case study questions class 12 biology chapter 11

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CBSE Class 12 Biology Case Study Questions PDF Download

Welcome to the comprehensive guide on CBSE Class 12 Biology Case Study Questions! In this article, we will delve into the world of Biology case studies, explore the importance of case study questions, and provide you with a PDF download that will assist you in mastering this subject. Whether you’re a student preparing for your examinations or a teacher looking to enhance your teaching materials, our detailed resource will help you excel in the realm of Biology.

Case studies are a widely used educational tool that involves the in-depth analysis of a particular situation, organism, process, or problem. In CBSE Class 12 Biology, case study questions are designed to evaluate students’ critical thinking, problem-solving abilities, and practical application of biological concepts. These questions require students to apply their theoretical knowledge to real-life scenarios, thus fostering a deeper understanding of the subject matter.

Class 12 Biology Case Study Questions

CBSE Class 12 Biology question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 12 Biology students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 12 Biology

Class 12 students should go through important Case Study problems for Biology before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Biology examinations. Our expert faculty for standard 12 Biology have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

Books for Class 12 Biology

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2022-23 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2022 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2022-23. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th.

Tips to Excel in CBSE Class 12 Biology Case Study Questions

Thoroughly grasp the concepts.

Before attempting case study questions, ensure you have a strong grasp of fundamental biological concepts. Familiarize yourself with topics such as genetics, ecology, human physiology, and evolution. A solid foundation will empower you to tackle complex case studies with confidence.

Practice Regularly

Practice makes perfect, and this is especially true for case study questions. Allocate time to work through a diverse range of case studies regularly. Familiarizing yourself with different scenarios will prepare you for the unpredictability of examination questions.

Collaborate and Discuss

Engaging in group discussions with fellow students or teachers can be highly beneficial. Sharing perspectives, analyzing cases together, and discussing possible solutions can broaden your understanding and offer fresh insights.

Seek Guidance from Teachers

Don’t hesitate to approach your teachers for assistance. They possess valuable experience and expertise to guide you through challenging case studies and provide valuable feedback.

Time Management

During examinations, time management is crucial. Practice solving case study questions under timed conditions to improve your efficiency and ensure you complete all questions within the allocated timeframe.

In conclusion, mastering case study questions in CBSE Class 12 Biology is essential for a holistic understanding of the subject and for excelling in examinations. By honing your analytical abilities, practicing regularly, and seeking guidance, you can conquer even the most challenging case studies. Remember that learning Biology is not just about memorizing facts but about applying knowledge to real-world situations. So, embrace the power of case studies and delve into the fascinating world of Biology.

With the provided CBSE Class 12 Biology Case Study Questions PDF, you have a valuable resource at your disposal. Practice diligently, and you’ll witness your confidence soar as you become a proficient problem solver in the realm of Biology. Good luck on your academic journey!

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Case Study Based Questions for CBSE Class 12 Biology Board Exam 2024: Read this article for Last Minute Revision

Cbse class 12 biology important case study questions : practise important case study based questions for class 12 biology board exam. these case study based questions are important for the upcoming cbse class 12 biology board exam 2024 on march 19, 2024..

CBSE Class 12th Biology Board Exam 2023-24 Pattern

The paper will be of 70 marks and the time duration for completing the paper will be 3 hours.

The paper will have 33 questions divided into 5 sections.

Section–A 16 questions of 1 mark each,

Section–B 5 questions of 2 marks each;

Section–C 7 questions of 3 marks each;

Section–D 2 case-based questions of 4 marks each,

and Section–E 3 questions of 5 marks each.

CBSE Class 12 Biology Important Case Study Based Questions

Case Study 1:  Nondisjunction is the failure of homologous chromosomes to disjoin correctly during meiosis. It leads to the formation of a new cell with an abnormal amount of genetic material. A number of clinical conditions are the result of this type of chromosomal mutation. This results in the production of gametes containing a greater or lesser chromosomal amount than normal ones. Consequently, the individual may develop a trisomy or monosomal syndrome. Nondisjunction can occur in both Meiosis I and Meiosis II of the cellular division. It is also the main cause of many genetic disorders; however, its origin and process remain vague. Although it results in the majority of cases from errors in maternal meiosis II, both paternal and maternal meiosis I do influence it. Maternal age is considered a risk factor for trisomy, as well as recombination alterations and many others that can affect chromosomal segregation.

• It is the presence of an extra chromosome in a diploid cell.
• An aneuploid cell differs from other cells only in size.
• It can be less number of chromosomes in a diploid cell.
• Aneuploidy always affects female individuals.
• both i and iii
• both ii and iii
• i, iii and iv
• Errors in meiosis I is the only cause of aneuploidy
• Aneuploidy always affects sex chromosomes.
• Most of the aneuploidy results from errors in cell division involved in egg formation.
• Nondisjunction in meiosis I can lead to more abnormal cells than disjunction in meiosis II.
• both I and iii
• both iii and iv
• I, iii and iv
• Aneuploidy is not influenced by the mother’s age.
• Delivery before 30 years of age can decrease the incidence of aneuploidy in most cases
• The chance of aneuploidy increases up to 22 years of age.
• There is a dramatic increase in aneuploidy if the maternal age exceeds 30

• both ii and iv
• Chromosomal disorders
• Mendelian disorders
• Incomplete dominance
• All the above

Q5: Assertion: All types of genetic disorders are caused by chromosomal nondisjunction.

• Both assertion and reason are correct and the reason is the correct explanation of assertion
• Both assertion and reason are correct but the reason is not the correct explanation of the assertion
• Assertion is correct but the reason is incorrect
• Both assertion and reason are incorrect

Case Study 2:  A Representative Diagram of the Human Genome Project:

• Biotechnology
• Biomonitoring
• Bioinformatics
• Biosystematics

Q2:  Name a free living, non-pathogenic nematode, the DNA of which has been completely sequenced.

Answer: Caenorhabditis elegans

Q3: Summarize the methodology adopted in the Human Genome Project.

Answer: Expressed Sequence Tags (ESTs) : The approach focused on identifying all the genes that are expressed as RNA.

Sequence Annotation : The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions.

Q4: What are SNPs’? How are they useful in human genomics?

• Identify disease-causing genes in humans
• Can be used to understand the molecular mechanisms of sequence evolution.

Q5: Mention at least four salient features of the Human Genome Project.

• Human genome contains 3164.7 million bp.
• Average gene consists of 3000 bases, but sizes vary greatly.
• Almost all (99.9 percent) nucleotide bases are exactly the same in all people.
• Less than 2 percent of the genome codes for proteins.

Case Study 3: Two blood samples of suspects ‘A’ and ‘B’ were sent to the Forensic Department along with sample ‘C’ from the crime scene. The Forensic Department was assigned the responsibility of running the samples and matching the samples of the suspects with that of the sample from the scene of the crime and thereby identifying the culprit.

• A radioactively labelled double stranded RNA molecule.
• A radioactively labelled double stranded DNA molecule.
• A radioactively labelled single stranded DNA molecule.
• A radioactively labelled single stranded RNA molecule.

Q3: What does ‘minisatellite’ and ‘microsatellite’ mean in relation to DNA Fingerprinting?

Answer:  Minisatellite: the repeating unit consists of 10-100 base pairs.

Microsatellite: the repeating unit consists of 2-6 base pairs.

Q3: How does polymorphism arise in a population?

Answer: Polymorphism (variation at the genetic level) arises due to mutations.

Q4: State the steps involved in DNA Fingerprinting in a sequential manner.

• DNA isolation
• DNA digestion with restriction enzymes.
• DNA fragment separation by electrophoresis.
• Hybridization
• DNA visualization under UV light.

Case Study 4: Bacteria like Streptococcus pneumoniae  and Haemophilus influenzae  are responsible for the disease pneumonia in humans which infects the alveoli (air-filled sacs) of the lungs. As a result of the infection, the alveoli get filled with fluid leading to severe problems in respiration. The symptoms of pneumonia include fever, chills, cough, and headache. In severe cases, the lips and fingernails may turn gray to bluish in colour. A healthy person acquires the infection by inhaling the droplets/aerosols released by an infected person or even by sharing glasses and utensils with an infected person. Dysentery, plague, diphtheria, etc., are some of the other bacterial diseases in man. Many viruses also cause diseases in human beings. Rhinoviruses represent one such group of viruses that cause one of the most infectious human ailments – the common cold. They infect the nose and respiratory passage but not the lungs.

The common cold is characterized by nasal congestion and discharge, sore throat, hoarseness, cough,

headache, tiredness, etc., which usually lasts for 3-7 days. Droplets resulting from the cough or sneezes of an infected person are either inhaled directly or transmitted through contaminated objects such as pens, books, cups, doorknobs, computer keyboards or mice, etc., and cause infection in a healthy person.

• By exhaling droplets of a non-infected person.
• By headache or leg pain.
• By eating fast food.
• By inhaling droplets of an infected person.

Q4:  How long does the common cold last?

