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CBSE Class 12 Biology Additional Practice Questions 2023

CBSE Class 12 Biology Additional Practice Questions

As CBSE Exam dates are coming close the Central Board of Secondary Education has released the CBSE Class 12 Biology Practice Questions 2023 on its official website at The candidates appearing in the CBSE Exam 2023 must solve the CBSE Class 12 Biology practice Questions 2023 given on this page to check their preparation for CBSE Exam 2023. 

CBSE Class 12 Additional Practice Question Paper 2023
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CBSE Class 12 Biology Additional Practice Questions 2023

The Central Board of Secondary Education is all set to conduct  CBSE Exam 2023 for Class 12. The  CBSE Exam 2023 for class 12 will begin in April 2023 with the Entrepreneurship, Beauty, and Wellness subjects and will be concluded on June 2023 with Psychology subject. The board will conduct CBSE Exam 2023 for Biology subject on 30th May 2023. The students appearing in CBSE Exam 2023 must bookmark this page to get all the latest updates from CBSE. 

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CBSE Class 12 Biology Additional Practice Questions 2023: Exam Pattern 

Here we have given the exam pattern of CBSE Class 12 Biology as per the sample paper released by the board. There will be three sections and 13 questions in the Class 12 Biology subject. The details of the following sections are given below: 

Section A: In this section, 6 questions of 2 marks each will be asked. 

Section B: In this section, 6 questions of 3 marks each will be asked. 

Section C: This section consists of case-based questions. Each question carries 5 marks. 

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CBSE Class 12 Biology Additional Practice Questions 2023: General Instructions


Max. Marks: 70

Time Allowed: 3 hours

General Instructions:
(i) All questions are compulsory.
(ii) The question paper has five sections and 33 questions. All questions are compulsory.
(iii) 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.
(iv) There is no overall choice. However, internal choices have been provided in some questions.
A student has to attempt only one of the alternatives in such questions.
(v) Wherever necessary, neat and properly labelled diagrams should be drawn.

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CBSE Class 12 Biology Additional Practice Questions 2023: SECTION A

Q1. Birth control tablets in females, popularly referred to as pills, prevent pregnancy by __________.
A. delaying menstruation
B. inhibiting ovulation and implantation
C. suppressing sperm motility and fertility
D. blocking the entry of sperms during coitus

Q2. Which of the following statements is/are correct about ZIFT and GIFT as methods of helping conception in cases of infertility?
P) ZIFT can help where the female is unable to form a viable ovum.
Q) ZIFT uses methods of in vitro fertilisation.
R) GIFT involves the injection of one’s own ovum into the body.
S) GIFT uses in vivo fertilisation method.
A. only P
B. only P and R
C. only Q, R and S
D. all – P, Q, R and S

Q3. A DNA sequence consisted of 20% adenine nucleotides. What would be the percentage of cytosine nucleotides in the same DNA sequence?
A. 20%
B. 30%
C. 60%
D. 80%

Q4. Comparative anatomy and morphology studies deepened the understanding of evolution. The presence of analogous and homologous structures provides important evidence in the favour of evolution. Which of the following is/are examples of HOMOLOGOUS structures found in plants?
A. only Q
B. only R
C. only P and Q
D. only Q and R

Q5. In certain diseased conditions such as pneumonia, the fingernails of an individual turn blue. What could be the reason for this?
A. blood does not reach fingernails due to an increase in fat content around the
B. nails become cold due to a decrease in fat content around the fingernails
C. reduced levels of oxygen in the blood
D. increase in oxygen levels in the blood

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Q6. In which of the following diseases is/are the parasites transmitted to a healthy individual through the bite of a female mosquito?
P) malaria
Q) ascariasis
R) filariasis
A. only P
B. only P and Q
C. only P and R
D. all – P, Q and R


Q7. A substrate is the surface on which an organism lives or survives on. Which of the following acts as the substrate that provides energy in a detritus food chain?
A. sunlight
B. green plants
C. decomposers
D. dead organic matter

Q8. Given below are two statements about polymerase chain reactions.
P) It mimics DNA replication that happens in a cell.
Q) It cannot be used to amplify RNA molecules.
Which of these is/are TRUE?
A. only P
B. only Q
C. both P and Q
D. neither P nor Q

Q9. Given below is a food web representative of the Arctic region. Increasing temperatures have been causing changes in the ocean ecosystem. These changes have caused the population of Arctic cod to decline rapidly.
Which of the following statement/s is/are most likely to be TRUE based on this information?
P) The population of arctic birds will increase.
Q) The ringed seal will slowly become extinct.
R) The harbour seal will be dependent on capelins alone.
A. only P
B. only R
C. only Q and R
D. all – P, Q and R

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Q10. In which of the following is competition MOST LIKELY to occur?
P) related species in the same environment
Q) related species in different environments
R) unrelated species in the same environment
S) unrelated species in different environments
A. only P and Q
B. only P and R
C. only Q and S
D. only P, Q and R

Q11. Which of the following is most likely to be true about the percentage of energy received by a horse and a crow from the producers in different food chains?
Horse Crow
P always the same always the same
Q always the same can be different
R can be different always the same
S can be different can be different
A. P
B. Q
C. R
D. S

Q12. A stable community is usually resistant to invasion by alien species. Which of the following would NOT be affected in a stable community due to this resistance?
A. species richness
B. productivity
C. co-extinction
D. total biomass

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.

