Autogamy is self-pollination within the same flower, requiring the flower to have both male (anther) and female (stigma) parts, i.e., be bisexual.


A unisexual flower possesses only one type of reproductive organ (either male or female).


Therefore, a unisexual flower cannot pollinate itself, which completely prevents autogamy.


Geitonogamy is the transfer of pollen from a flower to another flower on the same plant.


This is still possible if the plant is monoecious, bearing separate male and female unisexual flowers.


Thus, unisexuality prevents autogamy but not necessarily geitonogamy.
Quick Tip: Remember the definitions: Autogamy (self-pollination, same flower), Geitonogamy (different flower, same plant), and Xenogamy (different plant). Unisexuality is a natural barrier to autogamy.

The diagram illustrates the stages of sperm production (spermatogenesis).


'I' represents the initial diploid germ cells that undergo mitosis at puberty; these are the Spermatogonia.


Spermatogonia develop into primary spermatocytes, which then undergo Meiosis I to form haploid secondary spermatocytes, represented by 'II'.


The secondary spermatocytes ('II') undergo Meiosis II to produce four haploid cells called spermatids, represented by 'III'.


Finally, the spermatids ('III') mature into spermatozoa (sperm) through the process of spermiogenesis. These are represented by 'IV'.


Thus, the correct labels are: I-spermatogonia, II-sec. spermatocyte, III-spermatid, IV-spermatozoa.
Quick Tip: Memorize the sequence of spermatogenesis: Spermatogonium (2n) \(\rightarrow\) Primary Spermatocyte (2n) \(\rightarrow\) Secondary Spermatocytes (n) \(\rightarrow\) Spermatids (n) \(\rightarrow\) Spermatozoa (n). Meiosis I is the reductional division.

The evolutionary order is determined by the appearance of these plant groups in the fossil record, from oldest to newest.


(iii) Zosterophyllum is an extinct genus of early vascular plants from the Devonian period, representing one of the earliest land plant lineages.


(i) Ferns are pteridophytes that became abundant in the Carboniferous period, evolving after the primitive vascular plants like Zosterophyllum.


(ii) Ginkgo is a gymnosperm that appeared in the Permian period, representing the evolution of seed plants after the spore-bearing ferns.


(iv) Gnetales are considered a relatively advanced group of gymnosperms that evolved later than Ginkgo.


Therefore, the correct chronological order is Zosterophyllum \(\rightarrow\) Ferns \(\rightarrow\) Ginkgo \(\rightarrow\) Gnetales, which corresponds to (iii), (i), (ii), (iv).
Quick Tip: Recall the major steps in plant evolution: Early vascular plants (e.g., Psilophytes like Zosterophyllum) \(\rightarrow\) Pteridophytes (Ferns) \(\rightarrow\) Gymnosperms (e.g., Ginkgo, Gnetales) \(\rightarrow\) Angiosperms.

The concept that genetic information flows from DNA to RNA to protein is known as the "Central Dogma" of molecular biology.


This was proposed by Francis Crick in 1958.


The flow describes the processes of transcription (DNA to RNA) and translation (RNA to protein).


While Watson and Crick together discovered the DNA structure, the formulation of the Central Dogma is attributed solely to Francis Crick.


Hargobind Khorana and Marshall Nirenberg were key figures in deciphering the genetic code, not in proposing the Central Dogma.
Quick Tip: Associate scientists with their key contributions: Francis Crick is credited with the Central Dogma. Watson and Crick are credited with the DNA double helix structure. Nirenberg and Khorana helped crack the genetic code.

The diagram shows a primary transcript (hnRNA) before splicing.


Introns are the non-coding sequences that are looped out and removed during splicing. By counting the loops, we find there are 7 introns.


Exons are the coding sequences that are joined together to form the mature mRNA.


The exons are the linear segments. There is one at the beginning, one at the end, and one segment between each pair of adjacent introns.


With 7 introns, there are 6 segments between them. Adding the starting and ending segments gives 1 + 6 + 1 = 8 exons.


The question asks for the number of introns and exons, respectively.


Therefore, the count is 7 introns and 8 exons.
Quick Tip: For a linear gene transcript, the number of exons is always one more than the number of introns. If you count 'n' introns (loops), there will be 'n+1' exons (straight segments).

Let's analyze each statement regarding the haplodiploid sex-determination system in honey bees.


Statement (i) is incorrect. A zygote, formed by the fertilization of an egg by a sperm, develops into a diploid female (either a queen or a worker). Males (drones) develop from unfertilized eggs via parthenogenesis.


Statement (ii) is correct. Females are diploid (2n), and males are haploid (n). Therefore, males have half the number of chromosomes as females.


Statement (iii) is correct. Female honey bees are diploid and have a total of 32 chromosomes (2n=32). Males are haploid with 16 chromosomes (n=16).


Statement (iv) is incorrect. Since males develop from unfertilized eggs, they do not have a father. They only have a mother and a maternal grandfather.


Based on this analysis, only statements (ii) and (iii) are correct.
Quick Tip: In honey bees, sex is determined by the number of sets of chromosomes. Fertilized eggs (diploid, 2n=32) become females, while unfertilized eggs (haploid, n=16) become males. This is called haplodiploidy.

