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Bihar Board Class 12 Biology Question Paper 2026 with Solutions (Set I)

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Nidhi Bamnawat

| Updated On - Feb 12, 2026

Bihar Board Class 12 Biology Set I Question Paper 2026 with Solution pdf is available here for download. The exam was conducted by the Bihar School Examination Board on 3rd Feb 2026 in the first shift for a duration of 3 hours 15 min. Bihar Board Class 12 Biology exam is total of 70 Marks.

Bihar Board Class 12 Biology (Set I) Question Paper 2026 with Solution pdf

Bihar Board Class 12 Biology​ Question Paper 2026 with Solution pdf Download PDF Check Solutions
Bihar Board Class 12 Biology Question Paper 2026 with Solutions Set I

Section A

Question 1:

Who of the following scientists observed that restriction endonuclease can cut DNA strands in a particular shape?

  • (A) Herbert Boyer
  • (B) Stanley Cohen
  • (C) Watson
  • (D) Crick
Correct Answer: (A) Herbert Boyer
View Solution



Restriction endonucleases are enzymes that cut DNA at specific recognition sites.


Herbert Boyer, working with EcoRI, observed that this enzyme cuts DNA in a way that produces single-stranded overhanging ends known as ``sticky ends.''


This specific cutting shape allows DNA fragments from different sources to be joined together.


Stanley Cohen focused on plasmid isolation, while Watson and Crick proposed the double helix model.


Therefore, Herbert Boyer is the scientist credited with this specific observation regarding the shape of cut DNA strands.
Quick Tip: Boyer and Cohen collaborated to produce the first recombinant DNA in 1972, combining their expertise in enzymes and plasmids respectively.


Question 2:

Cutting DNA at a specific site is made possible by which enzyme?

  • (A) Restriction endonuclease
  • (B) Alkaline phosphatase
  • (C) DNA ligase
  • (D) Nuclease
Correct Answer: (A) Restriction endonuclease
View Solution



Restriction endonucleases are specific enzymes that recognize particular nucleotide sequences (palindromes) in DNA.


Upon recognition, they cut the sugar-phosphate backbone of both DNA strands at specific points.


Alkaline phosphatase removes phosphate groups, and DNA ligase joins DNA fragments together.


General nucleases cut DNA, but restriction endonucleases are required for cutting at \textit{specific sites.
Quick Tip: Restriction enzymes are often called ``molecular scissors'' because of their ability to cut DNA at precise locations.


Question 3:

In which of the following is the large volume of culture processed to produce appreciable quantities of products?

  • (A) Distillation unit
  • (B) Bioreactor
  • (C) PCR
  • (D) Electrophoresis unit
Correct Answer: (B) Bioreactor
View Solution



Biotechnological products (like enzymes, vaccines, proteins) often need to be produced on an industrial scale.


To achieve this, large volumes of culture (100-1000 litres) are processed in vessels called bioreactors.


Bioreactors provide optimal conditions (temperature, pH, oxygen, etc.) for the growth of the biological system to maximize product yield.


Distillation is a separation technique, PCR is for DNA amplification, and electrophoresis is for separating molecules.
Quick Tip: The most common type of bioreactor is the stirring type, which ensures mixing and oxygen availability throughout the reactor.


Question 4:

The base pillar of both strands of DNA double helix is made of

  • (A) Sugar -- Phosphate
  • (B) Sugar -- Nitrogenous base
  • (C) Nitrogenous base -- Phosphate
  • (D) Sugar -- Phosphate -- Nitrogenous base
Correct Answer: (A) Sugar -- Phosphate
View Solution



The DNA structure is often described as a twisted ladder or double helix.


The structural framework or ``backbone'' (base pillar) of the strands consists of alternating Deoxyribose sugar and Phosphate groups.


The nitrogenous bases project inwards from the sugar molecules to form the rungs or steps.


Therefore, the main structural support or pillar is the Sugar-Phosphate chain.
Quick Tip: Remember the backbone is hydrophilic (sugar-phosphate), while the core (bases) is hydrophobic.


Question 5:

What is essential for occurrence of the same kind of `sticky ends' of DNA fragments?

  • (A) To cut DNA by the same restriction endonuclease
  • (B) To cut DNA by different restriction endonucleases
  • (C) DNA amplification
  • (D) All of these
Correct Answer: (A) To cut DNA by the same restriction endonuclease
View Solution



To join two different DNA molecules (e.g., a vector and a source DNA), they must have complementary overhanging sequences known as sticky ends.


Restriction endonucleases cut at specific recognition sequences.


If both DNA molecules are cut with the \textit{same restriction endonuclease, they will produce identical, complementary sticky ends.


Cutting with different enzymes usually results in non-compatible ends that cannot easily pair.
Quick Tip: Compatible sticky ends allow hydrogen bonds to form between complementary bases, facilitating the action of DNA ligase.


Question 6:

During gel electrophoresis, fragments of DNA move towards which electrode?

  • (A) Cathode
  • (B) Anode
  • (C) Towards any electrode depending upon the size of DNA fragment
  • (D) None of these
Correct Answer: (B) Anode
View Solution



DNA molecules contain phosphate groups in their backbone, which carry a negative charge.


In an electric field established during gel electrophoresis, charged particles move towards the oppositely charged electrode.


Since DNA is negatively charged, it moves towards the positive electrode.


The positive electrode is called the Anode.
Quick Tip: Remember: \textbf{A}node \textbf{A}ttracts \textbf{A}nions (negative ions like DNA).


Question 7:

Which of the following is the sequence from where replication starts in cloning vector?

  • (A) \(tet^R\)
  • (B) \(amp^R\)
  • (C) ori
  • (D) more than one
Correct Answer: (C) ori
View Solution



Cloning vectors require a specific DNA sequence to initiate replication within the host cell.


This sequence is called the Origin of Replication, abbreviated as `ori'.


Any piece of DNA linked to this sequence can be replicated within the host cells.

\(tet^R\) and \(amp^R\) are selectable markers (antibiotic resistance genes), not the initiation site for replication.
Quick Tip: The copy number of the plasmid (how many copies exist in a cell) is also controlled by the `ori' sequence.


Question 8:

Which of the following cells are found in seminiferous tubules?

  • (A) Sertoli cells
  • (B) Interstitial cells
  • (C) Spermatogonia
  • (D) Both (A) and (C)
Correct Answer: (D) Both (A) and (C)
View Solution



The seminiferous tubules are the site of spermatogenesis in the testes.


The inner lining of each seminiferous tubule consists of two types of cells.


1. Male germ cells (Spermatogonia), which undergo division to form sperm.


2. Sertoli cells, which provide nutrition to the germ cells.


Interstitial cells (Leydig cells) are found in the interstitial spaces \textit{outside the seminiferous tubules.


Therefore, both Sertoli cells and Spermatogonia are found inside the tubules.
Quick Tip: Remember: "Inside" the tubule = Spermatogonia + Sertoli. "Outside" (Inter-stitial) = Leydig cells.


Question 9:

After how many months of pregnancy does the limbs and digits of foetus develop?

  • (A) One month
  • (B) Two months
  • (C) Three months
  • (D) Six months
Correct Answer: (B) Two months
View Solution



Human pregnancy development follows a specific timeline.


After one month of pregnancy, the embryo's heart is formed.


By the end of the second month of pregnancy, the foetus develops limbs and digits.


By the end of 12 weeks (first trimester), major organ systems are formed.


Therefore, limbs and digits are developed by the end of the second month.
Quick Tip: Sequence: 1st Month = Heart; 2nd Month = Limbs/Digits; 3rd Month = Major Organs/External Genitalia.


Question 10:

Which of the following placental hormones is produced during pregnancy?

  • (A) hCG
  • (B) hPL
  • (C) relaxin
  • (D) all of these
Correct Answer: (D) all of these
View Solution



The placenta acts as an endocrine tissue and produces several hormones.


It produces human chorionic gonadotropin (hCG) and human placental lactogen (hPL).


Relaxin is also produced in the later phase of pregnancy (primarily by the ovary, but the placenta also contributes).


In the context of NCERT biology, hCG, hPL, and relaxin are essentially produced/elevated specifically during pregnancy.


Therefore, all the options are correct.
Quick Tip: hCG, hPL, and relaxin are produced only during pregnancy, distinguishing them from other hormones like estrogen which are present otherwise but increase during pregnancy.


Question 11:

In which structure of sperm are the enzymes present which help in fertilisation of the ovum?

  • (A) Middle piece
  • (B) Neck
  • (C) Nucleus
  • (D) Acrosome
Correct Answer: (D) Acrosome
View Solution



The sperm head contains an elongated haploid nucleus.


The anterior portion of the sperm head is covered by a cap-like structure called the acrosome.


