Class 12 Biology Chapter-Wise Subjective Question Bank PDF — 15 Questions Per Chapter (Genetics & Reproduction)

(a) Development of Embryo Sac:
1. The Megaspore Mother Cell (MMC) undergoes meiosis to form a linear tetrad of four haploid megaspores.
2. Three of the megaspores, usually at the micropylar end, degenerate.
3. The one functional megaspore at the chalazal end enlarges.
4. The nucleus of the functional megaspore undergoes three successive free nuclear mitotic divisions, resulting in eight nuclei.
5. These eight nuclei arrange themselves: three move to the micropylar end to form the egg apparatus (one egg cell and two synergids), three move to the chalazal end to form antipodal cells, and the remaining two nuclei (polar nuclei) are situated in the large central cell below the egg apparatus.
6. Cell walls are then laid down, resulting in a 7-celled, 8-nucleate mature embryo sac.

(b) Monosporic Development:
It is called monosporic because the embryo sac develops from a single functional megaspore out of the four megaspores formed from the megaspore mother cell.

Difference between Endosperm and Embryo Development:
Endosperm Development: The Primary Endosperm Nucleus (PEN) undergoes repeated mitotic divisions to form a triploid endosperm tissue. Initially, these divisions are free-nuclear, creating a liquid endosperm. Later, cell wall formation occurs, making it cellular. Its primary function is to provide nutrition to the developing embryo.
Embryo Development: The diploid zygote divides mitotically to form the embryo. The development is more organized, proceeding through specific stages (proembryo, globular, heart-shaped, and mature embryo). It develops into the future plant.

Stages of Embryo Development (Dicot):
1. The zygote divides asymmetrically into a larger suspensor cell and a smaller embryonal cell.
2. The embryonal cell divides to form a proembryo.
3. Further divisions lead to a globular-shaped embryo.
4. The globular embryo then develops into a heart-shaped embryo due to the growth of two cotyledons. The suspensor pushes the embryo deep into the endosperm.
This development is called embryogeny.

(a) Development of Embryo Sac:
1. The Megaspore Mother Cell (MMC) is a large diploid cell in the nucellus of the ovule.
2. It undergoes meiosis to form a linear tetrad of four haploid megaspores.
3. Out of these four, three megaspores (usually towards the micropylar end) degenerate.
4. The one functional megaspore (usually towards the chalazal end) enlarges and becomes the first cell of the female gametophyte.
5. Its nucleus undergoes three successive free nuclear mitotic divisions, resulting in 8 nuclei within the same cytoplasm.
6. Cell walls then form, organizing these nuclei into a 7-celled, 8-nucleate embryo sac. This includes: one egg cell and two synergids at the micropylar end (egg apparatus), three antipodal cells at the chalazal end, and one large central cell with two polar nuclei.

(b) Role after fertilization:
* **Synergids:** One synergid guides the pollen tube entry into the embryo sac and degenerates upon its arrival. The other synergid also degenerates shortly after fertilization.
* **Antipodal cells:** These cells provide nutrition to the developing embryo sac and typically degenerate before or soon after fertilization.

(a) Sequence of events:
1. Pollen germination: The compatible pollen grain absorbs moisture and nutrients from the stigmatic surface and germinates. The tube cell emerges through a germ pore to form a pollen tube.
2. Pollen tube growth: The pollen tube grows down through the tissues of the stigma and style, guided by chemical signals.
3. Entry into ovule: The pollen tube enters the ovule, usually through the micropyle (porogamy). It then passes through the nucellus and penetrates the embryo sac, typically entering one of the synergids.
4. Discharge of male gametes: The tip of the pollen tube ruptures, releasing the two male gametes into the cytoplasm of the synergid.

(b) Pollen-Pistil Interaction:
It is a dynamic process involving pollen recognition followed by promotion or inhibition of the pollen. This interaction is mediated by chemical components of the pollen interacting with those of the pistil. If the pollen is compatible (of the right type), the pistil accepts the pollen and promotes post-pollination events. If the pollen is incompatible (of the wrong type), the pistil rejects the pollen by preventing pollen germination or pollen tube growth in the style. This recognition ability is the result of a continuous dialogue between pollen grain and the pistil.

(a) Double Fertilization:
1. The pollen tube, carrying two male gametes, enters the ovule and reaches the embryo sac, usually through a synergid.
2. The pollen tube releases the two male gametes into the cytoplasm of the synergid.
3. **Syngamy:** One of the male gametes fuses with the egg cell nucleus to form a diploid zygote (2n).
4. **Triple Fusion:** The other male gamete moves towards the central cell and fuses with the two polar nuclei (or the secondary nucleus) to form the triploid Primary Endosperm Nucleus (PEN) (3n).
Since two types of fusion, syngamy and triple fusion, occur in the embryo sac, the phenomenon is termed double fertilization.

(b) Significance:
* It leads to the formation of a zygote which develops into an embryo.
* It results in the formation of the Primary Endosperm Nucleus (PEN), which develops into the endosperm. The endosperm provides nourishment to the developing embryo, increasing the chances of seed viability and seedling survival.

(c) Structures formed:
* **Diploid (2n):** Zygote
* **Triploid (3n):** Primary Endosperm Nucleus (PEN) which later forms the Endosperm.

Parturition is induced by a complex neuro-endocrine mechanism involving cortisol, estrogens, and oxytocin.
**Mechanism of Parturition:**
1. **Initiation:** Signals for parturition originate from the fully developed fetus and the placenta, which induce mild uterine contractions called the fetal ejection reflex.
2. **Hormonal Role:** This triggers the release of oxytocin from the maternal pituitary.
3. **Positive Feedback Loop:** Oxytocin acts on the uterine muscles, causing stronger contractions. This, in turn, stimulates further secretion of oxytocin. The stimulatory reflex between the uterine contraction and oxytocin secretion continues, resulting in stronger and stronger contractions.
4. **Expulsion:** This leads to the expulsion of the baby out of the uterus through the birth canal.

**Initiation of Lactation:**
During pregnancy, the breasts undergo differentiation. After delivery, the sudden drop in progesterone and estrogen levels allows the hormone prolactin (from the pituitary) to act on the mammary glands, stimulating milk production (lactogenesis). The initial milk produced, called colostrum, is rich in antibodies.

(a) **Follicular Phase:** In the ovary, the primary follicles grow to become a fully mature Graafian follicle. In the uterus, the endometrium regenerates through proliferation.
**Luteal Phase:** In the ovary, the remaining parts of the Graafian follicle transform into the corpus luteum. In the uterus, the endometrium further thickens and its glands become secretory, preparing for implantation. If fertilization does not occur, the corpus luteum degenerates, causing the disintegration of the endometrium leading to menstruation.

(b) **Follicular Phase:** Gonadotropins (LH and FSH) increase, stimulating follicular development and secretion of estrogens by the growing follicles. Estrogen causes the proliferation of the endometrium.
**Luteal Phase:** The rapid secretion of LH (LH surge) induces rupture of the Graafian follicle and ovulation. After ovulation, the corpus luteum is formed which secretes large amounts of progesterone. Progesterone is essential for the maintenance of the endometrium. In the absence of fertilization, the decline in progesterone levels causes menstruation.