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Q13. Assertion (A): Apomixis and parthenocarpy are both asexual modes of reproduction.
Reason (R): Seeds are not produced in both apomixis and parthenocarpy.


Q14. Assertion (A): A colour-blind father will always have a colour-blind son.
Reason (R): Genes causing colour blindness are passed through a sex chromosome.


Q15. Assertion (A): Gene therapy is a method of treating a disorder but cannot cure it.
Reason (R): Cells are drawn from a patient and the functional gene is introduced into these cells and transferred back to the patient.

Q16. Assertion (A): In the absence of a predator, the prey population growth will always be exponential.
Reason (R): Exponential growth is when the resources and the environment allow an organism to realise fully its innate potential to grow in numbers.

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CBSE Class 12 Biology Additional Practice Questions 2023: SECTION B

Q17. Given below is the karyotype of an individual.
(a) What are the characteristic reproductive and physical features of such an individual?
(b) What is the category of such disorders called? How is it caused?

Q18. Nidhi performed gel electrophoresis after treating one vector with restriction enzymes. She added one mixture in well Q and another mixture in well R. Given below is an image of the results.
(a) What can be concluded about the mixtures loaded in wells P and Q?
(b) What is the likely reason that the fragments in wells Q and R are different?


Q19. In a patient, a mass of cells removed from the liver was found to be producing large amounts of the enzyme pepsin. In the same patient, a tumor was found in the stomach.
(a) What property of a tumor can be identified based on the statements above? Give a reason to support your answer.
(b) What are tumors exhibiting the property identified in (a) called?
(c) How will the tumors identified in (b) affect liver cells?

Q20. As reported by numerous medical sources, Reema Sandhu, is an account manager, lives in Bracknell, Berkshire, with husband and young son. She was diagnosed with multiple sclerosis in November 2015 after burning her face on a lamp. Multiple sclerosis is the most common demyelinating disease in which the insulating covers of nerve cells in the brain and spinal cord are damaged. This damage disrupts the ability of parts of the nervous system to transmit signals. As per reports, she regained much of her brain function including her vision through stem cell therapy.
(a) Which part of Reema’s body could these stem cells have been sourced from?
(b) Why would stem cell therapy have helped Reema?

Q21. Biomass is expressed in terms of dry weight and / or fresh weight. Which of the weights is more accurate as a unit of standing crop? Justify your answer with a reason.


Marshy areas often consist of hard outer coverings as detritus that are remains of organisms such as crabs. State TWO reasons why decomposition would be limited in such areas.

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CBSE Class 12 Biology Additional Practice Questions 2023: SECTION C

Q22. (a) Explain why non-occurrence of menstrual cycle could be indicative of pregnancy.
(b) The menstrual cycle can be divided into 4 phases: menstrual phase, follicular phase, ovulation phase, luteal phase. During the follicular phase, hormones like FSH and LH are released in good amounts. State TWO events that are triggered by LH.


Q23. Outbreeding helps in the maintenance of an organism’s ability to survive and perpetuate its genetic material. This is termed as biological fitness.
(a) What is the term used to signify reduction of such biological fitness?
(b) Explain one method of outbreeding devised by plants that requires a chemical intervention by the reproductive apparatus of a plant.

Q24. Hershey and Chase performed several experiments to find the chemical nature of the genetic material that is present in all organisms. The graph below shows the results of one such experiment. It tracks the amount of 32P and 35S found in the supernatant after the bacterial cell suspension was agitated in a blender. The Y-axis represents the percentage of radioactivity from 32P and 35S each as compared to all radioactivity detected in the supernatant.
[Source: Hershey A.D., Chase M. Independent functions of viral protein and nucleic acid in growth of bacteriophage. J Gen Physiol. 1952 May;36(1):39-56.]
(a) What did Hershey and Chase want to verify using this experiment?
(b) What do curves X and Y represent? Give a reason to support your answer.

Q25. In a population of 1000 individuals, 25% of individuals show the phenotype for sickle cell anaemia (genotype – ss).
(a) Assuming the population meets Hardy-Weinberg equilibrium, how many individuals would be carriers of the sickle cell allele but will not show the sickle cell phenotype?
(b) Can the Hardy-Weinberg principle be used to predict the frequency of the presence of the sickle cell allele in a sperm cell? Why or why not?

