CBSE Class 12 2025 Biology 57-1-2​ Question Paper Set-2 :Download Solutions with Answer Key

Step 1: Decomposition rates depend on the composition of organic matter in the soil. Sugar decomposes the fastest, followed by chitin, and lignin decomposes the slowest. Nitrogen content also enhances microbial activity, speeding up decomposition.

Step 2: Soil A has 40% lignin (slow to decompose), 0% sugar (fast to decompose), 45% chitin, and 5% nitrogen. The high lignin and low nitrogen suggest a slower decomposition rate.

Step 3: Soil B has 5% lignin, 35% sugar (fast to decompose), 15% chitin, and 45% nitrogen. The high sugar and nitrogen content indicate a faster decomposition rate.

Step 4: Comparing the two, Soil A, with more lignin and less sugar/nitrogen, will decompose more slowly than Soil B, which has more sugar and nitrogen to support faster microbial activity.

Thus, Soil A has a slower rate of decomposition than Soil B.
Quick Tip: In soil decomposition, high sugar and nitrogen content accelerate microbial activity, while high lignin content slows down the process.

Step 1: Pollen grains vary in size depending on the plant species, but their diameter typically ranges from 10 to 100 micrometers.

Step 2: For most plants, the average diameter of pollen grains is between 25 and 50 micrometers, which is a common range for many species.

Step 3: Comparing the given options, 25 to 50 micrometers aligns best with the typical size of pollen grains.

Thus, the diameter of the pollen grain is most accurately 25 to 50 micrometers.
Quick Tip: Pollen grain size can vary by species, but a typical range for many plants is 25 to 50 micrometers, making them visible under a light microscope.

Step 1: The menstrual cycle in human females consists of four main phases that occur in a specific order.

Step 2: The cycle begins with the Menstrual phase (shedding of the uterine lining), followed by the Follicular phase (follicle development), then the Ovulatory phase (release of the egg), and finally the Luteal phase (preparation for potential pregnancy).

Step 3: Comparing the options, option (B) correctly lists the sequence as Menstrual phase, Follicular phase, Ovulatory phase, and Luteal phase.

Thus, the correct sequence is Menstrual phase, Follicular phase, Ovulatory phase, Luteal phase.
Quick Tip: The menstrual cycle typically lasts 28 days and follows the order: Menstrual phase (days 1-5), Follicular phase (days 1-13), Ovulatory phase (day 14), and Luteal phase (days 15-28).

Step 1: A dihybrid cross involves two traits. Assuming seed color (Yellow YY vs. Green yy) and seed shape (Round RR vs. Wrinkled rr), pure line parents are YYRR and yyrr.

Step 2: The F\(_1\) generation from YYRR × yyrr is YyRr (yellow round, heterozygous). In the F\(_2\) generation (YyRr × YyRr), the phenotypic ratio is 9:3:3:1, giving 4 distinct phenotypes: yellow round, yellow wrinkled, green round, and green wrinkled.

Step 3: The genotypic ratio is 1:2:1:2:4:2:1:2:1, resulting in 9 distinct genotypes: YYRR, YYRr, YyRR, YyRr, YYrr, Yyrr, yyRR, yyRr, and yyrr.

Step 4: Comparing the options, option (D) correctly states 4 phenotypes and 9 genotypes.

Thus, in the F\(_2\) generation, there are 4 phenotypes and 9 genotypes.
Quick Tip: In a dihybrid cross, the F\(_2\) phenotypic ratio is 9:3:3:1 (4 phenotypes), and the genotypic ratio is 1:2:1:2:4:2:1:2:1 (9 genotypes), assuming independent assortment.

Step 1: RNA interference (RNAi) is a biological process where RNA molecules inhibit gene expression, typically by targeting mRNA.

Step 2: In the case of *Meloidogyne incognita*, RNAi is used to silence specific genes in the nematode by introducing double-stranded RNA that matches the target mRNA.

Step 3: This double-stranded RNA is processed into small interfering RNAs (siRNAs), which prevent the translation of the nematode's mRNA into proteins, effectively silencing the gene.

Thus, RNAi is achieved by preventing the translation of mRNA of the nematode.
Quick Tip: RNAi is a powerful tool in biotechnology, often used to silence harmful genes in pests like nematodes, by targeting their mRNA to prevent protein synthesis.

Step 1: The given DNA sequence, 5'-AATGCTAGGCAC-3', is the coding strand. The mRNA is transcribed from the template strand, which is complementary to the coding strand.

Step 2: The template strand (3' to 5') is 3'-TTACGATCCGTG-5'. The mRNA is synthesized 5' to 3', complementary to the template strand, replacing T with U: 5'-AAUGCUAGGCAC-3'.

Step 3: Comparing the options, option (D) matches the correct mRNA sequence: 5'-AAUGCUAGGCAC-3'.

Thus, the mRNA sequence is 5'-AAUGCUAGGCAC-3'.
Quick Tip: In transcription, mRNA is synthesized from the template strand of DNA, where A pairs with U (instead of T), and G pairs with C, following base-pairing rules.

