CUET 2023 Answer Key Biology- Download Slot-wise Answer Key with Solutions PDF

Step 1: Understand the organisms mentioned.


Yeast reproduces asexually by budding, a form of cell division.

Amoeba reproduces by binary fission, where one cell divides into two—this is purely cell division.

Paramecium reproduces asexually by binary fission too.

Fruitfly (Drosophila) reproduces sexually, so cell division is not the sole mode of reproduction.

Step 2: Eliminate incorrect options.


(D) is incorrect since Fruitfly doesn't reproduce by cell division alone. Quick Tip: In unicellular organisms like Amoeba and Paramecium, cell division is both a growth and a reproductive process. In multicellular organisms like Fruitfly, reproduction involves gametes and not simple cell division.

Step 1: Understand vegetative propagation in Agave.

Agave reproduces vegetatively through structures called bulbils, which grow from the inflorescence axis. These bulbils eventually fall to the ground and grow into new plants.

Step 2: Why not other options?

Leaf buds: Found in Bryophyllum, not Agave.

Offset: Common in water hyacinth, not Agave.

Rhizome: Found in ginger and turmeric.

Hence, the correct structure in Agave is a bulbil. Quick Tip: Bulbils are miniature vegetative buds that form on the floral axis, particularly in Agave, enabling asexual reproduction.

Step 1: Understand the structure of the mature embryo sac.

The mature embryo sac in angiosperms is formed by the development of the ovule. It contains 8 nuclei, with a total of 7 cells.

Step 2: Breakdown of the embryo sac structure.

The 8 nuclei in the embryo sac include:

1 egg cell,

2 synergid cells,

1 central cell (which has two polar nuclei),

3 antipodal cells.

These 8 nuclei are arranged in 7 cells: the egg cell, synergids, central cell, and antipodals.

Step 3: Conclusion.

Thus, the mature embryo sac is 8 nucleated and 7 celled.

The correct answer is: \[ \boxed{8 nucleated, 7 celled}. \] Quick Tip: In angiosperms, the mature embryo sac typically contains 8 nuclei, but only 7 cells due to the central cell containing two polar nuclei.

The pollen mother cell (PMC), also known as the microspore mother cell, undergoes meiosis to produce microspores. Each microspore develops into a pollen grain, which contains male gametes.

Step 1: Meiosis in PMC.

A pollen mother cell (PMC) is diploid (2n), and it undergoes meiosis to produce four haploid microspores. These microspores eventually develop into pollen grains.

Step 2: Pollen grain development.

Each microspore undergoes mitosis to form a mature pollen grain. In the case of male gametes, each pollen grain contains two male gametes (sperm cells).

Step 3: Number of male gametes.

Since one PMC produces four microspores (through meiosis), and each microspore develops into a pollen grain that contains two male gametes, the maximum number of male gametes produced from one pollen mother cell is:
\[ 4 \times 2 = 8. \]

Thus, the correct answer is:
\[ \boxed{8}. \] Quick Tip: Remember that meiosis in the pollen mother cell produces four microspores, and each microspore gives rise to two male gametes, leading to a total of eight male gametes per PMC.

Step 1: Identify the hormones secreted by the placenta.

The placenta is a temporary endocrine organ that secretes several hormones to maintain pregnancy. These include:

Estrogens: Help in maintaining pregnancy and preparing for delivery.

Progestogens (especially progesterone): Support the endometrium and suppress uterine contractions.

hCG (Human chorionic gonadotropin): Maintains the corpus luteum and progesterone secretion in early pregnancy.

Step 2: Evaluate Relaxin.

Relaxin is secreted by the corpus luteum, not the placenta. It helps in the relaxation of the pelvic ligaments and cervix during childbirth, but its source is not placental. Quick Tip: Placenta secretes estrogens, progesterone, hCG, and hPL, but not Relaxin — which comes from the corpus luteum.

Step 1: Recall Leydig cell function.

Leydig cells are located in the interstitial spaces between seminiferous tubules and are responsible for synthesizing and secreting androgens like testosterone.

Step 2: Analyze the given options.

Options (1) and (2) are true regarding Leydig cells.

Option (3) is unrelated; immune-competent cells are typically separate.

Option (4) describes a function more relevant to Sertoli cells, not Leydig cells. Quick Tip: Leydig cells secrete testosterone, essential for secondary male characteristics. Spermiogenesis is supported by Sertoli cells inside seminiferous tubules.

Step 1: Understand the popularity and practicality of contraceptive methods.

In India, Intrauterine Devices (IUDs) like Copper-T are the most widely accepted method due to their long-term effectiveness and affordability.

Step 2: Eliminate less popular methods.

(1) Vasectomy is permanent and used less frequently.

(2) Coitus interruptus is unreliable.

(4) Implants are less common due to cost/accessibility. Quick Tip: IUDs are a common and reversible method of contraception that are especially promoted in public health services in India.

