Biology Class 12 Sample Paper 2023 Solutions PDF Free Download

Biology Class 12 Sample Paper 2023 Solutions PDF Download Info

Biology Class 12 Sample Paper 2023 Solutions PDF Free Download, Answers PDF Solutions, 2022-23 PDF, Biology is a significant topic in the board test, along with other courses. CBSE has published a sample paper on its official website, www.cbseacademic.nic.in. You can do well in biology if you master the words and practise the properly-labeled diagram. To do well in the board test, all students are instructed to pull their socks up. You may practise by using the sample paper available on the official website. We have included direct download links for Physics Sample Paper in this post.

This page contains a Biology Sample Paper Class 12 2023 with Answers PDF. We also discussed the Biology Sample Paper Class 12 2023 Marking Scheme.

Biology Class 12 Sample Paper 2023 Solutions PDF

cbseacademic.nic.in currently has Biology Sample Paper Class 12 2023. Biology Sample Paper Class 12 2023 may be found here to help you achieve high grades in Biology. Sample Papers have been presented by the Central Board of Secondary Education for students in classes 10 and 12. These sample papers are available on the organization’s official website. Students may also obtain sample papers from the links provided below.

CBSE has provided sample papers for all courses based on the layout of the question paper that CBSE will use in this session 2022-23. CBSE has proposed just one term examination this time, which will be held in February 2023. Together with the Biology Sample Paper Class 12 2023, CBSE also publishes the Biology Sample Paper Class 12 2023 Marking System. With the aid of Biology Sample Paper Class 12 2023, students may obtain a good concept of the board exam pattern, question types, response possibilities, and other aspects.

We discussed the Biology Sample Paper Class 12 2023 in this post. Students must complete the Biology Sample Paper Class 12 2023 provided on this website and bookmark this page to get all CBSE Annual Exam updates.

Students searching for CBSE Sample Papers for Class 12 Biology 2022-23 may obtain the Sample question papers from this link. The CBSE Class 12 Sample Question Papers are provided here to help students prepare for their final year board exams. Let us inform you that sample papers, also known as model papers, are one of the finest resources for students preparing for their board examinations.

These CBSE Sample Papers for Class 12 Biology allow students to get experience before taking the final exam. They will also comprehend whether or not they are entirely prepared for the test. Students may test their understanding of the topic and gain confidence in their replies.

If there are any errors in the written responses, they might focus harder on those questions so that there are no errors in the final tests. As a result, here is a sample question paper for CBSE Biology class 12 with answers.

Sample CBSE Class 12 Biology Question Paper 2022-23: The CBSE board has issued Class 12 Biology Sample Paper, which is accessible on the Central Board of Secondary Education’s official website for all streams.

The board has issued the marking system for the next Class 10 and 12 board examinations in February, March, and April 2023, along with all other sample papers for 2022-23 board exams.

If you’re wondering how an example paper will help you prepare or if you’re not sure which themes to prioritise, you’ve come to the perfect spot. Sample papers might assist you figure out what kinds of questions are asked on certain themes. Meanwhile, the grading system might assist you in determining which sections to learn first based on your skills and shortcomings. We’ve included both of these for your convenience. You may prepare thoroughly and get more points by using CBSE sample question papers 2022-23 and the marking system.

Before trying to solve the example paper, ensure that you have covered all of the subjects listed in the syllabus. See the CBSE 12th Biology syllabus for 2022-23.

As a CBSE Class 12 Biology Board exam applicant, it is essential that you complete the CBSE Class 12 Biology Sample Question Paper 2022-23 to identify your blind spots and understand and review each idea from your curriculum.

Note that the sample paper is just for practise, and questions in the board examinations might come from any subject in the specified curriculum.

Sample Paper Class 12 Biology 2022-23 With Solutions

Section – A

Question1. An infertile couple was advised to undergo In vitro fertilization by the doctor. Out of the options given below, select the correct stage for transfer to the fallopian tube for successful results. 