The acrosome is filled with enzymes (like hyaluronidase and acrosin) that help dissolve the egg membranes (zona pellucida) for fertilization.


The middle piece contains mitochondria, the neck connects head and middle piece, and the nucleus carries genetic material.
Quick Tip: The acrosome is derived from the Golgi apparatus during spermiogenesis.


Question 12:

Which of the following cells is haploid?

  • (A) Oogonia
  • (B) Primary oocyte
  • (C) Secondary oocyte
  • (D) All of these
Correct Answer: (C) Secondary oocyte
View Solution



Ploidy refers to the number of sets of chromosomes in a cell.


Oogonia are diploid (2n) germ cells that undergo mitosis.


Primary oocytes are also diploid (2n), formed from oogonia, and are arrested in Prophase I of meiosis.


The primary oocyte completes Meiosis I to form a large Secondary oocyte and a small polar body. Meiosis I is a reduction division, reducing the chromosome number by half.


Therefore, the Secondary oocyte is haploid (n).
Quick Tip: Any cell named "primary" (primary spermatocyte/oocyte) is generally diploid. Any cell named "secondary" or "spermatid/ovum" is haploid.


Question 13:

What is the main role of large amount of progesterone secreted by corpus luteum?

  • (A) maintenance of endometrium of uterus
  • (B) maintenance of perimetrium of uterus
  • (C) maintenance of myometrium of uterus
  • (D) None of these
Correct Answer: (A) maintenance of endometrium of uterus
View Solution



After ovulation, the ruptured follicle transforms into the corpus luteum.


The corpus luteum secretes large amounts of progesterone.


Progesterone is essential for maintaining the endometrium (the inner lining of the uterus).


A maintained endometrium is necessary for the implantation of the fertilized ovum and subsequent pregnancy events.
Quick Tip: Progesterone is often called the "pregnancy hormone" because its decline leads to menstruation (shedding of the endometrium).


Question 14:

What is it called when an embryo having more than eight blastomeres is transferred to the uterus for development?

  • (A) IVF
  • (B) Intrauterine Transfer
  • (C) Zygote Intra-fallopian Transfer
  • (D) ICSI
Correct Answer: (B) Intrauterine Transfer
View Solution



In Assisted Reproductive Technologies (ART), embryos can be transferred at different stages.


If the embryo has up to 8 blastomeres (zygote or early embryo), it is transferred into the fallopian tube (ZIFT - Zygote Intra Fallopian Transfer).


If the embryo has \textit{more than 8 blastomeres, it is transferred directly into the uterus.


This specific procedure is called Intra Uterine Transfer (IUT).
Quick Tip: Remember the cutoff: \(\le\) 8 blastomeres \(\rightarrow\) Tube (ZIFT). \(>8\) blastomeres \(\rightarrow\) Uterus (IUT).


Question 15:

If a person's blood group is AB, what would be his/her genotype?

  • (A) \(I^A i\)
  • (B) \(I^B i\)
  • (C) \(ii\)
  • (D) \(I^A I^B\)
Correct Answer: (D) \(I^A I^B\)
View Solution



ABO blood grouping is controlled by the gene I, which has three alleles: \(I^A\), \(I^B\), and \(i\).


Alleles \(I^A\) and \(I^B\) are co-dominant, meaning both express themselves when present together.


Blood group A results from \(I^A I^A\) or \(I^A i\).


Blood group B results from \(I^B I^B\) or \(I^B i\).


Blood group O results from \(ii\).


Blood group AB results when both dominant alleles are present: \(I^A I^B\).
Quick Tip: Codominance is the phenomenon where F1 progeny resembles both parents (e.g., AB blood group has both A and B antigens).


Question 16:

How many types of gametes would be formed if the genotype of plant is AaBbCc?

  • (A) Two
  • (B) Four
  • (C) Eight
  • (D) Sixteen
Correct Answer: (C) Eight
View Solution



The number of types of gametes produced by an organism is given by the formula \(2^n\), where \(n\) is the number of heterozygous gene loci.


The given genotype is AaBbCc.


Aa is heterozygous (1).


Bb is heterozygous (2).


Cc is heterozygous (3).


So, \(n = 3\).


Number of gametes = \(2^3 = 2 \times 2 \times 2 = 8\).
Quick Tip: If the genotype was AABbCc, \(n\) would be 2 (only Bb and Cc are heterozygous), so \(2^2=4\) gametes.


Question 17:

Which of the following is not a chromosomal disorder?

  • (A) Klinefelter's syndrome
  • (B) Turner's syndrome
  • (C) Down's syndrome
  • (D) Haemophilia
Correct Answer: (D) Haemophilia
View Solution



Chromosomal disorders are caused by excess, absence, or abnormal arrangement of one or more chromosomes (aneuploidy).


Klinefelter's (XXY), Turner's (XO), and Down's (Trisomy 21) are all chromosomal disorders.


Haemophilia is a Mendelian disorder caused by a mutation in a single gene (specifically, an X-linked recessive gene).


Therefore, Haemophilia is a gene disorder, not a chromosomal disorder in the context of aneuploidy.
Quick Tip: Mendelian disorders (like Sickle-cell anemia, Haemophilia) follow Mendel's laws of inheritance, whereas chromosomal disorders involve whole chromosome changes.


Question 18:

In which of the following animals does sex determination depend on number of chromosome sets?

  • (A) Birds
  • (B) Honeybees
  • (C) Grasshoppers
  • (D) Mammals
Correct Answer: (B) Honeybees
View Solution



Birds use the ZW-ZZ system (chromosomal).


Grasshoppers use the XX-XO system (chromosomal).


Mammals use the XX-XY system (chromosomal).


Honeybees use the Haplodiploid sex-determination system.


In this system, sex is determined by the number of sets of chromosomes an individual receives.


Fertilized eggs (diploid, 2 sets) develop into females (queens or workers), while unfertilized eggs (haploid, 1 set) develop into males (drones).
Quick Tip: This means male honeybees have no father and cannot have sons, but have a grandfather and can have grandsons.


Question 19:

Which of the following laws is based on monohybrid cross?

  • (A) Law of dominance
  • (B) Law of segregation
  • (C) Law of independent assortment
  • (D) Both (A) and (B)
Correct Answer: (D) Both (A) and (B)
View Solution



Mendel conducted monohybrid crosses (involving one character) and dihybrid crosses (involving two characters).


Based on his monohybrid cross experiments, he proposed two general rules: The Law of Dominance and The Law of Segregation.


The Law of Independent Assortment is based on dihybrid cross observations (segregation of one pair of traits is independent of the other).


Therefore, both Law of Dominance and Law of Segregation are based on monohybrid crosses.
Quick Tip: The Law of Segregation is universally applicable (no exceptions), while Dominance and Independent Assortment have exceptions (Co-dominance, Linkage).


Question 20:

Fossils are found in which type of rocks?

  • (A) Sedimentary rock
  • (B) Metamorphic rock
  • (C) Igneous rock
  • (D) All of these
Correct Answer: (A) Sedimentary rock
View Solution



Fossils are the preserved remains or traces of ancient organisms.


Igneous rocks form from molten magma, which would destroy any organic remains.


Metamorphic rocks form under high heat and pressure, which usually destroys or severely deforms fossils.


Sedimentary rocks form by the accumulation of sediment (sand, mud) in layers over time at relatively low temperatures.


This process allows organisms to be buried and preserved as fossils.
Quick Tip: Paleontology is the study of fossils, which provides direct evidence for evolution.


Question 21:

Which of the following are examples of homologous organs in animals?

  • (A) Eyes of octopus and human
  • (B) Sweet potato and potato
  • (C) Flippers of penguins and dolphins
  • (D) Heart of reptiles and mammals
Correct Answer: (D) Heart of reptiles and mammals
View Solution



Homologous organs are those that have the same fundamental structure and origin but perform different functions.


The hearts of vertebrates (fishes, amphibians, reptiles, birds, and mammals) follow the same basic structural plan, indicating a common ancestry.


Eyes of octopus and human are analogous (different structure, same function).


Sweet potato (root) and potato (stem) are analogous.


Flippers of penguins (birds) and dolphins (mammals) are analogous.


Therefore, the heart of reptiles and mammals is a homologous organ.
Quick Tip: Homology indicates Divergent Evolution (common ancestor), while Analogy indicates Convergent Evolution (same habitat/need).


Question 22:

Which of the following is one of the most resistant organic materials?

  • (A) Pectin
  • (B) Cellulose
  • (C) Sporopollenin
  • (D) Chitin
Correct Answer: (C) Sporopollenin
View Solution



Sporopollenin is found in the exine (outer layer) of pollen grains.


It is considered one of the most resistant organic materials known.