(a) Ovulation is most likely to occur around the 14th day of a 28-day cycle.

(b) Events during the follicular phase:
In the Ovary: The primary follicles grow to become fully mature Graafian follicles. This is stimulated by FSH and LH.
In the Uterus: The endometrium regenerates through proliferation. This is stimulated by the increasing levels of ovarian hormones (estrogen).

(c) Hormonal Changes:
- Before Ovulation: FSH and LH levels gradually increase, stimulating follicular development and estrogen secretion. Estrogen levels rise, peaking just before ovulation. This high level of estrogen triggers a rapid secretion of LH (LH surge) which induces the rupture of the Graafian follicle, leading to ovulation.
- After Ovulation: The remaining parts of the Graafian follicle transform into the corpus luteum, which secretes large amounts of progesterone. Progesterone is essential for the maintenance of the endometrium.

(a) Sequence of events during fertilization:
1. Acrosomal reaction: The sperm's acrosome releases enzymes (like hyaluronidase) to penetrate the corona radiata and zona pellucida of the ovum.
2. Cortical reaction: Once a sperm fuses with the ovum's plasma membrane, it induces changes in the zona pellucida that block the entry of additional sperms (polyspermy block).
3. Completion of Meiosis II: The entry of sperm induces the completion of the meiotic division of the secondary oocyte, forming a second polar body and a haploid ovum (ootid).
4. Syngamy: The haploid nucleus of the sperm and that of the ovum fuse together to form a diploid zygote.

(b) Implantation is the process by which the blastocyst attaches to and embeds itself in the endometrium of the uterus. The trophoblast layer of the blastocyst gets attached to the endometrium. The uterine cells divide rapidly and cover the blastocyst, resulting in it becoming fully embedded in the endometrium.

(c) The three embryonic layers (germ layers) formed during gastrulation are: Ectoderm, Mesoderm, and Endoderm.

(a) Ovulation is most likely to occur around the 14th day of a 28-day cycle. The two pituitary hormones involved are Luteinizing Hormone (LH) and Follicle-Stimulating Hormone (FSH). A surge in LH level (LH surge) is the primary trigger for ovulation. (2 marks)
(b) After ovulation, the ruptured Graafian follicle transforms into a yellow glandular structure called the Corpus Luteum. The corpus luteum secretes large amounts of progesterone, which is essential for the maintenance of the endometrium, making it receptive to implantation of the fertilized ovum. (2 marks)
(c) If fertilization does not occur, the corpus luteum degenerates. This causes a fall in the level of progesterone, leading to the disintegration of the endometrium. This results in menstruation, marking the start of a new cycle. (1 mark)

As a health counsellor, I would suggest the following two categories of contraceptive methods for the couple:

1. **Barrier Methods (e.g., Condoms):**
* **Justification:** These are easily available, reversible, and offer the dual protection of preventing pregnancy and sexually transmitted infections (STIs). They have no hormonal side effects.
* **Mechanism of Action:** They act as a physical barrier, preventing the entry of ejaculated semen into the female reproductive tract, thus blocking the meeting of sperm and ovum.

2. **Hormonal Contraceptives (e.g., Oral Contraceptive Pills):**
* **Justification:** These are highly effective and reversible. Once the couple decides to have a child, they can simply stop taking the pills. Modern low-dose pills have minimal side effects for most healthy women.
* **Mechanism of Action:** These pills contain hormones (progestogen or a combination of progestogen-estrogen) that inhibit ovulation (release of the egg) and also alter the quality of cervical mucus to prevent the entry of sperm.

(a)
* **Disease:** Hepatitis-B
* **Causative Agent:** Hepatitis-B Virus (HBV)

(b) **Mode of Transmission:** Most STIs are transmitted through sexual contact with an infected person. Other modes include sharing of injection needles, surgical instruments with infected persons, transfusion of blood from an infected person, or from an infected mother to the foetus.

(c) **Three Preventive Measures:**
1. Always use condoms during coitus.
2. Avoid sex with unknown partners or multiple partners.
3. In case of doubt, one should go to a qualified doctor for early detection and get complete treatment if diagnosed with an infection.

(a) **Intra-Cytoplasmic Sperm Injection (ICSI)** would be the most appropriate technology for this couple.

(b) **Procedure of ICSI:** In this specialized procedure, a single healthy sperm is selected and directly injected into the cytoplasm of an ovum (collected from the wife/donor) in the laboratory. The resulting embryo is then transferred into the uterus of the female to achieve pregnancy. This bypasses the need for a high sperm count for natural fertilization.

(c) **Gamete Intra-Fallopian Transfer (GIFT):** This is another ART where fertilization occurs in vivo (inside the body). In this method, an ovum collected from a donor is transferred into the fallopian tube of another female who cannot produce one but can provide a suitable environment for fertilization and further development. Sperm is also introduced so that fertilization can occur naturally within the fallopian tube.

Suitable contraceptive methods for the couple would be Barrier Methods and Intra-Uterine Devices (IUDs) as they are effective and reversible.

1. Category: Barrier Method
- Method: Condoms (for male or female).
- Mode of Action: They prevent the physical meeting of sperm and ovum by creating a barrier.
- Advantage: Protects against Sexually Transmitted Infections (STIs) and is easily available.
- Disadvantage: May have issues with incorrect usage, leading to failure, or can tear.

2. Category: Intra-Uterine Devices (IUDs)
- Method: Copper T (CuT).
- Mode of Action: The copper ions released suppress sperm motility and their fertilizing capacity. It also causes phagocytosis of sperms within the uterus.
- Advantage: Highly effective, long-lasting (3-5 years), and reversible once removed.
- Disadvantage: Needs to be inserted by a doctor or trained nurse; can cause increased menstrual bleeding or cramping initially.

(Other valid examples like Oral Contraceptive Pills could also be explained).

a) The practice of determining the sex of the fetus is banned in India under the Pre-conception and Pre-natal Diagnostic Techniques (PCPNDT) Act to prevent female foeticide, which has led to a skewed sex ratio in many parts of the country. The technique they are referring to is Amniocentesis.
Principle of Amniocentesis: A small amount of amniotic fluid, which contains fetal cells, is collected from the uterus. These cells are cultured and their chromosomes are analyzed (karyotyping) to detect any genetic abnormalities.

b) This technique is legally permitted only to detect certain genetic disorders such as Down's syndrome, haemophilia, sickle-cell anemia, or other chromosomal abnormalities in the fetus. Its primary purpose is to diagnose serious congenital and genetic diseases, allowing for medical advice and preparation.

(a) F1 Generation:
Phenotype: All tall plants with violet flowers.
Genotype: All heterozygous for both traits (TtWw).

(b) F2 Generation Cross:
A dihybrid cross between TtWw x TtWw results in a phenotypic ratio of 9:3:3:1.
- 9 Tall, Violet
- 3 Tall, White
- 3 Dwarf, Violet
- 1 Dwarf, White
The genotypes would be in the ratio of 1:2:1:2:4:2:1:2:1 (e.g., TTWW, TtWW, ttWW, TTWw, TtWw, ttWw, TTww, Ttww, ttww).