Answer: 3-7 days

Q5:  Write any two symptoms of the common cold and pneumonia.

Answer: Cough and nasal congestion.

Case Study 5: When you insert a piece of alien DNA into a cloning vector and transfer it into a bacterial, plant, or animal cell, the alien DNA gets multiplied. In almost all recombinant technologies, the ultimate aim is to produce a desirable protein. Hence, there is a need for the recombinant DNA to be expressed. The foreign gene gets expressed under appropriate conditions. The expression of foreign genes in host cells involves understanding many technical details. After having cloned the gene of interest and having optimised the conditions to induce the expression of the target protein, one has to consider producing it on a large scale. Can you think of any reason why there is a need for large-scale production? If any protein encoding gene is expressed in a heterologous host, it is called a recombinant protein. The cells harbouring cloned genes of interest may be grown on a small scale in the laboratory. The cultures may be used for extracting the desired protein and then purifying it by using different separation techniques.

• A continuous culture system
• A stirred-tank bioreactor without in-lets and out-lets
• Laboratory flask of the largest capacity
• None of the above
• upstream processing
• downstream processing
• bioprocessing
• postproduction processing
• Human insulin
• Growth hormone
• cleaving and joining of DNA segments with endonuclease
• cleaving DNA segments with endonuclease and re-joining with ligase
• cleaving and re-joining DNA segments with ligase
• cleaving DNA segments with ligase and re-joining with endonuclease

Case Study 6: Gene Therapy

Read the following and answer the questions that follow:

• Replacing a disease-causing gene with a healthy copy of the gene
• Inactivating a disease-causing gene that is not functioning properly
• Introducing a new or modified gene into the body to help treat a disease
• Adenosine deaminase
• phenylketonuria
• Phenylalanine
• Bone marrow transplantation
• Southern blotting

Q4 Introduction of gene isolate from bone marrow producing ADA should be introduced at what age to

• acute diseases
• physiological diseases
• hereditary diseases
• infectious diseases
• CBSE Class 12 Biology Previous Year Question Papers with Solutions PDF Download
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• Important Questions for CBSE Class 12 Biology Chapter 11 Biotechnology: Principle and Process 2024-25

CBSE Class 12 Biology Chapter-11 Important Questions - Free PDF Download

Biotechnology principles and processes Class 12 important questions provide students a detailed insight into the topics covered in the chapter. These important questions and answers are designed by the subject experts at Vedantu. Students preparing for the Biology 12th board exam or the competitive exams must go through all the important questions for Class 12 biology Chapter 11 to get a deeper understanding of the Chapter of Biotechnology. These important questions on Class 12 Biology Chapter 11 are prepared following the updated CBSE guidelines so students can rely upon them for their exam preparation.

Download CBSE Class 12 Biology Important Questions 2024-25 PDF

Also, check CBSE Class 12 Biology Important Questions for other chapters:

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Study Important Questions for CBSE Class 12 Biology Chapter 11 - Biotechnology: Principles and Processes

Very short answer questions.  (1 mark).

1. A restriction enzyme digests DNA into fragments. Name the technique used to check the progression of this enzyme and separate DNA fragments.

Ans: Gel electrophoresis is the technique in which DNA fragments are separated after the digestion of DNA by a restriction enzyme that results in the formation of fragments.

2. Name two commonly used vectors in genetic engineering.

Ans: In genetic engineering, Plasmid and Bacteriophage are the two common vectors that are used.

3. Some enzymes are considered molecular scissors in genetic engineering. What is the name assigned to such enzymes?

Ans: In genetic engineering, restriction Enzymes are the enzymes that are responsible for the digestion of DNA strands resulting in the formation of fragments, thus they are called molecular scissors.

4. Write conventional nomenclature of EcoRI.

Ans: E. ​Escherichia is a bacterium where co stands for coli; R stands for ​the Name of Strain; I ​is the order in which the enzyme is isolated from the bacterial strain.

5. A linear DNA fragment and a plasmid have three restriction sites for EcoRI. How many fragments will be produced from linear DNA and plasmid respectively?

Ans: The linear DNA will produce 4 fragments while the plasmid produces 6 fragments after their digestion by a restriction enzyme.

6. An extra-chromosomal segment of circular DNA of a bacterium is used to carry the gene of interest into the host cell. What is the name given to it?

Ans: In bacterium, Plasmid is the extra-chromosomal segment of circular DNA that is useful in carrying the desired gene into the host cell.

7. Identify the recognition sites in the given sequences at which E. coli will be cut and make sticky ends.

5 ́-GAATTC-3 ́

3 ́-CTTAAG-5 ́

Ans:  

5 - G↓TAATTC3

3 - CTTAA↑G5 are the recognition sites that will make the sticky ends after they will be cut.       

8. Name the substance used as a medium in gel electrophoresis.

Ans: In the gel electrophoresis, Agarose will be a substance that will be used as a medium.

9. What is Bioconversion?

Ans: The process of conversion of raw materials into useful products with the help of various factors that include microbial, plant, or animal cells is called bioconversion.

10. Name the bacterium that yields thermostable DNA polymerase.

Ans: The thermostable DNA polymerase can be produced with the help of a bacterium which is named Thermusaquaticus.

11. Which enzymes are known as “molecular Scissors”?

Ans: The Restriction Endonuclease Enzymes are the enzymes that are responsible for the digestion of DNA strands resulting in the formation of fragments, thus they are called molecular scissors.

12. Name the commonly used vector for transformation in a plant cell?

Ans: The commonly used vector which is responsible for the transformation in plants cells is named Agrobacterium tumefacient.

13. Name the technique used for amplification of DNA?

Ans: The DNA amplification can be done by the technique named the Polymerase Chain Reaction.

14. Name the enzyme responsible for removal of 5 – phosphate group from nucleic acid?

Ans: To remove the 5 – phosphate group from nucleic acid, Alkaline Phosphates are the enzymes that are responsible.

15. Who isolated Restriction enzymes for the first time?

Ans: The Restriction enzymes were first isolated by Warner Arber & Hamilton-Smith.

16. Why do eukaryotic cells do not contain restriction enzymes?

Ans: In eukaryotic cells, the restriction enzymes are absent because the DNA is found to be methylated heavily.

17. Why does DNA moves towards the anode in the gel electrophoresis.

Ans: In the gel electrophoresis, the DNA is found to be moving towards the anode because the DNA is negatively charged due to the presence of a phosphate group that results in the movement of DNA towards the anode.

Short Answer Questions (2 Marks)

1. Name two main steps which are collectively referred to as the down streaming process. Why is this process significant?

Ans: Separation and Purification are the two main steps that are together referred to as the down streaming process.

This process is essential because the product needs to undergo clinical trial and quality control before it reaches the market.

2. How does plasmid differ from chromosomal DNA?

Ans: The differences between plasmid and chromosomal DNA are:

3. A bacterial cell is shown in the figure given below. Label the part ‘A’ and ‘B’. Also mention the use of part ‘A’ in rDNA technology.

Ans: Part A is labeled as Plasmid while part B is labeled as Nucleoid

The function of a plasmid is that, in rDNA technology, it acts as a vector that helps in transferring the desired gene into the host cell.

4. Mention two classes of restriction enzymes. Suggest their respective roles.

Ans: Exonucleases and endonucleases are two classes of restriction enzymes.

The function of Exonucleases is to remove the nucleotides from the ends of the DNA while the Endonucleases play a major role in cutting the DNA at specific sites between the ends of DNA.

5. In the given process of separation and isolation of DNA fragments, some of the steps are missing, Complete the missing steps –

A: Digestion of DNA fragments using restriction endonucleases

B: ..............................................................

C: Staining with ethidium bromide

D: Visualisation in U.V. light

E: .............................................................

F: Purification of DNA fragments.

Ans: At step B the process of Gel Electrophoresis takes place while at step E the process of Elution occurs.

6. Write any two properties of restriction endonuclease enzymes?

Ans: The properties of restriction endonuclease enzymes are:

(i) The Restriction endonuclease enzymes bind at the recognition sequence of the DNA after inspecting the length of the DNA sequence.

(ii) Its function is to cut the sugar-phosphate backbone at specific sites.

7. What are ‘Selectable markers’? What is their use in genetic engineering?

Ans: The host cells that contain the vector can be selected with the help of a gene called a selectable marker that results in the elimination of the non–transformant. It is used in genetic engineering for e.g. – the gene that encodes resistance towards the antibiotics are found to be useful selectable markers as they insert into a cell and result in the selective growth of transformants only.

8. How can the desired product formed after genetic engineering be produced on a commercial scale?

Ans: To make a commercial scale-out of the desired product that is formed after the process of genetic engineering there are a series of processes that need to be followed which are collectively called downstream processing and then it will lead to the final processes. The final processes that are involved in the downstream processing are Separation and purification.