In eukaryotes, there are three main types of RNA polymerase, each with specific functions.


(a) RNA polymerase I is responsible for transcribing ribosomal RNAs (rRNAs), specifically the 28S, 18S, and 5.8S rRNA genes. Therefore, (a) matches with (i).


(b) RNA polymerase II transcribes the precursor to messenger RNA (mRNA), which is called heterogeneous nuclear RNA (hnRNA). It also transcribes most small nuclear RNAs (snRNAs). From the options, the primary match is hnRNA. Therefore, (b) matches with (iii).


(c) RNA polymerase III transcribes transfer RNA (tRNA), 5S rRNA, and some small RNAs, including certain snRNAs (like U6 snRNA). In the context of this question, after matching the others, (c) correctly pairs with (ii).


The correct set of matches is (a)-(i), (b)-(iii), and (c)-(ii). This corresponds to option (D).
Quick Tip: Use the mnemonic "1, 2, 3 - r, m, t" to remember the primary products of RNA polymerases: Pol I makes rRNA, Pol II makes mRNA (via hnRNA), and Pol III makes tRNA.

First, determine the genotypes based on the given blood groups (phenotypes).


Mother has blood group AB, so her genotype must be \(I^{A}I^{B}\).


Father has blood group B, so his genotype can be either homozygous (\(I^{B}I^{B}\)) or heterozygous (\(I^{B}i\)).


Child has blood group A, so their genotype can be either homozygous (\(I^{A}I^{A}\)) or heterozygous (\(I^{A}i\)).


The child inherits one allele from each parent. The mother (\(I^{A}I^{B}\)) can contribute either an \(I^{A}\) or an \(I^{B}\) allele.


Since the child has blood group A, the child must have inherited the \(I^{A}\) allele from the mother.


To complete the genotype for blood group A, the child must inherit the other allele from the father. The father has blood group B and cannot provide an \(I^{A}\) allele.


Therefore, the child's genotype must be heterozygous (\(I^{A}i\)), meaning the child inherited the 'i' allele from the father.


For the father to provide an 'i' allele, his genotype must be heterozygous (\(I^{B}i\)).


Thus, the correct genotypes are: Father: \(I^{B}i\), Mother: \(I^{A}I^{B}\), Child: \(I^{A}i\).
Quick Tip: When solving blood group problems, always write down all possible genotypes for each parent's phenotype first. Then, use the child's phenotype to eliminate possibilities and find the only combination that works.

The immune system relies on the coordinated action of different cells.


T-lymphocytes, particularly helper T-cells (\(T_H\)), are central to the adaptive immune response.


Helper T-cells are required to activate B-lymphocytes (B-cells).


Once activated, B-cells differentiate into plasma cells, which are responsible for producing and secreting antibodies.


Therefore, a decrease in the number of T-lymphocytes leads to insufficient activation of B-cells.


This impairment directly results in a reduced or failed antibody production, weakening the humoral immune response.


This is the mechanism by which HIV, which targets helper T-cells, causes immunodeficiency (AIDS).
Quick Tip: Remember the chain of command in the immune system: Helper T-cells act as "managers" that activate other immune cells. Without them, B-cells (the "factories") cannot produce enough antibodies.

The question asks for the ratio of DNA types after the third generation, following the semi-conservative replication model. Let's trace the generations.


Generation 0: We start with bacteria grown in a heavy nitrogen isotope (\(N^{15}\)) medium. All DNA is heavy (\(N^{15}/N^{15}\)).


Generation 1: The bacteria are transferred to a light nitrogen (\(N^{14}\)) medium and allowed to replicate once. All the resulting DNA molecules are hybrid (\(N^{15}/N^{14}\)). The ratio of \(N^{15}/N^{15} : N^{15}/N^{14} : N^{14}/N^{14}\) is 0 : 100% : 0. Let's say we have 2 molecules. The ratio of molecule types is 0 : 2 : 0.


Generation 2: The bacteria replicate again in the \(N^{14}\) medium. The two hybrid molecules unwind. The two \(N^{15}\) strands form two new hybrid (\(N^{15}/N^{14}\)) molecules. The two \(N^{14}\) strands form two new light (\(N^{14}/N^{14}\)) molecules. We have 4 molecules in total. The ratio of molecule types is 0 (heavy) : 2 (hybrid) : 2 (light).


Generation 3: The bacteria replicate a third time in the \(N^{14}\) medium. We start with 2 hybrid and 2 light molecules.

- The 2 hybrid molecules produce 2 hybrid (\(N^{15}/N^{14}\)) and 2 light (\(N^{14}/N^{14}\)) molecules.

- The 2 light molecules produce 4 light (\(N^{14}/N^{14}\)) molecules.

- Total molecules = 8.

- Total heavy (\(N^{15}/N^{15}\)) = 0.

- Total hybrid (\(N^{15}/N^{14}\)) = 2.

- Total light (\(N^{14}/N^{14}\)) = 2 + 4 = 6.


The ratio of molecule types \(N^{15}/N^{15} : N^{15}/N^{14} : N^{14}/N^{14}\) is 0 : 2 : 6.