Q26. AIDS is a disease caused by the Human Immunodeficiency Virus (HIV) and, over time, this causes an individual to become immuno-deficient. The virus attaches itself to an animal cell where the viral genome replicates and produces more virus particles.
(a) Drugs that exist to treat AIDS are only partially effective. What process, after a virus has infected an animal cell, are these drugs most likely to target? Give a reason to support your answer.
(b) Why is the integration of the viral genome with the host genome important for the virus to form new virus particles?
(c) ELISA is a test commonly used in the detection of an HIV infection. State one situation in which a false negative result can be obtained.


The primary effluent in the treatment of sewage is sent to tanks for secondary treatment in the presence of aerobic bacteria.
(a) How would the BOD of the effluent be affected if anaerobic bacteria are used for secondary treatment?
(b) Name one condition that should be maintained in a sludge digester where biogas is produced.
(c) The slurry formed after biogas production is recommended as manure for plants. Which nutrients will the slurry be rich in and why?

Q27. Given below is the step-by-step process in the formation of yoghurt (curd) in a bioreactor.
(a) Why does the pH start decreasing a while after the mixture is incubated at 37-44°C?
(b) From the flowchart, identify two systems that the bioreactor would have. Give a reason to support your answer.


Q28. Latitudinal gradients have an impact on species diversity. While species diversity is highest at the tropics and lowest at the poles, loss of biodiversity also is highest in the tropics and lowest at the poles.
(a) Mention ONE possible reason for the low species diversity at the poles.
(b) Mention ONE possible reason for loss of biodiversity being higher in the tropics.
(c) How have humans used temperature conditions to conserve biodiversity in ex situ conditions?

CBSE Class 12 Biology Additional Practice Questions 2023: SECTION D 29 and 30 are case based questions. Each question has subparts with internal choice in one subpart.

Q29.  In maize, the trait for the purple kernel (P) is dominant over the yellow kernel (p). A plant with purple kernels is crossed with another plant with yellow kernels and produces 2 offspring with purple kernels and 2 offspring with yellow kernels.
(a) What is the genotype of the parental maize plants?
(b) Draw a Punnett square to depict the cross between the two offspring with purple kernels.
(c) Identify the genotypic and phenotypic ratios obtained from the cross in (b).


(c) Describe a method that can definitely help with the identification of an unknown genotype of a plant with purple kernels.

Q30. In the late 18th century, smallpox was a widely spreading disease causing the death of several affected individuals in Britain. Edward Jenner, who pioneered the concept of vaccination, inoculated matter from the cowpox lesions of a dairymaid into an 8-year-old boy. Post-inoculation, the boy developed a mild fever, loss of appetite and discomfort but was better after a few days. Next, he was inoculated with matter from a smallpox lesion and he did not develop any disease.
(a) What form of immunity, now known, did Edward Jenner provide the boy with? Give a reason to support your answer.
(b) Describe the form/s of immunity that is provided when an individual is vaccinated/immunised? Use an appropriate example/s to justify your answer.


(b) Which form of immunization does not generate a memory response? Give a reason to support your answer.

CBSE Class 12 Biology Additional Practice Questions 2023: SECTION E

Q31. Papaya is a widely cultivated crop in several regions. However, its production was limited by papaya ringspot disease which is caused by the Papaya ringspot virus (PRSV). Papaya plants infected by PRSV show symptoms of yellowing, discolouration of leaves and ‘ringspots’ on the fruit. PRSV belongs to the genus Potyvirus which has a single-stranded RNA as its genetic material.
(a) Explain the step-by-step process to inhibit the viral RNA from surviving in the papaya plants thus creating disease-resistance varieties of papaya.
(b) Name the biotechnological process described in (a) and give a reason why it is the appropriate process to be used in this case.


Growth hormone injection treatment is prescribed for children who have been diagnosed with growth hormone (GH) deficiency and other conditions causing short stature and insufficient growth. This hormone is produced by the pituitary gland in humans so the gene for this hormone was isolated from the pituitary gland and introduced into phGH407 vectors for production. However, a problem with this was that the protein so produced was 26 amino acids longer than the active growth hormone (24 amino acids long) and so this method could not be used.
(a) Given that the amino acid sequence of the active growth hormone was known, use a diagram to explain how human growth hormone could be produced outside the body.
(b) The vector consists of a lac gene which codes for the enzyme β-galactosidase. Describe how this gene can help with the selection of colonies containing the transgene.


Q32. Given below is a DNA sequence and the genetic code. Answer the questions based on these, assuming no post-transcriptional or post-translations modifications will take place.
(a) Write the nucleotide sequence that will be obtained on transcription of this DNA sequence.
(b) Will translation of this sequence take place? Give a reason to support your answer.
(c) What is the amino acid sequence that will be formed? Identify the sequence of the first tRNA.
(d) If the first guanine base in the DNA sequence gets replaced by thymine, how will the amino acid sequence change?
(e) Name and describe the mutation that occurred in (d).


The image below shows the DNA profile of four men, a mother and her child.
(a) Which man is most probably the father of the child? Give a reason to support your answer.
(b) Which technique, commonly used in forensic studies such as paternal testing, is depicted in the image?
(c) What is the basic principle that the technique identified in (b) is based on?
(d) What is the most likely relationship, if any, between men Q and S? Justify your answer.


