ISC Class 12 Biology Sample Paper 2022 with Solutions

ISC SEMESTER 2 EXAMINATION
SPECIMEN QUESTION PAPER
BIOLOGY PAPER 1 (THEORY)

Maximum Marks: 35
Time allowed: One and a half hour

SECTION A - 7 MARKS

Question 1

(i) Give the biological name of the causative agent of gonorrhoea. [1]

Answer: Neisseria gonorrhoea

Teacher's Note:
a) Always write biological binomial names with proper capitalization and underlining when handwritten, or in italics in print.
b) Students often misspell the genus name as Nesseria instead of Neisseria.

(ii) Sonora - 64 is a variety of: [1]
(A) Wheat
(B) Rice
(C) Maize
(D) Sugarcane

Answer: (A) Wheat

Teacher's Note:
a) Semi-dwarf wheat varieties like Sonora-64 and Lerma Rojo-64 were introduced during the Green Revolution in India.
b) Do not confuse wheat semi-dwarf varieties with high-yielding rice varieties like IR-8 and Jaya.

(iii) Assertion: PAN is a secondary pollutant.
Reason: The secondary pollutants are produced by the combination of primary emitted pollutants in the atmosphere. [1]
(A) Both assertion and reason are true, and reason is the correct explanation of assertion.
(B) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(C) Assertion is true but reason is false.
(D) Both assertion and reason are false.

Answer: (A) Both assertion and reason are true, and reason is the correct explanation of assertion.

Teacher's Note:
a) Peroxyacetyl nitrate (PAN) is formed photochemically in the atmosphere from nitrogen oxides and hydrocarbons.
b) Remember that primary pollutants are emitted directly from sources, whereas secondary pollutants form via chemical reactions in the atmosphere.

(iv) Give one significant contribution of Prof. R. Mishra. [1]

Answer: Prof. R. Misra formulated the first postgraduate course in ecology in India and is recognized as the father of ecology in India.

Teacher's Note:
a) Prof. Ramdeo Misra established the Department of Botany at Banaras Hindu University as a center of excellence in ecology.
b) Be precise with the scientist's name and their specific pioneering role in Indian environmental biology.

(v) Expand the term ADA. [1]

Answer: Adenosine deaminase

Teacher's Note:
a) ADA deficiency can cause severe combined immunodeficiency (SCID), which was the first disease treated using gene therapy.
b) Spelling of biochemical terms like deaminase must be accurate to avoid losing credit.

(vi) A substance that stimulates the production of antibodies is called __________. [1]

Answer: Antigen

Teacher's Note:
a) An antigen is any foreign molecule that triggers an immune response and the production of specific antibodies by B-lymphocytes.
b) Do not confuse antigen with antibody or allergen.

(vii) Why does a large ozone hole develop specifically over Antarctica? [1]

Answer: The unique atmospheric and chemical conditions of the Antarctic region, including extremely low temperatures that form polar stratospheric clouds and isolation of air during the polar vortex, facilitate rapid ozone destruction by reactive halogen gases.

Teacher's Note:
a) Polar stratospheric clouds convert inactive chlorine reservoirs into reactive chlorine atoms upon the return of sunlight in spring.
b) Mention polar stratospheric clouds and low winter temperatures to secure full marks.

SECTION B (16 MARKS)

Question 2
Briefly explain the role of the following in providing defense against infections in human body:

(i) B-cells [1]

Answer: B-cells get activated by binding to antigens, undergo proliferation, and differentiate into plasma cells that secrete a large army of specialized proteins called antibodies into the blood to fight pathogens.

Teacher's Note:
a) B-cells mediate humoral or antibody-mediated immune response.
b) Clearly state that B-cells produce antibodies, differentiating them from T-cells.

(ii) T-cells [1]

Answer: T-cells mature in the thymus, circulate in the body, and are responsible for cell-mediated immunity by recognizing antigens presented on antigen-presenting cells and helping B-cells produce antibodies or destroying infected cells directly.

Teacher's Note:
a) T-cells do not secrete antibodies themselves but regulate the overall immune response.
b) Mention cell-mediated immunity and the role of helper/cytotoxic T-cells for completeness.

