CUET 2026 May 20 Shift 1 Biology Question Paper is available for download here. NTA conducted the CUET 2026 exam from 11th May to 31st May.
- CUET 2026 Biology exam consists of 50 questions for 250 marks to be attempted in 60 minutes.
- As per the marking scheme, 5 marks are awarded for each correct answer, and 1 mark is deducted for incorrect answer.
Candidates can download CUET 2026 May 20 Shift 1 Biology Question Paper with Answer Key and Solution PDF from links provided below.
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CUET 2026 Biology May 20 Shift 1 Question Paper with Solution PDF
| CUET May 20 Shift 1 Biology Question Paper 2026 | Download PDF | Check Solutions |
Identify the chemicals produced by Calotropis
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Concept:
In terrestrial ecosystems, plants have evolved diverse chemical and morphological adaptations to protect themselves from herbivores, grazers, and browsers.
Secondary metabolites such as alkaloids, glycosides, tannins, and resins act as deterrents, toxins, or digestion inhibitors.
Step 1: Analysis of Chemical Defense in Calotropis:
Calotropis is a resilient weed commonly found growing wild in abandoned agricultural fields and arid wastelands.
The plant produces a thick, milky latex throughout its vegetative tissues containing poisonous cardiac glycosides.
Cardiac glycosides disrupt the cardiac function of animals by inhibiting the cellular sodium-potassium ATPase pump ($Na^+/K^+$-ATPase).
This leads to severe neuromuscular and cardiovascular distress in grazing mammals.
Consequently, cattle, goats, and wild browsers avoid eating or browsing this plant.
Step 2: Examination of the Other Chemicals:
- Strychnine: A highly poisonous alkaloid extracted from the seeds of Strychnos nux-vomica.
- Quinine: An alkaloid extracted from the dried bark of Cinchona trees, famously used as an antimalarial drug.
- Caffeine: A purine alkaloid produced by plants like Coffea arabica and Camellia sinensis to inhibit insect herbivory.
Step 3: Conclusion and Final Selection:
Among the given choices, the specific chemical produced by Calotropis for defense against herbivores is cardiac glycosides.
$\bullet$ Calotropis $\rightarrow$ Cardiac glycosides (toxic to heart).
$\bullet$ Cinchona $\rightarrow$ Quinine (antimalarial).
$\bullet$ Strychnos $\rightarrow$ Strychnine (neurotoxin).
$\bullet$ Coffea $\rightarrow$ Caffeine (stimulant/insect deterrent).
Match List-I with List-II:
Choose the correct answer from the options given below:
View Solution
Concept:
In an ecosystem, a trophic level represents the position occupied by an organism within a food chain based on its method of obtaining nourishment and energy.
Step 1: Identifying the First Trophic Level ($T_1$):
The first trophic level consists of primary producers (photoautotrophs) that fix solar energy into chemical energy.
In aquatic ecosystems, phytoplankton act as producers, while in terrestrial ecosystems, trees, grasses, and shrubs serve as producers.
Therefore, (C) Phytoplankton, trees corresponds to (II) First trophic level.
Step 2: Identifying the Second Trophic Level ($T_2$):
The second trophic level consists of primary consumers or herbivores that directly consume primary producers.
In aquatic ecosystems, zooplankton feed on phytoplankton, whereas cows and deer consume terrestrial vegetation.
Therefore, (B) Zooplankton, cow corresponds to (IV) Second trophic level.
Step 3: Identifying the Third Trophic Level ($T_3$):
The third trophic level comprises secondary consumers or primary carnivores that prey on herbivores.
Carnivorous birds, fishes, and wolves feed on primary consumers.
Therefore, (D) Birds, wolves corresponds to (I) Third trophic level.
Step 4: Identifying the Fourth Trophic Level ($T_4$):
The fourth trophic level contains tertiary consumers or top carnivores at the apex of the food chain.
Man and lions occupy apex predatory positions.
Therefore, (A) Man, lion corresponds to (III) Fourth trophic level.
Step 5: Compilation of Matches:
Matching combinations: (A)-(III), (B)-(IV), (C)-(II), (D)-(I).
This matches Option (C).
$T_1$ (Producers): Phytoplankton, Grass, Trees.
$T_2$ (Herbivores): Zooplankton, Cow, Deer.
$T_3$ (Primary Carnivores): Wolves, Small Birds, Fish.
$T_4$ (Top Carnivores): Lion, Man, Eagle.
The species confined to a particular region and not found anywhere else is:
View Solution
Concept:
Biogeographical distribution patterns describe how biological taxa are geographically distributed across different habitats, regions, and continents.
Step 1: Defining Endemism:
Endemism is the ecological and geographical state where a biological species is strictly restricted to a unique, defined geographical location (such as an island, mountain range, or specific biome) and does not occur naturally anywhere else in the world.
Such organisms are termed endemic species (e.g., the Western Ghats lion-tailed macaque, kangaroo in Australia).
Step 2: Evaluating the Alternative Ecological Terms:
- Cosmopolitan: Refers to species that have a nearly worldwide distribution and can thrive in diverse biomes across multiple continents (e.g., pigeons, houseflies).
- Hot-spot: A biodiversity hotspot is a biogeographical region characterized by exceptional levels of species richness and a high degree of endemism, while concurrently facing extreme habitat destruction.
- Alien species: Non-native or exotic species that have been introduced into an ecosystem outside their natural historical range, either intentionally or accidentally.
Step 3: Conclusion and Final Selection:
The phenomenon of being confined exclusively to a particular geographic zone is designated as Endemism.
$\bullet$ Endemic: Found only in one specific geographical region.
$\bullet$ Cosmopolitan: Found across the globe.
$\bullet$ Biodiversity Hotspot: Region with high species richness and high endemism.
Identify the correct reasons for cutting and clearing of Amazon rain forest?
(A) Cultivating soyabeans
(B) Growing wheat and maize
(C) Setting up industries
(D) Conversion to grasslands for raising beef cattle
Choose the correct answer from the options given below:
View Solution
Concept:
Habitat loss and fragmentation is considered the most significant cause of the global extinction of animal and plant species, critically impacting tropical rainforest biomes.
Step 1: Environmental Drivers of Amazonian Deforestation:
The Amazon rainforest in South America is often referred to as the 'lungs of the planet' because it accounts for a massive proportion of global terrestrial photosynthesis and oxygen release.
Despite harbouring millions of species, vast tracts of this rainforest are cleared and fragmented at alarming rates.
According to ecological studies and NCERT records, the two main anthropogenic drivers for clearing the Amazon rainforest are:
1. Clearing lands for the large-scale commercial cultivation of soybeans (Statement A).
2. Converting the cleared forested land into grasslands and pastures for raising beef cattle (Statement D).
Step 2: Evaluating the Given Statements:
- Statement (A) is correct: Intensive soybean farming is a primary reason for deforestation.
- Statement (B) is incorrect: Large-scale wheat and maize agriculture is not the primary driver cited for clearing the tropical rainforest.
- Statement (C) is incorrect: Setting up industrial plants is not the principal direct cause of the massive clearing of the Amazon basin compared to commercial agriculture and livestock ranching.
- Statement (D) is correct: Pasture conversion for cattle ranching is the largest direct driver of deforestation in the Amazon.
Step 3: Final Selection:
The valid pair of reasons is (A) and (D).
Hence, Option (B) is correct.
The Amazon rainforest is cleared mainly for:
1. Cultivating soybeans.
2. Conversion to grasslands for raising beef cattle.
Which one of the following is a inhibitor genes in lac
View Solution
Concept:
The lac operon in Escherichia coli, elucidated by François Jacob and Jacques Monod, is a polycistronic transcriptional unit controlled by a shared promoter and regulatory elements.
Step 1: Genetic Organization of the lac Operon:
The lac operon comprises a regulatory gene along with three adjacent structural genes:
1. Regulatory Gene ($i$ gene): The letter '$i$' is derived from the word inhibitor. It codes constitutively for the lac repressor protein.
2. Structural Gene $z$: Transcribes and codes for the enzyme $\beta$-galactosidase, which hydrolyzes lactose into glucose and galactose.
3. Structural Gene $y$: Codes for $\beta$-galactoside permease, which increases membrane permeability to lactose.
4. Structural Gene $a$: Codes for $\beta$-galactoside transacetylase.
Step 2: Mechanism of Action of the Inhibitor Gene:
The product of the $i$ gene (the active repressor monomer that forms a tetramer) binds to the operator ($O$) region in the absence of an inducer (such as lactose or allolactose).
Binding to the operator physically impedes RNA polymerase from initiating transcription of the downstream structural genes $z, y,$ and $a$.
Because it inhibits transcription under uninduced conditions, gene $i$ is called the inhibitor gene.
Step 3: Final Answer Formulation:
Therefore, the inhibitor gene in the lac operon is Gene $i$.
$\bullet$ In lac operon, '$i$' stands for inhibitor (NOT inducer).
$\bullet$ Structural genes: $z$ ($\beta$-galactosidase), $y$ (permease), $a$ (transacetylase).
The scientist who developed a method for sequencing the DNA fragments was:
View Solution
Concept:
The development of biochemical methods for determining the exact sequence of nucleotide bases in DNA laid the foundation for modern genomics and genetic engineering.
Step 1: Historical Contributions of Frederick Sanger:
Frederick Sanger developed the dideoxy chain-termination technique, commonly known as the Sanger sequencing method.
This method uses 2',3'-dideoxynucleoside triphosphates (ddNTPs) that terminate DNA strand synthesis when incorporated by DNA polymerase.
Automated sequencing machines were subsequently designed based on Sanger's principles and were instrumental in completing the Human Genome Project (HGP).
Sanger also developed the method for sequencing amino acids in proteins (specifically insulin).
Step 2: Evaluating Other Scientists' Contributions:
- Frederick Griffith (1928): Discovered the 'Transforming Principle' using virulent and avirulent strains of Streptococcus pneumoniae in mice.