It can withstand high temperatures, strong acids, and strong alkalis.


No enzyme that degrades sporopollenin is so far known.


Pectin, Cellulose, and Chitin are degradable by specific enzymes (pectinase, cellulase, chitinase).
Quick Tip: The presence of sporopollenin allows pollen grains to be preserved as fossils for thousands of years.


Question 23:

Which of the following cells is present in pollen grain of angiospermic plants?

  • (A) Vegetative cell
  • (B) Generative cell
  • (C) Antipodal cell
  • (D) Both (A) and (B)
Correct Answer: (D) Both (A) and (B)
View Solution



A mature pollen grain in angiosperms typically contains two cells.


1. The Vegetative cell: It is bigger, has abundant food reserve, and a large irregularly shaped nucleus.


2. The Generative cell: It is small and floats in the cytoplasm of the vegetative cell.


Antipodal cells are found in the embryo sac (female gametophyte), not the pollen grain (male gametophyte).


Therefore, both vegetative and generative cells are present.
Quick Tip: In 60% of angiosperms, pollen is shed at this 2-celled stage. In others, the generative cell divides to form two male gametes (3-celled stage) before shedding.


Question 24:

In which of the following seeds is perisperm not found?

  • (A) Beet
  • (B) Black pepper
  • (C) Wheat
  • (D) Both (A) and (B)
Correct Answer: (C) Wheat
View Solution



Perisperm is the persistent, residual nucellus in the seed.


In most seeds, the nucellus is completely consumed during embryo development.


However, in some seeds like black pepper and beet, remnants of the nucellus persist as perisperm.


Wheat is a caryopsis where the nucellus is consumed; it has endosperm but no perisperm.


Therefore, perisperm is not found in wheat.
Quick Tip: Don't confuse Perisperm (nucellus remnant) with Pericarp (fruit wall derived from ovary wall).


Question 25:

In which of the following seeds more than one embryo is present?

  • (A) Citrus
  • (B) Mango
  • (C) Orange
  • (D) All of these
Correct Answer: (D) All of these
View Solution



The occurrence of more than one embryo in a seed is called polyembryony.


It is commonly observed in many Citrus species (like Orange, Lemon) and Mango varieties.


In these plants, the nucellar cells surrounding the embryo sac start dividing, protrude into the embryo sac, and develop into embryos.


Since Citrus (which includes Orange) and Mango both exhibit this, the correct answer covers all options.
Quick Tip: Nucellar polyembryony produces clones of the mother plant (apomixis), which is useful in horticulture to maintain hybrid characters.


Question 26:

What is cotyledon called in grass family?

  • (A) Plumule
  • (B) Radicle
  • (C) Hypocotyl
  • (D) Scutellum
Correct Answer: (D) Scutellum
View Solution



In the embryo of the grass family (monocots), there is only one cotyledon.


This single shield-shaped cotyledon is situated towards one side (lateral) of the embryonal axis.


It is called the Scutellum.


Plumule is the shoot tip, Radicle is the root tip, and Hypocotyl is the part of the axis below the cotyledon.
Quick Tip: The protective sheath covering the plumule is the Coleoptile, and the one covering the radicle is the Coleorhiza.


Question 27:

Which cell of the embryo sac becomes primary endosperm cell after triple fusion?

  • (A) Central cell
  • (B) Synergids
  • (C) Antipodals
  • (D) Egg cell
Correct Answer: (A) Central cell
View Solution



Double fertilization involves two fusion events: Syngamy and Triple Fusion.


In Triple Fusion, the second male gamete fuses with the two polar nuclei located in the large Central cell.


This fusion results in a triploid nucleus (Primary Endosperm Nucleus - PEN).


Following this fusion, the Central cell becomes the Primary Endosperm Cell (PEC) and develops into the endosperm.


The Egg cell becomes the Zygote. Synergids and Antipodals degenerate.
Quick Tip: Triple fusion involves three haploid nuclei (1 male gamete + 2 polar nuclei), resulting in a triploid (3n) tissue.


Question 28:

Which of the following plays an important role in guiding the pollen tube into the synergids?

  • (A) Antipodals
  • (B) Filiform apparatus
  • (C) Central cell
  • (D) Micropyle
Correct Answer: (B) Filiform apparatus
View Solution



The synergids have special cellular thickenings at the micropylar tip called the Filiform apparatus.


The Filiform apparatus secretes chemical substances that attract and guide the pollen tube into the synergid.


This ensures the male gametes are released in the vicinity of the egg cell and central cell.
Quick Tip: Remember: "Filiform" means thread-like. These finger-like projections increase surface area for secretion/absorption.


Question 29:

How many mega-biodiversity countries are there in the world?

  • (A) 25
  • (B) 34
  • (C) 14
  • (D) 17
Correct Answer: (D) 17
View Solution



Mega-biodiversity countries are a group of nations that harbor the majority of the Earth's species and high numbers of endemic species.


According to the World Conservation Monitoring Centre, there are 17 recognized mega-diverse countries.


These include countries like India, Brazil, China, Australia, etc.


34 is the number of Biodiversity Hotspots (initially 25, then increased).
Quick Tip: India is one of the 17 mega-biodiversity countries and has 2.4% of the world's land area but 8.1% of its species diversity.


Question 30:

Which region got comparatively shorter span of time for species evolution and diversification due to frequent glaciation?

  • (A) temperate region
  • (B) tropical region
  • (C) polar region
  • (D) glaciation occurred in all regions
Correct Answer: (A) temperate region
View Solution



Speciation is generally a function of time.


Tropical regions have remained relatively undisturbed for millions of years.


In contrast, temperate regions were subjected to frequent glaciations (Ice Ages) in the past.


This frequent disruption provided less evolutionary time for species diversification in temperate regions compared to the tropics.
Quick Tip: This is one of the three main hypotheses explaining why the tropics have greater biodiversity than temperate zones.


Question 31:

Which is called the lungs of the earth?

  • (A) Forest
  • (B) Temperate region
  • (C) Amazon rainforest
  • (D) Tropical Savannah
Correct Answer: (C) Amazon rainforest
View Solution



The Amazon rainforest is the largest tropical rainforest in South America.


It is estimated to produce about 20% of the total oxygen in the Earth's atmosphere through photosynthesis.


Because of this massive contribution to the planet's oxygen supply, it is popularly known as the "Lungs of the Planet" or "Lungs of the Earth".
Quick Tip: Despite being the "lungs", the Amazon is currently facing a huge threat from deforestation for soybean cultivation and cattle ranching.


Question 32:

Which of the following species became extinct due to over-exploitation?

  • (A) Passenger pigeon
  • (B) Giant Panda
  • (C) Nile Perch
  • (D) 200 species of cichlid fish
Correct Answer: (A) Passenger pigeon
View Solution



Humans have always depended on nature for food and shelter, but when need turns to greed, it leads to over-exploitation.


The Passenger pigeon (once very abundant in North America) and Steller's sea cow became extinct in the last 500 years due to over-exploitation by humans.


Nile Perch is an invasive species that caused the extinction of Cichlid fish.


Giant Panda is an endangered species, not extinct.
Quick Tip: "The Evil Quartet" of biodiversity loss: Habitat loss, Over-exploitation, Alien species invasions, and Co-extinctions.


Question 33:

Deriving countless direct economic benefits from nature by humans is .......... argument for conservation of biodiversity.

  • (A) narrowly utilitarian
  • (B) broadly utilitarian
  • (C) ethical
  • (D) all of these
Correct Answer: (A) narrowly utilitarian
View Solution



Arguments for conserving biodiversity are grouped into three categories:


1. Narrowly utilitarian: Focuses on direct economic benefits like food, firewood, fibre, tannin, medicines, etc.


2. Broadly utilitarian: Focuses on ecosystem services like oxygen production, pollination, and aesthetic pleasure.


3. Ethical: We have a moral duty to care for other species.


Since the question mentions "direct economic benefits", it corresponds to the narrowly utilitarian argument.
Quick Tip: "Narrow" = Tangible, economic goods. "Broad" = Intangible, ecosystem services.


Question 34:

Which of the following is not the primary consumer of an ecosystem?

  • (A) Zooplankton
  • (B) Cow
  • (C) Grasshopper
  • (D) Wolf
Correct Answer: (D) Wolf
View Solution



Primary consumers are herbivores that feed directly on producers (plants/algae).


Zooplankton feed on phytoplankton (producers) in aquatic ecosystems.


Cows feed on grass (producers).


Grasshoppers feed on plant parts (producers).


Wolves are carnivores; they feed on herbivores (primary consumers). Therefore, a wolf is a Secondary Consumer (or higher).
Quick Tip: Trophic levels: Producers (\(1^{st}\)) \(\rightarrow\) Primary Consumers (\(2^{nd}\)) \(\rightarrow\) Secondary Consumers (\(3^{rd}\)).


