CBSE Class 12 – Biology Question Paper 2023

SECTION A

Questions no. 1 to 16 are Multiple Choice (MCQ) type Questions, carrying 1 mark each.

1. Which one of the following processes results in the production of recombinants in future generations?
(i) Mutation
(ii) Independent assortment during meiosis I
(iii) Independent assortment during meiosis II
(iv) Crossing over of bivalents

(a) (iv) only
(b) (ii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (ii), (iii) and (iv)

Answer: (a) (iv) only

2.Identify the region ‘X’, the factor ‘Y’ and the enzyme ‘Z’ involved in the process of transcription in prokaryote as shown in the schematic representation given below.

Region Factor Y Enzyme Z
(a) Terminator Sigma ( ) RNA polymerase
(b) Promoter Rho ( ) RNA polymerase
(c) Promoter Sigma ( ) RNA polymerase
(d) Promoter Sigma ( ) DNA polymerase

Answer :

Region X: Represents the Promoter on the DNA, where transcription initiation begins.
Factor Y: Is the Sigma (σ) factor, a protein that helps RNA polymerase recognize and bind to the promoter.
Enzyme Z: Is RNA polymerase, the enzyme responsible for synthesizing RNA using the DNA template.

3. The status of the human population reflected in the human age pyramid
given below is

(a) Declining population (b) Stable population
(c) Expanding population (d) Extinct populatio

Answer

(a) Declining population.

Shape: The pyramid is narrower at the base (pre-reproductive) and wider at the top (post-reproductive). This indicates that there are fewer younger individuals compared to older individuals.
Interpretation: This shape suggests a declining population. In a declining population, there are fewer births than deaths, leading to a smaller and older population over time.

4. The graph plotted below is based on the data collected by biology
students with respect to the levels of oxygen at the specific points in the
river flowing outside their city. Which point in the graph indicates the
entry of untreated sewage in the river ?

(a) Point (i) (b) Point (ii)
(c) Point (iii) (d) Point (iv

Answer

(b) Point (ii).

Sewage Impact: When untreated sewage enters a river, it brings with it a high amount of organic matter (like food waste).
Oxygen Consumption: Bacteria in the river start decomposing this organic matter. This process uses up a lot of dissolved oxygen in the water.
Mineral Increase: The decomposition also releases minerals and nutrients into the water, causing an increase in mineral ion levels.

Looking at the graph:

Point (ii): Shows a sharp drop in dissolved oxygen levels and a corresponding increase in mineral ion levels. This pattern is characteristic of sewage pollution entering the river.

5. Given below are two columns. In Column I is the list of four enzymes and in Column II is the list of functions of the given enzymes. Which one of the following options shows the enzymes matched with their respective functions correctly?

Column I (Enzyme) | Column II (Function)
P. DNA Ligase | i. Removes nucleotides from ends of DNA
Q. Restriction exonuclease | ii. Extends primer on a DNA template
R. Taq polymerase | iii. Joins the DNA fragments
S. Restriction endonuclease | iv. Cuts DNA at a specific position

(a) P-i, Q-ii, R-iv, S-iii
(b) P-iv, Q-iii, R-ii, S-i
(c) P-i, Q-iv, R-iii, S-ii
(d) P-iii, Q-i, R-ii, S-iv

Answer: (d) P-iii, Q-i, R-ii, S-iv

6. Select the option that gives the correct description of the process of
Natural Selection with respect to the length of the neck of giraffe.

Answer

Directional Selection: This type of natural selection favors one extreme phenotype over the other. In this case, giraffes with longer necks would have an advantage in reaching food sources higher up in trees.
Graph Interpretation: The graph shows a shift in the population towards longer necks. This indicates that giraffes with longer necks are more likely to survive and reproduce, passing their genes for longer necks to the next generation.

8. Choose the option that gives the correct number of pollen grains that will be formed after 325 microspore mother cells undergo microsporogenesis.

