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SOLUTIONS

2023 BOARD EXAM

CBSE CLASS 12-PCB BIOLOGY Board Paper 2023 — Set 1

2023

BIOLOGY

CLASS 12-PCB

CBSE EXAMINATION SOLVED PAPER-2023 BIOLOGY

CBSE EXAMINATION PAPER-2023

BIOLOGY

(Solved)

Time allowed : 3 hours

Maximum Marks : 75

General Instructions :

Read the following instructions carefully and follow them :

This question paper contains 32 questions. All questions are compulsory.
This question paper is divided into 5 sections.
Section A – questions number 1 to 2 are case based questions
Section B – questions number 3 to 13 are multiple choice questions
Section C – questions number 14 to 18 are very short answer
Section D – questions number 19 to 26 are short answer
Section E – questions number 27 to 32 are long answer
There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
Use of calculator is NOT allowed.

Section A

Question 1.

Question 2.

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

(1)

Name the region in the world that records the greatest biodiversity and mention why.

[1 Marks]

Answer: The region that records the greatest biodiversity on Earth is the Amazonian rain forest in South America. This rainforest is home to more than 40,000 species of plants, 3,000 species of fishes, 1,300 species of birds, 427 species of mammals, 427 species of amphibians, 378 species of reptiles, and over 125,000 species of invertebrates. Its unique climatic conditions, diverse habitats, and geographical isolation contribute to this extraordinary level of biodiversity, making it a crucial area for conservation efforts.

Key Points: Amazonian rain forest; home to 40,000+ plant species, 3,000 fish species; high species richness and endemism; vital for global ecology.

(2)

Write the observations made regarding the species diversity when moving from region 'A' to region 'B'. Give two reasons also

[3 Marks]

Answer: When moving from region 'A' to region 'B', it is observed that species diversity decreases significantly. In region 'A', located near the equator, there is a higher variety of plant and animal species due to the favorable climatic conditions that support various ecosystems. As we transition to region 'B', which is further away from the equator, there is a marked decrease in species richness. This decline in diversity can be attributed to two primary reasons: first, the temperature and climatic variations that occur with increasing latitude make survival more challenging for many species. Second, a reduction in habitat types and the overall area available for different species as one moves towards the poles limits ecological niches, leading to decreased biodiversity.

Key Points: Species diversity decreases from region 'A' to 'B'-Higher diversity in equatorial regions due to favorable climate-Reduced habitat types and ecological niches limit biodiversity in polar regions

(3)

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

[1 Marks]

Answer: India is home to over 1,200 species of birds, which can be attributed to its diverse habitats and climatic conditions. The country's location in tropical latitudes, along with various ecosystems such as forests, wetlands, and mountains, provides a suitable environment for a rich avifauna. This diversity in habitats allows for a wide array of bird species to thrive in the region, thereby contributing to India's status as a biodiversity hotspot.

Key Points: India has over 1,200 bird species - Tropical latitudes contribute to diversity - Diverse habitats support various species

Section B

Question 3.

Identify the region of transcription in prokaryote as shown in the schematic representation given below.

[1 Marks]

(A) Promoter Rho (p) RNA polymerase
(B) Promoter Sigma (σ) DNA polymerase
(C) Promoter Sigma (σ) RNA polymerase
(D) Terminator Sigma (σ) RNA polymerase

Explanation: The correct option is 'Promoter Sigma (σ) RNA polymerase' because the promoter is the region where RNA polymerase binds to initiate transcription in prokaryotes. The sigma (σ) factor is an initiation factor that assists RNA polymerase in recognizing the promoter region, thereby facilitating the start of transcription.

Question 4.

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?