Question 35:

How much percentage of photo-synthetically active radiation is captured by plants?

  • (A) 50%
  • (B) 2 - 10%
  • (C) 10%
  • (D) 20%
Correct Answer: (B) 2 - 10%
View Solution



Photosynthetically Active Radiation (PAR) is the range of light wavelength (400-700 nm) used for photosynthesis.


Less than 50% of the total solar radiation incident on earth is PAR.


Plants capture only 2-10% of the PAR to perform photosynthesis and produce biomass.


This small amount of energy sustains the entire living world.
Quick Tip: Don't confuse: Plants capture 2-10% of PAR, which is roughly 1-5% of total incident solar radiation.


Question 36:

Which of the following is/are responsible for pneumonia in humans?

  • (A) Streptococcus pneumoniae
  • (B) Salmonella
  • (C) Haemophilus influenzae
  • (D) Both (A) and (C)
Correct Answer: (D) Both (A) and (C)
View Solution



Pneumonia is a bacterial disease that affects the alveoli (air sacs) of the lungs.


The most common causative agents are \textit{Streptococcus pneumoniae and \textit{Haemophilus influenzae.


Salmonella causes Typhoid.


Therefore, both (A) and (C) are correct.
Quick Tip: Symptoms include fever, chills, cough, and headache. In severe cases, lips and fingernails may turn gray to bluish (cyanosis).


Question 37:

What type of organism is the pathogen responsible for malaria?

  • (A) Virus
  • (B) Bacteria
  • (C) Protozoa
  • (D) Nematode
Correct Answer: (C) Protozoa
View Solution



Malaria is caused by a tiny parasite of the genus Plasmodium.


\textit{Plasmodium (e.g., \textit{P. vivax, P. falciparum) belongs to the kingdom Protista and is a Protozoan.


Viruses cause Dengue/Chikungunya, Bacteria cause Typhoid/Pneumonia, Nematodes cause Filariasis/Ascariasis.
Quick Tip: The vector for malaria is the female \textit{Anopheles mosquito, but the pathogen itself is a Protozoan.


Question 38:

Which of the following diseases is/are transmitted by insect vectors?

  • (A) Typhoid
  • (B) Malaria
  • (C) Filaria
  • (D) Both (B) and (C)
Correct Answer: (D) Both (B) and (C)
View Solution



Vectors are carriers that transmit pathogens from one host to another.


Malaria is transmitted by the female Anopheles mosquito.


Filariasis (Elephantiasis) is transmitted by the female \textit{Culex mosquito.


Typhoid is typically transmitted through contaminated food and water (fecal-oral route), though houseflies can act as mechanical carriers, they are not biological vectors in the same sense. The grouping of Malaria and Filaria is the standard example of vector-borne diseases.


Therefore, both (B) and (C) are correct.
Quick Tip: Another common vector-borne disease is Dengue, transmitted by the \textit{Aedes mosquito.


Question 39:

What kind of barrier is the interferon secreted by virus infected cells?

  • (A) Cytokine barrier
  • (B) Physical barrier
  • (C) Cellular barrier
  • (D) Physiological barrier
Correct Answer: (A) Cytokine barrier
View Solution



Innate immunity consists of four types of barriers: Physical, Physiological, Cellular, and Cytokine.


Virus-infected cells secrete proteins called Interferons.


Interferons protect non-infected cells from further viral infection.


This type of defense falls under Cytokine barriers.


Physical barriers are skin/mucus. Physiological are acid/tears. Cellular are WBCs like PMNL-neutrophils/macrophages.
Quick Tip: "Cyto-kine" means cell-movement/signaling. These are chemical messengers used for communication between cells.


Question 40:

Which of the following cells are responsible for humoral immune response?

  • (A) B-lymphocytes
  • (B) T-lymphocytes
  • (C) Killer lymphocytes
  • (D) Macrophages
Correct Answer: (A) B-lymphocytes
View Solution



The immune system has two main branches: Humoral (Antibody-mediated) and Cell-mediated.


Humoral immune response is mediated by antibodies present in body fluids (humors).


B-lymphocytes produce these antibodies.


T-lymphocytes are responsible for Cell-mediated immunity (CMI).


Therefore, B-lymphocytes are responsible for the humoral response.
Quick Tip: T-cells do not produce antibodies themselves but help B-cells to produce them.


Question 41:

What is taken by the patient to stop graft rejection?

  • (A) Immunosuppressant
  • (B) Vaccine
  • (C) To introduce readymade antibodies in the body
  • (D) All of these
Correct Answer: (A) Immunosuppressant
View Solution



Organ transplantation (grafting) often leads to rejection because the body's immune system recognizes the graft as "non-self".


This rejection is primarily mediated by Cell-Mediated Immunity (T-lymphocytes).


To prevent this, patients must take drugs that suppress the immune response for the rest of their lives.


These drugs are called Immunosuppressants (e.g., Cyclosporin A).
Quick Tip: Cyclosporin A is a bioactive molecule produced by the fungus Trichoderma polysporum.


Question 42:

Which of the following are primary lymphoid organs?

  • (A) Bone marrow and tonsils
  • (B) Bone marrow and thymus
  • (C) Lymph nodes
  • (D) Thymus and lymph nodes
Correct Answer: (B) Bone marrow and thymus
View Solution



Lymphoid organs are classified into Primary and Secondary.


Primary lymphoid organs are where immature lymphocytes differentiate into antigen-sensitive lymphocytes.


The two primary lymphoid organs in humans are the Bone Marrow and the Thymus.


Secondary lymphoid organs (Spleen, Lymph nodes, Tonsils, Peyer's patches) are sites where lymphocytes interact with antigens.
Quick Tip: Bone marrow is the site of origin for all blood cells, including both B and T lymphocytes. T-cells migrate to the Thymus to mature.


Question 43:

Which of the following viruses has reverse transcriptase enzyme?

  • (A) TMV
  • (B) Bacteriophage
  • (C) HIV
  • (D) All of these
Correct Answer: (C) HIV
View Solution



Reverse transcriptase is an enzyme that catalyzes the formation of DNA from an RNA template (Reverse Transcription).


This enzyme is characteristic of Retroviruses.


Human Immunodeficiency Virus (HIV) is a retrovirus containing RNA as genetic material.


Inside the host, HIV uses reverse transcriptase to make viral DNA from its RNA.


TMV is an RNA virus but not a retrovirus. Bacteriophages are usually DNA viruses.
Quick Tip: The process: RNA \(\xrightarrow{Reverse Transcriptase}\) DNA \(\xrightarrow{Transcription}\) RNA \(\xrightarrow{Translation}\) Protein. This is the central dogma modified for retroviruses.


Question 44:

Which of the following drugs interferes with transport of dopamine?

  • (A) Nicotine
  • (B) Hashish
  • (C) Morphine
  • (D) Cocaine
Correct Answer: (D) Cocaine
View Solution



Coca alkaloid or cocaine is obtained from the coca plant \textit{Erythroxylum coca.


It interferes with the transport of the neuro-transmitter dopamine.


Cocaine blocks the reuptake of dopamine in the synapse, leading to its accumulation and potent stimulating action on the CNS.


Nicotine stimulates adrenal glands. Hashish is a cannabinoid. Morphine is an opioid.
Quick Tip: Dopamine is the "feel-good" neurotransmitter. Excess dopamine causes a sense of euphoria and increased energy.


Question 45:

Which vitamin content increases due to conversion of milk into curd?

  • (A) Vitamin A
  • (B) Vitamin \(B_2\)
  • (C) Vitamin C
  • (D) Vitamin \(B_{12}\)
Correct Answer: (D) Vitamin \(B_{12}\)
View Solution



Microorganisms such as Lactobacillus (LAB - Lactic Acid Bacteria) grow in milk and convert it to curd.


During this process, the LAB produce acids that coagulate and partially digest the milk proteins.


It also improves the nutritional quality of the curd by increasing the content of Vitamin \(B_{12\).
Quick Tip: In our stomach, LAB also plays a very beneficial role in checking disease-causing microbes.


Question 46:

Which of the following is the blood cholesterol lowering agent?

  • (A) Statin
  • (B) Cyclosporin-A
  • (C) Antibiotic
  • (D) Streptokinase
Correct Answer: (A) Statin
View Solution



Statins are bioactive molecules produced by the yeast \textit{Monascus purpureus.


Statins act as blood-cholesterol lowering agents.


They function by competitively inhibiting the enzyme responsible for the synthesis of cholesterol.


Cyclosporin-A is an immunosuppressant. Streptokinase is a clot buster.
Quick Tip: Remember: "S" for Statin, "S" for Synthesis inhibition (of cholesterol).