Q33. Angiosperms such as pea plants undergo double fertilisation. The male gametophyte has a simple structure while the female gametophyte has a much more complex structure with multiple supporting cells in it. (a) What is/are the product/s of double fertilisation?
(b) If you are given a pea pod, how can you identify the product/s of double fertilisation in it?
(c) How does a structural difference help each gametophyte perform its functions better?


Gametogenesis is the process of production of gametes. In males, it is spermatogenesis and in females it is oogenesis. The cells in the germline that undergo meiosis, primary spermatocytes or primary oocytes, are derived from the zygote by a long series of mitosis before the onset of the two meiotic cycles to form the mature gametes. Testosterone is an androgen that plays an important role in the formation and release of sperm from the seminiferous tubules.
(a) What is the count of chromosomes after the first and second meiotic divisions in the formation of sperms? Give a reason to support your answer.
(b) In an individual with low testosterone levels –
(i) which process in spermatogenesis is likely to not happen?
(ii) if the semen sample of such an individual is collected, what is likely to be observed?
(c) What is likely to happen to the polar bodies formed after each meiotic cycle in oogenesis? Give a reason to support your answer.


CBSE Class 12 Biology Additional Practice Questions 2023 pdf download link

Below is the CBSE Class 12 Biology Additional Practice Questions 2023 pdf download link:

CBSE Class 12 Biology Additional Practice Questions 2023 pdf download

Class 12 Biology Additional Practice Questions: Previous Year


1. Malaria, caused by Plasmodium is often characterized by a high fever recurring every three to four days. Explain why this happens.

Ans. Malaria caused by Plasmodium is often characterized by a high fever recurring every three to four days, this happens because the parasite enters the bloodstream and infects the red blood cells. the parasite multiplies inside the red blood cells causing the cell to break within 48 to 72 hours, which infects more red blood cells.

2. Rural energy shortage has been largely mitigated by biogas plants.
(a) What is the primary flammable gas present in biogas?
(b) Why is cattle dung used to produce biogas?

Ans. (a) Methane is the primary flammable gas present in biogas.

(b) Biogas is produced from cattle dung because it lowers the amount of waste and lowers the amount of methane released into the atmosphere by covering it with carbon dioxide. carbon dioxide is 25 percent less damaging than methane.


Nicotine stimulates the adrenal gland and increases blood pressure and heart rate.
(a) Despite higher heart rate leading to faster pumping of the blood, why would a smoker show oxygen deficiency in the body?
(b) Why does a chronic smoker experience withdrawal symptoms if he abruptly discontinues smoking?

Ans. (a) The rise in blood pressure is due both to an increase in cardiac output and total peripheral vascular resistance. The blood pressure rise appears immediately and occurs before any increase in circulating catecholamines. It is a paradox that while smoking acutely increases blood pressure, a slightly lower blood pressure level has been found among smokers than nonsmokers in larger epidemiological studies. Because blood pressure may increase after the cessation of smoking, a smoke quitting program should not postpone the initiation of antihypertensive treatment in patients otherwise in need of such treatment.

(b) A study of 496 adult marijuana smokers who tried to quit found that 95.5% of them experienced at least one withdrawal symptom while 43.1% experienced more than one symptom. The number of symptoms the participants experienced was significantly linked to how often and how much the subjects smoked before trying to quit.

3. The population growth can be exponential (unimpeded) or logistic (competitive). These growth curves can be represented by the graphs as shown below:

CBSE Class 12 Biology Additional Practice Questions 2023_130.1

(a) Define carrying capacity.

Ans. The carrying capacity of an environment is the maximum population size of a biological species that can be sustained by that specific environment, given the food, habitat, water, and other resources available.
(b) What limits the growth of organisms as seen in the graphs above?

Ans. The logistic growth limits the growth of the organisms. the growth in population limits the resources.

4. While studying the human population data of a particular geographical area it was noticed that the area has an age pyramid as shown below.

CBSE Class 12 Biology Additional Practice Questions 2023_140.1
(a) What kind of natality and mortality rate of the pre-reproductive population contributes to the kind of shorter base of the age pyramid as shown above?

(b) Why do we usually have an age pyramid with a broader base?

Ans. The reason the pyramid is broader at the bottom and narrow at the apex in context with food components: Each food group is depicted by a band or level. Narrow bands at the apex signify lower quantities, while wider bands at the base mean that more from that food group needs to be used up.

5. For an individual ‘X’ with a history of lung cancer in the parents, doctors advised certain genetic testing processes that help in detecting the inheritance of mutations.
(a) Malignant tumors spread rapidly and avoid detection. α-interferon is a biological response modifier and can target specific disease-causing mechanisms.
How does α-interferon help in the treatment of malignant tumors?
(b) For cancer caused by the inheritance of genetic mutations, how will the malignancy spread internally?

