CBSE Class 12 Science Half Yearly Sample Paper 2025 with Solution

Sections

No.of Questions 

Marks 

A

16

16

B

5

10

C

7

21

D

2

08

E

3

15

Total

33

70

Q.No

Answer Key

1.

(a) common set of characters in groups of different ancestry

2.

A. 23% According to Chargaff’s rules, in DNA,

 A =T and G=C; 

Thus, A +T+G+C =100 

Given T = 27% so A=T =27% 

Thus A+T = 27 +27 =54% 

Thus, G+C =100 – 54 = 46% 

Since G = C so G = 46/2 =23%

3.

(c) a) - iii, b) - iv, c) - ii, d) - i 

4.

A. present in the medium, and it binds to the repressor. 

5.

(a) Lysine and Arginine

6.

B. CO2

7.

(c) Preformed Antibodies, Passive

8.

A. Reduction in BOD

9.

(b) commensalism

10.

(b) No growth, growth

11. 

D. Rapid divergence of traits among populations inhabiting a given geographical area.

12. 

A. 1& 5; 5 &1

Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below: 

A. Both A and R are true and R is the correct explanation of A 

B. Both A and R are true and R is not the correct explanation of A 

C. A is true but R is false

D. A is False but R is true

13.

A. Both A and R are true and R is correct explanation of A.

14. 

(a) Both A and R are true and R is the correct explanation of A

15.

C. A is true but R is false. 

16.

(a) Both A and R are true and R is the correct explanation of A

17.

(a) Palindromic sequences (0.5), endonuclease enzyme (½ Mark) 

(b) Restriction enzymes can make complementary cut counterparts forming sticky ends for recombination DNA / RDNA technology/ to facilitate ligation of vector and foreign DNA.(1 Mark)

18.

• The variation in colour of colonies is due to the principle of insertional inactivation. (0.5) 

•  In this, a recombinant DNA is inserted within the coding sequence of an enzyme, β-galactosidase. This results into inactivation of the enzyme, which is referred to as insertional inactivation. (0.5) 

•  The presence of a chromogenic substrate gives blue-coloured colonies if the plasmid in the bacteria does not have an insert. (0.5) 

•  Presence of insert results into insertional inactivation of the β - galactosidase and the colonies do not produce any colour, these are identified as recombinant colonies. (0.5)

19.

(a) Inverted pyramids of biomass are seen in aquatic conditions where a small standing crop of phytoplankton supports a large standing crop of zooplankton/fish/In terrestrial ecosystem where a large number of insects are feeding on the leaves of a tree. (1 Mark) (b) No, the Pyramid of energy is always upright, and can never be inverted because when energy flows from one trophic level to the next trophic level some amount of energy is always lost as heat at each step. (1 Mark) 

(a) Inverted pyramid because a large number of insects feed on one tree. 

(b) No, the Pyramid of energy is always upright, and can never be inverted because when energy flows from one trophic level to the next trophic level some amount of energy is always lost as heat at each step. (1 x 2 = 2 marks)

20.

(a) A –Pituitary gland; B: Ovary(½ x 2 =1 Mark) (b) Endometrium of the uterus regenerates through proliferation. (1 Mark) 

21.

Making the correct punnett square (1 mark) 

Phenotype - All Inflated green pods (½ mark) 

Genotype –FfGg (½ mark)

22.

(a) Corn: Wind. Numerous flowers are packed in an inflorescence; the tassels seen in the corn cob are the stigma and style which wave in the wind to trap pollen grains. 

(b) Water hyacinth: Insects or wind. In water hyacinth the flowers emerge above the level of water and are pollinated by insects or wind as in most of the land plants. 

(c) Vallisneria: Water, In Vallisneria - the female flower reaches the surface of water by the long stalk and the male flowers or pollen grains are released onto the surface of water. They are carried passively by water currents; some of them eventually reach the female flowers and the stigma. 

23.

An antibody molecule consists of four polypeptide chains, two are long called heavy (H) chains while other two are short called light (L) chains. Both are arranged in the shape of Y. Hence, the antibody is represented as H2 L2. 

(Diagram with Labels – 

Light chain (½ mark), Heavy Chain (½ Mark) 

Types of Antibody – IgA, IgM, IgE,IgG (1mark awarded when all 4 types are stated) 

IgA – Lactating Mother to protect their infant(½ Mark) 

Ig E – To protect from allergen(½ Mark) 

OR 

(a) When a female Anopheles mosquito bites an infected person, the parasites enter the mosquito’s body as gametocytes(½ mark). It leads to fertilization and development in the gut (½Mark)of the mosquito and undergoes further development to form sporozoites that are stored in salivary glands (½ Mark) until their transfer to human body. 

In the human body – the sporozoites reach the liver and reproduce asexually (½ Mark), bursting the cells and releasing them into the RBCs as gametocytes (½ Mark). 

(Labeled diagram explaining the mentioned stages can also be considered) 

(b) The rupture of RBCs releases a toxic substance called haemozoin, (1/2 Mark) which is responsible for the chill and high fever.