- George Gamow: A physicist who proposed that genetic code is a triplet code made up of three bases to specify 20 amino acids ($4^3 = 64$ codons).
- Har Gobind Khorana: Successfully synthesized defined poly-ribonucleotides and co-polymers to help decipher the triplet genetic code.
Step 3: Conclusion and Final Selection:
Thus, the scientist credited with developing the fundamental method for sequencing DNA fragments is Frederick Sanger.
$\bullet$ Frederick Sanger: DNA sequencing (Chain termination method) \& Protein sequencing.
$\bullet$ Griffith: Transformation experiment.
$\bullet$ Gamow: Triplet nature of genetic code.
$\bullet$ Khorana: Chemical synthesis of defined RNA homopolymers.
Which one of the followings is not a terminator codon?
View Solution
Concept:
During messenger RNA (mRNA) translation, triplet codons specify amino acids. Three of the 64 possible codons do not code for any amino acid and instead act as stop/termination signals.
Step 1: Understanding Terminator (Stop) Codons:
In the standard universal genetic code:
- 61 codons code for the 20 standard amino acids (sense codons).
- 3 codons act as stop/nonsense/terminator codons that signal the termination of polypeptide synthesis.
The three standard terminator codons are:
1. UAA (Ochre)
2. UAG (Amber)
3. UGA (Opal)
When a ribosome encounters any of these three codons in the A-site, release factors bind to them, hydrolyzing the bond between the polypeptide and tRNA, thereby completing translation.
Step 2: Identification of UUC:
The codon UUC is a sense codon.
Both UUU and UUC specify the amino acid Phenylalanine (Phe).
Hence, UUC is an amino-acid coding codon and does not serve as a terminator codon.
Step 3: Final Conclusion:
Among the given options, UUC is not a terminator codon.
$\bullet$ UAA, UAG, UGA (All three start with 'U' and end translation).
$\bullet$ Initiator Codon: AUG (codes for Methionine).
$\bullet$ UUU and UUC code for Phenylalanine (Phe).
Select the correct statements regarding Hardy-Weinberg principle.
(A) Sum total of all the allelic frequencies is 1.
(B) Genetic drift affect Hardy-Weinberg principle.
(C) Allele frequencies in a population are stable and is constant from generation to generation.
(D) The gene pool continues changing generation after generation.
Choose the correct answer from the options given below:
View Solution
Concept:
The Hardy-Weinberg principle states that allele and genotype frequencies in a large, randomly mating population remain constant and stable from generation to generation in the absence of evolutionary disturbances.
Step 1: Detailed Statement Analysis:
- Statement (A): "Sum total of all the allelic frequencies is 1."
If a gene has two alleles $A$ and $a$ with frequencies $p$ and $q$, then $p + q = 1$. The sum total of allelic frequencies in a population is always 1.0 (or 100\%). This statement is correct.
- Statement (B): "Genetic drift affect Hardy-Weinberg principle."
Five specific evolutionary factors disrupt Hardy-Weinberg equilibrium: gene migration/gene flow, genetic drift, mutation, genetic recombination, and natural selection. Thus, genetic drift alters allele frequencies and disturbs equilibrium. This statement is correct.
- Statement (C): "Allele frequencies in a population are stable and is constant from generation to generation."
This represents the fundamental definition of genetic equilibrium (Hardy-Weinberg equilibrium). This statement is correct.
- Statement (D): "The gene pool continues changing generation after generation."
Under Hardy-Weinberg equilibrium, the gene pool (total genes and their alleles in a population) remains constant. A constantly changing gene pool indicates evolution, which violates equilibrium. This statement is incorrect.
Step 2: Conclusion:
Statements (A), (B), and (C) are correct.
Therefore, Option (B) is the correct choice.
$\bullet$ Allele frequencies: $p + q = 1$
$\bullet$ Genotype frequencies: $p^2 + 2pq + q^2 = 1$
Disrupted by: Genetic drift, Mutation, Gene flow, Recombination, and Natural selection.
Identify the agencies which are involved in developing the technology of biogas production.
(A) Khadi and Village Industries Commission
(B) Indian Farmers Welfare Association
(C) Indian Biogas Production Institute
(D) Indian Agricultural Research Institute
Choose the correct answer from the options given below:
View Solution
Concept:
Biogas is a clean, combustible gaseous fuel generated via the anaerobic microbial digestion of cattle dung and biomass, spearheaded by national research and rural development agencies in India.
Step 1: Background of Biogas Technology in India:
In rural India, cattle dung (commonly called 'gobar') is available in large quantities.
Dung is rich in methanogens (Methanobacterium), which digest cellulosic material anaerobically inside biogas plants to produce biogas (rich in methane, $CH_4$, along with $CO_2$ and $H_2$).
Step 2: Identifying the Developing Agencies:
The practical technology and design of standardized biogas plants (such as the floating gas-holder type model) were developed in India through the combined and collaborative research efforts of:
1. IARI - Indian Agricultural Research Institute (Statement D)
2. KVIC - Khadi and Village Industries Commission (Statement A)
Step 3: Verification of Options:
- Statements (A) and (D) accurately represent the authorized institutes mentioned in standard biological literature.
- Organizations listed under (B) and (C) are fictitious or uninvolved in the primary development of this technology.
Therefore, (A) and (D) are correct.
$\bullet$ IARI: Indian Agricultural Research Institute
$\bullet$ KVIC: Khadi and Village Industries Commission
Both developed rural gobar gas plants.
Identify the bacteria which can fix atmospheric nitrogen and are free-living in the soil.
(A) Azotobacter
(B) Rhizobium
(C) Azospirillum
(D) Anabaena
View Solution
Concept:
Nitrogen-fixing microorganisms convert molecular atmospheric dinitrogen ($N_2$) into usable ammonia ($NH_3$). They are classified as symbiotic or free-living organisms.
Step 1: Classifying the Nitrogen-Fixing Microorganisms:
- Azotobacter: An aerobic, free-living soil bacterium capable of fixing atmospheric nitrogen independently without forming symbiotic plant associations. (Statement A is free-living bacterium).
- Azospirillum: A free-living/associative nitrogen-fixing bacterium found living freely in the rhizosphere and soil, enriching the soil nitrogen content. (Statement C is free-living bacterium).
- Rhizobium: A symbiotic nitrogen-fixing bacterium that forms root nodules specifically with leguminous plants; it cannot fix nitrogen when living freely in soil. (Statement B is symbiotic).
- Anabaena: A cyanobacterium (blue-green alga) that fixes nitrogen symbiotically (e.g., in Azolla and Cycas coralloid roots) or in aquatic environments, but is classified as a photosynthetic cyanobacterium rather than a typical free-living soil bacterium.
Step 2: Selecting the Correct Combination:
The free-living nitrogen-fixing bacteria in the soil explicitly recognized together in textbook literature are Azotobacter and Azospirillum.
Hence, statements (A) and (C) are the correct choices.
Step 3: Final Answer:
The matching choice is Option (A).
$\bullet$ Free-living bacteria: Azotobacter, Azospirillum, Beijerinckia.
$\bullet$ Symbiotic bacteria: Rhizobium (legumes), Frankia (non-legumes like Alnus).
$\bullet$ Cyanobacteria: Anabaena, Nostoc, Oscillatoria.
Arrange the following steps involved in PCR in the correct sequence.
(A) Amplification
(B) Extension of primers
(C) Primer annealing
(D) Denaturation
Choose the correct answer from the options given below:
View Solution
Concept:
Polymerase Chain Reaction (PCR), developed by Kary Mullis, is an in vitro enzymatic method used to amplify a specific segment of DNA into billions of copies through cyclic thermal reactions.
Step 1: Step-by-Step Mechanism of PCR:
A single cycle of PCR involves three consecutive temperature-dependent steps followed by overall amplification:
1. Denaturation (D): The double-stranded target DNA is heated to high temperature (approx. $94^\circ\text{C}-96^\circ\text{C}$), which breaks the hydrogen bonds between complementary base pairs, separating it into two single strands.
2. Annealing of Primers (C): The reaction temperature is lowered to around $50^\circ\text{C}-60^\circ\text{C}$ to allow two sets of synthetic oligonucleotide primers to hybridize (anneal) to their complementary sequences at the 3' end of each single-stranded DNA template.
3. Extension of Primers (B): The temperature is raised to optimal $72^\circ\text{C}$ where a thermostable enzyme, Taq DNA polymerase (isolated from Thermus aquaticus), extends the primers using free deoxyribonucleoside triphosphates (dNTPs).
4. Amplification (A): Repeating these three steps across 30 cycles results in the exponential amplification of the target DNA sequence to approximately one billion ($10^9$) times.
Step 2: Ordering the Steps:
The correct chronological order is:
Denaturation (D) $\rightarrow$ Primer annealing (C) $\rightarrow$ Extension of primers (B) $\rightarrow$ Amplification (A).
This corresponds to sequence (D), (C), (B), (A).
Step 3: Final Answer:
Thus, the correct option is (D).
1. Denaturation ($\sim 94^\circ\text{C}$)
2. Annealing ($\sim 54^\circ\text{C}$)
3. Extension ($\sim 72^\circ\text{C}$)
Leads to Amplification after repeated cycles.
Which one of the following microorganisms is used as a cloning vector to deliver the genes of interest into a variety of plants?
View Solution
Concept:
Plant genetic engineering utilizes natural vectors capable of stably transferring exogenous DNA into plant host genomes.
Step 1: Role of Agrobacterium tumefaciens as a Cloning Vector:
Agrobacterium tumefaciens is a soil-borne plant pathogen known as 'nature's genetic engineer'.
It naturally infects dicotyledonous plants and transfers a specific segment of DNA, termed T-DNA (transfer DNA), located on its tumor-inducing (Ti) plasmid into the plant nuclear genome.
In recombinant DNA technology, the pathogenic tumor-inducing genes in the Ti-plasmid are disarmed and replaced with desired genes of interest.
The modified Ti-plasmid delivers the foreign gene into the plant host genome without causing crown gall disease.