Simplifying this ratio by dividing by 2 gives 0 : 1 : 3.
Quick Tip: In Meselson-Stahl problems, remember that the number of hybrid (\(N^{15}/N^{14}\)) molecules will always be 2 after the first generation. The total number of molecules after 'n' generations is \(2^n\). The number of light molecules is \(2^n - 2\).

Let's evaluate each statement.


(A) This statement is incorrect. The enzyme that joins the gene of interest and the vector DNA is DNA ligase, not a polymerase.


(B) This statement is incorrect. Gel electrophoresis is a technique used to separate DNA fragments based on their size, not for amplification.


(C) This statement is incorrect. PCR (Polymerase Chain Reaction) is used for the amplification (making multiple copies) of a specific DNA segment, not for its initial isolation and separation.


(D) This statement is correct. A vector is a DNA molecule used as a vehicle to carry foreign genetic material into another cell. Plasmids are commonly used as vectors for this purpose in genetic engineering.
Quick Tip: Associate each biotechnology tool with its primary function: PCR \(\rightarrow\) Amplification; Gel Electrophoresis \(\rightarrow\) Separation; DNA Ligase \(\rightarrow\) Joining/Ligation; Plasmid/Vector \(\rightarrow\) Transfer.

Different microbes are used for the commercial production of various organic acids and other products.


(A) Aspergillus niger, a fungus, is widely used for the industrial production of citric acid.


(B) Lactobacillus species are bacteria used in the production of lactic acid, such as in the making of curd from milk.


(C) Clostridium butylicum is a bacterium used for the production of butyric acid.


(D) Saccharomyces cerevisiae, a yeast, is used in baking (as baker's yeast) and in breweries for producing ethanol.


Therefore, the correct microbe for citric acid production is Aspergillus niger.
Quick Tip: Create a table to memorize important microbes and their commercial products: Aspergillus niger (citric acid), Acetobacter aceti (acetic acid), Lactobacillus (lactic acid), Saccharomyces cerevisiae (ethanol).

The Assertion (A) states that the corpus luteum secretes progesterone. This is a correct statement. After ovulation, the remnant of the Graafian follicle transforms into the corpus luteum, which acts as a temporary endocrine gland.


The Reason (R) states that progesterone is essential for the maintenance of the endometrium (the inner lining of the uterus). This is also a correct statement. Progesterone makes the endometrium receptive to implantation and supports the early stages of pregnancy.


To check if (R) explains (A), we ask: "Why does the corpus luteum secrete progesterone?" The biological significance of this secretion is precisely to maintain the endometrium for a potential pregnancy.


Therefore, the Reason provides the correct functional explanation for the Assertion.
Quick Tip: For Assertion-Reason questions, follow a three-step process: 1. Check if Assertion is true. 2. Check if Reason is true. 3. If both are true, check if the Reason correctly explains the Assertion by linking them with the word "because".

The Assertion (A) describes the phenomenon of industrial melanism observed in the peppered moth (Biston betularia). Before industrialization, light-colored (white-winged) moths were camouflaged against lichen-covered trees. After industrialization, soot darkened the tree trunks, making the white moths highly visible to predators, which caused their numbers to decrease. So, the Assertion is true.


The Reason (R) states that the effects of industrialization (like pollution and soot deposition) were more marked in rural areas. This is incorrect. The effects of industrialization were most prominent in and around industrial cities and towns, not in the less-polluted rural areas.


Since the Assertion is true and the Reason is false, the correct option is (C).
Quick Tip: Industrial melanism in peppered moths is a classic example of natural selection in action. Remember that the environmental change (soot-covered trees) occurred in industrial areas, providing a selective advantage to the dark-colored (melanic) moths there.

The Assertion (A) states that Streptococcus pneumoniae and Haemophilus influenzae cause infectious diseases in humans. This is true; both are well-known pathogens responsible for causing pneumonia and other respiratory infections.


The Reason (R) describes the mode of transmission for these pathogens, stating that infection is acquired by inhaling aerosols (droplets) from an infected person. This is the primary way these respiratory diseases spread. Hence, the Reason is also true.


Now, we check if the Reason explains the Assertion. The fact that the disease is "infectious" (as stated in A) is explained by its mode of transmission from one person to another (as stated in R). The reason directly explains how the disease caused by these organisms spreads, confirming its infectious nature.


Therefore, both statements are true, and (R) is the correct explanation for (A).
Quick Tip: For infectious diseases, always link the pathogen, the disease it causes, and its mode of transmission. Here: (S. pneumoniae, H. influenzae) \(\rightarrow\) Pneumonia \(\rightarrow\) Inhaling aerosols/droplets.

The Assertion (A) states that restriction endonucleases recognize and cut DNA at specific palindromic sequences. This is the fundamental definition of how these enzymes function. A palindromic sequence in DNA reads the same on both strands when read in the same orientation (e.g., 5' to 3'). This statement is true.


The Reason (R) claims that a palindromic sequence has two unique recognition sites, PstI and PvuI. This statement is incorrect and nonsensical. PstI and PvuI are two different types of restriction enzymes, and they each recognize their own unique palindromic sequence. A single palindromic sequence is recognized by a single type of restriction enzyme, not multiple different ones like PstI and PvuI.