Question 47:

Which of the following pairs is incorrect?

  • (A) Cyclosporin-A -- Produced by fungus
  • (B) Statin -- Produced by yeast
  • (C) Streptokinase -- Produced by bacteria
  • (D) Butyric acid -- Produced by fungus
Correct Answer: (D) Butyric acid -- Produced by fungus
View Solution



Let's verify each pair:

(A) Cyclosporin-A is produced by Trichoderma polysporum, which is a fungus. (Correct)

(B) Statin is produced by \textit{Monascus purpureus, which is a yeast (fungus). (Correct)

(C) Streptokinase is produced by \textit{Streptococcus, which is a bacterium. (Correct)

(D) Butyric acid is produced by \textit{Clostridium butyricum, which is a bacterium, NOT a fungus. (Incorrect)


Therefore, pair (D) is incorrect.
Quick Tip: Other acids: Citric acid (Fungus - \textit{Aspergillus), Acetic acid (Bacterium - Acetobacter), Lactic acid (Bacterium - Lactobacillus).


Question 48:

Which of the following cyanobacteria fixes atmospheric nitrogen?

  • (A) Nostoc
  • (B) Anabaena
  • (C) Oscillatoria
  • (D) All of these
Correct Answer: (D) All of these
View Solution



Cyanobacteria (blue-green algae) are autotrophic microbes.


Many of them can fix atmospheric nitrogen into organic forms.


\textit{Nostoc and \textit{Anabaena are the most common examples cited in textbooks, possessing specialized cells called heterocysts for N-fixation.


\textit{Oscillatoria is also a nitrogen-fixing cyanobacterium (though it lacks heterocysts, it fixes nitrogen under specific conditions).


Therefore, all the given examples are capable of nitrogen fixation.
Quick Tip: These biofertilizers are crucial in paddy fields (rice) to maintain soil fertility without chemical fertilizers.


Question 49:

With the help of what are butterfly caterpillars controlled?

  • (A) Pseudomonas
  • (B) Trichoderma
  • (C) Bacillus thuringiensis
  • (D) Baculovirus
Correct Answer: (C) Bacillus thuringiensis
View Solution



To control butterfly caterpillars, the bacteria \textit{Bacillus thuringiensis (Bt) is used.


Bt produces a protein crystal containing a toxic insecticidal protein (Bt toxin).


When the caterpillar eats the spores (sprayed on plants), the toxin is released in its gut, killing the larva.


\textit{Trichoderma controls root pathogens. Baculovirus attacks insects but is usually noted for species-specific control. Bt is the classic example for caterpillars.
Quick Tip: This microbial biocontrol agent was the basis for developing genetically modified Bt-crops (like Bt Cotton).


Question 50:

DNA cannot pass through cell membrane because DNA is .......... molecule.

  • (A) hydrophobic
  • (B) hydrophilic
  • (C) amphipathic
  • (D) neutral
Correct Answer: (B) hydrophilic
View Solution



The cell membrane is composed of a phospholipid bilayer, which has a hydrophobic (water-hating) interior.


Hydrophilic (water-loving) molecules cannot easily pass through this hydrophobic barrier.


DNA is a hydrophilic molecule due to its negatively charged sugar-phosphate backbone.


Therefore, it cannot pass through the cell membrane without assistance (like bacterial transformation treatments).
Quick Tip: This is why cells must be made "competent" (e.g., using Calcium ions) to take up DNA during transformation experiments.


Question 51:

Themrostable DNA polymerase used in PCR is isolated from which bacteria?

  • (A) Thermus aquaticus
  • (B) E. coli
  • (C) Salmonella
  • (D) Eubacteria
Correct Answer: (A) Thermus aquaticus
View Solution



PCR (Polymerase Chain Reaction) involves a denaturation step at very high temperatures (\(94^{\circ}\)C).


Normal DNA polymerase enzymes would denature and become inactive at this temperature.


Therefore, a thermostable DNA polymerase is required.


This enzyme, known as Taq polymerase, is isolated from the thermophilic bacterium Thermus aquaticus, which survives in hot springs.
Quick Tip: "Taq" is an abbreviation of \textbf{Thermus \textbf{aq}uaticus.


Question 52:

Why does Bt toxin protein not kill the Bacillus?

  • (A) Bt toxin protein exists as inactive protoxins
  • (B) Bt toxin protein is converted into harmless form by Bacillus
  • (C) One enzyme of Bacillus inactivates Bt toxin protein
  • (D) All of these
Correct Answer: (A) Bt toxin protein exists as inactive protoxins
View Solution



The Bt toxin protein produced by the bacterium \textit{Bacillus thuringiensis exists as an inactive crystal form.


This inactive form is called a protoxin.


Because it is inactive inside the bacterium, it does not kill the host \textit{Bacillus.


The protoxin is converted into the active toxic form only when it enters the alkaline pH of the insect gut.
Quick Tip: The activated toxin binds to the surface of midgut epithelial cells, creating pores that cause cell swelling and lysis.


Question 53:

Which technique was used to prevent infection of tobacco plants from Meloidegyne incognitia tapeworm?

  • (A) RNA interference
  • (B) Cry protein
  • (C) Somatic hybridisation
  • (D) Pesticide
Correct Answer: (A) RNA interference
View Solution



The nematode \textit{Meloidegyne incognita infects the roots of tobacco plants, reducing yield.


(Note: The question refers to it as a "tapeworm" but it is biologically a nematode/roundworm).


RNA interference (RNAi) was used to develop resistance in tobacco plants.


Specific genes were introduced into the plant that produced double-stranded RNA (dsRNA), which silenced the specific mRNA of the nematode, preventing its survival.
Quick Tip: RNAi is a natural cellular defense mechanism in all eukaryotic organisms.


Question 54:

Gene therapy was first used to remove the deficiency of which of the following enzymes?

  • (A) DNA polymerase
  • (B) Uracil DNA glycosylase
  • (C) Adenosine deaminase
  • (D) Beta galactosidase
Correct Answer: (C) Adenosine deaminase
View Solution



The first clinical gene therapy was given in 1990 to a 4-year-old girl.


She was suffering from Severe Combined Immunodeficiency (SCID).


This condition was caused by the deficiency of the enzyme Adenosine Deaminase (ADA), which is crucial for the immune system to function.
Quick Tip: ADA deficiency results in the dysfunction of T-lymphocytes.


Question 55:

In 1997 an American company got the patent rights for a new variety of Basmati developed by hybridizing Indian Basmati rice with which variety of rice?

  • (A) IR-36
  • (B) Pusa 1121
  • (C) Semi-dwarf variety of rice
  • (D) Indica rice
Correct Answer: (C) Semi-dwarf variety of rice
View Solution



In 1997, an American company (RiceTec) was granted a patent for "Basmati rice lines and grains".


They claimed to have invented a new variety.


This variety was actually derived by crossing Indian Basmati varieties with semi-dwarf varieties.


This is often cited as a case of biopiracy.
Quick Tip: Biopiracy is the use of bio-resources by multinational companies without proper authorization from the countries concerned.


Question 56:

Which human protein is found in milk of the first transgenic cow- Rosie?

  • (A) Alpha-1 antitrypsin
  • (B) Alpha lactalbumin
  • (C) Albumin
  • (D) p53
Correct Answer: (B) Alpha lactalbumin
View Solution



Rosie, the first transgenic cow, was produced in 1997.


She produced human protein-enriched milk (2.4 grams per litre).


The milk contained the human protein alpha-lactalbumin.


This made the milk a more nutritionally balanced product for human babies than natural cow milk.
Quick Tip: Alpha-1 antitrypsin is a transgenic product used to treat emphysema.


Question 57:

Which of the following was more like ape?

  • (A) Dryopithecus
  • (B) Ramapithecus
  • (C) Homo erectus
  • (D) Australopithecus
Correct Answer: (A) Dryopithecus
View Solution



About 15 million years ago, primates called \textit{Dryopithecus and \textit{Ramapithecus existed.


Fossil evidence suggests that they were hairy and walked like gorillas and chimpanzees.


\textit{Ramapithecus was more man-like in its dentition and posture.


\textit{Dryopithecus was more ape-like.
Quick Tip: Evolutionary sequence: Dryopithecus \(\rightarrow\) Ramapithecus \(\rightarrow\) Australopithecus \(\rightarrow\) Homo habilis.


Question 58:

Which of the following enzymes causes lactose to enter the bacterial cell?

  • (A) Beta galactosidase
  • (B) Permease
  • (C) Transacetylase
  • (D) Lactase
Correct Answer: (B) Permease
View Solution



The Lac operon in \textit{E. coli consists of three structural genes: \(z\), \(y\), and \(a\).