Step 1: Colour blindness is an X-linked recessive disorder. A man inherits his X chromosome from his mother and Y chromosome from his father.

Step 2: Since the man's father was colour-blind, and the disorder is X-linked, the man himself is not colour-blind (he received the normal X from his mother). His genotype is \( XY \) with a normal X.

Step 3: The woman had a colour-blind mother and normal father, so she must be a carrier (\( X^C X \)).

Step 4: Cross between man (\( XY \)) and woman (\( X^C X \)) gives:

Male offspring: 50% \( X Y \) (normal), 50% \( X^C Y \) (colour-blind)

Thus, 50% of male children will be colour-blind.
Quick Tip: In X-linked disorders, sons get their X chromosome from their mother. Carrier mothers can pass the defective gene to 50% of sons.

Step 1: Human evolution follows a specific sequence of primate ancestors leading to modern humans (*Homo sapiens*).

Step 2: The correct sequence begins with early primates like *Dryopithecus* (not listed in the correct option but part of the broader timeline), followed by *Ramapithecus*, then *Australopithecines* (early hominids), *Homo erectus*, Neanderthals (*Homo neanderthalensis*), and finally *Homo sapiens*.

Step 3: Option (B) lists Australopithecines \rightarrow Homo erectus \rightarrow Neanderthal \rightarrow Homo sapiens,
which aligns with the accepted evolutionary timeline.

Thus, the correct series is Australopithecines \rightarrow Homo erectus \rightarrow Neanderthal \rightarrow Homo sapiens.
Quick Tip: Human evolution progressed from early primates like Australopithecines to Homo erectus, then to Neanderthals, before the emergence of modern Homo sapiens.

Step 1: GEAC is an acronym related to biotechnology and genetic engineering in India.

Step 2: It is a regulatory body that oversees the approval of genetically modified organisms and products.

Step 3: The correct full form of GEAC is Genetic Engineering Approval Committee, which matches option (D).

Thus, the answer is (D) Genetic Engineering Approval Committee. Quick Tip: GEAC is a key regulatory body in India for approving GMOs and ensuring biosafety.

Step 1: Transplantation rejection occurs when the recipient's immune system identifies the transplanted tissue/organ as foreign.

Step 2: The immune system has two main types of responses: humoral (antibody-mediated) and cell-mediated (T-cell mediated).

Step 3: Rejection of transplants is primarily due to the action of T-cells, which recognize foreign antigens on the transplanted tissue and attack it. This is a cell-mediated immune response.

Step 4: Autoimmune response (A) targets the body’s own tissues, humoral response (B) involves antibodies, and physiological immune response (C) is not a standard term in this context.

Thus, the correct answer is (D) Cell mediated immune response. Quick Tip: Cell-mediated immunity, driven by T-cells, is critical in transplant rejection, which is why immunosuppressive drugs are often used post-transplant.

Step 1: Amplification of a gene of interest is typically done using the Polymerase Chain Reaction (PCR) technique, which relies on DNA polymerase.

Step 2: In PCR, the DNA is doubled in each cycle. A standard PCR run often includes 30–40 cycles.

Step 3: Mathematically, after \( n \) cycles, the amplification factor is \( 2^n \). For 30 cycles, \( 2^{30} \approx 1.07 \times 10^9 \), which is roughly 1 billion.

Step 4: Options (A), (B), and (D) are either too low or too high for typical PCR amplification.

Thus, the correct answer is (C) 1 billion times. Quick Tip: PCR amplification is exponential: after 30 cycles, a single DNA molecule can theoretically be amplified to over 1 billion copies.

Step 1: The development of a dicot embryo begins with the zygote, which is formed after fertilization (P).

Step 2: The zygote divides to form a proembryo, where the suspensor develops as a structure that supports the embryo by anchoring it and providing nutrients (Q).

Step 3: As the embryo develops into a globular and then heart-shaped stage, the cotyledons (seed leaves) form, which are visible in the heart-shaped embryo (R).

Step 4: In the mature embryo, the plumule (the shoot tip that will develop into the shoot system) is visible between the cotyledons (S).

Step 5: The radicle (future root) is also part of the mature embryo but is not labeled as S in the correct option. Comparing options, (B) correctly labels P as Zygote, Q as Suspensor, R as Cotyledon, and S as Plumule.

Thus, the correct answer is (B). Quick Tip: In dicot embryo development, the suspensor supports the early embryo, while the cotyledons and plumule are key structures in the mature embryo.

Step 1: Perisperm is the remnant of the nucellus, a maternal tissue in the seed, which surrounds the embryo in some seeds (e.g., in black pepper). Thus, Reason (R) is true.

Step 2: The nucellus is a diploid tissue (2n) since it is part of the maternal sporophyte. Hence, perisperm, being the remains of the nucellus, is also diploid. Assertion (A) is true.

Step 3: While both statements are true, the reason does not explain why perisperm is diploid; it only describes what perisperm is. The ploidy explanation requires understanding the maternal origin of the nucellus, which is not directly stated in (R).