(a) Zygote only 

(b) Zygote or early embryo up to 8 blastomeres 

(c) Embryos with more than 8 blastomeres 

(d) Blastocyst Stage 

Solution: (b) Zygote or early embryo upto 8 blastomeres

Question2. Given below are four contraceptive methods and their modes of action. Select the correct match:

(a) a)–(i) b)–(ii) c)– (iii) d)–(iv) 

(b) a)–(ii) b)–(iii) c)–(iii) d) – (I) 

(c) a)–(iii) b)–(iv) c)–(ii) d)–(I) 

(d) a)–(iv) b)–(i) c)– (iii) d)–(ii)

Solution:(c) a)–(iii) b)–(iv) c)–(ii) d)–(I) 

Question3. Which of the following amino acid residues will constitute the histone core?

(a) Lysine and Arginine

(b) Asparagine and Arginine

(c) Glutamine and Lysine

(d) Asparagine and Glutamine

Solution:(a) Lysine and Arginine

Question4. Evolutionary convergence is the development of an

(a) the common set of functions in groups of different ancestry.

(b) dissimilar set of functions in closely related groups.

(c) the common set of structures in closely related groups.

(d) dissimilar set of functions in unrelated groups.

Solution:(a) the common set of functions in groups of different ancestry.

Question5. Apis mellifera are killer bees possessing toxic bee venom. Identify the treatment and the type of immunity developed from the given table to treat a person against the venom of this bee.         

Solution:(c) Preformed Antibodies, Passive

Question6.  Interferons are most effective in making non-infected cells resistant to the spread of which of the following diseases in humans? 

(a) ascariasis 

(b) ringworm 

(c) amoebiasis 

(d) AIDS      

Solution:(d) AIDS

Question7. Which of the following water samples in the table given below, will have a higher concentration of organic matter? 

Solution:(a) High, High 

Question8. The figure below shows the structure of a plasmid. 

 A foreign DNA was ligated at BamH1. The transformants were then grown in a medium containing antibiotics tetracycline and ampicillin. Choose the correct observation for the growth of bacterial colonies from the given table

Solution:(b) No growth, growth

Question9. Swathi was growing a bacterial colony in a culture flask under ideal laboratory conditions where the resources are replenished. Which of the following equations will represent the growth in this case? (Where population size is N, the birth rate is b, the death rate is d, the unit time period is t, and the carrying capacity is K).

(a) dN/dt = KN

(b) dN/dt = r N

(c) dN/dt = r N(K-N/K)

(d) dN/dt = r N(K+N/K)

Solution:(b) dN/dt = r N

Question10. Sea Anemone gets attached to the surface of the hermit crab. The kind of population interaction exhibited in this case is

(a) amensalism.

(b) commensalism.

(c) mutualism.

(d) parasitism.

Solution: (b) commensalism 

Question11.  Which of the following food chains is the major conduit for energy flow in terrestrial and aquatic ecosystems respectively?

Solution:(c) Detritus; Grazing food chain respectively

Question12.  Which of the following is an example of ex-situ conservation?

(a) Sacred Groves

(b) National Park

(c) Biosphere Reserve

(d) Seed Bank   

Solution:  (d) Seed Bank  

Questions No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer these questions by 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.

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

Question13. Assertion: Apomictic embryos are genetically identical to the parent plant.

Reason: Apomixis is the production of seeds without fertilization.

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

Question14. Assertion: When white-eyed, yellow-bodied Drosophila females were hybridized with red-eyed, brown-bodied males; and F1 progeny was intercrossed, the F2 ratio deviated from 9 : 3 : 3: 1.

Reason: When two genes in a dihybrid are on the same chromosome, the proportion of parental gene combinations is much higher than in the non-parental type.

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

Question15. Assertion: Functional ADA cDNA genes must be inserted in the lymphocytes at the early embryonic stage.

Reason: Cells in the embryonic stage are mortal, differentiated, and easy to manipulate.