(c) Law of Independent Assortment. This law states that when two pairs of traits are combined in a hybrid, the segregation of one pair of characters is independent of the other pair of characters.

(a) Let C be the allele for normal vision and c be the allele for colour blindness.
Man's genotype: XcY
Woman's genotype: XCXc
Cross:
XcY x XCXc
Progeny:
- Daughter (XCXc): Carrier, normal vision (25%)
- Daughter (XcXc): Colour-blind (25%)
- Son (XCY): Normal vision (25%)
- Son (XcY): Colour-blind (25%)
Probability: 50% of daughters will be colour-blind. 50% of sons will be colour-blind. Overall, 50% of children will be colour-blind.

(b) Colour blindness is an X-linked recessive trait. Males have only one X chromosome (XY). If they inherit one recessive allele (Xc) on their X chromosome from their mother, they will express the trait. Females have two X chromosomes (XX). They need to inherit the recessive allele from both parents (XcXc) to be colour-blind. Having one recessive allele (XCXc) makes them a carrier with normal vision.

(c) Haemophilia.

(a) Aneuploidy is the phenomenon of gain or loss of one or more chromosomes, which results from the failure of segregation of chromatids during cell division. Example: Down's syndrome (Trisomy of chromosome 21) or Turner's syndrome (Monosomy of X chromosome).

(b) Klinefelter's syndrome is caused by the presence of an additional copy of the X-chromosome, resulting in a karyotype of 47, XXY. This is a trisomy of the sex chromosomes.

(c) Characteristic symptoms of Klinefelter's syndrome include:
1. The individual is male but shows feminine development (gynaecomastia, i.e., development of breasts).
2. Overall masculine development is seen, but individuals are sterile.
3. They are often tall with long limbs and have poor beard growth.

a) Man's genotype: Iᴬi
Woman's genotype: Iᴮi
Child's genotype: ii
b) Other possible blood groups are A (Iᴬi), B (Iᴮi), and AB (IᴬIᴮ).
Cross:
Parents: Iᴬi (Man) × Iᴮi (Woman)
Gametes: (Iᴬ, i) and (Iᴮ, i)
Progeny: IᴬIᴮ (AB group), Iᴬi (A group), Iᴮi (B group), ii (O group). The probability for each blood group is 25%.
c) This demonstrates Co-dominance and Multiple Allelism.
Co-dominance: Both alleles Iᴬ and Iᴮ express themselves completely when present together, resulting in the AB blood group. Neither allele is dominant or recessive to the other.
Multiple Allelism: The gene 'I' for blood group exists in more than two allelic forms: Iᴬ, Iᴮ, and i.

a) The phenotype of the F1 generation will be Round and Yellow seeds, as these are the dominant traits. The genotype will be RrYy.
b) The phenotypic ratio observed in the F2 generation will be 9 (Round, Yellow) : 3 (Round, Green) : 3 (Wrinkled, Yellow) : 1 (Wrinkled, Green).
c) This cross demonstrates Mendel's Law of Independent Assortment. It states that when two pairs of traits are combined in a hybrid, the segregation of one pair of characters is independent of the other pair of characters.
d) The gametes from F1 (RrYy) would be RY, Ry, rY, and ry. The Punnett square for the F2 generation would be a 16-square grid showing the combination of these gametes, resulting in the 9:3:3:1 phenotypic ratio. (A correctly drawn Punnett square is expected).

(a) Meselson and Stahl grew E. coli in a medium containing ¹⁵NH₄Cl (a heavy isotope of Nitrogen) for many generations. This resulted in the incorporation of ¹⁵N into the newly synthesized DNA. This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a CsCl density gradient. Then they transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various definite time intervals as the cells multiplied. The DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of ‘light’ DNA.
(b) CsCl density gradient centrifugation was used because it separates DNA molecules based on their densities. The heavy DNA with ¹⁵N settled lower down the tube than the light DNA with ¹⁴N. The hybrid DNA (¹⁵N-¹⁴N) settled at a level intermediate between the heavy and light DNA, which was crucial for their conclusion.

(a) Initiation: Involves the charging or aminoacylation of tRNA, where a specific amino acid is activated and linked to its cognate tRNA. The small ribosomal subunit binds to the mRNA at the start codon (AUG). Then, the initiator tRNA (carrying methionine) binds to the start codon. Finally, the large ribosomal subunit binds to this complex, forming the complete initiation complex.
(b) Elongation: A second charged tRNA with an appropriate anticodon binds to the A-site of the ribosome. A peptide bond is formed between the amino acid on the initiator tRNA (at the P-site) and the amino acid on the tRNA at the A-site. The ribosome then translocates along the mRNA, moving the tRNA from the A-site to the P-site, and the uncharged tRNA from the P-site to the E-site, from where it exits. This cycle repeats.
(c) Termination: Elongation continues until a stop codon (UAA, UAG, or UGA) on the mRNA reaches the A-site. Release factors bind to the stop codon, causing the hydrolysis of the bond between the polypeptide chain and the tRNA at the P-site. The completed polypeptide is released, and the ribosomal subunits dissociate from the mRNA.

(a) When lactose is absent: The regulator gene (i) synthesizes a repressor protein. This repressor protein binds to the operator region (o) of the operon. This binding prevents RNA polymerase from transcribing the structural genes (z, y, and a). As a result, the enzymes required for lactose metabolism are not synthesized.
(b) When lactose is present: Lactose (or its metabolite allolactose) acts as an inducer. It binds to the repressor protein, causing a conformational change that inactivates it. The inactivated repressor cannot bind to the operator region. This allows RNA polymerase to access the promoter and transcribe the structural genes z, y, and a. The transcription leads to the synthesis of enzymes (β-galactosidase, permease, and transacetylase) necessary for lactose metabolism.

(a) According to Chargaff's rule, A=T and G=C.
Number of Adenine (A) = 520. Therefore, Number of Thymine (T) = 520.
Total number of A+T = 520 + 520 = 1040.
Total nucleotides = 2000.
Number of G+C = 2000 - 1040 = 960.
Since G=C, Number of Guanine (G) = 480 and Number of Cytosine (C) = 480.
Pyrimidine bases are Cytosine (C) and Thymine (T).
Total pyrimidines = C + T = 480 + 520 = 1000.

(b) DNA replication in E. coli is semi-conservative. The main enzyme is DNA-dependent DNA polymerase. It polymerises nucleotides in the 5'→3' direction. Since the two strands of the DNA helix are anti-parallel, one strand with polarity 3'→5' acts as a template for continuous synthesis (leading strand). The other strand with polarity 5'→3' acts as a template for discontinuous synthesis in the form of short fragments called Okazaki fragments (lagging strand). These fragments are later joined together by the enzyme DNA ligase.

(a) Template Strand: It has polarity 3'→5'. It acts as a template for the synthesis of mRNA during transcription. The sequence of mRNA is complementary to this strand (with U in place of T).
Coding Strand: It has polarity 5'→3'. It does not code for anything. Its sequence is the same as that of the mRNA transcribed (except that T is present instead of U).

