ISC Class 12 Biology Sample Paper 2023 with Solutions

SECTION A - 20 MARKS

Question 1
Answer the following questions briefly.

(i) Name the type of bioreactor which provides greater surface area for oxygen transfer. [1 Mark]

Answer:
Sparged stirred-tank bioreactor.

Teacher's Note:
a) A sparged stirred-tank bioreactor allows sterile air bubbles to be bubbled through the culture medium, significantly increasing the surface area for oxygen mass transfer.
b) Students often confuse this with simple stirred-tank bioreactors; remember that sparging introduces gas directly into the liquid.

(ii) Name the causative agent of genital warts. [1 Mark]

Answer:
Human Papillomavirus (HPV).

Teacher's Note:
a) Genital warts are a common sexually transmitted infection caused by specific strains of the Human Papillomavirus.
b) Ensure the full name is written or the standard abbreviation HPV is used; do not write generic terms like bacteria or fungus.

(iii) If a segment of double-stranded DNA has 18% thymine, calculate the percentage of cytosine in the DNA. [1 Mark]

Answer:
32%

Teacher's Note:
a) According to Chargaff's rules, the percentage of thymine equals adenine (\( \% T = \% A = 18\% \)), and cytosine equals guanine (\( \% C = \% G \)). Total purines and pyrimidines sum to \( 100\% \), so \( \% A + \% T + \% C + \% G = 100\% \), meaning \( 36\% + 2\text{C} = 100\% \), leading to \( \% C = 32\% \).
b) Always show the intermediate step using Chargaff's equivalence rules to ensure full credit.

(iv) A woman has normal vision, but her father is colourblind. If she marries a colourblind man, what is the probability of her son being colourblind? [1 Mark]

Answer:
50% (or 1/2)

Teacher's Note:
a) The woman's father is colourblind (\( X^cY \)), so she must be a carrier for colourblindness (\( XX^c \)). Her husband is colourblind (\( X^cY \)). Their sons inherit the X chromosome from the mother; thus, there is a 50% chance the son receives the affected X chromosome.
b) Pay close attention to whether the question asks for sons, daughters, or total offspring.

(v) What are Ramsar sites? [1 Mark]

Answer:
Ramsar sites are wetlands of international importance declared under the Ramsar Convention for the conservation and sustainable utilization of wetlands.

Teacher's Note:
a) The Ramsar Convention is an international treaty for the conservation and sustainable use of wetlands, signed in Ramsar, Iran in 1971.
b) Keywords like "international importance" and "wetlands" must be included in the definition.

(vi) Define standing state in an ecosystem. [1 Mark]

Answer:
The amount of nutrient elements, such as nitrogen, phosphorus, and calcium, present in the non-living (abiotic) component of an ecosystem at any given time is known as the standing state.

Teacher's Note:
a) Standing state refers to inorganic nutrients in the environment, which should not be confused with standing crop, which refers to the mass of living organisms.
b) Mentioning "abiotic component" or "reservoir pool" is essential for scoring full marks.

(vii) Name the toxin which is responsible for causing the symptoms of malaria. [1 Mark]

Answer:
Haemozoin.

Teacher's Note:
a) Haemozoin is a toxic byproduct formed from the digestion of hemoglobin by the malaria parasite (\( Plasmodium \)) inside red blood cells.
b) The release of haemozoin is responsible for the chill and high fever recurring every three to four days.

(viii) Name the bond which exists between chain-A and chain-B of human insulin. [1 Mark]

Answer:
Disulphide bond (or disulphide bridges).

Teacher's Note:
a) Mature human insulin consists of two short polypeptide chains - chain A and chain B, linked together by interchain disulphide bridges.
b) Specify "disulphide bond" clearly rather than general covalent bonds.

