

Step 2: Enter alveoli, bloodstream.

Step 3: Cause asthma, COPD, heart disease → highest risk. Quick Tip: PM10 → upper airways; PM2.5 → deep lungs.


Question 24:

Maintenance of constant internal environment is called as

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

Step 1: Homeostasis → stable internal conditions.

Step 2: Includes temperature, pH, ion balance.

Step 3: Osmoregulation, thermoregulation → components of homeostasis. Quick Tip: Walter Cannon coined "homeostasis".


Question 25:

The animals which are active during day time:

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

Step 1: Diurnal → active in daylight (e.g., humans, birds).

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

Step 3: Vesporal, auroral → not standard. Quick Tip: Diurnal animals have good color vision.


Question 26:

Which of the following statement is incorrect related to biomes?

  • (A) High temperature and minimum rainfall help to form grasslands.
  • (B) Variation in temperature and mean precipitation accounts for the major biomes.
  • (C) Low temperature and less rainfall is a characteristic of Tundra 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, <25 cm/year rainfall.

Step 2: Option (D) says more rainfall + low temperature → incorrect.

Step 3:
- Tundra: cold, low rain
- Grasslands: moderate rain, warm
- Biomes depend on temp + precipitation Quick Tip: Tropical rainforests → high rain + high temp.


Question 27:

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

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

Step 1: Arthropods show diverse respiratory organs:

- Gills (crustaceans)

- Book gills (horseshoe crabs)

- Book lungs (scorpions, spiders)

- Trachea (insects)

Step 2: No other phylum has all these.

Molluscs → ctenidia; Annelids → skin; Sponges → no organs. Quick Tip: Tracheae → air-filled tubes in insects.


Question 28:

Which of the following plant is used to extract Colchicine?

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

Step 1: Colchicine → alkaloid from Colchicum autumnale (autumn crocus).

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

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


Question 29:

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

Step 1: Setae → locomotion in earthworm.

Step 2: Absent in:

- 1st segment (peristomium)

- Last segment (anal)

- Clitellum (14–16 segments, glandular)

Step 3: Option (A) correct. Quick Tip: Clitellum → cocoon secretion.


Question 30:

Cell theory was formulated by

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

Step 1: M.J. Schleiden (1838) → plants made of cells.

Step 2: T. Schwann (1839) → animals made of cells.

Step 3: Cell theory (1839): all organisms cellular.

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


Question 31:

The type of Polysaccharide present in a cotton fibre

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

Step 1: Cotton → cell wall of seed hairs.

Step 2: Plant cell wall → cellulose (β-1,4-glucose).

Step 3: Cotton ≈ 95% cellulose → textile fiber.

Starch (storage), glycogen (animal), insulin (hormone). Quick Tip: Humans lack cellulase → can't digest cellulose.


Question 32:

Enzyme involved in crossing over

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

Step 1: Crossing over → pachytene of meiosis.

Step 2: Recombinase (e.g., DMC1, RAD51) → strand exchange.

Step 3: Forms chiasmata → recombination.

Endonuclease → cuts DNA; ligase → seals. Quick Tip: Synaptonemal complex → holds homologues.


Question 33:

Kranz anatomy can be seen in

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

Step 1: Kranz anatomy → C4 plants.

Step 2: Bundle sheath cells with chloroplasts.

Step 3: Maize → C4; Pea, tomato, potato → C3. Quick Tip: C4 → Hatch-Slack pathway → high efficiency.


Question 34:

Respiratory quotient of glucose is

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

Step 1: RQ = CO₂ / O₂

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 respiration.


Question 35:

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

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

Step 1: Epiglottis → cartilage flap over trachea.

Step 2: Closes during swallowing → prevents food entry.

Step 3: Improper closure → food in windpipe → cough reflex. Quick Tip: Choking → Heimlich maneuver.


Question 36:

Binomial nomenclature is introduced by

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

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

Step 2: Binomial: Genus + species.

Step 3: Universal system → e.g., Homo sapiens. Quick Tip: ICBN → plants; ICZN → animals.


Question 37:

Bovine spongiform encephalopathy is caused by

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

Step 1: BSE (Mad Cow) → neurodegenerative.

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

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


Question 38:

Phycoerythrin and Floridean starch is found in

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

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

Step 2: Store floridean starch (α-1,4-glucan).

Step 3: Deep water → absorb blue light. Quick Tip: Agar → Gelidium, Gracilaria.


Question 39:

Menstrual cycle is exhibited by:

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

Step 1: Menstrual cycle → shedding of uterine lining (bleeding).

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

Step 3: Cow, tiger, rat → estrous cycle (no menstruation). Quick Tip: Human cycle ≈ 28 days; apes similar.


Question 40:

An example of dioecious plant:

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

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

Step 2: Papaya → male and female trees.

Step 3: Mango, cucurbita, coconut → monoecious. Quick Tip: Date palm → dioecious.


Question 41:

Stalk of the Stamen is:

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

Step 1: Stamen = anther + filament.

Step 2: Filament → stalk supporting anther.

Step 3: Peduncle (inflorescence), pedicel (flower), petiole (leaf). Quick Tip: Androecium → stamens.


Question 42:

The ovule of angiosperm is technically known as:

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

Step 1: Ovule = integumented megasporangium.

Step 2: Nucellus = megasporangium tissue.

Step 3: Produces megaspore from MMC. Quick Tip: Ovule → seed.


