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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

myCBSEguide: An app as intriguing as biology

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Case Study Questions for Class 11 Biology PDF Download

We have provided here Case Study questions for Class 11 Biology for final board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your exams.

CBSE 11th Standard CBSE Biology question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Biology books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Biology Case Study Questions

CBSE Class 11 Biology question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 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 11 Biology

• Chapter 1 : The Living World
• Chapter 2 : Biological Classification
• Chapter 3 : Plant Kingdom
• Chapter 4 : Animal Kingdom
• Chapter 5 : Morphology of Flowering Plants
• Chapter 6 : Anatomy of Flowering Plants
• Chapter 7 : Structural Organisation in Animals
• Chapter 8 : Cell : The Unit of Life
• Chapter 9 : Biomolecules
• Chapter 10 : Cell Cycle and Cell Division
• Chapter 11 : Transport in Plants
• Chapter 12 : Mineral Nutrition
• Chapter 13 : Photosynthesis in Higher Plants
• Chapter 14 : Respiration in Plants
• Chapter 15 : Plant Growth and Development
• Chapter 16 : Digestion and Absorption
• Chapter 17 : Breathing and Exchange of Gases
• Chapter 18 : Body Fluids and Circulation
• Chapter 19 : Excretory Products and their Elimination
• Chapter 20 : Locomotion and Movement
• Chapter 21 : Neural Control and Coordination
• Chapter 22 : Chemical Coordination and Integration

Class 11 MCQ Questions

Class 11 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 11 Biology examinations. Our expert faculty for standard 11 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 11 students understand the concepts and also easy to learn solutions.

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Case Study Questions for Class 11 Biology

Chapter 1: The Living World Chapter 2: Biological Classification Chapter 3: Plant Kingdom Chapter 4: Animal Kingdom Chapter 5: Morphology of Flowering Plants Chapter 6: Anatomy of Flowering Plants Chapter 7: Structural Organisation in Animals Chapter 8: The Unit of Life Chapter 9: Biomolecules Chapter 10: Cell Cycle and Cell Division Chapter 11: Transport in Plants Chapter 12: Mineral Nutrition Chapter 13: Photosynthesis in Higher Plants Chapter 14: Respiration in Plants Chapter 15: Plant Growth and Development Chapter 16: Digestion and Absorption Chapter 17: Breathing and Exchange of Gases Chapter 18: Body Fluids and Circulation Chapter 19: Excretory Products and Their Elimination Chapter 20: Locomotion and Movement Chapter 21: Neural Control and Coordination Chapter 22: Chemical Coordination and Integration

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CBSE Important Questions for Class 11 Biology are available in Printable format for Free Download.Here you may find NCERT Important Questions and Extra Questions for Class 11 Biology chapter wise with answers also. These questions will act as chapter wise test papers for Class 11 Biology. These Important Questions for Class 11 Biology are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Board Examinations

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Biology Topics to be covered for Class 11 Science

• The Living World
• Biological Classification
• Plant Kingdom
• Animal Kingdom
• Morphology of Flowering Plants
• Anatomy of Flowering Plants
• Structural Organisation in Animals
• Cell-The Unit of Life
• Biomolecules
• Cell Cycle and Cell Division
• Transport in Plants
• Mineral Nutrition
• Photosynthesis in Higher Plants
• Respiration in Plants
• Plant - Growth and Development
• Digestion and Absorption
• Breathing and Exchange of Gases
• Body Fluids and Circulation
• Excretory Products and Their Elimination
• Locomotion and Movement
• Neural Control and Coordination
• Chemical Coordination and Integration

Structure of CBSE Biology Sample Paper for Class 12 Science is

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CBSE Class 11 Biology Sample Papers

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Importance of Question Bank for Exam Preparation?

There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows

• Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
• CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.
• Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.
• Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.

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Class 11 Biology: Case Study of Chapter 19 Excretory Products and their Elimination PDF Download

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In Class 11 Final Exams 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 11 Biology Chapter 19 Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Biology Excretory Products and their Elimination to know their preparation level.

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In CBSE Class 11 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.

