ISC Class 12 Biology Board Exam Question Paper 2024 with Solutions

CISCE Class 12 Biology Board Exam Question Paper 2024 with Solutions

SECTION A - 20 MARKS

Q1. Answer the following questions briefly. [1 Mark each]

1.1. In human plasma, five different types of immunoglobulins are found. Which type of immunoglobulin is responsible for allergic reactions? [1 Mark]

Answer:
Immunoglobulin E (IgE).

Teacher's Note:
a) IgE mediates type I hypersensitivity reactions which manifest as common allergies.
b) Remember that IgE binds to mast cells and basophils, triggering degranulation.

1.2. Some orchids live on the branches of mango trees. Name the type of interaction that exists between the mango tree and the orchid. [1 Mark]

Answer:
Commensalism.

Teacher's Note:
a) In commensalism, one species benefits while the other is unaffected (+, 0 interaction).
b) The orchid gets sunlight and mechanical support, while the mango tree suffers no harm or benefit.

1.3. Four triplet codons code for the amino acid valine. Three of them are given below. GUU GUC GUA. Write the fourth codon. [1 Mark]

Answer:
GUG.

Teacher's Note:
a) The genetic code is degenerate, meaning multiple codons can specify the same amino acid.
b) Ensure you write RNA codons using Uracil (U) instead of Thymine (T).

1.4. A haemophilic man marries a carrier woman and they have a daughter. What is the probability of their daughter being haemophilic? [1 Mark]

Answer:
50%

[Figure: Cross showing X^hY (haemophilic man) crossed with XX^h (carrier woman) producing XX^h (carrier daughter) and X^hX^h (haemophilic daughter)]

Teacher's Note:
a) Haemophilia is an X-linked recessive disorder.
b) Out of the female offspring, half receive two recessive alleles and thus express the disease.

1.5. Home-made fruit juices are turbid, while the bottled fruit juices purchased from the market are clear. Give a reason for this difference. [1 Mark]

Answer:
Bottled juices are filtered and clarified using pectinases and proteases, whereas home-made juices contain natural pulp and fibres.

Teacher's Note:
a) Commercial fruit juices are treated with microbial pectinases and proteases to clear the juice.
b) Home-made juices retain suspended plant cell wall fragments and fibres.

1.6. The number of lily plants in a pond was found to be 50. After one year, the number increased to 65. Calculate the natality of lily plants. [1 Mark]

Answer:
Natality rate = 0.3 per plant per year.

Teacher's Note:
a) Natality is calculated as the increase in population divided by the original population over a time period (\(15 \div 50 = 0.3\)).
b) Always state the correct unit, which is offspring/individuals per unit time per existing individual.

1.7. Based on the table given below, identify the type of natural selection taking place. [1 Mark]

Answer:
Disruptive selection.

[Figure: Graph showing two peaks at extremes and a trough in the middle representing disruptive selection]

Teacher's Note:
a) Disruptive selection favors individuals at both extremes of a phenotypic range over intermediate forms.
b) Here, small and large seeds have high germination rates while medium-sized seeds have low rates.

1.8. Give the name of the target pest of gene cry 1 Ac. [1 Mark]

Answer:
Cotton bollworms and tobacco budworm.

Teacher's Note:
a) The cry1Ac gene codes for crystal proteins that are toxic to lepidopteran insects.
b) Be precise with the specific insect orders targeted by different cry genes.

1.9. If a person shows the production of interferons in his body, then he is suffering from __________. [1 Mark]
1. Malaria
2. Ring worm
3. Dengue
4. Typhoid

Answer:
3. Dengue

Teacher's Note:
a) Interferons are antiviral proteins produced by virus-infected cells to protect non-infected cells.
b) Among the choices, only Dengue is caused by a virus.