9. What is “Insertional Inactivation”?

Ans: The insertional inactivation is the process of insertion of the recombinant DNA into the coding Sequence of enzyme B– galactosidase leading to the inactivation of the enzyme. An example is when the insert is absent in the plasmid of bacteria then it will lead to the insertional inactivation leading to the production of colorless colonies instead of blue-colored colonies due to the presence of chromogenic substrate.

10. What are the two basic techniques involved in modern Biotechnology?

Ans: The two basic techniques involved in modern Biotechnology are:

a) Genetic Engineering - the technique which involves the introduction of the genome into another host organism or results in the alternation of the nature of genetic material that leads to change in its phenotype.

b) Techniques that are performed under sterile conditions are for the manufacturing of a large number of the desired microbes or cells by the process of multiplication and growth.

11. Represent diagrammatically the E. coli. Cloning vector PBR 322.

Ans: The diagram of E. coli. Cloning vector PBR 322 is represented as:

12. Differentiate between plasmid DNA and chromosomal DNA? 

Ans: The difference between plasmid DNA and chromosomal DNA are:

13. What is the role of enzyme “Ligase” in genetic Engineering?

Ans: In genetic engineering, the enzyme “Ligase” acts as a molecular Suture that helps in binding the two DNA pieces together. This process of joining the two DNA pieces requires energy in the form of ATP and results in the formation of a phosphodiester bond between the two cohesive ends of DNA.

14. Name the components a bioreactor must possess to achieve the desired product?

Ans: The bioreactors are the devices in the form of the vessel which contains various organisms or chemical substances that undergo the chemical processes and result in the formation of the biologically active substances.  A bioreactor must consist of the following components that result in the formation of the desired product. These components include temperature, substrate, pH, oxygen, vitamins, and salts. 

15. The following proteins of given molecular weight are Subjected to Get electrophoresis. Write the order of Sequence in which these proteins are isolated in a gel?

Ans: The sequence of proteins obtained from top to bottom in a gel:

Myosin > Insulin >Haemoglobin> Ribozyme > Keratin > Albumin.

16. How is gene Z used as a marker?

Ans: The Lac Z Gene Is responsible for the coding of Β-galactosidase enzyme this results in the inactivation of the enzyme due to the insertion of recombinant DNA in the coding sequence of an enzyme Β-galactosidase. The bacterial colony normally produces blue-colored colonies by they get the insert into their plasmid then it will result in the production of colonies that are colorless.

17. What is Bioreactor? What are the advantages of Stirred tank Bioreactor over Shake flask? Show diagrammatically a simple Stirred tank Bioreactor?

Ans: The bioreactors are the devices in the form of the vessel which contains various organisms or chemical substances that undergo the chemical processes and result in the formation of the biologically active substances. They consist of large vessels where the raw materials using microbial, plant, animal, or human cells are converted biologically into specific proteins. The advantages of Bioreactor over shake flask are:

a) To produce the optimum growth of the desired product, it provides the optimal conditions e.g., temp, pH, etc.

b) For testing the sample, a small volume of cultures can be withdrawn periodically from the bioreactor.

c) It has an agitation system, temp control system, from control system & pH control system.

Short Answer Question (3 Marks)

1. Since DNA is a hydrophilic molecule, it cannot pass through cell membranes. Name and explain the technique with which the DNA is forced into 

(i) a bacterial cell 

Ans: In the bacterial cell, the DNA can enter by the technique of Chemical treatment and may be exposed to cold and high temp (42°C) alternatively.

(ii) a plant cell 

Ans: The DNA in the plant cell can enter through the technique called Biolistic or gene gun.

(iii) an animal cell.

Ans: The DNA in the animal cell can enter through the technique called Micro-injection. 

2. How will you obtain purified DNA from a cell?

Ans: Various enzymes are used to treat cells that will result in the release of DNA. Various enzymes are cellulose (plant cells), Lysozyme (bacteria), and chitinase (fungus) while ribonuclease and protease enzymes are used for the treatment of and removal of RNA and proteins respectively.

3. In recombinant DNA technology, vectors are used to transfer a gene of interest in the host cells. Mention any three features of vectors that are most suitable for this purpose.

Ans: The vectors in the recombinant DNA technology are utilized in transferring the gene of interest in the host cells due to the following features of vectors:

(i) Vectors constitute of origin of replication (Ori) where the gene of interest attaches.

(ii) They constitute various selectable markers for the various genes of interest.

(iii) They are also composed of at least one recognition site.

4. Why is Agrobacterium-mediated genetic engineering transformation​ in plants considered natural genetic engineering?

Ans: Agrobacterium tumefaciens is a pathogen in many dicot plants due to its ability to deliver a piece of DNA (T​DNA) that results in the transformation of normal plant cells into a tumor cell leading to the production of chemicals by these tumor cells that are required by the pathogen.

5. Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following question​

5 ́ - CAGAATTCTTA - 3 ́

3 ́ - GTCTTAAGAAT - 5 ́

(a) Name a restriction enzyme that can recognize this DNA sequence.

Ans: EcoRI is the name of a restriction enzyme that helps in the recognition of various sequences of DNA.

(b) Write the sequence after digestion.

(c) Why are the ends generated after digestion called sticky ends?

Ans: The ends that are obtained after the digestion of the DNA sequences are called the sticky ends because they result in the formation of hydrogen bonds with their complementary cut parts.

6. A selectable marker is used in the section of recombinants on the basis of their ability to produce color in presence of chromogenic substrate.

(a) Mention the name of the mechanism involved.

Ans: The insertional inactivation is the process of insertion of the recombinant DNA into the coding Sequence of enzyme B– galactosidase leading to the inactivation of the enzyme.

(b) Which enzyme is involved in the production of color?

Ans: The b-galactosidase is the enzyme that helps in the production of color in presence of chromogenic substrate.

(c) How is it advantageous to overuse antibiotic-resistant genes as a selectable marker?

Ans: The use of antibiotic-resistant genes as a selectable marker is more beneficial because due to the inactivation of antibiotics the selection of recombinants requires simultaneous plating on two plates that are having different antibiotics.

7. Mention the important properties which a good vector must possess?

Ans: The important properties which a good vector must possess are -

i) Size - The size of the vector must be small so that it helps them in purification and isolation easily.

ii) Origin of Replication – A site where replication starts and is made up of the sequence of base pairs. When the DNA is attached to this sequence then it will result in the replication within its host cell & thus, controls the number of copies of linked DNA.

iii) Selectable Marker - The host cells that contain the vector can be selected with the help of a gene called a selectable marker that results in the elimination of the non–transformant. It is used in genetic engineering.

iv) Cloning Sites - The site in the vector where the foreign/alien DNA attaches is also called the unique recognition site. A particular restriction enzyme will cut the vector-only at a particular recognition site.

8. Describe any three vectors less method of introducing the rDNA into a competent host cell?

Ans: The three vectors less method of introducing the rDNA into a competent host cell are:

i) Transformation: The bacterial cell is first treated with the specific concentration of divalent cation e.g., Ca2+ so that they can be competent enough to take up the DNA in the plasmid. The calcium ions increase the efficiency of DNA to enter into a bacterium through pores in its cell wall. By the process of incubating the cells with recombinant DNA on ice, then placing them at 420 C, and then again putting them back into ice the recombinant DNA can then be forced into cells. This helps the bacteria in taking up the recombinant DNA.

ii) Microinjection: With the help of a microneedle of the tip with a diameter (~ 4mm), the recombinant DNA can be injected directly into the nucleus of an animal cell.

iii) Biolistic / Gene gun: The cells of the DNA that are coated with particles of gold or tungsten are bombarded with high-velocity micro-gun.

9. Why is Agrobacterium-mediated genetic transformation described as Natural Genetic engineering in plants?

Ans: The Agrobacterium tumefaciens-interceded plant genetic transformation measure requires the presence of two genetic segments situated on the bacterial Ti-plasmid. Basically, the main basic part is the T-DNA, which is characterized by conserved 25-base pair imperfect repeats at the closures of the T-region known as a border sequence. The tumor-inducing (Ti) plasmid is used as a cloning vector to transfer the desired gene into plants as they insert a part of their DNA in plants during the infection. The second is the virulence (vir) region, which is made out of in any event seven significant loci (virA, virB, virC, virD, virE, virF, and virG) encoding parts of the bacterial protein machinery which is mediating the T-DNA processing and transfer. The gene of interest is attached to the T-DNA so that it automatically gets transformed into plant cells thus, Agrobacterium tumefacien is known as the “Natural Genetic Engineer” of plants.

10. Mention the important tools required for genetic engineering technology?

Ans: The process of genetic engineering is accomplished only when we have the following key tools:

a) Restriction Enzymes: In genetic engineering, restriction Enzymes are the enzymes that are responsible for the digestion of DNA strands resulting in the formation of fragments, thus they are called molecular scissors. They are of types: Endonucleases and Exonucleases.

b) Cloning Vector: The cloning Vector is the part of the DNA molecule that is attached to the DNA Segment of an organism that is desired and then transfers into the cell or DNA of another organism.

c) Desired Foreign DNA: The desired foreign DNA segment is the part of DNA that consists of genes having desired characters that are being transferred into the genome of another cell with the help of the cloning vector.