[1 Marks]

(A) P-i, Q-iv, R-iii, S-ii
(B) P-iii, Q-i, R-ii, S-iv
(C) P-i, Q-ii, R-iv, S-iii
(D) P-iv, Q-iii, R-ii, S-i

Explanation: The correct answer is 'P-i, Q-iv, R-iii, S-ii' because each enzyme is matched to its specific function based on the classification of enzymes mentioned in the context provided. For example, oxidoreductases catalyze oxidoreduction reactions, which corresponds to the correct function listed in Column II.

Question 5.

Study the DNA profiles obtained as a result of DNA fingerprinting of a child 'X' and three individuals 1, 2 and 3. Which one of the following options shows the possible parents of the child 'X' ?

[1 Marks]

(A) 2 and 3
(B) 1 and 2
(C) 1 and 3
(D) Only individual 3

Explanation: The correct answer is '1 and 3', because child 'X' must inherit DNA bands from both parents. Therefore, the banding patterns of child 'X' should closely match the patterns of individual 1 and individual 3, indicating they are likely the parents.

Question 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.

[1 Marks]

(A) Disruptive selection as giraffes with smaller and longer neck lengths are selected.
(B) Stabilising selection as giraffes with longer neck lengths are selected further.
(C) Directional selection as giraffes with longer neck lengths are selected.
(D) Stabilising selection as giraffes with medium neck lengths are selected.

Explanation: The correct option is 'Directional selection as giraffes with longer neck lengths are selected.' This is because, as stated in the context, giraffes evolved longer necks to better reach leaves on tall trees, indicating a shift in the average neck length over generations towards longer necks due to natural selection favoring this trait.

Question 7.

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

[1 Marks]

(A) 650
(B) 1300
(C) 975
(D) 325

Explanation: During microsporogenesis, each microspore mother cell undergoes meiosis to produce four microspores. Therefore, if 325 microspore mother cells undergo this process, the total number of pollen grains formed will be 325 x 4 = 1300. Thus, the correct option is 1300.

Question 8.

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.

[1 Marks]

(A) P-iv, Q-iii, R-ii, S-i
(B) P-i, Q-ii, R-iii, S-iv
(C) P-iii, Q-i, R-iv, S-ii
(D) P-ii, Q-iv, R-iii, S-i

Explanation: The correct option is P-iii, Q-i, R-iv, S-ii. This option correctly correlates each contraceptive device with its mode of action based on the context provided. For instance, IUDs (P) prevent implantation by making the uterus unsuitable for it (iii), while progestogen pills (Q) work by suppressing ovulation (i). Additionally, barrier methods like condoms (R) prevent sperm from reaching the egg (iv), and hormonal contraceptives like implants (S) are designed to alter the hormonal environment to prevent ovulation (ii).

Question 9.

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

[1 Marks]

(A) Sertoli cells Testosterone Development of secondary sexual characteristics
(B) Leydig cells Androgen Initiates the production of sperms
(C) Ovary FSH Stimulates follicular development
(D) Placenta Estrogen Initiates secretion of milk

Explanation: The correct option is 'Leydig cells Androgen Initiates the production of sperms.' Leydig cells in the testes produce androgens (like testosterone), which are essential for the process of spermatogenesis, i.e., the production of sperm.

Question 10.

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

[1 Marks]

(A) Dr. Venugopalan
(B) Ms. Vandana Shiva
(C) Mr. Ajay Phadke
(D) Dr. R.A. Mashelkar

Explanation: Ms. Vandana Shiva is known for her activism in protecting traditional knowledge and challenging unjust patent claims on indigenous resources, like turmeric. Her efforts align with the concerns mentioned in the context about safeguarding traditional herbal medicines from patent exploitation.

Question 11.

Assertion (A): A patient of ADA deficiency undergoing treatment for gene therapy requires periodic infusion of genetically engineered lymphocytes.

Reason (R): Lymphocytes are immortal.

[1 Marks]

(A) Assertion (A) is false, but Reason (R) is true.
(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) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation: Assertion (A) is true, but Reason (R) is false. The patient indeed needs periodic infusion of genetically engineered lymphocytes for treatment; however, lymphocytes are not immortal, which means they cannot survive indefinitely without periodic replacement.