Since the Assertion is true and the Reason is false, the correct option is (C).
Quick Tip: Remember that each type of restriction enzyme (like EcoRI, PstI, HindIII) is highly specific and recognizes only one particular DNA sequence (its recognition site). The concept of a single site being recognized by two different enzymes is incorrect.

(A) Pollination in Vallisneria and Zostera:


Both plants exhibit hydrophily (pollination by water), but through different mechanisms.


1. Vallisneria: It shows epihydrophily (pollination on the surface of the water).

The female flower reaches the water surface with its long stalk.

The male flowers are released from the male plant and float on the water surface.

Water currents carry the male flowers passively towards the female flowers, leading to pollination.


2. Zostera (Seagrass): It shows hypohydrophily (pollination beneath the water surface).

The female flowers remain submerged in water.

The male plant releases long, ribbon-like pollen grains into the water.

These pollen grains are carried by water currents and come into contact with the submerged stigma of the female flower to effect pollination.


OR


(B) Four strategies to prevent self-pollination (Outbreeding devices):


1. Unisexuality (Dicliny): The plant bears either male or female flowers, but not both. This ensures cross-pollination. If male and female flowers are on different plants, it is called dioecy (e.g., Papaya).


2. Dichogamy: The anthers and stigma mature at different times in a bisexual flower.

- \textit{Protandry: Anthers mature earlier than the stigma (e.g., Sunflower).

- \textit{Protogyny: Stigma matures earlier than the anthers (e.g., Ficus).


3. Self-incompatibility: This is a genetic mechanism that prevents self-pollen (from the same flower or other flowers of the same plant) from fertilizing the ovules by inhibiting pollen germination or pollen tube growth in the pistil.


4. Herkogamy (Spatial Separation): Anthers and stigma are placed at different positions so that pollen from the same flower cannot land on its stigma (e.g., in Gloriosa and Hibiscus).
Quick Tip: Hydrophily is a rare form of pollination. Distinguish between epihydrophily (on water surface, e.g., Vallisneria) and hypohydrophily (under water, e.g., Zostera). For outbreeding devices, associate the term with its mechanism (e.g., Dichogamy = different timing).

(a) Identification and Inheritance of the Trait:


The trait is an autosomal recessive trait.


Explanation of Inheritance:

1. The trait appears in the offspring (Generation I) even though the parents are unaffected. This phenomenon, where a trait "skips a generation," is a hallmark of a recessive disorder.

2. If the trait were dominant, at least one parent would have to be affected to have affected offspring.

3. Since the parents are unaffected but have affected children, both parents must be heterozygous carriers for the trait (e.g., genotype Aa).

4. The trait affects both male (shaded square) and female (shaded circle) offspring, suggesting it is autosomal rather than sex-linked.

5. The affected children have the genotype (aa), having inherited one recessive allele (a) from each of the carrier parents (Aa).


(b) Example of such a trait in humans:


One example of an autosomal recessive trait in humans is Sickle-cell anemia or Phenylketonuria (PKU).
Quick Tip: Key clues for pedigree analysis: 1. Trait skips a generation \(\rightarrow\) Recessive. 2. Affected parents have an unaffected child \(\rightarrow\) Dominant. 3. Affects males and females roughly equally \(\rightarrow\) Autosomal.

(A) Situations requiring pre-formed antibodies (Passive Immunity):


A doctor would recommend injecting pre-formed antibodies when a quick, immediate immune response is required, as the body does not have time to mount its own primary immune response.


1. Snakebite: If a person is bitten by a venomous snake, an injection of antivenom is given. Antivenom contains pre-formed antibodies against the snake venom, which neutralize the toxin immediately.


2. Tetanus Infection: If a person with a deep, dirty wound is not immunized against tetanus, they may be given an injection of Tetanus Antitoxin (ATS). This provides immediate protection against the toxin produced by the bacterium Clostridium tetani.


OR


(B) Delayed appearance of Malaria symptoms:


The symptoms of malaria do not appear immediately after a mosquito bite due to the life cycle of the parasite (Plasmodium) in the human body.


1. Entry and Liver Stage (Pre-erythrocytic cycle): When an infected female Anopheles mosquito bites a human, it injects sporozoites into the bloodstream. These sporozoites do not cause symptoms. They travel to the liver.


2. Asexual Reproduction in Liver: Inside the liver cells, the sporozoites multiply asexually for a period (the incubation period), producing thousands of merozoites. This stage is asymptomatic.


3. RBC Stage (Erythrocytic cycle): The liver cells eventually rupture, releasing the merozoites into the bloodstream. These merozoites then invade red blood cells (RBCs) and multiply asexually.


4. Symptom Onset: The symptoms of malaria, such as recurring high fever and chills, appear only when the infected RBCs rupture. This releases new merozoites along with a toxic substance called hemozoin, which is responsible for the characteristic symptoms.
Quick Tip: Distinguish between passive and active immunity. Passive immunity (pre-formed antibodies) is for emergencies (e.g., snakebite), providing fast but temporary protection. Active immunity (vaccination or infection) is slower but provides long-term memory.

(a) Source Organism:


The scientific name of the source organism is Thermus aquaticus.