Long Answer Questions (5 Marks)

1. The development of bioreactors is required to produce large quantities of products.

(a) Give optimum growth conditions used in bioreactors.

Ans: The bioreactors are the devices in the form of the vessel which contains various organisms or chemical substances that undergo the chemical processes and result in the formation of the biologically active substances. To produce the optimum growth of the desired product the optimum temperature, pH, substrates, salts, vitamins, and oxygen are required.

(b) Draw a well-labeled diagram of a simply stirred ​ tank bioreactor.

Ans: A simply stirred​ tank bioreactor is shown below:

(c) How does a simply stirred tank​ bioreactor to differ from sparged stirred – tank’ bioreactor?

Ans: In the simply​ stirred tank bioreactor the stirrer facilitates the even mixing and the oxygen availability throughout the process, whereas for proper mixing throughout the reactor in the case of sparged stirred-tank bioreactor the air is found to be bubbled.

2. In the given figure, one cycle of polymerase chain reaction (PCR) is shown-

(a) Name the steps A, B, and C.

Ans: A: Denaturation, B: Annealing, and C: Extension.

(b) Give the purpose of each of these steps.

Ans: The purpose of each step is:

(i) Denaturation: ​ Due to high temperature, the heat will break or denature the two complementary strands of DNA and results in their separation.

(ii) Annealing: The hybridization of the denatured DNA strands takes place with the help of the primers.

(iii) Extension: The target DNA sequence will synthesize its copies by the process of the extension of the primers.

(c) State the contribution of the bacterium Thermus Aquaticus in this process.

Ans: The Taq polymerase enzyme is found to be isolated from the bacterium Thermus Aquaticus which functions at a very high temperature and results in the denaturation of double-stranded DNA.

3. Study the figure of vector pBR322 given below in which foreign DNA is ligated at the Bam H1 site of the tetracycline resistance gene.

Answer the following questions:

(a) Mention the function of rop.

Ans: The ​rop is responsible for the​ coding of the proteins that are involved in the replication of plasmids.

(b) What will be the selectable marker for this recombinant plasmid and why?

Ans: The selectable marker for this recombinant plasmid will be the ampicillin resistance gene. They after placing them on an ampicillin-containing medium will undergo the process of plating that will help in the differentiation between the trAns.formants from non-trAns.formants.

(c) Explain transformation.

Ans: Transformation is the process of transferring DNA from one cell and then placing them into the other cell which will result in the formation of the recombinant cell consisting of the properties of both the cells. 

4. Describe the various steps involved in Recombinant DNA technology with the help of a well labeled. Diagram?

Ans: The steps involved in the recombinant DNA are:

i) Identification of DNA with Desirable Genes: The addition of chilled ethanol will result in obtaining the purified DNA while by using various other appropriate techniques the other molecules in the target cell can also be removed.

ii) Cutting the DNA at a Specific Location: The source and vector DNAS after being cut will be introduced together having a gene of interest and specific restriction site and will be joined together with the help of an enzyme called ligase. 

iii) Insertion of Recombinant DNA into Host Cell: The bacterial cell is first treated with the specific concentration of divalent cation e.g., Ca2+ so that they can be competent enough to take up the DNA in the plasmid. The calcium ions increase the efficiency of DNA to enter into a bacterium through pores in its cell wall. It is a vector-less method.

iv) Selection & Screening: The selectable marker for this recombinant plasmid will be the ampicillin resistance gene. They after placing them on an ampicillin-containing medium will undergo the process of plating that will help in the differentiation between the trAns.formants from non-trAns.formants.

v) Obtaining the Foreign Gene product: The gene of interest after its cloning along with the presence of optimum growth conditions will result in the expression of target proteins which then needs to be produced on a large scale.

5. Expand PCR? Describe the different steps involved in this technique?

Ans: The Polymerase Chain Reaction is the method of making millions of DNA copies from a DNA sample. It has two main reagents: primers (short single-stranded DNA fragments that are a complementary sequence to the target DNA), and DNA polymerase. The DNA polymerase is heat stable, that is Taq polymerase which is extracted from the bacteria Thermus aquaticus. Each cycle has three steps:

a) DENATURATION: In the first step, the two strands of the DNA helix are physically separated at a heat during a process called macromolecule denaturation.

b) RENATURATION / ANNEALING: In the second step, the temperature is lowered so that the primers can bind to the complementary sequences of DNA.

c) EXTENSION: The third step is the target DNA sequence will synthesize its copies by the process of the extension of the primers.  The temperature is raised to 750c. At this temperature, Taq – polymerase initiates DNA Synthesis at the 3-OH end of the primer.

6. What are Restriction enzymes? Why do bacteria have these restriction enzymes? Show diagrammatically a restriction enzyme its recognition & the product it produces?

Ans: Restriction Enzymes are the endonuclease enzymes that are responsible for the digestion of DNA strands resulting in the formation of fragments, thus they are called molecular scissors. They are found in bacteria cells as they help in cutting the foreign DNA and results in the modification of the restriction system and thus results in the improving of the immunity in the bacterial cell. 

Name of Restriction enzyme- EcoRI Substrate DNA on which it acts

Download Important Question for Class 12 Biology Chapter 11 PDF

Some of the important topics covered in Class 12 Biology Chapter 11 important questions are - Biotechnology Principle and Processes, Tools of Recombinant DNA Technology, Processes of Recombinant DNA Technology.

Chapter 11 Biotechnology Principles and Its Processes Summary

Biotechnology is the broader field of research and development, which uses the technology and the application of living organisms to develop and produce products useful for human welfare. The term ‘Biotechnology’ was first coined by Karoly Ereky, so he is known as the father of biotechnology.

Principles of Biotechnology - According to modern biotechnology, some of the main principles are genetic engineering and bioprocess engineering. Genetic engineering is used to modify the DNA of the target organisms by changing the phenotype of the organism. Bioprocess engineering maintains the sterile condition to support the growth of large quantities of desired microbes and other eukaryotic cells. Which is used for the production of new or modified biotechnology products like antibiotics, enzymes, and vaccines.

Genetic engineering primarily includes the isolation of DNA fragments from the donor organism. Then it is inserted into the vector DNA, later it is transferred into an appropriate host. Then cloning of recombinant DNA in the host organism.

Recombinant DNA Technology - It is also known as genetic engineering, it is the process of joining two DNA molecules from two different organisms. Some of the important steps involved in the processes of recombinant DNA technology are - Isolation of DNA, DNA fragmentation using restriction endonucleases. Then ligation of the desired DNA fragment into the vector and transfer of the recombinant DNA into the host. Later culture of the transformed cells in a nutrient medium and extraction of the desired product.

DNA Cloning - It is the process of making multiple, identical copies of a piece of DNA, this process requires cloning vectors. With some of the properties like - It should be smaller in size but should be able to carry a large DNA insert. The cloning vector should have the origin of replication so that it can autonomously replicate in the host organism. It should have a restriction site with a selectable marker to screen recombinant organisms. And It should also possess multiple cloning sites.

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By going through all the important questions for Class 12 Biology Chapter 11 , students can prepare and revise the concepts for their 12th boards and competitive exams. It will also help them to understand the fundamental topics covered in the chapter Biotechnology Principle and Its Processes. These important questions are designed in such a way that it not only covers the conceptual parts but also the problem-solving and applicational approach too. The answers of Biotechnology Principles and processes Class 12 important questions are detailed and connected to the theory so students can get a clear idea about the topics by referring to these questions and answers.

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FAQs on Important Questions for CBSE Class 12 Biology Chapter 11 Biotechnology: Principle and Process 2024-25

1. What is the role of Agrobacterium tumefaciens in plant transformation, according to Chapter 11 of Class 12 Biology?

Agrobacterium tumefaciens negatively infects plants. They are the pathogens that are responsible for the plant's death. They infect plants such as tomatoes, sunflowers, etc. It has a tumor-inducing plasmid which helps this pathogen in making a plant infected. This plasmid is also known as Ti plasmid. It also has T DNA which eventually causes the death of the plant. This DNA is also responsible for tumor formation in the host plant.

2. What is a bioreactor? Explain different types of bioreactors, according to Chapter 11 of Class 12 Biology.

Biological reactions are carried out in a bioreactor. It is also useful in curating aerobic cells, which then conduct cellular and enzymatic immobilities. Here are the different types of bioreactors:

Stirred tank 

Bubble column

3. What is a polymerase chain reaction? What are the steps involved? Mention its applications, according to Chapter 11 of Class 12 Biology.

It is a concept of molecular biology in which several copies of a specific segment of DNA are produced. The polymerase chain reaction is used to perform this task. The steps which are followed in this process are denaturation, annealing, and extension. The polymerase chain reaction is used in various fields. These fields include research in genetics, medicine, etc. The important questions of Chapter 11 of Class 12 Biology are available free of cost on the Vedantu website and the Vedantu app.