Question 12.

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.

[1 Marks]

(A) Assertion (A) is true, but Reason (R) is false.
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(C) Assertion (A) is false, but Reason (R) is true.
(D) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Explanation: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). The context provided clearly states that the cattle egret benefits from the presence of grazing cattle, which help to flush out insects, making it easier for the egrets to forage. This exemplifies the commensalism interaction described in Assertion (A).

Question 13.

Assertion (A): Birds like pigeon have heterogametic females whereas the males are homogametic.

Reason (R): In pigeons, females have Z and W sex chromosomes whereas males have ZZ sex chromosomes.

[1 Marks]

(A) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is 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.

Explanation: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). In birds, including pigeons, females are heterogametic with ZW chromosomes, while males are homogametic with ZZ chromosomes, thus explaining the assertion.

Section C

Question 14.

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.

[2 Marks]

Answer: The human embryonic stage shown is the blastocyst, which appears approximately 5 to 6 days after fertilization. During normal pregnancy, it is located in the uterus as it implants into the uterine wall. Part 'X' represents the inner cell mass, which develops into the embryo, while part 'Y' is the trophoblast, responsible for forming the placenta and facilitating nutrient exchange between the mother and embryo.

Question 15.

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

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

[2 Marks]

Answer: The pollen tube enters the embryo sac of a Hibiscus flower from the micropylar end of the ovule. It grows towards the ovule and penetrates one of the synergids, allowing the male gametes to be delivered. After entry, one of the male nuclei fuses with the egg cell, leading to fertilization, while the other male nucleus fuses with the central cell to form the triploid endosperm, crucial for seed development.

Question 16.

(i) Identify and name the structures 'A' and 'B' marked in the image given below:

(ii) State their importance in various biotechnology experiments.

[2 Marks]

Answer: Structure 'A' is identified as red blood cells, which are round and biconcave in shape, facilitating efficient oxygen transport. Structure 'B' is white blood cells, which are amoeboid, allowing them to move and respond to infections. These cells are crucial in biotechnology as red blood cells can be used in blood transfusions and white blood cells are essential for immunological studies and vaccine development.

Question 17.

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.

[2 Marks]

Answer: Bacterial cells can be made competent by using divalent cations like calcium chloride, which facilitates DNA uptake. The process begins by treating the bacterial cells with calcium ions, increasing membrane permeability. The cells are then incubated with foreign DNA on ice, followed by a brief heat shock at 42°C. This temperature change creates pores in the cell membrane, allowing the DNA to enter. Finally, the cells are returned to ice to stabilize them, completing the transformation.

Question 18.

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

[2 Marks]

Answer: Ecological pyramids, while useful, have limitations. Firstly, they can overlook organisms that exist at multiple trophic levels, which skews the energy dynamics. Secondly, they assume a simplistic food chain while real ecosystems function as complex food webs. This complexity means that interactions between species are not accurately represented. Thus, ecological pyramids may misrepresent energy flow and biomass distribution in ecosystems.

Section D

Question 19.

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.

[3 Marks]

Answer: The formation of recombinant DNA using restriction endonuclease EcoRI and DNA ligase can be illustrated in three major steps. First, EcoRI recognizes specific palindromic sequences in both the source DNA and the plasmid vector, cutting both at these sites to generate sticky ends. Next, the desired DNA fragment with complementary sticky ends is mixed with the open plasmid, allowing the sticky ends to hybridize. Finally, DNA ligase is added to facilitate the joining of the DNA strands, forming stable recombinant DNA. This recombinant vector can now be introduced into host cells for replication and protein expression.

Question 20.

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.

