
CBSE Class 12 Biology Set 1 Question Paper PDF (57/1/1) is now available for download. CBSE conducted the Class 12 Biology examination on March 19, 2024, from 10:30 AM to 1:30 PM. The question paper consists of 33 questions carrying a total of 70 marks. Section A includes 16 MCQs for 1 mark each, Section B contains 5 very short-answer questions for 2 marks each, Section C comprises 7 short-answer questions for 3 marks each, Section D comprises 2 Case-based questions carries 4 marks each and Section E comprises 3 long-answer questions carries 5 marks each.
Candidates can use the link below to download the CBSE Class 12 Biology Set 1 Question Paper with detailed solutions.
| CBSE Class 12 2024 Biology Question Paper with Answer Key | Check Solution |
In which of the following plants are both male and female flowers born on the same plant, and the mode of pollination can be geitonogamy or xenogamy?
Plants that bear both male and female flowers on the same plant are called monoecious plants. Maize is an example of a monoecious plant.
Geitonogamy refers to the transfer of pollen between flowers on the same plant.
Xenogamy refers to the transfer of pollen between flowers of different plants. Maize exhibits both types of pollination. Quick Tip: Keep in mind:
Papaya and Date Palm are dioecious (with separate male and female plants).
Monoecious plants, like maize, have both male and female flowers on the same plant, which facilitates both geitonogamy and xenogamy.
Which one of the following hormones is secreted by the human placenta that helps in the maintenance of pregnancy?
Human Chorionic Gonadotropin (hCG) is a hormone secreted by the placenta, playing a crucial role in sustaining pregnancy by supporting the corpus luteum and progesterone production during the early stages.
Relaxin is produced later in pregnancy to help relax the pelvic muscles in preparation for delivery.
Oxytocin triggers uterine contractions during labor.
Human Placental Lactogen aids in fetal growth and metabolism, though it doesn't directly contribute to maintaining pregnancy.
Quick Tip: Key hormones and their roles:
hCG: Ensures the continuation of pregnancy.
Oxytocin: Induces labor.
Relaxin: Prepares the body for childbirth.
Placental Lactogen: Supports fetal development and metabolism.
The periodic abstinence by a couple for family planning should be from:
Periodic abstinence is a natural family planning method in which couples refrain from sexual intercourse during the fertile window of the menstrual cycle to avoid pregnancy. This approach requires an understanding of the woman's menstrual cycle and ovulation patterns.
Ovulation, the process of releasing an egg from the ovary, usually occurs on Day 14 of a standard 28-day cycle, though this timing can vary based on individual cycle length.
The fertile window, when pregnancy is most likely, includes the days leading up to and immediately following ovulation. This is because sperm can survive up to 5 days in the female reproductive system, and the egg is viable for 12 to 24 hours after ovulation.
In a typical 28-day cycle, ovulation happens around Day 14, so it is recommended to abstain from Day 10 to Day 17 to account for potential cycle variations and to reduce the chance of pregnancy.
For women with shorter or longer cycles than 28 days, ovulation may occur earlier or later than Day 14, requiring careful tracking of the cycle to adjust the abstinence period. Quick Tip: Key concept:
Ovulation usually takes place around Day 14 in a 28-day cycle.
The fertile window is from Day 10 to Day 17, when conception is most likely.
Avoiding sexual intercourse outside of this period reduces the risk of pregnancy.
Select the incorrect match from the following:
| Human Karyotype | Characters |
|---|---|
| 45 + XX | Broad palm with characteristic palm crease |
| 44 + XXY | Overall feminine development |
| 44 + XO | Sterile females as ovaries are rudimentary |
| 44 + XY | Normal male |
(A) 45 + XX: This refers to Down Syndrome, which is characterized by distinct physical traits such as broad palms and palm creases.
(B) 44 + XXY: Incorrect. This karyotype represents Klinefelter Syndrome, which affects males and results in partial masculine development, not feminine characteristics.
(C) 44 + XO: This refers to Turner Syndrome, where females are infertile due to underdeveloped ovaries.
(D) 44 + XY: This corresponds to normal males.