1.10. Match the columns I and II with reference to weeks of pregnancy and the development of a human embryo. Select the correct option from the choices given below: [1 Mark]

1. I - (P), II - (Q), III - (R), IV - (S)
2. I - (Q), II - (P), III - (R), IV - (S)
3. I - (R), II - (S), III - (P), IV - (Q)
4. I - (S), II - (R), III - (Q), IV - (P)

Answer:
2. I - (Q), II - (P), III - (R), IV - (S)

Teacher's Note:
a) Embryonic development milestones must be memorized thoroughly.
b) At 12 weeks major organ systems are formed (limbs and external genitalia), and digits develop earlier around 8 weeks.

1.11. Assertion: In a bioreactor, it is not necessary to maintain sterile ambience.
Reason: Sterile conditions promote the growth of unwanted microbes in the culture medium. [1 Mark]
1. Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
2. Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
3. Assertion is true and Reason is false.
4. Both Assertion and Reason are false.

Answer:
4. Both Assertion and Reason are false.

Teacher's Note:
a) Maintaining a sterile environment is absolutely crucial in a bioreactor.
b) Unwanted microbes contaminate the culture, ruining the desired product yield.

1.12. Assertion: Lymphocytes originate and proliferate in primary lymphoid organs.
Reason: Spleen is a secondary lymphoid organ. [1 Mark]
1. Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
2. Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
3. Assertion is true and Reason is false.
4. Both Assertion and Reason are false.

Answer:
2. Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.

Teacher's Note:
a) Primary lymphoid organs provide sites for maturation of lymphocytes (bone marrow and thymus).
b) Spleen is indeed a secondary lymphoid organ, but it does not explain why lymphocytes originate in primary organs.

1.13. Name the chemical used to visualise the movement of DNA fragments in the gel. [1 Mark]

Answer:
Ethidium bromide (EtBr).

[Figure: Agarose gel electrophoresis setup showing wells, DNA bands moving from negative to positive electrode, indicating largest to smallest fragments]

Teacher's Note:
a) Ethidium bromide intercalates with DNA base pairs.
b) Stained DNA fluoresces bright orange under UV light radiation.

1.14. In humans, somatic gene therapy was carried out to correct an immunodeficiency disease. Name this disease. [1 Mark]

Answer:
SCID (Severe Combined Immunodeficiency).

Teacher's Note:
a) SCID is caused by a deficiency of the enzyme adenosine deaminase (ADA).
b) It was the first disease treated using gene therapy in 1990.

1.15. What could be the genotype of the affected male? [1 Mark]

[Figure: Pedigree chart showing autosomal recessive inheritance pattern with affected males and females]

Answer:
\(aa\) (homozygous recessive).

Teacher's Note:
a) Thalassemia is an autosomal recessive disorder.
b) Affected individuals must possess two copies of the mutant allele to show symptoms.

1.16. Answer the following questions: [1 Mark]

1.16. (a) In a karyotype analysis, X and Y chromosomes represent sex chromosomes. Name the scientist who discovered the X chromosome. [1 Mark]

Answer:
Hermann Henking.

Teacher's Note:
a) Hermann Henking identified the X body in insects in 1891.
b) Later, it was recognized as a sex chromosome.

1.16. (b) Expand to its full form: NACO [1 Mark]

Answer:
National AIDS Control Organisation.

Teacher's Note:
a) NACO coordinates policies and programs for HIV/AIDS prevention and control in India.
b) Memorize all key medical and biological abbreviations.

1.17. Name the hypodermal cell labelled '1' which divides periclinally. [1 Mark]

[Figure: Diagram of transverse section of early microsporangium showing hypodermal cell layer labeled 1]

Answer:
Archesporial cell.

Teacher's Note:
a) Archesporial cells divide periclinally to form primary parietal cells and primary sporogenous cells.
b) They initiate the formation of pollen sacs.

1.18. Give a reason for each of the following: [2 Marks]

1.18. (a) The second half of the menstrual cycle is called the luteal phase as well as the secretory phase. [1 Mark]

Answer:
It is called the luteal phase because the remaining ruptured follicle transforms into the corpus luteum, and the secretory phase because the endometrium secretes large amounts of progesterone.