(ix) Which row is correct with respect to the features of Neutrophils and B-lymphocytes? [1 Mark]
Neutrophils B-lymphocytes
I can change shape. get activated by contact with antigens.
II are found in organs rather than in blood. kill virus-infected cells.
III may be long-lived cells. are always short-lived cells.
IV their lysosomal enzymes digest bacteria. secrete cytokines.
(a) I
(b) II
(c) III
(d) IV

Answer: (a) I

Neutrophils are phagocytic leukocytes that can change shape (ameboid movement) to engulf pathogens, while B-lymphocytes get activated upon direct contact with specific antigens.

Teacher's Note:
a) Neutrophils exhibit ameboid movement and diapedesis, allowing them to change shape and squeeze through capillary walls.
b) T-lymphocytes kill virus-infected cells (cytotoxic T cells), whereas B-lymphocytes differentiate into plasma cells to secrete antibodies.

(x) How many ova and sperms would be produced from 50 primary oocytes and 50 primary spermatocytes during gametogenesis? [1 Mark]
(a) 100 ova, 50 sperms
(b) 100 ova, 200 sperms
(c) 50 ova, 200 sperms
(d) 50 ova, 100 sperms

Answer: (c) 50 ova, 200 sperms

One primary oocyte undergoes meiosis to form 1 ovum (and 3 polar bodies), so 50 primary oocytes form 50 ova. One primary spermatocyte undergoes meiosis to form 4 functional sperms, so 50 primary spermatocytes form \( 50 \times 4 = 200 \) sperms.

Teacher's Note:
a) Remember that oogenesis produces only one functional ovum per meiotic division, whereas spermatogenesis yields four functional spermatozoa.
b) Read the question carefully to check whether primary or secondary oocytes/spermatocytes are specified.

(xi) Which one of the following is a palindromic sequence? [1 Mark]
(a) 5’-CGTATG-3’ / 3’-CGAATG-5’
(b) 5’-CGAATG-3’ / 3’-GCATAC-5’
(c) 5’-GAATTC-3’ / 3’-CTTAAG-5’
(d) 5’-CGAATG-3’ / 3’-CTTAAG-5’

Answer: (c) 5’-GAATTC-3’ / 3’-CTTAAG-5’

A palindromic DNA sequence is a sequence of base pairs that reads same on the two strands when reading direction is kept the same (5prime to 3prime). 5prime-GAATTC-3prime matches 3prime-CTTAAG-5prime when read backwards.

Teacher's Note:
a) EcoRI recognition site (GAATTC) is the classic example of a palindrome in molecular biology.
b) Always check the 5prime to 3prime orientation of both strands when verifying palindromic sequences.

(xii) Assertion: Energy value of biogas is lower than that of organic matter.
Reason: Biogas minimises the chances of spread of faecal pathogens. [1 Mark]
(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: (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

Biogas is a cleaner fuel with a lower calorific value compared to pure organic matter, and anaerobic sludge digestion destroys pathogens, but the reason does not explain why its energy value is lower.

Teacher's Note:
a) Biogas is primarily composed of methane and carbon dioxide, giving it a moderate energy density.
b) In assertion-reason questions, verify if the reason statement is factually correct on its own before testing causality.

(xiii) Give one significant contribution of each of the following scientists: [2 Marks]
(a) S. Cohen
(b) H. Boyer

Answer:
(a) S. Cohen: Along with Herbert Boyer, constructed the first recombinant DNA organism by combining an antibiotic resistance gene with a plasmid vector in 1972.
(b) H. Boyer: Isolated the restriction enzyme EcoRI and co-developed recombinant DNA technology with Stanley Cohen.

Teacher's Note:
a) Stanley Cohen and Herbert Boyer are pioneers of genetic engineering.
b) Keep contributions specific to rDNA technology and restriction enzymes.

(xiv) Give a term for the following: [2 Marks]
(a) The technique used to amplify a gene.
(b) The technique used for early diagnosis of HIV infection.

Answer:
(a) Polymerase Chain Reaction (PCR).
(b) Polymerase Chain Reaction (PCR) / ELISA.

Teacher's Note:
a) PCR is used for both gene amplification and early detection of pathogens like HIV when viral titers are low.
b) ELISA is also a standard diagnostic test for HIV, but PCR is specifically noted for early molecular diagnosis.