Quick Tip: Karyotype tips: 44 + XO: Turner Syndrome (Infertile females). 44 + XXY: Klinefelter Syndrome (Affects males with partial masculine development). 45 + XX/XY: Down Syndrome.
You know that there are twenty different types of naturally occurring amino acids and four different types of bases in the DNA. A combination of 3 such bases codes for a specific amino acid. If instead there are 96 different amino acids and 12 different bases in the DNA, then the minimum number of combination of bases required to form a codon is:
To find the minimum number of base combinations needed to code for 96 distinct amino acids, we can apply the formula for the number of possible combinations:
\[ Number of combinations = (Number of bases)^{Number of bases per codon} \]
In this case:
- There are 12 different base types in DNA.
- We need to determine how many bases per codon are necessary to produce at least 96 different amino acids.
Let \( n \) represent the number of bases required per codon. The total number of possible codons will be \( 12^n \), and we want this value to be at least 96.
\[ 12^n \geq 96 \]
Now, calculating the powers of 12:
- \( 12^1 = 12 \)
- \( 12^2 = 144 \)
Since \( 12^2 = 144 \) exceeds 96, the minimum number of bases per codon required is 2.
Thus, the correct answer is \( \boxed{2} \).
Option (A): Incorrect. \( 12^6 \) would result in a much higher number than necessary.
Option (B): Incorrect. \( 12^3 \) provides more than 96 codons, and 3 bases per codon is unnecessary.
Option (C): Correct. The minimum number of bases per codon is 2, as \( 12^2 = 144 \) exceeds 96.
Option (D): Incorrect. Using 4 bases per codon would create far more combinations than needed, but 2 bases is the minimum required.
Quick Tip: To determine the minimum number of bases per codon needed to encode a given number of amino acids, use the formula \( Number of combinations = (Number of bases)^{Number of bases per codon} \), and solve for \( n \).
The type of bond represented by the dotted line ‘-----’ in a schematic polynucleotide chain is:

Explanation: Phosphodiester bonds connect the phosphate group of one nucleotide to the sugar of another, forming the backbone of a polynucleotide chain.
This bond is known as an N-glycosidic linkage. It connects the 1' carbon of the pentose sugar to the nitrogen atom of the nitrogenous base.
Other bonds in the polynucleotide structure:
Phosphodiester bond: Links the sugar molecules of adjacent nucleotides.
Hydrogen bond: Occurs between complementary nitrogenous bases in double-stranded DNA.
Peptide bond: Found in proteins, not in polynucleotides.
Quick Tip: Identify bonds in biomolecules: N-glycosidic linkage: Between sugar and base in nucleotides. Phosphodiester bond: Backbone of DNA/RNA. Hydrogen bond: Holds DNA strands together. Peptide bond: Found in proteins.
In which of the following conditions/diseases is there a substantial increase in the activity of mast cells observed in the human body?
Mast cells are crucial in the body’s allergic reactions. They release histamine and other substances during an allergic response, which results in symptoms such as itching, swelling, and redness.
Conditions like typhoid, ascariasis, or AIDS do not involve mast cell activation. Quick Tip: Keep in mind:
Mast cells = Essential for allergic reactions.
Histamine release causes common allergy symptoms like sneezing, rash, and swelling.
Lactobacillus that sets milk into curd is categorised as:
Lactobacillus is a heterotrophic bacterium that ferments lactose in milk to produce lactic acid, leading to curd formation.
It ferments lactose in milk, producing lactic acid and causing the milk to curdle. Quick Tip: Bacteria classification: - Heterotrophic bacteria: Depend on organic matter. - Cyanobacteria: Photosynthetic. - Archaebacteria: Survive extreme environments.
Which one of the following transgenic animals is being used to test the safety of the polio vaccine?
Transgenic mice are widely used in biomedical research, including testing the safety and efficacy of vaccines like the polio vaccine.
. Quick Tip: Transgenic mice = Common in vaccine research due to their genetic similarity and ease of modification.
Restriction Endonuclease – Hind II always cuts DNA molecules at a particular point by recognising a specific sequence of:
Hind II recognises specific palindromic sequences in DNA consisting of 6 base pairs and cleaves precisely at this point.