(b) Given template strand: 3'–ATGCATGCATGC–5'
(i) Complementary coding strand: 5'–TACGTACGTACG–3'
(ii) mRNA transcribed: 5'–UACGUACGUACG–3'

(c) In bacteria, a single DNA-dependent RNA polymerase catalyses the transcription of all types of RNA. RNA polymerase binds to the promoter and initiates transcription (initiation). It uses nucleoside triphosphates as a substrate and polymerises in a template-dependent fashion following the rule of complementarity. It facilitates the opening of the helix and continues the elongation of the RNA chain. When it reaches the terminator region, the nascent RNA falls off, as does the RNA polymerase, terminating the process.

(a) Frequency of brown beetles (genotype bb), q² = 360/1000 = 0.36. Therefore, the frequency of allele 'b' (q) = √0.36 = 0.6. Since p + q = 1, the frequency of allele 'B' (p) = 1 - 0.6 = 0.4.
(b) Frequency of genotype BB (p²) = (0.4)² = 0.16. Frequency of genotype Bb (2pq) = 2 * 0.4 * 0.6 = 0.48.
(c) The number of heterozygous individuals (Bb) = Frequency of Bb × Total population = 0.48 × 1000 = 480 beetles.
(d) Two factors that can disrupt Hardy-Weinberg equilibrium are: Gene flow (or gene migration) and Genetic drift. (Also acceptable: Mutation, Natural selection, Non-random mating).

(a) The type of speciation is Allopatric Speciation. This occurs when a population is divided by a geographical barrier (like a river), preventing gene flow. The isolated populations then evolve independently due to different selective pressures, genetic drift, and mutations, eventually leading to reproductive isolation and the formation of new species.
(b) Three pre-zygotic isolating mechanisms could be:
1. **Ecological/Habitat Isolation:** The two populations may have adapted to different habitats (e.g., one on the riverbank, one in the forest) and thus do not meet to mate.
2. **Temporal Isolation:** The mating seasons of the two populations might have shifted to different times of the year.
3. **Behavioral/Mating Isolation:** The courtship rituals or mating signals (e.g., specific pheromones or dance patterns) may have diverged, so individuals of one population are no longer attracted to the other.

(a) The fossil distribution suggests a direct evolutionary lineage where organism X likely evolved into Y, and Y evolved into Z over geological time. The appearance of more complex forms in younger rock layers (higher up) indicates progressive evolution. The structural similarities support the idea of descent with modification from a common ancestor.
(b) This evidence supports evolution by demonstrating that: (i) Life forms have changed over time (are not static). (ii) Different life forms appeared at different times in Earth's history. (iii) It shows transitional forms, providing a record of the sequence of evolutionary change from simpler to more complex organisms.
(c) Two other types of evidence for evolution are:
1. **Comparative Anatomy and Morphology:** Evidence from homologous and analogous structures, as well as vestigial organs.
2. **Biochemical Evidence:** Similarities in proteins, DNA sequences, and metabolic pathways among different species.

(a) Homologous structures have a common embryonic origin and similar basic structure but perform different functions (e.g., forelimbs of humans, cheetahs, whales, and bats). Analogous structures have different origins and structures but perform similar functions (e.g., wings of a butterfly and a bird).
(b) Homologous structures suggest a common ancestry and divergent evolution, where related species adapt to different needs. Analogous structures suggest convergent evolution, where unrelated species evolve similar traits independently to adapt to similar environments or ecological niches.
(c) The wings of a bat and a butterfly are analogous structures. This represents convergent evolution. It shows that different organisms, without a recent common ancestor, can independently evolve similar features as adaptations to a similar way of life (in this case, flying).

(a) The aim was to simulate the conditions thought to be present on the early Earth and to test the hypothesis that organic molecules could be synthesized from inorganic precursors.
(b) Gaseous mixture: Methane (CH₄), Ammonia (NH₃), Hydrogen (H₂), and Water vapour (H₂O) in a closed flask. Conditions: High temperature (800°C) and continuous electric sparks to simulate lightning.
(c) After a week, they observed the formation of a collection of organic molecules in the condensed water, including amino acids like glycine, alanine, and aspartic acid.
(d) The significant conclusion was that organic molecules, the building blocks of life, could indeed form abiotically under the primitive Earth's conditions. This provided strong evidence for the theory of chemical evolution.

Parturition is induced by a complex neuro-endocrine mechanism involving cortisol, estrogens, and oxytocin.
**Mechanism of Parturition:**
1. **Initiation:** Signals for parturition originate from the fully developed fetus and the placenta, which induce mild uterine contractions called the fetal ejection reflex.
2. **Hormonal Role:** This triggers the release of oxytocin from the maternal pituitary.
3. **Positive Feedback Loop:** Oxytocin acts on the uterine muscles, causing stronger contractions. This, in turn, stimulates further secretion of oxytocin. The stimulatory reflex between the uterine contraction and oxytocin secretion continues, resulting in stronger and stronger contractions.
4. **Expulsion:** This leads to the expulsion of the baby out of the uterus through the birth canal.

**Initiation of Lactation:**
During pregnancy, the breasts undergo differentiation. After delivery, the sudden drop in progesterone and estrogen levels allows the hormone prolactin (from the pituitary) to act on the mammary glands, stimulating milk production (lactogenesis). The initial milk produced, called colostrum, is rich in antibodies.

(a) **Follicular Phase:** In the ovary, the primary follicles grow to become a fully mature Graafian follicle. In the uterus, the endometrium regenerates through proliferation.
**Luteal Phase:** In the ovary, the remaining parts of the Graafian follicle transform into the corpus luteum. In the uterus, the endometrium further thickens and its glands become secretory, preparing for implantation. If fertilization does not occur, the corpus luteum degenerates, causing the disintegration of the endometrium leading to menstruation.

(b) **Follicular Phase:** Gonadotropins (LH and FSH) increase, stimulating follicular development and secretion of estrogens by the growing follicles. Estrogen causes the proliferation of the endometrium.
**Luteal Phase:** The rapid secretion of LH (LH surge) induces rupture of the Graafian follicle and ovulation. After ovulation, the corpus luteum is formed which secretes large amounts of progesterone. Progesterone is essential for the maintenance of the endometrium. In the absence of fertilization, the decline in progesterone levels causes menstruation.

(a) Ovulation is most likely to occur around the 14th day of a 28-day cycle.

(b) Events during the follicular phase:
In the Ovary: The primary follicles grow to become fully mature Graafian follicles. This is stimulated by FSH and LH.
In the Uterus: The endometrium regenerates through proliferation. This is stimulated by the increasing levels of ovarian hormones (estrogen).

(c) Hormonal Changes:
- Before Ovulation: FSH and LH levels gradually increase, stimulating follicular development and estrogen secretion. Estrogen levels rise, peaking just before ovulation. This high level of estrogen triggers a rapid secretion of LH (LH surge) which induces the rupture of the Graafian follicle, leading to ovulation.
- After Ovulation: The remaining parts of the Graafian follicle transform into the corpus luteum, which secretes large amounts of progesterone. Progesterone is essential for the maintenance of the endometrium.