(xv) Expand the following abbreviations: [2 Marks]
(a) ICSI
(b) IUCD

Answer:
(a) ICSI - Intracytoplasmic Sperm Injection.
(b) IUCD - Intrauterine Contraceptive Device.

Teacher's Note:
a) Spellings in expansions must be precise; marks are deducted for spelling errors in medical terms.
b) ICSI is an assisted reproductive technology (ART) used for severe male infertility.

(xvi) Give a reason for each of the following: [2 Marks]
(a) A person with cuts and bruises following an accident is administered tetanus anti-toxin.
(b) Origin of life is not possible under the present atmospheric conditions.

Answer:
(a) Tetanus anti-toxin provides preformed antibodies that confer passive immunity for immediate neutralization of tetanus toxins.
(b) The present oxidizing atmosphere (presence of free oxygen) destroys complex organic molecules, unlike the primitive reducing atmosphere that favoured abiogenesis.

Teacher's Note:
a) Anti-toxin is an example of passive immunization used for quick action against fast-acting toxins.
b) Oxygen is reactive and oxidizes organic compounds, preventing spontaneous synthesis of life today.

SECTION B - 14 MARKS

Question 2 [2 Marks]
(i) How does the Reproductive and Child Health Care Programme run by the government benefit the society?

Answer:
1. Creates awareness among people about reproduction-related aspects and sexually transmitted infections (STIs).
2. Reduces maternal and infant mortality rates through proper pre-natal and post-natal care.

Teacher's Note:
a) RCH programs focus on reproductive health and family welfare.
b) Mentioning healthcare awareness and mortality reduction secures full marks.

OR

(ii) Write any four causes of infertility in males. [2 Marks]

Answer:
1. Oligospermia (low sperm count) or azoospermia (absence of sperms).
2. Poor sperm motility (asthenozoospermia).
3. Hormonal disorders (such as low testosterone levels).
4. Blockage of the vas deferens or erectile dysfunction.

Teacher's Note:
a) Male infertility factors can be pre-testicular, testicular, or post-testicular.
b) Listing any four clear physiological or anatomical causes is sufficient.

Question 3 [2 Marks]
What is meant by lactational amenorrhoea? Discuss the physiological mechanism which makes lactational amenorrhoea a natural contraceptive method.

Answer:
Lactational amenorrhoea refers to the temporary absence of menstruation during intense lactation following childbirth. High levels of prolactin secreted during breastfeeding inhibit gonadotropin-releasing hormone (GnRH) release from the hypothalamus, thereby preventing ovulation.

Teacher's Note:
a) This natural method is effective only up to six months following parturition.
b) Mentioning the role of prolactin in suppressing GnRH/ovulation is vital.

Question 4 [2 Marks]
Study the pedigree chart given below and answer the questions that follow.

[Figure: Pedigree chart showing unaffected parents having affected offspring, skipping generations, with both male and female affected individuals.]

(i) Is the trait recessive or dominant? Give a reason for your answer. [1 Mark]
(ii) Is the trait sex-linked or autosomal? Give a reason for your answer. [1 Mark]

Answer:
(i) Recessive, because unaffected parents can produce affected children (carriers transmit the trait).
(ii) Autosomal, because both male and female offspring are affected equally and unaffected parents transmit it to both sexes.

Teacher's Note:
a) Skips in generations indicate a recessive inheritance pattern.
b) Equal distribution among male and female children rules out X-linked dominant or recessive inheritance.

Question 5 [2 Marks]
The NPP of a terrestrial ecosystem is 1500 Kg per meter square per year and the respiratory loss of the ecosystem is 1200 Kg per meter square per year. Calculate the GPP of the given ecosystem.

Answer:
Formula: Gross Primary Productivity (GPP) = Net Primary Productivity (NPP) + Respiration loss (R)
GPP = 1500 Kg / m² / year + 1200 Kg / m² / year = 2700 Kg / m² / year.

Teacher's Note:
a) Gross primary production is the total rate of organic matter synthesis, part of which is respired away.
b) Always state the formula and include appropriate units in the final answer.