Step 1: Understand mechanism of oral pills.

Oral contraceptives contain synthetic estrogen and progesterone, which inhibit ovulation and prevent implantation by altering the uterine lining.

Step 2: Eliminate other options.

(2) Lippes loop is a physical barrier in the uterus.

(3) Condoms are external barriers.

(4) Tubectomy is surgical sterilization. Quick Tip: Oral pills prevent both ovulation and implantation by manipulating hormone levels.

Step 1: Understand chromosomal abnormality.

An extra copy of chromosome 21 results in Trisomy 21, which causes Down's syndrome.

Step 2: Eliminate incorrect options.

(2) Klinefelter: XXY male genotype

(3) Turner's: XO female genotype

(4) Haemophilia: X-linked disorder, not due to chromosome number Quick Tip: Down’s syndrome is caused by Trisomy 21 — having 3 copies of chromosome 21 instead of 2.

Step 1: Define genotypes.

Let \( A \) = axial (dominant), \( a \) = terminal (recessive). Heterozygous plants have genotype \( Aa \).

Step 2: Perform cross of Aa × Aa.
\[ Genotypic ratio: AA : Aa : aa = 1 : 2 : 1 \]

Step 3: Identify phenotypes.

Only \( aa \) produces terminal flowers. So, 1 out of 4 or 25% will show terminal flowers. Quick Tip: Use Punnett squares for Mendelian monohybrid crosses to easily determine phenotypic ratios.

Step 1: Understanding the mutation.

Sickle cell anaemia is a genetic disorder caused by a mutation in a single base pair in the hemoglobin gene. This leads to a change in the shape of red blood cells.

Step 2: Analyzing the options.

Sickle cell anaemia is caused by a single nucleotide polymorphism (SNP) where a substitution of adenine for thymine occurs, changing glutamic acid to valine.

Thalassemia, colour blindness, and cystic fibrosis are also genetic disorders, but they are not primarily caused by a single base pair change.

Step 3: Conclusion.

Thus, the correct answer is: \[ \boxed{Sickle cell anaemia}. \] Quick Tip: Sickle cell anaemia is a classical example of a disease caused by a single base pair mutation in the hemoglobin gene.

Step 1: Understand the DNA fingerprinting process.

The steps involved in DNA fingerprinting are:

1. Isolation and digestion of DNA (C): First, the DNA is isolated from the sample and digested by restriction enzymes.

2. Separation of DNA fragments by electrophoresis (A): The digested DNA fragments are separated by size using gel electrophoresis.

3. Transferring of separated DNA fragments to synthetic membranes (E): The separated DNA fragments are then transferred to a membrane.

4. Hybridisation using labelled VNTR probe (B): A labelled probe is used to hybridize with specific regions of the DNA.

5. Detection of hybridised DNA fragments by autoradiography (D): Finally, the hybridized fragments are detected using autoradiography.


Step 2: Conclusion.

Thus, the correct order of steps is: \[ \boxed{(C), (A), (E), (B), (D)}. \] Quick Tip: DNA fingerprinting involves several key steps including isolation, digestion, separation, hybridisation, and detection.

Step 1: Understand the functions of each polymerase.

RNA Polymerase II (A) transcribes hn RNA, which includes mRNA.

RNA Polymerase I (B) is responsible for the transcription of 28 s rRNA.

Ribosome (C) consists of 23 s rRNA in bacteria and 5 s rRNA in eukaryotes.

RNA Polymerase III (D) transcribes 5 s rRNA.


Step 2: Match List-I with List-II.

(A)-(II) \( \Rightarrow \) RNA Polymerase II transcribes hn RNA.

(B)-(I) \( \Rightarrow \) RNA Polymerase I transcribes 28 s rRNA.

(C)-(IV) \( \Rightarrow \) Ribosomes are composed of 23 s rRNA in bacteria.

(D)-(III) \( \Rightarrow \) RNA Polymerase III transcribes 5 s rRNA.


Thus, the correct answer is: \[ \boxed{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}. \] Quick Tip: RNA polymerases transcribe different types of RNA. Be familiar with the functions of each RNA polymerase and the RNA they transcribe.

Step 1: Understand VNTR.

VNTR (Variable Number Tandem Repeat) refers to regions in DNA where a short nucleotide sequence is repeated. The size of these repeats can vary.

Step 2: Analyze the options.

VNTRs typically vary in size from 0.1 to 20 kb (kilobases), making Option (1) the correct answer.

The other options are not correct as VNTRs do not typically range from only a few base pairs or kilobases.

Thus, the correct answer is: \[ \boxed{0.1 to 20 kb}. \] Quick Tip: VNTRs are highly variable in size and are used in genetic fingerprinting for individual identification.

Step 1: Understand the brain capacity of Homo erectus.

Homo erectus is known for having a significantly larger brain compared to earlier hominids. Their brain size ranged from about 600 cc to 1,100 cc.