Excretory Products and their Elimination Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Biology  Chapter 19 Excretory Products and their Elimination

Case Study/Passage-Based Questions

Case Study 1:

In humans, the excretory system consists of a pair of kidneys, one pair of ureters, a urinary bladder and a urethra. Kidneys are reddish brown, bean shaped structures situated between the levels of last thoracic and third lumbar vertebra close to the dorsal inner wall of the abdominal cavity. Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120- 170 g. towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter. Inner to the hilum is a broad funnel shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and an inner medulla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces. The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Each kidney has nearly one million complex tubular structures called nephrons, which are the functional units. Each nephron has two parts – the glomerulus and the renal tubule. Glomerulus is a tuft of capillaries formed by the afferent arteriole – a fine branch of renal artery. Blood from the glomerulus is carried away by an efferent arteriole.

The renal tubule begins with a double walled cup-like structure called Bowman’s capsule, which encloses the glomerulus. Glomerulus alongwith Bowman’s capsule, is called the Malpighian body or renal corpuscle. The tubule continues further to form a highly coiled network – proximal convoluted tubule (PCT). A hairpin shaped Henle’s loop is the next part of the tubule which has a descending and an ascending limb. The ascending limb continues as another highly coiled tubular region called distal convoluted tubule (DCT). The DCTs of many nephrons open into a straight tube called collecting duct, many of which converge and open into the renal pelvis through medullary pyramids in the calyces.

The Malpighian corpuscle, PCT and DCT of the nephron are situated in the cortical region of the kidney whereas the loop of Henle dips into the medulla. In majority of nephrons, the loop of Henle is too short and extends only very little into the medulla. Such nephrons are called cortical nephrons. In some of the nephrons, the loop of Henle is very long and runs deep into the medulla. These nephrons are called juxta medullary nephrons.

The efferent arteriole emerging from the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries. A minute vessel of this network runs parallel to the Henle’s loop forming a ‘U’ shaped vasa recta. Vasa recta is absent or highly reduced in cortical nephrons.

1.) Inner side to the hilum is a broad funnel shaped cavity space is present, is known as___________

• a) Renal pelvic
• b) Renal pelvis
• c) Renal tubule

Ans: b) Renal pelvis

2.) Blood from turf of capillaries is carried away by ____________

• a) Afferent arteriole
• b) Branch of renal artery
• c) Efferent arteriole
• d) None of the above

Ans: c) Efferent arteriole

3.) How Glomerulus formed?

Ans: Glomerulus is a tuft of capillaries. Glomerulus are formed by a very fine branch of renal artery i.e. afferent arteriole.

4.) What are cortical nephrons?

Ans: Nephrons in which the loop of Henle is too short and extends only very little into the medulla, such nephrons are named as cortical nephrons.

5.) What is juxta medullary nephrons?

Ans: Nephrons in which the loop of Henle is very long and runs deep into the medulla. These nephrons are called juxta medullary nephrons.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Biology Chapter 19 Excretory Products and their Elimination with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 11 Biology Excretory Products and their Elimination 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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Case Study Questions Class 11 Biology Biomolecules

Case study questions class 11 biology chapter 9 biomolecules.

CBSE Class 11 Case Study Questions Biology Biomolecules. 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 Biomolecules.

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 Biomolecules

Enzymes are composed of one or several polypeptide chains. However, there are a number of cases in which non-protein constituents called co-factors are bound to the enzyme to make the enzyme catalytically active. In these instances, the protein portion of the enzymes is called the apoenzyme. Three kinds of cofactors may be identified: prosthetic groups, co-enzymes and metal ions. Prosthetic groups are organic compounds and are distinguished from other cofactors in that they are tightly bound to the apoenzyme. For example, in peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is a part of the active site of the enzyme. Co-enzymes are also organic compounds but their association with the apoenzyme is only transient, usually occurring during the course of catalysis. Furthermore, co-enzymes serve as co-factors in a number of different enzyme catalyzed reactions. The essential chemical components of many coenzymes are vitamins, e.g., coenzyme nicotinamide adenine dinucleotide (NAD) and NADP contain the vitamin niacin. A number of enzymes require metal ions for their activity which form coordination bonds with side chains at the active site and at the same time form one or more cordination bonds with the substrate, e.g., zinc is a cofactor for the proteolytic enzyme carboxypeptidase. Catalytic activity is lost when the co-factor is removed from the enzyme which testifies that they play a crucial role in the catalytic activity of the enzyme.