Teacher's Note:
a) Progesterone maintains the uterine endometrium for implantation.
b) Without fertilization, the corpus luteum degenerates, leading to menstruation.

1.18. (b) Streptokinase is administered to the patients having myocardial infarction. [1 Mark]

Answer:
Streptokinase acts as a clot buster, dissolving blood clots in blood vessels of patients suffering from myocardial infarction.

Teacher's Note:
a) It is produced by the bacterium Streptococcus and modified through genetic engineering.
b) It prevents extensive heart muscle damage by restoring blood flow.

SECTION B - 14 MARKS

Q2. [2 Marks]

2. (a) Name any two Cu-ions releasing IUDs. [1 Mark]

Answer:
Cu-T and Cu-7.

Teacher's Note:
a) Copper-releasing IUDs increase phagocytosis of sperms within the uterus.
b) Other examples include Multiload 375.

2. (b) Explain any two ways by which IUDs devices act as contraceptives. [1 Mark]

Answer:
1. They suppress sperm motility and the fertilising capacity of sperms.
2. They make the uterus unsuitable for implantation.

Teacher's Note:
a) Copper ions released suppress sperm motility.
b) Phagocytosis of sperm increases inside the uterine cavity.

Q3. A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following: [2 Marks]

3.1. Frequency of the allele (A). [0.5 Marks]

Answer:
\(p = 1 - q = 1 - 0.4 = 0.6\).

Teacher's Note:
a) According to Hardy-Weinberg equilibrium, \(p + q = 1\).
b) Here q represents the recessive allele frequency and p represents the dominant allele frequency.

3.2. The number of homozygous dominant fruit flies. [0.5 Marks]

Answer:
\(p^{2} \times 200 = (0.6)^{2} \times 200 = 0.36 \times 200 = 72\).

Teacher's Note:
a) Homozygous dominant genotype frequency is given by \(p^{2}\).
b) Multiply by the total population size to get the absolute number.

3.3. The number of homozygous recessive fruit flies. [0.5 Marks]

Answer:
\(q^{2} \times 200 = (0.4)^{2} \times 200 = 0.16 \times 200 = 32\).

Teacher's Note:
a) Homozygous recessive genotype frequency is represented by \(q^{2}\).
b) Ensure proper squaring of decimals before multiplying.

3.4. The number of carrier fruit flies. [0.5 Marks]

Answer:
\(2pq \times 200 = 2 \times 0.6 \times 0.4 \times 200 = 0.48 \times 200 = 96\).

Teacher's Note:
a) Heterozygous individuals act as carriers and their frequency is \(2pq\).
b) Sum of all numbers (72 + 32 + 96) equals the total population of 200, verifying calculations.

Q4. Jacob is genetically a carrier of the disorder that affects the shape of the RBCs, as shown in the diagram below. His son James suffers from the same disorder. [2 Marks]

[Figure: Sickle-shaped red blood cells compared to normal biconcave red blood cells]

i. Give the biochemical reason for the disorder that changes the shape of the RBCs, as shown above. [1 Mark]

Answer:
Deoxygenated sickle-cell haemoglobin undergoes polymerisation under low oxygen tension, changing the RBC shape into elongated sickle-like structures.

Teacher's Note:
a) Glutamic acid is substituted by valine at the 6th position of the beta globin chain.
b) This amino acid substitution drastically alters protein solubility and polymerisation.

ii. Draw a Punnett square to show the genotype of the mother of James. [0.5 Marks]

Answer:
Mother's genotype is either Hb^AHb^S (carrier) or Hb^SHb^S (affected).

Teacher's Note:
a) Since James has sickle cell anemia (Hb^SHb^S), he must receive one Hb^S allele from each parent.
b) Since Jacob is a carrier, the mother must also carry or possess the Hb^S allele.

iii. Name and define the type of 'point mutation' responsible for this disorder. [0.5 Marks]

Answer:
Substitution mutation (specifically missense mutation), where a single base pair change alters the resultant amino acid.

Teacher's Note:
a) A single nucleotide substitution from GAG to GUG leads to valine incorporation instead of glutamic acid.
b) This is a classic example of qualitative genetic defects in humans.