This bacterium was isolated from hot springs, and the enzyme is commonly known as Taq polymerase.


(b) Advantage of use:


The primary advantage of using a thermostable DNA polymerase is that it can withstand the high temperatures required during the PCR cycle.

Specifically, it remains active and functional during the denaturation step, which occurs at a high temperature (around 95°C).

This stability eliminates the need to add fresh DNA polymerase after each cycle, which was necessary with earlier, non-thermostable enzymes.

This makes the automation of the entire PCR process possible.
Quick Tip: The key to PCR's success is the thermostable Taq polymerase from Thermus aquaticus. Its ability to survive high heat allows for repeated cycles of denaturation, annealing, and extension without adding new enzyme each time, making the process efficient and automated.

David Tilman's long-term ecosystem experiments using outdoor plots led to two main conclusions regarding the relationship between biodiversity and ecosystem stability:


1. Increased Stability: Tilman found that plots with more species (higher biodiversity) showed less year-to-year variation in their total biomass. This indicates that increased diversity leads to increased ecosystem stability and resilience.


2. Increased Productivity: His experiments also demonstrated that, in his experimental plots, increased diversity contributed to higher overall productivity. Plots with a greater variety of species tended to produce more total plant biomass.
Quick Tip: Remember the key takeaway from Tilman's experiments: "More Biodiversity = More Stability and Productivity". This is a fundamental concept in ecology that provides strong evidence for the importance of conserving biodiversity.

(a) Reasons for effectiveness of Copper releasing IUDs (Intra Uterine Devices):


1. The released copper ions (\(Cu^{2+}\)) suppress the motility of sperm, reducing their ability to swim towards the egg.


2. The copper ions also reduce the fertilizing capacity of the sperm.


(b) Mechanism of Oral Contraceptive Pills:


Oral contraceptive pills contain either a combination of progesterone and estrogen, or progesterone alone.


They prevent pregnancy through the following mechanisms:


1. They inhibit ovulation by suppressing the release of gonadotropins (FSH and LH) from the pituitary gland.


2. They alter the quality of the cervical mucus, making it thick and viscous, which prevents or retards the entry of sperm into the uterus.


3. They can also make the uterine endometrium unsuitable for implantation.
Quick Tip: Remember the primary modes of action for contraceptives. Copper IUDs are spermicidal. Hormonal pills primarily work by preventing ovulation.

(a) Calculation of male gametophytes:


1. A bilobed, dithecous anther has two lobes, and each lobe has two thecae (microsporangia). Therefore, the total number of microsporangia in the anther is 4.


2. Each microsporangium has 200 Microspore Mother Cells (MMCs).


3. Total number of MMCs in the anther = Number of microsporangia \(\times\) MMCs per microsporangium = 4 \(\times\) 200 = 800 MMCs.


4. Each diploid MMC undergoes meiosis to produce four haploid microspores (pollen grains).


5. Total number of microspores produced = Total MMCs \(\times\) 4 = 800 \(\times\) 4 = 3200.


6. Each microspore (pollen grain) develops into a male gametophyte.


Therefore, 3200 male gametophytes can be produced by this anther.


(b) Composition of Intine and Exine:


Exine: The hard outer layer of the pollen grain. It is composed of sporopollenin, one of the most resistant organic materials known.


Intine: The thin inner wall of the pollen grain. It is composed of cellulose and pectin.
Quick Tip: Key formulas for anther calculations: Total microsporangia in a dithecous anther = 4. Each MMC produces 4 microspores via meiosis. Remember the tough, protective exine is made of sporopollenin.

The Hardy-Weinberg equilibrium states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.


Natural selection is one of these evolutionary influences and disrupts the Hardy-Weinberg equilibrium by favoring certain alleles and phenotypes over others.


This leads to a change in allele frequencies in the population over time.


Natural selection can affect the equilibrium in three ways, as shown by graphs of a phenotypic trait:


1. Stabilizing Selection: In this type, individuals with the mean (average) phenotype are favored, while individuals with extreme phenotypes are selected against. This reduces variation but does not change the mean value of the trait. The peak of the graph gets narrower and taller.



2. Directional Selection: This type favors individuals with one extreme of the phenotypic range. Over time, this causes the population's average phenotype to shift in that direction. The peak of the graph shifts to one side (left or right).



3. Disruptive Selection: Here, individuals at both extremes of the phenotypic range are favored over intermediate phenotypes. This can lead to the formation of two distinct phenotypes in the population. The graph shows two peaks with a valley in the middle.
Quick Tip: Remember the five factors that disrupt Hardy-Weinberg equilibrium: Gene Flow, Genetic Drift, Mutation, Non-random Mating, and Natural Selection. Natural selection is the only one that leads to adaptive evolution.

Let's define the alleles for the autosomal feature. Let 'A' be the dominant allele and 'a' be the recessive allele.


Parental Genotypes:

- Homozygous female: Her genotype could be homozygous dominant (AA) or homozygous recessive (aa). Let's assume she is homozygous recessive (aa).

- Heterozygous male: His genotype is (Aa).


Cross: (aa) Female \(\times\) (Aa) Male


Gametes produced:

- Female (aa) produces only one type of gamete: (a).