[3 Marks]

Answer: The trait depicted in the pedigree chart is likely recessive, as it appears to skip generations, indicating that unaffected individuals can carry the gene. If both parents express the trait and their child does not, it further supports the recessive nature. Regarding whether the trait is autosomal or sex-linked, it appears to be autosomal because the trait is transmitted equally to both male and female offspring without a clear gender bias. This pattern is typical for autosomal inheritance. For the children numbered '1' and '3', if we consider possible parental genotypes, '1' could have the genotype 'Aa' while '3' could be 'aa', suggesting varying carrier statuses among the siblings.

Question 21.

(a) (a) Write the characteristics of 'stem cells'.

(b) From where can one obtain 'stem cells' in humans?

[3 Marks]

Answer: Stem cells are unique cells characterized by their ability to differentiate into various cell types and self-renewal. They are undifferentiated cells, capable of giving rise to specialized cells for all tissues in the body. Stem cells can be obtained from various sources in humans, primarily from embryonic tissues, umbilical cord blood, and adult bone marrow. Applications of stem cells include treatment of blood disorders like leukemia, and regenerative medicine, such as repairing heart tissue after a heart attack.

Question 22.

(a) Differentiate between malignant and benign tumours.

(b) Name and explain the most feared property of a malignant tumour.

[3 Marks]

Answer: Benign tumours are non-cancerous growths that remain localized and do not invade surrounding tissues or spread to other parts of the body. They usually cause minimal damage and can often be surgically removed without significant harm to the body. In contrast, malignant tumours are cancerous; they grow rapidly, invade nearby tissues, and can metastasize, meaning they can spread to distant sites through the bloodstream, forming new tumours. The most feared property of malignant tumors is their ability to metastasize, which makes them particularly dangerous as they compromise the function of multiple organs.

Question 23.

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

[3 Marks]

Answer: The treatment of wastewater involves various stages crucial for reducing its polluting potential. (a) Primary sludge is obtained during the primary treatment stage when larger particles settle out as sediment from the sewage, representing a combination of solid waste and biologically active microorganisms. (b) Activated sludge is formed when primary effluent enters aeration tanks, where it is agitated and aerated. This process encourages the growth of microbial flocs that digest organic matter, significantly lowering BOD levels. A portion of this sludge is recycled back to maintain microbial activity. (c) Anaerobic sludge digesters are large, oxygen-free tanks where the accumulated activated sludge is treated further. Here, anaerobic bacteria decompose the organic matter, releasing biogas, which mainly consists of methane, a valuable renewable energy source. This sequenced approach effectively cleans wastewater before its release.

Question 24.

(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 ?

[3 Marks]

Answer: (a) The two primate ancestors of modern humans that existed approximately 15 million years ago are Dryopithecus and Ramapithecus. Dryopithecus is considered to be more ape-like, while Ramapithecus displays more human-like characteristics. (b) According to geological records, Australopithecines lived around 2 million years ago in the East African grasslands. They were known for their bipedal locomotion and likely used stone tools. (c) Two differences between Homo habilis and Homo erectus are: Homo habilis had a smaller brain capacity of 650-800cc, whereas Homo erectus had a larger brain, typically ranging from 600-1,100cc. Additionally, Homo habilis is believed to have primarily consumed fruits, while Homo erectus had a more varied diet, including meat.

Question 25.

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

(1) ZIFT

(2) ICSI

(3) IUT

(4) GIFT

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

[3 Marks]

Answer: The expansions of the abbreviations are as follows: ZIFT stands for Zygote Intra Fallopian Transfer, ICSI stands for Intra Cytoplasmic Sperm Injection, IUT stands for Intra Uterine Transfer, and GIFT stands for Gamete Intra Fallopian Transfer. Among these, GIFT is not considered a procedure of IVF because it does not involve fertilization taking place in the laboratory. Instead, GIFT transfers the collected ovum and sperm into the fallopian tube for natural fertilization, while IVF involves fertilizing the egg outside the body and then transferring the embryo.

Question 26.

