students must start practicing the questions from CBSE Sample Papers for Class 12 Biology with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

  1. All questions are compulsory.
  2. The question paper has five sections and 33 questions. ALL questions are compulsory.
  3. Section-A has 16 questions of 1 mark each: Section-B has 5 questions of 2 marks each; Section- C has 7 questions of 3 marks each; Section- D has 2 case-based questions of 4 marks each: and Section-E has 3 questions of 5 marks each.
  4. There is no overall choice. However, internat choices have been provided in some questions. A student has to attempt onty one of the alternatives ¡n such questions.
  5. Wherever necessary, neat and properly Labeled diagrams should be drawn.

Section – A 

Question 1.
…… A ……… enzyme is used to remove dots from the blood vessels of patients who have undergone …… B ……. leading to heart attack.
(a) A-Streptokinase; B-Myocardial infarction
(b) A-Lipases; B-Arteriosclerosis
(c) A-Proteases; B-Myocardial infarction
(d) A-Pectinases; B-Atherosclerosis
Answer:
(a) A-Streptokinase; B-Myocardial infarction
Explanation:
An enzyme and thrombolytic drug is streptokinase. In some cases of myocardial infarction (heart attack), it is used as a medicine to dissolve clots.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

Question 2.
Identify components and enzymes for the given figure
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 1

(a) X- Repressor protein, Y- Inducer (lactose), E1 -Permease, E1-Transacetylase, E3– galactosidase.
(b) X-Repressor protein, Y-Inducer (lactose), E1galactosidase, E2 -Transacetylase, E3 – Permease.
(c) X- Inducer (lactose), Y- Repressor protein, E1-galactosidase, E2-Permease, E3– Transacetylase.
(d) X- Repressor protein, Y- Inducer (lactose), E1-galactosidase, E2-Permease, E3– Transacetylase.
Answer:
(d) X-Repressor protein, Y-Inducer (lactose),
E1-galactosidase,
E2-Permease,
E3– Transacetylase
Explanation:
‘X’ is a repressor that binds to the inducer region in the absence of lactose; in the presence of lactose(Y), the inducer protein binds to the repressor. The lactose operon gene in E Coli is a polycistronic gene. Lactose is converted to glucose and galactose by the enzyme E1, a-galactosidase. A transporter protein called E2-permease moves lactose from the media into the cell. E3 is a transacetylase that converts acetyl CoA into galactosides.

Question 3.
Which will be able to cope with global warming?
(a) Birds
(b) Mammals
(c) Amphibians
(d) Fishes
Answer:
(a) Birds
Explanation:
Birds will be able to cope with global warming because they are homeotherms as their body temperature remains constant irrespective to change in surroundings.

Question 4.
A tough tissue layer that covers the egg and prevents it from desiccation and pathogen infection. However, the same tissue also shields the egg and sperm nuclei from one another. However, sperm is known to release enzymes that break down this resilient sheet. Which sperm component is it?
(a) Acrosome
(b)Tail
(c) Mitochondria
(d) Sperm nuclei
Answer:
(a) Acrosome
Explanation:
An acrosome is a cap-like structure that can be seen on the sperm head’s anterior-most end. It contains many enzymes, including hyaluronic acid, which are released when an egg comes into contact with the female oviduct. These enzymes work by breaking down the membrane, allowing the head to pass through and the subsequent release of male nuclei.

Question 5.
The trophic level occupied by fishes, zooplankton, man and trees are respectively.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 2
(a) B, A, D, C
(b) C, D, B, A
(c) C, B, D, A
(d) C, B, A, D
Answer:
(c) C, B, D, A
Explanation:
An ecosystem contains a wide range of creatures. According to the trophic levels, the creatures in the environment should be in the following order:
Fishes- C (Secondary consumer) Zooplanktons- B (Primary consumer)
Man- D (Tertiary consumer)
Trees- A (Primary producer)

Question 6.
Rahul observed a plant in his garden. He hypothesized that the stem height exhibited incomplete dominance. To check for his, he created true-breeding lines of tall and short plants. He then crossed these and sampled 1000 progeny. Which of the following cases, matches his hypothesis?
(a) 500 tall plants, 250 intermediate plants, and 250 small plants
(b) 250 tall plants, 500 intermediate plants, and 250 small plants
(c) 250 tall plants, 250 intermediate plants, and 500 small plants
(d) 125 tall plants, 750 intermediate plants, and 125 small plants
Answer:
(b) 250 tall plants, 500 intermediate plants, and 250 small plants.
Explanation:
The phenotypic ratio of the progeny that exhibits incomplete dominance is 1:2:1. Therefore, of 1000 plants sampled from the progeny, 250 should be tall, 250 should be small, and 500 intermediate plants if incomplete dominance holds.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

Question 7.
Match the columns:

CoLumn-I CoLumn-II
(A)Leaching (i) Break down of detritus into smaller particle.
(B) Fragmentation (ii) Precipitation of water-soluble nutrients as unavailable salt Jes.
(C) CataboLism (iii) Degradation of detritus by bacterial and fungal enzymes.