Solution:(c) A is true but R is false

Question16. Given below is the Age Pyramid of the population in one of the states in India as per the 2011 census. It depicts the male population on the left-hand side, the female population on the right-hand side, newborns towards the base and gradually increasing age groups as we move from the base to the top, with the oldest population at the top. Study this pyramid and comment upon the appropriateness of the Assertion and the reason.  

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

Section – B

Question17.  In the figure given below, parts A and B show the level of hormones that influence the menstrual cycle. Study the figure and answer the questions that follow: 

(a) Name the organs which secrete the hormones represented in parts A and B.

(b) State the impact of the hormones in part B on the uterus of the human female during 6 to 15 days of the menstrual cycle?

Solution:(a) A –Pituitary gland; B: Ovary

(b) Endometrium of the uterus regenerates through proliferation.

Question18. A true breeding pea plant, homozygous dominant for inflated green pods is crossed with another pea plant with constricted yellow pods (ffgg). With the help of Punnett square show the above cross and mention the results obtained phenotypically and genotypically in the F1 generation. 

Solution: Making the correct Punnett square

Phenotype – All Inflated green pods

Genotype –FfGg 

Question19. During a field trip, one of your friends in the group suddenly became unwell, she started sneezing and had trouble breathing. Name and explain the term associated with such sudden responses. What would the doctor recommend for relief?

Solution:(a) Allergy, the exaggerated response of the immune response to certain antigens present in the environment is called allergy. 

(b) Doctors would administer drugs like antihistamines, adrenaline, and steroids (anyone) to reduce the symptoms

Question20.  CTTAAG  

GAATTC  

(a) What are such sequences called? Name the enzyme used that recognizes such nucleotide sequences.

 (b) What is their significance in biotechnology?

Solution:(a) Palindromic sequences (0.5), endonuclease enzyme

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

Question21. (a) Given below is a pyramid of biomass in an ecosystem where each bar represents the standing crop available at the trophic level. With the help of an example explain the conditions where this kind of pyramid is possible in nature. 

(b) Will the pyramid of energy be also of the same shape in this situation? Give a reason for your response.

Solution:(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 a terrestrial ecosystem where a large number of insects are feeding on the leaves of a 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.

OR 

(a) Draw a pyramid of numbers where a large number of insects are feeding on the leaves of a tree. What is the shape of this pyramid? 

(b) Will the pyramid of energy be also of the same shape in this situation? Give a reason for your response. 

Solution:(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.

Section – C

Question22.  Explain the functions of the following structures in the human male reproductive system.

(a) Scrotum

(b) Leydig cells

(c) Male accessory glands

Solution:(a) Scrotum: The testes are situated outside the abdominal cavity within a pouch called the scrotum. The scrotum helps in maintaining the low temperature of the testes (2–2.5 degree celsius lower than the normal internal body temperature) necessary for spermatogenesis. 

(b) Leydig cells: The regions outside the seminiferous tubules called interstitial spaces, contain small blood vessels and interstitial cells or Leydig cells. Leydig cells synthesize and secrete testicular hormones called androgens. 

(c) Male accessory glands: The male accessory glands include paired seminal vesicles, prostate, and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose, calcium, and certain enzymes. The secretions of bulbourethral glands also help in the lubrication of the penis. 

Question23.  State the agent(s) which help in pollinating in the following plants. Explain the adaptations in these plants to ensure pollination:

(a) Corn

(b) Water hyacinth

(c) Vallisneria

Solution:(a) Corn: Wind. Numerous flowers are packed in an inflorescence; the tassels seen in the corn cob are the stigma and style that 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 land plants. 

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

Question24.  (a) Identify the polarity of x to X in the diagram below and mention how many more amino acids are expected to be added to this polypeptide chain.

(b) Mention the codon and anticodon for alanine.

(c) Why are some untranslated sequences of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?

Solution:(a) x to X is 5′———– > 3′No more amino acids will be added

(b) GCA

 Anticodon is CGU

(c) The untranslated regions are required for an efficient translation process. 

They are present before the initiation codon at the 5’ – end and after the stop/termination codon, at the 3’ – end 

Question25.(a) How is Hardy-Weinberg’s expression “(p²+ 2pq+q²) = 1”derived?