Thus, the correct answer is (B). Quick Tip: Perisperm (diploid) is different from endosperm (triploid). Perisperm is maternal tissue (nucellus), while endosperm results from double fertilization.

Step 1: Alexander Fleming, while working on Staphylococci, observed that a mold, Penicillium notatum, inhibited bacterial growth, leading to the discovery of penicillin. Assertion (A) is true.

Step 2: However, Fleming did not commercially extract penicillin or establish its full potential. This was done later by Howard Florey and Ernst Chain, who developed penicillin into a usable drug. Reason (R) is false.

Step 3: Since (A) is true and (R) is false, the correct option is (C).

Thus, the correct answer is (C). Quick Tip: Fleming discovered penicillin, but Florey and Chain were key in its commercial production and application, earning them a shared Nobel Prize with Fleming in 1945.

Step 1: Degeneracy is a property of the genetic code, meaning that multiple codons can code for the same amino acid. Assertion (A) is true.

Step 2: For example, amino acids like leucine are coded by multiple codons (e.g., CUU, CUC, CUA, CUG). This is exactly what degeneracy means. Reason (R) is true.

Step 3: Since (R) directly explains why the genetic code is degenerate, the correct option is (A).

Thus, the correct answer is (A). Quick Tip: Degeneracy in the genetic code reduces the impact of mutations, as a change in a codon might still code for the same amino acid.

Step 1: A bioreactor does provide optimal conditions (e.g., temperature, pH, oxygen levels) for microorganisms to grow and produce the desired product, such as enzymes or antibiotics. Assertion (A) is true.

Step 2: Stirring-type bioreactors (stirred-tank bioreactors) are indeed the most commonly used due to their efficiency in mixing and aeration. Reason (R) is true.

Step 3: However, the type of bioreactor (stirring type) does not explain why a bioreactor provides optimal conditions; it’s just a common design. The explanation for (A) lies in the control of growth parameters, not the bioreactor type.

Thus, the correct answer is (B). Quick Tip: Bioreactors optimize growth by controlling factors like pH, temperature, and oxygen, regardless of their design (e.g., stirred-tank or air-lift).

Offspring 1: Turner’s Syndrome (45, XO)

Caused due to absence of one X chromosome (Monosomy).
Genotype: 44 + XO (total 45 chromosomes).
Phenotype: Female with underdeveloped secondary sexual characters, short stature, and sterility.


Offspring 2: Klinefelter’s Syndrome (47, XXY)

Caused due to the presence of an extra X chromosome (Trisomy).
Genotype: 44 + XXY (total 47 chromosomes).
Phenotype: Male with feminine body features, gynecomastia, and sterility. Quick Tip: Turner’s syndrome (XO) results in 45 chromosomes (female), while Klinefelter’s syndrome (XXY) results in 47 chromosomes (male). These are chromosomal disorders caused by non-disjunction during gamete formation.

[(a)]

Individual 1 (B): Possible genotype = \( I^B I^O \)
Individual 2 (A): Possible genotype = \( I^A I^O \)

\textit{(Since their child (individual 8) has blood group O, both parents must carry the \( I^O \) allele.)


[(b)]

Individual 5 (AB): Antigens present = A and B antigens
Individual 8 (O): Antigens present = None (no A or B antigen present on RBC membrane) Quick Tip: Blood group inheritance follows codominance (A and B alleles) and simple recessiveness (O allele). Genotype must be inferred from both phenotype and offspring data.

Step 1: ADA (Adenosine Deaminase) deficiency leads to severe combined immunodeficiency (SCID), where lymphocytes are non-functional.

Step 2: Periodic infusion of genetically engineered lymphocytes is needed because these modified cells have a limited lifespan and cannot multiply to maintain the therapeutic effect.

Step 3: Since these engineered cells are not incorporated into the patient's genome permanently, repeated infusions are necessary.

Step 4: A permanent cure can be achieved by gene therapy where a functional ADA gene is introduced into embryonic stem cells or early bone marrow cells of the patient, allowing lifelong expression. Quick Tip: Gene therapy offers a long-term solution for ADA deficiency by incorporating the correct gene into the patient's genome—eliminating the need for repeated infusions.

Step 1: Microinjection: The recombinant DNA is directly injected into the nucleus of an animal cell using a fine micro-needle under a microscope.

Step 2: Gene Gun or Biolistics: High-velocity micro-particles coated with DNA are bombarded into plant cells to deliver the recombinant DNA directly.

Step 3: These physical methods bypass the need for a vector, allowing direct DNA transfer into host cells. Quick Tip: Microinjection and gene gun are physical, vector-free methods for direct DNA delivery—commonly used in animal and plant biotechnology.

Step 1: Morphine and heroin are both opioids derived from the opium poppy plant, Papaver somniferum.

Step 2: Morphine is a natural alkaloid extracted directly from the opium poppy, while heroin is a semi-synthetic drug produced by chemically modifying morphine (diacetylmorphine).

Step 3: Both act on the central nervous system by binding to opioid receptors, leading to pain relief, euphoria, and sedation. However, they also cause side effects like drowsiness, respiratory depression, and constipation.