(a) (A)-(ii), (B)-(i), (C)-(iii)
(b) (A)-(ii), (B)-(iii), (C)-(i)
(c) (A)-(iii), (B)-(ii), (C)-(i)
(d) (A)-(iii), (B)-(i), (C)-(ii)
Answer:
(a) (A)-(ii), (B)-(i). (C)-(iii)
Explanation:
The process that breaks down trash into smaller particles is known as fragmentation. It is mostly caused by the behavior of invertebrates that feed on detritus (detritivores, e.g., termites, earthworms, etc.). As it moves through an animal’s digestive system, the debris is crushed. Detritus particles’ surface area has significantly expanded as a result of fragmentation. Catabolism, or the enzymatic conversion of the disintegrating debris into simpler and inorganic compounds, is carried out by the extracellular enzymes secreted by bacteria and fungi. Leaching is the procedure by which dissolved nutrients are transferred to lower soil or groundwater layers to create inaccessible salts.

Question 8.
identify the correct option:

Conformers Regulators
(a) Less active More active
(b) Homeostasis is very Little Possess homeostasis
(c) Can maintain their body temperature Changes according to  environment
(d) Consume Less amount of energy Consume more amount of energy

Answer:
(c) Conformers: Can maintain their body temperature: Regulators: Changes according to environment

Question 9.
How is nascent oxygen toxic to aerobic living organisms?
(a) Cause mutation in DNA
(b) Can degrade proteins
(c) Can degrade enzymes
(d) All of these
Answer:
(d) All of these
Explanation:
All types of biomolecules, including DNA, proteins, and enzymes, that are present in living things can undergo reactions due to nascent oxygen’s high reactivity. It is hazardous to aerobic life forms because it can alter DNA and break down proteins and enzymes through reactions.

Question 10.
The pie chart shows for the number of species of major taxa of invertebrates:
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 3

identify A, B, C, and D in this figure

(a) (b) (c) (d)
A-In sects

B-Crus taceans

C-Mot Luscs

D-Other animat groups

A-Othe  animal groups

B-Mot Luscs

C-Crus tcicea ns

D-In sects

A-Mot tuscs

B-Insects

C-Other animal groups

D-Crusto ceans

A-In sects

B-Mot Luscs

C-Crus tacea ns

D-Other animal groups

Answer:
(b) A-Other animal groups, B-Molluscs, C-Crustaceans, D-lnsects ,
Explanation:
The pie chart shows the global biodiversity of invertebrates.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 14

From that, it is observed that A represents Other animal groups, B represents Molluscs, C represents Crustaceans, and D represents Insects.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

Question 11.
Identify the mendelian disorders.
(a) Down’s Syndrome
(b) Albinism
(c) Turner’s Syndrome
(d) Klinefelter’s Syndrome
Answer:
(b) Albinism
Explanation:
Albinism is the mendelian disorders while rest are chromosomal disorders.

Question 12.
Which of the following is not the source of variation?
(a) Recombination
(b) Genetic drift
(c) Migration
(d) Adaptations
Answer:
(d) Adaptations
Explanation:
There can be various sources of variations like recombination, genetic drift, gene flow etc. But adaptation is another phenomenon, it is not related to variation.

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.

Question 13.
Depending on where the reading begins, the genetic code can be interpreted in a variety of ways. For instance, reading could begin with the letter G in the base sequence GGGAAACCC, and there would be three codons: GGG, AAA, and CCC. The string will contain the codons GGA and AAC if reading begins at G in the second position.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 4

Assertion (A): UAA, UAG and UGA terminate protein synthesis.
Reason (R): They are not recognised by RNA,
(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.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
Synthesis of polypeptide terminates when a nonsense codon of mRNA reaches the A site, lac mRNA There are three nonsense codons UAA, UAG and UGA. These codons are not recognized by any of the RNAs. Therefore, no more aminoacyltRNA reaches the A-site. The P- site tRNA is hydrolyzed and the completed polypeptide is released in the presence of release factor. Thus termination occurs.