Q5. [2 Marks]

5.1. The diagram given below shows the three types of endosperms in angiosperms. [2 Marks]

[Figure: Diagrams of three types of endosperms: Type 1 Free nuclei, Type 2 Cellular, Type 3 Helobial]

1. Identify the three types of endosperms shown above. [1 Mark]

Answer:
Type I: Nuclear endosperm, Type II: Cellular endosperm, Type III: Helobial endosperm.

Teacher's Note:
a) Nuclear endosperm involves repeated free nuclear divisions without immediate wall formation.
b) Cellular endosperm has wall formation after every division.

2. Name the type of endosperm which commonly occurs in polypetalous dicots. [1 Mark]

Answer:
Nuclear endosperm.

Teacher's Note:
a) Nuclear endosperm is the most common type found in angiosperms.
b) Coconut water is a familiar example of free-nuclear endosperm.

OR

5.2. The diagram given below shows the various steps in spermatogenesis. [2 Marks]

[Figure: Spermatogenesis schematic diagram showing spermatogonia, primary spermatocytes (1), secondary spermatocytes (2), spermatids (3), and spermatozoa (4)]

a. Name the parts labelled '1', '2' and '3'. [1.5 Marks]

Answer:
1 - Primary spermatocytes
2 - Secondary spermatocytes
3 - Spermatids

Teacher's Note:
a) Primary spermatocytes undergo the first meiotic division.
b) Spermatids are haploid round cells that differentiate into spermatozoa.

b. Name the process by which part '3' changes to part '4'. [0.5 Marks]

Answer:
Spermiogenesis.

Teacher's Note:
a) Spermiogenesis is the transformation of non-motile spermatids into motile spermatozoa.
b) Do not confuse spermiogenesis with spermiation, which is the release of sperms from Sertoli cells.

Q6. [2 Marks]

6.1. (a) Write the scientific name of the filarial worm that causes filariasis. [1 Mark]

Answer:
Wuchereria bancrofti (or Wuchereria malayi).

Teacher's Note:
a) Binomial nomenclature rules must be followed (genus capitalized, species lowercase, both underlined or italicized).
b) It resides in the lymphatic vessels of humans.

6.1. (b) Write the mode of transmission for the following diseases: Filariasis [1 Mark]

Answer:
Transmitted to humans through the bite of infected female Culex mosquitoes.

Teacher's Note:
a) Vectors transmit infective larvae (microfilariae) during biting.
b) Chronic infection leads to severe swelling known as elephantiasis.

6.2. (a) Write the scientific name of the causative agent of the following diseases: Typhoid [1 Mark]

Answer:
Salmonella typhi.

Teacher's Note:
a) This pathogenic bacterium enters through contaminated food and water.
b) The Widal test is used for diagnosis.

6.2. (b) Write the mode of transmission for the following diseases: Typhoid [1 Mark]

Answer:
Ingestion of food and water contaminated with the bacteria.

Teacher's Note:
a) Typhoid is a classic water-borne disease.
b) Poor sanitation practices facilitate its spread.

Q7. [2 Marks]

7.1. A male plant bearing red flowers was crossed with a female plant bearing yellow flowers. In the F₁ generation, all the flowers were orange in colour. Give a reason to explain the change of colours in F₁ generation. [1 Mark]

Answer:
Incomplete dominance, where neither allele is completely dominant over the other, resulting in an intermediate phenotype.

Teacher's Note:
a) Incomplete dominance deviates from classical Mendelian dominance.
b) The heterozygous condition expresses a blended or intermediate trait.

7.2. A male plant bearing red flowers was crossed with a female plant bearing yellow flowers. In the F₁ generation, all the flowers were orange in colour. Mention the ratio of red flowers, yellow flowers and orange flowers in the F₂ generation. [1 Mark]

Answer:
1 (Red) : 2 (Orange) : 1 (Yellow).