(a) 325
(b) 650
(c) 1300
(d) 975

Answer: (c) 1300

Explanation:
During microsporogenesis, each microspore mother cell undergoes meiosis to produce 4 pollen grains. So, if 325 microspore mother cells undergo meiosis, the total number of pollen grains produced will be:
325 × 4 = 1300

9. Given below are two columns. In Column I the names of four contraceptive devices are given and in Column II the modes of action of the contraceptives are given. Select the option where the contraceptive devices are correctly matched with their respective modes of action.

Column I (Contraceptive devices) | Column II (Modes of action)
P. Lippes loop | i. Inhibition of ovulation
Q. Multiload 375 | ii. Phagocytosis of sperms in uterus
R. Subcutaneous Norplant | iii. Causes thickening of cervical mucous
S. Saheli | iv. Makes cervix hostile to sperms

(a) P-ii, Q-iv, R-iii, S-i
(b) P-i, Q-ii, R-iii, S-iv
(c) P-iii, Q-i, R-iv, S-ii
(d) P-iv, Q-iii, R-ii, S-i

Answer: (d) P-iv, Q-iii, R-ii, S-i

10. In which one of the following options does the endocrine gland correctly match with its hormonal secretion and its function?

Endocrine Gland	Hormone	Function
(a) Sertoli cells	Testosterone	Development of secondary sexual characteristics
(b) Placenta	Estrogen	Initiates secretion of milk
(c) Leydig cells	Androgen	Initiates the production of sperms
(d) Ovary	FSH	Stimulates follicular development

Answer: (c) Leydig cells – Androgen – Initiates the production of sperms

11. The organism used in the construction of the first artificial recombinant DNA by Cohen and Boyer in 1972 was:

(a) E. coli
(b) Salmonella typhimurium
(c) Agrobacterium tumefaciens
(d) Bacillus thuringiensis

Answer: (a) E. coli

Explanation:
Cohen and Boyer used E. coli as a host organism to construct the first artificial recombinant DNA in 1972. They inserted a foreign gene into the E. coli, which successfully expressed the gene, marking the birth of recombinant DNA technology.

12. Who among the following challenged the patent right granted to the University of Mississippi Medical Centre for ‘use of turmeric in wound healing’?

(a) Mr. Ajay Phadke
(b) Ms. Vandana Shiva
(c) Dr. Venugopalan
(d) Dr. R.A. Mashelkar

Answer: (b) Ms. Vandana Shiva

Explanation:
Ms. Vandana Shiva, an environmental activist and scholar, challenged the patent rights granted to the University of Mississippi Medical Centre for the use of turmeric in wound healing, arguing that turmeric’s medicinal properties were a part of traditional knowledge and should not be patented.

For Questions 13 to 16, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c), and (d) as given below.

13. Assertion (A): A patient of ADA deficiency undergoing treatment for gene therapy requires periodic infusion of genetically engineered lymphocytes.
Reason (R): Lymphocytes are immortal.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Answer: (c) Assertion (A) is true, but Reason (R) is false.

Explanation:

Assertion (A) is true because patients with ADA (Adenosine Deaminase) deficiency require periodic infusion of genetically engineered lymphocytes as part of their gene therapy.
Reason (R) is false because lymphocytes are not immortal. While they can be engineered in the lab to have a longer lifespan or be modified for gene therapy purposes, lymphocytes are not naturally immortal.

14. Assertion (A): A cattle egret and grazing cattle in close association is a classic example of commensalism.
Reason (R): As grazing cattle move through the field, they stir up and flush out insects from the vegetation that otherwise might be difficult for egrets to find and catch.

Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

Assertion (A) is true because the relationship between cattle egrets and grazing cattle is a classic example of commensalism. In commensalism, one species benefits, while the other is neither helped nor harmed.
Reason (R) is also true because as the cattle graze and move through the field, they stir up insects from the vegetation. The egrets take advantage of this by feeding on the insects that are now exposed. Therefore, Reason (R) correctly explains the commensalistic relationship.