Question 14.
Assertion (A): PCR, ELISA, DNA hybridization are confirmatory tests for sexually transmitted diseases.
Reason (R): incidence of STDs are very high in persons above the age of 30.

(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.
Answer:
(c) A is true but R is false.
Explanation:
STDs are a major threat to healthy society. Incidence of STDs is very high in persons who are 15-24 years of age.

Question 15.
Assertion (A): Interferons are natural proteins produced by the cells of the immune systems of most animals in response to challenges by foreign agents. Reason (R): Interferons stimulate inflammation at the site of injury.
(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.
Answer:
(c) A is true but R is false.
Explanation:
Interferons are natural proteins produced by the cells of the immune systems of most animals in response to challenges by foreign agents such as viruses, bacteria, parasites and tumor cells. They stimulate both macrophages and NK (Natural Killer) cells, smack is more powerful analgesic than morphine. They both are also active against tumors. Interferon-gamma is involved in the regulation of the immune and inflammatory response.

Related Theory:
Interferons belong to the large class of glycoproteins known as cytokines. In human body there are three types of interferon-a, p and y. Interferon alpha and beta are produced by many cell types, including T-cells and B-cells and are important components of the antiviral response.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

Question 16.
Assertion (A): A woman can go to the court if a man refuses to own his child. He has blood group B and women has A, while child has O.
Reason (R): Genetically she is right, he can be father of the child.
(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.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
The B blood type male has the genotype (IBi) The blood type A female carries the genotype ( IAi )The kid has genotype ii, which causes the blood type O to induce. This shows that since both parents are heterozygous for their respective blood groups, the woman and the man can have a child with blood group O.

Related Theory:
The ABO blood group in human blood involves two antigens and two antibodies which are antigen A and antigen B, antibody A and antibody B respectively. Thus we can classify human blood into 4 groups:

  • Antigen A with antibody B
  • Antigen B with antibody A
  • Antigen AB with no antibodies
  • Antigen nil with antibody A and B.

Caution:
Students usually make mistakes whiile choosing the correct option. According to the figure given below, the man might be the child’s father in this situation. -As a result, if he declines to take the child, the lady has the right to file a lawsuit. However, it doesn’t provide conclusive evidence that he is the child’s father.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 15

Section – B

Question 17.
Biotechnology is a field of engineering that makes uke of biological systems, living organisms, or fragments thereof, to develop or produce various goods.Processes that fall under the category of biotechnology include brewing and baking bread. List two critical research areas of biotechnology.
Answer:
Two critical research areas of biotechnology are:

  1. Providing the best catalyst in the form of an improved organism usually a microbe or pure enzyme.
  2. Creating optimal conditions through . engineering for a catalyst to act.

Question 18.
The diagram refers to the chromosome complement of each sex of a fruit fly.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 5
Identify the unlabelled parts.
Answer:
One pair of sex chromosomes are present in Drosophila. in Drosophila, gender ” determination is homogametic for females and heterogametic for males, much like in humans. The female Drosophila is represented by the first figure, and the male Drosophila is represented by second figure. X-chromosomes are indicated by ‘A’ in female Drosophila. ‘B’ denotes Autosome. In male Drosophila, ‘C’ denotes X-chromosomes. ‘D’ denotes a male Drosophila’s Y chromosome.

Question 19.
Give one reason why a mother’s breastfeeding of her child acts as a natural form of contraception.
OR
Choose and write the appropriate device from the list below in the table:
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 6

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 7

Method of Birth Control Device
Barrier
IUD
Surgical technique.
Administering hormones

Answer:
Lactation or by interfering with the hormone prolactin, breastfeeding the baby delays the .onset or return of the menstrual cycle and the ovulation cycle. As a result, during this time—up to six months—there is no chance of conception. Therefore, breastfeeding the child may serve as a natural form of contraception for the mother (lactational amenorrhea).

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions
OR

Method of Birth Control Device
Barrier
IUD
Surgical technique.
Hormonal administrations

Question 20.
Which of the two cloning vectors, bacteriophages or plasmids, will be selected by biotechnologists? Justify with reason.
Answer:
Because they have many copies of their genome within the bacterial cells, bacteriophages are preferred by biotechnologists for cloning over plasmids. Some plasmids may only have one or two copies per cell, while others may haye 15-100 copies. Plasmids are less effective than phage vectors for cloning substantial DNA segments.