[Figure: Punnett square showing RR (Red), Ry (Orange), Ry (Orange), yy (Yellow)]

Teacher's Note:
a) Both genotypic and phenotypic ratios in incomplete dominance are identical (1:2:1) in the F₂ generation.
b) This confirms that alleles remain distinct and do not mix permanently.

Q8. Microbes are useful to human beings in diverse ways. Give the biological name of the following microbe: [2 Marks]

8.1. Lactic acid producing bacterium. [0.5 Marks]

Answer:
Lactobacillus.

Teacher's Note:
a) LAB convert milk into curd by producing acids that coagulate milk proteins.
b) They also improve nutritional quality by increasing vitamin B12.

8.2. Microbe known as Baker's yeast. [0.5 Marks]

Answer:
Saccharomyces cerevisiae.

Teacher's Note:
a) Used extensively in baking and brewing industries.
b) Ferments sugars to produce carbon dioxide and ethanol.

8.3. Fungus which helps in the production of cyclosporin-A. [0.5 Marks]

Answer:
Trichoderma polysporum.

Teacher's Note:
a) Cyclosporin-A is used as an immunosuppressive agent in organ transplant patients.
b) It prevents tissue rejection by suppressing cell-mediated immunity.

8.4. Microbe used in the production of statins. [0.5 Marks]

Answer:
Monascus purpureus.

Teacher's Note:
a) Statins act as blood-cholesterol lowering agents.
b) They function by competitively inhibiting the enzyme responsible for cholesterol synthesis.

SECTION C - 21 MARKS

Q9. The diagram given below is the L.S. of a typical fruit. [3 Marks]

[Figure: L.S. of a fruit showing 1 - Epicarp, 2 - Mesocarp, 3 - Endocarp]

i. Identify the parts labelled '1', '2' and '3'. [1 Mark]

Answer:
1 - Epicarp, 2 - Mesocarp, 3 - Endocarp.

Teacher's Note:
a) These three layers form the pericarp of a true fruit.
b) They develop from the ovary wall after fertilization.

ii. State the difference between a true fruit and a false fruit. [1 Mark]

Answer:
True fruits develop exclusively from the ovary, whereas false fruits develop from other floral parts like the thalamus along with the ovary.

Teacher's Note:
a) Apple and strawberry are classic examples of false fruits.
b) In false fruits, the edible portion is usually the swollen thalamus.

iii. What is the significance of the formation of fruit in angiosperms? [1 Mark]

Answer:
Fruits protect seeds from harsh environmental conditions and assist in seed dispersal over long distances.

Teacher's Note:
a) Fleshy fruits attract animals, which aid in zoochory (seed dispersal).
b) Dry fruits have specialized mechanisms for wind or explosive dispersal.

Q10. Suneeta is planning an experiment to clone a gene in a vector. So, she has to choose a good cloning vector. Which one of the vectors shown below should she choose? Justify your answer by giving two reasons. [3 Marks]

[Figure: Vector A, Vector B, Vector C showing antibiotic resistance genes, origin of replication, and restriction sites]

Answer:
Vector C. Reasons: 1. It contains selectable markers with two antibiotic resistance genes (amp^R and tet^R) for screening recombinants. 2. It possesses a suitable origin of replication (ori) and unique restriction sites.

Teacher's Note:
a) A good cloning vector must have selectable markers to differentiate transformants from non-transformants.
b) Insertion of foreign DNA at a restriction site inactivates one of the marker genes, aiding insertional inactivation screening.

Q11. Study the two figures shown below that represent two growth models. [3 Marks]

[Figure: Figure A showing J-shaped exponential growth curve, Figure B showing S-shaped logistic growth curve]

i. Which one of the two figures represents an unlimited supply of nutrients? Give a reason. [1 Mark]

Answer:
Figure A represents an unlimited supply of nutrients because it shows exponential (J-shaped) growth where resources are never limiting.

Teacher's Note:
a) Exponential growth occurs when resources are abundant.
b) The population grows at a maximum intrinsic rate of natural increase.

ii. Which figure depicts a challenge to population growth? [0.5 Marks]

Answer:
Figure B.