SECTION B

17. The diagram given below shows a developmental stage of human embryo.
Answer the following questions with reference to it :

(a) Identify and name the human embryonic stage shown.

(b) Mention its exact location in the normal pregnancy of a woman.

Answer

a) Identify and name the human embryonic stage shown.

The human embryonic stage shown in the diagram is a Blastocyst.

b) Mention its exact location in the normal pregnancy of a woman.

In a normal pregnancy, the blastocyst would be located in the uterus, specifically implanted in the endometrium (the inner lining of the uterus).

c) Write one function of each of the structures labeled X and Y.

X: This structure represents the trophoblast. Its primary function is to:
- Implant the blastocyst into the endometrium.
- Form the placenta, which provides nutrients and oxygen to the developing embryo and removes waste products.
Y: This structure is the inner cell mass. It is crucial for the development of the embryo itself. It will eventually differentiate into various tissues and organs of the developing fetus.

18. (a) From which end of the ovule, and how does the pollen tube gain its entry into the embryo sac of a Hibiscus flower?

Answer:
The pollen tube enters the embryo sac of the Hibiscus flower through the micropyle at the micropylar end of the ovule.

The pollen tube grows down the style after pollination and enters the micropyle, which is a small opening in the integuments of the ovule.
This allows the pollen tube to reach the embryo sac, where fertilization will occur.

(b) State the fate of the male nuclei present in the pollen tube.

Answer:

The pollen tube contains two male nuclei (sperm cells) in the Hibiscus flower.
One male nucleus fuses with the egg cell to form the zygote (fertilization).
The other male nucleus fuses with the two polar nuclei in the central cell of the embryo sac to form the triploid primary endosperm nucleus, which will develop into the endosperm (providing nutrition to the developing embryo).

19. (a) (i) Identify and name the structures ‘A’ and ‘B’ marked in the image given below:

(ii) State their importance in various biotechnology experiments.

Answer

19. (a) (i) Identify and name the structures ‘A’ and ‘B’ marked in the image given below:

A: Plasmid
B: Bacteriophage

19. (a) (ii) State their importance in various biotechnology experiments.

Plasmids:
- Cloning vectors
- Gene libraries
- Protein production
- Gene therapy
Bacteriophages:
- Phage display
- Phage therapy
- Gene delivery

(b) Explain the process by which a bacterial cell can be made ‘competent’ to take up foreign DNA from its surroundings, using divalent cations and temperature treatment.

Answer:
To make a bacterial cell competent:

The cells are treated with divalent cations like CaCl₂ to neutralize the charges on the cell membrane and DNA, making it more permeable.
The cells are subjected to a heat shock (e.g., 42°C for 30-60 seconds) to temporarily increase the membrane’s permeability, allowing the foreign DNA to enter.
Afterward, the cells are cooled on ice and incubated in a recovery medium to express the new DNA.

20. Ecological pyramids give important information about the ecological system, but do have some limitations. List any two limitations of ecological pyramids.

Answer:

Does not account for energy flow in decomposers: Ecological pyramids do not include the role of decomposers (bacteria, fungi, etc.), which are crucial for nutrient recycling in ecosystems.
Ignores the complexity of food webs: Ecological pyramids typically represent simple food chains, but ecosystems often have complex food webs that the pyramids do not reflect.

21. India is the seventh largest country in the world in terms of total land area including land and water. Write the value of the land area of our country (in terms of percentage) of the world. Mention then, what makes India one of the 12 mega-diversity countries of the world.

Answer:

India’s land area is 2.4% of the world’s total land area.
India as a mega-diversity country: India is one of the 12 mega-diverse countries of the world because it has rich biodiversity, including a wide variety of ecosystems, species, and endemic species that contribute significantly to global biodiversity.