4.  What are the properties of a good vector, according to Chapter 11 of Class 12 Biology?

The properties of the good vector are:

To isolate and purify it is required that the vector should be small in size.

A base or pair sequence is required where the replication starts, the origin should be the replication.

A selectable marker is required which then helps in the selection process of transformed host cells.

A good vector should have the property that it has a special recognition site that can bind the foreign DNA to it.

5. Mention any three vector-less methods that are used to introduce recombinant DNA into a competent host cell, according to Chapter 11 of Class 12 Biology.

Microinjection, gene gun method, and transformation are three vectors fewer methods used to introduce recombinant DNA into a competent host cell. Microinjection is a method in which the DNA is directly infused into an animal nucleus. In the gene gun method, the gold and tungsten-coated DNA is injected with force in the cells; in the transformation method, the process is natural as the bacteria naturally takes the genetic material from the surroundings.

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CBSE 12th Standard Biology Subject Biotechnology: Principles and Processes Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 Biology, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

12th Standard CBSE

Final Semester - June 2015

Case Study Questions

Read the following and answer any four questions from (i) to (v) given below: Gene manipulation is a fast emerging science. It started with development of recombinant DNA molecule. It is named variously as DNA manipulation biotechnology, recombinant DNA technology and genetic engineering. This technology, that mostly involves cutting and pasting of desired DNA fragments, is based on two important discoveries in bacteria, i.e., presence of plasmid in bacteria and restriction endonucleases. Paul Berg was able to introduce a gene of SV-40 into a bacterium. The science of recombinant DNA technology took birth when Cohen and Boyer (1973) were able to introduce a piece of gene containing foreign DNA into plasmid of E.coli. (i) Biotechnology is also known as

(ii) A bacterial plasmid is a/an

(iii) Father of genetic engineering is

(iv) Which of the following is used by Paul Berg to introduce a gene of SV-40 in a bacterium?

(v) Read the given statements and select the correct option. Assertion : Biotechnology started with the development of recombinant DNA molecule. Reason : Biotechnology mostly involves cutting and pasting of desired DNA fragments.

Read the following and answer any four questions from (i) to (v) given below: The foundations of recombinant DNA (rDNA) were laid by the discovery of restriction enzymes. These enzymes are present in many bacterias where they function as a part of their defense mechanism called the Restriction Modification system (RM system). Molecular basis of this system was explained first by Werner Arber in 1962. The Restriction t-'l0dification system consists of two components: 1. A restriction enzyme (called restriction endonuclease) identifies the introduced foreign DNA and cuts it into pieces. 2.The second component is a modification enzyme (methylase) that adds a methyl group to DNA at specific site to protect it from the restriction enzyme cleavage. (i) Restriction endonucleases are enzymes present in (i) where they function as a part of (ii) mechanism.

(ii) Which of the following statements regarding modification enzyme is correct?

(iii) Which of the following is a type II restriction enzyme?

(iv) Which of the following is the first discovered restriction endonuclease?

(v) Components of Restriction Modification System include

Read the following and answer any four questions from (i) to(v) given below: In recombinant DNA technology, the fragments of DNA generated after cutting the DNA by restriction enzymes are separated according to their size or length by gel electrophoresis. Gel electrophoresis is performed in a gel matrix so that molecules of similar electric charges can be separated on the basis of size. Most commonly used matrix in gel electrophoresis is agarose. The fragments are separated under the influence of electric field. The separated DNA fragments can be seen only after staining the DNA with compound known as ethidium bromide (EtBr) followed by exposure to UV radiation as bright orange band. (i) Gel electrophoresis is used for the separation of

(ii) Most commonly used matrix is (i) which is a (ii) extracted from (iii).

(iii) A DNA molecule was treated with a restriction endonuclease and three fragments of size (i) 426 kb, (ii) 129 kb and (iii) 46 kb were obtained. Identify the order in which these bands will arrange themselves in the gel plate after gel electrophoresis is completed. (Assuming that negative part of electrode is towards the well)

(iv) Which of the following statements regarding gel electrophoresis is incorrect?

(v) The factor that will not affect the rate of DNA migration in gel electrophoresis is

(ii) During cDNA formation, what would happen if DNA formed by reverse transcriptase is not treated with the alkali?

(iii) Enzyme that helps in the formation of double stranded cDNA is

(iv) DNA polymerase can be obtained form

(v) DNA synthesised without a template is referred to as

*****************************************

Cbse 12th standard biology subject biotechnology: principles and processes case study questions with solution 2021 answer keys.

(a) A - Denaturation; D - Extension. (b) B - Primers; they are oligonucleotides. (c) C - Taq polymerase; its source organism is Thermus aquaticus.

(i) (d) (ii) (b) : Plasmid in a bacterial cell is an extra chromosomal material that undergo replication with or without chromosomal DNA. (iii) (a)  (iv) (c) : In 1972, Paul Berg was able to introduce a gene of SV-40 virus into a bacterium with the help of lambda phage. (v) (b)

(i) (d) (ii) (a) (iii) (d) : Different examples of Type II restriction endonuclease are Alu I, EcoR I, BamH I, etc. (iv) (c) (v) (d) : The restriction modification system consists of two components (i) A restriction enzyme called restriction endonucleases which identifies the introduced foreign DNA and cuts it into pieces and (ii) A modification enzyme (methylase) that adds a methyl group to DNA at a specific site to protect the site from restriction endonuclease cleavage.

(i) (d) : Gel electrophoresis is a technique used to separate fragments of molecules, i.e., DNA, RNA and protein. (ii) (a) (iii) (b) (iv) (c) : Under the influence of electric field positively charged molecules move towards the cathode and negatively charged molecules move towards the anode. (v) (b) : Concentration of DNA will not affect the migration of DNA molecule in a gel electrophoresis. As in gel electrophoresis, molecules separate according to their size therefore DNA size will affect migration. Increased voltage supply will increase rate of migration and the more concentrated gel will reduce rate of migration.

(i) (a) (ii) (b) : The cDNA formation involves the alkaline denaturation of the mRNA-cDNA hybrid. The double stranded DNA molecule formed after the activity of reverse transcriptase is treated with alkali to digest mRNA. (iii) (c) : A cDNA strand is formed on the separated single stranded DNA template with the help of DNA polymerase enzyme. (iv) (d) (v) (c)

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Class 11 Biology Case Study Questions

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If you are finding it difficult to solve Class 11 Biology Case Study Questions, you are not alone. Many students face difficulties in solving such questions, as they require in-depth knowledge of the subject. However, with the right resources and guidance, it is possible to overcome these difficulties. One of the best resources for Class 11 Biology Case Study Questions is the myCBSEguide app.

myCBSEguide provides detailed information and Class 11 Biology Case Study Questions that can help you understand the concepts better. Class 11 Biology students can also find several practice questions at the end of each chapter that can help Class 11 Biology students in understanding the concept better.

Biology: The study of living organisms

Biology is the study of life and all living things. It is a natural science that covers a wide range of topics, from the structure and function of the human body to the behavior of plants and animals. Class 11 biology students learn about the various branches of biology, such as anatomy, physiology, ecology, and evolution. In addition, they also study the cell, the building block of all living things. By understanding how cells work, students can better understand how the body works as a whole.

Class 11 students entering the world of Biology

For Class 11 students, biology is the foundation for Class 12 CBSE students. It is a vital topic that helps students grasp the fundamental notions of life and living beings. Cell structure and function, genetics, evolution, ecology, and plant and animal physiology are all themes addressed in biology. Biology is a fascinating topic that teaches students about the natural world around them. Biology is an excellent foundation for Class 11 CBSE students who want to pursue a career in medicine. Biology is critical for understanding the human body and its processes, as well as developing medical remedies.

Significance of Biology for class 11 students

• Biology encourages students to learn the fundamentals of biology.
• It promotes a rational/scientific attitude toward issues such as population, environment, and development by encouraging the acquisition of new information and its application to individuals and society.
• It raises public knowledge of environmental issues, problems, and remedies.
• It raises students’ understanding of the diversity of living species and fosters respect for other living beings.
• It understands that even the most complicated biological phenomena are based on fundamentally simple processes.

Case study questions in Class 11 Biology

Case studies are a part of to Class 11 biology examination paper pattern. These case studies can be used to assess a student’s understanding of a subject as well as their ability to apply that understanding in a real-world context. Incorporating case study questions into Class 11 Biology can provide students with a more hands-on and realistic experience with the subject. Class 11 Biology students can better learn how chemical concepts are utilized in the real world by going through real-life problems. Class 11 Biology Case study questions can also aid in the development of critical thinking and problem-solving abilities.

Examples of Class 11 Biology Case Study Questions

We must solve a range of Class 11 Biology case study questions in order to achieve high grades in Biology. Class 11 Biology students must be seeking some examples of case study questions in order to improve their grades. myCBSEguide has prepared a variety of Class 11 Biology case study questions that will undoubtedly assist all students studying the subject. We have compiled a selection of Class 11 Biology case study questions for you. Have a look at the following Class 11 Biology case study question examples.