24.

• isolation of DNA,  

• digestion of DNA by restriction endonucleases,  

• separation of DNA fragments by electrophoresis,  

• transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,  

• hybridisation using labelled VNTR probe, and  

• detection of hybridised DNA fragments by autoradiography. (0.5 x 6 =3)

25.

Impacts of loss of biodiversity on the ecosystem: 

(a) (i) Decline in plant production 

(ii) Lowered resistance to environmental perturbations such as drought 

(iii) Increased variability in certain ecosystems – processes such as plant productivity, water use, pest and disease cycles. (½ x 3 = 1 ½ marks) 

(b) (i) Habitat loss and fragmentation 

(ii) Over-exploitation 

(iii) Alien invasive species (iv) Co-extinctions. (Any three - ½ x 3 = 1½)

26.

(a) x to x’ is 5′———– > 3′ (½ Mark) No more amino acids will be added(½ Mark) 

(b) GCA(½ Mark) Anticodon is CGU (½ Mark) 

(c) The untranslated regions are required for an efficient translation process. (½Mark) They are present before the initiation codon at the 5’ – end and after the stop/termination codon, at the 3’ – end (½ Mark)

27.

A. Seed X- 3 embryos; 1embryo sac; 1ovule; (0.5 x 3=1.5) B. The nucellar cells grow mitotically and develop into the embryos by asexual reproduction. (0.5) C. The plants growing from seed X will have to share the resources/endosperm so there is a possibility of some plant being undernourished/; only one plant in seed Y will use the entire endosperm for its growth or as the plants of seed X are clones they will not show variation and may succumb to environmental stress;/ plants from seed Y will have genetic variation and so can show greater adaptability. (1)

For visually impaired students 

A. Seed X- 3 embryos; 1embryo sac; 1ovule; (0.5 x 3=1.5) 

B. The nucellar cells grow mitotically and develop into the embryos by asexual reproduction. (0.5) 

C. The plants growing from seed X will have to share the resources/endosperm so there is a possibility of some plant being undernourished/; only one plant in seed Y will use the entire endosperm for its growth or as the plants of seed X are clones they will not show variation and may succumb to environmental stress;/ plants from seed Y will have genetic variation and so can show greater adaptability. (1)

28.

PCR stands for Polymerase Chain Reaction. In this reaction, multiple copies of the gene (or DNA) of interest are synthesised in vitro using two sets of primers (small chemically synthesised oligonucleotides that are complementary to the regions of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. (1) 

If the process of replication of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times, i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of a thermostable DNA polymerase (isolated from a bacterium, Thermus aquaticus), which remains active during the high temperature induced denaturation of double stranded DNA. The amplified fragment if desired can now be used to ligate with a vector for further cloning. 

(1) Each cycle has three steps:

 (i) Denaturation,

 (ii) Annealing and

 (iii) Extensions. 

29.

A. The rupture of RBCs associated with the release of toxic substance haemozoin is responsible for the chills and fever/recurring every 3 - 4 days. (1) 

B. - The parasite reproduces asexually in liver cells, bursting the cell and releasing into the blood. (1) - Parasite further reproduces asexually in red blood cells. Released parasite infects new red blood cells. Sexual stages (gametocytes) develop in red blood cells. (1) 

Student to attempt either subpart C or D. 

C. The infection is caused by the bite of the female Anopheles mosquito which introduces the sporozoites in the human body. (1) 

OR 

D. Fertilisation and development take place in the mosquito’s gut. 

For visually impaired students 

A. The rupture of RBCs associated with the release of toxic substance haemozoin is responsible for the chills and fever/recurring every 3-4 days. (1) 

B. - 

The parasite reproduces asexually in liver cells, bursting the cell and releasing into the blood. (1) 

Parasite further reproduces asexually in red blood cells. Released parasite infects new red blood cells. Sexual stages (gametocytes) develop in red blood cells. (1) 

Student to attempt either subpart C or D. 

C. The infection is caused by the bite of the female Anopheles mosquito which introduces the sporozoites in the human body. (1) 

OR 

D. Fertilisation and development take place in the mosquito’s gut. (1)

30.

(a) Concentration of nicotine is maximum at 10 minutes/ conc. of nicotine increases steadily in the blood to reach 45mg/cm3 (1 Mark) 

(b) The Concentration of CO will increase resulting in reduced concentration of haemboundoxygen.(1 Mark) 

(c) Nicotine results in stimulating the adrenal gland which results in release of adrenaline / nor - adrenaline in the blood resulting in increase of blood pressure and heart rate. (2 Marks) 

31.

