KCET 2022 Biology A1 Question Paper With Answer Key And Solutions PDF

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Shivam Yadav

Updated on - Nov 16, 2025

KCET 2022 Biology A-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 (A-1) Question Paper with Answer Key

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

KCET 2022 Biology A1 Question Paper With Answer Key And Solutions PDF


Question 1:

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: Urine formation begins with filtration of blood plasma.
Step 2: This occurs in the glomerulus, a tuft of capillaries in the nephron.
Step 3: High pressure forces water, ions, urea, glucose into Bowman's capsule → glomerular filtrate.
Step 4: PCT, DCT, collecting duct handle reabsorption/secretion, not filtration.
Hence, filtration takes place only in the glomerulus. Quick Tip: Urine formation steps: 1. Glomerular filtration (Glomerulus) 2. Tubular reabsorption (PCT, Loop of Henle, DCT) 3. Tubular secretion (DCT, Collecting duct)


Question 2:

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: The brain has left and right cerebral hemispheres.
Step 2: Corpus callosum is a broad band of ~200 million nerve fibers.
Step 3: It connects left ↔ right cerebral hemispheres for coordination.
Step 4: Cerebellum is connected via cerebellar peduncles; spinal cord via medulla oblongata.
Thus, only (D) is correct. Quick Tip: - Corpus Callosum → Lateral integration (left-right brain) - Cerebellar Peduncles → Cerebrum ↔ Cerebellum - Brainstem → Brain ↔ Spinal cord


Question 3:

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 (menstruation).
Step 2: Occurs in primates (humans, apes, monkeys).
Step 3: Other mammals (cow, tiger, rat) have estrous cycle → endometrium reabsorbed, no bleeding.
Hence, only apes exhibit menstrual cycle. Quick Tip: - Menstrual cycle → Primates (visible bleeding) - Estrous cycle → Most mammals (no bleeding)


Question 4:

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, female, hermaphrodite plants → dioecious.
- Mango, Cucurbita, Coconut: Both flowers on same plant → monoecious.
Hence, only papaya is dioecious. Quick Tip: - Dioecious: ♂ and ♀ on different plants - Monoecious: ♂ and ♀ on same plant


Question 5:

Stalk of the Stamen is:

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

Step 1: Stamen = anther (pollen) + stalk.
Step 2: Stalk is called filament.
Step 3:
- Petiole → leaf stalk
- Peduncle → inflorescence axis
- Pedicel → flower stalk
Only filament is part of stamen. Quick Tip: - Stamen: Anther + Filament - Flower stalk: Pedicel - Inflorescence stalk: Peduncle


Question 6:

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 is the female reproductive unit in angiosperms.
Step 2: Contains nucellus (tissue), integuments, and megaspore mother cell (MMC).
Step 3: MMC → meiosis → megaspore → embryo sac.
Step 4: Entire ovule is homologous to megasporangium in lower plants.
Hence, ovule = megasporangium. Quick Tip: - Ovule = Megasporangium + Integuments - MMC → inside nucellus → forms megaspore


Question 7:

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 (common) embryo sac:
- Megaspore → 3 mitotic divisions → 8 nuclei.
Step 2: Arrangement:
- 3 antipodals
- 2 synergids
- 1 egg cell
- 1 central cell (2 polar nuclei → binucleate)
Step 3: Total: 7 cells, 8 nuclei.
Hence, (D) is correct. Quick Tip: Embryo sac (Polygonum type): - 7 cells: 3 antipodal + 2 synergid + 1 egg + 1 central (2 nuclei) - 8 nuclei total


Question 8:

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

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

Step 1: Excavation at Masada (King Herod’s fortress), near Dead Sea.
Step 2: Date palm seed (~2000 years old) found.
Step 3: Germinated in 2005 → named “Methuselah”.
Step 4: Scientific name: Phoenix dactylifera.
Hence, (D) is correct. Quick Tip: Oldest viable seed: Phoenix dactylifera (Date palm) → ~2000 years Location: Masada, Israel


Question 9:

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

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

Step 1: Spermatogenesis needs 2–3°C below body temperature (37°C).
Step 2: Scrotum keeps testes outside abdomen → cooler environment.
Step 3:
- Sperm storage → epididymis
- Hormone secretion → Leydig cells
- Release → vas deferens
Only temperature regulation is the primary reason. Quick Tip: - Scrotum → Thermoregulation for sperm production - Pampiniform plexus → Cools arterial blood


Question 10:

Identify the odd one from the following:

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

Step 1: Parts of fallopian tube:
- Infundibulum → funnel end
- Fimbriae → finger-like projections
- Isthmus → narrow part near uterus
Step 2: Labia minora → part of external genitalia (vulva).
Step 3: (A), (B), (C) → internal; (D) → external.
Hence, labia minora is the odd one. Quick Tip: Fallopian tube parts: Infundibulum → Fimbriae → Ampulla → Isthmus Labia minora → Vulva (external)


Question 11:

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

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

Step 1: Fetal development timeline:
- 4th month: Limbs fully formed, heartbeat detectable.
- 5th month: Quickening (first fetal movements felt by mother), lanugo (fine hair) appears on head and body.
- 8th month: Subcutaneous fat increases, hair thickens.
Hence, first movements + hair on head → 5th month. Quick Tip: - Quickening: 18–20 weeks (~5th month) - Lanugo: Appears at ~5th month, sheds by 7–8th month


Question 12:

The most abundant type of WBC cells

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

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


Question 13:

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 bud → Agave (correct).
- Conidia: Asexual spores → Penicillium, not Hydra.
- Spores: Sexual/asexual → Ferns, not sponges.
- Gemmules: Internal buds → Sponges, not Amoeba.
Hence, only (A) is correct. Quick Tip: - Bulbil: Agave, Garlic - Gemmule: Sponges - Conidia: Fungi - Budding: Hydra


Question 14:

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:
- RCH: Reproductive & Child Health (program)
- MTP: Medical Termination of Pregnancy
- RTI: Reproductive Tract Infection
- ART: Assisted Reproductive Technology (IVF, IUI, ZIFT, GIFT)
Step 2: To treat infertility → ART.
Hence, (D) is correct. Quick Tip: - Infertility treatment: ART (IVF, ICSI, etc.) - Contraception: MTP, IUD - Health programs: RCH


Question 15:

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: Sampling of amniotic fluid (16–18 weeks).
Step 2: Contains fetal cells → karyotyping → sex determination, genetic disorders (Down’s, thalassemia).
Step 3:
- Cell culture → secondary use
- Heart/brain disease → ultrasound
Hence, primary use → sex & genetic diagnosis. Quick Tip: - Amniocentesis: 16–18 weeks → Fetal cells → Karyotype - Chorionic Villus Sampling (CVS): 10–12 weeks - Banned in India for sex determination


Question 16:

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: Human evolution:
- Australopithecus → Homo habilis (2.4–1.4 mya)
- Homo habilis: "Handy man", used stone tools, first in Homo genus.
- Homo erectus: Later, used fire.
- Homo sapiens: Modern humans.
Hence, first human-like → Homo habilis. Quick Tip: Human evolution order: H. habilis → H. erectus → H. sapiens


Question 17:

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

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

Step 1:
- XY system (humans): Male = XY (heterogametic), Female = XX (homogametic)
- XO system (grasshopper): Male = XO (heterogametic), Female = XX
Step 2: In both, male produces two types of gametes → male heterogamety.
Hence, (B) is correct. Quick Tip: - Male heterogamety: XY, XO, XY (male produces X & Y or X & O) - Female heterogamety: ZW (birds)


Question 18:

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 disorders: Gene mutations → single gene (thalassemia, haemophilia, cystic fibrosis).
Step 2: Non-Mendelian: Chromosomal abnormalities → Down’s syndrome (trisomy 21).
Hence, (D) is non-Mendelian. Quick Tip: - Mendelian: Sickle cell, Haemophilia - Non-Mendelian: Down’s, Turner’s, Klinefelter’s (aneuploidy)


Question 19:

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 syndrome: XXY male.
Step 2: Symptoms:
- Tall, underdeveloped testes
- Gynecomastia (breast development due to estrogen imbalance)
Step 3: Turner’s (XO) → female, no gynecomastia.
Hence, (D) is correct. Quick Tip: - Klinefelter’s (XXY): Male + gynecomastia, sterile - Turner’s (XO): Female, short, infertile


Question 20:

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
- Shaded → Affected
- Unshaded → Normal
Step 2: Affected male → shaded square.
Hence, (A) is correct. Quick Tip: Pedigree: - Male: □ (normal), ■ (affected) - Female: ○ (normal), ● (affected)


Question 21:

The given diagram represents:

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

Step 1: The diagram typically shows DNA wrapped around histone octamer with a linker.
Step 2: This structure is "beads on a string" → nucleosome.
Step 3:
- Chromosome: Condensed chromatin
- Mesosome: Bacterial membrane infolding
- Ribosome: Protein synthesis organelle
Hence, the diagram is nucleosome. Quick Tip: - Nucleosome = DNA + Histone octamer (H2A, H2B, H3, H4) × 2 - "Beads on a string" → 11 nm fiber


Question 22:

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 (maintains corpus luteum)
- hPL (human placental lactogen)
- Progestogens (progesterone)
- Relaxin (late pregnancy)
Step 2: FSH → secreted by anterior pituitary, not placenta.
Hence, (C) is correct. Quick Tip: - Placental hormones: hCG, hPL, Progesterone, Estrogen, Relaxin - Pituitary: FSH, LH, TSH, ACTH


Question 23:

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

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

Step 1: Griffith used S (smooth) and R (rough) strains of Streptococcus pneumoniae.
Step 2: Heat-killed S + live R → mice died → transformation.
Step 3: Led to discovery of DNA as genetic material.
Hence, (C) is correct. Quick Tip: - Griffith (1928): S. pneumoniae (R + heat-killed S → virulent) - Avery, MacLeod, McCarty (1944): DNA is transforming principle


Question 24:

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 = \(4^3 = 64\)
Step 2:
- 3 stop codons: UAA, UAG, UGA → no amino acid
- 61 sense codons → code for 20 amino acids (degenerate code)
Hence, 61 codons code for amino acids. Quick Tip: - 64 codons total - 61 code amino acids - 3 stop codons - AUG: Start + Methionine


Question 25:

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).
Step 2: These are satellite DNA → highly repetitive, non-coding.
Step 3: Number of repeats varies between individuals → unique pattern.
Step 4: Purine/pyrimidine ratio, DNA amount, fingerprint ridges → not used.
Hence, (A) is correct. Quick Tip: - DNA Fingerprinting: VNTRs (minisatellites) - Alec Jeffreys (1984) - Probe + Southern blot + Autoradiography


Question 26:

Identify the most infectious and fatal type of malarial parasite:

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

Step 1:
- P. falciparum: Cerebral malaria, high parasitemia, most fatal.
- P. vivax: Benign tertian, relapses
- P. malariae: Quartan, chronic
- P. ovale: Mild, rare
Step 2: Causes blackwater fever, multi-organ failure.
Hence, most infectious & fatal → P. falciparum. Quick Tip: - P. falciparum: Malignant tertian, cerebral, fatal - P. vivax: Benign tertian, Duffy antigen


Question 27:

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: Allergen → binds IgE on mast cells/basophils → histamine release.
Step 3:
- IgG, IgM: Immunity
- IgA: Mucosal
Hence, allergy → IgE. Quick Tip: - Allergy: IgE → Mast cell degranulation → Histamine - Anaphylaxis: Severe IgE reaction


Question 28:

One of the side-effect 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 → androgenic effects in females.
Step 2: Causes:
- Deep voice
- Facial hair
- Enlarged clitoris → masculinisation
Step 3: Liver damage possible, but masculinisation is classic in females.
Hence, (C) is correct. Quick Tip: - Anabolic steroids in females: Virilization (masculinisation) - In males: Gynecomastia, testicular atrophy


Question 29:

Which one of the following is a opiate narcotics?