Teacher's Note:
a) Logistic growth accounts for environmental resistance and carrying capacity (K).
b) As population density increases, competition for resources intensifies.

iii. Explain the term reproductive fitness. [0.5 Marks]

Answer:
Reproductive fitness refers to an organism's ability to survive, reproduce, and pass on its genes to subsequent generations relative to other members of the population.

Teacher's Note:
a) Also referred to as Darwinian fitness.
b) High fitness leads to greater representation of traits in future gene pools.

iv. Give the mathematical expressions for Figure A and Figure B. [1 Mark]

Answer:
For Figure A: dN/dt = rN.
For Figure B: dN/dt = rN((K - N)/K).

Teacher's Note:
a) N represents population density, r is intrinsic rate of increase, and K is carrying capacity.
b) Memorize both differential equations and the terms involved.

Q12. The diagram given below represents the schematic structure of proinsulin, which undergoes certain modifications before it becomes a fully functional insulin. Study the diagram carefully and answer the questions that follow: [3 Marks]

[Figure: Schematic structure of proinsulin showing A chain, B chain, and C-peptide connected by disulfide bonds]

i. State the change the proinsulin undergoes to become fully functional. [1 Mark]

Answer:
The inactive C-peptide is removed by enzymatic cleavage, leaving the A and B chains connected by disulfide bonds to form mature insulin.

Teacher's Note:
a) Proinsulin is a precursor molecule synthesized in pancreatic beta cells.
b) Removal of the C-peptide is essential for biological activity.

ii. Name the modern scientific technique used for the production of human insulin. [1 Mark]

Answer:
Recombinant DNA (rDNA) technology.

Teacher's Note:
a) Human insulin genes are inserted into bacterial plasmids (like E. coli) for mass production.
b) This avoids allergic reactions caused by animal-derived insulin.

iii. How are the two polypeptide chains of the fully functional insulin held together? [1 Mark]

Answer:
By disulfide bonds (disulfide bridges).

Teacher's Note:
a) Disulfide bonds provide structural stability to the active hormone.
b) Chain A and Chain B are linked together by these covalent sulfur-sulfur bonds.

Q13. [3 Marks]

13.1. (a) A patient was given an anti-retroviral drug by the doctor. Which disease was the patient diagnosed with? [0.5 Marks]

Answer:
AIDS (HIV infection).

Teacher's Note:
a) Antiretroviral drugs target retroviruses like HIV.
b) They help suppress viral replication and prolong patient survival.

13.1. (b) Mention any one symptom of HIV disease. [0.5 Marks]

Answer:
Swollen lymph glands, persistent fever, weight loss, or chronic diarrhea.

Teacher's Note:
a) Symptoms appear after a prolonged latency period.
b) Immune system collapse makes patients susceptible to opportunistic infections.

13.2. A patient was given an anti-retroviral drug by the doctor. Give the scientific name of the causative agent of HIV disease. [1 Mark]

Answer:
Human Immunodeficiency Virus (HIV).

Teacher's Note:
a) HIV is a retrovirus containing RNA as its genetic material.
b) It specifically attacks helper T-lymphocytes (CD4+ T cells).

13.3. A patient was given an anti-retroviral drug by the doctor. Which method was used to diagnose HIV disease? [0.5 Marks]

Answer:
ELISA (Enzyme-Linked Immunosorbent Assay).

Teacher's Note:
a) ELISA is the primary screening test for HIV.
b) Western blot is used as a confirmatory test.

13.4. A patient was given an anti-retroviral drug by the doctor. What is the role of Reverse Transcriptase and Integrase in the life cycle of a retrovirus? [0.5 Marks]

Answer:
Reverse transcriptase converts viral RNA into complementary DNA (cDNA), and integrase integrates this viral DNA into the host cell genome.

Teacher's Note:
a) Reverse transcription defies the central dogma by going from RNA to DNA.
b) Integration allows the virus to hijack host cell machinery permanently.