Step 4: Heroin is more potent and addictive than morphine, crossing the blood-brain barrier faster, leading to a higher risk of dependency and overdose. Long-term use of both can result in tolerance, addiction, and severe withdrawal symptoms. Quick Tip: Morphine and heroin are structurally related as opioids, but heroin’s faster action makes it more addictive and dangerous.

Step 1: (i) Alcoholic drinks produced by microbes:

(1) \textit{With distillation: Whisky is produced by fermenting grains (e.g., barley) using yeast (Saccharomyces cerevisiae), followed by distillation to increase alcohol content.

(2) \textit{Without distillation: Beer is produced by fermenting barley malt with yeast, without distillation, resulting in a lower alcohol content.

Step 2: (ii) Cyanobacteria as bio-fertilizer: Cyanobacteria, such as Anabaena and Nostoc, can fix atmospheric nitrogen into ammonia due to the presence of heterocysts (specialized cells).

Step 3: This ammonia is converted into nitrates, enriching the soil with nitrogen, which is essential for plant growth. Cyanobacteria are often used in paddy fields to naturally enhance soil fertility.

Step 4: Additionally, cyanobacteria produce growth-promoting substances and improve soil structure, making them effective bio-fertilizers. Quick Tip: Cyanobacteria are eco-friendly bio-fertilizers, especially in rice fields, as they fix nitrogen naturally and reduce the need for chemical fertilizers.

Step 1: Primary productivity refers to the rate at which biomass is produced by photosynthetic organisms (producers) in an ecosystem, measured as energy or biomass per unit area per unit time.

Step 2: A young forest: A young forest has rapidly growing plants with high photosynthetic rates, leading to high primary productivity due to less competition and more resources (light, nutrients). However, the total biomass is lower than in an old forest.

Step 3: A natural old forest: An old forest has mature trees with high biomass but lower growth rates. Primary productivity is moderate because, while photosynthesis occurs, much energy is used for maintenance rather than new growth.

Step 4: A shallow polluted lake: A shallow polluted lake often suffers from eutrophication (nutrient overload), leading to algal blooms that increase primary productivity initially. However, pollution can cause oxygen depletion, harming aquatic life and reducing overall productivity over time.

Step 5: \textit{Comparison: The young forest is likely the most productive in terms of primary productivity because of its rapid growth and high photosynthetic rate. A shallow polluted lake may have high productivity during algal blooms but is unsustainable. An old forest, while stable, has lower productivity due to slower growth rates.

Thus, the young forest will be the most productive in terms of primary productivity. Quick Tip: Young ecosystems often have higher primary productivity due to rapid growth, while old ecosystems have more biomass but slower growth rates.

Step 1: Gross Primary Productivity (GPP): GPP is the total amount of energy or biomass produced by photosynthetic organisms (producers) through photosynthesis per unit area per unit time in an ecosystem.

Step 2: Net Primary Productivity (NPP): NPP is the energy or biomass that remains after accounting for the energy used by producers for respiration. It is the energy available for consumption by herbivores and decomposers.

Step 3: Mathematical Relationship: NPP = GPP - Respiration (R), where respiration is the energy used by plants for metabolic activities like growth and maintenance.

Step 4: Key Difference: GPP represents the total energy fixed by photosynthesis, while NPP is the energy left after plants’ respiratory losses, which supports the rest of the ecosystem (e.g., herbivores).

For example, in a forest, GPP might be high due to photosynthesis, but NPP is lower because mature trees use a significant portion of energy for respiration. Quick Tip: NPP is what sustains ecosystems as it represents the energy available for higher trophic levels, while GPP includes energy plants use themselves.

Step 1: ‘ori’ (origin of replication) is a DNA sequence that initiates the replication of the plasmid within the host \textit{E. coli cell.

Step 2: The presence of ‘ori’ ensures that the plasmid can replicate independently of the host chromosome, allowing multiple copies of the recombinant DNA to be produced per cell.

Step 3: ‘rop’ (repressor of primer) is a gene that encodes a protein regulating the copy number of the plasmid by controlling replication at the ‘ori’.

Step 4: The ‘rop’ protein helps maintain a stable, moderate copy number of the plasmid, preventing over-replication, which could be toxic to the host cell.

Thus, ‘ori’ enables replication, while ‘rop’ ensures controlled replication for efficient cloning. Quick Tip: ‘ori’ is essential for plasmid replication, and ‘rop’ fine-tunes the process to avoid overburdening the host cell.

Step 1: Exonuclease: An exonuclease is an enzyme that removes nucleotides from the ends (either 5’ or 3’) of a DNA or RNA strand.

Step 2: Endonuclease: An endonuclease is an enzyme that cleaves the phosphodiester bonds within the DNA or RNA strand, cutting at specific internal sites.

Step 3: Key Difference: Exonucleases act on the terminal ends of the nucleic acid, while endonucleases cut within the strand. For example, restriction enzymes like EcoRI (seen in the diagram) are endonucleases, cutting at specific recognition sites.