SECTION C

22. With the help of a schematic diagram only, show in three steps, the formation of recombinant DNA by the action of restriction endonuclease EcoRI and DNA ligase.

Answer:

Step 1: Action of EcoRI Restriction Endonuclease

EcoRI cuts the DNA at specific sequences, producing sticky ends (overhangs).

Example: DNA sequence:
5′ – GAATTC – 3′
3′ – CTTAAG – 5′

After EcoRI cuts:
5′ – G AATTC – 3′
3′ – CTTAA G – 5′

The cut produces sticky ends with single-stranded overhangs.

Step 2: Insertion of Foreign DNA

The foreign DNA is also cut with EcoRI, producing complementary sticky ends.

Example foreign DNA:
5′ – GAATTC – 3′
3′ – CTTAAG – 5′

After EcoRI cuts the foreign DNA, the sticky ends will match the sticky ends of the plasmid DNA.

Step 3: DNA Ligase Action

DNA ligase joins the sticky ends of the plasmid and the foreign DNA by forming phosphodiester bonds, creating the recombinant DNA molecule.

Result:
The foreign gene is now inserted into the plasmid vector, and the recombinant DNA is formed.

23. Study the given pedigree chart and answer the questions that follow.

(a) Is the trait given in the chart dominant or recessive? Give reason in support of your answer.

(b) Is this trait autosomal or sex-linked? Give reason in support of your answer.

Answer

a) Is the trait given in the chart dominant or recessive? Give reason in support of your answer.

Recessive.

In the first generation, both parents are unaffected, but they have an affected child. This indicates that the trait is recessive. If it were dominant, at least one parent would need to exhibit the trait.

b) Is this trait autosomal or sex-linked? Give reason in support of your answer.

Autosomal.

The trait appears equally in both males and females in the pedigree, suggesting it is not linked to a sex chromosome (X or Y).

c) Write the possible genotypes of the children numbers ‘1’ and ‘3’ of the second generation.

Child 1 (unaffected): Possible genotypes are Aa or AA (where “A” represents the dominant allele and “a” represents the recessive allele). Since the trait is recessive, they must have at least one dominant allele.
Child 3 (affected): Genotype must be aa to exhibit the recessive trait.

24.
(a) Where can one obtain stem cells in humans?
(b) State any two applications of stem cells in curing human diseases.

Answer:
(a) Stem cells in humans can be obtained from the following sources:

Embryonic stem cells: These are obtained from the inner cell mass of the blastocyst (early-stage embryo).
Adult stem cells: These are found in various tissues such as bone marrow, brain, skin, and liver.
Induced pluripotent stem cells (iPSCs): These are created by reprogramming somatic cells (such as skin cells) to become pluripotent.

(b) Two applications of stem cells in curing human diseases:

Regeneration of damaged tissues: Stem cells can be used to regenerate tissues such as heart muscle after a heart attack or nerve cells in cases of spinal cord injury.
Blood-related diseases: Stem cells from bone marrow can be used in treatments for diseases like leukemia through bone marrow transplants.

25.
(a) Differentiate between malignant and benign tumors.
(b) Name and explain the most feared property of a malignant tumor.

Answer:
(a) Difference between malignant and benign tumors:

Feature	Malignant Tumors	Benign Tumors
Growth	Rapid and uncontrolled growth.	Slow and localized growth.
Metastasis	Can spread to other parts of the body (invasive).	Do not spread to other body parts.
Cell Appearance	Cells are abnormal and poorly differentiated.	Cells appear similar to normal cells.
Effect on Body	Can be life-threatening.	Generally not life-threatening.

(b) The most feared property of a malignant tumor is metastasis. This refers to the ability of cancer cells to spread from their original site to other parts of the body through the bloodstream or lymphatic system, leading to secondary tumors and making the cancer more difficult to treat.