Hind II recognizes and cuts specific sequences of six base pairs, following its recognition site in the DNA sequence.
Quick Tip: Restriction enzymes:
Recognise specific palindromic sequences in DNA.
Hind II = Recognises 6 base pairs.
The improved trait found in the genetically modified transgenic crop – Golden rice is:
Golden rice has been genetically engineered to produce beta-carotene in the endosperm, which is a precursor to vitamin A, thereby helping to combat vitamin A deficiency. Quick Tip: Keep in mind:
Golden rice = A source of Vitamin A.
Helps to address health issues caused by vitamin A deficiency.
The rate of formation of new organic matter by consumers, and the biomass available for consumption of herbivores as well as decomposers, are referred to as:
Gross primary productivity (GPP) refers to the total energy captured by autotrophs.
Secondary productivity is the biomass created by consumers from the organic matter they consume. Quick Tip: Productivity explained:
GPP: Energy captured by producers.
NPP: GPP minus the energy used in respiration.
Secondary productivity: Biomass generated by consumers.
Assertion (A): The laws of our country permit legal adoption, and it is as yet, one of the best methods for childless couples looking for parenthood.
Reason (R): Emotional, religious, and social factors are no deterrents to the legal adoption of orphaned and destitute children in India.
- The laws of India permit legal adoption, making it one of the best options for childless couples.
- However, emotional, religious, and social factors often act as deterrents to the adoption of children, especially in conservative settings. Quick Tip: When solving assertion-reason questions: - Check the truth of each statement individually. - Ensure that the reason explains the assertion directly if both are true.
Assertion (A): Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1.
Reason (R): Linked genes do not undergo independent assortment.
Solution: Linked genes are inherited together because they are located on the same chromosome. They do not follow Mendel's law of independent assortment, which is why the dihybrid F2 ratio 9:3:3:1 is not observed.
Assertion (A): Agrobacterium tumefaciens is a pathogen of several monocot plants.
Reason (R): It is able to deliver a piece of DNA known as ‘T-DNA' to transform normal plant cells into a tumor.
Solution: Agrobacterium tumefaciens is primarily a pathogen of dicot plants, not monocots. However, it does deliver T-DNA into the host plant genome, causing the formation of crown gall tumors.
Assertion (A): Indian Government has set up an organisation known as GEAC to decide the validity of GM research.
Reason (R): Genetic modification of organisms has no effect when such organisms are introduced into the ecosystem.
Solution: The Genetic Engineering Appraisal Committee (GEAC) evaluates and monitors genetically modified organisms (GMOs) for research and release. However, genetic modification can significantly affect ecosystems when GMOs are introduced due to possible crossbreeding or ecological imbalance.
Identify A, B, C, and D in the table given below:
| Terms | Part of the plant it represents |
|---|---|
| Pericarp | A |
| B | Cotyledon in seed of grass family |
| Embryonal axis | C |
| D | Remains of nucellus in a seed |
Solution:
• Pericarp (A): The fruit wall derived from the ovary wall.
• Scutellum (B): A specialized cotyledon found in monocot seeds (grasses).
• Embryonal axis (C): Includes the plumule (shoot) and radicle (root).
• Remains of nucellus (D): Called the perisperm in mature seeds.
Observe the picture given below.

Name the naturalist and write the explanation given by him that evolution of life forms had occurred on the basis of this example.
Solution:
Naturalist: Jean-Baptiste Lamarck
Solution: Lamarck explained evolution through the theory of Use and Disuse and Inheritance of Acquired Characteristics. He suggested that giraffes evolved long necks by stretching to reach tall trees for food. This acquired trait was passed on to subsequent generations.
Write the basic steps followed in the Assisted Reproductive Technologies (ART) programme to help childless couples. Why is it also known as test tube baby programme?
Solution:
Basic Steps in ART Programme:
1. Retrieval of eggs from the female and collection of sperm from the male.
2. Fertilization of the egg and sperm outside the body in a laboratory dish (in vitro fertilization).
3. Development of the fertilized egg (embryo) in a controlled environment.
4. Transfer of the embryo into the uterus for implantation and pregnancy.
Reason for Test Tube Baby Programme: The term “test tube baby" refers to the process of fertilization occurring outside the human body, typically in a laboratory setup, rather than a test tube.