Class 11 Biology case study questions 1

Read the following and answer any four questions: The detailed structure of the membrane was studied only after the advent of the electron microscope in the 1950s. Meanwhile, chemical studies on the cell membrane, especially in human red blood cells (RBCs), enabled the scientists to deduce the possible structure of the plasma membrane. These studies showed that the cell membrane is composed of lipids, proteins and carbohydrates.

• Nucleic acid
• Carbohydrate
• Phosphoglycerides
• Glycoproteins
• Both assertion and reason are true, and reason is the correct explanation of the assertion.
• Both assertion and reason are true, and reason is not the correct explanation of the assertion.
• Assertion is true but reason is false.
• Both assertion and reason are false.

Answer Key:

• (b) The lipids are arranged in a bilayer in the plasma membrane with the polar head towards the outer sides and the hydrophobic tails towards the inner part.
• (a) The lipid component of the membrane mainly consists of phosphoglycerides.
• (c) In human beings, the membrane of the erythrocyte has approximately 52 percent protein and 40 percent lipids.
• (b) Depending on the ease of extraction, membrane proteins can be classified into two types – integral or peripheral.
• (b) The plasma membrane is selectively permeable to some molecules present on either side of it. Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient, i.e., from higher concentration to the lower. Hence, both assertion and reason are true, but reason is not the correct explanation of the assertion.

Class 11 Biology case study questions 2

Read the following and answer any four questions: Plastids are found in all plant cells and in euglenoids. These are easily observed under the microscope as they are large. They bear some specific pigments, thus imparting specific colours to the plants. Plastids consist of numerous membrane layers embedded in a material called the stroma. They have their own genome and ribosomes.

• Leucoplasts
• Chloroplasts
• Chromoplasts
• Carotenoids
• Amyloplasts
• Aleuroplasts
• Elaioplasts
• (a) The leucoplasts are the colourless plastids of varied shapes and sizes with stored nutrients.
• (b) The aleuroplasts store proteins in grains.
• (a) Amyloplasts are leucoplasts store carbohydrates (starch) in potato.
• (c) The space limited by the inner membrane of the chloroplast is called the stroma.
• (c) The chloroplasts contain chlorophyll and carotenoid pigments which are responsible for trapping light energy essential for photosynthesis. The chromoplasts impart colours to the parts of the plant as yellow, orange or red colour. Hence, Assertion is true but reason is false.

Class 11 Biology case study questions 3

Read the following and answer any four questions: In human beings, the lungs are situated in the thoracic chamber which is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs, and on the lower side by the dome-shaped diaphragm. The anatomical setup of the lungs in the thorax is such that any change in the volume of the thoracic cavity will be reflected in the lung (pulmonary) cavity. Such an arrangement is essential for breathing. Breathing involves two stages – inspiration and expiration. During inspiration, the atmospheric air is drawn in and during expiration, the alveolar air is released out.

• 12 – 16
• 70 – 72
• Ribs lift up
• Diaphragm flattens
• Ribs flatten
• Both ribs lift up and diaphragm flattens
• Tidal volume
• Inspiratory Reserve Volume
• Residual Volume
• Vital Capacity
• 6000 to 8000 mL
• 2500 mL to 3000 mL
• 1000 mL to 1100 mL
• 1100 mL to 1200 mL
• The movement of air into and out of the lungs is carried out by creating a pressure gradient.
• Expiration can occur if the pressure within the lungs (intra-pulmonary pressure) is less than the atmospheric pressure.
• The diaphragm and a specialised set of muscles help in generation of pressure gradients.
• Expiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber in the antero-posterior axis.

Choose from below the correct alternative.

• a. Only I is true
• b. I and IV are true
• c. III and II are true
• d. I and III are true
• (a) On an average, a healthy human breathes 12-16 times/minute.
• (d) When we breathe in, the ribs are lifted up and the diaphragm flattens which increases the size of the chest cavity. Because of this, the air is sucked into the lungs and fills the expanded alveoli.
• (a) Volume of air inspired or expired during normal respiration is called tidal volume.
• (d) Residual volume of air is the remaining air in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL.
• (d) The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere. Inspiration can occur if the pressure within the lungs (intra-pulmonary pressure) is less than the atmospheric pressure. The diaphragm and a specialized set of muscles – external and internal intercostals between the ribs, help in the generation of pressure gradients. Inspiration is initiated by the contraction of the diaphragm which increases the volume of thoracic chamber in the antero-posterior axis. Hence, statements I and III are true.

Class 11 Biology case study questions 4

Read the following and answer any four questions: Exchange of gases also occurs between blood and tissues. O 2  and CO 2  are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient. The solubility of the gases, as well as the thickness of the membranes involved in diffusion, are also some important factors that can affect the rate of diffusion.

• Atmospheric pressure
• Partial pressure
• Differential pressure
• Capillary pressure
• pO 2  – 104 mm Hg, pCO 2  – 40 mm Hg
• pO 2  – 104 mm Hg, pCO 2  – 140 mm Hg
• pO 2  – 95 mm Hg, pCO 2  – 40 mm Hg
• pO 2  – 40 mm Hg, pCO 2  – 45 mm Hg
• The given diagram represents the exchange of gases at the alveolus and the body tissues with blood and the transport of oxygen and carbon dioxide.
• The amount of CO 2  that can diffuse through the diffusion membrane per unit difference in partial pressure is much lesser compared to that of O 2 .
• All the factors in our body are favourable for the diffusion of O 2  from tissues to alveoli and that of CO 2  from alveoli to tissues.
• The total thickness of the diffusion membrane is much less than a millimetre.
• Only I is true
• I and IV are true
• III and II are true
• I and III are true
• (b) Pressure contributed by an individual gas in a mixture of gases is called partial pressure.
• (a) Alveoli are the primary sites of exchange of gases.
• (c) The diffusion membrane is made up of three major layers.
• (d) The values of pO 2  and pCO 2  in the body tissues is: pO 2  – 104 mm Hg, pCO 2  – 40 mm Hg.
• (b) The given diagram represents the exchange of gases at the alveolus and the body tissues with blood and the transport of oxygen and carbon dioxide. The amount of CO 2  that can diffuse through the diffusion membrane per unit difference in partial pressure is much higher compared to that of O 2 . All the factors in our body are favourable for the diffusion of O 2  from alveoli to tissues and that of CO 2  from tissues to alveoli. The total thickness of diffusion membrane is much less than a millimetre.

Dealing with Class 11 Biology case study questions

There are a number of different ways to approach Class 11 Biology case study questions, but the most important thing is to make sure that Class 11 Biology students answer all parts of the question and provide as much detail as possible. In some cases, you may need to research the topic further in order to be able to answer the Class 11 Biology case study questions fully.

When dealing with Class 11 Biology case study questions, it is also important to think about the different perspectives that might be involved. For example, if you are asked to evaluate a particular decision made by a scientist, you will need to consider the impact of that decision from both the scientist’s perspective and the perspective of those affected by the decision.

Answering case study questions can be challenging, but it is an important skill to develop if you want to pursue a career in fields such as business or law. With practice, Class 11 Biology students will be able to approach these questions with confidence and provide well-reasoned, detailed answers.

Class 11 Biology curriculum: As fascinating as Biology itself

The current curriculum of Class 11 Biology provides students with up-to-date principles as well as more extensive exposure to current topics in the discipline. Class 11 Biology curriculum also strives to emphasize the basic concepts that are shared by animals, plants, and microbes, as well as the link between Biology and other fields of study. Class 11 Biology structure provides for a straightforward, sequential flow of ideas. It connects the science of biology to actual life through technological advancements. It connects biological discoveries and breakthroughs to everyday issues including the environment, industry, health, and agriculture. The new curriculum of Class 11 Biology also emphasizes scientific ideas and their application, while ensuring that enough chances and opportunities for mastering and recognizing fundamental concepts remain within its framework.

CBSE Class 11 Biology (Code No. 044)

COURSE STRUCTURE

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Class 12 Biology Case Study Questions Chapter 1 The Living World

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Biology Chapter 1 The Living World Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve NCERT Case Study Questions The Living World to know their preparation level.

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In CBSE Class 12 Biology Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The Living World Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Biology  Chapter 1 The Living World

Case Study/Passage-Based Questions

Case Study 1: There are millions of plants and animals in the world and they are known by their local names in their area. The local names do vary from place to place, even within a country. Hence, scientists have established procedures to assign a scientific name to each organism, which is acceptable to biologists all over the world.

(a) Name the system of naming organisms given by Linnaeus.

Answer: The system of naming organisms given by Linnaeus is called “binomial nomenclature.”

(b) Mention the two components in each scientific name.

Answer: The two components in each scientific name are the genus and species names.

(c) Give the scientific name of: (i) human beings (ii) wheat.