Question 43:

Typical mature embryosac of angiosperm is

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

Step 1: Polygonum type → 7 cells: egg, 2 synergids, 3 antipodals, 1 central cell.

Step 2: Central cell → 2 polar nuclei → 8 nuclei total.

Step 3: 8-nucleated, 7-celled. Quick Tip: Female gametophyte = embryosac.


Question 44:

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

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

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

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

Step 3: Oldest viable seed. Quick Tip: "Judean date palm" → extinct variety revived.


Question 45:

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

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

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

Step 2: Scrotum → external sac.

Step 3: Pampiniform plexus → heat exchange. Quick Tip: Cryptorchidism → infertility.


Question 46:

Identify the odd one from the following:

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

Step 1: Infundibulum, fimbriae, isthmus → fallopian tube.

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

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


Question 47:

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

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

Step 1: Quickening → 5th month (mother feels movement).

Step 2: Lanugo (head hair) → 5th month.

Step 3: Heartbeat → 1st month; limbs → 8th week. Quick Tip: Gestation = 280 days ≈ 40 weeks.


Question 48:

The most abundant type of WBC cells

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

Step 1: Neutrophils → 60–65% of WBCs.

Step 2: Phagocytic, first responders.

Step 3: Multilobed nucleus. Quick Tip: WBC count: 4,000–11,000/μL.


Question 49:

Filtration of blood during urine formation takes place in

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

Step 1: Ultrafiltration → glomerulus.

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

Step 3: Forms glomerular filtrate. Quick Tip: GFR ≈ 125 mL/min.


Question 50:

Corpus Callosum connects the

  • (A) Cerebrum and cerebellum
  • (B) Two lobes of cerebellum
  • (C) Spinal cord with the brain
  • (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: Enables inter-hemispheric communication. Quick Tip: Split-brain → corpus callosum sectioned.


Question 51:

Amniocentesis is a process to:

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

Step 1: Amniocentesis → 15–20 weeks; amniotic fluid extracted.

Step 2: Fetal cells cultured → karyotyping.

Step 3: Detects sex (XX/XY), chromosomal disorders.

Banned for sex determination in India (PCPNDT Act). Quick Tip: Amniocentesis → diagnostic, not therapeutic.


Question 52:

The first human like being is

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

Step 1: Homo habilis (2.4–1.4 mya) → earliest Homo.

Step 2: Brain ~650 cm³, Oldowan tools.

Step 3: "Handy man" → first human-like.

H. erectus → later (1.8 mya). Quick Tip: Evolution: Australopithecus → H. habilis → H. erectus → H. sapiens.


Question 53:

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: (D) Female Heterogamety
View Solution

Step 1: XO-type (grasshopper): female XO → heterogametic.

Step 2: XY-type: male XY → heterogametic.

Step 3: Question focuses on XO → female heterogamety. Quick Tip: ZW (birds) → female heterogamety.


Question 54:

Example for Non-Mendelian disorder:

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

Step 1: Mendelian → single gene (thalassemia, haemophilia, CF).

Step 2: Down’s syndrome → trisomy 21 (nondisjunction).

Step 3: Chromosomal → non-Mendelian. Quick Tip: Non-Mendelian: chromosomal, polygenic, mitochondrial.


Question 55:

Gynecomastia is a symptom of

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

Step 1: Klinefelter’s (47, XXY) → male.

Step 2: Extra X → low testosterone, high estrogen → gynecomastia (breast development).

Step 3: Turner’s → female (XO). Quick Tip: Klinefelter: tall, sterile, gynecomastia.


Question 56:

The affected male in the pedigree chart is symbolized by:

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

Step 1: Pedigree symbols:

- Square → male

- Circle → female

- Filled → affected

Step 2: Affected male → filled square (■). Quick Tip: Carrier → half-filled.


Question 57:

The given diagram represents:


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

Step 1: Nucleosome → DNA wrapped around histone octamer.

Step 2: H1 histone → linker between nucleosomes.

Step 3: Diagram shows DNA + H1 → nucleosome. Quick Tip: Beads-on-string → 11 nm fiber.


Question 58:

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

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

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

Step 2: FSH → anterior pituitary.

Step 3: Not placental. Quick Tip: hCG → pregnancy test.


Question 59:

Which of the following is correctly matched?

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

Step 1: Bulbil → vegetative propagule in Agave.

Step 2: Gemmules → sponge; conidia → fungi.

Step 3: Correct → Bulbil – Agave. Quick Tip: Runner → grass; offset → Eichhornia.


Question 60:

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

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

Step 1: ART (Assisted Reproductive Technology) → IVF, IUI, ZIFT, GIFT.

Step 2: Treats infertility.

Step 3: MTP → abortion; RTI → infection; RCH → program. Quick Tip: First IVF baby: Louise Brown (1978).

KCET Questions

  • 1.
    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

    • 2.
      Which of the following statements are correct with reference to prokaryotic genome?

        • Monocistronic structural genes
        • Introns absent in structural genes
        • Transcription and translation are coupled processes
        • Primary transcript undergoes splicing

      • 3.
        In which of the following, HIV replicates and produces its progeny viruses?

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

        • 4.
          When single gene exhibits multiple phenotypic expression, the phenomenon is called ____

            • Pleiotropy
            • Co-dominance
            • Polygenic inheritance
            • Incomplete dominance

          • 5.

            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

            • 6.
              When a change in the gene frequency of population occurs by chance, it is called

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

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