The \(y\) gene codes for the enzyme permease.


Permease increases the permeability of the bacterial cell membrane to \(\beta\)-galactosides (like lactose).


This allows lactose to enter the cell to be utilized.
Quick Tip: A very low level of expression of the lac operon must be present in the cell all the time, otherwise, lactose cannot enter.


Question 59:

Who stated that in double stranded DNA sum of purine nitrogenous bases is equal to the sum of pyrimidine nitrogenous bases?

  • (A) Miescher
  • (B) Watson and Crick
  • (C) Wilkins
  • (D) Chargaff
Correct Answer: (D) Chargaff
View Solution



Erwin Chargaff proposed a set of rules for the base composition of DNA.


One of Chargaff's rules states that in a double-stranded DNA, the ratio between Adenine and Thymine and between Guanine and Cytosine is constant and equals one.


This implies that the total amount of Purines (A + G) equals the total amount of Pyrimidines (T + C).
Quick Tip: Recall: A pairs with T (2 H-bonds) and G pairs with C (3 H-bonds).


Question 60:

A typical nucleosome is composed of

  • (A) 200 base pairs of DNA
  • (B) Histone octamer
  • (C) Both (A) and (B)
  • (D) Acidic protein
Correct Answer: (C) Both (A) and (B)
View Solution



In eukaryotes, DNA packaging involves the wrapping of DNA around histone proteins.


A unit of this packaging is called a nucleosome.


It consists of a core of eight histone molecules (histone octamer) charged positively.


Negatively charged DNA (approximately 200 base pairs) is wrapped around this octamer.


Therefore, a nucleosome is composed of both the DNA and the histone octamer.
Quick Tip: Nucleosomes constitute the repeating unit of a structure in the nucleus called chromatin, appearing as "beads-on-string".


Question 61:

What did Hershey and Chase conclude on the basis of their experiments?

  • (A) Protein from the virus enters the bacteria
  • (B) DNA from the virus enters the bacteria
  • (C) RNA from the virus enters the bacteria
  • (D) Transforming principle
Correct Answer: (B) DNA from the virus enters the bacteria
View Solution



Hershey and Chase (1952) conducted experiments using bacteriophages labeled with radioactive Phosphorus (\(^{32}P\), labels DNA) and radioactive Sulfur (\(^{35}S\), labels Protein).


They found that only \(^{32}P\) entered the bacterial cells, while \(^{35}S\) remained outside.


This proved that DNA is the genetic material that enters the bacteria from the virus, not protein.
Quick Tip: This experiment provided unequivocal proof that DNA is the genetic material.


Question 62:

What is the process of removal of introns and mutual joining of exons called?

  • (A) Splicing
  • (B) Tailing
  • (C) Capping
  • (D) Transcription
Correct Answer: (A) Splicing
View Solution



In eukaryotic transcription, the primary transcript (hnRNA) contains both coding sequences (exons) and non-coding sequences (introns).


Before it becomes functional mRNA, the introns must be removed.


The process where introns are removed and exons are joined together in a defined order is called Splicing.


Capping and Tailing are modifications at the 5' and 3' ends respectively.
Quick Tip: Splicing represents the dominance of the RNA world in ancient times.


Question 63:

Which of the following statements is incorrect about the genetic code?

  • (A) The codon is triplet
  • (B) It is unambiguous and specific
  • (C) It is degenerate
  • (D) All 64 codons code for amino acids
Correct Answer: (D) All 64 codons code for amino acids
View Solution



The genetic code consists of 64 codons (triplets of bases).


However, 3 of these codons (UAA, UAG, UGA) are Stop Codons or nonsense codons.


These stop codons do not code for any amino acid; they signal the end of translation.


Therefore, only 61 codons code for amino acids. Statement (D) is incorrect.
Quick Tip: The Start Codon is usually AUG, which codes for Methionine.


Question 64:

What is the raw material for decomposition called?

  • (A) Humus
  • (B) Inorganic substance
  • (C) Detritus
  • (D) Organic substance
Correct Answer: (C) Detritus
View Solution



Decomposition is the breakdown of complex organic matter.


The input or raw material for this process consists of dead plant parts (leaves, bark, flowers) and dead animal remains, including fecal matter.


This collective dead organic matter is technically called Detritus.


Humus is a product of decomposition, not the raw material.
Quick Tip: Detritus is the beginning of the Detritus Food Chain (DFC).


Question 65:

What is secondary consumer called?

  • (A) Herbivores
  • (B) Primary carnivores
  • (C) Secondary carnivores
  • (D) Consumers
Correct Answer: (B) Primary carnivores
View Solution



In a food chain, trophic levels are arranged as follows:


1. Producers (Plants)


2. Primary Consumers (Herbivores that eat plants)


3. Secondary Consumers (Carnivores that eat herbivores)


Since Secondary Consumers eat herbivores, they are the first level of carnivores, hence called Primary Carnivores.
Quick Tip: A Tertiary consumer would be a Secondary Carnivore.


Question 66:

What are those called which obtain energy and nutrition from breaking dead organic matter or detritus?

  • (A) Saprotrophs
  • (B) Heterotrophs
  • (C) Detritivores
  • (D) Parasite
Correct Answer: (C) Detritivores
View Solution



The question refers to organisms that feed on detritus (dead organic matter).


While Saprotrophs (decomposers) also break down dead matter, the specific terminology for organisms that ingest and break down detritus (fragmentation) is Detritivores (e.g., earthworms).


The Hindi text specifically uses the word "Khandan" (Fragmentation), which links directly to the function of Detritivores in the NCERT syllabus.


Therefore, Detritivores is the specific answer for organisms breaking down detritus.
Quick Tip: Detritivores (like earthworms) physically break down litter, increasing the surface area for bacterial/fungal (saprotrophic) action.


Question 67:

Which of the following is the pollinating agent of Ophrys?

  • (A) Wasp
  • (B) Humming bird
  • (C) A species of bee
  • (D) Bat
Correct Answer: (C) A species of bee
View Solution



The Mediterranean orchid \textit{Ophrys employs a mechanism called "sexual deceit" to ensure pollination.


One petal of its flower bears an uncanny resemblance to the female of the bee in size, color, and markings.


The male bee is attracted to the flower, mistaking it for a female, and attempts to copulate (pseudocopulation), picking up pollen in the process.


NCERT explicitly states "female of the bee" and "male bee".
Quick Tip: This is a classic example of co-evolution, where the flower's appearance has evolved to match the pollinator.


Question 68:

Which of the following processes decreases population density?

  • (A) Immigration
  • (B) Death
  • (C) Emigration
  • (D) Both (B) and (C)
Correct Answer: (D) Both (B) and (C)
View Solution



Population density is influenced by four basic processes:


1. Natality (Births): Increases density.


2. Mortality (Deaths): Decreases density.


3. Immigration: Individuals coming in. Increases density.


4. Emigration: Individuals leaving. Decreases density.


Therefore, both Death (Mortality) and Emigration contribute to a decrease in population density.
Quick Tip: Formula: \(N_{t+1} = N_t + [(B+I) - (D+E)]\).


Question 69:

Which of the following was used to control the cactus that had been introduced in Australia?

  • (A) Moth
  • (B) Calotropis
  • (C) Phytophagous insects
  • (D) Virus
Correct Answer: (A) Moth
View Solution



In the 1920s, the prickly pear cactus was introduced into Australia and became invasive, spreading rapidly.


The biological control method used to check this invasion was the introduction of a cactus-feeding predator.


This predator was a moth named \textit{Cactoblastis cactorum.


The moth larvae feed on the cactus, successfully controlling its population.
Quick Tip: Biological control uses natural predators/parasites to control pests, avoiding chemical pesticides.


Question 70:

What is lichen an example of?

  • (A) Mutualism
  • (B) Competition
  • (C) Parasitism
  • (D) Commensalism
Correct Answer: (A) Mutualism
View Solution



Lichens represent an intimate symbiotic relationship between a fungus and a photosynthesizing partner (algae or cyanobacteria).


The fungus provides shelter and absorbs water/minerals for the algae.


The algae prepares food (via photosynthesis) for the fungus.


Since both partners benefit from this association, it is a classic example of Mutualism.
Quick Tip: Lichens are excellent pollution indicators; they do not grow in areas with high air pollution (\(SO_2\)).

Section B

Question 1:

What is RNA interference?

Correct Answer:
View Solution



RNA interference (RNAi) is a biological process in which RNA molecules inhibit gene expression or translation.


It takes place in all eukaryotic organisms as a method of cellular defense.