(a) Sequence of events during fertilization:
1. Acrosomal reaction: The sperm's acrosome releases enzymes (like hyaluronidase) to penetrate the corona radiata and zona pellucida of the ovum.
2. Cortical reaction: Once a sperm fuses with the ovum's plasma membrane, it induces changes in the zona pellucida that block the entry of additional sperms (polyspermy block).
3. Completion of Meiosis II: The entry of sperm induces the completion of the meiotic division of the secondary oocyte, forming a second polar body and a haploid ovum (ootid).
4. Syngamy: The haploid nucleus of the sperm and that of the ovum fuse together to form a diploid zygote.

(b) Implantation is the process by which the blastocyst attaches to and embeds itself in the endometrium of the uterus. The trophoblast layer of the blastocyst gets attached to the endometrium. The uterine cells divide rapidly and cover the blastocyst, resulting in it becoming fully embedded in the endometrium.

(c) The three embryonic layers (germ layers) formed during gastrulation are: Ectoderm, Mesoderm, and Endoderm.

(a) Ovulation is most likely to occur around the 14th day of a 28-day cycle. The two pituitary hormones involved are Luteinizing Hormone (LH) and Follicle-Stimulating Hormone (FSH). A surge in LH level (LH surge) is the primary trigger for ovulation. (2 marks)
(b) After ovulation, the ruptured Graafian follicle transforms into a yellow glandular structure called the Corpus Luteum. The corpus luteum secretes large amounts of progesterone, which is essential for the maintenance of the endometrium, making it receptive to implantation of the fertilized ovum. (2 marks)
(c) If fertilization does not occur, the corpus luteum degenerates. This causes a fall in the level of progesterone, leading to the disintegration of the endometrium. This results in menstruation, marking the start of a new cycle. (1 mark)

(a) The bacterium is *Bacillus thuringiensis* (Bt).
(b) Spores of *Bacillus thuringiensis* are available in sachets as a dry powder. These are mixed with water and sprayed onto vulnerable plants like cotton. When the insect larvae (bollworms) ingest the plant parts, the toxin is released in their gut, killing them.
(c) The toxin exists as an inactive protoxin (crystal protein) in the bacterium. It gets activated only in the alkaline pH of an insect's gut, which solubilizes the crystals. This activated toxin binds to the surface of midgut epithelial cells, creating pores that cause cell swelling and lysis, leading to the death of the insect. Since the bacterium's internal environment and the gut of other non-target insects (like bees) are not alkaline, the protoxin remains inactive and harmless.

Two possible reasons for low gas yield:
1. Lack of methanogenic bacteria: The initial slurry might not have a sufficient population of methanogens. Adding a small amount of slurry from a previously working biogas plant can act as an inoculum.
2. Incorrect temperature/pH: The digester tank may not be at the optimal temperature (around 35-40°C) or pH for methanogens to function efficiently.

The three stages of biogas production are:
(i) Solubilisation/Hydrolysis: Complex organic polymers like cellulose, proteins, and fats are broken down into simpler soluble monomers by hydrolytic enzymes produced by fermentative bacteria.
(ii) Acidogenesis: The simple monomers are further converted by acidogenic bacteria into organic acids, primarily acetic acid.
(iii) Methanogenesis: This is the final stage where methanogenic bacteria act on the organic acids to produce a mixture of gases, primarily methane (CH₄), along with carbon dioxide (CO₂) and traces of H₂S. This stage is strictly anaerobic.

(a) Microbial Sources:
Cyclosporin A: Fungus *Trichoderma polysporum*.
Statins: Yeast *Monascus purpureus*.
(b) Mode of Action:
Cyclosporin A: It acts as an immunosuppressive agent by inhibiting the activation of T-lymphocytes, which are responsible for cell-mediated immunity that causes graft rejection.
Statins: They act by competitively inhibiting the enzyme HMG-CoA reductase, which is responsible for the synthesis of cholesterol in the body. This lowers blood cholesterol levels.
(c) Cyclosporin A is essential because the body's immune system recognizes the transplanted organ as foreign and mounts an immune response (cell-mediated) to reject it. By suppressing the immune system, particularly T-cells, Cyclosporin A prevents this rejection, allowing the organ to function in the recipient's body.

(a)
- **Swiss Cheese:** The bacterium *Propionibacterium shermanii*.
- **Yoghurt:** *Lactobacillus* species (e.g., *Lactobacillus bulgaricus*) and *Streptococcus thermophilus*.

(b)
- **Swiss Cheese Formation:** *Propionibacterium shermanii* is used for the ripening of Swiss cheese. It grows on the curd and carries out fermentation, producing a large amount of carbon dioxide (CO2). The CO2 gas gets trapped within the cheese matrix, leading to the formation of the characteristic large holes or 'eyes' in Swiss cheese.
- **Yoghurt Formation:** Lactic Acid Bacteria (LAB) like *Lactobacillus* are added to milk. They grow and ferment the lactose (milk sugar) into lactic acid. The production of lactic acid causes the coagulation and partial digestion of milk proteins (casein), leading to the formation of a thick, semi-solid curd with a characteristic tangy flavour. This process converts milk into yoghurt.

In secondary treatment, the primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows for the vigorous growth of useful aerobic microbes into 'flocs' (masses of bacteria associated with fungal filaments to form mesh-like structures).

Roles of microbes:
1. **Organic Matter Consumption:** These microbes consume the major part of the organic matter in the effluent. This significantly reduces the Biochemical Oxygen Demand (BOD) of the effluent. BOD is a measure of the organic matter present in the water.
2. **Formation of Activated Sludge:** Once the BOD is reduced, the effluent is passed into a settling tank where the bacterial 'flocs' are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum.

This stage is crucial because it removes about 90-95% of the organic pollutants, which are the main cause of water pollution. By drastically reducing the BOD, it prevents the depletion of dissolved oxygen in natural water bodies, thus protecting aquatic life.

(a) The circular plasmid is 4000 bp. The restriction sites are at BamHI (1000 bp), EcoRI (2000 bp), and EcoRI (3500 bp).
Double digestion with both enzymes will cleave the plasmid at all three sites.
The fragments obtained will be:
1. Fragment between BamHI (1000 bp) and EcoRI (2000 bp) = 2000 - 1000 = 1000 bp.
2. Fragment between EcoRI (2000 bp) and EcoRI (3500 bp) = 3500 - 2000 = 1500 bp.
3. Fragment between EcoRI (3500 bp) and BamHI (1000 bp, crossing the origin) = (4000 - 3500) + 1000 = 500 + 1000 = 1500 bp.
So, three fragments will be produced with sizes 1000 bp, 1500 bp, and 1500 bp.

(b) A selectable marker is essential to help in identifying and eliminating non-transformants and selectively permitting the growth of the transformants. It helps to differentiate host cells which have taken up the vector from those which have not. Example: Ampicillin resistance gene (ampR) or tetracycline resistance gene (tetR) in pBR322.

(a) The technique is Polymerase Chain Reaction (PCR). The steps are:
1. Denaturation: The double-stranded DNA is heated to 94-96°C to separate the two strands.
2. Annealing: The temperature is lowered to 50-65°C to allow the primers to bind to their complementary sequences on the single-stranded DNA templates.
3. Extension: The temperature is raised to 72°C, the optimal temperature for Taq polymerase to synthesize new DNA strands, starting from the primers.
(b) Role of Taq polymerase: It is a thermostable DNA polymerase that synthesizes the new DNA strand using the original strand as a template and dNTPs. It is preferred because it can withstand the high temperatures of the denaturation step, eliminating the need to add fresh polymerase in each cycle.
(c) Primers are short, chemically synthesized oligonucleotides that are complementary to the regions of DNA at the 3' end of the template strands. Their function is to provide a starting point (a free 3'-OH group) for the DNA polymerase to begin synthesis.