Q14. Draw a neat and well labelled diagram of T.S. of anther. [3 Marks]

Answer:
[Figure: Diagram of T.S. of an anther showing Epidermis, Endothecium, Middle layers, Tapetum, Sporogenous tissue, and Connective tissue]

Teacher's Note:
a) Label all four wall layers clearly from outside to inside.
b) Tapetum is the innermost nutritive layer nourishing developing microspores.

OR

14.2. Draw a neat and well labelled diagram of T.S. of the mammalian ovary. [3 Marks]

Answer:
[Figure: Diagram of T.S. of mammalian ovary showing Germinal epithelium, Primary follicle, Graafian follicle, Antral fluid, Ovum, Corpus luteum, and Corpus albicans]

Teacher's Note:
a) Clearly show follicles at various stages of development.
b) Include annotations for corpus luteum and secondary oocyte.

Q15. The table given below shows the Area, Y-intercept and regression coefficient of the continents namely, Africa and Europe. Study the table carefully and answer the questions that follow: [3 Marks]

i. Calculate the species richness (S) of each continent. [1 Mark]

Answer:
Africa: S = 327.01; Europe: S = 450.33.

Teacher's Note:
a) Use the species-area relationship formula: log S = log C + Z log A.
b) Substitute the values of C, Z, and A provided in the table and calculate antilog/exponential values.

ii. Which of these continents shows a higher biodiversity? [1 Mark]

Answer:
Europe (based on the calculated species richness values).

Teacher's Note:
a) Higher species richness indicates higher biodiversity.
b) Compare the final calculated S values for both regions.

iii. State any two factors that cause an increase in biodiversity. [1 Mark]

Answer:
1. Habitat diversity and stability.
2. Favorable climatic conditions (such as high solar energy and stable temperature in tropical regions).

Teacher's Note:
a) Tropical regions have high biodiversity due to undisturbed environments over long evolutionary time frames.
b) Increased niche specialization promotes species coexistence.

SECTION D - 15 MARKS

Q16. Meena had grown Rose and China-rose plants in her garden. She collected pollen grains from China-rose plants and sprinkled them on the stigma of the Rose flowers, as she wanted to grow a hybrid variety of Rose. [5 Marks]

a. Will this pollination give the desired results? Give a reason for your answer. [2 Marks]

Answer:
No. Reason: Rose and China-rose belong to different species/genera, and pollination is species-specific due to pollen-pistil interaction mechanisms.

Teacher's Note:
a) Successful pollination requires pollen recognition by the compatible stigma.
b) Incompatible pollen fails to germinate or grow pollen tubes.

b. What is geitonogamy? Why is it considered equivalent to cross-pollination in ecological context and self-pollination in genetic context? [3 Marks]

Answer:
Geitonogamy is the transfer of pollen grains from the anther to the stigma of another flower on the same plant. Ecological context: It requires a pollinating agent. Genetic context: It is genetically similar to self-pollination since pollen comes from the same plant.

Teacher's Note:
a) Geitonogamy occurs in monoecious plants.
b) Even though it involves pollinators (ecological cross-pollination), the gametes originate from the same genetically identical plant (genetic self-pollination).

OR

16.2. Fertilisation is the key process in sexually reproducing organisms and it acts as a vital link between two generations. Flowering plants adopt a unique pattern of sexual reproduction as compared to other organisms. [5 Marks]

a. Explain the process of fertilisation in angiosperms. [2 Marks]

Answer:
Angiosperms exhibit double fertilisation. One male gamete fuses with the egg cell to form a zygote (syngamy), and the other male gamete fuses with the two polar nuclei to form the primary endosperm nucleus (triple fusion).

[Figure: Diagram showing double fertilisation with synergids, polar nuclei, and egg cell]

Teacher's Note:
a) Double fertilisation is a unique characteristic feature of angiosperms.
b) It results in the formation of a diploid zygote and a triploid primary endosperm nucleus.

b. What is the precise location and function of filiform apparatus in the embryo sac of angiosperms? [2 Marks]

Answer:
Location: At the micropylar end of the synergids. Function: It guides the entry of the pollen tube into the synergid.