Question 21.
What is a food chain? List the types of food chains.
Answer:
Food Chain: A nutritive interaction among biotic communities (organisms) involving a producer, various levels of consumers, and a decomposer forms a food chain. Each step in a food chain is called a trophic level. Kinds of the food chain. There are 3 kinds of food chains: predator, parasitic and saprophytic chains.

Section – C

Question 22.
Natural selection is directional because it occurs in the context of adaptation, whereas creation and the presence of variation are directionless. Comment. 3
Answer:
Because variation occurs spontaneously and randomly, it is regarded as having no direction. It can be observed in organisms that reproduce sexually and results from gamete fusion or crossing over during meiosis. Generations after generation inherit the variations that aid humans in adjusting to their surroundings. Because it only leads down one road, i.e., selection, natural selection is directional. The survival of the fittest and the eradication of the unfit is the result of an evolutionary transformation.

Question 23.
Differentiate between the functions of anaerobic sludge digesters and floes in the treatment of sewage.
Answer:
In the process of treating sewage, floes consume a significant portion of organic waste, turning it into microbial biomass and liberating a large amount of minerals. While many anaerobic bacteria are present in anaerobic sludge digesters and break down the organic content, it lowers the BOD of sewage. Methane, C02, and other gases are created during this digestion.

Question 24.
Describe a gene library.
OR
Why are probes seen to be superior to other types of diagnostic equipment for diagnosing diseases? 3
Answer:
Gene Library: The phrase “gene library” refers to the collection of several cell clones that reflect nearly all of an organism’s genes and each contain one or a few foreign genes. Finding a clone with the desired gene can be done using that gene library. The genome of an organism must first be divided into smaller DNA fragments containing one or a few genes. These fragments can then be cloned in a cell, resulting in a group of cells with identical foreign DNA, which are referred to as clones.
OR
Probes superior than other types of diagnostic equipment due to the following reasons:

  1. They are a lot easier, quite focused, and relatively quick.
  2. They are incredibly effective, especially when used in conjunction with PCR: even the smallest amount of the test sample’s constituent molecules can be recognised.
  3. The risk of accidental infection of laboratory staff is eliminated because the culture of microbes is not necessary.

Question 25.
In a Dihybrid cross 556 seeds were collected in F2 generation. Count them shape-wise and color-wise as round yellow, round green, wrinkled yellow, wrinkled green seeds separately and write their numbers. Also, explain your answer.
Answer:
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 16
In F2 generation Mendel obtained pLants in the phenotypic ratio:

Round Yellow Round green Wrinkled yellow Wrinkled green
9/16 3/16 3/16 1/16

The number of plants with Round yellow seeds
= (9/16)x556 = 312.75 or 312
The number of plants with Round green seeds
= (3/16)x556= 104.25 or 104
The number of plants with wrinkled yellow
seeds = (3/16)x556 = 104.25 or 104
The number of plants with wrinkled green
seeds = (1/I6)x556 = 34.75 or 34

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

Question 26.
Sometimes the presence of foreign species causes native species to become extinct. Provide any two examples that will support this claim.
Answer:
Indigenous species are driven to extinction as a result of invasion by alien species, which also threaten native species.

  1. More than 200 species of cichlid fish in Lake Victoria go extinct as a result of the introduction of Nile perch.
  2. Introduced plants like Lantana and carrot grass (Parthenium) have become invasive and harmed the ecology. They endanger the indigenous plant species that live in our woodlands.

Question 27.
The above diagram shows a part of the human female reproductive system.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 8

(A) Identify (i), (ii), (iii), (iv).
(B) What is the other name for the fallopian tube?
(C) Where will you find the fimbriae? What is its function?
Answer:
(A) (i) Ovary,
(ii) isthmus,
(iii) Ampulla,
(iv) Infundibulum.
(B) Oviduct.
(C) Fimbriae are present at the edges of the Infundibulum. They help in collection of ovum after ovulation.

Question 28.
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 9

Answer the following:
(A) Which pond among the two mentioned in the above graph might have a greater amount of organic matter?
(B) Describe the process of primary and secondary treatment of sewage.
Answer:
(A) Pond 1, as the biological oxygen demand is increasing with time. Greater the BOD, greater will be the amount of organic matter.