(a) The technique is Polymerase Chain Reaction (PCR). The three steps are:
1. Denaturation: The double-stranded DNA is heated to a high temperature (around 94-96°C) to separate the two strands.
2. Annealing: The temperature is lowered (around 50-65°C) to allow short, synthetic DNA primers to bind (anneal) to their complementary sequences on the single-stranded DNA templates.
3. Extension: The temperature is raised again (around 72°C), allowing Taq polymerase to synthesize new DNA strands by adding nucleotides, starting from the primers.

(b) Role of Taq polymerase: Taq polymerase is a thermostable DNA polymerase that synthesizes new DNA strands. It is used because it can withstand the high temperatures of the denaturation step, which would denature most other polymerases. This stability eliminates the need to add fresh enzyme after each cycle.

(c) Other essential components: Two other essential components are primers (short synthetic oligonucleotides) and deoxynucleoside triphosphates (dNTPs - dATP, dGTP, dCTP, dTTP).

(a) Vessel: Bioreactor.
A simple stirred-tank bioreactor diagram should be drawn, labelling the following: Agitator system, Acid/base for pH control, Steam for sterilization, Culture broth, Sterile air, Foam breaker, Motor, and Flat bladed impeller.

(b) Functions:
- Agitator system: Ensures uniform mixing of the culture medium and even distribution of nutrients and oxygen throughout the bioreactor.
- Oxygen delivery system: Supplies sterile air/oxygen to the culture, which is essential for the aerobic growth of the recombinant cells.
- Foam control system: Prevents foam formation, which can interfere with the process and lead to contamination.

(c) Downstream Processing: It is a series of processes that follow the biosynthetic stage (fermentation) to separate and purify the desired product from the culture medium. Two processes include: Separation (e.g., centrifugation or filtration) and Purification (e.g., chromatography).

(a) The equipment used is a Bioreactor (or Fermenter).

(b) Four essential components and their functions are:
1. Agitator System: Consists of blades or impellers for uniform mixing of the contents and ensuring that the microbial cells are evenly exposed to nutrients and oxygen.
2. Oxygen Delivery System: A sparger or tube that bubbles sterile air or oxygen through the culture medium to meet the aerobic requirements of the cells for growth and product formation.
3. Foam Control System: A sensor and addition port to add anti-foaming agents, as excessive foam can interfere with the process and contaminate the system.
4. Temperature Control System: A jacket or internal coils through which water circulates to maintain the optimal temperature for enzymatic activity and cell growth, removing excess heat generated during metabolism.
(Other valid points: pH control system, sampling ports)

(c) Continuous culture is more productive because it maintains the cells in their physiologically most active log/exponential phase. This is achieved by continuously draining used medium and adding fresh medium, leading to a prolonged period of high growth rate and product formation, unlike batch culture where growth is limited by nutrient depletion and waste accumulation.

(a) The circular plasmid is 4000 bp. The restriction sites are at BamHI (1000 bp), EcoRI (2000 bp), and EcoRI (3500 bp).
Double digestion with both enzymes will cleave the plasmid at all three sites.
The fragments obtained will be:
1. Fragment between BamHI (1000 bp) and EcoRI (2000 bp) = 2000 - 1000 = 1000 bp.
2. Fragment between EcoRI (2000 bp) and EcoRI (3500 bp) = 3500 - 2000 = 1500 bp.
3. Fragment between EcoRI (3500 bp) and BamHI (1000 bp, crossing the origin) = (4000 - 3500) + 1000 = 500 + 1000 = 1500 bp.
So, three fragments will be produced with sizes 1000 bp, 1500 bp, and 1500 bp.

(b) A selectable marker is essential to help in identifying and eliminating non-transformants and selectively permitting the growth of the transformants. It helps to differentiate host cells which have taken up the vector from those which have not. Example: Ampicillin resistance gene (ampR) or tetracycline resistance gene (tetR) in pBR322.

(a) The technique is Polymerase Chain Reaction (PCR). The steps are:
1. Denaturation: The double-stranded DNA is heated to 94-96°C to separate the two strands.
2. Annealing: The temperature is lowered to 50-65°C to allow the primers to bind to their complementary sequences on the single-stranded DNA templates.
3. Extension: The temperature is raised to 72°C, the optimal temperature for Taq polymerase to synthesize new DNA strands, starting from the primers.
(b) Role of Taq polymerase: It is a thermostable DNA polymerase that synthesizes the new DNA strand using the original strand as a template and dNTPs. It is preferred because it can withstand the high temperatures of the denaturation step, eliminating the need to add fresh polymerase in each cycle.
(c) Primers are short, chemically synthesized oligonucleotides that are complementary to the regions of DNA at the 3' end of the template strands. Their function is to provide a starting point (a free 3'-OH group) for the DNA polymerase to begin synthesis.

(a) The technique is Polymerase Chain Reaction (PCR). The three steps are:
1. Denaturation: The double-stranded DNA is heated to a high temperature (around 94-96°C) to separate the two strands.
2. Annealing: The temperature is lowered (around 50-65°C) to allow short, synthetic DNA primers to bind (anneal) to their complementary sequences on the single-stranded DNA templates.
3. Extension: The temperature is raised again (around 72°C), allowing Taq polymerase to synthesize new DNA strands by adding nucleotides, starting from the primers.

(b) Role of Taq polymerase: Taq polymerase is a thermostable DNA polymerase that synthesizes new DNA strands. It is used because it can withstand the high temperatures of the denaturation step, which would denature most other polymerases. This stability eliminates the need to add fresh enzyme after each cycle.

(c) Other essential components: Two other essential components are primers (short synthetic oligonucleotides) and deoxynucleoside triphosphates (dNTPs - dATP, dGTP, dCTP, dTTP).

(a) Vessel: Bioreactor.
A simple stirred-tank bioreactor diagram should be drawn, labelling the following: Agitator system, Acid/base for pH control, Steam for sterilization, Culture broth, Sterile air, Foam breaker, Motor, and Flat bladed impeller.

(b) Functions:
- Agitator system: Ensures uniform mixing of the culture medium and even distribution of nutrients and oxygen throughout the bioreactor.
- Oxygen delivery system: Supplies sterile air/oxygen to the culture, which is essential for the aerobic growth of the recombinant cells.
- Foam control system: Prevents foam formation, which can interfere with the process and lead to contamination.

(c) Downstream Processing: It is a series of processes that follow the biosynthetic stage (fermentation) to separate and purify the desired product from the culture medium. Two processes include: Separation (e.g., centrifugation or filtration) and Purification (e.g., chromatography).

(a) The equipment used is a Bioreactor (or Fermenter).