(b) List any two factors that can disturb the genetic equilibrium.

Solution: (a) Sum Total of All the Allele Frequencies is 1: Let there be two alleles A and an in a population. The frequencies of alleles A and a are ‘p’ and ‘q’ respectively. The frequency of AA individuals in a population is p² and it can be explained that the probability that an allele A with a frequency of p would appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p². Similarly, the frequency of aa is q² and that of Aa is 2pq. 

p² + 2pq+q²  ) = 1, where p²  represents the frequency of the homozygous dominant genotype, 2pq represents the frequency of the heterozygous genotype and q²  represents the frequency of the homozygous recessive.

(b) Factors that affect Hardy–Weinberg equilibrium: (i) Gene migration or gene flow (ii) Genetic drift (iii) Mutation (iv)Genetic recombination (v) Natural Selection 

Question26. Highlight the structural importance of an antibody molecule with a diagram. Name the four types of antibodies found to give a humoral immune response, mentioning the functions of two of them you have studied.

Solution: An antibody molecule consists of four polypeptide chains, two are long called heavy (H) chains while the 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, Heavy Chain 

Types of Antibody –

IgA, IgM, IgE,IgG 

IgA – Lactating Mothers to protect their infant

Ig E – To protect from allergen

OR

(a) Explain the Life cycle of Plasmodium starting from its entry into the body of female Anopheles till the completion of its life cycle in humans.

(b) Explain the cause of periodic recurrence of chill and high fever during a malarial attack in humans.

Solution: (a) When a female Anopheles mosquito bites an infected person, the parasites enter the mosquito’s body as gametocytes. It leads to fertilization and development in the gut of the mosquito and undergoes further development to form sporozoites that are stored in salivary glands until their transfer to the human body. In the human body – the sporozoites reach the liver and reproduce asexually, bursting the cells and releasing them into the RBCs as gametocytes. (Labeled diagram explaining the mentioned stages can also be considered) 

(b) The rupture of RBCs releases a toxic substance called hemozoin, which is responsible for the chill and high fever

Question27.  Carefully observe the given picture. A mixture of DNA with fragments ranging from 200 base pairs to 2500 base pairs was electrophoresed on agarose gel with the following arrangement.  

(a) What result will be obtained on staining with ethidium bromide? Explain with reason.  

(b) The above setup was modified and a band with 250 base pairs was obtained at X. 

What change(s) were made to the previous design to obtain a band at X? Why did the band appear at position X? 

Solution:(a) No bands will be obtained as/All DNA will be seen in the well only; DNA fragments being negatively charged will not move towards the -ive end/ cathode. DNA that is negatively charged will remain stationed at the positive end/ anode end of the agar block; 

 (b) (a) Position of the positive terminal/ end/ anode and the negative terminal/ end/ cathode was inter-changed 

(ii) The fragment with the least base pairs will get separated faster and move faster to the anode end.

Question28.  (a) There was a loss of biodiversity in the ecosystem due to a new construction project in that area. What would be its impact on the ecosystem? State any three. (b) List any three major causes of loss of biodiversity. 

Solution: 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.

(b)(i) Habitat loss and fragmentation

(ii) Over-exploitation

(iii) Alien invasive species

(iv) Co-extinctions. 

Section – D

Question29.  Study the Pedigree chart given below and answer the questions that follow

(a) On the basis of the inheritance pattern exhibited in this pedigree chart, what conclusion can you draw about the pattern of inheritance? 

(b) If the female is homozygous for the affected trait in this pedigree chart, then what percentage of her sons will be affected?

(c) Give the genotype of offspring 1,2,3 and 4 in III generations. 

Solution:(a) X- linked , Recessive trait 

(b) 100% (1 Mark) 

(c) XY OR XY, 2. X X, 3. XY, 4. XX 

OR 

(c) In this type of inheritance pattern, out of male and female children one has less probability of receiving the trait from the parents. Give a reason

Solution: The possibility of the female getting the trait is less. (1 Mark) The female will get the trait only if the mother is at least a carrier and the father is affected.