Step 4: Application: Exonucleases are used to remove unwanted nucleotides (e.g., in DNA polishing), while endonucleases are crucial in cloning for cutting DNA at specific sites to insert foreign DNA.

Thus, the primary difference lies in their site of action—ends versus internal regions. Quick Tip: Endonucleases like restriction enzymes are widely used in genetic engineering to create sticky or blunt ends for DNA ligation.

Step 1: Alexander Von Humboldt’s observation describes the species-area relationship, where species richness (\( S \)) increases with area (\( A \)) but eventually levels off due to habitat saturation or other limiting factors.

Step 2: Graph Construction: Plot species richness (\( S \)) on the y-axis and area (\( A \)) on the x-axis. The graph starts with a steep increase in \( S \) as \( A \) increases, then gradually levels off, forming a curve that approaches a maximum value (asymptote). This is typically a logarithmic or power-law relationship that plateaus.

Step 3: Equation: The species-area relationship is often expressed as a power law: \( S = cA^Z \), where \( S \) is the number of species, \( A \) is the area, \( c \) is a constant, and \( Z \) is the regression coefficient. However, since the increase levels off, a more appropriate model might be a logarithmic form: \( S = k \log(A) + b \), where \( k \) and \( b \) are constants, reflecting the slowing growth rate of species richness.

Step 4: In practice, the graph would show a curve that rises quickly at first (when \( A \) is small) and then flattens as \( A \) becomes very large, indicating that new species are harder to find in larger areas due to ecological limits. Quick Tip: The species-area relationship often follows a power law (\( S = cA^Z \)) or logarithmic curve, but always levels off due to ecological constraints like habitat diversity.

Step 1: The regression coefficient \( Z \) in the species-area relationship equation \( S = cA^Z \) represents the slope of the relationship on a log-log scale, indicating how quickly species richness increases with area.

Step 2: \( Z \) values typically range between 0.1 and 0.3 for most ecosystems, depending on the type of habitat and the taxonomic group studied.

Step 3: For small areas or islands (e.g., island biogeography), \( Z \) is often between 0.2 and 0.3, as seen in studies of birds or insects on islands. For larger continental areas, \( Z \) can be lower, around 0.1 to 0.2, because species diversity increases more slowly due to greater habitat homogeneity.

Step 4: In extreme cases, such as highly isolated or species-poor regions, \( Z \) might approach 0.4 or higher, but this is less common. For example, in Humboldt’s observations of tropical regions, \( Z \) might be around 0.15 to 0.25, reflecting high biodiversity but with a plateau effect.

Thus, typical \( Z \) values range from 0.1 to 0.3, with higher values in more isolated or diverse ecosystems. Quick Tip: \( Z \) values are higher in isolated ecosystems like islands (0.2–0.3) and lower in continental areas (0.1–0.2) due to differences in habitat diversity and species dispersal.

Step 1: To test for genetic disorders like cystic fibrosis and Down’s syndrome, prenatal genetic testing is recommended.

Step 2: One common method is amniocentesis, which can be performed between 15–20 weeks of pregnancy, but since Radha is in her second month (around 8 weeks), a more suitable option is Chorionic Villus Sampling (CVS). CVS can be done between 10–13 weeks.

Step 3: In CVS, a small sample of placental tissue (chorionic villi) is taken and analyzed for genetic abnormalities. For Down’s syndrome, a karyotype analysis can detect an extra chromosome 21 (trisomy 21). For cystic fibrosis, DNA testing can identify mutations in the CFTR gene.

Step 4: Additionally, non-invasive prenatal testing (NIPT) can be done as early as 10 weeks by analyzing fetal DNA in the mother’s blood to screen for Down’s syndrome, though it’s less definitive for cystic fibrosis.

Thus, CVS is a suitable method at this stage to test for both conditions, followed by specific genetic analysis. Quick Tip: CVS and NIPT are early prenatal tests for genetic disorders; CVS is invasive but diagnostic, while NIPT is non-invasive but a screening tool.

Step 1: If the fetus is diagnosed with cystic fibrosis or Down’s syndrome and poses a risk to Radha’s health (e.g., due to complications like severe stress or related medical issues), one option is medical termination of pregnancy (MTP).

Step 2: Radha is in her second month (around 8 weeks) of pregnancy. MTP is generally considered safe in the first trimester (up to 12 weeks) using methods like medication (e.g., mifepristone and misoprostol) or a minor surgical procedure (e.g., aspiration).

Step 3: At 8 weeks, MTP is safe for Radha as the fetus is small, and the procedure carries lower risks compared to later stages. Complications like infection or heavy bleeding are rare when performed by a medical professional.

Step 4: However, Radha’s overall health, emotional well-being, and local legal regulations must be considered before proceeding with MTP.

Thus, MTP is a viable and safe option at 8 weeks, provided it’s medically supervised. Quick Tip: MTP is safest in the first trimester; early intervention reduces risks to the mother compared to later stages of pregnancy.

Step 1: The process mentioned in (b) is medical termination of pregnancy (MTP). Its legality depends on the country’s laws and specific conditions.