(b) Four essential components and their functions are:
1. Agitator System: Consists of blades or impellers for uniform mixing of the contents and ensuring that the microbial cells are evenly exposed to nutrients and oxygen.
2. Oxygen Delivery System: A sparger or tube that bubbles sterile air or oxygen through the culture medium to meet the aerobic requirements of the cells for growth and product formation.
3. Foam Control System: A sensor and addition port to add anti-foaming agents, as excessive foam can interfere with the process and contaminate the system.
4. Temperature Control System: A jacket or internal coils through which water circulates to maintain the optimal temperature for enzymatic activity and cell growth, removing excess heat generated during metabolism.
(Other valid points: pH control system, sampling ports)

(c) Continuous culture is more productive because it maintains the cells in their physiologically most active log/exponential phase. This is achieved by continuously draining used medium and adding fresh medium, leading to a prolonged period of high growth rate and product formation, unlike batch culture where growth is limited by nutrient depletion and waste accumulation.

The farmer should use vegetative propagation for both plants.

1. **For Banana:**
* **Method:** Propagation through Rhizome.
* **Propagule:** The vegetative propagule is the rhizome. Small sections of the underground stem (rhizome), each containing at least one bud ('eye'), are cut and planted.
* **Process:** These rhizome pieces, when planted in moist soil, develop roots and shoots from the buds, growing into new, genetically identical banana plants.

2. **For Ginger:**
* **Method:** Propagation through Rhizome.
* **Propagule:** The vegetative propagule is the rhizome.
* **Process:** Ginger also has an underground, modified stem called a rhizome. The rhizome has distinct nodes and internodes. Small pieces of the rhizome, each with a few buds, are planted. These buds sprout to form new plants.

**Advantages for the farmer:**
* **Genetic Uniformity:** All new plants are genetically identical to the parent, ensuring desirable traits like fruit quality and yield are maintained.
* **Faster Growth:** Plants grown from vegetative propagules mature and bear fruit much earlier than those grown from seeds.
* **Propagation of Seedless Varieties:** This is the only method to propagate plants like bananas that do not produce viable seeds.

The farmer should use the following vegetative propagules:
1. **Potato:** Eyes of the tuber. The 'eyes' are axillary buds that can sprout to form new plants.
2. **Ginger:** Rhizome. The rhizome is an underground modified stem with nodes and internodes. Buds at the nodes can give rise to new plants.
3. **Water Hyacinth:** Offset. The offset is a lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots, capable of forming a new plant.

Two major advantages for the farmer are:
* **Genetic Uniformity (Clones):** It is a method of producing genetically identical individuals (clones). This ensures that desirable traits like high yield, disease resistance, or specific fruit quality of the parent plant are preserved.
* **Faster Propagation:** Plants raised by vegetative propagation reach maturity and start bearing flowers and fruits much earlier than those produced from seeds. This leads to a quicker harvest.

(a)
For Banana: The suggested vegetative propagule is the Rhizome (or suckers arising from it). Justification: The underground stem (rhizome) of banana possesses nodes and buds which can give rise to new plants. This ensures rapid and uniform crop production.
For Ginger: The suggested vegetative propagule is the Rhizome. Justification: Ginger has a modified underground stem called a rhizome which stores food and has buds. These buds can sprout to form new plants, making it an effective method for propagation.

(b)
Vegetative propagation is considered a type of asexual reproduction because it involves the formation of new individuals from a single parent, without the formation and fusion of gametes. The offspring are genetically identical to the parent plant.
Advantages for farmers:
1. It is a faster, easier, and cheaper method of propagation for many plants.
2. It helps in preserving the parental characters (like superior fruit quality, disease resistance) for successive generations.
Disadvantages for farmers:
1. Since there is no genetic variation, the entire crop is susceptible to the same diseases and pests. A single disease can wipe out the whole population.

(a) Embryogenesis is the process of development of an embryo from the zygote. During embryogenesis, the zygote undergoes two major events:
1. Cell Division (Mitosis): The zygote divides repeatedly by mitosis to increase the number of cells in the developing embryo. This is also called cleavage.
2. Cell Differentiation: The cells formed through division undergo differentiation to form specialized tissues and organs, which ultimately organize to form a complete organism.

(b) Analysis of difference:
In oviparous animals, the development of the zygote takes place outside the female's body, within a calcareous shell. The embryo is vulnerable to environmental threats and predators.
In viviparous animals, the zygote develops into a young one inside the body of the female organism.
Better embryonic care and protection: Viviparity provides better care and protection because the developing embryo remains safe inside the mother's body, receiving continuous nourishment and protection from predators and harsh environmental conditions. After a certain stage of growth, the young ones are delivered out of the mother's body. Because of this proper embryonic care and protection, the chances of survival of young ones are greater in viviparous organisms.

(a) All of them will sprout and form new plants. This is because all these parts are modified vegetative parts of the plant that contain nodes with adventitious buds. Under favorable conditions (like moist soil), these buds can grow into new plants.

(b) The vegetative propagules are:
- Potato tuber: Eyes (axillary buds)
- Ginger: Rhizome
- Bryophyllum leaf: Leaf buds (adventitious buds on leaf margins)
- Water hyacinth: Offset

(c) Vegetative propagation is considered a type of asexual reproduction because it involves only a single parent, and the offspring produced are genetically identical to the parent (clones). There is no formation or fusion of gametes involved.

(a) The pyramid of energy is always upright because energy flow in a food chain is unidirectional. At each successive trophic level, there is a significant loss of energy (about 90%) as heat, according to the 10% law. Thus, the energy available at the producer level is always maximum, and it decreases at each subsequent trophic level (primary consumer, secondary consumer, etc.).

(b) The pyramid of biomass in a large tree ecosystem is inverted (spindle-shaped).
Justification: A single large tree (producer) has a very large biomass. It supports a large number of herbivorous birds (primary consumers), which have a smaller total biomass. These birds, in turn, support a very small number of parasites or predators (secondary consumers), which have an even smaller biomass. Thus, the producer biomass is large, consumer biomass is less, and top consumer biomass is least.

(c) In a terrestrial ecosystem (e.g., grassland), the pyramid of biomass is typically upright because the total biomass of producers is greater than that of primary consumers, which is greater than that of secondary consumers. However, in an aquatic ecosystem (e.g., a pond), the pyramid of biomass is often inverted. This is because the producers are phytoplanktons, which have a very short life span and a small standing crop biomass at any given time. They are consumed rapidly by zooplanktons (primary consumers), which have a larger biomass.

(a) The key steps in decomposition are: Fragmentation, Leaching, and Catabolism. Humification and Mineralisation occur simultaneously during decomposition.

(b) Temperature and soil moisture are crucial climatic factors that regulate decomposition. The rate of decomposition is higher in warm and moist environments. Low temperature and anaerobiosis inhibit decomposition, resulting in the build-up of organic materials. For example, decomposition is faster in tropical rainforests compared to temperate or arctic regions.

(c) The rate of decomposition is largely controlled by the chemical composition of detritus. Decomposition is slower if the detritus is rich in lignin and chitin, as these are complex polymers that are difficult for microbes to break down. Conversely, decomposition is quicker if the detritus is rich in nitrogen and water-soluble substances like sugars, as these provide an easily accessible source of energy and nutrients for the decomposers.