Answer: (Scientific names of the organisms: (i) Human beings: Homo sapiens (ii) Wheat: Triticum aestivum

Case Study 2: The Living World provides a fundamental understanding of the diversity of life forms on Earth. It begins by defining life and living organisms, moving on to explain the concept of biodiversity, its types, and the importance of this diversity. The chapter further introduces the hierarchal organization of life forms in the biosphere, starting from individuals and progressing to species, genus, families, and so on, up to the kingdom level. It also emphasizes the system of binomial nomenclature developed by Carl Linnaeus and the importance of classification, taxonomy, and systematics in biology. Additionally, it provides insights into herbariums, botanical gardens, zoological parks, and museums, which serve as repositories of collected plant and animal specimens.

What is Biodiversity?

A) The variety of life forms present on Earth

B) The system of naming organisms

C) The study of plant and animal specimens

D) The process of classifying organisms into various categories

Answer: A) The variety of life forms present on Earth

What does the hierarchal organization of life forms begin with?

Answer: C) Species

Who developed the system of binomial nomenclature?

A) Charles Darwin

B) Gregor Mendel

C) Carl Linnaeus

D) Jean-Baptiste Lamarck

Answer: C) Carl Linnaeus

What is the importance of classification in biology?

A) It helps in understanding the interrelationships among different groups of organisms

B) It helps in defining life

C) It helps in identifying the number of species on Earth

D) It helps in understanding the concept of biodiversity

Answer: A) It helps in understanding the interrelationships among different groups of organisms

What is the purpose of botanical gardens and zoological parks?

A) They serve as tourist attractions

B) They serve as places for scientific research

C) They serve as repositories of plant and animal specimens

D) They serve as places for breeding endangered species

Answer: C) They serve as repositories of plant and animal specimens

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Biology Chapter 1 The Living World with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Biology The Living World Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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• Biology Important Questions
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CBSE Important Questions for Class 12 Biology

CBSE Important Questions for Class 12 Biology are prepared by the subject experts and are extensively helpful for students in their exam preparations. These important questions provide a strategy to prepare for various board examinations and other competitive exams.

Biology is the most important subject for students preparing for medical entrance exams, including  AIIMS, NEET, etc. Students aspiring to excel in the board and entrance examinations require deep knowledge and a thorough understanding of this subject. Therefore, the important questions for CBSE Class 12 Biology are essential for students to prepare for their exams.

For competitive examinations, students require an excellent level of preparation. Therefore, we cover all the CBSE Important Questions which are necessary for students preparing for both the Class 12 board and medical entrance examinations. All the questions in these materials are framed according to the NCERT textbook.

Class 12 Biology Important Questions

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Case Study Questions for Class 12 Biology Chapter 12 Biotechnology and Its Applications

• Last modified on: 2 years ago
• Reading Time: 5 Minutes

Question 1:

Transgenic cows have extra gene or genes inserted into their DNA. Firstly the genes for the desired product is identified and sequenced. Then a gene construct containing this desired gene is introduced into female cow cells. Transgenic bovine cells are selected and fused with bovine oocytes that have had all of their chromosomes removed. Once fused with the oocyte, the transgenic cells chromosomes are reprogrammed to direct development which can be implanted into a recipient cow. The resulting transgenic cow only express the transgene in her milk. This is because expression of the transgene is controlled by a promoter specific to lactating mammary cells.

(i) The gene construct with desired gene is introduced into female cow cells by (a) transformation (b) transduction (c) transfection (d) transplantation.

(ii) Production of transgenic cow fulfill the objective of (a) increased milk production (b) increased meat production (c) molecular farming (d) all of these.

(iii) The name of first transgenic cow is (a) Tracy (b) Dolly (c) Rosie (d) ANDI.

(iv) Transgenic cow is produced through the implantation of ______ containing transgene into recipient cow. (a) ova (b) embryo (c) mammary cell (d) both (a) and (b)

(v) Read the given statements and select the correct option. Statement I : Transgenes only express in the mammary glands of transgenic cow. Statement II : Transgenes are present in chromosomes of every cell in transgenic cow. (a) Both statements I and II are true. (b) Both statements I and II are false. (c) Statement I is true but statement II is false. (d) Statement I is false but statement II is true.

Case Study and Passage Based Questions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

Last modified on:2 years agoReading Time:2MinutesCase Study/Passage Based Questions: Question 1: Given below is a flowchart for the formation of somatic hybrid, Pomato. (i) A certain tissue, of a plant, infected with TMV was used to obtain a new plant using tissueculture technique. Identify the technique used and reason out the possibility of obtaininga new healthy…

Case Study and Passage Based Questions for Class 12 Biology Chapter 7 Evolution

Last modified on:2 years agoReading Time:2MinutesCase Study/Passage Based Questions: Question 1: Darwin found the varieties of finches that in travelled to Galapagos Islands and observed variations in them. (i) What role does an individual organism play as per Darwin’s theory of natural selection?(ii) How did Darwin explain the existence of different varieties of finches on Galapagos…

Case Study and Passage Based Questions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Last modified on:2 years agoReading Time:2MinutesCase Study/Passage Based Questions: Question 1: Given below is the diagram of a tRNA molecule. Answer the questions based on the above diagram:(i) Why is charging of tRNA essential in translation?(ii) Where does peptide bond formation occur in a bacterial ribosome?(iii) Name the scientist who called tRNA an adaptor molecule. Answer…

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Case Study Questions Class 11 Biology Cell : The Unit of Life

Case study questions class 11 biology chapter 8 cell : the unit of life.

CBSE Class 11 Case Study Questions Biology Cell : The Unit of Life. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Cell : The Unit of Life.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Biology Cell : The Unit of Life

In 1838, Matthias Schleiden, a German botanist, examined a large number of plants and observed that all plants are composed of different kinds of cells which form the tissues of the plant. At about the same time, Theodore Schwann (1839), a British Zoologist, studied different types of animal cells and reported that cells had a thin outer layer which is today known as the ‘plasma membrane’. He also concluded, based on his studies on plant tissues, that the presence of cell wall is a unique character of the plant cells. On the basis of this, Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells.

Schleiden and Schwann together formulated the cell theory. This theory however, did not explain as to how new cells were formed. Rudolf Virchow (1855) first explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape. Cell theory as understood today is: (i) all living organisms are composed of cells and products of cells. (ii) All cells arise from pre-existing cells.

1.) Identify the incorrect statement

Statement 1 – Theodore Schwann reported the presence of cell membrane

Statement 2 – Rudolph Virchowgive the cell theory a final shape.

Statement 3 – New cells arise from pre-existing cells.

Statement 4 – Living organisms are composed of cells and products of cells.

a.) Statement 1 and 3 are incorrect

b.) Statement 2 and 3 are incorrect

c.) Statement 4 is incorrect

d.) All statement are correct

2.) ____________ proposed the theory, which states that the bodies of animals and plants are composed of cells and products of cells.

a) Antony Von Leeuwenhoek

b) Matthias Schleiden (1838)

c) Rudolph Virchow (1855)

d.) Theodore Schwann (1839)

3.) Give the name of scientist who stated that the animal cell had a thin outer layer.

4.) Who was the first to explain that new cells arise from pre-existing cells and gave final shape to cell theory?

5.) What is mean by “Omnis cellula-e cellula”?

6.) Explain cell theory and give its postulates.

3.) British Zoologist Theodore Schwann (1839) studied different types of animal cells and stated that animal cells had a thin outer layer which is today known as the ‘plasma membrane’.

4.) Rudolf Virchow (1855) explained that cells divided and new cells are formed from pre-existing cells.

5.) “Omnis cellula-e cellula” – means new cells develop by cell division of pre-existing cell.

6.) Schleiden and Schwann together formulated the cell theory. This theory however, did not explain as to how new cells were formed. Rudolf Virchow (1855) first explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e cellulae). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape.

Cell theory postulates,

• All living organisms are composed of cells and products of cells.
• All cells arise from pre-existing cells.

The prokaryotic cells are represented by bacteria, blue-green algae, mycoplasma and PPLO (Pleuro Pneumonia like Organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells. They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (spiral).

The organisation of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane except in mycoplasma. The fluid matrix filling the cell is the cytoplasm. There is no well-defined nucleus. The genetic material is basically naked, not enveloped by a nuclear membrane. In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA are called plasmids. The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. Nuclear membrane is found in eukaryotes. No organelles, like the ones in eukaryotes, are found in prokaryotic cells except for ribosomes. Prokaryotes have something unique in the form of inclusions. A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of cell membrane.

1.) ______________ is the fluid matrix, which fills the prokaryotic cell.

a.) Cell sap

b) Cytoplasm

d.) Both a & b

2.) Identify incorrect statement

Statement 1 – In prokaryotic cell nucleus is absent.

Statement 2 – In prokaryotic cells genetic material appears naked.

Statement 3 – In prokaryotic cells genetic material not enveloped by a nuclear membrane.