A farmer while working on his farm was bitten by a poisonous snake. He was rushed to a nearby health centre where the doctor gave him an injection to save his life.
(i) What did the doctor inject and why?
(ii) Name the kind of immunity provided by this injection.
Solution:
(i) The doctor injected anti-venom serum. This contains pre-formed antibodies that neutralize the snake venom, providing immediate relief and preventing further damage.
(ii) The immunity provided by this injection is passive immunity, as the antibodies are directly administered rather than being produced by the body.
Why do organic farmers not recommend complete eradication of insect pests? Explain giving reason.
Solution: Organic farmers discourage complete eradication of insect pests because:
1.Ecological Balance:
Insects play a role in the food chain and biodiversity. Their complete eradication can disrupt the balance.
2.Natural Pest Control:
Some pests attract their predators, which naturally control pest populations, reducing the need for chemical interventions.
Study the diagram of a pyramid of biomass given below.

Name the two standing crops that could be occupying level 'A' and level ‘B' in it. Name this type of pyramid and the ecosystem in which it is found.
Solution:
Level A: Zooplankton (Primary Consumers)
Level B: Phytoplankton (Primary Producers)
Type of Pyramid: Inverted Pyramid of Biomass.
Ecosystem: Found in an aquatic ecosystem.
In such ecosystems, the biomass of primary producers (phytoplankton) is lower than that of primary consumers (zooplankton) at any given time. This is because phytoplankton reproduce rapidly and are consumed quickly by zooplankton.
Explain the mode of action of contraceptive pills taken by human females. Mention the schedule to be followed for an effective outcome.
Solution:
Mode of Action:
1. Contraceptive pills contain synthetic hormones (estrogen and progesterone) that prevent ovulation by inhibiting the secretion of FSH and LH from the pituitary gland.
2. They thicken cervical mucus, making it difficult for sperm to enter the uterus.
3. They alter the endometrial lining, preventing implantation of a fertilized egg.
Schedule for Effective Outcome:
Pills should be taken daily for 21 days starting from the 5th day of the menstrual cycle, followed by a 7-day gap during which withdrawal bleeding occurs.
Name and write two characteristics of the type of DNA that forms the basis of DNA fingerprinting technique.
Solution:
• Type of DNA: Short Tandem Repeats (STRs) or Variable Number Tandem Repeats (VNTRs).
• Characteristics:
1. They are non-coding, repetitive sequences of DNA.
2. They show high polymorphism, making them unique to each individual (except identical twins).
Mention any two applications of this technique.
Solution:
1. Forensic Science: Used for identification in criminal investigations.
2. Paternity Testing: Determines biological relationships.
Explain the significance of the experiment carried out by S.L. Miller. Name the scientists whose hypothesis prompted him to carry out this experiment.
Solution:
Significance of Miller's Experiment:
• Demonstrated that organic molecules like amino acids could be synthesized abiotically under conditions resembling early Earth's atmosphere.
• Supported the theory that life originated from simple organic compounds.
• Scientists: The experiment was based on the hypothesis of Alexander Oparin and J.B.S. Haldane.
How does meteorite analysis favour this hypothesis?
Solution: Meteorite analysis revealed the presence of simple organic compounds, such as amino acids, on extraterrestrial objects. This suggests that organic molecules can form naturally, supporting the idea of abiotic synthesis.
Identify A, B, C, D, E, and F in the table given below:
| Name of Human Disease | Causative Organism | Symptoms |
|---|---|---|
| Pneumonia | Streptococcus | A |
| Typhoid | B | High fever, weakness, headache, stomach pain |
| Common Cold | Rhino virus | C |
| Ringworm | D | Dry scaly lesions on body parts, redness, itching |
| Ascariasis | Ascaris | E |
| F | Entamoeba histolytica | Constipation, cramps, stools with mucous and blood clots |
Solution:
• Pneumonia symptoms: A include breathing difficulty, cough, and chest pain.
• Typhoid causative organism: B is Salmonella typhi.
• Common cold symptoms: C include nasal congestion, sore throat, and cough.