Question 3. What is the role of a probe in recombinant DNA technology? [2]

Answer:
1. A probe is a single-stranded DNA or RNA tagged with a radioactive molecule.
2. It is used to detect and locate specific complementary gene sequences in a sample genome through hybridization.

Teacher's Note:
a) Probes are extensively used in screening recombinant clones and genetic fingerprinting.
b) Always mention that the probe is single-stranded and hybridizes with its complementary sequence.

Question 4. The level of air pollution is very high in a city. Suggest any four measures which should be taken by the government to control it. [2]

Answer:
1. Shifting public transport vehicles like buses and auto-rickshaws to clean fuels like CNG.
2. Implementing schemes like odd-even vehicle rotation during high pollution periods.
3. Mandating and strictly enforcing electrostatic precipitators in industrial plants.
4. Relocating highly polluting industries to designated industrial zones outside city limits.

Teacher's Note:
a) Give practical, actionable governance measures such as vehicular emission control and industrial regulation.
b) Bullet points should be concise and direct since the question carries 2 marks.

Question 5

(i) If 20000 K Cal energy is available at the level of producers, what will be the amount of energy available at the level of secondary consumers. [2]

Answer:
1. According to the 10 percent law of energy transfer, only 10 percent of energy is transferred from one trophic level to the next.
2. Energy at producer level = \( 20000 \) K Cal.
Energy at primary consumer level = \( 20000 \times 10\% = 2000 \) K Cal.
Energy at secondary consumer level = \( 2000 \times 10\% = 200 \) K Cal.

Teacher's Note:
a) Show step-by-step calculation across trophic levels from producer to primary consumer, then to secondary consumer.
b) Always include the correct unit (K Cal) in the final answer.

OR

(ii) The number of mice in a laboratory was 100 on a particular day. After one year their number increased to 120. Calculate the growth rate in the population. [2]

Answer:
1. Formula: Growth rate = \(\frac{\text{Current population} - \text{Previous population}}{\text{Previous population}} \times 100\)
2. Calculation: \(\frac{120 - 100}{100} \times 100 = \frac{20}{100} \times 100 = 20\%\)

Teacher's Note:
a) State the percentage growth rate clearly with proper units.
b) Ensure the formula is explicitly written before substituting the values.

Question 6. Give any four advantages of mutation breeding. [2]

Answer:
1. It helps in eliminating undesirable characters from a crop variety.
2. It improves desirable morphological and agronomic characters of crops.
3. It induces resistance against specific diseases and pathogens.
4. It enhances specific traits in well-adapted high-yielding crop varieties.

Teacher's Note:
a) Mutation breeding uses physical or chemical mutagens to create genetic variability.
b) Keep points brief since 2 marks require concise statements.

Question 7. Name and describe the technique that helps in separation of DNA fragments. [2]

Answer:
1. Name of technique: Gel electrophoresis.
2. Description: Negatively charged DNA fragments are forced to move through an agarose gel matrix under an electric field, separating them based on their size where smaller fragments move faster and farther than larger ones.

Teacher's Note:
a) Mention agarose gel matrix, electrical field, and size-based separation.
b) Visualization is typically done using ethidium bromide under UV light.

Question 8. Which molecules are called molecular scissors? Why are they called so? [2]

Answer:
1. Restriction enzymes (restriction endonucleases) are called molecular scissors.
2. They are called so because they inspect DNA sequences and make specific cuts or incisions at specific recognition sites on DNA strands.

Teacher's Note:
a) Specify restriction endonucleases as the exact biological term.
b) Mention that they recognize specific palindrome nucleotide sequences.

Question 9. Differentiate between primary succession and secondary succession with appropriate examples. [2]

Answer:

Teacher's Note:
a) Use a tabular format for clear differentiation as expected in board marking schemes.
b) Include accurate examples for both types of ecological succession.

SECTION C (12 MARKS)

Question 10
Study the graph given below and answer the following questions:
[Figure: Species richness versus Area graph on a log-log scale showing a straight line with equation log S = log C + Z log A, where line 'a' is a steep curve and line 'b' is a less steep straight line]

(i) Explain the ecological principle represented by the graph and write a mathematical expression for the same. [2]

Answer:
1. Principle: Within a region, species richness increases with increasing explored area, but only up to a limit (Alexander Von Humboldt).
2. Mathematical expression (on logarithmic scale): \(\log S = \log C + Z \log A\) (where \(S\) = species richness, \(A\) = area, \(Z\) = regression coefficient/slope, \(C\) = Y-intercept).