6. Eichhornia crassipes (commonly known as water hyacinth) is an aquatic plant, native to the Amazon basin. It was introduced to the water bodies of India and other SouthEast Asian countries for industrial use of its fibers in making bags and footwear, as a substrate for biogas production, and for its ability to uptake heavy metals from the water bodies. However, Eicchornia has been named the ‘terror of Bengal’ due to its prolific, invasive growth.

(a) How does the growth of water hyacinth affect the growth of other native species?
(b) There are a host of algae and fungi that form lichens in freshwater lakes. What would be the fate of freshwater snails that feed on such lichens if the algae and fungi are destroyed by the growth of water hyacinth Mention the scientific term used to denote such threats to biodiversity.

Conservation of biodiversity is vital to maintaining the balance in the ecosystem.
(a) Mention ONE condition when ex-situ conservation is a more viable process to conserve a species.
(b) What are the TWO most important parameters that help in naming an area as a biodiversity hotspot?

7. A pest control program needed to be developed for the Sunderbans mangrove ecosystem and biocontrol methods were being explored.
(a) Why would biocontrol methods be the best method for a pest control program in a biome like Sunderbans?

Ans. In a biome like Sundarbans where the reaches of creation species are more the chemical pest control pesticides may affect the quality of water bodies and may interfere in the ecosystem and may disturb the balance of the ecosystem, the biocontrol methods especially target the best organisms and will completely eradicate the organism may keep the best population in proper control by checks and balances
(b) What is the difference in the action of biocontrol agents against the action of chemical pesticides?

Ans. Biocontrol can be defined as the control of one type of living organism with the application of another. The use of Biocontrol agents is also called biological control and is mainly used for the reduction of pest population and to produce yields that are free of any pests. The biocontrol methods are a long-term method and most importantly, a self-sustaining one that helps in the control of invasive species of plants. Weed infestation and pests are controlled with the help of insects, pathogens, and grazing animals. The natural animals in the form of parasitism, predation along with other similar mechanisms perform the role of biocontrol agents which help in maintaining the balance.
(c) Mention any two points of vital information about the pests that the designers of the pest control program need to be aware of?


Lymphocytes are an integral part of our immune system and help in the humoral and cell-mediated immune response process.
(a) Specify the type of lymphocytes that mediate humoral immune response and the ones that mediate cell-mediated immunity.

Ans. B cells are involved in humoral immunity and T cells are the primary mediator of cell-mediated immunity
(b) What is the chemical nature of antibodies?

Ans. Antibodies are proteins in nature. Antibodies are usually beneficial to the host in that they increase resistance toward a subsequent challenge by antigens.
(c) How is an organ rejected by the body due to an ‘unmatched’ transplant?

Ans. Hyperacute rejection occurs a few minutes after the transplant when the antigens are completely unmatched. The tissue must be removed right away so the recipient does not die. This type of rejection is seen when a recipient is given the wrong type of blood. For example, when a person has been given type A blood when he or she is type B.

8. The image below represents the replication of a retrovirus. In the image, steps 1- 5 depict different stages in the invasion of the retrovirus into the host cell, and steps 6 – 9 show the invasion of the host DNA and the processes resulting from it.

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(a) Why does the retrovirus need to use reverse transcriptase to infect the host genome?

Ans. Retroviruses also have the enzyme reverse transcriptase, which allows them to copy RNA into DNA and use that DNA “copy” to infect human, or host, cells. When HIV infects a cell, it first attaches to and fuses with the host cell. The new copies of HIV then leave the host cell and move on to infect other cells.

(b) What is the significance of steps 7 and 8 (after the viral genome enters the host nucleus) as shown in the diagram? 

Ans.  New viral RNA is used as genomic RNA to make viral proteins
Viral DNA is transported across the nucleus and integrated into the host DNA

9. Diversity is seen in the living world at various levels. The distribution of biodiversity shows specific patterns that account for the species richness or paucity across the globe.

(a) Explain, with reasons, how species diversity changes with changing latitudes.

Ans. Diversity increases as you move away from the polar regions and towards the equator. This idea is also referred to as the latitudinal diversity gradient, meaning that as you move from the equator towards the poles, diversity lessens.

(b) The graph below represents the species-area relationship. What does such a graph signify?

CBSE Class 12 Biology Additional Practice Questions 2023_160.1

Ans. The graph shows that within a region, species richness increased with an increase in the explored area but only up to a limit.

10. In recombinant DNA technology, the endonuclease cuts the DNA into fragments.

(a) Where does a restriction endonuclease cut the DNA strand?

Ans. Restriction endonuclease enzyme cut the strand of DNA from the center of palindromic sequence these sequences have same bases two DNA strand from 5 prime to 3 Prime.

(b) How is the restriction endonuclease able to cut the DNA strand as mentioned in (a)?

Ans. By recognizing a specific sequence, restriction enzymes cut DNA molecules at a precise location. Each restriction endonuclease examines the length of a DNA sequence. It will connect to the DNA and cut each of the two strands of the double helix at precise spots in its sugar-phosphate backbone after it has found its specific recognition sequence.