Differentiate between the following :

(i) Perisperm and Pericarp

(ii) Syncarpous pistil and Apocarpous pistil

(iii) Plumule and Radicle

[3 Marks]

Answer: Perisperm refers to the nutritive tissue that surrounds the embryonic sac in some seeds, while pericarp is the outer layer of the fruit that develops from the ovary wall after fertilization. A syncarpous pistil consists of fused carpels, whereas an apocarpous pistil features separate, distinct carpels. Lastly, the plumule is the part of the seed that develops into the shoot or stem of the plant, while the radicle is the embryonic root that develops downward into the soil.

Section E

Question 27.

Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment and answer the following questions.

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

(ii) Why did they take samples of E. coli at definite time intervals for their observation ?

(iii) State the role of caesium chloride density gradient in their experiment. (iv) Write the conclusions they arrived at.

[5 Marks]

Answer: In their landmark experiment, Meselson and Stahl used two types of nitrogen isotopes: 15N, a heavy isotope, and 14N, the lighter, more common isotope. The purpose of using these isotopes was to track the incorporation of nitrogen into the DNA of E. coli, thereby providing evidence for the mode of DNA replication. By isolating pure DNA and subjecting it to caesium chloride density gradient centrifugation, they could distinguish between the densities of DNA molecules after replication. Samples were taken at specific intervals to observe the transition from the heavy 15N DNA to the hybrid DNA, thus demonstrating semiconservative replication. The gradient allowed them to analyze the density differences, confirming that each daughter DNA molecule consisted of one strand from the parent and one newly synthesized strand.

Question 28.

(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.

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

[5 Marks]

Answer: To determine the inheritance from a true-breeding tall pea plant (genotype TT, RR) crossed with a dwarf pea plant (genotype tt, rr), we first examine the F1 generation. All offspring will be tall with round seeds (TtRr). When the F1 generation is self-pollinated, the F2 generation emerges. Using a Punnett square for both traits, we find the phenotypic ratio: 9 Tall Round (T_R_), 3 Tall Wrinkled (T_rr), 3 Dwarf Round (ttR_), and 1 Dwarf Wrinkled (ttrr). This totals to 3:1:3:1, showing that both characters assort independently. The derived Mendelian principle from this cross is the Law of Independent Assortment, which states that alleles of different genes segregate independently of each other during gamete formation.

Question 29.

(i) What is the chemical name of 'smack'? Why is the consumption of smack considered as an abuse?

(ii) Name the source plant and one effect of the following drugs on the human body:

(1)Marijuana

(2) Cocaine

(3) Morphine

[5 Marks]

Answer: The chemical name of 'smack' is diacetylmorphine, which is derived from morphine obtained from the latex of the opium poppy plant, Papaver somniferum. Heroin acts as a depressant, slowing down bodily functions and, when abused, can lead to serious health consequences, including addiction, respiratory failure, and death. Consumption is considered abuse when taken for non-medicinal purposes or in excessive amounts that impair physical or psychological functions. (1) Marijuana is sourced from the Cannabis sativa plant and can cause relaxation and altered perceptions. (2) Cocaine, derived from coca leaves, is a powerful stimulant that increases energy and alertness. (3) Morphine also comes from the opium poppy and is primarily used as a pain reliever, but can lead to physical dependence and addiction.

Question 30.

Study the diagram given below showing the modes of pollination. Answer the questions that follow.

(i) The given diagram shows three methods of pollen transfer in plants. What are the technical terms used for pollen transfer methods '1', '2' and '3'?

(ii) How do the following plants achieve pollination successfully ?

(1) Water lily

(2) Vallisneria

(iii) Flowering plants have developed many devices to avoid inbreeding depression. Explain one hereditary and one physiological device which helps plants to achieve this target.