Teacher's Note:
a) The filiform apparatus consists of cellular thickenings.
b) It plays a critical chemotactic role during pollen tube guidance.

c. Fruits and seeds are generally formed due to fertilisation. Name the process involved in the production of the following without fertilisation: 1. Fruits 2. Seeds [1 Mark]

Answer:
1. Fruits: Parthenocarpy.
2. Seeds: Apomixis.

Teacher's Note:
a) Parthenocarpy produces seedless fruits naturally or artificially (e.g., using growth hormones).
b) Apomixis mimics sexual reproduction but produces seeds without fertilization.

Q17. The diagram given below shows the process of decomposition in the forest ecosystem. Observe the diagram carefully and answer the questions that follow: [5 Marks]

[Figure: Diagram of decomposition in a forest ecosystem showing falling leaves, earthworms, and centipedes]

i. Why is the breaking down of complex organic matter an important event in the ecosystem? [1 Mark]

Answer:
It releases trapped inorganic nutrients back into the soil, making them available to producers for nutrient recycling.

Teacher's Note:
a) Decomposition prevents nutrient depletion in ecosystems.
b) Mineralization converts organic matter into inorganic ions.

ii. The forest soil has a higher humus content than the desert soil. Give a reason to justify this statement. [1 Mark]

Answer:
Forests have abundant vegetation and moisture supporting rapid litter accumulation and slow partial decomposition, leading to humus formation, whereas deserts lack adequate organic matter and moisture.

Teacher's Note:
a) Humus is a dark-colored, amorphous substance resistant to microbial action.
b) It undergoes extremely slow decomposition, enriching soil fertility.

iii. Earthworms and centipedes play an important role in the decomposition process of forest ecosystems. At which stage of the decomposition are these organisms involved? [1 Mark]

Answer:
Fragmentation.

Teacher's Note:
a) Detritivores like earthworms break down detritus into smaller particles.
b) This increases the surface area for microbial action.

iv. The net annual primary productivity of a particular wetland ecosystem is found to be 8,000 kcal/m² per year. If respiration by the aquatic producers is 11,000 kcal/m² per year, calculate the gross primary productivity for this ecosystem. [2 Marks]

Answer:
GPP = 19,000 kcal/m²/year.

Teacher's Note:
a) Formula: NPP = GPP - R, therefore GPP = NPP + R.
b) Calculation: 8000 + 11000 = 19000 kcal/m²/year.

Q18. Griffith conducted a series of experiments on mice with two different strains of the bacterium Diplococcus pneumoniae. [5 Marks]

[Figure: Griffith's transformation experiment showing S-strain (virulent), R-strain (non-virulent), heat-killed S-strain, and mixture of heat-killed S and live R-strain causing mouse death]

i. Describe the entire procedure for this experiment. [2 Marks]

Answer:
Griffith injected mice with live R-strain (mice lived), live S-strain (mice died), heat-killed S-strain (mice lived), and a mixture of live R-strain and heat-killed S-strain (mice died).

Teacher's Note:
a) S-strain bacteria possess a mucous polysaccharide coat making them virulent.
b) Heat-killed S-strain alone is non-pathogenic.

ii. Write the conclusion of this experiment. [2 Marks]

Answer:
The live non-virulent R-strain bacteria were transformed into virulent S-strain bacteria by a 'transforming principle' derived from the heat-killed S-strain bacteria.

Teacher's Note:
a) This experiment laid the foundation for discovering DNA as the genetic material.
b) Avery, MacLeod, and McCarty later identified this transforming principle as DNA.

iii. What would have been the result of the experiment if both strains of bacteria were first heat-killed, mixed and then injected in the mice? [1 Mark]

Answer:
The mice would have survived.

Teacher's Note:
a) Heat-killing destroys the viability and pathogenicity of both bacterial strains.
b) Without living R-cells present to undergo transformation, no lethal infection can develop.