(c) DNA fragments formed by the action of endonuclease can be separated by gel electrophoresis. What is the principle on which gel electrophoresis works? 

Ans. Gel electrophoresis is a technique used to separate macromolecules such as DNA, RNA, and proteins. Both DNA and RNA molecules are separated based on their size while proteins are separated based on both size and charge. Agarose gel electrophoresis is the technique used to separate both DNA and RNA. From 100 bp to 25 kb DNA fragments can be separated by agarose gel electrophoresis. Generally, DNA is a positively-charged molecule since they possess negative charges in its phosphate groups. Thus, DNA migrates towards the positive electrode during gel electrophoresis.

11. Polymerase Chain Reaction (PCR) is often used to detect the presence of pathogens like bacteria and viruses, even though they may be in low concentration.

(a) What is the technique that allows PCR to detect low concentrations of pathogens?


In a diagnostic PCR test, the machine can detect the presence of a pathogen after replicating the genetic material. However, certain viruses, including SARS-CoV-2, consist of RNA rather than DNA. For these viruses, the RNA undergoes a process called reverse transcription PCR (rtPCR)Trusted Source. This turns the RNA into DNA before copying it. The time it takes to get results from a PCR test can vary from a few minutes to several days. With an onsite analyzer, the results are rapid. It can take longer for results to come back when doctors send samples to an off-site lab, due to processing delays.

(b) The image below shows the steps involved in a PCR. Step 1 is denaturation, step 2 is annealing, and step 3 is an extension. During a PCR, if step 2 is bypassed, what would be the implication of the process? 

CBSE Class 12 Biology Additional Practice Questions 2023_170.1

Ans. PCR occurs in three steps:

  1. Denaturation: It maintains a temperature at 94°C. The high temperature melts the hydrogen bonds and separates the two strands of DNA.
  2. Annealing: It occurs at 55°C. The two oligonucleotide primers anneal with separate strands.
  3. Extension: It carries out at 72°C. Finally, DNA polymerase from the complementary strand.

Hence if the “Annealing” step is missed the primers will not attach. DNA polymerase cannot function without a primer and therefore DNA will not extend.

12. Populations of organisms respond differently to the abiotic factors of the environment. Organisms that can maintain physiological homeostasis ensuring constant body temperature are called regulators while others that cannot, are termed conformers. Crocodiles are often seen basking in the sun more during certain periods of the year.

CBSE Class 12 Biology Additional Practice Questions 2023_180.1

(a) Identify, with a reason, if the species described above is an environmental conformer or regulator.

Ans. Crocodiles can not maintain body temperature, as they are cold-blooded, and therefore they are environmental conformers.
(b) Crocodiles are usually sluggish and show minimum movement. How does such behavior aid in the conservation of body heat in a crocodile?

Ans. Crocodilians are ectotherms, producing relatively little heat internally and relying on external sources to raise their body temperatures. The sun’s heat is the main means of warming for any crocodilian, while water immersion may either raise its temperature by conduction or cool the animal in hot weather. The main method for regulating its temperature is behavioral. For example, an alligator in temperate regions may start the day by basking in the sun on land. A bulky animal warms up slowly, but at some time later in the day, it moves into the water, still exposing its dorsal surface to the sun. At night it remains submerged, and its temperature slowly falls. The basking period is extended in winter and reduced in summer. For crocodiles in the tropics, avoiding overheating is generally the main problem. They may bask briefly in the morning but then move into the shade, remaining there for the rest of the day, or submerge themselves in water to keep cool. Gaping with the mouth can provide cooling by evaporation from the mouth lining
(c) If the geographical location of a population of crocodiles was to receive harsh winter conditions, what could be the TWO most viable survival methods for the animals?



Consider the information given below to answer the questions that follow: When scientists look at improving food production and crop value, they have looked at the possibility of using arid conditions for increasing available farmlands. Plants growing in low water content areas can have three modes of adaptation: – drought escape: where the plants complete the life cycle before the dry season comes in – drought avoidance: where the plants naturally have adaptations like reduced leaves, lower stomatal presence to reduce water loss – drought tolerance: where the plants inherently have low water requirement and can grow in dry conditions When any of these is genetically incorporated in a plant that otherwise grows in moderate water availability, it may impact physiological processes like mobilization and storage of minerals, maturation of flowers and fruits, etc. The National Academies Press published a report titled Transgenic Plants and World Agriculture (2000). In the chapter named Examples of GM crops that can benefit World Agriculture, the report speaks about techniques for developing pest resistance in GM crops and its advantages. “There is a benefit to farmers if transgenic plants are developed that are resistant to a specific pest. For example, papaya-ringspot-virus-resistant papaya has been commercialized and grown in Hawaii since 1996. (Gonsalves 1998). Developments resulting in commercially produced varieties in countries such as the United States and Canada have centered on increasing the shelf life of fruits and vegetables, conferring resistance to insect pests or viruses, and producing tolerance to specific herbicides. While these traits have had benefits for farmers, it has been difficult for the consumers to see any benefit other than, in limited cases, a decreased price owing to reduced cost and increased ease of production (Nelson et al. 1999; Falck-Zepeda et al. 1999). A possible exception is the development of GM technology that delays ripening of fruit and vegetables, thus allowing an increased length of storage.” 