Question 6 [2 Marks]
Give any two differences between normal body cells and cancer cells.

Answer:

Teacher's Note:
a) Contact inhibition and metastasis are key defining characteristics of cancer cells.
b) Presenting the differences in a tabular format ensures clarity and full credit.

Question 7 [2 Marks]
(i) Name the first human-like hominid ancestor. What was its cranial capacity? [1 Mark]
(ii) Name the hominid ancestor that existed about 1.5 mya. What was its cranial capacity? [1 Mark]

Answer:
(i) Homo habilis; cranial capacity was about 650 - 800 cc.
(ii) Homo erectus; cranial capacity was about 900 cc.

Teacher's Note:
a) Homo habilis is considered the first tool-maker (handman).
b) Homo erectus lived around 1.5 million years ago and probably ate meat.

Question 8 [2 Marks]
A person in good health visited a garden where flowers were in full bloom. While returning from the garden he suddenly started sneezing and wheezing.
(i) Name and define the response of the person’s immune system in the above-mentioned case. [1 Mark]
(ii) Name the cell of the immune system and the type of antibody involved in this kind of response. [1 Mark]

Answer:
(i) Allergy (Hypersensitivity) - An exaggerated immune response to certain antigens present in the environment.
(ii) Mast cells and IgE antibodies.

Teacher's Note:
a) Allergies involve the release of histamines and seretonin from mast cells due to allergen binding on IgE.
b) Correct identification of both mast cells and IgE is mandatory for the second part.

SECTION C - 21 MARKS

Question 9 [3 Marks]
Draw a neatly labelled diagram of a microspore.

Answer:
[Figure: Diagram of a microspore or pollen grain showing vacuole, nucleus, exine with germ pore, and intine.]
Labels: Exine, Intine, Germ pore, Nucleus, Vacuole.

Teacher's Note:
a) Diagrams must be neat, cleanly drawn with a sharp pencil, and correctly labelled.
b) Exine (with germ pore) and intine layers should be clearly differentiated.

Question 10 [3 Marks]
(i)

[Figure: Diagram of a stirred-tank bioreactor showing motor at top, flat bladed impeller, nutrient broth, and sterile air bubbler at base with label 'A' pointing to the agitator/impeller.]

(a) Identify the diagram given above. [½ Mark]
(b) What is the role of the part labelled ‘A’? [½ Mark]
(c) Redraw the diagram and label any three parts. [2 Marks]

Answer:
(a) Simple stirred-tank bioreactor.
(b) The agitator (impeller) mixes the contents and facilitates even mixing and oxygen availability throughout the bioreactor.
(c) [Figure: Redrawn outline of stirred-tank bioreactor with labels Motor, Agitator, Culture broth, Foam breaker.]

Teacher's Note:
a) Stirred-tank bioreactors are widely used for large-scale production of recombinant proteins and enzymes.
b) Ensure all three labels are clearly pointed to distinct parts in the sketch.

OR

(ii) A bacterial culture was grown on a specific culture medium containing a chromogenic substrate. After sometime, it was observed that some colonies developed a blue-coloured appearance while some remained colourless. Briefly explain the phenomenon responsible for this observation. [3 Marks]

Answer:
1. This phenomenon is known as insertional inactivation used in recombinant selection.
2. The gene for alpha-galactosidase / beta-galactosidase is inactivated due to insertion of a recombinant DNA fragment.
3. Non-recombinants produce active enzyme which converts the chromogenic substrate into blue colonies, whereas recombinants remain colourless.

Teacher's Note:
a) Insertional inactivation is a reliable method to distinguish recombinant colonies from non-recombinant ones.
b) Blue colonies indicate non-transformants or plasmids without inserts, while white/colourless colonies indicate successful insertion.

Question 11 [3 Marks]
What is meant by ‘biocontrol’? Explain how Trichoderma and Baculovirus act as biocontrol agents.