• Ringworm causative organism: D is Trichophyton.
• Ascariasis symptoms: E include internal bleeding, muscular pain, and anemia.
• F: Disease caused by Entamoeba histolytica is Amoebiasis.
In a family, the father, the daughter, and the son are colour blind, whereas the mother has normal vision. Do you think the son and daughter have inherited the disease from their father? Work out a cross to justify your answer.
Solution: Colour blindness is an X-linked recessive disorder. Males inherit the X chromosome from their mother, while females inherit one X chromosome from each parent. The daughter inherits the Xc chromosome from her father and XC (normal) from her mother, making her a carrier. The son cannot inherit Xc from the father as he receives only the Y chromosome from him.
Cross:
Father (XcY) × Mother (XCXC)
Gametes: Xc, Y (Father) and XC, XC (Mother)
Offspring:
Daughter (XCXc) (Carrier)
Son (XCY) (Normal).
Hence, the son and daughter cannot inherit colour blindness from the father as it is X-linked.
What are transgenic animals?
Solution: Transgenic animals are those that have had foreign genes deliberately inserted into their genome. These genes are introduced to study gene functions, improve livestock traits, or produce biologically important substances.
Name the first transgenic cow and state its importance.
Solution: The first transgenic cow is Rosie. Importance: Rosie produced milk enriched with the human protein alpha-lactalbumin, making it nutritionally more balanced for infants.
(i) Explain the convention for naming EcoRI.
(ii) With the help of an illustration only, show the action of EcoRI on a DNA Polynucleotide.
Solution: (i) EcoRI:
1. "E" = Genus (Escherichia).
2. "co" = Species (coli).
3. "R" = Strain (RY13).
4. "I" = First identified enzyme from this strain.
(ii)
Explain how it is ensured that the orchid Ophrys is pollinated by a specific species of bee.
Solution: The orchid Ophrys ensures pollination by mimicking the appearance and scent of a female bee. Male bees, attracted to the orchid, attempt to mate with it (pseudocopulation) and inadvertently carry pollen to other flowers.
Describe co-evolution with the help of this example.
Solution: Co-evolution is the process where two or more species influence each other's evolution. In the case of Ophrys and its specific bee species, the orchid evolved to mimic the female bee, while the bee adapted to the orchid's mimicry.
Questions No. 29 and 30 are case-based questions. Each question has 3 sub-questions with internal choice in one sub-question. 29. Read the following passage and answer the questions that follow.
Isn't it incredible that India's land area is only 2.4 percent of the world's total land area whereas its share of the global species diversity is an impressive 8-1 percent! However, in these estimates of species, prokaryotes do not figure anywhere. Biologists are always keen on collecting data with respect to species diversity observed in different regions of the world. The data collected based on the survey conducted for species richness of groups of mammals in three different regions of the world is shown in the bar graph below.
(a) Why is the species richness maximum in Region III in the bar graph?
Solution: Region III has maximum species richness due to: 1. Favourable climatic conditions such as tropical climate with high rainfall and temperature. 2. High resource availability and ecological niches supporting a large number of species.
(b) Why is the species richness minimum in Region I in the bar graph?
Solution: Region I has minimum species richness because: 1. It likely represents a cold or arid climate, which is less suitable for supporting diverse life forms. 2. Limited resource availability and harsh environmental conditions restrict the number of species.
(c) Plants and animals do not have uniform diversity in the world but show rather uneven distribution. Mention what this kind of diversity is referred to as.
Solution: This uneven distribution of species diversity is referred to as Beta Diversity. It reflects the variation in species composition between different ecosystems or regions.
(d) Why is it that prokaryotes do not have an estimated number of their species diversity as seen in plants and animals? Explain.
Solution:
1. Prokaryotes lack morphological diversity, making species identification challenging.
2. Many prokaryotic species cannot be cultured in laboratories, limiting data collection.
3. Molecular techniques like DNA sequencing are required to identify them, which is time-consuming and resource-intensive.
Study the diagram given below that shows the steps involved in the procedure of selecting transformed bacteria and answer the questions that follow:

(a) Identify the colony that has got transformed. Justify your answer.