Step 2: In many countries, MTP may be illegal if performed beyond a certain gestational age (e.g., after 20–24 weeks) unless the mother’s life is at risk or the fetus has severe abnormalities, as per local laws (e.g., in India under the MTP Act).

Step 3: It may also be illegal if done without medical supervision, by unqualified practitioners, or for reasons not permitted by law, such as sex-selective abortion (e.g., banned in India under the PCPNDT Act).

Step 4: Additionally, if the procedure is performed without the mother’s informed consent or in a region where abortion is entirely prohibited (e.g., in some countries with strict anti-abortion laws), it would be illegal.

Thus, MTP can be illegal if it violates gestational limits, lacks medical oversight, or is performed for unapproved reasons like sex selection. Quick Tip: MTP laws vary globally; always check local regulations, especially regarding gestational age and approved medical reasons.

Step 1: A palindromic sequence in DNA reads the same on both strands in the 5’ to 3’ direction when considering complementary base pairing.

Step 2: The given sequence is 5’–GAATTC–3’. Its complementary strand (3’ to 5’) is determined by base pairing: G pairs with C, A with T, T with A, etc.

Step 3: Complementary strand (3’ to 5’): 3’–CTTAAG–5’. When read in the 5’ to 3’ direction, the complementary strand is 5’–GAATTC–3’, which matches the original sequence.

Step 4: Thus, the palindromic sequence is:

5’–GAATTC–3’

3’–CTTAAG–5’

This is a palindromic sequence because both strands read 5’–GAATTC–3’ in their respective directions. Quick Tip: Palindromic sequences are targets for restriction enzymes; they read the same on both strands in the 5’ to 3’ direction.

Step 1: The sequence 5’–GAATTC–3’ is a well-known palindromic sequence recognized by a specific restriction endonuclease.

Step 2: The restriction endonuclease that recognizes this sequence is EcoRI, derived from the bacterium \textit{Escherichia coli RY13.

Step 3: EcoRI cuts between the G and A on both strands, producing sticky ends. This makes it widely used in genetic engineering for cloning.

Thus, the restriction endonuclease is EcoRI. Quick Tip: EcoRI is a commonly used restriction enzyme in biotechnology, producing sticky ends that facilitate DNA ligation.

Step 1: Sticky ends are produced when a restriction endonuclease like EcoRI cuts a palindromic DNA sequence asymmetrically, leaving overhanging single-stranded regions.

Step 2: For the sequence 5’–GAATTC–3’, EcoRI cuts between G and A on both strands:

5’–G\(\downarrow\)AATTC–3’

3’–CTTAA\(\uparrow\)G–5’

This results in:

5’–G \quad \quad AATTC–3’

3’–CTTAA \quad \quad G–5’

The overhanging sequences (5’–AATT–3’ and 3’–TTAA–5’) are the sticky ends.

Step 3: Role of Sticky Ends: Sticky ends can base-pair with complementary overhangs on other DNA fragments cut by the same enzyme, facilitating ligation during cloning.

Step 4: This base-pairing ensures precise joining of DNA fragments, making sticky ends crucial for constructing recombinant DNA in genetic engineering.

Thus, sticky ends are produced by staggered cuts and are essential for efficient DNA ligation in cloning. Quick Tip: Sticky ends allow specific and efficient joining of DNA fragments, unlike blunt ends, which are harder to ligate accurately.

Step 1: The life cycle of Plasmodium, the parasite causing malaria, involves multiple stages, and the female Anopheles mosquito acts as the vector.

Step 2: The infective stage of Plasmodium that is stored in the salivary glands of the female Anopheles mosquito is the \textit{sporozoite.

Step 3: When the mosquito bites a human, sporozoites are injected into the bloodstream, initiating infection.

Thus, the infective stage is the sporozoite. Quick Tip: Sporozoites are the motile, infective forms of Plasmodium that target the liver cells of the human host to begin the infection cycle.

Step 1: In the life cycle of Plasmodium, fertilization and development occur within the mosquito vector, as shown in the diagram at stage Q.

Step 2: Fertilization occurs when male and female gametocytes, taken up by the mosquito during a blood meal, fuse in the mosquito’s gut to form a zygote.

Step 3: The zygote develops into an ookinete, which then forms an oocyst in the mosquito’s gut wall. The oocyst eventually ruptures, releasing sporozoites that migrate to the salivary glands.

Thus, fertilization and development of the parasite take place in the gut of the female Anopheles mosquito. Quick Tip: The mosquito’s gut provides the environment for Plasmodium’s sexual reproduction, while the human host supports asexual reproduction.

Step 1: The diagram shows a part of the Plasmodium life cycle, with labels P and Q indicating specific stages.

Step 2: Label P represents the stage where sporozoites are present in the mosquito’s salivary glands, ready to infect a human host. This stage is the \textit{sporozoite stage.

Step 3: Label Q represents the process of fertilization and development, where gametocytes fuse and develop into sporozoites within the mosquito’s gut. This is the \textit{fertilization and development stage.

Thus, P is the sporozoite stage, and Q is the fertilization and development stage. Quick Tip: Understanding the stages of Plasmodium’s life cycle helps in developing strategies to interrupt malaria transmission.