Step 2: Evaluating Alternative Options:
- Bacillus thuringiensis: A bacterium producing Cry insecticidal crystal proteins; it is not used as a cloning vector.
- Rhino viruses: Viruses causing common cold in humans.
- Nucleopolyhedrovirus: Baculoviruses used as narrow-spectrum biological control agents against insect pests.
Step 3: Final Answer:
Thus, the microorganism widely utilized as a disarmed plant cloning vector is Agrobacterium tumefaciens (Option C).
$\bullet$ Plants $\rightarrow$ Disarmed Ti-plasmid of Agrobacterium tumefaciens.
$\bullet$ Animals $\rightarrow$ Disarmed Retroviruses.
Which one of the following organisms has been disarmed and used to deliver desirable genes into animal cells?
View Solution
Concept:
Viruses with natural mechanisms for integrating their genetic material into host cells can be genetically engineered (disarmed) to serve as delivery vehicles (vectors) for therapeutic or desired genes in mammalian systems.
Step 1: Retroviruses as Animal Cloning Vectors:
In their wild form, retroviruses are RNA viruses that infect animal cells and integrate their reverse-transcribed viral DNA stably into the host cell chromosomes, sometimes transforming normal cells into cancerous cells.
In genetic engineering, the pathogenic and oncogenic genes of retroviruses are deleted or disarmed.
These disarmed retroviruses retain their efficient cell-entry and host-genome integration capabilities, making them optimal vectors for delivering functional therapeutic genes into animal cells (e.g., in gene therapy for ADA deficiency).
Step 2: Examining the Other Options:
- Cyanobacteria: Photosynthetic prokaryotes, not used as animal transformation vectors.
- Yeast: Eukaryotic unicellular organisms used as expression hosts or cloning hosts (e.g., YAC vectors), but not as delivery viruses.
- Baculoviruses: Pathogens belonging to the genus Nucleopolyhedrovirus that attack insects and other arthropods; used as biocontrol agents.
Step 3: Final Selection:
Therefore, disarmed Retroviruses are used to deliver desirable genes into animal cells.
$\bullet$ Plants: Agrobacterium tumefaciens (Ti-plasmid).
$\bullet$ Animals: Retroviruses (convert normal animal cells to transgenic cells without pathogenesis).
Arrange the following steps which are involved in action of Bt toxin proteins on insects in correct sequence.
(A) Conversion of inactive toxin into active one by alkaline pH of the gut of insect
(B) Creation of pores that cause cell swelling and lysis which leads to death of insect
(C) Inactive protoxins is ingested by insect
(D) Activated toxin binds to the surface of midgut epithelial cells of insect
Choose the correct answer from the options given below:
View Solution
Concept:
Bacillus thuringiensis produces crystal endotoxin proteins (Bt toxins) that act as insecticidal agents following ingestion and activation within susceptible insect larvae.
Step 1: Mechanism of Action of Bt Toxin:
1. Ingestion (C): The insect feeds on the transgenic plant parts and ingests the insecticidal crystalline protein in its inactive protoxin form.
2. Solubilization and Activation (A): Inside the insect's alimentary canal, the alkaline pH of the midgut solubilizes the protein crystals. Proteolytic cleavage converts the inactive protoxin into an active form of the toxin.
3. Receptor Binding (D): The activated toxin binds specifically to receptors present on the apical brush border membrane of the insect midgut epithelial cells.
4. Pore Formation and Lysis (B): The bound toxin oligomerizes and inserts into the membrane, creating trans-membrane lytic pores. This causes osmotic swelling, cell lysis, gut perforation, and ultimately results in the death of the insect.
Step 2: Determining the Correct Sequence:
The correct order of events is:
Ingestion of inactive protoxins (C) $\rightarrow$ Activation by alkaline pH (A) $\rightarrow$ Binding to midgut epithelium (D) $\rightarrow$ Pore formation and cell lysis (B).
This forms the sequence (C), (A), (D), (B).
Step 3: Final Answer:
Hence, Option (C) is the correct answer.
$\text{Protoxin Ingestion (C)} \xrightarrow{\text{Alkaline pH}} \text{Active Toxin (A)} \xrightarrow{\text{Midgut Epithelium}} \text{Binding (D)} \xrightarrow{\text{Pores}} \text{Lysis \& Death (B)}$.
Which one of the following techniques serve the purpose of early diagnosis of diseases?
View Solution
Concept:
Early detection of disease allows timely medical intervention. Molecular diagnostic tools can detect pathogens even when their concentration is extremely low in the body.
Step 1: Limitations of Conventional Diagnostics:
Conventional diagnostic methods (such as serum analysis, urine analysis, and stool analysis) generally detect diseases only when the pathogen has multiplied significantly and clinical symptoms appear.
Consequently, they do not serve the purpose of early disease diagnosis.
Step 2: Principles of Molecular Diagnostics for Early Detection:
Advanced recombinant DNA technology, Polymerase Chain Reaction (PCR), and Enzyme-Linked Immunosorbent Assay (ELISA) allow early detection of diseases:
- ELISA (Enzyme-Linked Immunosorbent Assay): Based on the principle of antigen-antibody interaction. Infection by pathogens can be detected by identifying very low concentrations of antigens (proteins, glycoproteins) or antibodies synthesized against the pathogen.
- PCR: Amplifies trace amounts of microbial nucleic acids (DNA/RNA) to detect pathogens before symptoms develop.
Step 3: Conclusion:
Among the given options, Enzyme linked immuno-sorbent assay (ELISA) is the recognized molecular tool used for early diagnosis of diseases.
$\bullet$ Recombinant DNA technology
$\bullet$ Polymerase Chain Reaction (PCR)
$\bullet$ ELISA (Antigen-Antibody interaction)
Conventional methods (Serum, Urine, Stool analysis) only detect infections at later stages.
When the pistils are fused together, this state is called:
View Solution
Concept:
The gynoecium represents the female reproductive organ of a flower and may consist of a single pistil/carpel or multiple pistils/carpels that can either be free or fused together.
Step 1: Botanical Terms Describing Carpel Fusion:
- Syncarpous: When a flower possesses a multicarpellary gynoecium in which the individual carpels/pistils are fused together into a single compound structure (e.g., Papaver, Hibiscus).
- Apocarpous: When a multicarpellary gynoecium has carpels/pistils that remain completely free and separate from each other (e.g., Michelia, Lotus, Rose).
- Monocarpellary: A condition where the gynoecium consists of only one single pistil/carpel (e.g., pea, bean).
- Multicarpellary: A condition where the gynoecium contains more than one carpel (can be either syncarpous or apocarpous).
Step 2: Conclusion:
When the pistils are fused together, the morphological condition is termed Syncarpous.
$\bullet$ Fused pistils = Syncarpous (Example: Papaver)
$\bullet$ Free pistils = Apocarpous (Example: Michelia)
Pollination is not carried out by water in -
View Solution
Concept:
Hydrophily (pollination by water) is restricted to a small number of genera (mostly monocotyledons). Many aquatic plants with emergent flowers rely on wind (anemophily) or insects (entomophily).
Step 1: Examining Pollination in the Aquatic Plants:
- Water lily (Nymphaea) and Water hyacinth (\textit{Eichhornia):} Although they grow in aquatic environments, their flowers emerge above the surface of the water and are pollinated by insects (entomophily) or wind (anemophily), not by water.
- Vallisneria: An aquatic plant where the female flower reaches the surface of water by a long stalk, and pollen grains are released onto the surface of water to reach the stigma (epihydrophily).
- Hydrilla: A submerged freshwater plant that utilizes water currents for the transport of pollen grains to stigmas.
- Zostera (Marine sea grasses): Submerged marine angiosperms where female flowers remain submerged and long, ribbon-like pollen grains are carried passively inside the water (hypohydrophily).
Step 2: Conclusion:
Pollination is not carried out by water in Water lily.
Therefore, Option (B) is the correct answer.
$\bullet$ Water lily and Water hyacinth $\rightarrow$ Flowers emerge above water level and are pollinated by insects or wind.
$\bullet$ Vallisneria \& Zostera $\rightarrow$ True water pollination (Hydrophily).
Identify the structures which have the potency to give rise to all the tissues and organs.
View Solution
Concept:
During human embryonic development, cellular differentiation leads to the formation of specialized tissues and organs from pluripotent progenitor cells.
Step 1: Role and Potency of Stem Cells:
The blastocyst stage of embryonic development contains an outer cellular layer called the trophoblast and an inner group of cells attached to one pole called the inner cell mass (ICM).
The inner cell mass contains specialized, undifferentiated cells termed stem cells.
Stem cells possess pluripotency, which gives them the developmental capacity to differentiate into all the specialized embryonic germ layers (ectoderm, mesoderm, and endoderm) and subsequently form all tissues and organs of the adult organism.
Step 2: Evaluating the Alternative Structures:
- Trophoblast: The outer epithelial layer of the blastocyst that attaches to the endometrium and helps form extraembryonic membranes and the fetal portion of the placenta.
- Chorionic villi: Finger-like projections produced by the trophoblast that interdigitate with maternal uterine tissue to form the placenta.
- Placenta: The structural and functional organ facilitating nutrient, respiratory gas, and waste exchange between maternal and fetal blood.
Step 3: Conclusion:
The cells that possess the potency to give rise to all tissues and organs are Stem cells.
$\bullet$ Inner Cell Mass (ICM) / Stem Cells: Pluripotent cells giving rise to all three germ layers and all body tissues/organs.
$\bullet$ Trophoblast: Forms extra-embryonic membranes and placenta.
Which one of the following structures is not essential for maturation and motility of sperms?
View Solution
Concept:
After spermatogenesis in the seminiferous tubules, spermatozoa undergo physiological maturation and acquire progressive motility through secretions of male accessory ducts and accessory glands.
Step 1: Role of Male Reproductive Ducts and Glands:
- Epididymis: Stores spermatozoa temporarily, allowing them to undergo functional biochemical maturation and acquire forward motility.