Step 2: Analyze the options.

Option (1), 900 cc, falls within this range and is the correct answer.

The other options are either too low or do not match the generally accepted size for Homo erectus.

Thus, the correct answer is: \[ \boxed{900 \, cc}. \] Quick Tip: When studying human evolution, consider the brain size as a critical factor in distinguishing species like Homo erectus from earlier hominids.

In Miller's famous experiment (1953), he tested the hypothesis of chemical evolution by simulating the conditions of early Earth in a laboratory setting.

Step 1: Understand the experimental setup.

Miller filled a closed flask with gases that were thought to be present in Earth's early atmosphere. He then introduced electric sparks to simulate lightning and observed the formation of amino acids and other organic molecules.

Step 2: Identify the gases used in the experiment.

The gases that Miller used in his experiment were methane (CH\(_4\)), hydrogen (H\(_2\)), ammonia (NH\(_3\)), and water vapor (H\(_2\)O). This mixture was believed to represent the early Earth's atmosphere.

Step 3: Correct answer.

The correct mixture of gases used in the experiment is: CH\(_4\), H\(_2\), NH\(_3\) and H\(_2\)O.

Thus, the correct answer is:

CH\(_4\), H\(_2\), NH\(_3\) and H\(_2\)O
Quick Tip: Miller’s experiment used gases like methane, hydrogen, and ammonia to simulate early Earth conditions.

Hardy-Weinberg equilibrium describes a population that is not evolving. There are five conditions that must be met for Hardy-Weinberg equilibrium:
1. No mutation: There are no changes in the genetic material.
2. No migration: No gene flow between populations.
3. Large population size: No genetic drift due to a large enough population.
4. Random mating: All individuals have an equal chance to mate with others.
5. No natural selection: All individuals have an equal chance of surviving and reproducing.

Step 1: Understand the factors affecting Hardy-Weinberg equilibrium.

Genetic drift: A random change in allele frequencies, especially in small populations, which can affect Hardy-Weinberg equilibrium.

Mutation: Changes in the DNA sequence, introducing new alleles, can affect the equilibrium.

Random mating: This condition ensures that every individual has an equal chance of mating with any other individual in the population, and it is necessary for maintaining Hardy-Weinberg equilibrium.


However, random mating is one of the conditions needed to maintain Hardy-Weinberg equilibrium. Thus, the correct answer, based on the phrasing of the question, is not an affecting factor since it’s a required condition.

Thus, the correct answer is:
\[ \boxed{Random mating}. \] Quick Tip: To maintain Hardy-Weinberg equilibrium, random mating is a condition that must be satisfied, rather than a factor that disrupts it.

We are given diseases in List - I and their pathogens in List - II. We need to match each disease with its causative organism.

Step 1: Match Pneumonia with its pathogen.

- Pneumonia is caused by Haemophilus influenzae (IV).

Step 2: Match Typhoid with its pathogen.

- Typhoid is caused by Salmonella typhi (I).

Step 3: Match Malaria with its pathogen.

- Malaria is caused by Plasmodium falciparum (II).

Step 4: Match Filariasis with its pathogen.

- Filariasis is caused by Wuchereria bancrofti (III).

Thus, the correct matching is:
\[ \boxed{(A)-(IV), (B)-(I), (C)-(II), (D)-(III)}. \] Quick Tip: Be familiar with common diseases and their causative organisms to effectively match them in questions like these.

Elephantiasis is a disease caused by parasitic infections, primarily Wuchereria bancrofti, which affect the lymphatic system. The symptoms include extreme swelling and deformities due to the obstruction of lymph flow.

Step 1: Understand the symptoms of Elephantiasis.

Elephantiasis leads to:

(A) Inflammation of the lower limb: This is a common symptom caused by the blockage of lymphatic vessels, leading to the accumulation of fluid in the lower limbs.

(D) Gross deformities of genital organs: The accumulation of fluid can cause swelling and deformities in the genital organs due to lymphatic damage.


Step 2: Analyze other options.

(B) Scaly lesions on the scalp: This is not a typical symptom of elephantiasis. It may be seen in other skin conditions, but not specifically in elephantiasis.

(C) Anaemia and blockage of intestinal passage: These symptoms are also not directly associated with elephantiasis.

Thus, the correct answer is:
\[ \boxed{(A) \, and \, (D) \, Only}. \] Quick Tip: Elephantiasis typically affects the limbs and genital organs, causing severe swelling and deformities due to the lymphatic obstruction.

Antibodies, also known as immunoglobulins, are protein molecules that play a crucial role in the immune system by recognizing and neutralizing foreign invaders like bacteria and viruses.

Step 1: Structure of Antibodies.

An antibody molecule consists of four peptide chains:

Two heavy chains (larger in size).

Two light chains (smaller in size).


Step 2: How the chains are connected.