Solution: The transformed colony is the one that grows on Plate M (with ampicillin) but does not grow on Plate N (with both ampicillin and tetracycline). Justification: The recombinant DNA disrupts the tetracycline resistance gene, making the bacteria sensitive to tetracycline, but retains ampicillin resistance, allowing growth on Plate M.
(b) What are the sites in a plasmid called where ampicillin and tetracycline resistance genes are inserted? State their role in genetic engineering.
Solution: The sites are called selectable marker sites. Role: These genes help in identifying and selecting transformed cells: Ampicillin resistance gene ensures growth on ampicillin-containing media. Tetracycline resistance gene helps distinguish recombinant bacteria (tetracycline-sensitive) from non-recombinant ones.
(c) Name two enzymes playing an important role in genetic engineering.
Solution:
1. Restriction endonucleases: Cut DNA at specific sequences.
2. DNA ligase: Joins DNA fragments to form recombinant DNA.
OR
(c) State the role of β-galactosidase in insertional inactivation.
Solution: Insertional inactivation of the β-galactosidase gene prevents the production of the enzyme, resulting in white colonies (instead of blue) on media with X-gal. This helps identify recombinant bacteria.
Explain the development of male gametophyte in an angiosperm.
Solution: The male gametophyte in angiosperms develops within the pollen grain:
1. Microspore Formation: Each microspore undergoes mitotic division to form a small generative cell and a larger vegetative cell.
2. Mature Pollen Grain: The generative cell divides mitotically to produce two male gametes, resulting in a three-celled structure (two male gametes + one vegetative cell).
Draw a labelled diagram of a three-celled male gametophyte. (Diagram skipped as requested.)
Solution:


Draw a diagrammatic sectional view of the ovary of a human female and label the following:
Solution:


At which stage of life are primary follicles formed in a human female?
Solution: Primary follicles are formed during the fetal stage of development in a female.
Explain the events (both hormonal and structural) that occur at the time of ovulation till the onset of the next menstrual cycle.
Solution:
1. At Ovulation: Surge in LH (Luteinizing Hormone) leads to the rupture of the Graafian follicle and release of the ovum.
2. Post-Ovulation: The ruptured follicle transforms into the corpus luteum, which secretes progesterone. Progesterone prepares the endometrium for implantation.
3. If Fertilization Does Not Occur: The corpus luteum degenerates, leading to a drop in progesterone levels. The endometrium breaks down, resulting in menstruation.
Stability, as one of the properties of genetic material, was very evident in one of the very early experiments in genetics. Name the scientist and describe his experiment. State the conclusion he arrived at.
Solution:
• Scientist: Frederick Griffith
• Experiment:
1. He worked on Streptococcus pneumoniae (S strain = virulent, R strain = non-virulent).
2. Heat-killed S strain mixed with live R strain transformed R into a virulent S strain.
• Conclusion: A "transforming principle" (later identified as DNA) is responsible for genetic stability and transfer of traits.
A tall pea plant bearing violet flowers with unknown genotype is given. Find the genotype by working out different crosses by selfing the plants. Write the genotypic and phenotypic ratios of each cross shown by you.
Solution:
1. Case 1: Homozygous (TTVV)
• Cross: TTVV × TTVV
• Genotypic ratio: 100% TTVV
• Phenotypic ratio: 100% Tall Violet
2. Case 2: Heterozygous (TtVv)
• Cross: TtVv × TtVv
• Genotypic ratio: 1:2:1 (TT:Tt:tt) for height, 1:2:1 (VV:Vv:vv) for color.
• Phenotypic ratio: 9:3:3:1 (Tall violet: Tall white: Dwarf violet: Dwarf white).
Name and explain the property present in normal cells but lost in cancer cells.
Solution: The property present in normal cells but lost in cancer cells is contact inhibition. Contact inhibition refers to the process where normal cells stop dividing when they come into contact with neighboring cells. This mechanism ensures that cells grow in an organized manner and maintain proper tissue structure. It prevents overgrowth and helps to regulate tissue size. In cancer cells, this property is lost, and they continue to divide uncontrollably, even when they come into contact with other cells. This uncontrolled division results in tumor formation and cancer progression.