- Seminal Vesicles: Secrete seminal plasma rich in fructose, prostaglandins, and clotting factors that nourish sperm and enhance motility.
- Prostate Gland: Secretes a slightly alkaline milky fluid containing citrate and enzymes that contribute to seminal fluid volume and maintain sperm motility.
Secretions of the epididymis, vas deferens, seminal vesicles, and prostate are essential for the maturation and motility of sperms.
Step 2: Role of Cervix:
The cervix is a structural component of the female reproductive tract that forms the cervical canal connecting the uterus to the vagina.
It acts as a passage during insemination and childbirth (part of the birth canal), but is not a male accessory structure responsible for sperm maturation and motility.
Step 3: Conclusion:
Hence, the cervix is not an essential structure for the maturation and motility of sperms.
$\bullet$ Epididymis
$\bullet$ Vas deferens
$\bullet$ Seminal vesicles
$\bullet$ Prostate gland
The cervix belongs to the female reproductive system.
Match List-I with List-II:
Choose the correct answer from the options given below:
View Solution
Concept:
Contraceptive methods prevent unwanted pregnancies and are categorized into natural methods, barrier methods, intra-uterine devices (IUDs), oral contraceptives, and surgical sterilization methods.
Step 1: Matching Intra-Uterine Devices and Barriers:
1. Non-medicated IUDs: Inert plastic/polyethylene devices that promote phagocytosis of sperms within the uterus without releasing chemical agents. The primary example is Lippes loop. Thus, (A) matches with (IV).
2. Copper-releasing IUDs: Devices that slowly release cupric ions ($Cu^{2+}$) to suppress sperm motility and fertilizing capacity. Examples include CuT, Cu7, and Multiload 375. Thus, (B) matches with (I).
3. Hormone-releasing IUDs: Devices that release synthetic progestogens, making the uterus unsuitable for implantation and the cervix hostile to sperms. Examples include Progestasert and LNG-20. Thus, (C) matches with (II).
4. Barrier methods: Mechanical devices made of rubber/latex that prevent physical contact between sperm and ovum. Examples include diaphragms, cervical caps, and Vaults. Thus, (D) matches with (III).
Step 2: Compilation of Pairs:
(A)-(IV), (B)-(I), (C)-(II), (D)-(III).
This corresponds directly to Option (D).
$\bullet$ Non-medicated: Lippes loop
$\bullet$ Copper releasing: CuT, Cu7, Multiload 375
$\bullet$ Hormone releasing: Progestasert, LNG-20
$\bullet$ Barriers: Condoms, Diaphragms, Cervical caps, Vaults
Which one of the following diseases is completely curable if detected early and treated properly?
View Solution
Concept:
Sexually Transmitted Infections (STIs) or venereal diseases (VD) are infections transmitted through sexual contact. Bacterial STIs are generally curable with antibiotics, whereas certain viral STIs remain incurable.
Step 1: Analyzing Curable and Incurable STIs:
According to NCERT medical guidelines on reproductive health:
- Except for Hepatitis-B, Genital herpes, and HIV infections, other sexually transmitted diseases are completely curable if detected early and treated properly with appropriate therapeutics.
- Gonorrhoea is a bacterial STI caused by the bacterium Neisseria gonorrhoeae. Because it is bacterial, early detection allows effective treatment and complete eradication using antibiotics (such as cephalosporins/penicillins).
Step 2: Evaluating the Incurable Viral STIs:
- Hepatitis-B: Caused by Hepatitis B virus (HBV); cannot be completely cured once established in a chronic state.
- Genital herpes: Caused by Herpes simplex virus (HSV); resides latently in neural ganglia and remains non-curable.
- HIV: Human Immunodeficiency Virus causes AIDS; current antiretroviral therapy only prolongs life without providing a complete cure.
Step 3: Final Selection:
Gonorrhoea is completely curable when diagnosed early and treated appropriately.
$\bullet$ Incurable STIs: Hepatitis-B, Genital herpes, HIV.
$\bullet$ Curable STIs: Gonorrhoea, Syphilis, Chlamydiasis, Trichomoniasis (if detected early).
"Two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually." This statement was given by:
View Solution
Concept:
Interspecific competition occurs when individuals of different species compete for shared, limiting environmental resources such as food or space.
Step 1: Defining Gause's Competitive Exclusion Principle:
Russian ecologist G.F. Gause conducted laboratory competition experiments using species of Paramecium (P. aurelia and P. caudatum).
Based on his observations, he formulated Gause's Competitive Exclusion Principle, which states that:
"Two closely related species competing for the same limiting resources cannot co-exist indefinitely, and the competitively inferior species will eventually be eliminated by the competitively superior one."
This principle holds true particularly when resources are limited.
Step 2: Evaluating Other Ecological and Evolutionary Principles:
- Malthus theory: Addressed geometric population growth outstripping arithmetic food supply.
- Verhulst-Pearl logistic growth: Described resource-limited population growth mathematically using the formula $\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)$.
- Darwin's theory of natural selection: Focused on survival of the fittest based on heritable variations.
Step 3: Conclusion:
The given statement defines Gause's competitive exclusion principle (Option B).
$\bullet$ Gause's Principle: Competitive exclusion of inferior competitor.
$\bullet$ MacArthur: Resource partitioning promotes co-existence (e.g., Warblers).
$\bullet$ Verhulst-Pearl: Logistic growth model ($K$ = carrying capacity).
The trophic level that has a certain mass of living material at a particular time is called as:
View Solution
Concept:
Ecosystem structure is characterized by quantitative parameters that measure the accumulation of living organic matter and energy across different trophic levels at a specific moment in time.
Step 1: Understanding Standing Crop:
Each trophic level has a certain mass of living material at a particular time, which is referred to as the standing crop.
The standing crop is measured either as:
1. The biomass (mass of living organisms) per unit area.
2. The total number of individual organisms per unit area.
Measurement of biomass in terms of dry weight is considered more accurate than fresh weight because fresh weight fluctuates with water content.
Step 2: Evaluating Other Ecological Concepts:
- Productivity: The rate of biomass or organic matter production per unit area over a specified time period (e.g., $g/m^2/yr$).
- Stratification: The vertical distribution and layering of different species occupying different physical levels in an ecosystem (e.g., trees form top layer, shrubs middle, herbs/grasses bottom).
- Respiration loss ($R$): The amount of fixed energy consumed by autotrophs or heterotrophs for their cellular metabolic processes.
Step 3: Conclusion:
The mass of living material present at a particular time at a trophic level is called the standing crop.
$\bullet$ Standing crop: Total mass of living matter at a given time (measured as dry biomass).
$\bullet$ Standing state: Total amount of non-living inorganic nutrients ($C, N, P, Ca$) in soil at a given time.
Match List-I with List-II:
Choose the correct answer from the options given below:
View Solution
Concept:
Sacred groves represent traditional, community-protected forest patches conserved due to religious, cultural, and spiritual beliefs, serving as vital repositories for in-situ biodiversity conservation.
Step 1: Identifying Locations of Sacred Groves in India:
According to NCERT conservation records:
1. Khasi and Jaintia Hills: Located in the state of Meghalaya (famous for protecting many rare and threatened endemic species). Thus, (A) matches with (IV).
2. Western Ghat regions: Span across Karnataka and Maharashtra. Thus, (B) matches with (II).
3. Sarguja, Chanda, and Bastar areas: Located in the central forested belts of Madhya Pradesh / Chhattisgarh. Thus, (C) matches with (III).
4. Aravalli Hills: Located in the arid and semi-arid landscape of Rajasthan. Thus, (D) matches with (I).
Step 2: Compiling the Matches:
- (A) $\rightarrow$ (IV)
- (B) $\rightarrow$ (II)
- (C) $\rightarrow$ (III)
- (D) $\rightarrow$ (I)
This set corresponds directly to Option (D).
Step 3: Final Answer:
Hence, the correct option is (D).
$\bullet$ Khasi and Jaintia Hills $\rightarrow$ Meghalaya
$\bullet$ Aravalli Hills $\rightarrow$ Rajasthan
$\bullet$ Western Ghats $\rightarrow$ Karnataka \& Maharashtra
$\bullet$ Sarguja, Chanda, Bastar $\rightarrow$ Madhya Pradesh
Arrange the following steps involved in the regulation of gene expression in correct sequence.
(A) Processing level
(B) Translational level
(C) Transcriptional level
(D) Transport of mRNA from the nucleus to the cytoplasm
Choose the correct answer from the options given below:
View Solution
Concept:
In eukaryotic organisms, gene expression is a multi-step process leading to the synthesis of a functional protein. Regulation can occur at several sequential stages along this biological pathway.
Step 1: Sequential Levels of Gene Regulation in Eukaryotes:
In eukaryotes, gene expression is regulated in the following chronological sequence from the gene in the nucleus to the finished protein in the cytoplasm:
1. Transcriptional level (C): Formation of the primary RNA transcript (hnRNA) from the DNA template in the nucleus.
2. Processing level (A): Post-transcriptional modification of primary transcripts, including 5'-capping, 3'-polyadenylation, and splicing of non-coding introns to form mature mRNA.
3. Transport level (D): Transport of the processed, mature mRNA through nuclear pore complexes from the nucleus into the cytoplasm.
4. Translational level (B): Translation of mRNA into a polypeptide chain by ribosomes and tRNAs in the cytoplasm.
Step 2: Compiling the Sequence:
The chronological order of gene regulation steps is:
Transcriptional level (C) $\rightarrow$ Processing level (A) $\rightarrow$ Transport of mRNA (D) $\rightarrow$ Translational level (B).
This matches sequence (C), (A), (D), (B).
Step 3: Final Answer:
Thus, the correct option is (C).
1. Transcription (hnRNA synthesis)
2. Processing (Splicing, Capping, Tailing)
3. Nuclear Transport (Nucleus to Cytoplasm)
4. Translation (Polypeptide synthesis)
Identify the organism whose genome has not been sequenced.