26.
Treatment of wastewater is done in a sewage treatment plant to make it less polluting. Explain the following with reference to this treatment process:
(a) Primary sludge
(b) Activated sludge
(c) Anaerobic sludge digesters

Answer:
(a) Primary sludge:
Primary sludge is the solid waste removed during the primary treatment phase of sewage treatment. It consists of large particles like debris and organic matter that settle at the bottom of the primary settling tank.

(b) Activated sludge:
Activated sludge refers to the sludge formed during the secondary treatment of sewage, where microorganisms are added to the sewage to break down organic pollutants. This leads to the formation of microbial biomass, which is called activated sludge.

(c) Anaerobic sludge digesters:
Anaerobic sludge digesters are used to treat sludge in the absence of oxygen. Bacteria break down organic matter in the sludge, producing biogas (mainly methane) and reducing the volume of the sludge.

27.
(a) Name the two primate ancestors of the present-day humans, who existed approximately about 15 million years ago.
(b) According to geological records, when and where did Australopithecines live?
(c) Give two differences between Homo habilis and Homo erectus.

Answer:
(a) The two primate ancestors of present-day humans who existed approximately 15 million years ago are:

Proconsul
Dryopithecus

(b) According to geological records, Australopithecines lived around 4 million to 1.4 million years ago in Africa. They were early hominins that are considered to be significant in human evolution.

(c) Two differences between Homo habilis and Homo erectus:

Brain Size:
- Homo habilis had a smaller brain size (about 510–600 cc), whereas Homo erectus had a larger brain size (about 600–1,100 cc).
Body Proportions:
- Homo habilis had more ape-like features with shorter limbs, while Homo erectus had longer limbs and a more human-like body proportion, which helped in bipedal walking.

28. (a) (i) Expand the abbreviations given below, used for different modes of assisted reproductive technologies:

ZIFT – Zygote Intrafallopian Transfer
ICSI – Intracytoplasmic Sperm Injection
IUT – Intrauterine Transfusion
GIFT – Gamete Intrafallopian Transfer

(ii) Which one of them cannot be considered as a procedure of IVF? Give reasons in support of your answer.

Answer:
IUT (Intrauterine Transfusion) cannot be considered a procedure of IVF because it involves the transfusion of blood or blood products into the uterus, which is unrelated to fertilization or embryo implantation. IVF specifically involves the fertilization of eggs outside the body and the transfer of embryos into the uterus.

(b) Differentiate between the following:

(i) Perisperm and Pericarp

Answer:

Feature	Perisperm	Pericarp
Definition	Nutritive tissue surrounding the embryo in the seed.	Outer covering of the fruit.
Location	Inside the seed around the embryo.	Surrounds the seed and fruit.
Function	Provides nourishment to the developing embryo.	Protects the seed and aids in seed dispersal.

(ii) Syncarpous pistil and Apocarpous pistil

Answer:

Feature	Syncarpous Pistil	Apocarpous Pistil
Definition	A pistil with fused carpels.	A pistil with free carpels.
Example	Example: Tomato, Peach.	Example: Buttercup, Magnolia.
Ovary Structure	The ovary consists of multiple fused carpels.	The ovary consists of separate carpels.

(iii) Plumule and Radicle

Answer:

Feature	Plumule	Radicle
Location	Embryonic shoot part of the seed.	Embryonic root part of the seed.
Function	Develops into the stem and leaves.	Develops into the primary root.
Appearance	Located above the cotyledons.	Located below the cotyledons.

SECTION D

29. Gene expresses itself in a cell system as a protein/enzyme. How does an expression of gene occur in a cell system and when does it need to occur, and how the gene expression is regulated in a prokaryote cell system was studied by the combined efforts of Jacque Monod, the biochemist, and Francois Jacob, the geneticist. For their work on lactose metabolism in E. coli and the lac operon, they were awarded the Nobel Prize in 1965.