Question30. The data below shows the concentration of nicotine smoked by a smoker taking 10 puffs/ minute. 

(a) With reference to the above graph explain the concentration of nicotine in blood at 10 minutes.

(b) How will this affect the concentration of carbon monoxide and homebound oxygen at 10 minutes?

(c) How does cigarette smoking result in high blood pressure and an increase in heart rate?

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

(b) The Concentration of CO will increase resulting in reduced.

(c)the concentration of haemboundoxygen.  Nicotine results in stimulating the adrenal gland which results in the release of adrenaline/noradrenaline in the blood resulting in an increase in blood pressure and heart rate. 

OR

(c) How does cigarette smoking result in lung cancer and emphysema? 

Solution: (c) Chemical carcinogens present in tobacco smoke are the major cause of lung cancer. Cigarette smoke irritates the air passages of the lungs causing them to produce mucus which causes cough resulting in enlarging air spaces/ reduce surface area/loss their elasticity (any point can be mentioned) thus difficulty in breathing causing emphysema. 

Section- E 

Question31. Trace the events from copulation to zygote formation in a human female.

Solution: 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 fertilization and pregnancy.

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

vi) During fertilization, 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 fertilize 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. 

OR

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

Trace the development of a megaspore mother cell to the formation of a 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 organized 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 the egg apparatus. Thus, a typical angiosperm embryo sac, at maturity, though 8-nucleate is 7-celled.

Solution:

Question32. Observe the segment of mRNA given below. 

(a) Explain and illustrate the steps involved to make fully processed hnRNA? 

(b) Gene encoding RNA Polymerase I and III have been affected by mutation in a cell. Explain its impact on the synthesis of a polypeptide, stating the reasons. 

Solution:(a) The hnRNA undergoes processes called capping and tailing followed by splicing. In capping, an unusual nucleotide is added to the 5¢-end of hnRNA methyl guanosine triphosphate. In tailing, adenylate residues (about 200–300) are added at 3¢-end in a template-independent manner. Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA called splicing.

(b) The process of translation will not happen, thus the polypeptide synthesis is stopped/ hampered. 

The reason for the above is: RNA polymerase I transcribe rRNAs which is the cellular factory for protein synthesis. 

RNA polymerase III helps in the transcription of tRNA which is the adaptor molecule/ that transfers amino acids to the site of protein synthesis. 

OR 

Study the schematic representation of the genes involved in the lac operon given below and answer the questions that follow

(a) The active site of enzyme permease present in the cell membrane of a bacterium has been blocked by an inhibitor, how will it affect the lac operon?

(b) The protein produced by the I gene has become abnormal due to unknown reasons. Explain its impact on lactose metabolism stating the reason.

(c) If the nutrient medium for the bacteria contains only galactose; will operon be expressed? Justify your answer. 

Solution: (a) When the active site of enzyme permease present in the cell membrane of a bacterium has been blocked by an inhibitor, the lactose is not transported into the cell. As lactose is the inducer, the lac operon will not be switched on. 

(b) Since the repressor protein synthesized by the I gene is abnormal, it will not bind to the operator region of the operon (1 Mark), resulting in a continuous state of the transcription process.

 (c) No because galactose is not an inducer/ it is a product of lactose metabolism 

Question33. The oil spill is a major environmental issue. It has been found that different strains of Pseudomonas bacteria have genes to break down the four major groups of hydrocarbons in oil. Trials are underway to use different biotechnological tools to incorporate these genes and create a genetically engineered strain of Pseudomonas – a ‘super-bug’, to break down the four major groups of hydrocarbons in oil. Such bacteria might be sprayed onto surfaces polluted with oil to clean thin films of oil.  

(a)  List two advantages of using bacteria for such biotechnological studies? 

(b) For amplification of the gene of interest PCR was carried out. The PCR was run with the help of polymerase which was functional only at a very low temperature. How will this impact the efficiency of the PCR? Justify. 