The four chains are held together by disulfide bonds, which are covalent bonds formed between the sulfur atoms of cysteine residues in the chains. These bonds provide structural stability and are crucial for the correct 3D folding of the antibody molecule.

Step 3: Eliminate other bond types.

Peptide bonds: These link individual amino acids together in a protein, but they do not link different chains of the antibody.

Phosphodiester bonds: These bonds link nucleotides in DNA or RNA, not in protein molecules.

Glycosidic bonds: These bonds link sugar molecules, not involved in the structure of antibodies.

Thus, the correct answer is:
\[ \boxed{Disulfide bond}. \] Quick Tip: Disulfide bonds help maintain the structural integrity of proteins, including antibodies, by linking different chains.

Step 1: Understand the varieties of crops.

Wheat (A): Pusa Shubhra is a variety of wheat.

Cauliflower (B): Pusa Shubhra is known as a cauliflower variety.

Cowpea (C): Pusa Komal is the correct variety for cowpea.

Chilli (D): Pusa Sadabahar is a variety of chili.


Thus, the correct answer is: \[ \boxed{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}. \] Quick Tip: Familiarize yourself with popular crop varieties and their associated traits. Crop varieties are essential for breeding programs.

Step 1: Understand the wheat varieties.

Atlas-66 is a wheat variety that is known for its high protein content. It is commonly used for breeding programs aiming to improve the protein quality of wheat.

Step 2: Analyze the options.

Sonalika and Jaya are popular wheat varieties, but they are not specifically known for high protein content.

Ratna is another wheat variety, but it does not have higher protein content compared to Atlas-66.


Step 3: Conclusion.
Thus, the correct answer is: \[ \boxed{Atlas-66}. \] Quick Tip: High-protein wheat varieties like Atlas-66 are often bred to improve the nutritional value of wheat-based products.

Step 1: Understand the process of inducing resistance.

Mutation is the process used to introduce changes in the genetic makeup of an organism, resulting in resistance to diseases like yellow mosaic virus and powdery mildew in mung bean. This is achieved by inducing genetic changes that enhance disease resistance.

Step 2: Analyze the options.

Hybridisation involves crossing two different varieties or species, but it is not the method used to induce disease resistance in mung beans.

Recombination refers to genetic material exchange, which is related but not the main method to induce resistance in this context.

Micropropagation is a method for plant propagation, not specifically used to induce disease resistance.


Step 3: Conclusion.
Thus, the correct answer is: \[ \boxed{Mutation}. \] Quick Tip: Inducing resistance to diseases through mutation is a common method in plant breeding for improving crop protection.

Step 1: Understand the products associated with each microorganism.

Aspergillus niger (A) is known for the production of Citric acid (III).

Trichoderma polysporum (B) is used for producing Ethanol (II).

Lactobacillus (C) is used for producing Lactic acid (IV).

Saccharomyces cerevisiae (D) is used for producing Cyclosporin A (I).

Step 2: Conclusion.

Thus, the correct matching is: \[ \boxed{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}. \] Quick Tip: Microorganisms are utilized for the production of various industrially important substances like citric acid, ethanol, lactic acid, and cyclosporin.

Step 1: Understand the role of bio-fertilizers.

Azospirillum is a genus of bacteria that acts as a bio-fertilizer in paddy fields by fixing nitrogen from the air and making it available to plants.

Step 2: Analyze the options.

Nostoc is another nitrogen-fixing organism, but Azospirillum is widely used for bio-fertilization in paddy fields.

Rhizobium is mainly used with legumes and Azotobacter is used in other crops, but Azospirillum is the best answer for paddy fields.

Step 3: Conclusion.
Thus, the correct answer is: \[ \boxed{Azospirillum}. \] Quick Tip: Azospirillum is used as a bio-fertilizer in paddy fields due to its ability to fix nitrogen, which is essential for rice growth.

Step 1: Understand the process of dough fermentation.

When dough is left to ferment (e.g., while making bread), yeast cells present in the dough metabolize sugars. This is an anaerobic process called fermentation.

Step 2: Products of yeast fermentation.

Yeast fermentation produces: \[ Glucose \longrightarrow Ethanol (Alcohol) + Carbon dioxide gas \]

Step 3: Identify which product causes puffing.

It is the carbon dioxide gas that forms bubbles and gets trapped in the dough, causing it to rise and appear puffed.

Step 4: Eliminate other options.

(2) Alcohol is also produced but evaporates during baking and does not cause puffing.

(3) Acetic acid is not a typical product here.

(4) Oxygen gas is not produced in anaerobic fermentation. Quick Tip: In yeast-based fermentation, carbon dioxide causes the dough to rise by forming gas bubbles, giving a puffed-up appearance.

Step 1: Understand gel electrophoresis.

Gel electrophoresis separates DNA fragments based on size. Negatively charged DNA moves through an agarose gel when electric field is applied.

Step 2: Sequence of events.

(B) The DNA fragments move toward the anode under electric field.