It involves the silencing of a specific mRNA due to a complementary double-stranded RNA (dsRNA) molecule that binds to and prevents translation of the mRNA (silencing).
Quick Tip: The source of this complementary RNA could be from an infection by viruses having RNA genomes or mobile genetic elements (transposons).


Question 2:

What is explant?

Correct Answer:
View Solution



An explant is any part of a plant (such as a piece of root, shoot, leaf, or stem) that is excised (cut out) from the original plant.


It is then grown in a test tube under sterile conditions in a special nutrient medium to regenerate a whole new plant.


This process relies on the property of cellular totipotency.
Quick Tip: The capacity to generate a whole plant from any cell/explant is called totipotency.


Question 3:

What is latitudinal gradient in biodiversity?

Correct Answer:
View Solution



The latitudinal gradient in biodiversity refers to the well-established pattern where species diversity decreases as we move away from the equator towards the poles.


The tropics (low latitudes) harbor more species than temperate or polar regions (high latitudes).


This is due to higher solar energy availability, less seasonal variation, and a longer evolutionary time without major disruptions (like glaciations) in the tropics.
Quick Tip: Colombia (near the equator) has nearly 1400 species of birds, while New York (\(41^{\circ}\)N) has 105.


Question 4:

What do you mean by `the Evil Quartet'?

Correct Answer:
View Solution



'The Evil Quartet' is a term used to describe the four major causes of biodiversity loss.


These causes are:


1. Habitat loss and fragmentation (the most important cause).


2. Over-exploitation (harvesting species faster than they can reproduce).


3. Alien species invasions (introduction of non-native species).


4. Co-extinctions (when a species becomes extinct, the plant and animal species associated with it also become extinct).
Quick Tip: Remember the acronym HOAC for the four causes.


Question 5:

Write three advantages of predation.

Correct Answer:
View Solution



1. Energy Transfer: Predators act as conduits for energy transfer across trophic levels (from prey to higher consumers).


2. Population Control: Predators keep prey populations under control, preventing them from becoming unstable or destroying the ecosystem.


3. Species Diversity: Predators help maintain species diversity in a community by reducing the intensity of competition among competing prey species.
Quick Tip: Example: The starfish Pisaster maintains diversity in rocky intertidal communities; its removal leads to extinction of invertebrates due to competition.


Question 6:

What is humus and what effect does it have on soil?

Correct Answer:
View Solution



Humus is a dark-coloured, amorphous, organic substance formed by the decomposition of dead organic matter (detritus).


Effects on Soil:


1. It is highly resistant to microbial action and undergoes decomposition at an extremely slow rate.


2. Being colloidal in nature, it serves as a reservoir of nutrients.


3. It improves soil texture and aeration.
Quick Tip: Humification leads to accumulation of humus; mineralization further releases inorganic nutrients from humus.


Question 7:

Define recognition sequence.

Correct Answer:
View Solution



A recognition sequence is a specific base sequence of DNA that is recognized by a restriction endonuclease enzyme.


The enzyme binds to the DNA at this specific site and cuts the two strands.


These sequences are typically palindromic, meaning they read the same on the two strands when orientation is kept the same (e.g., 5' GAATTC 3').
Quick Tip: For EcoRI, the recognition sequence is 5'-GAATTC-3'.


Question 8:

Define transformation.

Correct Answer:
View Solution



Transformation is a process by which a cell takes up naked DNA fragments (genetic material) from its surrounding medium and incorporates them into its own genome.


This results in the alteration of the phenotype of the recipient cell.


This phenomenon was first discovered by Frederick Griffith in \textit{Streptococcus pneumoniae bacteria.
Quick Tip: In biotechnology, cells are made "competent" (e.g., using Calcium chloride) to facilitate transformation.


Question 9:

Write four salient features of Human Genome Project.

Correct Answer:
View Solution



1. The human genome contains 3164.7 million nucleotide bases.


2. The average gene consists of 3000 bases, but sizes vary greatly (largest is Dystrophin).


3. The total number of genes is estimated at 30,000.


4. Almost all (99.9%) nucleotide bases are exactly the same in all people.


(Additional: Less than 2% of the genome codes for proteins; Chromosome 1 has the most genes (2968) and Y has the fewest (231)).
Quick Tip: The project was a 13-year effort coordinated by the US Department of Energy and the National Institutes of Health.


Question 10:

What is the example of divergent evolution?

Correct Answer:
View Solution



Divergent evolution occurs when the same structure develops along different directions due to adaptations to different needs (Homologous organs).


Examples:


1. Forelimbs of mammals: Whales (swimming), Bats (flying), Cheetahs (running), and Humans (grasping) share the same anatomical structure of bones (humerus, radius, ulna, etc.) but function differently.


2. Thorns of \textit{Bougainvillea and tendrils of \textit{Cucurbita (both are modified axillary buds).
Quick Tip: Remember: Homology = Common Ancestry = Divergent Evolution.


Question 11:

What is clot bluster? What is its source?

(Note: The question text has a typo "bluster", it should be "buster".)

Correct Answer:
View Solution



A clot buster is an enzyme used in medicine to dissolve blood clots that have formed in the blood vessels, particularly to treat myocardial infarction (heart attack).


The specific enzyme is Streptokinase.


Source: It is produced by the bacterium \textit{Streptococcus and modified by genetic engineering.
Quick Tip: It works by activating plasminogen to plasmin, which breaks down the fibrin mesh of the clot.


Question 12:

What are cannabinoids?

Correct Answer:
View Solution



Cannabinoids are a group of chemicals that interact with specific cannabinoid receptors present principally in the brain.


They are obtained chiefly from the inflorescences of the plant \textit{Cannabis sativa (Hemp plant).


Examples include marijuana, hashish, charas, and ganja.


They primarily affect the cardiovascular system of the body.
Quick Tip: Natural cannabinoids are obtained from the flower tops, leaves, and resin of the plant.


Question 13:

What is autogamy? What adaptations are there in flower for it?

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Autogamy is a type of self-pollination where the transfer of pollen grains takes place from the anther to the stigma of the \textit{same flower.


Adaptations for Autogamy:


1. Synchrony: Pollen release and stigma receptivity must happen simultaneously.


2. Proximity: The anther and stigma should lie close to each other so that self-pollination can occur.


3. Cleistogamy: Production of flowers which do not open at all (e.g., \textit{Viola, \textit{Oxalis), ensuring complete autogamy.
Quick Tip: Autogamy results in genetically identical offspring (pure lines) but can lead to inbreeding depression.


Question 14:

What is emasculation?

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Emasculation is the removal of anthers from the flower bud of a bisexual flower before the anther dehisces (releases pollen).


This is done using a pair of forceps.


It is a crucial step in artificial hybridization experiments to prevent self-pollination in the female parent, ensuring that only the desired pollen is used for fertilization.
Quick Tip: Emasculated flowers must be "bagged" immediately to prevent contamination by unwanted pollen.


Question 15:

What are sex-chromosomes and autosomes?

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Sex-chromosomes (Allosomes): These are the chromosomes involved in the determination of the sex of an individual. In humans, these are the X and Y chromosomes (the \(23^{rd}\) pair).


Autosomes: These are the chromosomes that determine somatic (body) characters and are not involved in sex determination. In humans, there are 22 pairs of autosomes, which are the same in both males and females.
Quick Tip: Human Karyotype: 46 chromosomes = 44 Autosomes + 2 Sex Chromosomes (XX for female, XY for male).


Question 16:

What is Turner's syndrome?

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Turner's syndrome is a chromosomal disorder in females caused by the absence of one of the X chromosomes.


Karyotype: 45, XO (Monosomy of sex chromosome).


Symptoms: Such females are sterile as ovaries are rudimentary. They lack secondary sexual characters, have short stature, and may have a webbed neck.
Quick Tip: This is an example of Aneuploidy arising from non-disjunction of chromosomes during gamete formation.


Question 17:

What is the use of Widal test?

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The Widal test is a serological diagnostic test.


It is specifically used for the diagnosis of Typhoid fever.


It works by detecting the presence of antibodies in the patient's serum that react against the antigens of the bacterium Salmonella typhi.
Quick Tip: Typhoid is characterized by sustained high fever (\(39^{\circ\) to \(40^{\circ}\)C), stomach pain, and constipation.


Question 18:

What are cellular barriers in innate immunity?

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Cellular barriers constitute the internal defense line of the body's innate immunity.


They consist of specialized cells capable of phagocytosing (eating) and destroying microbes.


These include:


1. Leukocytes (WBCs): Polymorphonuclear leukocytes (PMNL-neutrophils) and Monocytes in the blood.


2. Natural Killer (NK) cells: A type of lymphocyte.


3. Macrophages: Found in tissues.
Quick Tip: Phagocytosis involves engulfing the pathogen and destroying it using lysosomal enzymes.