(a) You can easily grow a large quantity of the bacteria/no ethical issues/have plasmids/ can easily transform (any 1) (b) PCR will not amplify the gene. (½ Mark) If the polymerase enzyme denatures at low temp, it will not be able to withstand high temperature which is essential for separating/opening/unwinding/ denaturing DNA strand to open. Thus subsequent step of extending the primers using the nucleotides provided in the reaction and the genomic DNA as template will not occur.(1½ Marks) (c) Positive effect: oil spills can be treated and the environment becomes better/ cleaner/ water becomes more potable/ safe for aquatic forms/ safe for water birds like sea gulls. (any one 1)

Negative effect: the bacteria can mutate/ can harm other organisms/ can conjugate with other non-virulent forms and make them super bugs with detrimental effect/ unpredictable/ for a longer duration it may reduce the dissolved oxygen and leading to mortality of aquatic organisms (any one 1) 

OR

(a) Species III is least susceptible (1 Mark) 

(b) Bt toxin protoxins are converted into an active form in the gut which solubilises the toxin crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of the insect (2 Marks) 

(c) Insect species I and II have alkaline gut pH which solubilises the insecticidal protein crystals of protoxin and makes it active. Species III has an acidic and the protoxin continues to remain in an inactive form doing no harm to insect species III (2 Marks)

32.

Student to attempt either option A or B. 

A. 

(i)  Sperm count decreases, spermatogenesis is impaired;  Spermatids do not get nourishment to develop into spermatozoa thus spermiogenesis will be affected;  Leydig cells synthesize and secrete androgen hormones (like testosterone) so secretion of androgens will be affected. (0.5 x 3 = 1.5) 

(ii) Spremiation (0.5) 

(iii) Artificial insemination (AI) technique. In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI – intra – uterine insemination) of the female. 

OR 

Intra cytoplasmic sperm injection (ICSI) is another specialised procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum. (1) (iv) The zygote or early embryos (with upto 8 blastomeres) could be transferred into the fallopian tube (ZIFT–zygote intra fallopian transfer); embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer), to complete its further development. (1+1)

For visually impaired students 

(iv) The zygote or early embryos (with upto 8 blastomeres) could then be transferred into the fallopian tube (ZIFT–zygote intra fallopian transfer); embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer), to complete its further development. (1+1+0.5 mark for ZIFT with full form) 

OR 

B. 

(I) 

(i) In rose – bay plant, the time of maturation of stamen and pistil is not same, the pollen will not be able to germinate on the stigma. This prevents autogamy in rose-bay. (1) 

(ii) Different position and incompatible placement of the reproductive structure prevent successful pollination and thus autogamy in primrose. 

(1) 

(iii) Pollen pistil interaction for same species is not possible; this is a genetic mechanism which prevent the pollen grain from forming pollen tube on the pistil of the same flower. (1) 

(II) The male and female flowers are present in the same plant but are not in proximity preventing self-fertilization in castor. In papaya, the male flower and female flowers are in different plants, it prevents autogamy. (2)

33.

i) During copulation (coitus) semen is released by the penis into the vagina (insemination). 

ii) The motile sperms swim rapidly, pass through the cervix, enter into the uterus and finally reach the ampullary region of the fallopian tube. 

iii) The ovum released by the ovary is also transported to the ampullary region where fertilization takes place. 

iv) Fertilisation can only occur if the ovum and sperms are transported simultaneously to the ampullary region. This is the reason why not all copulations lead to fertilisation and pregnancy. 

v) The process of fusion of a sperm with an ovum is called fertilisation. 

vi) During fertilisation, a sperm comes in contact with the zona pellucida layer of the ovum and induces changes in the membrane that block the entry of additional sperms. Thus, it ensures that only one sperm can fertilise an ovum. 

vii) The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane.

viii) This induces the completion of the meiotic division of the secondary oocyte.

ix) The second meiotic division is also unequal and results in the formation of a second polar body and a haploid ovum (ootid). 

x) Soon the haploid nucleus of the sperms and that of the ovum fuse together to form a diploid zygote. (½ x 10 = 5)

OR

Trace the development of a megaspore mother cell to the formation of mature embryo sac in a flowering plant. 

The process of formation of megaspores from the megaspore mother cell is called megasporogenesis.

 i) Ovules generally differentiate a single megaspore mother cell (MMC) in the micropylar region of the nucellus. It is a large cell containing dense cytoplasm and a prominent nucleus. The MMC undergoes meiotic division to form megaspores. 

ii) In a majority of flowering plants, one of the megaspores is functional while the other three degenerate. Only the functional megaspore develops into the female gametophyte (embryo sac). This method of embryo sac formation from a single megaspore is termed monosporic development. 

iii) The nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac.

iv) Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac. 

v) These mitotic divisions are strictly free nuclear, that is, nuclear divisions are not followed immediately by cell wall formation. 

vi) After the 8-nucleate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo sac. 

vii) Six of the eight nuclei are surrounded by cell walls and organised into cells; the remaining two nuclei, called polar nuclei are situated in the large central cell. 

viii) Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus, in turn, consists of two synergids and one egg cell. The synergids have special cellular thickenings at the micropylar tip called filiform apparatus. 

ix) Three cells are at the chalazal end and are called the antipodals. 

x) The large central cell, as mentioned earlier, has two polar nuclei. Which come to lie below egg apparatus. Thus, a typical angiosperm embryo sac, at maturity, though 8-nucleate is 7-celled. (½ x 10 = 5)