(D) Smaller fragments move faster, leading to size-based separation.

(A) DNA fragments are stained with ethidium bromide (EtBr), a dye that intercalates with DNA.

(C) The gel is exposed to UV light to visualize the DNA bands.

Step 3: Verify the correct order.

(B), (D), (A), (C) matches the actual process. Quick Tip: Always apply electric field first, then stain the DNA and visualize under UV. The separation is based on size—smaller fragments move farther.

Step 1: Purpose of recombinant DNA technology.

To produce many copies of the target DNA, it is inserted into a cloning vector (e.g., a plasmid).

Step 2: Role of origin of replication (ori).

The ori is a sequence in the plasmid from where DNA replication starts. A high copy number ori ensures the plasmid replicates multiple times within the host cell, leading to many copies of target DNA.

Step 3: Eliminate wrong options.

(1) Reducing cloning sites limits flexibility.

(3) Selectable markers help in identifying transformed cells but don't increase copy number.

(4) More cloning sites allow multiple insertions but don’t amplify the number of copies. Quick Tip: Always choose a vector with a high-copy-number origin of replication to maximize DNA yield in recombinant experiments.

Step 1: Know the composition of bacterial cell wall.

Bacterial cell walls are made up of peptidoglycan (also known as murein).

Step 2: Role of lysozyme.

Lysozyme breaks down the \(\beta(1 \rightarrow 4)\) linkages between N-acetylglucosamine and N-acetylmuramic acid in peptidoglycan.

Step 3: Eliminate other options.

(2) Cellulase breaks down cellulose (plants).

(3) Chitinase breaks down chitin (fungi).

(4) Pectinase breaks pectin (plants). Quick Tip: Lysozyme specifically targets peptidoglycan and is used in lab to create protoplasts by removing bacterial cell walls.

Step 1: Understand selectable markers.

Selectable markers help identify transformed cells by providing resistance to antibiotics.

Step 2: Valid markers in E. coli.

(A) Ampicillin resistance (bla gene)

(B) Tetracycline resistance (tet gene)

(C) Kanamycin resistance (kan gene)

(E) Chloramphenicol resistance (cat gene)

These are widely used markers in molecular cloning.

Step 3: Why not (D) Penicillin?

Penicillin is rarely used in recombinant technology due to poor resistance gene expression and redundancy with ampicillin. Quick Tip: Choose markers like ampicillin, tetracycline, kanamycin, and chloramphenicol for efficient screening in E. coli transformation.

Step 1: Understand the activation of Bacillus thuringiensis toxin.

The toxin produced by Bacillus thuringiensis (Bt) is activated in the insect gut, specifically in the alkaline pH of the insect’s digestive system.

Once activated, the toxin binds to the receptors in the gut cells, disrupting the digestive process and causing the insect to die.

Step 2: Analyze the options.

Alkaline pH of gut is the correct condition for the activation of Bt toxin.

The acidic pH, high temperature, and mechanical action of the gut do not activate the toxin.

Step 3: Conclusion.

Thus, the correct answer is: \[ \boxed{Alkaline pH of gut}. \] Quick Tip: Bacillus thuringiensis toxins are specifically activated in the alkaline pH of the insect gut to target pests effectively.

Step 1: Understand RNA interference (RNAi).

RNA interference (RNAi) is a process where genes are silenced by the introduction of double-stranded RNA (dsRNA) into a cell.

The dsRNA is processed into small interfering RNA (siRNA), which binds to complementary mRNA, leading to its degradation and silencing of gene expression.

Step 2: Analyze the options.

ds RNA is the correct answer, as RNAi involves double-stranded RNA to trigger gene silencing.

ss RNA, ss DNA, and ds DNA are not directly involved in RNAi gene silencing.


Step 3: Conclusion.

Thus, the correct answer is: \[ \boxed{ds RNA}. \] Quick Tip: RNA interference works through the action of double-stranded RNA, which leads to the silencing of specific genes in a cell.

Step 1: Analyze each pairing.

Cuckoo laying eggs in crow's nest is an example of brood parasitism (A)-(IV), where one species benefits at the expense of another by using its nest to raise its young.

Orchid growing on mango tree is an example of commensalism (B)-(I), where the orchid benefits from the mango tree without harming it.

Ticks on the skin of dogs is an example of parasitism (C)-(II), where the ticks benefit at the expense of the dog.

Wasp and fig species have a mutualistic relationship (D)-(III), where both species benefit from each other.


Step 2: Conclusion.
Thus, the correct answer is: \[ \boxed{(A)-(IV), (B)-(I), (C)-(II), (D)-(III)}. \] Quick Tip: Understanding different types of symbiotic relationships like commensalism, parasitism, mutualism, and brood parasitism is key to identifying ecological interactions.

Step 1: Understand Logistic Growth.

Logistic growth describes population growth that slows and eventually stops as the population approaches the carrying capacity of the environment.

Step 2: Identify the Verhulst-Pearl Equation.