View Solution
Concept:
The Human Genome Project (HGP) not only sequenced human nuclear DNA but also sequenced several non-human model organisms to understand basic biological processes, comparative genomics, and evolutionary relationships.
Step 1: Genome Projects of Non-Human Model Organisms in NCERT:
During the Human Genome Project, several non-human model organisms had their complete genomes mapped and sequenced due to their immense utility in genetic, physiological, and developmental research.
The prominent non-human organisms listed in canonical biological curricula whose genomes were successfully sequenced include:
1. Bacteria (e.g., Escherichia coli)
2. Yeast (Saccharomyces cerevisiae)
3. Caenorhabditis elegans (a free-living non-pathogenic nematode whose genome was completely sequenced)
4. Drosophila melanogaster (the fruit fly)
5. Plants such as Arabidopsis thaliana and Oryza sativa (rice).
Step 2: Evaluating the Non-Sequenced Organism:
- Option (A) Caenorhabditis elegans: Genome fully sequenced as an essential model for developmental biology and cell lineage.
- Option (B) Arabidopsis: Genome fully sequenced as a classical plant genetic model.
- Option (C) Drosophila: Genome fully sequenced as a classical model organism for eukaryotic genetics.
- Option (D) Sea urchin: Although sea urchin is studied in classical embryology, it is not among the benchmark model organisms sequenced and highlighted as part of the Human Genome Project curriculum.
Step 3: Final Answer:
Thus, among the given choices, the organism whose genome is not cited as sequenced under the HGP model organisms curriculum is Sea urchin.
$\bullet$ Nematode: Caenorhabditis elegans
$\bullet$ Insect: Drosophila melanogaster
$\bullet$ Plants: Arabidopsis thaliana, Rice (Oryza sativa)
$\bullet$ Microbes: Yeast (S. cerevisiae), E. coli
Which one of the following statements is the correct fact about human genome project?
View Solution
Concept:
The Human Genome Project (HGP) provided quantitative insights into the structure, organization, size, coding capacity, and distribution of genes in the human genome.
Step 1: Evaluation of Salient Features of Human Genome:
- Option (A): The human genome contains approximately $3.1647 \times 10^9$ base pairs. The average gene consists of 3,000 bases (not 30,000 bases), although sizes vary greatly (with dystrophin being the largest at 2.4 million bases). Thus, (A) is incorrect.
- Option (B): The functions are unknown for over 50 percent (more than half) of the discovered genes, not less than 20 percent. Thus, (B) is incorrect.
- Option (C): Less than 2 percent of the entire human genome sequence codes for functional proteins; the remaining over 98 percent consists of non-coding, repetitive sequences. Thus, (C) is correct.
- Option (D): Chromosome 1 contains the highest number of genes (2,968 genes), whereas the Y chromosome has the fewest genes (231 genes). The statement has these values reversed. Thus, (D) is incorrect.
Step 2: Final Answer:
The only true statement regarding the human genome is that less than 2 percent of the genome codes for proteins.
$\bullet$ Total bases $\approx 3.164$ billion bp.
$\bullet$ Average gene size = 3,000 bases.
$\bullet$ Coding region $<$ 2\% of total genome.
$\bullet$ Chromosome 1 has 2968 genes (maximum).
$\bullet$ Chromosome Y has 231 genes (minimum).
Arrange the following stages of human evolution in correct chronological order.
(A) Australopithecines
(B) Homo erectus
(C) Ramapithecus
(D) Homo sapiens
Choose the correct answer from the options given below:
View Solution
Concept:
Human evolution is characterized by anatomical, morphological, and cranial developments occurring across fossil hominid taxa over geological time periods.
Step 1: Tracing the Chronological Timeline of Human Ancestors:
1. Ramapithecus (C): Lived about 15 million years ago (mya). They were ape-like, hairy, and walked more like modern hominids.
2. Australopithecines (A): Existed around 2 to 4 mya in East African grasslands. They hunted with stone weapons and possessed a brain capacity around 400--500 cc.
3. Homo habilis: First hominid human-like ancestor with brain capacity 650--800 cc (lived around 2 mya).
4. Homo erectus (B): Emerged around 1.5 mya. Fossils discovered in Java (Java man) had a cranial capacity of approximately 900 cc and probably ate meat.
5. Neanderthal man: Lived near east and central Asia between 100,000 to 40,000 years ago with a cranial capacity of 1400 cc.
6. Homo sapiens (D): Modern humans arose in Africa between 75,000 to 10,000 years ago during the Ice Age.
Step 2: Sequential Ordering:
Arranging the given taxa chronologically:
Ramapithecus (C) $\rightarrow$ Australopithecines (A) $\rightarrow$ Homo erectus (B) $\rightarrow$ Homo sapiens (D).
This corresponds to sequence (C), (A), (B), (D).
Step 3: Final Answer:
Hence, Option (A) is the correct answer.
Dryopithecus $\rightarrow$ Ramapithecus (15 mya) $\rightarrow$ Australopithecus (2-4 mya) $\rightarrow$ Homo habilis (650-800 cc) $\rightarrow$ Homo erectus (900 cc, 1.5 mya) $\rightarrow$ Neanderthal (1400 cc) $\rightarrow$ Homo sapiens.
Which one of the given terms states that mutations caused speciation?
View Solution
Concept:
Hugo de Vries formulated the Mutation Theory of Evolution based on his breeding experiments with the evening primrose (Oenothera lamarckiana), contrasting with Darwin's view of gradual evolution.
Step 1: Mechanism of Speciation according to Hugo de Vries:
Hugo de Vries believed that large, sudden, discontinuous, random, and directionless genetic changes called mutations are the primary cause of evolution and the origin of new species.
He proposed that mutation causes speciation in a single step, which he termed saltation (single-step large mutation).
This differed from Darwinian evolution, which emphasized minor, continuous, directional variations accumulating gradually over long periods.
Step 2: Evaluating the Other Terms:
- Variation: Morphological or physiological differences among individuals of a species, described by Darwin as small and continuous.
- Genetic recombination: Exchange of genetic material during crossing over in meiosis leading to new allelic combinations.
- Gene migration: The movement of alleles into or out of a population due to individual immigration or emigration.
Step 3: Final Answer:
The term that states that single-step large mutations cause speciation is Saltation (Option D).
$\bullet$ Darwin: Small, gradual, continuous, directional variations.
$\bullet$ Hugo de Vries: Large, sudden, discontinuous, directionless mutations = Saltation (Single-step large mutation causing speciation).
Match List-I with List-II:
Choose the correct answer from the options given below:
View Solution
Concept:
Biocontrol refers to the use of natural predators, parasites, or biological agents to control plant diseases and agricultural pests without relying on synthetic chemicals.
Step 1: Identifying the Target Organism for Each Biocontrol Agent:
1. Ladybird: A familiar beetle with red and black markings that preys on and controls populations of **aphids**. Thus, (A) matches with (II).
2. Bacillus thuringiensis (Bt): A bacterium whose spores produce insecticidal Cry proteins that kill insect larvae such as **butterfly caterpillars**. Thus, (B) matches with (IV).
3. Dragonflies: Aerial predatory insects that feed on and control populations of **mosquitoes**. Thus, (C) matches with (I).
4. Trichoderma: A free-living, soil-dwelling fungus found in root ecosystems that acts as an effective biocontrol agent against several **plant pathogens**. Thus, (D) matches with (III).
Step 2: Compilation of Pairs:
Matching set:
(A)-(II), (B)-(IV), (C)-(I), (D)-(III).
This corresponds directly to Option (C).
$\bullet$ Ladybird beetle $\rightarrow$ Aphids
$\bullet$ Dragonflies $\rightarrow$ Mosquitoes
$\bullet$ Bacillus thuringiensis $\rightarrow$ Butterfly caterpillars
$\bullet$ Trichoderma $\rightarrow$ Root-borne fungal plant pathogens
Match List-I with List-II:
Choose the correct answer from the options given below:
View Solution
Concept:
Microorganisms produce diverse bioactive molecules, industrial metabolites, and perform ecological services such as nitrogen fixation.
Step 1: Identifying Microorganisms and Their Specific Bioactive Roles:
1. Immunosuppressive agent: Cyclosporin A is a bioactive molecule used as an immunosuppressive agent in organ-transplant patients, commercially produced by the fungus Trichoderma polysporum. Thus, (A) matches with (II).
2. Blood-cholesterol lowering agents: Statins competitively inhibit HMG-CoA reductase (the enzyme responsible for cholesterol synthesis) and are produced by the yeast Monascus purpureus. Thus, (B) matches with (III).
3. Nitrogen fixation: Oscillatoria is an autotrophic cyanobacterium that enriches organic matter and fixes atmospheric nitrogen in aquatic and paddy soil habitats. Thus, (C) matches with (IV).
4. Holes in 'Swiss cheese': The large holes in Swiss cheese are due to the production of large amounts of $CO_2$ gas by the bacterium Propionibacterium sharmanii. Thus, (D) matches with (I).
Step 2: Summary of Matches:
(A)-(II), (B)-(III), (C)-(IV), (D)-(I).
This matches Option (B).
$\bullet$ Cyclosporin A $\rightarrow$ Trichoderma polysporum (Immunosuppressant).
$\bullet$ Statins $\rightarrow$ Monascus purpureus (Lowers blood cholesterol).
$\bullet$ Swiss cheese holes $\rightarrow$ Propionibacterium sharmanii ($CO_2$ release).
$\bullet$ Biofertilizer cyanobacteria $\rightarrow$ Oscillatoria, Nostoc, Anabaena.
Select the correct methods to introduce the recombinant DNA into the host cells.
(A) Biolistics or gene gun for plant cells
(B) Micro-injection for an animal cell
(C) 'Disarmed pathogen' vectors
(D) By heating the host cell at 100°C and then cooling to 70°C
Choose the correct answer from the options given below:
View Solution
Concept:
Since DNA is a hydrophilic molecule, it cannot pass directly across lipid cellular membranes. Recombinant biotechnology employs physical, chemical, and biological methods to introduce recombinant DNA into competent host cells.