Teacher's Note:
a) Clearly define each term in the species-area relationship equation.
b) Mention Alexander Von Humboldt as the naturalist who discovered this relationship.

(ii) What will happen if the slope of the line 'b' becomes steeper? [1]

Answer: If the slope (Z value) becomes steeper, the species-richness will increase much faster with an increase in area, typically observed when large areas like entire continents are analyzed.

Teacher's Note:
a) Steeper slopes indicate a higher Z value, which ranges between \(0.6\) and \(1.2\) for very large geographic areas.
b) Ensure the concept of regression coefficient is correctly linked to area scale.

Question 11. Explain the steps involved in the process of plant breeding. [3]

Answer:
1. Collection of variability: Collecting and preserving all wild varieties, species, and relatives of the crop having desired alleles (germplasm collection).
2. Evaluation and selection of parents: Evaluating the collected germplasm to identify plants with desirable traits and selecting pure lines for hybridization.
3. Cross hybridization among selected parents: Crossing the chosen parents to combine desired characters into a single progeny.
4. Selection and testing of superior recombinants: Selecting plants with superior desired character combinations and self-pollinating them until uniformity is achieved.
5. Testing, release, and commercialization of new cultivars: Evaluating the newly selected lines in research fields for at least three growing seasons before commercial release.

Teacher's Note:
a) List the classical plant breeding steps sequentially.
b) Mentioning germplasm collection, hybridization, and field testing is mandatory for full marks.

Question 12. The rDNA technology has provided a method to control the nematode parasite Meloidogyne incognita. Explain the principle involved in this technique. [3]

Answer:
1. RNA interference (RNAi) is used, which involves silencing a specific mRNA of the nematode due to complementary single-stranded RNA binding.
2. Using Agrobacterium vectors, nematode-specific genes were introduced into the host tobacco plant host to produce both sense and anti-sense RNA in plant cells.
3. These two RNA strands form double-stranded RNA (dsRNA) that triggers RNAi, silencing the specific mRNA of the parasite, thereby protecting the transgenic plant from infection.

Teacher's Note:
a) Emphasize RNA interference (RNAi) and gene silencing mechanisms.
b) Mention sense and anti-sense RNA formation and double-stranded RNA production clearly.

Question 13

(i) Consider the amount of organic matter at the different trophic levels in an ecosystem given below:

On the basis of the data provided above, construct an ecological pyramid. Comment upon its nature giving at least one example of the type of organism occupying each of the above mentioned trophic levels. [3]

Answer:
1. Nature of pyramid: It is an inverted pyramid of biomass (commonly found in aquatic ecosystems where biomass of producers is smaller than consumers).
2. Trophic level examples:
- Tertiary consumers: Orcas / Apex predators
- Secondary consumers: Fishes / Carnivores
- Primary consumers: Zooplanktons / Copepods
- Primary producers: Phytoplanktons
[Figure: Inverted ecological pyramid with Phytoplankton at the base, followed upward by primary consumers, secondary consumers, and tertiary consumers at the top]

Teacher's Note:
a) Analyze the given organic matter values carefully to determine that the biomass increases from producers to top consumers, making the pyramid inverted.
b) Give correct marine ecosystem examples for each trophic level.

OR

(ii) On the basis of the demographic data of a country given below, construct an age pyramid and explain whether the population is stable, declining or growing. [3]

Answer:
1. Nature of population: The population is growing (expanding population).
2. Reason: The number of pre-reproductive individuals (\(20,000\)) is significantly larger than the reproductive (\(15,000\)) and post-reproductive (\(10,000\)) individuals, forming a broad-based triangular pyramid.
[Figure: Upright triangular age pyramid with wide base for pre-reproductive group, narrower middle for reproductive group, and narrowest top for post-reproductive group]

Teacher's Note:
a) Explain that a broad base indicates a high proportion of young individuals, pointing towards rapid population growth.
b) State the demographic status clearly as 'growing' or 'expanding'.