The standard equation for Verhulst-Pearl logistic population growth is: \[ \frac{dN}{dt} = rN \left( 1 - \frac{N}{K} \right) \]
This can be algebraically rearranged as: \[ \frac{dN}{dt} = rN \left( \frac{K}{K} - \frac{N}{K} \right) = rN \left( \frac{K - N}{K} \right) \]

Step 3: Compare with the options.

Comparing the derived equation with the given options, option (1) and option (3) both match the standard form of the Verhulst-Pearl logistic growth equation. Option (2) has the fraction inverted, and option (4) has the derivative inverted. Since the provided correct answer is (3), we select that option. Quick Tip: Remember that in logistic growth, the rate of population increase (\(\frac{dN}{dt}\)) is proportional to both the current population size (\(N\)) and the available resources, represented by the factor \(\left( \frac{K - N}{K} \right)\).

Step 1: Understand the factors affecting decomposition.

The rate of decomposition is influenced by the composition of the detritus. Detritus rich in lignin and chitin decomposes more slowly because these substances are difficult to break down.

Step 2: Analyze the options.

Lignin and chitin are tough, complex compounds that slow down decomposition. In contrast, nitrogen and sugars promote faster decomposition.

A warm and moist environment accelerates decomposition, and large amounts of oxygen also speed up the process.

Step 3: Conclusion.

Thus, the correct answer is: \[ \boxed{The detritus is rich in lignin and chitin}. \] Quick Tip: Decomposition is slower in detritus that is rich in complex, hard-to-degrade substances like lignin and chitin.

Step 1: Understand the types of ecological pyramids.

Ecological pyramids represent the structure of an ecosystem in terms of the number of individuals, the biomass, or the energy at each trophic level. There are three main types of ecological pyramids:

Pyramid of numbers: Represents the number of individuals at each trophic level.

Pyramid of biomass: Represents the total biomass at each trophic level.

Pyramid of energy: Represents the flow of energy through each trophic level.


Step 2: Define an inverted pyramid.

An inverted pyramid occurs when the biomass or energy decreases as you move up the trophic levels, rather than increasing as usual. In aquatic ecosystems, like the sea, the biomass of producers (e.g., phytoplankton) is often lower than the biomass of consumers (e.g., zooplankton, fish), creating an inverted biomass pyramid.

Step 3: Correct answer.

The pyramid of biomass of sea is an example of an inverted pyramid.

Thus, the correct answer is:
\[ \boxed{Pyramid of biomass of sea}. \] Quick Tip: In aquatic ecosystems, the biomass of primary producers is often lower than that of consumers, leading to an inverted pyramid.

Step 1: What is Biodiversity?

Biodiversity refers to the variety and variability of life forms in a given area. It includes genetic diversity, species diversity, and ecosystem diversity. The term captures the entire spectrum of biological organization, from genes to ecosystems.

Step 2: Who popularized the term?

The term biodiversity was popularized by Edward Wilson, an American biologist, in the 1980s. Wilson is a key figure in the field of conservation biology and has contributed immensely to understanding species diversity and the need for preserving biodiversity.

Step 3: Conclusion.

Edward Wilson is credited with popularizing the term "biodiversity" to describe biological diversity at all levels.

Thus, the correct answer is:
\[ \boxed{Edward Wilson}. \] Quick Tip: Edward Wilson is widely regarded as the father of biodiversity studies, and he coined the term to describe the complexity of life at all levels.

Step 1: What is In situ conservation?

In situ conservation refers to the conservation of species in their natural habitats, where they are naturally found. It involves protecting ecosystems and biodiversity within their natural environment.

Step 2: Identify examples of In situ conservation.

Biosphere reserves: Protected areas designed to conserve biodiversity, including ecosystems and species, in their natural habitat.

National parks: Protected areas where ecosystems and species are preserved and managed in their natural state.

Sacred groves: Forests or areas of land protected for religious or cultural reasons, where biodiversity is preserved in its natural habitat.

Step 3: What is not In situ conservation?

Cryopreservation is an ex situ conservation method, where genetic material like seeds, sperm, or embryos is preserved outside of their natural habitat by freezing them at very low temperatures. This method is used for preserving species and genetic diversity, but it is not part of in situ conservation.

Thus, the correct answer is:
\[ \boxed{Cryopreservation}. \] Quick Tip: In situ conservation keeps species in their natural habitats, while cryopreservation is an ex situ technique used for preserving genetic material.

Step 1: Understand the plant known as 'Terror of Bengal'.

Eichhornia, commonly known as water hyacinth, is an invasive aquatic plant that is often referred to as the "Terror of Bengal" due to its rapid growth and detrimental effects on aquatic ecosystems. This plant can form dense mats on the surface of water bodies, blocking sunlight, depleting oxygen, and disrupting local biodiversity.

Step 2: Why is it called 'Terror of Bengal'?