Step 1: Evaluating Gene Transfer Methods:
- Statement (A): Biolistics or gene gun is a direct physical transformation method suitable for plant cells, where cells are bombarded with high-velocity microscopic gold or tungsten particles coated with recombinant DNA. This statement is correct.
- Statement (B): Micro-injection is a direct method used for animal cells, where recombinant DNA is injected directly into the nucleus using a glass micropipette. This statement is correct.
- Statement (C): Disarmed pathogen vectors (e.g., disarmed Ti-plasmid of Agrobacterium for plants, and disarmed retroviruses for animals) transfer foreign genes upon infecting host cells without causing disease. This statement is correct.
- Statement (D): Chemical transformation of bacterial cells using divalent cations ($Ca^{2+}$) involves heat shock at $42^\circ\text{C}$ followed by rapid cooling on ice ($0^\circ\text{C}$). Heating cells to $100^\circ\text{C}$ would denature proteins, melt cellular membranes, and kill the host cells. Thus, (D) is incorrect.
Step 2: Conclusion:
Statements (A), (B), and (C) are correct.
Hence, Option (A) is the correct choice.
$\bullet$ Gene gun / Biolistics: Plant cells (Gold/Tungsten microparticles).
$\bullet$ Micro-injection: Animal cells (Direct injection into nucleus).
$\bullet$ Heat Shock: $Ca^{2+}$ treated bacteria placed on ice, then $42^\circ\text{C}$, then back to ice.
Arrange the following steps which are involved in the process of recombinant DNA technology in correct sequence.
(A) Fragmentation of DNA by restriction endonucleases
(B) Transferring the recombinant DNA into the host
(C) Ligation of desired DNA fragment into a vector
(D) Isolation of DNA
Choose the correct answer from the options given below:
View Solution
Concept:
Recombinant DNA technology comprises sequential genetic and molecular manipulations aimed at isolating a target gene and producing transgenic organisms or products.
Step 1: Tracing the Core Sequence of rDNA Technology:
1. Isolation of DNA (D): The host and donor cells are lysed enzymatically to release and purify high-molecular-weight genomic DNA free of other macromolecules.
2. Fragmentation of DNA (A): The isolated purified DNA is cleaved into fragments using specific restriction endonucleases, followed by agarose gel electrophoresis to isolate the gene of interest.
3. Ligation into a Vector (C): The isolated gene of interest is ligated into an enzymatically cleaved cloning vector (plasmid/bacteriophage) using the enzyme DNA ligase to construct recombinant DNA.
4. Transferring rDNA into the Host (B): The recombinant vector is introduced into a competent recipient host cell (e.g., bacterial transformation) for multiplication and expression.
Step 2: Formulating the Chronological Sequence:
The correct chronological order of the given steps is:
Isolation of DNA (D) $\rightarrow$ Fragmentation by restriction enzymes (A) $\rightarrow$ Ligation into a vector (C) $\rightarrow$ Transferring rDNA into the host (B).
This corresponds to sequence (D), (A), (C), (B).
Step 3: Final Answer:
Thus, Option (D) is the correct answer.
1. DNA Isolation $\rightarrow$ 2. Restriction Digestion $\rightarrow$ 3. Gel Isolation $\rightarrow$ 4. Ligation into Vector $\rightarrow$ 5. Transformation into Host $\rightarrow$ 6. Culturing in Bioreactors $\rightarrow$ 7. Downstream Processing.
Identify the Bt toxin gene which controls the corn borer.
View Solution
Concept:
Bacillus thuringiensis produces specific crystalline endotoxins encoded by different cry genes, each targeting specific insect orders and agricultural pests.
Step 1: Specificity of Bt cry Genes in Crops:
The insecticidal crystal proteins produced by Bacillus thuringiensis are coded by genes called cry genes.
Different cry genes have distinct pest-specificity:
- cryIAc and cryIIAb genes: Code for delta-endotoxin proteins that effectively control the cotton bollworm (Helicoverpa armigera).
- cryIAb gene: Specifically codes for an endotoxin that controls the **corn borer** (Ostrinia nubilalis).
Step 2: Conclusion:
The specific gene utilized in biotechnology to control the corn borer pest in maize crops is cryIAb.
Therefore, Option (C) is the correct answer.
$\bullet$ Cotton bollworms $\rightarrow$ cryIAc and cryIIAb
$\bullet$ Corn borer $\rightarrow$ cryIAb
Select the incorrect statement about the insulin hormone produced by rDNA technique.
View Solution
Concept:
Human insulin consists of two short polypeptide chains: Chain A (21 amino acids) and Chain B (30 amino acids) linked together by disulfide bonds. Synthesizing active human insulin via rDNA required bypassing proinsulin processing.
Step 1: Biotechnology of Humulin Production (Eli Lilly, 1983):
In humans, insulin is synthesized as a pro-hormone containing an extra C-peptide (connecting peptide) of 33 amino acids that is cleaved during maturation to yield mature insulin.
Because E. coli cannot process eukaryotic proinsulin properly, the American pharmaceutical company Eli Lilly (1983) devised an innovative strategy:
1. They chemically synthesized two separate DNA oligonucleotide sequences corresponding to Chain A and Chain B of human insulin.
2. These sequences were inserted into plasmids and introduced into E. coli hosts to express Chain A and Chain B separately.
3. The individual A and B chains were extracted, purified, and then combined in vitro by creating disulfide bonds to produce mature, functional human insulin (Humulin).
4. The C chain is not added during the synthesis of mature recombinant insulin; mature insulin completely lacks the C-peptide.
Step 2: Analysis of the Statements:
- Statements (A), (B), and (D) are factually correct.
- Statement (C) is incorrect because chains A and B are linked by disulfide bonds, not combined with the C chain.
Step 3: Final Answer:
Hence, Option (C) is the incorrect statement.
$\bullet$ Proinsulin = A chain + B chain + C peptide.
$\bullet$ Mature functional insulin = A chain + B chain (joined by disulfide bonds, NO C-peptide).
$\bullet$ Eli Lilly produced Chains A and B separately in E. coli and joined them via disulfide bridges.
Intine of the pollen grain is made up of:
View Solution
Concept:
The microspore or pollen grain wall (sporoderm) in angiosperms has a two-layered protective wall architecture: an outer exine and an inner intine.
Step 1: Biochemical Composition of Pollen Wall Layers:
- Exine (Outer Wall): Composed of sporopollenin, one of the most resistant organic biopolymers known, which resists degradation by high temperatures, strong acids, alkalis, and enzymes. It exhibits apertures called germ pores where sporopollenin is absent.
- Intine (Inner Wall): The thin, continuous, flexible inner layer situated beneath the exine. It is composed biochemically of cellulose and pectin (pectocellulosic in nature). During pollen germination, the pollen tube emerges as an extension of the intine through a germ pore.
Step 2: Evaluating the Given Options:
- Option (A) Starch and agarose: Incorrect, agarose is derived from red seaweed.
- Option (B) Heparin and lignin: Incorrect, heparin is an animal anticoagulant.
- Option (C) Cellulose and pectin: Correct chemical composition of intine.
- Option (D) Glycogen and collagen: Incorrect, these are animal storage and structural macromolecules.
Step 3: Final Answer:
Thus, the intine of the pollen grain is made up of Cellulose and pectin (Option C).
$\bullet$ Exine: Outer, thick, sculpted layer made of sporopollenin.
$\bullet$ Intine: Inner, thin, continuous layer made of cellulose and pectin.
At maturity, a typical angiosperm embryo sac is:
View Solution
Concept:
In angiosperms, monosporic female gametophyte development (Polygonum type) involves three consecutive mitotic nuclear divisions of the functional megaspore, followed by cellular organization into an embryo sac.
Step 1: Cellular and Nuclear Organization of Mature Embryo Sac:
The functional megaspore undergoes three free-nuclear mitotic divisions, producing 8 nuclei.
Subsequent cytokinesis and cell wall formation result in the following organization:
1. Egg Apparatus (Micropylar end): Composed of 3 cells---one central egg cell and two flanking synergids (total = 3 cells, 3 nuclei).
2. Antipodal Cells (Chalazal end): Composed of 3 distinct cells that degenerate after fertilization (total = 3 cells, 3 nuclei).
3. Central Cell: A large single cell located in the middle containing two free polar nuclei (total = 1 cell, 2 nuclei).
Step 2: Counting Nuclei and Cells:
- Total number of cells = $3 \text{ (egg apparatus)} + 3 \text{ (antipodals)} + 1 \text{ (central cell)} = 7\text{ cells}$.
- Total number of nuclei = $3 \text{ (egg apparatus)} + 3 \text{ (antipodals)} + 2 \text{ (polar nuclei)} = 8\text{ nuclei}$.
Therefore, a typical mature female gametophyte (embryo sac) is 8-nucleate and 7-celled.
Step 3: Final Answer:
Hence, the correct option is (D).
$\bullet$ 7 Cells: 3 antipodals + 1 central cell + 2 synergids + 1 egg cell.
$\bullet$ 8 Nuclei: 3 in antipodals + 2 polar nuclei in central cell + 3 in egg apparatus.
Structure = 8-nucleate, 7-celled.
Which one of the following hormones is not involved in the process of parturition?
View Solution
Concept:
Parturition (the process of delivering the baby) is induced by a complex neuroendocrine mechanism initiated by signals originating from the fully developed fetus and the placenta.
Step 1: Endocrine Mechanism of Parturition:
- The fully developed fetus and placenta induce mild uterine contractions called the fetal ejection reflex.
- Fetal adrenal glands release cortisol, which stimulates placental production of estrogens.
- An increased estrogen-to-progesterone ratio enhances uterine sensitivity to contractile stimuli and stimulates oxytocin receptor expression on myometrial cells.
- The fetal ejection reflex triggers the maternal posterior pituitary gland to secrete oxytocin.