All normal human cells have genes that may become cancerous under certain conditions. Name them and explain how.
Solution: The genes that may become cancerous under certain conditions are oncogenes and tumor suppressor genes.
1. Oncogenes: Oncogenes are mutated forms of normal genes called proto-oncogenes. Proto-oncogenes normally promote cell growth and division. However, when mutated or overexpressed, they become oncogenes, leading to uncontrolled cell division and potentially causing cancer. Example: The Ras gene, when mutated, continuously signals the cell to divide, even when not needed, leading to uncontrolled cell proliferation.
2. Tumor Suppressor Genes: Tumor suppressor genes are responsible for inhibiting cell growth and promoting the repair of damaged DNA or apoptosis (programmed cell death). If these genes are mutated or inactivated, cells can bypass growth control mechanisms and continue to divide. Example: The p53 gene is a critical tumor suppressor gene. When it is mutated, it fails to prevent the division of cells with damaged DNA, which can lead to cancer development.
State the role of the following techniques in the detection and diagnosis of cancer:
1. Biopsy and Histopathology
2. Magnetic Resonance Imaging (MRI)
Solution:
1. Biopsy and Histopathology: A biopsy involves removing a small sample of tissue from a suspected tumor or abnormal growth. The tissue is then examined under a microscope through histopathology. The role of biopsy and histopathology is essential for diagnosing cancer. They help identify the presence of cancerous cells, determine the type of cancer, and assess its stage and grade. This also provides critical information on the malignancy of the tumor and helps plan appropriate treatment options.
2. Magnetic Resonance Imaging (MRI): MRI is a non-invasive imaging technique that uses magnetic fields and radio waves to produce detailed images of the inside of the body. It is especially useful in detecting soft tissue abnormalities such as brain, spinal cord, and breast cancers. The role of MRI in cancer diagnosis includes identifying and locating tumors, determining their size, and assessing whether cancer has spread to other parts of the body (metastasis). It is also used to monitor treatment progress and plan surgeries or radiation therapy.
Large quantities of sewage are generated every day in cities as well as in towns and are treated in Sewage Treatment Plants (STPs) to make them less polluting. Given below is the flow diagram of stages of STP.
Study the flow diagram and answer the questions that follow:

(i) 1. Why is primary effluent passed into large aeration tanks?
Solution: Primary effluent is passed into large aeration tanks to increase the oxygen supply to the effluent. This helps in the growth of aerobic microorganisms that break down the organic matter in the sewage, converting it into simpler substances. The process of aeration encourages the growth of these microorganisms, which helps in reducing the biochemical oxygen demand (BOD) of the water, making it less polluting.
2. What is the ‘sediment’ formed referred to? Mention its significance.
Solution: The sediment formed is called sludge. It consists of organic and inorganic matter that settles at the bottom of the settling tank. The significance of sludge is that it contains a high concentration of organic matter, which can be further processed to extract biogas or treated for use as fertilizer. Proper disposal of sludge is crucial for preventing environmental pollution.
3. Explain the final step in the settling tank before the treated effluent is released into water bodies.
Solution: The final step involves the treatment of the effluent in the settling tank, where the remaining suspended particles, including microorganisms, are allowed to settle down. This is followed by the removal of the clear water (supernatant), which is relatively free from pollutants. This treated water is then released into water bodies. Further treatment may be performed, depending on the quality of the effluent.
(ii) Name any two organisms commonly used as biofertilizers, belonging to different kingdoms. Write how each one acts as a biofertilizer.
Solution:
Rhizobium (Kingdom: Bacteria): Rhizobium is a nitrogen-fixing bacterium that forms a symbiotic relationship with leguminous plants. It helps in fixing atmospheric nitrogen into a form that the plants can use for growth, thereby acting as a natural fertilizer.
Azotobacter (Kingdom: Bacteria): Azotobacter is a free-living nitrogen-fixing bacterium found in soil. It helps in fixing nitrogen from the atmosphere and making it available to plants, improving soil fertility. Unlike Rhizobium, Azotobacter does not form symbiotic relationships with plants but independently fixes nitrogen.
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