13. Bt cotton is known to be a pest-resistant GM crop. How does Bt cotton get its insect resistance?

Ans. Bt cotton was created through the addition of genes encoding toxin crystals in the Cry group of endotoxin. When insects attack and eat the cotton plant the Cry toxins or crystal protein are dissolved due to the high pH level of the insect’s stomach. The dissolved and activated Cry molecules bond to cadherin-like proteins on cells comprising the brush border molecules The epithelium of the brush border membranes separates the body cavity from the gut while allowing access to nutrients. The Cry toxin molecules attach themselves to specific locations on the cadherin-like proteins present on the epithelial cells of the midgut and ion channels are formed which allow the flow of potassium. Regulation of potassium concentration is essential and, if left unchecked, causes the death of cells. Due to the formation of Cry ion channels sufficient regulation of potassium ions is lost and results in the death of epithelial cells. The death of such cells creates gaps in the brush border membrane.

14. One advantage of incorporating pest resistance in GM crops is that it helps in pest control. Traditionally, chemical pesticides have been used as pest control methods in agricultural farmlands.
(a) What is the environmental advantage of developing GM crops with pest resistance?

Ans.  GM crops with plant resistance to pests are that it is environmentally preferable to the heavy use of insecticides it replaces. Again, there is nothing special here about GM; the advantage could equally be claimed for traditionally bred resistant varieties.
(b) Name and explain the cellular defense mechanism that has been used to develop such pest resistance against specific nematodes.


  • Many plants have impenetrable barriers, such as bark and waxy cuticles, or adaptations, such as thorns and spines, to protect them from herbivores.
  • If herbivores breach a plant’s barriers, the plant can respond with secondary metabolites, which are often toxic compounds, such as glycol cyanide, that may harm the herbivore.
  • When attacked by a predator, damaged plant tissue releases jasmonate hormones that promote the release of volatile compounds, attracting parasitoids, which use, and eventually kill, the predators as host insects.

15. For a fruit plant naturally growing in moderate water availability, GM techniques can help in incorporating drought escape characteristics. This will reduce their life cycle duration and can have adverse impacts. State ONE such possible impact on the commercial value of the fruit.

Ans. Powerful scientific techniques have caused a dramatic expansion of genetically modified crops leading to altered agricultural practices posing direct and indirect environmental implications. Despite the enhanced yield potential, risks and biosafety concerns associated with such GM crops are the fundamental issues to be addressed. An increasing interest can be noted among the researchers and policymakers in exploring the unintended effects of transgenes associated with gene flow, the flow of naked DNA, weediness, and chemical toxicity. The current state of knowledge reveals that GM crops impart damaging impacts on the environment such as modification in crop pervasiveness or invasiveness, the emergence of herbicide and insecticide tolerance, transgene stacking, and disturbed biodiversity, but these impacts require a more in-depth view and critical research to unveil further facts. Most of the reviewed scientific resources provide similar conclusions and currently, there is an insufficient amount of data available up until today, the consumption of GM plant products is safe for consumption to a greater extent with few exceptions. This paper updates the undesirable impacts of GM crops and their products on target and non-target species and attempts to shed light on the emerging challenges and threats associated with it. Underpinning research also realizes the influence of GM crops on a disturbance in biodiversity, development of resistance and evolution slightly resembles the effects of non-GM cultivation

Answer the following questions about the diagram and information on the pBR322 plasmid:
1. The ability of plasmids and bacteriophages to replicate inside a bacterial cell, independent of the chromosomal DNA control is used in the technique of cloning vectors.
2. E.coli cloning vector pBR322 has restriction sites, site of origin, and antibiotic-resistant genes that make it perfect as a cloning vehicle.

CBSE Class 12 Biology Additional Practice Questions 2023_190.1

Q. pBR322 and normal E. coli genes were incorporated into the DNA of two test plant specimens A and B respectively. Later, the antibiotic ampicillin was administered to the two plant specimen to combat bacterial infections.
What would MOST LIKELY be the fate of each of the two test specimens and why?

16. The sequences amp marker and tet marker are the selectable markers in the plasmid and help in transformation and can help in differentiating the non-recombinants from the transformants. This is done using ‘chromogenic substrates’.
(a) What is the chemical basis on which the chromogenic substance act?