- Oxytocin acts on the myometrium, causing strong, rhythmic uterine contractions, driving the expulsion of the baby.
Step 2: Role of Prolactin:
- Prolactin is synthesized and secreted by the anterior pituitary gland. Its physiological function is the synthesis and production of milk within the mammary alveoli during pregnancy and lactation.
- It is not involved in the physical process or hormonal cascade of uterine contractions during parturition.
Step 3: Final Answer:
Thus, Prolactin is not involved in the process of parturition (Option D).
$\bullet$ Parturition: Oxytocin (strong uterine contractions), Estrogen, Fetal Cortisol, Relaxin.
$\bullet$ Lactation (Milk production): Prolactin.
$\bullet$ Milk ejection / letdown: Oxytocin.
Which one of the following structures provide energy for sperm movement?
View Solution
Concept:
A human spermatozoon is a microscopic, flagellated motile gamete composed of four distinct anatomical regions: head, neck, middle piece, and tail.
Step 1: Structure and Energetics of Sperm Parts:
- Head: Contains an elongated haploid nucleus housing the paternal genetic material, capped anteriorly by the acrosome.
- Acrosome: A cap-like structure derived from the Golgi complex containing hydrolytic enzymes (hyaluronidase, acrosin) that aid in penetrating the egg ovum coverings during fertilization.
- Middle Piece: Contains numerous spirally arranged mitochondria (known as Nebenkern) wrapped around the axial filament. These mitochondria undergo oxidative phosphorylation to produce abundant ATP (cellular energy). This energy powers the lashing movements of the flagellar tail, facilitating forward sperm motility essential for fertilization.
- Tail: A long flagellar axoneme that executes undulating movements powered by ATP supplied by the middle piece.
Step 2: Conclusion:
The middle piece houses the mitochondria that produce energy for sperm motility.
Hence, Option (C) is the correct answer.
$\bullet$ Acrosome: Lytic enzymes for ovum penetration.
$\bullet$ Head: Haploid paternal nucleus.
$\bullet$ Middle Piece: Mitochondria ('powerhouse') generating ATP for motility.
$\bullet$ Tail: Flagellum responsible for movement.
Which one of the following hormones is at its peak during ovulation?
View Solution
Concept:
The human menstrual cycle is regulated by gonadotropins (LH and FSH) from the anterior pituitary and ovarian steroids (estrogen and progesterone).
Step 1: Hormonal Dynamics during Ovulation (Mid-cycle, $\sim$14th Day):
During the proliferative (follicular) phase, growing ovarian follicles secrete estrogen. High levels of estrogen exert positive feedback on the pituitary gland.
Around the midpoint of the 28-day menstrual cycle (day 14), both LH and FSH attain peak secretion levels.
The rapid, dramatic surge in Luteinising Hormone (LH) concentration—commonly known as the LH surge—reaches its maximum peak.
This LH surge induces the rupture of the mature Graafian follicle, leading to the release of the secondary oocyte into the peritoneal cavity (ovulation).
Step 2: Evaluating the Other Hormones:
- Progesterone: Remains low during the follicular and ovulatory phases; it reaches its peak later, during the luteal (secretory) phase when secreted by the corpus luteum.
- Estrogen: Peaks just prior to ovulation, but the sharpest physiological peak directly causing ovulation is the LH surge.
- Luteinising Hormone (LH): The definitive hormone whose mid-cycle surge and maximal peak trigger ovulation.
Step 3: Final Selection:
Therefore, the hormone at its decisive peak during ovulation is Luteinising hormone (Option C).
$\bullet$ Day 14 (Ovulation): Peak LH (LH surge $\rightarrow$ follicle rupture) and peak FSH.
$\bullet$ Late Follicular Phase: Peak Estrogen.
$\bullet$ Mid-Luteal Phase (Day 21--22): Peak Progesterone (secreted by Corpus Luteum).
Drugs and Drug Abuse
Read the following passage carefully and answer the given questions.
Use of drugs has been on the rise, especially among the youth. This is really a cause of concern
as it could result in many harmful effects. Proper education and guidance would enable young
people to safeguard themselves against these dangerous behaviour patterns and follow healthy
lifestyles. The drugs which are commonly abused are opioids, cannabinoids and coca alkaloids.
The majority of these are obtained from flowering plants. Some are obtained from fungi. Various
plants, fruits and seeds having hallucinogenic properties have been used for hundreds of years
in folk-medicine, religious ceremonies and rituals all over the globe. When these are taken for a
purpose other than medicinal use or in amounts/frequency that impairs one’s physical, physiological
or psychological functions, it constitutes drug abuse. These days, some drugs are also being abused
by some sports-persons.
Question 41:
Chemically, heroin is:
View Solution
Concept:
Opioids are a class of psychoactive drugs that bind to specific opioid receptors present in the central nervous system and gastrointestinal tract.
Step 1: Chemical Nature of Heroin:
Heroin, commonly known on the street as 'smack', is a white, odorless, bitter crystalline compound.
Chemically, heroin is synthesized by the acetylation of morphine (extracted from the latex of the opium poppy, Papaver somniferum).
Acetylation of the two hydroxyl ($-OH$) groups of morphine with acetic anhydride yields diacetylmorphine.
Heroin acts as a depressant that slows down bodily functions.
Step 2: Conclusion:
Chemically, heroin is Diacetylmorphine.
Hence, Option (B) is the correct answer.
$\text{Morphine} \xrightarrow{\text{Acetylation}} \text{Diacetylmorphine (Heroin / Smack)}$.
Source: Latex of Papaver somniferum.
Which one of the following drugs is not produced from Cannabis
View Solution
Concept:
Cannabinoids are a group of chemicals that interact with cannabinoid receptors located primarily in the brain. They are derived from the hemp plant Cannabis sativa.
Step 1: Products Derived from Cannabis sativa:
The inflorescences, flower tops, leaves, and resins of the plant Cannabis sativa are processed in various combinations to produce:
- Marijuana
- Hashish
- Charas
- Ganja
These cannabinoids are typically inhaled or ingested orally and are known for affecting the cardiovascular system.
Step 2: Evaluating Smack:
- Smack is the street name for heroin (diacetylmorphine), which is an opioid derived from the opium poppy (Papaver somniferum), not from the Cannabis plant.
Step 3: Final Answer:
Therefore, Smack is not produced from the Cannabis plant (Option C).
$\bullet$ Cannabis sativa $\rightarrow$ Marijuana, Hashish, Charas, Ganja.
$\bullet$ Papaver somniferum $\rightarrow$ Morphine, Codeine, Smack (Heroin).
$\bullet$ Erythroxylum coca $\rightarrow$ Cocaine (Crack).
A very effective sedative and painkiller, which is very useful in patients who have undergone surgery is:
View Solution
Concept:
Certain opioid alkaloids obtained from medicinal plants possess strong central analgesic and sedative properties, making them medically valuable when prescribed under strict clinical control.
Step 1: Therapeutic Role of Morphine:
Morphine is a principal phenanthrene alkaloid obtained directly from the dried latex of unripened capsules of Papaver somniferum.
It acts as a potent central nervous system analgesic and sedative.
In clinical medicine, morphine is widely administered as an effective painkiller and sedative to alleviate severe, intractable post-operative pain in patients who have undergone major surgery.
Step 2: Evaluating Other Options:
- Heroin: A highly addictive illicit drug with no standard clinical use in post-operative recovery due to high dependence liability.
- Cocaine: A central nervous system stimulant that interferes with dopamine reuptake, producing euphoria; not used as a general sedative.
- Kanamycin: An aminoglycoside antibiotic used to treat bacterial infections, lacking analgesic or sedative properties.
Step 3: Final Answer:
Hence, Morphine is the correct therapeutic agent (Option A).
"Morphine is a very effective sedative and painkiller, and is very useful in patients who have undergone surgery."
Which one of the following drugs is not normally used as medicine to help patients cope with mental illnesses?
View Solution
Concept:
Psychotropic drugs modify mental activity, mood, and behaviour, and are legitimately prescribed to manage psychological disorders such as depression, anxiety, and insomnia.
Step 1: Identifying Psychotropic Medicines:
Medications that are legitimately prescribed to help patients cope with mental illness, depression, and insomnia include:
- Barbiturates: CNS depressants that act as hypnotics and sedatives.
- Benzodiazepines: Anxiolytics that treat anxiety, panic disorders, and sleeplessness.
- Amphetamines: Central stimulants used to treat attention deficit hyperactivity disorder (ADHD) and narcolepsy.
When abused without clinical indication, these substances impair normal mental and physical health.
Step 2: Evaluating Chloramphenicol:
Chloramphenicol is a broad-spectrum antibiotic that inhibits bacterial protein synthesis by targeting the 50S ribosomal subunit.
It is used to treat serious bacterial infections such as typhoid fever, meningitis, and cholera, and has no therapeutic application in treating mental illnesses.
Step 3: Final Answer:
Thus, Chloramphenicol is not used as a medicine for mental illnesses (Option A).
$\bullet$ Mental illness medications (frequently abused): Barbiturates, Amphetamines, Benzodiazepines, Lysergic acid diethylamides (LSD).
$\bullet$ Chloramphenicol: Broad-spectrum antibiotic for bacterial infections (e.g., typhoid).
Which one of the following plants does not have hallucinogenic properties?
View Solution
Concept:
Hallucinogens (psychedelics) are substances that alter sensory perception, mood, and thought processes, causing hallucinations. Many plants contain hallucinogenic tropane alkaloids.
Step 1: Evaluating Plants with Hallucinogenic Properties:
- Atropa belladonna: Contains tropane alkaloids (atropine, scopolamine) and possesses potent hallucinogenic properties.
- Datura: Contains hyoscyamine and scopolamine; well known in botanical toxicology for its strong hallucinogenic effects.
- Erythroxylum coca: Produces cocaine, an alkaloid that at higher dosages induces hallucinations and severe mental agitation.