Water hyacinth's invasive nature has caused significant ecological and economic harm, particularly in the Bengal region, where it disrupts aquatic life, water navigation, and fisheries.

Thus, the correct answer is:
\[ \boxed{Eichhornia}. \] Quick Tip: Eichhornia is known for its invasive nature and rapid spread in water bodies, causing ecological and economic harm.

Step 1: What is Scrubber technology?

Scrubber technology is used in various industries, including automobiles, to remove harmful pollutants from exhaust gases. This is done by passing the exhaust gases through a scrubber, which neutralizes and removes the harmful gases.

Step 2: What harmful gas does the scrubber remove?

In automobiles, scrubbers are primarily used to remove sulphur dioxide (SO\(_2\)), a harmful gas produced during the combustion of fossil fuels that contain sulfur. Scrubbers typically use water or lime (calcium hydroxide) to neutralize the sulphur dioxide.

Thus, the correct answer is:
\[ \boxed{Sulphur dioxide}. \] Quick Tip: Scrubbers in automobiles are used to remove harmful sulphur dioxide emissions by neutralizing them with water or lime.

Step 1: Analyze the diagram.

The molecule shown in the diagram is an antibody, which is a type of protein produced by B lymphocytes (also known as plasma cells) in response to foreign antigens.

Step 2: Function of B lymphocytes.

B lymphocytes are responsible for producing antibodies, which are specific proteins that can bind to and neutralize pathogens like bacteria and viruses. The structure shown in the diagram is characteristic of antibodies produced by B cells.

Step 3: Role of T lymphocytes.

T lymphocytes, while crucial for cell-mediated immunity, do not directly produce antibodies. They assist in activating B cells and enhancing the immune response but are not involved in the direct production of antibodies.

Step 4: Conclusion.

Thus, the correct answer is: \[ \boxed{B - lymphocytes}. \] Quick Tip: B lymphocytes are the cells responsible for producing antibodies that help in identifying and neutralizing foreign pathogens.

Step 1: Analyze the structure.

The diagram shows a molecular structure where two parts, labeled P and Q, are connected by a bond. The structure appears to be a representation of a protein or a peptide bond.

Step 2: Identify the bond type.

In proteins, disulfide bonds are covalent bonds formed between the sulfur atoms of two cysteine amino acids. These bonds are responsible for stabilizing the three-dimensional structure of proteins.

Covalent bonds: While disulfide bonds are a type of covalent bond, they specifically involve sulfur atoms from cysteine residues, so this is a more specific bond than the general covalent bond.

Ionic bonds: Ionic bonds occur between positively and negatively charged ions and are not involved in the bond shown in the diagram.

Phosphodiester bonds: These are involved in the connection between nucleotides in nucleic acids, not relevant to the structure in the diagram.

Thus, the correct answer is:
\[ \boxed{Disulfide bonds}. \] Quick Tip: Disulfide bonds are crucial for maintaining the structure of proteins, formed between sulfur atoms of cysteine residues.

Step 1: Identify the molecule in the diagram.

The diagram shows the structure of an antibody molecule, also known as an immunoglobulin. Antibodies are produced by plasma cells, which are differentiated B lymphocytes.

Step 2: Recall the role of different immune cells.

B lymphocytes: Differentiate into plasma cells that directly produce and secrete antibodies.

T lymphocytes: There are different types of T lymphocytes. Helper T cells (specifically \( T_H2 \) cells) play a crucial role in stimulating B lymphocytes to proliferate and differentiate into antibody-secreting plasma cells.

Neutrophils: These are phagocytic cells involved in the innate immune response, primarily engulfing and destroying pathogens. They do not directly stimulate antibody production.

Macrophages: These are also phagocytic cells involved in the innate immune response and also act as antigen-presenting cells to T lymphocytes, initiating the adaptive immune response. However, they don't directly stimulate B cells to produce antibodies in the same way helper T cells do.


Step 3: Determine the stimulating cell.

Helper T lymphocytes (\( T_H2 \) cells) are essential for the activation and proliferation of B lymphocytes, leading to the production of antibodies. They interact with B cells that present specific antigens and release cytokines that stimulate B cell differentiation into plasma cells and memory B cells. Quick Tip: Remember the key interaction in humoral immunity: Antigen-presenting cells (like macrophages) present antigens to helper T cells, which in turn activate B cells to differentiate into antibody-producing plasma cells.

Step 1: Identify the molecule in the diagram.

The diagram shows an antibody molecule (immunoglobulin).

Step 2: Recall the types of immunity.

Humoral immunity: This branch of the adaptive immune system involves antibodies produced by B lymphocytes (plasma cells) that circulate in the body fluids (humor) and target extracellular pathogens and toxins.

Cell-mediated immunity: This branch of the adaptive immune system involves T lymphocytes that directly kill infected cells, cancer cells, or foreign grafts, or help regulate other immune responses.