- Male (Aa) produces two types of gametes: (A) and (a).


Punnett Square for the F1 Generation:

\begin{tabular{c|c|c
& A & a
\hline
a & Aa & aa
\hline
a & Aa & aa

\end{tabular


Distribution of Genotypes and Phenotypes in F1 Generation:

- Genotypic Ratio: The resulting genotypes are Aa and aa. The ratio is 2 Aa : 2 aa, which simplifies to 1 Aa : 1 aa.

- Phenotypic Ratio: Individuals with genotype 'Aa' will show the dominant phenotype. Individuals with genotype 'aa' will show the recessive phenotype. The ratio is 1 (Dominant phenotype) : 1 (Recessive phenotype).


This means that 50% of the offspring will be heterozygous and show the dominant trait, while 50% will be homozygous recessive and show the recessive trait.
Quick Tip: A cross between a heterozygous individual (Aa) and a homozygous recessive individual (aa) is called a test cross. The resulting phenotypic ratio in the offspring is always 1:1.

(a) Need for drug tests:

Drug tests are conducted to prevent athletes from using performance-enhancing drugs (PEDs). The primary reasons are:

1. To ensure fair competition by preventing any athlete from gaining an unfair advantage.

2. To protect the health of the athletes, as many PEDs have serious and harmful side effects.


(b) Drugs usually tested for:

Authorities usually look for a range of substances, including:

1. Anabolic steroids

2. Stimulants (e.g., amphetamines)

3. Narcotic analgesics (painkillers)

4. Cannabinoids

5. Diuretics (masking agents)


(c) Scientific names of two source plants:

1. Erythroxylum coca: The source of cocaine, which is a potent stimulant.

2. Cannabis sativa: The source of cannabinoids like marijuana and hashish.

(Other examples include: Papaver somniferum for opioids like morphine, and Atropa belladonna for atropine).
Quick Tip: Drug abuse in sports is a major issue. Remember the main categories of banned substances (like steroids and stimulants) and associate them with their source plants and effects.

(a) Specific Bt. toxin gene for corn borer:

The protein coded by the gene cryIAb controls the corn borer.


(b) Mechanism of action of Bt. toxin:

1. Bacillus thuringiensis produces the toxin as an inactive crystalline protein, called a protoxin.


2. When an insect (like the corn borer) ingests the plant containing this protoxin, it reaches the insect's midgut.


3. The alkaline pH of the insect's gut solubilizes the protein crystal and activates the protoxin into its toxic form.


4. The activated toxin binds to the surface of the midgut epithelial cells.


5. This binding creates pores in the cell membranes, leading to cell swelling and lysis (bursting).


6. This process causes paralysis of the digestive system, the insect stops feeding, and eventually dies.
Quick Tip: Remember that Bt toxin is a protoxin that requires the alkaline pH of an insect's gut for activation. This specificity is why it is harmless to humans and other animals with acidic stomach pH.

(a) Identification of 'X' and 'Y':

The pie chart shows the relative diversity of invertebrates. The largest group of animals on Earth is insects.

- X represents Insects, which is the most species-rich group among invertebrates.

- Y represents Molluscs, which is the second-largest invertebrate phylum.


(b) Most species-rich group:

- 'X' (Insects) is the most species-rich taxonomic group.

- Insects account for over 70% of all animal species recorded. Looking at the chart, group 'X' clearly represents the vast majority of the invertebrate section.


(c) Levels of Biodiversity:

(i) Estuaries and alpine meadows in India: This represents Ecological diversity. It describes the variety of different ecosystems, habitats, and ecological communities within a geographical area.


(ii) The medicinal plant Rauwolfia vomitoria: This example, particularly if referring to variations in its genetic makeup (e.g., differences in the concentration of the active chemical reserpine found in plants growing in different Himalayan ranges), represents Genetic diversity. It refers to the total number of genetic characteristics in the genetic makeup of a species.
Quick Tip: Remember the three levels of biodiversity: Genetic (variation within a species), Species (variety of species in an area), and Ecological (variety of ecosystems). Insects are the most diverse group of organisms on the planet.

(a) Specialised Cells for Immune Response:

The two types of specialised cells are B-lymphocytes and T-lymphocytes.

Both primary and secondary responses are carried out by these cells, with the secondary response specifically involving memory B and memory T cells.


(b) Humoral Immune Response:

Antibody-mediated immunity is called the humoral immune response because the antibodies are found circulating in the body's fluids or 'humors', such as blood plasma and lymph.


(c) Graft Rejection:

(i) Characteristic of Immune System:

The ability of the immune system to differentiate between 'self' and 'non-self' is responsible for graft rejection. The recipient's body recognizes the transplanted organ as foreign.


(ii) Type of Immune Response and Cell:

The type of immune response is Cell-Mediated Immunity (CMI).

The cells primarily involved are the T-lymphocytes (specifically cytotoxic T-cells).


OR


(d) Difference between Active and Passive Immunity:

\begin{tabular{|l|l|l|
\hline
Feature & Active Immunity & Passive Immunity
\hline
Source of Antibodies & Body produces its own antibodies. & Receives pre-formed antibodies.