(B) Primary treatment: It basically involves physical removal of particles from the sewage through filtration and sedimentation. These are removed in stages; initially, floating debris is removed by sequential filtration. Then the grit are removed by sedimentation. All solids that settle form the primary sludge, and the supernatant forms the effluent.

Secondary treatment:
The primary effluent is passed into large aeration tanks where it is constantly agitated and air is pumped into it allowing vigorous growth of useful aerobic microbes into floes. Here, other kinds of anaerobic bacteria, digest the bacteria and the fungi in the sludge. The effluent from the secondary treatment plant is generally released into natural water bodies.

Section – D

(Q. No. 29 and 30 are case based questions. Each question has subparts with internal choice in one subpart.)

Question 29.
A functional mammary gland is characteristic of all female mammals. Mammary glands are paired structures that contain glandular tissue and variable amount of fat. Refer to the given figure of a mammary gland and answer the following questions.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 10
(A) Identify part Labelled as T and ‘R’
(B) Identify part labelled as ‘P’ and ‘S’
(C) Why is nursing advised during the early stages of an infant’s development?

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions
OR
(C) How are mammary ampullae formed? What is areola?
Answer:
(A) T is mammary ampulla and ‘R’ is mammary alveoli.

(B) ‘P’ is areola and ‘S’ is mammary duct.

(C) Female mammary glands begin generating milk around the end of pregnancy through the lactation process, aiding the mother in feeding the baby. The milk produced in the first few days is called colostrum. Colostrum contains antibodies that are essential for helping babies establish resistance, thus doctors advise expecting parents to have a healthy baby.
OR
Near the nipple, mammary ducts expand to form mammary ampullae where some milk may be stored before going to lactiferous ducts. Areola is the fibrous (connective) tissue that supports the alveoli and the ducts.

Caution:
Students usually get confused with the labellings and write wrong answers. The following figure shows the labelled structure of human female breast

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 17

Question 30.
The given graphs show fluctuations in blood sugar of person X and Y during a 24 hour time period
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 11

Based on the above information, answer the following questions.
(A) Person Y is suffering from which disease?
(B) What causes weakness in person Y?
(C) What do you understand about person X from the above graph?
OR
What is the difference between the disease in person Y and type II diabetes?
Answer:
(A) Elevated blood sugar levels in person Y indicate that he is suffering from diabetes mellitus.

(B) Excretion of glucose in urine and excessive urination caused weakness in person Y.

(C) Blood sugar level fluctuations in person X indicate that sugar level never exceeds the normal limit and sufficient s secretion of insulin at required times removes any extra sugar from blood and converts it into glycogen for future use This implies that person X is normal and healthy.
OR
Type I diabetes (disease in person ‘y’) or insulin-dependent diabetes mellitus or juvenile diabetes is an autoimmune disorder caused by failure of beta cells to produce adequate amounts of insulin. Type II diabetes or noninsulin-dependent diabetes mellitus involves failure of insulin to facilitate the movement of glucose into body the cells.

Section – E

Question 31.
(A) Molecular marker genotyping with gel electrophoresis is frequently used in plant breeding and genomics, but there are many other applications as well. For instance, the polymerase chain reaction (PCR) is used to amplify specific DMA fragments that are used as markers and isolated from individual plants. The amplified DNA fragments are then loaded onto a gel Describe the theory that underlies the separation of DNA fragments during the following process.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 12

(B) How and why does recombinant DNA technology use the bacterium Thermus aquaticus? Explain.
(C) Identify the benefits of Taq polymerase and its origin.
OR
(A) Define recombinant proteins. How do bioreactors assist in their manufacture?

(B) By joining a gene to a plasmid vector, you have produced a recombinant DNA molecule. Your friend accidentally adds an exonuclease enzyme to the tube that contains the recombinant DNA. How would your experiment be impacted if you proceed with the shift right away?

(C) Why and how do bioreactors sustain a continuous culture system?
Answer:
(A) Get electrophoresis is a method of moving molecules like DNA, RNA, and protein in the direction of an electrode with the opposite charge based on their size while being influenced by an electric field. Molecules with positive charges travel in the direction of the cathode (-ve electrode), and vice versa. These molecules can be divided . based on their size and pass through a medium or matrix.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

(B) Thermus aquaticus bacteria is used in recombinant DNA technology because it contains thermostable DNA polymerase (Taq Polymerase), which keeps working even when a PCR step is denaturized by high temperatures.
When a gene is amplified via PCR. this enzyme is used (Polymerase Chain ’ Reaction). For additional cloning, the amplified region can be ligated to a vector.