1.) In complex of protein and co-factors, protein is referred as ___________________.

a) Proenzyme

b) Coenzyme

c) Apoenzyme

d.) Proteinase enzyme.

2.) ________________ Co-factor are found very tightly bound to the apoenzyme.

a.) Co-enzyme

b.) Proenzyme

c,) Proteinase

d.) Prosthetic

3.) Enlist the type of co-factor with examples.

4.) Define co-factors.

5.) What result come off if co-factor is removed from the enzyme?

3.) Three kinds of cofactors

• Prosthetic groups – e.g. Haem
• Co-enzymes – e.g. Niacin
• Metal ions – e.g. zinc

4.) Co-factor are the non-protein constituents are bound to the enzyme to make the enzyme catalytically active. Enzymes are composed of one or several polypeptide chains. However, there are a number of cases in which non-protein constituents are bound to the enzyme to make the enzyme catalytically active.

5.) Catalytic activity is lost when the co-factor is removed from the enzyme which testifies that they play a crucial role in the catalytic activity of the enzyme.

The activity of an enzyme can be affected by a change in the conditions which can alter the tertiary structure of the protein. These include temperature, pH, and change in substrate concentration or binding of specific chemicals that regulate its activity. Temperature and pH Enzymes generally function in a narrow range of temperature and pH. Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH. Activity declines both below and above the optimum value. Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.

Concentration of Substrate With the increase in substrate concentration, the velocity of the enzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (Vmax) which is not exceeded by any further rise in concentration of the substrate. This is because the enzyme molecules are fewer than the substrate molecules and after saturation of these molecules, there are no free enzyme molecules to bind with the additional substrate molecules.

The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme. When the binding of the chemical shuts off enzyme activity, the process is called inhibition and the chemical is called an inhibitor.

When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. Due to its close structural similarity with the substrate, the inhibitor competes with the substrate for the substrate binding site of the enzyme. Consequently, the substrate cannot bind and as a result, the enzyme action declines, e.g., inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure. Such competitive inhibitors are often used in the control of bacterial pathogens.

1.) _______________ is a chemical compound or molecule which is responsible for decrease or stop the enzyme activity by binding to an enzyme.

a.) Catalyser

b) Inhibitor

c) Regulator

d) Controller

2.) _______________ preserve the enzyme and keep them in temporarily inactive state.

a.) Optimum pH

c) Optimum temperature

d) Low temperature

3.) Give reason – why most of the enzymes destroyed in high temperature condition?

4.) Explain the relation between substrate concentration and enzymatic activity?

5.) Explain competitive inhibition and inhibitor.

3.) Enzymes are composed of one or several polypeptide chains. Almost all enzymes are protein. High temperature condition destroys enzymatic activity because proteins are denatured by heat.

4.) Concentration of Substrate With the increase in substrate concentration, the velocity of the enzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (Vmax) which is not exceeded by any further rise in concentration of the substrate. This is because the enzyme molecules are fewer than the substrate molecules and after saturation of these molecules, there are no free enzyme molecules to bind with the additional substrate molecules.

5.) When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. Due to its close structural similarity with the substrate, the inhibitor competes with the substrate for the substrate binding site of the enzyme. Consequently, the substrate cannot bind and as a result, the enzyme action declines. This phenomenon is called as competitive inhibition.