(c) If such bacteria are sprayed on water bodies with oil spills, how will this have a positive or negative effect on the environment? Discuss. 

Solution:(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. If the polymerase enzyme denatures at low temp, it will not be able to

withstand high temperatures which is essential for separating/opening/unwinding/ denaturing DNA strands to open. Thus subsequent step of extending the primers using the nucleotides provided in the reaction and the genomic DNA as the template will not occur.

(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 seagulls. Negative effect: the bacteria can mutate/ can harm other organisms/ can conjugate with other non-virulent forms and make them superbugs with detrimental effects/ unpredictable/ for a longer duration it may reduce the dissolved oxygen and lead to mortality of aquatic organisms

OR 

Insects in the Lepidopteran group lay eggs on maize crops. The larvae on hatching feed on maize leaf and tender cob. In order to arrest the spread of three such Lepidopteran pests, Bt maize crops were introduced in an experimental field. A study was carried out to see which of the three species of lepidopteran pests was most susceptible to Bt genes and its product. The lepidopteran pests were allowed to feed on the same Bt-maize crops grown on 5 fields (A-E). The graph below shows the leaf area damaged by these three pests after feeding on maize leaves for five days. 

Insect gut pH was recorded as 10, 8, and 6 respectively for Species I, II, and III respectively. 

(a) Evaluate the efficacy of the Bt crop on the feeding habits of the three species of stem borer and suggest which species is least susceptible to Bt toxin. 

(b) Which species is most susceptible to Bt-maize, explain why? (c) Using the given information, suggest why a similar effect was not seen in the three insect species. 

Solution:(a) Species III is the least susceptible 

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

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

Biology Sample Paper Class 12 Term 1

Cbse Board Exam 2022: The Cbse Class 12 Term 1 Board Examinations For Minor Papers Will Commence On November 16. Moreover, The Cbse Will Begin Conducting Board Exams For Main Papers On November 30.

According To The Schedule, The Cbse Class 12 Biology Term 1 Board Test Will Be Held On December 18.

The Cbse Term 1 Test Will Be 90 Minutes Long And Mcq-based For Each Topic. Students Will Have 20 Minutes Instead Of 15 Minutes To Read The Paper.

For Medical Students, Biology Is The Most Significant Subject. The Cbse Class 12 Biology Curriculum Is Broad, And Students May Find It Difficult To Finish The Whole Curriculum While Also Making Time For Practise And Review. This Can Only Be Done If Their Education Is Well Planned. These Preparation Suggestions Are Intended To Assist Students In Preparing For Board Examinations By Ensuring That Students Have Conceptual Clarity In All Areas And May Get The Highest Possible Mark On The Test.

First And Foremost, It Is Vital To Grasp The Arrangement Of The Cbse Class 12 Biology Exam. Every Year, The Board Releases The Pattern To Help Students Get Acquainted With It And Appropriately Prepare. Understanding The Pattern Might Also Aid In Time Management Throughout The Test.

Despite The Fact That The Board Exam Gives Options, The Majority Of The Choice-based Questions Come From The Same Topic.

As A Consequence, Students Must Prepare The Whole Curriculum Completely And Be Confident In Their Ability To Answer Any Question On The Test. Since The Questions Are Selected From The Curriculum, Students Should Confine Their Board Exam Preparation To The Syllabus.

Biology Is Both A Theoretical And A Conceptual Discipline. Students Must Understand Critical Words. In Addition To Terms And Definitions, The Cbse Sample Exam Class 12 Biology 2022-2023 Offers A Number Of Diagram-related Problems.

Students May Utilise Flashcards Before The Test To Assist Them Recall And Refresh Important Language And Pictures. Students Must Also Practise Drawing The Schematics Multiple Times So That They Can Draw It Rapidly In The Test.