Response Time & Slow, takes time to develop. & Fast and provides immediate relief.

Immune Memory & Develops memory cells; is long-lasting. & No memory is formed; is temporary.

Example & Immunity after infection or vaccination. & Antivenom, antibodies from mother to foetus.

\hline
\end{tabular
Quick Tip: Associate Cell-Mediated Immunity (CMI) with T-cells and graft rejection. Associate Humoral Immunity with B-cells, antibodies, and body fluids ('humors'). Active immunity is what you "make," while passive immunity is what you are "given."

(a) Main Enzyme of Transcription:

The main enzyme involved is DNA-dependent RNA polymerase.


(b) Identification of DNA Strands:

The enzyme RNA polymerase reads the DNA strand with 3' \(\rightarrow\) 5' polarity to synthesize RNA in the 5' \(\rightarrow\) 3' direction.

- Template Strand: The strand with polarity 3' \(\rightarrow\) 5'. It acts as the template for RNA synthesis.

- Coding Strand: The strand with polarity 5' \(\rightarrow\) 3'. It has a sequence similar to the RNA transcribed (with T instead of U) and does not code for anything directly.


(c) Identification and Significance of (C) and (D):

- (C) is the Promoter: It is the DNA sequence where RNA polymerase binds to start the process of transcription. It defines the start site of transcription.

- (D) is the Terminator: It is the DNA sequence that signals the end of transcription. When RNA polymerase reaches this site, it stops transcription and dissociates from the DNA.


OR


(d) Location of (C) and (D):

The transcription unit is defined by the promoter, the structural gene, and the terminator.

- Location of (C) - Promoter: The promoter is located upstream (towards the 5'-end of the coding strand) of the structural gene.

- Location of (D) - Terminator: The terminator is located downstream (towards the 3'-end of the coding strand) of the structural gene.
Quick Tip: Remember that the template strand is read 3' to 5', and the new RNA is built 5' to 3'. The coding strand has the same polarity and sequence (with T for U) as the new RNA. The promoter is the "start" signal and the terminator is the "stop" signal for transcription.

(A) Oogenesis in Human Females


(i) Schematic Representation of Oogenesis:


Oogonium (2n, 46 chromosomes)
\(\downarrow\) Mitosis \& Differentiation

Primary oocyte (2n, 46 chromosomes) - Arrested at Prophase-I
\(\downarrow\) Meiosis-I (completed prior to ovulation)

Secondary oocyte (n, 23 chromosomes) + First polar body (n, 23 chromosomes)

- Secondary oocyte is arrested at Metaphase-II
\(\downarrow\) Meiosis-II (completed after fertilization)

Ovum (n, 23 chromosomes) + Second polar body (n, 23 chromosomes)



(ii) Correlation with Life Phases:

- Fetal Life: Oogonia are formed and multiply by mitosis. They start meiosis and get arrested at the prophase-I stage, becoming primary oocytes. No more oogonia are formed after birth.


- Childhood to Puberty: The primary oocytes remain arrested in Prophase-I.


- At Puberty: The primary oocyte within a maturing follicle completes its first meiotic division just before ovulation to form a large secondary oocyte and a tiny first polar body.


- During Ovulation: The secondary oocyte, arrested in metaphase-II, is released from the ovary.


- After Fertilization: If a sperm enters the secondary oocyte, it completes the second meiotic division, resulting in the formation of a haploid ovum and a second polar body.


OR


(B) Pollination in Chasmogamous Flowers


(i) Three Types of Pollination:

Chasmogamous flowers are open flowers with exposed anthers and stigma. A chasmogamous bisexual flower can undergo three types of pollination:


1. Autogamy: This is self-pollination where pollen grains are transferred from the anther to the stigma of the \textit{same flower. It leads to inbreeding.


2. Geitonogamy: This is the transfer of pollen grains from the anther of one flower to the stigma of \textit{another flower on the same plant. Genetically, it is similar to autogamy since the pollen comes from the same plant.


3. Xenogamy: This is cross-pollination where pollen grains are transferred from the anther of a flower on one plant to the stigma of a flower on a \textit{different plant of the same species. It is the only type of pollination that brings genetically different types of pollen to the stigma.


(ii) Diagram of a Mature Pollen Grain (2-celled stage):


Labels:

1. Exine (outer layer)

2. Intine (inner layer)

3. Vegetative cell (or Tube cell)

4. Generative cell
Quick Tip: For oogenesis, remember the two key arrest points: Prophase I (from birth to puberty) and Metaphase II (at ovulation, completed only upon fertilization). For pollination types, focus on the source and destination: same flower (auto-), different flower/same plant (geitono-), different plant (xeno-).

(A) Recombinant DNA Technology


(i) Making a Bacterial Cell Competent:

DNA is a hydrophilic molecule, so it cannot readily pass through the cell membrane. To make a bacterial cell competent to take up DNA, the following procedure is used:


1. Chemical Treatment: The bacterial cells are treated with a specific concentration of a divalent cation, such as calcium chloride (\(CaCl_2\)). This increases the efficiency with which DNA enters the bacterium through pores in its cell wall.