Step 2: Evaluating Rauwolfia:
- Rauwolfia serpentina: A medicinal plant found in the Himalayan ranges that produces the active alkaloid reserpine.
- Reserpine is used medically to treat hypertension (high blood pressure) and psychotic agitation, but it does not possess recreational hallucinogenic properties.
Step 3: Final Answer:
Therefore, Rauwolfia is the plant that lacks hallucinogenic properties (Option D).
$\bullet$ Atropa belladonna
$\bullet$ Datura
$\bullet$ Erythroxylum coca (at high doses)
Rauwolfia serpentina yields reserpine (used for blood pressure control, non-hallucinogenic).
Mendelian Disorders
Read the following passage carefully and answer the given questions.
Genetic disorders may be grouped into two categories – Mendelian disorders and Chromosomal
disorders. Mendelian disorders are mainly determined by alteration or mutation in a single gene.
These disorders are transmitted to the offspring on the same lines as in the principle of inheritance.
The pattern of inheritance of such Mendelian disorders can be traced in a family by pedigree analy-
sis. Most common and prevalent Mendelian disorders are Haemophilia, Cystic fibrosis, Sickle-cell
anaemia, Colour blindness, Phenylketonuria, Thalassemia, etc. The Mendelian disorders may be
dominant or recessive. By pedigree analysis, one can easily understand whether the trait in question
is dominant or recessive. Similarly, the trait may also be linked to the sex chromosome or an autosome.
Question 46:
Which one of the following diseases is an autosomal dominant disorder?
View Solution
Concept:
Mendelian disorders follow predictable inheritance patterns (autosomal dominant, autosomal recessive, X-linked recessive, or X-linked dominant) depending on the chromosomal location of the gene and the dominance relationship of the mutant allele.
Step 1: Classifying the Inheritance Patterns of the Listed Disorders:
- Myotonic dystrophy: An autosomal dominant disorder characterized by progressive muscle wasting, weakness, and delayed muscle relaxation after contraction. Because it is dominant, a single copy of the mutant allele on an autosome is sufficient to manifest the disease phenotype.
- Sickle-cell anaemia: An autosomal recessive disorder caused by a mutation in the $HBB$ gene on chromosome 11.
- Phenylketonuria (PKU): An autosomal recessive metabolic disorder caused by a mutation in the phenylalanine hydroxylase gene on chromosome 12.
- Thalassemia: An autosomal recessive blood disorder causing reduced synthesis of globin polypeptide chains.
Step 2: Final Selection:
Among the given options, Myotonic dystrophy is the autosomal dominant disorder.
Hence, Option (B) is correct.
$\bullet$ Autosomal Dominant: Myotonic dystrophy, Huntington's chorea.
$\bullet$ Autosomal Recessive: Sickle-cell anaemia, Phenylketonuria, Thalassemia, Cystic fibrosis.
$\bullet$ X-linked Recessive: Haemophilia, Colour blindness.
Which one of the following substitutions is a cause of sickle-cell anaemia?
View Solution
Concept:
Sickle-cell anaemia is a classical example of a molecular point mutation (single base substitution) that alters the primary structure of a protein and changes its quaternary behavior under hypoxic conditions.
Step 1: Molecular Basis of Sickle-Cell Anaemia:
- In the normal $\beta$-globin gene ($Hb^A$), the sixth codon on the sense strand of DNA is GAG, which transcribes into mRNA as GAG, coding for the hydrophilic amino acid Glutamic acid (Glu).
- In the mutant sickle-cell gene ($Hb^S$), a point mutation substitutes a single base pair: adenine ($A$) is replaced by thymine ($T$) in DNA ($GAG \rightarrow GTG$).
- The resulting mutant mRNA codon becomes GUG, which translates into the hydrophobic amino acid Valine (Val) at the sixth position of the $\beta$-globin polypeptide chain.
- Under low oxygen tension, this substitution causes mutant hemoglobin ($Hb^S$) molecules to polymerize into long insoluble crystalline fibers, distorting biconcave red blood cells into rigid, sickle-shaped erythrocytes.
Step 2: Conclusion:
Sickle-cell anaemia is caused by the substitution of Glutamic acid by Valine at the sixth position of the beta-globin chain.
Hence, Option (C) is the correct statement.
$\bullet$ DNA: $GAG \rightarrow GTG$
$\bullet$ mRNA: $GAG \rightarrow GUG$
$\bullet$ Amino Acid (Position 6 of $\beta$-chain): $\text{Glutamic acid (Glu)} \rightarrow \text{Valine (Val)}$.
Which one of the followings is an example of an inborn error of metabolism and inherited as autosomal recessive trait?
View Solution
Concept:
Inborn errors of metabolism are congenital metabolic disorders caused by single-gene defects that result in deficient or non-functional enzymes along essential metabolic pathways.
Step 1: Biochemical Pathology of Phenylketonuria (PKU):
Phenylketonuria is an autosomal recessive inborn error of metabolism.
The affected individual lacks the functional hepatic enzyme phenylalanine hydroxylase.
Normally, this enzyme converts the dietary essential amino acid phenylalanine into tyrosine.
Due to enzyme deficiency, phenylalanine accumulates in the body and is converted into phenylpyruvic acid and related keto-derivatives.
Accumulation of these toxic metabolites in the brain causes severe mental retardation and impaired neurological development.
Excess phenylpyruvic acid is also excreted in the urine because of poor renal reabsorption.
Step 2: Evaluating the Other Disorders:
- Thalassemia: An autosomal recessive quantitative disorder of globin chain synthesis, not an enzymatic inborn error of metabolism.
- Haemophilia: An X-linked recessive blood clotting factor deficiency.
- Colour Blindness: An X-linked recessive defect in red/green retinal cone photoreceptors.
Step 3: Final Answer:
Hence, Phenylketonuria is the correct answer (Option A).
$\bullet$ Autosomal recessive inborn error of metabolism.
$\bullet$ Enzyme deficiency: Phenylalanine hydroxylase.
$\bullet$ Metabolic block: Phenylalanine $\not\rightarrow$ Tyrosine.
$\bullet$ Results in: Phenylpyruvate accumulation, mental retardation, excretion in urine.
α-thalassemia is controlled by the genes:
View Solution
Concept:
Thalassemia is an autosomal recessive blood disorder categorized into $\alpha$-thalassemia and $\beta$-thalassemia based on which globin chain of the hemoglobin tetramer ($\alpha_2\beta_2$) has reduced synthesis.
Step 1: Genetics of $\alpha$-Thalassemia vs $\beta$-Thalassemia:
- $\alpha$-Thalassemia: The synthesis of $\alpha$-globin chains is impaired. It is controlled by two closely linked genes, HBA1 and HBA2, located on **chromosome 16** of each parent (providing four alleles in a diploid cell). The condition is caused by the deletion or mutation of one or more of these four alleles; severity increases with the number of affected genes.
- $\beta$-Thalassemia: The synthesis of $\beta$-globin chains is impaired. It is controlled by a single gene, HBB, located on **chromosome 11** of each parent.
Step 2: Evaluating the Given Options:
- Option (A) & (D): Mention chromosome 11 and gene HBB, which controls $\beta$-thalassemia.
- Option (B): Uses incorrect gene designations (HAA1/HAB1).
- Option (C): Correctly specifies genes HBA1 and HBA2 on chromosome 16.
Step 3: Final Answer:
Therefore, $\alpha$-thalassemia is controlled by the genes HBA1 and HBA2 on chromosome 16 of each parent (Option C).
$\bullet$ $\alpha$-Thalassemia: Genes HBA1 and HBA2 on Chromosome 16 (4 alleles total).
$\bullet$ $\beta$-Thalassemia: Gene HBB on Chromosome 11 (2 alleles total).
Failure of segregation of chromatids during cell division results in:
View Solution
Concept:
Chromosomal aberrations and numerical abnormalities arise due to mitotic or meiotic non-disjunction during cell division cycles.
Step 1: Mechanism of Aneuploidy vs Polyploidy:
- Aneuploidy: The failure of sister chromatids (or homologous chromosomes) to segregate properly during anaphase of cell division (non-disjunction) results in the gain or loss of one or a few individual chromosomes in daughter cells ($2n+1, 2n-1, 2n+2$, etc.). Examples include Down syndrome (trisomy 21), Turner syndrome ($45, XO$), and Klinefelter syndrome ($47, XXY$).
- Polyploidy: The failure of cytokinesis (cytoplasmic division) after telophase results in an increase in a whole set of chromosomes ($3n, 4n$, etc.), a phenomenon frequently observed in plants.
Step 2: Evaluating the Other Terms:
- Apomixis: A form of asexual reproduction in flowering plants that mimics sexual reproduction by producing seeds without fertilization.
- Parthenocarpy: The development of fruit without fertilization, resulting in seedless fruits (e.g., banana).
Step 3: Final Answer:
The failure of segregation of chromatids during cell division results in Aneuploidy.
Hence, Option (B) is the correct answer.
$\bullet$ Failure of chromatid segregation $\rightarrow$ Aneuploidy (gain/loss of individual chromosomes: $2n \pm 1$).
$\bullet$ Failure of cytokinesis after telophase $\rightarrow$ Polyploidy (gain of entire sets of chromosomes: $3n, 4n$).
CUET UG 2026 Exam Pattern
| Parameter | Details |
|---|---|
| Exam Name | Common University Entrance Test (CUET UG) 2026 |
| Conducting Body | National Testing Agency (NTA) |
| Exam Mode | Computer-Based Test (CBT) |
| Exam Duration | 60 minutes per test |
| Total Sections | 3 (Languages, Domain Subjects, General Test) |
| Question Type | Multiple Choice Questions (MCQs) |
| Questions per Test | 50 questions (all compulsory) |
| Marking Scheme | +5 for correct, -1 for incorrect |
| Maximum Marks | 250 marks per test |
| Maximum Subject Choices | 5 subjects in total |
| Syllabus Base | Class 12 NCERT (mainly for Domain Subjects) |








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