Students Should Answer Previous Year’s Exams And Cbse Sample Paper Class 12 Biology 2022-2023 To Understand The Kind And Structure Of Questions. Solving Multiple Questions Will Offer Students With Experience For Dealing With Any Kind Of Question On The Test. Students May Also Be Able To Finish The Task Within The Time Span Specified. Students May Utilise A Cbse Sample Paper Class 12 Biology 2022-2023 To Assess Their Readiness For The Actual Exam. They Discover Their Own Talents And Shortcomings. Students Who Work On Them Will Do Better On The Test.

The Last Days Before The Board Test Should Be Spent Mostly Revising. Continuous Review Helps To Recall All Of The Important Subjects, And One May Successfully Remember The Ideas During The Test. Therefore, It Is The Time To Go Through The Text And Diagrams Completely. It’s Also A Good Idea To Devote Special Attention To The More Important Issues.

Here Are Some Of The Biology Test Preparation Tips. In Addition To These Preparation Suggestions, Many Effective Test-taking Strategies Are Offered Below To Help With Last-minute Preparation For The Class 12 Biology Exam.

Biology Is A Topic That Has A Particular Place In The Hearts Of Students Because Of The Fascinating Facts And Details They Study. The Pupils Are Interested In Learning About The Globe As Well As Our Body. As A Result, In Order To Assist Students In Developing A Precise Knowledge And Earning High Scores In Tests, Vedantu Ensures That The Students Are Provided With The Greatest Study Resources And Advice Available. Vedantu’s Sample Questions Are Also Quite Useful And Enlightening For Students Studying For Board Exams.

The Biology Example Questions Are Good Practise Material For Pupils. They May Go Through The Curriculum And Determine Which Areas Need Their Attention. The Sample Questions Will Also Assist Students Understand How The Questions Will Look On The Real Test. Students May Simply Download This From Vedantu’s Website And Save It In Pdf Format For Future Use. Let Us Have A Look At The Key Concepts From The Chapters That Will Be Relevant For The Tests.

CBSE Term 1 board exam 2022 Biology sample paper: Details

Time: 90 minutes

General instructions:

1. The Question Paper contains three sections.
2. Section A has 24 questions. Attempt any 20 questions.
3. Section B has 24 questions. Attempt any 20 questions.
4. Section C has 12 questions. Attempt any 10 questions.
5. All questions carry equal marks.
6. There will be no negative markings.

ANSWER PRESENTATION STRATEGIES:

• Use a sharp pencil to write properly and make diagrams.

• Label the diagram with a ruler, and label the diagrams on the right side of the paper if possible.

• Highlight crucial words and phrases in the response.

• New parts should begin on a new page.

• Even if the question does not specifically ask for diagrams, include them.

• Students should review the Solution PDF for previous year Biology examinations, which has been issued by CBSE. This will aid students in comprehending response writing skills and the procedure of assigning marks in a gradual manner.

TIME MANAGEMENT STRATEGIES:

• Finish section A in 25 minutes due to the short answer type questions.

• Section B should take 40 minutes to complete because there are seven questions.

• Section C has 12 questions and should be completed in under an hour.

• Section D consists of long-answer questions, and it should take no more than 45 minutes to complete.

• Use the remaining 10 to 15 minutes to revise and double-check that all of the questions have been answered correctly.

Steps to download CBSE Class 12 Biology Sample Paper 2023

Download the Sample Paper Class 12 Biology Exam 2023 with solutions for the exam by following the below-mentioned steps :

Step I- Visit the official website of CBSE Academic @ www.cbseacademic.nic.in or Click on the CBSE Class 12 Biology Sample Paper with Solutions for the Exam mentioned above.  

Step II- Click on the notification appearing in the academic section- “Sample Question Papers of Classes XII Exams 2022-23”.

Step III- Click on the link under “Sample Papers Class XII”.

Step IV- The list of all subjects “Class XII Sample Question Paper & Marking Scheme for Exam 2022-23” appears on the screen.  

Step V- Click on “SQP” for “Biology” and download CBSE Class 12 Physics Sample Paper 2022-23 pdf along with the Solution.  

Step VI- Check the marking scheme after attempting each subject CBSE Class 12 Biology Sample Paper 2022-23.