Ans. Enzymes are proteins that catalyze most of the chemical reactions that take place in the body. They make it possible for chemical reactions to occur at neutral pH and body temperature. The chemical compound upon which the enzyme exerts its catalytic activity is called a substrate. Proteolytic enzymes act on their natural substrates, proteins, and peptides by hydrolyzing one or more peptide bonds (s). This process is usually highly specific in the sense that only peptide bonds adjacent to certain amino acids are cleaved. Chromogenic substrates are peptides that react with proteolytic enzymes under the formation of color. They are made synthetically and are designed to possess a selectivity similar to that of the natural substrate for the enzyme. Attached to the peptide part of the chromogenic substrate is a chemical group that when released after the enzyme cleavage gives rise to color. The color change can be followed spectrophotometrically and is proportional to the proteolytic activity. Chromogenic substrate technology was developed in the early 1970s and has since then become a tool of substantial importance in basic research. The majority of chromogenic substrate applications are found in various clinical fields. In particular, they have been used to generate fundamental knowledge of the mechanisms regulating blood coagulation and fibrinolysis. Furthermore, products based on chromogenic substrate technology have brought a new generation of diagnostics into the clinical laboratory.

(b) Explain how the substrates help in the differentiation process.


Cellular differentiation, or simply cell differentiation, is the process through which a cell changes gene expression to become a more specific type of cell. The process of cell differentiation allows multi-cellular organisms to create uniquely functional cell types and body plans. The process of cell differentiation is driven by genetics and their interaction with the environment. All organisms begin from a single cell. This single-cell carries the DNA coding for all the proteins the adult organism will use. However, if this cell-expressed all of these proteins at once it would not be functional. This cell must divide repeatedly, and the cells must begin the process of cell differentiation as they divide. The cell lines begin to emerge, and the cells get more and more specific. Eventually, an entire organism is formed with hundreds of different cell types from this process of cell differentiation. The original mass of cells, which have not undergone differentiation, are known as stem cells. Unlike normal cell division, which creates two identical daughter cells, the division of stem cells is asymmetric cell division. In this case, one of the cells remains identical to the parent stem cell. In the other cell, chemical triggers activate the process of cell differentiation, and the cell will start to express the DNA of a specific cell type. Stem cells that can differentiate into entire organisms are known as embryonic stem cells and are said to be totipotent. By contrast, the body also has many only pluripotent cells. These cells have already undergone some cell differentiation. These stem cells can only divide into a narrow range of cell types. Bone marrow, for instance, contains somatic stem cells which can only become red blood cells. These cells are necessary for the constant replenishment of blood cells, which are mostly inactive besides their oxygen-carrying ability.

(c) Why are such chromogenic substrates advantageous in the separation of nonrecombinants and transformants?


First, the biggest feature of MAOS is that the maximum absorption wavelength of its oxidized product is far more than that of ordinary chromogenic reagents. Even in the new Trinder’s reagent, its product UV absorption wavelength is 630nm, which is quite high. The maximum absorption wavelength of products of many chromogenic reagents is in the visible light region. If testing human blood or other body fluids, due to the complex components in the sample to be tested, the sample itself contains a small number of components that absorb wavelengths in the visible light region or absorb with chromogenic products. The wavelengths are close, which will make the detection result high. If you use something like MAOS, the maximum absorption wavelength of the product is in the ultraviolet region and it is relatively high, which will greatly reduce interference. Therefore, it is recommended to use MAOS as a chromogenic substrate for some biochemical testing items that require highly accurate values. Secondly, the chromogenic reaction of MAOS chromogenic substrate has wider adaptability to the pH of the reaction system, which greatly improves its adaptability. The reaction of some chromogenic reagents is often performed under acidic conditions, but biochemical detection usually requires the participation of enzymes. We know that the catalytic effect of enzymes is relatively sensitive to the pH of the reaction environment, and it may not match the chromogenic reagent Inactivation will be greatly restricted in use, and MAOS reagents do not have this problem. Another point is that the MAOS chromogenic reaction has a larger molar absorption intensity, so its sensitivity is relatively high. It should be noted that MAOS will fade when the detection time is too long, so the MAOS detection needs to be completed in time and cannot be interrupted. In addition, MAOS is also relatively safe. Unlike DAB and other biphenyl chromogenic reagents, although the price is relatively low, it has certain carcinogenicity or mutagenicity.

17. Mention any TWO characteristics that are important for a cloning vector.


1. Presence of Origin of replication (ori)- The vector should have an ori which is a sequence from where replication starts. Any piece of DNA which is linked to the ori can be made to replicate within the host cells The ori site also controls the copy number so cloning vectors having ori that support a high copy number are chosen

2. Selectable marker – The vector should possess a selectable marker that helps to distinguish and identify the non-transformants from transformants and selectively permit the growth of transformants Genes encoding for antibiotics like ampicillin, tetracycline, chloramphenicol, or kanamycin are considered to be useful selectable markers for E.coli.

3. Cloning Sites – The cloning vector should contain a single recognition site for the restriction enzymes to link the alien DNA The presence of more than one recognition site generates several fragments and complicates the cloning process.

4. It should have a high copy number so that we obtain many copies of the DNA linked to it

5. They should be able to replicate independently.

6. They should help easy linking alien DNA

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