Step 1: Plasmodium’s life cycle involves two hosts: a primary host (where asexual reproduction occurs) and a secondary host (where sexual reproduction occurs).

Step 2: The asexual phase occurs in the human host, where sporozoites develop into merozoites in the liver and blood, causing malaria symptoms.

Step 3: The sexual phase occurs in the female Anopheles mosquito, where gametocytes undergo fertilization and development to produce sporozoites.

Thus, the asexual phase occurs in the human, and the sexual phase occurs in the female Anopheles mosquito. Quick Tip: The dual-host life cycle of Plasmodium makes malaria control challenging, requiring interventions targeting both humans and mosquitoes.

Step 1: Convergent and divergent evolution are two patterns of evolution that describe how species adapt and evolve over time.

Step 2: \textit{Convergent Evolution: This occurs when unrelated species develop similar traits due to similar environmental pressures, despite not sharing a common ancestor. For example, the wings of birds and insects both serve the purpose of flight, but they evolved independently.

Step 3: \textit{Divergent Evolution: This occurs when related species or populations diverge and develop distinct traits due to different environmental pressures or adaptations. For example, Darwin’s finches evolved different beak shapes from a common ancestor to adapt to different food sources.

Step 4: \textit{Comparison: In convergent evolution, species become more similar in function or structure (analogous structures), while in divergent evolution, species become more distinct (homologous structures). Convergent evolution reflects adaptation to similar environments, while divergent evolution reflects adaptation to different environments.

Thus, convergent evolution leads to similar traits in unrelated species, while divergent evolution leads to distinct traits in related species. Quick Tip: Convergent evolution results in analogous structures (e.g., wings of birds and insects), while divergent evolution results in homologous structures (e.g., forelimbs of mammals).

Step 1: The diagram shows a structure with a DNA double helix wrapped around a core, which is a characteristic feature of eukaryotic chromosome organization.

Step 2: This structure is known as a \textit{nucleosome, where DNA is wound around a histone protein core to compact the genetic material.

Step 3: The nucleosome is the fundamental unit of chromatin, aiding in DNA packaging within the nucleus.

Thus, the structure shown is a nucleosome. Quick Tip: Nucleosomes help in compacting DNA into the nucleus and play a role in regulating gene expression by controlling access to DNA.

Step 1: In the diagram of the nucleosome, label P points to the core around which the DNA is wrapped, and label Q points to the DNA strand itself.

Step 2: The core (P) consists of histone proteins, specifically a histone octamer made of two copies each of histones H2A, H2B, H3, and H4.

Step 3: The strand (Q) is the DNA double helix that wraps around the histone core, approximately 147 base pairs in length.

Thus, P is the histone core, and Q is the DNA double helix. Quick Tip: The histone core and DNA together form the nucleosome, with linker DNA connecting adjacent nucleosomes in chromatin.

Step 1: Histone proteins are a key component of the nucleosome, facilitating DNA packaging in eukaryotic cells.

Step 2: Histones are small, positively charged proteins due to their high content of basic amino acids like lysine and arginine.

Step 3: This positive charge allows histones to interact strongly with the negatively charged phosphate backbone of DNA, stabilizing the nucleosome structure.

Thus, histone proteins are basic and positively charged. Quick Tip: The basic nature of histones enables tight DNA binding, while their modifications (like acetylation) regulate gene expression.

Step 1: Chromatin in eukaryotic cells exists in two forms: euchromatin and heterochromatin, differing in structure and function.

Step 2: Euchromatin is loosely packed, transcriptionally active, and rich in gene expression, as it allows access to DNA for transcription machinery.

Step 3: Heterochromatin is tightly packed, transcriptionally inactive, and often found in regions like centromeres and telomeres, where it helps maintain structural integrity.

Step 4: Structurally, euchromatin has a more open conformation, while heterochromatin is condensed, often with more histone modifications that silence genes.

Thus, euchromatin is loosely packed and transcriptionally active, while heterochromatin is tightly packed and transcriptionally inactive. Quick Tip: Euchromatin and heterochromatin states can switch during the cell cycle or development, influencing gene regulation and genome stability.

Step 1 (a): Monarch butterflies acquire the chemical by feeding on poisonous milkweed plants during their larval stage. These chemicals accumulate in their body making them unpalatable to predators.

Step 2 (b):

Population at \( t + 1 = 800 \)

Using formula:
\[ N_{t+1} = N_t + Natality + Immigration - Mortality - Emigration \] \[ 800 = N_t + 200 + 200 - 150 - 100 \] \[ 800 = N_t + 150 \Rightarrow N_t = 650 \]

Since natality and immigration are higher than mortality and emigration, it indicates a growing population. Therefore, the type of age pyramid formed would be **expanding**.

Step 3 (c):

For carrot grass (uniform, small individuals), population density is best measured in terms of **number of individuals per unit area**.

For a huge banyan tree (single large organism with many offshoots), population is better measured in terms of **biomass** or **canopy cover**.