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

Step 1: Opiates → derived from opium poppy (Papaver somniferum).
Step 2:
- Morphine, Heroin, Codeine → opiates
- Amphetamines: Stimulants
- Barbiturates: Sedatives
- LSD: Hallucinogen
Hence, morphine is opiate narcotic. Quick Tip: - Opiates: Morphine, Heroin (CNS depressant, analgesic) - Cannabinoids: Marijuana - Cocaine: Stimulant


Question 30:

The large holes in 'Swiss - Cheese' are made by a

  • (A) Bacterium producing a large amount of \(CO_{2}\)
  • (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 \(CO_{2}\)
View Solution

Step 1: Swiss cheese → made with Propionibacterium shermanii.
Step 2: Ferments lactic acid → propionic acid + CO₂.
Step 3: CO₂ forms large bubbles → holes (eyes) in cheese.
Step 4: Not methane, not fungus, not machine.
Hence, (A) is correct. Quick Tip: - Swiss cheese holes: CO₂ by Propionibacterium - Roquefort: Penicillium roqueforti (blue veins)


Question 31:

Which vitamin is increased by 'LAB' in curd ?

  • (A) Vitamin \(B_{12}\)
  • (B) Vitamin C
  • (C) Vitamin E
  • (D) Vitamin B
Correct Answer: (A) Vitamin \(B_{12}\)
View Solution

Step 1: LAB → Lactic Acid Bacteria (e.g., Lactobacillus) used in curd formation.
Step 2: During fermentation, LAB synthesizes Vitamin \(B_{12}\) (cobalamin).
Step 3:
- Vitamin C → not produced
- Vitamin E → fat-soluble, not microbial
- Vitamin B → vague; \(B_{12}\) is specific
Hence, Vitamin \(B_{12}\) increases in curd. Quick Tip: - Curd (Yogurt): LAB → ↑ Vitamin \(B_{12}\), folate - Sauerkraut, Kimchi: Also ↑ \(B_{12}\)


Question 32:

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 → lipids/fats.
Step 2: Lipase → hydrolyzes triglycerides → glycerol + fatty acids.
Step 3:
- Amylase: Starch
- Renin: Milk clotting (not in laundry)
- Protease: Proteins
Hence, lipase removes oily stains. Quick Tip: - Detergent enzymes: – Lipase → Oil – Protease → Protein (blood, egg) – Amylase → Starch


Question 33:

DNA replicates semiconservatively was first shown in :

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

Step 1: Meselson & Stahl (1958) used E. coli.
Step 2: Grew in \(^{15}\)NH₄Cl → heavy DNA → transferred to \(^{14}\)N medium.
Step 3: Density gradient centrifugation → hybrid band after one generation → semiconservative replication.
Hence, first shown in E. coli. Quick Tip: - Meselson-Stahl: \(^{15}\)N → \(^{14}\)N → Hybrid DNA - Conservative: HH + LL - Dispersive: All intermediate


Question 34:

What does the sample of given base sequence represent? \(5^{\prime}\)-GAATTC-\(3^{\prime}\) \(3^{\prime}\)-CTTAAG-\(5^{\prime}\)

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

Step 1: Sequence: \(5^{\prime}\)-GAATTC-\(3^{\prime}\) \(3^{\prime}\)-CTTAAG-\(5^{\prime}\)
Step 2: Reads same on complementary strand in opposite direction → palindrome.
Step 3: Recognized by EcoRI (restriction enzyme).
Step 4: Not replication, codon, or mutation.
Hence, palindromic sequence. Quick Tip: - Palindrome: 5'-GAATTC-3' = 3'-CTTAAG-5' (reverse complement) - EcoRI: Cuts between G and A


Question 35:

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 electrophoresis → DNA moves toward anode (negative charge).
Step 2: Smaller fragments move faster → separate by size.
Step 3:
- Ligation → recombinant DNA
- Restriction enzyme → cutting
- PCR/Extraction → isolation
Hence, separation by size. Quick Tip: - Gel electrophoresis: – Small DNA → travels far – Large DNA → stays near well


Question 36:

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: Vector has ampicillin resistance gene (ampᵣ).
Step 2: Only transformed cells (took up plasmid) grow on ampicillin plate.
Step 3: Non-recombinants also grow → use insertional inactivation (e.g., lacZ).
Step 4: First selection → transformed cells.
Hence, (D) is correct. Quick Tip: - Selectable marker: ampᵣ, tetᵣ → selects transformants - Blue-white screening: X-gal → recombinants (white)


Question 37:

Silencing of specific mRNA in RNAi is by

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

Step 1: RNAi → dsRNA introduced.
Step 2: Dicer cuts dsRNA → siRNA (21–23 nt).
Step 3: RISC uses siRNA → degrades complementary mRNA.
Step 4: ssDNA, ssRNA, dsDNA → not used.
Hence, dsRNA triggers RNAi. Quick Tip: - RNAi pathway: dsRNA → Dicer → siRNA → RISC → mRNA cleavage


Question 38:

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: Specific to lepidopteran larvae (butterflies/moths).
Step 3: Cotton bollworm (Helicoverpa armigera) → lepidopteran → controlled by Cry-IAC.
Step 4:
- Cry-IIAB: Nematodes
- Corn borer: Cry-IA (different)
Hence, cotton bollworms. Quick Tip: - Cry-I: Lepidoptera (bollworm) - Cry-II: Diptera + Lepidoptera - Cry-III: Coleoptera (beetle)


Question 39:

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 → Adenosine Deaminase → needed in T-lymphocytes.
Step 2: Deficiency → SCID (no immunity).
Step 3: Bone marrow → source of hematopoietic stem cells → produce healthy lymphocytes.
Step 4: First gene therapy → ADA gene into lymphocytes.
Hence, bone-marrow transplantation. Quick Tip: - ADA deficiency: SCID - Treatment: Bone marrow transplant or Gene therapy


Question 40:

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: 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 41:

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: Molecular scissors → cut DNA at specific sites.
Step 2: Restriction enzymes (e.g., EcoRI, HindIII) recognize palindromic sequences and cleave DNA.
Step 3: Used in rDNA technology for cloning.
Step 4: Polymerases → synthesize DNA/RNA, not cut.
Hence, restriction enzyme. Quick Tip: - Restriction enzyme → "Molecular scissors" - Ligase → "Molecular glue"


Question 42:

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


Question 43:

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 characteristics 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 + very low rainfall (<25 cm/year).
Step 2: (D) says more rainfall + low temperature → incorrect.
Step 3:
- (A): Grasslands → moderate rain, warm
- (B): Temperature + precipitation → biome distribution
- (C): Tundra → cold, low precipitation
Hence, (D) is incorrect. Quick Tip: - Desert: Hot, <25 cm rain - Tundra: Cold, <25 cm rain - Rainforest: >200 cm rain


Question 44:

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 (400–700 nm) → used in photosynthesis.
Step 2: Of total solar radiation:
- ~50% is PAR
- Plants capture 2–10% of PAR for gross primary productivity.
Step 3: Most energy lost as heat, reflection, transmission.
Hence, 2–10%. Quick Tip: - PAR: 50% of solar radiation - Plants use: 2–10% of PAR - NPP: ~1–2% of solar energy


Question 45:

The given graph represents \(S=CA^{Z}\)

  • (A) Population growth
  • (B) Enzyme activity
  • (C) Species area relationship
  • (D) Growth of organisms
Correct Answer: (C) 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 46:

Cuscuta is an example of

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

Step 1: Cuscuta → yellow, leafless vine.
Step 2: Attaches to host stem externally → penetrates via haustoria → absorbs water/nutrients.
Step 3:
- Ectoparasite: Lives outside host (e.g., lice, Cuscuta)
- Endoparasite: Lives inside (e.g., tapeworm)
Hence, ectoparasitism. Quick Tip: - Cuscuta: Stem holoparasite → ectoparasite - Rafflesia: Root endoparasite


Question 47:

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


Question 48:

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 → maintaining stable internal conditions (pH, temperature, ions).
Step 2:
- Thermoregulation: Temperature
- Osmoregulation: Water/salt
- Metastasis: Cancer spread
Step 3: Homeostasis includes all.
Hence, homeostasis. Quick Tip: - Homeostasis → "Same standing" (Claude Bernard) - Feedback loops: Negative (most), Positive (rare)