Question 19:

Draw well labelled diagram of sperm.

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(Description of the diagram components):


A well-labelled diagram of a human sperm should show the following four main parts enveloped by a plasma membrane:


1. Head: Contains an elongated haploid Nucleus and is capped by the Acrosome (containing enzymes).


2. Neck: Short region connecting head and middle piece.


3. Middle Piece: Contains numerous Mitochondria (powerhouse) to provide energy for motility.


4. Tail: A long flagellum that facilitates sperm movement.
Quick Tip: The mitochondria in the middle piece are spirally arranged and are sometimes called the Nebenkern.


Question 20:

Name male accessory glands with their roles.

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The male accessory glands include:


1. Seminal Vesicles (A pair).


2. Prostate Gland (Single).


3. Bulbourethral Glands (A pair, also called Cowper's glands).


Roles:


- Their secretions constitute the Seminal Plasma, which is rich in fructose, calcium, and certain enzymes.


- This fluid provides nutrition and a medium for the transport and motility of sperms.


- Secretions of the bulbourethral glands specifically help in the lubrication of the penis.
Quick Tip: Fructose acts as the fuel source for sperm motility.


Question 21:

Compare endosperm and perisperm. Describe the development of endosperm.

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Comparison between Endosperm and Perisperm:


1. Origin: Endosperm is formed by the fusion of the second male gamete with the secondary nucleus (Triple Fusion). Perisperm is the persistent remnant of the nucellus.


2. Ploidy: Endosperm is generally triploid (3n). Perisperm is diploid (2n).


3. Occurrence: Endosperm is found in almost all angiosperm seeds (consumed in non-albuminous, persistent in albuminous). Perisperm is found in only a few seeds like black pepper and beet.


Development of Endosperm:


1. The Primary Endosperm Cell (PEC) undergoes repeated nuclear divisions to give rise to free nuclei. This stage is called Free-nuclear endosperm.


2. Subsequently, cell wall formation occurs, and the endosperm becomes cellular.


3. Example: In coconut, the water is free-nuclear endosperm (made of thousands of nuclei) and the surrounding white kernel is cellular endosperm.
Quick Tip: Endosperm development precedes embryo development to ensure a food supply is ready for the developing embryo.


Question 22:

Write short notes on any two of the following:

(a) Intra-uterine devices

(b) Structure of testis

(c) Incomplete dominance

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(a) Intra-uterine Devices (IUDs):

These are contraceptive devices inserted by doctors or expert nurses into the uterus through the vagina.

They are of three types:

1. Non-medicated IUDs (e.g., Lippes loop): They increase phagocytosis of sperms within the uterus.

2. Copper releasing IUDs (e.g., CuT, Cu7): Copper ions suppress sperm motility and fertilizing capacity.

3. Hormone releasing IUDs (e.g., Progestasert, LNG-20): They make the uterus unsuitable for implantation and the cervix hostile to sperms.


(b) Structure of Testis:

Testes are the primary sex organs in males, situated within the scrotum.

Each testis has about 250 compartments called testicular lobules.

Each lobule contains 1-3 highly coiled seminiferous tubules where sperms are produced.

The tubule is lined by Spermatogonia (germ cells) and Sertoli cells (nurse cells).

The regions outside the tubules (interstitial spaces) contain Leydig cells, which secrete androgens.


(c) Incomplete Dominance:

It is a phenomenon where the \(F_1\) hybrid exhibits a phenotype intermediate between the two parents, rather than resembling the dominant parent.

A classic example is the flower colour in \textit{Antirrhinum (Snapdragon).

When a true-breeding red flower (RR) is crossed with a white flower (rr), the \(F_1\) offspring (Rr) are Pink.

The phenotypic and genotypic ratio in the \(F_2\) generation is identical: 1:2:1 (Red:Pink:White).
Quick Tip: Incomplete dominance disproves the blending theory because the parental traits (Red and White) reappear in the \(F_2\) generation.


Question 23:

Write short notes on any two of the following:

(a) Chromosomal theory of inheritance

(b) Nucleosome

(c) Transfer RNA

Correct Answer:
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(a) Chromosomal Theory of Inheritance:

Proposed by Sutton and Boveri (1902).

They noted that the behavior of chromosomes was parallel to the behavior of genes (Mendelian factors).

Key points:

1. Chromosomes occur in pairs (homologous pairs), just like alleles.

2. They segregate during gamete formation (meiosis) such that only one of each pair is transmitted to a gamete.

3. Independent pairs of chromosomes segregate independently of each other.

Sutton united chromosomal segregation with Mendelian principles to formulate this theory.


(b) Nucleosome:

It is the structural unit of chromatin in eukaryotes.

It consists of a core of eight histone proteins (Histone Octamer: two each of H2A, H2B, H3, and H4).

Negatively charged DNA is wrapped around this positively charged octamer (about 200 base pairs of DNA per nucleosome).

Under an electron microscope, chromatin looks like ``beads-on-string'' due to these nucleosomes.


(c) Transfer RNA (tRNA):

tRNA acts as an adapter molecule that reads the genetic code and brings the specific amino acids to the ribosome during translation.

Structure:

1. Anticodon loop: Has bases complementary to the code on mRNA.

2. Amino acid acceptor end: Binds to the specific amino acid at the 3' end.

In 2D, it looks like a clover-leaf; in 3D, it looks like an inverted 'L'.
Quick Tip: tRNA was specifically termed "sRNA" (soluble RNA) before its function was fully understood.


Question 24:

Describe sewage treatment.

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Sewage treatment is carried out in Sewage Treatment Plants (STPs) in two main stages:


1. Primary Treatment (Physical):

It involves the physical removal of particles (large and small) through filtration and sedimentation.

- Filtration: Floating debris is removed by sequential filtration.

- Sedimentation: The grit (soil and small pebbles) is removed by settling. The solid that settles is the Primary Sludge, and the supernatant is the effluent.


2. Secondary Treatment (Biological):

- The primary effluent is passed into large aeration tanks where it is agitated and air is pumped in.

- This allows the vigorous growth of aerobic microbes into flocs (bacteria associated with fungal filaments).

- These microbes consume the major part of the organic matter, significantly reducing the BOD (Biochemical Oxygen Demand).

- Once BOD is reduced, the effluent is passed into a settling tank where the flocs settle as Activated Sludge.

- A small part of this sludge is pumped back to the aeration tank (inoculum), and the rest goes to Anaerobic Sludge Digesters.

- Here, anaerobic bacteria digest the bacteria and fungi, producing a mixture of gases (Biogas - Methane, \(H_2S\), \(CO_2\)).
Quick Tip: The lower the BOD of wastewater, the lower is its polluting potential.


Question 25:

Explain various components of ecosystem.

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An ecosystem consists of two main categories of components:


1. Abiotic Components (Non-living):

- Physical factors: Temperature, Light, Water, Soil, pH, Topography.

- Chemical factors: Inorganic nutrients (C, N, P, K) and organic substances (proteins, carbohydrates) present in the soil/water.


2. Biotic Components (Living):

- Producers (Autotrophs): Organisms that synthesize their own food using sunlight (e.g., Green plants, Algae, Phytoplankton).

- Consumers (Heterotrophs): Organisms that depend on others for food.

(a) Primary Consumers (Herbivores): Feed on producers (e.g., Insects, Deer).

(b) Secondary Consumers (Carnivores): Feed on herbivores (e.g., Frogs).

(c) Tertiary Consumers: Top carnivores (e.g., Lion, Tiger).

- Decomposers (Saprotrophs): Microorganisms (Bacteria, Fungi) that break down dead organic matter and release nutrients back into the soil, closing the cycle.
Quick Tip: Functional aspects of an ecosystem include Productivity, Decomposition, Energy Flow, and Nutrient Cycling.


Question 26:

Describe in brief the features of cloning vector.

Correct Answer:
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A cloning vector is a DNA molecule used to carry foreign genetic material into another cell. Key features include:
1. Origin of Replication (ori):

This is a sequence from where replication starts. It controls the copy number of the linked DNA. A vector with a high copy number ori is preferred for making many copies.


2. Selectable Marker:

It helps in identifying and eliminating non-transformants and selectively permitting the growth of transformants. Common examples are antibiotic resistance genes (e.g., \(amp^R\), \(tet^R\)).


3. Cloning Sites (Restriction Sites):

The vector needs a single recognition site for a specific restriction enzyme to insert the foreign DNA. Multiple sites would generate several fragments, complicating the process.


4. Size:

The vector should ideally be small in size to facilitate easy isolation and minimize the chance of breaking during purification.
Quick Tip: Plasmids (like pBR322) and Bacteriophages are the most commonly used vectors.

*The article might have information for the previous academic years, please refer the official website of the exam.

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