2. Incubation on Ice: The cells are incubated with the recombinant DNA on ice.


3. Heat Shock: The cells are then briefly placed at a higher temperature (e.g., 42°C) and then put back on ice. This heat shock creates transient pores in the bacterial cell membrane, allowing the recombinant DNA to enter the cell.


(ii) Steps of PCR (Polymerase Chain Reaction):

PCR is used to amplify a gene of interest into millions of copies. It involves three main steps repeated in cycles:


1. Denaturation: The double-stranded DNA is heated to a high temperature (94-96°C). This breaks the hydrogen bonds between the two strands, separating them into single strands to act as templates.


2. Annealing: The temperature is lowered (50-65°C), which allows short, synthetic DNA primers to bind (anneal) to their specific complementary sequences on the single-stranded DNA templates.


3. Extension (Elongation): The temperature is raised to 72°C, the optimal temperature for the thermostable DNA polymerase (like Taq polymerase). The polymerase adds nucleotides to the 3' end of the primers, synthesizing a new complementary strand of DNA.

This three-step cycle is repeated 25-35 times, leading to an exponential amplification of the target DNA segment.


OR


(B) Transgenic Animals


(i) Definition:

Transgenic animals are animals that have had their DNA manipulated to possess and express an extra, foreign gene from another species.


(ii) Four Reasons for Producing Transgenic Animals:

1. Study of Disease: Many transgenic animals are designed to serve as models for human diseases. This allows scientists to study how genes contribute to the development of a disease and to test new treatments. Examples include models for cancer and Alzheimer's disease.


2. Biological Products: Transgenic animals can be created to produce useful biological products, a field known as molecular farming. For example, transgenic sheep can produce human proteins like alpha-1-antitrypsin in their milk, which is used to treat emphysema.


3. Vaccine Safety Testing: Before being used on humans, vaccines must be tested for safety. Transgenic mice are being developed for use in testing the safety of vaccines, such as the polio vaccine.


4. Normal Physiology and Development: Transgenic animals can be used to study how genes are regulated and how they affect the normal functions of the body and its development. For example, studying the effects of growth factors like insulin-like growth factor.
Quick Tip: Remember the PCR steps with the mnemonic "DAE": Denature, Anneal, Extend. For transgenic animals, focus on their applications in medicine: understanding disease, producing drugs, and testing safety.

(A) Tropical Biodiversity and Species-Area Relationship


(i) Reasons for High Tropical Species Diversity:

1. Longer Evolutionary Time: Tropical latitudes have remained relatively undisturbed by glaciations for millions of years. This long period of stability has allowed for uninterrupted evolution and speciation, leading to greater species diversity.


2. Constant and Predictable Environment: The tropical environment is less seasonal and more predictable than temperate regions. This constancy promotes niche specialisation and allows a greater number of species to coexist.


3. Higher Solar Energy and Productivity: The tropics receive more intense solar energy, which leads to higher primary productivity. This greater availability of food resources can support a larger number of species at different trophic levels.


(ii) Species-Area Relationship:


- Naturalist: The relationship was first observed and studied by the German naturalist and geographer Alexander von Humboldt.


- Observation: He observed that within a region, species richness increases with increasing explored area, but only up to a certain limit. The relationship is a rectangular hyperbola, described by the equation S = cA\(^z\), where S is species richness, A is area, and c and z are constants.


OR


(B) Biodiversity Loss and Conservation


(i) Three Major Causes of Biodiversity Loss (The 'Evil Quartet'):

1. Habitat Loss and Fragmentation: This is the single most important cause. Deforestation for agriculture, mining, and urbanization destroys the natural habitats of countless species. When large habitats are broken up into small, isolated fragments, it can lead to population declines and extinctions.


2. Over-exploitation: Humans have always depended on nature for food and shelter, but when the need turns to greed, it leads to the over-exploitation of natural resources. Over-harvesting of species (e.g., over-fishing) has endangered many commercially important species and led to extinctions (e.g., Steller's sea cow, passenger pigeon).


3. Alien Species Invasions: When new (exotic) species are introduced into a habitat, either intentionally or unintentionally, they may turn invasive and cause the decline or extinction of indigenous species. They can outcompete native species for resources or introduce new diseases. (e.g., The introduction of Nile Perch into Lake Victoria in east Africa led to the extinction of more than 200 species of cichlid fish).


(ii) 'Ex situ' Conservation Approach:

'Ex situ' (off-site) conservation is the approach of protecting and conserving threatened plants and animals \textit{outside their natural habitats. It is used when a species is critically endangered and its survival in its natural habitat is not possible.


Two Examples of Ex situ Conservation:

1. Zoological Parks (Zoos) and Botanical Gardens: These places maintain and breed threatened animal and plant species under protected conditions with the aim of eventually reintroducing them into the wild.


2. Cryopreservation and Gene Banks: This involves preserving the gametes, seeds, or tissues of threatened species in a viable condition at very low temperatures (e.g., in liquid nitrogen at -196°C). These can be used in the future to revive the species.
Quick Tip: For biodiversity, remember the 'Evil Quartet' as the main causes of loss: Habitat Loss, Over-exploitation, Alien Species Invasion, and Co-extinctions. Conservation strategies are either In situ (in nature) or Ex situ (outside nature).