Question 49:

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 Disease.
Step 2: Caused by prions → misfolded PrP proteins.
Step 3: No nucleic acid → resistant to heat, radiation.
Step 4: Human form → vCJD.
Hence, prions. Quick Tip: - Prions: Proteinaceous infectious particles - Diseases: BSE, CJD, Kuru, Scrapie


Question 50:

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 (Rhodophyta):
- Pigment: Phycoerythrin (red)
- Storage: Floridean starch (similar to glycogen)
Step 2:
- Brown algae: Fucoxanthin, laminarin
- BGA: Phycocyanin, glycogen
- Green algae: Chlorophyll, starch
Hence, red algae. Quick Tip: - Red algae: – Phycoerythrin (red) – Floridean starch – Agar, carrageenan


Question 51:

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 adaptations:
- Gills: Aquatic (crabs, prawns)
- Book gills: Limulus (horseshoe crab)
- Book lungs: Scorpions, spiders
- Trachea: Insects
Step 2:
- Sponges → no organs
- Annelids → moist skin
- Molluscs → gills, ctenidia
Hence, only arthropods have all four. Quick Tip: - Arthropod respiration: – Aquatic: Gills – Terrestrial: Trachea, book lungs, book gills


Question 52:

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 used in polyploidy induction.
Step 2: Extracted from corms of Colchicum autumnale (autumn crocus).
Step 3: Inhibits spindle formation → arrests mitosis at metaphase.
Hence, Colchicum. Quick Tip: - Colchicine source: Colchicum autumnale - Use: Polyploidy, gout treatment


Question 53:

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: Present in all segments except:
- 1st segment (peristomium)
- Last segment (anus)
- Clitellar segments (14–16) → no setae
Step 3: Clitellum → gland for cocoon.
Hence, first, last, and clitellum. Quick Tip: - Earthworm setae: – Absent in 1st, last, 14–16 (clitellum) – S-shaped, 4 pairs/segment


Question 54:

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


Question 55:

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 fiber → seed hair of Gossypium.
Step 2: Composed of >90% cellulose (β-1,4 glucose polymer).
Step 3:
- Starch → storage (plants)
- Glycogen → animal storage
- Insulin → hormone
Hence, cellulose. Quick Tip: - Cotton: Cellulose (95%) - Jute: Lignin + cellulose - Silk: Protein


Question 56:

Enzyme involved in crossing over

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

Step 1: Crossing over → exchange of genetic material in pachytene.
Step 2: Recombinase (e.g., Spo11, RAD51) → forms Holliday junction.
Step 3:
- Endonuclease → cuts DNA
- Ligase → joins
- Polymerase → fills gaps
But recombinase initiates recombination.
Hence, recombinase. Quick Tip: - Crossing over: – Synaptonemal complex → Recombinase → Chiasma


Question 57:

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: Features:
- Bundle sheath cells with chloroplasts
- Mesophyll → radial chloroplasts
Step 3: Maize → C4 plant.
Pea, tomato, potato → C3.
Hence, maize. Quick Tip: - Kranz anatomy: – C4 plants (maize, sugarcane) – Bundle sheath → CO₂ fixation (PEP carboxylase)


Question 58:

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 = \(\frac{CO_2 produced}{O_2 consumed}\)
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 values: – Carbohydrate: 1.0 – Fat: 0.7 – Protein: 0.9 – Anaerobic: ∞


Question 59:

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.
Step 2: Closes larynx during swallowing → food enters esophagus.
Step 3: If fails → food enters trachea → choking reflex (cough).
Hence, epiglottis. Quick Tip: - Swallowing: – Epiglottis closes larynx – Soft palate closes nasopharynx


Question 60:

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: Binomial nomenclature → Genus + species.
Step 2: Introduced by Carolus Linnaeus in "Systema Naturae" (1758) and "Species Plantarum" (1753).
Step 3:
- Bentham & Hooker → classification
- John Ray → early binomial
Hence, Linnaeus. Quick Tip: - Binomial system: Linnaeus - ICBN/ICZN: Rules for naming - Homo sapiens: Linnaeus

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