(C) Thermostable bacteria, specifically Thermus aquaticus, are used to make Taq Polymerase. It is employed for DNA denaturation during PCR and stays active at higher temperatures.

OR
(A) Recombinant proteins are created whenever a gene encoding a protein is produced in a heterologous host. Large-scale recombinant protein manufacturing is facilitated by bioreactors. The ideal circumstances for producing the necessary recombinant protein via biological processes are offered by a bioreactor.

(B) Since the recombinant DNA molecule is circular and closed with no free ends, the experiment is not likely to be impacted. As a result, it won’t serve as a substrate for the exonuclease enzyme, which cuts nucleotides from DNA’s free ends.

(C) The used medium is drained out of one side of the bioreactor and the fresh medium is introduced from that side in order to maintain a continuous culture system. This kind of cultivating technique generates more biomass, which increases yields of the desired product

Question 32.
The female gametophyte’s embryo sac is a structure with seven cells and eight nuclei. Justify the statement using a labeled diagram.
OR
(A) Can pollen from a flowering plant in Pune pollinate a plant in Noida that is of the same species? Give details to support your response.
(B) Create a diagram of a successful pollination-affected pistil. Label the components necessary to deliver the male gametes to their final location.
Answer:
The egg apparatus is made up of three cells that are typically grouped together at the micropylar end of an embryo sac The egg apparatus itself is made up of one egg cell and two synergids. At the chalazal end, three cells are cast as antipodals. There are two polar nuclei in the large central cell. A typical angiosperm embryo sac has seven cells and eight nuclei when it is mature.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 20

OR
(A) Yes, it is possible through artificial hybridization, which involves introducing pollen grains from one flower artificially – onto the stigma of another flower. Nevertheless, self-incompatibility is not acceptable.
To protect the stigma from tampering, it’s crucialto use only the desired pollen grains when pollinating (from unwanted pollen). Emasculation and bagging techniques are used to achieve this.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 18

Question 33.
(A) Why is it impossible for an alien DNA to join a chromosome anywhere along its length and carry out regular replication?
(B) A structural gene has two DNA strands X and Y shown along side.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 13

Identify the tempLate strand.
(C) Describe the function of the DNA ligase enzyme In DNA replication.
(D) In a typical nucleus, some chromatin areas are stained Lightly white others are stained darkly. What does It mean and why is ¡tthat way?
OR
(A) ExpLain where and how each of the foUowing components fits into a transcription unit using a schemotic
diagram: structural gene, promoter, and terminator.

(B) How would the Lac opel-on fl.rnction if E. coU were to grow in a culture medium containing Lactose as a source of sugar?

(C) Where and how do peptide bonds arise in o bacterial ribosome?
Answer:
(A) Because there is no origin of replication, where the replication process is started, alien DNA cannot join a chromosome anywhere along its length and replicate correctly.

CBSE Sample Papers for Class 12 Biology Set 6 with Solutions

(B) ‘X’ is template strand. It is because the template strand has the polarity in 3′ → 5′ direction.

(C) By catalysing the creation of phosphodiester bonds, DNA ligase makes it easier for Okazaki fragments in lagging DNA strands to be joined together. Additionally, it aids in the restoration of duplex DNA single-strand breaks.

(D) In a normal nucleus, some chromatin regions are stained light because of loose packing, whereas other chromatin regions are stained dark because of dense packing. While heterochromatin (darkly stained) is transcriptionaLly inactive, euchromatin (lightly stained) is transcriptionally active chromatin.
OR
(A) Structure of a transcription unit: Transcription start site
CBSE Sample Papers for Class 12 Biology Set 4 with Solutions 19

The structural gene is flanked by the promoter and terminator in a transcription unit. The promoter aids in the initiation of transcription by interacting with RNA polymerase and is situated at the 5’end (upstream) of the structural gene. The terminator, which is situated near the 3’end (downstream) of the coding strand, often indicates the conclusion of the transcription process. Between the promoter and the terminator is the structural gene. For structural purposes, it codes for an enzyme or protein.

(B) In a medium containing E coli, lactose acts as a substrate for the enzyme p-galactosidase, which activates the operon. As a result, it is also known as an inducer. By attaching to the repressor, it deactivates it and grants the RNA polymerase access to the promoter.

(C) The amino group (-NH2) of the amino acid at the A-site and the carboxyl group (-COOH) at the P-site combine to create a peptide bond. A bacterial ribosome’s peptidyltransferase enzyme creates it.