OR



Step 3 (d):

Two methods to measure tiger population density:

1. **Pugmark method** – tracking and counting footprints.

2. **Camera trapping** – using motion-sensor cameras to photograph and identify individual tigers. Quick Tip: Population density is dynamic and varies with birth, death, immigration, and emigration. Different species require different techniques to estimate their population effectively.

Step 1 (a): The hormone responsible for the peak in both Case 1 and Case 2 is **Luteinizing Hormone (LH)**.

\hspace*{0.5cm Function: LH stimulates ovulation and formation of corpus luteum.

Step 2 (b): During the follicular phase, in the ovary, a primary follicle develops into a Graafian follicle.

In the uterus, the endometrium regenerates under the influence of estrogen.

Step 3 (c): Hormone Q in Case 2 is **Progesterone**.

\hspace*{0.5cm Function: Maintains the endometrial lining for implantation and supports early pregnancy.


OR


Step 3 (d): The functional structure in Case 2 is the **Corpus luteum**.

\hspace*{0.5cm It is formed from the Graafian follicle after ovulation under the influence of LH. Quick Tip: LH surge is essential for ovulation. Progesterone from the corpus luteum helps maintain pregnancy until the placenta takes over.

Step 1 (A):

In a mature male gametophyte of an angiosperm, two cells are enclosed:

Generative cell: Divides to form two male gametes (sperms).
Vegetative cell: Forms the pollen tube for gamete transfer.



Technical terms:

1 — Autogamy
2 — Geitonogamy
3 — Xenogamy

Pollination modes:

Water lily: By insects (entomophily)
Vallisneria: By water (hydrophily)

Advantages of Xenogamy (method ‘3’):

Introduces genetic variation.
Helps in evolution and adaptability.




Step 2 (B):

Fate of sperms:

P: Fails to penetrate the ovum.
Q: Successfully fertilizes the ovum.
R: Degenerates.

Role of Zona pellucida:

It prevents polyspermy by blocking entry of additional sperms once one sperm has penetrated.

Changes after sperm entry:

Completion of meiosis II.
Formation of female pronucleus.
Zona reaction to block polyspermy.

Role of acrosome and middle piece:

Acrosome: Contains enzymes to digest egg membranes.
Middle piece: Packed with mitochondria providing energy for sperm motility. Quick Tip: Remember — pollen transfer can occur within the same flower, between flowers of the same plant, or different plants. Similarly, acrosomal enzymes like hyaluronidase and zona lysin aid sperm entry during fertilization.

Step 1 (A):

MALT (Mucosa-Associated Lymphoid Tissue) is a component of the immune system found in mucosal linings such as the respiratory tract, gut, and urogenital tract. It provides localized immunity at the mucosal surfaces.
Cytokine barriers refer to proteins like interferons secreted by virus-infected cells that protect non-infected cells from further viral infection.
The diagnostic test for AIDS is the ELISA test. It works on the principle of antigen-antibody interaction, detecting HIV antibodies in the blood.
Bone marrow produces lymphocytes while the thymus is where T-lymphocytes mature. Together, they generate a functional immune system capable of defending the body.


Step 2 (B):

See the filled table above in the question section.
Baculoviruses are used as biological control agents because they are species-specific, do not harm non-target organisms, and are safe for the environment. They infect only specific insect pests. Quick Tip: Interferons and cytokines are key natural defense mechanisms against viral infections. Also, microbes provide a wide range of bioactive compounds used in medical and industrial applications.

Step 1 (A):

Haemophilia and red-green colourblindness are X-linked recessive disorders. Since men have only one X chromosome, the presence of a single defective gene leads to expression of the disorder. Women, having two X chromosomes, must inherit two defective copies to express the condition, which is less likely.
For equal production of haemophilic daughters and sons, the cross must be between a carrier female (X\textsuperscript{HX\textsuperscript{h) and a haemophilic male (X\textsuperscript{hY):
[4pt]

\begin{tabular{c|c c
& X\textsuperscript{H & X\textsuperscript{h

\hline
X\textsuperscript{h & X\textsuperscript{HX\textsuperscript{h (Carrier Daughter) & X\textsuperscript{hX\textsuperscript{h (Haemophilic Daughter)

Y & X\textsuperscript{HY (Normal Son) & X\textsuperscript{hY (Haemophilic Son)
\end{tabular



This cross results in 1 haemophilic daughter and 1 haemophilic son out of 4 children.


Step 2 (B):

In bacteria, both transcription and translation occur in the cytoplasm. In eukaryotes, transcription occurs in the nucleus, and translation occurs in the cytoplasm.
\textit{[Insert labelled diagram of DNA replication fork here]
DNA has base pairing: Adenine pairs with Thymine, and Guanine pairs with Cytosine. So, 240 Adenine means 240 Thymine. That’s 480 purines.

Total nucleotides = 1000, so remaining 520 are pyrimidines (240 Thymine + 280 Cytosine). Quick Tip: X-linked disorders affect males more due to the presence of a single X chromosome. Always use complementary base pairing logic when calculating nucleotide distribution in DNA.