Statement 4 – prokaryotic cells do not have a cell wall.

c.) Both 2 & 3

d.) All of the above

3.) Give reason – why genetic material in prokaryotic cell is not enveloped in nuclear membrane?

4.) Define mesosome.

5.) Give the any two characteristic of prokaryotic cells.

6.) Enlist the shapes of bacteria are generally occurs.

3.) In prokaryotic cell well-defined nucleus is absent. Hence all genetic material appears naked and not enveloped by a nuclear membrane.

4.) Mesosome – A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of cell membrane

5.) Characteristic of prokaryotic cells;

• There is no well-defined nucleus.
• The genetic material is basically naked, not enveloped by a nuclear membrane.
• Cellulose is the fluid matrix, which fills the prokaryotic cell.
• The prokaryotic cells are generally smaller and multiply more rapidly than the eukaryotic cells.
• All prokaryotes have a cell wall surrounding the cell membrane except in mycoplasma.

6.) The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (spiral).

Studies showed that the cell membrane is mainly composed of lipids and proteins. The major lipids are phospholipids that are arranged in a bilayer. Also, the lipids are arranged within the membrane with the polar head towards the outer sides and the hydrophobic tails towards the inner part. This ensures that the nonpolar tail of saturated hydrocarbons is protected from the aqueous environment. In addition to phospholipids membrane also contains cholesterol.

The ratio of protein and lipid varies considerably in different cell types. In human beings, the membrane of the erythrocyte has approximately 52 per cent protein and 40 per cent lipids.

Depending on the ease of extraction, membrane proteins can be classified as integral and peripheral. Peripheral proteins lie on the surface of membrane while the integral proteins are partially or totally buried in the membrane.

An improved model of the structure of cell membrane was proposed by Singer and Nicolson (1972) widely accepted as fluid mosaic model. According to this, the quasi-fluid nature of lipid enables lateral movement of proteins within the overall bilayer. This ability to move within the membrane is measured as its fluidity.

One of the most important functions of the plasma membrane is the transport of the molecules across it. The membrane is selectively permeable to some molecules present on either side of it. Many molecules can move briefly across the membrane without any requirement of energy and this is called the passive transport. Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient, i.e., from higher concentration to the lower. Water may also move across this membrane from higher to lower concentration. Movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the nonpolar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane. A few ions or molecules are transported across the membrane against their concentration gradient, i.e., from lower to the higher concentration. Such a transport is an energy dependent process, in which ATP is utilised and is called active transport, e.g., Na+/K+ Pump.

1.) _______________ is the major constituent of the cell membrane.

a) Phospholipids

b) Lipoproteins

c) Proteins

d) Carbohydrates

2.) The lipids arranged within membrane in such way, that the polar head oflipids towards the ___________ sides and the hydrophobic tails towards the ________side.

a) inner ,   outer

b) outer ,   inner

c) outer ,   outer

d) inner ,  inner

3.) In cell membrane, the tail of phospholipid is _________________.

a.) Polar and hydrophilic

b) Non-polar and hydrophilic

c) Polar and hydrophobic

d) Non-polar and hydrophobic

4.) Define simple diffusion.

5.) Give the classification of membrane protein.

6.) What is the ratio of protein and lipid present in cell membrane of erythrocyte?

4.) Simple diffusion it is defined as the movement of the neutral solute across the membrane along with the concentration gradient, i.e., from higher concentration to the lower.

5.) Depending on the ease of extraction, membrane proteins can be classified as integral and peripheral. Peripheral proteins lie on the surface of membrane while the integral proteins are partially or totally buried in the membrane.

6.) The ratio of protein and lipid varies considerably in different cell types. In human beings, the membrane of the erythrocyte has approximately 52 per cent protein and 40 per cent lipids.

The membranous organelles is distinct in terms of its structure and function, many of these are considered together as an endomembrane system because their functions are coordinated. The endomembrane system include endoplasmic reticulum (ER), Golgi complex, lysosomes and vacuoles. Since the functions of the mitochondria, chloroplast and peroxisomes are not coordinated with the above components, these are not considered as part of the endomembrane system.

Electron microscopic studies of eukaryotic cells reveal the presence of a network or reticulum of tiny tubular structures scattered in the cytoplasm that is called the endoplasmic reticulum (ER). Hence, ER divides the intracellular space into two distinct compartments, i.e., luminal (inside ER) and extra luminal (cytoplasm) compartments. The ER often shows ribosomes attached to their outer surface. The endoplasmic reticulum bearing ribosomes on their surface is called rough endoplasmic reticulum (RER). In the absence of ribosomes they appear smooth and are called smooth endoplasmic reticulum (SER). RER is frequently observed in the cells actively involved in protein synthesis and secretion. They are extensive and continuous with the outer membrane of the nucleus. The smooth endoplasmic reticulum is the major site for synthesis of lipid. In animal cells lipid-like steroidal hormones are synthesised in SER.

1.) Endomembrane system facilitate the ______________

a) Coordinated functions of cell organelles with in the cell

b) Coordinated functions of membranous organelles with in the cell

c) Coordinated functions of plasma membrane with in the cell

d) Both a and b

2.) Endoplasmic reticulum is the ____________________

a) Tiny tubular secretion scattered in the cytoplasm

b) tiny tubular structures embedded in membrane

c) Tiny tubular structures scattered in the cytoplasm

d) Tiny tubular structures scattered in the vacuoles

3) Define Rough endoplasmic reticulum and Smooth endoplasmic reticulum.

4) Give the functions of Endoplasmic reticulum.

5) Enlist the name of membranous organelles which comes under endomembrane system and why?

3.) Rough endoplasmic reticulum: The ER which shows ribosomes attached to their outer surface which gives rough appearance, suchribosomebearing ER is called as rough endoplasmic reticulum (RER). In the absence of ribosomes they appear smooth and are called smooth endoplasmic reticulum (SER).

4.) Functions of endoplasmic reticulum:

• Protein synthesis
• Synthesis of lipids
• Synthesis of steroids
• Synthesis of hormones

5.) Membranous organelles are distinct in terms of structure and function, many of these are considered together as an endomembrane system because their functions are coordinated. The endomembrane system include endoplasmic reticulum (ER), Golgi complex, lysosomes and vacuoles

Camillo Golgi (1898) first observed densely stained reticular structures near the nucleus. These were later named Golgi bodies after him. They consist of many flat, disc-shaped sacs or cisternae of 0.5µm to 1.0µm diameter. These are stacked parallel to each other. Varied number of cisternae are present in a Golgi complex. The Golgi cisternae are concentrically arranged near the nucleus with distinct convex cis or the forming face and concave trans or the maturing face. The cis and the trans faces of the organelle are entirely different, but interconnected. The Golgi apparatus principally performs the function of packaging materials, to be delivered either to the intra-cellular targets or secreted outside the cell. Materials to be packaged in the form of vesicles from the ER fuse with the cis face of the Golgi apparatus and move towards the maturing face. A number of proteins synthesised by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus before they are released from its Trans face. Golgi apparatus is the important site of formation of glycoproteins and glycolipids. It is also involved in the formation of lysosomes.

Mitochondria unless specifically stained, are not easily visible under the microscope. The number of mitochondria per cell is variable depending on the physiological activity of the cells. In terms of shape and size also, considerable degree of variability is observed. Typically it is sausage-shaped or cylindrical having a diameter of 0.2-1.0µm (average 0.5µm) and length 1.0-4.1µm. Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e., the outer compartment and the inner compartment. The inner compartment is filled with a dense homogeneous substance called the matrix. The outer membrane forms the continuous limiting boundary of the organelle. The inner membrane forms a number of infoldings called the cristae towards the matrix. The cristae increase the surface area. The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP, hence they are called ‘power houses’ of the cell. The matrix also possesses single circular DNA molecule, a few RNA molecules, ribosomes (70S) and the components required for the synthesis of proteins. The mitochondria divide by fission.

1.) Golgi cisternae formed faces of the Golgi apparatus are ____________

b) Different

c) Interconnected

d) Both b and c

2.) In mitochondria, on the inner membrane number of infoldings present, are called as __________

b) Follicles

d) Cisternae

3.) Name the method by which mitochondria divides to form new mitochondria.

4.) Name site which releases the proteins, synthesised by ribosomes on the endoplasmic reticulum.

5.) Give the functions of Golgi apparatus.

6.) Why the Golgi apparatus remains in close association with the ribosomes?

3.) The mitochondria divide by fission.

4.) Proteins synthesised by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus and then they are released from its Trans face.

5.) Functions of Golgi apparatus:

• Storage, modification and packaging of products
• Involved in formation of lysosomes
• Golgi apparatus is the important site of formation of glycoproteins and glycolipids.
• performs the function of packaging materials, to be delivered either to the intra-cellular targets

6.) A number of proteins synthesised by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus before they are released from its Trans face. Golgi apparatus is the important site of formation of glycoproteins and glycolipids.

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