Answer:
Biocontrol refers to the use of biological methods for controlling plant diseases and pests. Trichoderma species are free-living fungi effective against several plant pathogens as they produce antifungal metabolites and parasitize pathogens. Baculoviruses (primarily Nucleopolyhedrovirus) are pathogens that attack insects and other arthropods, serving as species-specific narrow-spectrum bio-pesticides.

Teacher's Note:
a) Biocontrol reduces reliance on toxic chemical pesticides and preserves beneficial non-target organisms.
b) Mentioning the specific roles of both Trichoderma and Baculovirus is required for full marks.

Question 12 [3 Marks]
What are linked genes? Give a schematic representation of a test cross between a white-eyed female Drosophila and a red-eyed male Drosophila.

Answer:
Linked genes are genes located close together on the same chromosome that tend to be inherited together during meiosis due to minimal crossing over.
Schematic Cross:
Parents: White-eyed female (\( X^w X^w \)) × Red-eyed male (\( X^W Y \))
F1 generation: Red-eyed females (\( X^W X^w \)) and White-eyed males (\( X^w Y \)).
Test cross: F1 red-eyed female (\( X^W X^w \)) × White-eyed male (\( X^w Y \))
Offspring: 1 Red-eyed female (\( X^W X^w \)), 1 White-eyed female (\( X^w X^w \)), 1 Red-eyed male (\( X^W Y \)), 1 White-eyed male (\( X^w Y \)).

Teacher's Note:
a) Linkage violates Mendel's law of independent assortment when genes are closely situated on chromosomes.
b) Use standard superscript notations for eye color alleles in Drosophila (\( w \) for white, \( W \) or \( + \) for red).

Question 13 [3 Marks]
Alcoholic drinks are produced by fermentation, but some beverages are produced through an additional process of distillation. How do the distilled and undistilled alcoholic beverages differ in their quality and composition? Explain by giving one example each for a distilled alcoholic beverage and an undistilled alcoholic beverage.

Answer:
Distilled beverages have a higher alcohol concentration and are purified through distillation, whereas undistilled beverages have lower alcohol content. Examples: Whisky, Brandy, or Rum are distilled; Wine and Beer are undistilled.

Teacher's Note:
a) Distillation increases the percentage of alcohol by volume in the fermented wash.
b) Provide one clear example for each category to ensure full marks.

Question 14 [3 Marks]
Consider the situation, where a variety of birds depend upon a big tree for their survival. The birds in turn are hosts for the different parasites surviving on them. Draw a pyramid of number to represent the above situation.

Answer:
[Figure: Inverted pyramid of numbers: Tree (1) at base, Birds (many) in middle, Parasites (very large number) at top.]
Description: The pyramid is inverted because a single tree supports many birds, which in turn support numerous parasites.

Teacher's Note:
a) A pyramid of numbers can be inverted in parasitic or tree ecosystems.
b) Clearly label trophic levels: Tree (producer) -> Birds (herbivore/primary consumer) -> Parasites (secondary consumer/parasite).

Question 15 [3 Marks]
Explain any three ex situ methods of conservation of biodiversity.

Answer:
1. Botanical Gardens and Zoological Parks: Maintaining live specimens of rare and threatened plant and animal species outside their natural habitats.
2. Cryopreservation: Storing gametes, embryos, or tissues at ultra-low temperatures (-196 °C in liquid nitrogen) for long-term conservation.
3. Seed Banks / Gene Banks: Preserving viable seeds under low temperature and moisture conditions for genetic conservation.

Teacher's Note:
a) Ex situ conservation involves protecting endangered species outside their natural habitats.
b) Mentioning cryopreservation or seed banks along with botanical gardens covers all standard methods.

SECTION D - 15 MARKS

Question 16
(i)

[Figure: Diagram of a typical angiosperm embryo sac showing 1 (antipodals), 2 (central cell polar nuclei), 3 (egg cell), 4 (synergids with filiform apparatus).]

Observe the given diagram of a typical embryo sac in angiosperm and answer the following questions.