(a) Why is lac operon said to be a transcriptionally regulated system?
Answer:
The lac operon is said to be a transcriptionally regulated system because the expression of genes involved in lactose metabolism is controlled by the transcription of the structural genes (lacZ, lacY, lacA) into mRNA, which is subsequently translated into enzymes. The operon is only transcribed when lactose (or its inducer, allolactose) is present, allowing the cell to save energy by only producing the enzymes when needed.

(b) It is said that the lac operon has to be operational at a very low level in the bacterial cell all the time. Justify.
Answer:
The lac operon needs to be operational at a very low level all the time because even in the absence of lactose, a small amount of glucose is metabolized and some basal level of the lac enzymes (like β-galactosidase) is required for the cell to be ready for lactose metabolism when lactose becomes available. This basal expression ensures the operon can quickly respond when lactose enters the environment.

(c) Why is the regulator gene in lac operon necessary?
Answer:
The regulator gene in the lac operon (lacI) is necessary because it produces the lac repressor protein, which controls the operon. When lactose is not present, the lac repressor binds to the operator region of the operon, preventing transcription. In the presence of lactose, the repressor is inactivated, allowing transcription to occur and enabling the cell to metabolize lactose.

30. Study the diagrammatic representation given below of the Earth with regions marked ‘A’ and ‘B’ respectively. Answer the questions that follow.

a) Write the observations made regarding the species diversity when moving from the equator (region A) towards the poles (region B). Give two reasons also.

Observation:

Species diversity generally decreases as we move from the equator (region A) towards the poles (region B).

Reasons:

Climate and Temperature: The equatorial regions experience relatively stable and warm temperatures throughout the year, which supports a greater variety of habitats and niches for different species. As we move towards the poles, the climate becomes colder and more extreme, with harsher conditions that limit the diversity of life.
Solar Radiation: The equator receives the most direct sunlight, leading to higher temperatures, increased rainfall, and greater primary productivity. This abundant energy supports a more complex and diverse food web, leading to higher species richness.

b) Stating the reason, mention the approximate number of bird species recorded in India.

India, located in the tropical and subtropical regions, is known for its rich biodiversity. It is estimated that India is home to around 1,300 – 1,400 bird species.

The reason for this high diversity is India’s diverse geographical features, ranging from the Himalayas to the Gangetic plains and the Western Ghats. This diverse landscape provides a wide range of habitats, supporting a large variety of bird species.

SECTION E

31. (a) Meselson and Stahl carried out an experiment to prove the nature of DNA replication.

(i) Which two types of nitrogen were used by them in their experiment and why?
Answer:

^14N (nitrogen-14) and ^15N (nitrogen-15).
^15N was used to label the DNA, and ^14N was used to track new DNA synthesis.

(ii) Why did they take samples of E. coli at definite time intervals for their observation?
Answer:
To observe changes in DNA density and track the replication process.

(iii) State the role of caesium chloride density gradient in their experiment.
Answer:
It was used to separate DNA based on density, helping distinguish between ^15N-labeled and ^14N-labeled DNA.

(iv) Write the conclusions they arrived at.
Answer:
DNA replication is semi-conservative, with each new DNA molecule consisting of one old strand and one new strand.

(b) (i) A true breeding tall pea plant with round seeds is crossed with a recessive dwarf pea plant having wrinkled seeds. Work out the cross up to F2 generation giving the phenotypic ratios of F1 and F2 generation respectively.

Answer:

Parental Generation (P):
Tall (TT) and Round (RR) × Dwarf (tt) and Wrinkled (rr)
F1 Generation:
Genotype: Tt Rr (All tall with round seeds)
F2 Generation:
Phenotypic ratio:
- 9 Tall and Round : 3 Tall and Wrinkled : 3 Dwarf and Round : 1 Dwarf and Wrinkled

(ii) State the Mendelian principle that can be derived only with the help of such a cross.

Answer:
Law of Independent Assortment.