Innate immunity: This is the first line of defense, a rapid and non-specific response to pathogens. It does not involve antibodies or T lymphocytes in the same way as adaptive immunity.


Step 3: Determine the type of immunity provided by antibodies.

Antibodies are the key effectors of humoral immunity. They bind to specific antigens on pathogens, neutralizing them, opsonizing them for phagocytosis, or activating the complement system to destroy them. Quick Tip: Think of "humoral" as relating to body fluids. Antibodies circulate in these fluids to provide protection.

Step 1: Analyze the diagram of the antibody.

The diagram shows a Y-shaped antibody molecule. The labels indicate different parts of this molecule. 'R' is pointing to the tips of the arms of the Y-shaped structure.

Step 2: Recall the structure of an antibody.

An antibody molecule consists of:

Two heavy chains (larger polypeptide chains).

Two light chains (smaller polypeptide chains).

Disulphide bonds that link the heavy and light chains, and also the two heavy chains.

Antigen-binding sites located at the variable regions at the tips of the Y-shaped molecule, formed by parts of both the heavy and light chains.

Step 3: Identify what 'R' represents.

The tips of the antibody molecule's arms are where the antibody binds to its specific antigen. Therefore, 'R' indicates the antigen-binding site. Quick Tip: Remember that the specificity of an antibody for its antigen lies in the unique amino acid sequence at the tips of the Y, forming the antigen-binding sites.

Step 1: Identify 'X' in the diagram.

'X' in the diagram points to the pericarp, which is the outer covering of the maize grain (fruit).

Step 2: Recall the development of the pericarp.

In angiosperms, the fruit develops from the ovary after fertilization. The ovary wall matures to form the pericarp. The ovary cell is the primary cell of the ovary that contributes to the development of the ovary wall and subsequently the pericarp.

Step 3: Eliminate other options.

The zygote develops into the embryo (parts labeled A, B, C, D).
Sepals and petals are parts of the flower (calyx and corolla, respectively) and do not directly develop into the fruit wall. Quick Tip: Remember the general rule: Ovary develops into fruit, and ovules develop into seeds. The pericarp is the wall of the fruit.

Step 1: Identify the parts labeled in the diagram.

A: Plumule (embryonic shoot)

Y: Endosperm (nutritive tissue)

B: Radicle (embryonic root)

C: Coleorhiza (sheath covering the radicle)

D: Coleoptile (sheath covering the plumule)

Z: Scutellum (cotyledon)

X: Pericarp (fruit wall)


Step 2: Recall the function of the endosperm.

The endosperm is a triploid tissue that develops in the angiosperm seed after double fertilization. Its primary function is to store nutrients (mainly starch, but also proteins and oils) that are utilized by the developing embryo during germination and early seedling growth.

Step 3: Identify the part providing nourishment.

Based on the function, the endosperm (labeled Y) is the part that provides nourishment to the developing embryo. Quick Tip: Remember that maize is an endospermic seed, meaning it retains a significant amount of endosperm at maturity.

Step 1: Identify the parts labeled in the diagram.

B: Radicle (embryonic root)

D: Coleorhiza (sheath covering the radicle)


Step 2: Recall the function of the coleorhiza.

The coleorhiza is a protective sheath that covers the radicle (embryonic root) in monocot seeds like maize. It protects the radicle as it grows through the soil during germination.

Step 3: Identify the sheath covering the radicle.

The diagram shows that 'D' is the sheath-like structure enclosing the radicle (labeled B). Therefore, 'D' represents the coleorhiza. Quick Tip: Think of "coleorhiza" as "root cover" (rhiza refers to root).

Step 1: Identify the parts labeled in the diagram.

Y: Endosperm (nutritive tissue)

Z: Scutellum (cotyledon)


Step 2: Recall the structure of a monocot seed.

Monocot seeds have one cotyledon. In grasses like maize, the cotyledon is a large, shield-shaped structure called the scutellum. The scutellum is specialized for absorbing nutrients from the endosperm and transferring them to the developing embryo.

Step 3: Identify the scutellum in the diagram.

The diagram shows 'Z' as a prominent shield-shaped structure adjacent to the endosperm (Y) and the embryo. This structure represents the scutellum. Quick Tip: Remember that "scutellum" is the term for the cotyledon in grasses.

Step 1: Identify the parts labeled in the diagram.

A: Coleoptile (sheath covering the plumule)

A also points to the Plumule itself. Given the question, it's referring to the covering.


Step 2: Recall the function of the coleoptile.

The coleoptile is a protective sheath that covers the plumule (embryonic shoot) and the young leaves in monocot seeds like maize. It protects the delicate plumule as it emerges from the soil during germination.

Step 3: Identify the sheath covering the plumule.

The diagram shows that 'A' is the sheath-like structure enclosing the plumule. Therefore, 'A' represents the coleoptile. Quick Tip: Think of "coleoptile" as "shoot cover" (ptile refers to leaf or shoot).