[5 Marks]

Answer: Pollination in flowering plants can occur through various mechanisms. The technical terms for the methods shown in the diagram are as follows: (1) Autogamy, which refers to self-pollination within the same flower; (2) Geitonogamy, where pollen is transferred between different flowers on the same plant; and (3) Allogamy, indicating cross-pollination between flowers of different plants. Water lilies achieve pollination through water, utilizing hydrophily, as their flowers float on the surface, while Vallisneria employs a similar method, where flowers emerge only when pollination is needed. To prevent inbreeding depression, hereditary devices can include self-incompatibility mechanisms that prevent self-fertilization, while physiological devices might involve temporal separation of male and female flower maturation, ensuring cross-pollination occurs.

Question 31.

Observe the diagram given below showing the menstrual cycle of a normal human female and answer the questions that follow :

(i) What are the suitable technical terms used for the following ?

(1) Days 1-7

(2) Days 8-12

(3) Days 16-28

(4) Days 13-15

(ii) Explain the role of ovarian and pituitary hormones during the following time periods :

(1) Days 8-12

(2) Days 13-15

(3) Days 16-28

[5 Marks]

Answer: The menstrual cycle consists of four main phases. (1) Days 1-7 are called the Menstrual Phase, characterized by menstrual flow. (2) Days 8-12 are termed the Follicular Phase, where follicles in the ovaries mature under FSH influence. (3) Days 16-28 are known as the Luteal Phase, marked by the development of the corpus luteum. (4) Days 13-15 correspond to Ovulation, when an ovum is released. During Days 8-12, estrogen levels rise due to follicular development. Days 13-15 see a surge in LH, triggering ovulation. From Days 16-28, progesterone from the corpus luteum prepares the endometrium for potential implantation.

Question 32.

A time-bound vaccination programme is followed for the children in our country from their birth up to ten years of age. A graph plotted below shows the effect of the vaccination followed by infection by the same pathogen, and the antibody concentration in the blood of the child.

(i) Explain why the administration of a vaccine causes an increase in the antibody concentration.

(ii) If the child is infected with the same pathogen almost four months later, the antibody concentration in his/her blood increases very fast. Explain why.

(iii) A table given below gives information about different types of immunity and how they are attained. Identify P , Q , R , S and T in the table.

[5 Marks]

Answer:

(i) A vaccine contains harmless antigens (inactivated pathogen or protein fragments) that stimulate your B-cells to divide and differentiate into plasma cells, which secrete large amounts of specific antibodies—so you get a rise in antibody concentration without having to suffer the actual disease.

(ii) When you were vaccinated, you not only made antibodies but also produced long-lived memory B-cells. On re-exposure to the same pathogen those memory cells immediately reactivate, proliferate and become plasma cells, so antibody levels soar much faster than on the first encounter.

(iii) Filling in P, Q, R, S and T in the immunity table

Row	Type of immunity	Production of Abs	Memory cells	Mode attained
(1)	Natural, active	Yes	P = Yes	Q = Infection by pathogen
(2)	Natural, passive	No	R = No	Across the placenta…
(3)	Acquired, active	S = Yes	Yes	Getting a vaccine
(4)	Acquired, passive	T = No	No	Injection of antibodies

So the missing entries are:

P = Yes
Q = Infection by (exposure to) the live pathogen
R = No
S = Yes
T = No

Paper Details

CBSE Board Exam 2023

Class

CLASS 12-PCB

Subject

BIOLOGY

Year

2023

Set

Set 1

Other Years — BIOLOGY

2022 Set-1

Paper

Solutions

2022 Set-2

Paper

Solutions

2022 Set-3

Paper

Solutions

2022 Set-4

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2022 Set-5

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Solutions

2023 Set-1

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Solutions

2023 Set-3

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2023 Set-5

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2024 Set-1

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2024 Set-2

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2024 Set-3

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2024 Set-4

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2024 Set-5

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2025 Set-1

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2025 Set-2

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2025 Set-3

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2025 Set-4

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2025 Set-5

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2025 Set-6

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