(a) Identify the parts labelled 1, 2, 3 and 4. [2 Marks]
(b) Define Syngamy and Triple fusion. [1 Mark]
(c) How many nuclei and cells constitute an embryo sac? [1 Mark]
(d) Give one point on the significance of double fertilization. [1 Mark]

Answer:
(a) 1 - Antipodal cells, 2 - Polar nuclei, 3 - Egg cell, 4 - Synergids.
(b) Syngamy is the fusion of a male gamete with the egg cell to form a zygote. Triple fusion is the fusion of the second male gamete with the diploid secondary nucleus to form the primary endosperm nucleus (PEN).
(c) 8 nuclei and 7 cells.
(d) It ensures that endosperm formation starts only after fertilization, preventing wastage of energy.

Teacher's Note:
a) Double fertilization is a unique characteristic feature of angiosperms.
b) Clearly distinguish between the cellular and nuclear counts of a mature embryo sac.

OR

(ii) A couple was expecting their child and visited a doctor for routine check-up. They came to know that the foetus was suffering from an incurable disorder. The doctor advised them to go for MTP.

(a) What is the full form of MTP? [1 Mark]
(b) In what way has the technique of MTP been misused? [1 Mark]
(c) Which diagnostic technique helped the doctor to detect the disorder in the embryo? [1 Mark]
(d) Give one similarity and one difference between Cu7 and LNG-20. [2 Marks]

Answer:
(a) Medical Termination of Pregnancy.
(b) Misused for illegal female foeticide based on sex determination.
(c) Amniocentesis (or ultrasonography).
(d) Similarity: Both are intrauterine devices (IUDs) used for contraception. Difference: Cu7 is a copper-releasing IUD that suppresses sperm motility, whereas LNG-20 is a hormone-releasing IUD that makes the uterus unsuitable for implantation and cervix hostile to sperms.

Teacher's Note:
a) Amniocentesis checks chromosomal abnormalities but has been misused for female foeticide.
b) Copper IUDs release copper ions, whereas hormonal IUDs release progestogens.

Question 17 [5 Marks]
(i) Explain the ‘Species-Area Relationship’ with the help of a graph. Give its mathematical expression also.
(ii) Explain ‘Rivet Popper Hypothesis’.

Answer:
(i) Alexander von Humboldt observed that within a region, species richness increases with increasing explored area, but only up to a limit. The rectangular hyperbola is represented by the equation \( S = CA^Z \) or on a logarithmic scale as \( \log S = \log C + Z \log A \).
[Figure: Graph showing species richness (S) on y-axis and area (A) on x-axis, showing a rectangular hyperbola curve and a straight line on logarithmic scale.]
(ii) Paul Ehrlich's Rivet Popper Hypothesis likens an ecosystem to an airplane where species are rivets. Removing rivets (extinction of species) one by one initially may not affect flight safety, but removing key species (keystone species) can cause catastrophic ecosystem collapse.

Teacher's Note:
a) The slope of the regression line (\( Z \)) generally lies between 0.1 and 0.2, but becomes steeper (0.6 to 1.2) for very large areas like entire continents.
b) The rivet popper analogy emphasizes the importance of every species in maintaining ecosystem stability.

Question 18 [5 Marks]
Describe the process of transcription in prokaryotes.

Answer:
Transcription in prokaryotes involves three major steps catalyzed by DNA-dependent RNA polymerase:
1. Initiation: RNA polymerase binds to the promoter site on DNA with the help of sigma factor (\( \sigma \) factor), initiating transcription.
2. Elongation: The enzyme uses nucleotide triphosphates as substrates and polymerizes them in a template-dependent fashion following complementary base pairing rules, opening the DNA helix.
3. Termination: When the polymerase reaches the terminator region, the rho factor (\( \rho \) factor) associates with the enzyme, causing nascent RNA and RNA polymerase to fall off, terminating transcription.

Teacher's Note:
a) In bacteria, mRNA does not require extensive processing and can be translated even while transcription is ongoing (coupled transcription-translation).
b) Mentioning the roles of sigma factor and rho factor is essential for full marks.