(a) Primary Succession: It begins in an area where no living organisms ever existed, e.g., bare rock, cooled lava.
Secondary Succession: It begins in an area that has lost all the living organisms that existed there, but where soil or sediment is present, e.g., abandoned farmland, burned or cut forests.

(b) During succession:
- Species diversity: Increases as succession proceeds. The number of species and organisms increases.
- Biomass: The total amount of living organic matter (biomass) increases.
- Humus content: Decomposers also increase, leading to an increase in the humus content of the soil.

(c) Secondary succession is faster than primary succession for two main reasons:
1. Presence of Soil: Secondary succession starts on well-developed soil or sediment. This soil already contains nutrients and organic matter, eliminating the long process of soil formation which is a necessary first step in primary succession.
2. Presence of Propagules: Some seeds, spores, roots, or other propagules of previous organisms may remain in the soil, allowing for quicker re-colonization compared to the arrival of pioneer species on a sterile surface.

(a) In a marine ecosystem, the pyramid of biomass is inverted. The biomass of producers (phytoplankton) at any given time is much lower than the biomass of the primary consumers (zooplankton) that feed on them. This happens because phytoplankton have a very short life span and a high turnover rate; they are consumed rapidly by zooplankton. The zooplankton, in turn, are consumed by small fish, and so on, with biomass increasing at each successive trophic level.

Pyramid Structure:
Base: Phytoplankton (Producers) - Smallest biomass
Above Base: Zooplankton (Primary Consumers)
Above that: Small Fish (Secondary Consumers)
Top: Large Fish (Tertiary Consumers) - Largest biomass

(b) Three limitations of ecological pyramids are:
1. They do not account for the same species belonging to two or more trophic levels (e.g., omnivores).
2. They assume a simple food chain, which almost never exists in nature; they do not accommodate a food web.
3. Decomposers (saprophytes) are not given any place in ecological pyramids even though they play a vital role in the ecosystem.

The key steps of the phosphorus cycle that make it available to plants are:
1. **Weathering of Rocks:** The natural reservoir of phosphorus is rock, which contains phosphorus in the form of phosphates. Through weathering and erosion, these phosphates are dissolved in soil solution and absorbed by the roots of plants.
2. **Decomposition:** When organisms die, decomposers like bacteria and fungi break down the organic matter, returning phosphates to the soil (mineralization), making it available for producers again.

Two ways human activities disrupt the phosphorus cycle:
1. **Use of Fertilizers:** The extensive use of phosphate fertilizers in agriculture significantly alters the natural cycle. The excess phosphorus runs off from fields into water bodies, causing eutrophication (algal blooms) which depletes oxygen and harms aquatic life.
2. **Mining:** Humans mine phosphate rock to produce fertilizers. This depletes the natural geological reserves of phosphorus, which are non-renewable on a human timescale.

(a) In-situ (on-site) conservation involves protecting species within their natural habitats. This allows the entire ecosystem to be protected, and the species can continue to evolve in its natural environment. Example: National Parks, Sanctuaries.
Ex-situ (off-site) conservation involves protecting species outside their natural habitats in special settings. This is used for rare or endangered species that cannot be safely protected in the wild. Example: Zoological parks, Botanical gardens, Cryopreservation.
(b) For the Royal Bengal Tiger, the in-situ conservation strategy is highly recommended.
Reasons:
1. Ecosystem Protection: Conserving the tiger in its natural habitat (like a Tiger Reserve) also ensures the protection of its entire ecosystem, including its prey base (deer, wild boar) and forest habitat. This maintains ecological balance.
2. Natural Behavior and Evolution: In-situ conservation allows tigers to live, hunt, and reproduce in their natural environment, preserving their natural behaviors and allowing them to continue adapting and evolving. This is not possible in captivity (ex-situ).

(a) The equation for the species-area relationship on a logarithmic scale is:
log S = log C + Z log A
Where S = Species richness, A = Area, Z = slope of the line (regression coefficient), and C = Y-intercept.

(b) The slope of the line ('Z') will be steeper for Site A (the island) compared to Site B (the continental mainland).
**Ecological Reason:** On islands (and other isolated habitats), species are more susceptible to extinction, and immigration from other areas is less frequent. This leads to a situation where an increase in area results in a more rapid accumulation of species compared to a continental area of the same size, where species are more widespread and can easily move between adjacent areas. Therefore, the species turnover rate is higher on islands, resulting in a steeper slope.

(c) The typical range for the 'Z' value for small or continental areas is 0.1 to 0.2. A much steeper slope (Z values in the range of 0.6 to 1.2) signifies that species richness increases very rapidly with an increase in area. This is characteristic of very large areas like entire continents or for specific taxonomic groups like frugivorous birds and mammals in tropical forests.

(a) Biodiversity hotspots are regions with very high levels of species richness and a high degree of endemism (species confined to that region and not found anywhere else).

(b) Two biodiversity hotspots in India are: The Western Ghats and Sri Lanka, and the Indo-Burma region. (Himalaya is also a valid answer).

(c) The three criteria for a region to be a biodiversity hotspot are:
1. **High Species Richness:** It must contain a large number of different species.
2. **High Degree of Endemism:** It must have a significant number of endemic species (species that are found nowhere else).
3. **High Degree of Threat:** It must have lost at least 70% of its original habitat.

(d) The hotspot concept is a practical approach because it allows conservation efforts and resources to be focused on areas where they can have the greatest impact. By protecting these hotspots, a very large number of species can be protected in a cost-effective manner, especially those that are most vulnerable to extinction.

(a) **In-situ conservation:** This approach involves protecting species in their natural habitat. The entire ecosystem is protected. It is the on-site conservation of genetic resources in natural populations.
Examples: National Parks, Wildlife Sanctuaries, Biosphere Reserves, Sacred Groves.

**Ex-situ conservation:** This approach involves placing threatened animals and plants in special care units for their protection, away from their natural habitat. It is the conservation of biological diversity components outside their natural habitats.
Examples: Zoological parks, Botanical gardens, Cryopreservation (gene banks, seed banks).

(b) The ex-situ approach is a secondary option because:
1. It removes the species from its natural evolutionary pressures and ecological interactions.
2. It can be expensive and may not be feasible for all species.
3. The population size that can be maintained is limited due to space and resource constraints.
4. Reintroduction into the wild can be difficult and unsuccessful. Therefore, protecting the entire ecosystem (in-situ) is always the preferred and more holistic approach.

(a) The latitudinal gradient pattern shows that species diversity generally decreases as we move away from the equator towards the poles. The tropics (latitudinal range of 23.5° N to 23.5° S) harbor more species than temperate or polar areas. For example, Colombia, located near the equator, has nearly 1,400 species of birds, while New York (41° N) has 105 species, and Greenland (71° N) only has 56 species.

(b) Three hypotheses to explain this pattern are:
1. **Speciation is a function of time:** Temperate regions were subjected to frequent glaciations in the past, while tropical latitudes have remained relatively undisturbed for millions of years, thus allowing a long evolutionary time for species diversification.
2. **Tropical environments are less seasonal:** Tropical environments are more constant and predictable than temperate ones. This promotes niche specialization and leads to a greater species diversity.
3. **More solar energy is available in the tropics:** This contributes to higher productivity, which in turn might support a greater diversity of species indirectly.