Almost all enzymes are proteins. There are some nucleic acids that behave like enzymes. These are called ribozymes. An enzyme like any protein has a primary structure, i.e., amino acid sequence of the protein. An enzyme like any protein has the secondary and the tertiary structure. When you look at a tertiary structure you will notice that the backbone of the protein chain folds upon itself, the chain criss-crosses itself and hence, many crevices or pockets are made. One such pocket is the ‘active site’. An active site of an enzyme is a crevice or pocket into which the substrate fits. Thus enzymes, through their active site, catalyse reactions at a high rate. Enzyme catalysts differ from inorganic catalysts in many ways, but one major difference needs mention. Inorganic catalysts work efficiently at high temperatures and high pressures, while enzymes get damaged at high temperatures (say above 40°C). However, enzymes isolated from organisms who normally live under extremely high temperatures (e.g., hot vents and sulphur springs), are stable and retain their catalytic power even at high temperatures (upto 80°-90°C). Thermal stability is thus an important quality of such enzymes isolated from thermophilic organisms.

1.) _____________ is the pocket like region of an enzyme into which substrate molecules bind.

a) Protein site

b) Co-factors

c) Coenzyme

d) Active site

2.) Identify incorrect statement

Statement 1 – Nucleic acids which behave like enzymes are commonly termed as nucliozymes.

Statement 2 – An enzyme like any protein has a primary, secondary and the tertiary structure.

Statement 3 – Enzyme catalysts differ from inorganic catalysts in many ways.

Statement 4 – All enzymes are proteins.

b) Both 1 & 3

d.) None of the above

3.) How active site of enzymes are formed?

4.) Explain how Enzyme catalysts differ from inorganic catalysts?

5.) What is ribozymes?

3.) Enzyme have primary, secondary and tertiary structure like proteins. In tertiary structure, backbone of the protein chain folds upon itself, the chain criss-crosses itself and leads to the formation of many crevices or pockets are made. These pockets are referred as active site of enzyme. An active site of an enzyme is a crevice or pocket into which the substrate fits.

4.) Enzyme catalysts differ from inorganic catalysts in many ways. Inorganic catalysts work efficiently at high temperatures and high pressures, while enzymes get damaged at high temperatures (above 40°C). There are some exceptions such as enzyme isolated from thermophilic organisms.

5.) There are some nucleic acid behave like an enzymes, these nucleic acid is termed as ribozymes.

Metabolic pathways can lead to a more complex structure from a simpler structure (for example, acetic acid becomes cholesterol) or lead to a simpler structure from a complex structure (for example, glucose becomes lactic acid in our skeletal muscle). The former cases are called biosynthetic pathways or anabolic pathways. The latter constitute degradation and hence are called catabolic pathways. Anabolic pathways, as expected, consume energy. Assembly of a protein from amino acids requires energy input. On the other hand, catabolic pathways lead to the release of energy. For example, when glucose is degraded to lactic acid in our skeletal muscle, energy is liberated. This metabolic pathway from glucose to lactic acid which occurs in 10 metabolic steps is called glycolysis. Living organisms have learnt to trap this energy liberated during degradation and store it in the form of chemical bonds. As and when needed, this bond energy is utilised for biosynthetic, osmotic and mechanical work that we perform. The most important form of energy currency in living systems is the bond energy in a chemical called adenosine triphosphate (ATP).

There are thousands of chemical compounds in a living organism, otherwise called as metabolites or biomolecules, are present at concentrations characteristic of each of them. For example, the blood concentration of glucose in a normal healthy individual is 4.2 mmol/L– 6.1 mmol/L, while that of hormones would be nanograms/mL. The most important fact of biological systems is that all living organisms exist in a steady-state characterised by concentrations of each of these biomolecules. These biomolecules are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium. The steady state is a non-equilibrium state. Systems at equilibrium cannot perform work. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady state to be able to perform work; living process is a constant effort to prevent falling into equilibrium. This is achieved by energy input. Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous. Without metabolism there cannot be a living state.

1.) ________________ is the destructive process, which involves complex structure breakdown into simple form.

a) Amphibolic pathway

b) Anabolic pathway

c) Catabolic pathway

d) None of the above

2.) ______________ is the normal glucose concentration in normal healthy individual.

a) 9 mmol/L– 6.8 mmol/L

b) 5 mmol/L– 6.5 mmol/L

c) 0 mmol/L– 7.1 mmol/L

d) 2 mmol/L– 6.1 mmol/L

3.) Give any one example of catabolic reaction that take place in human body.

4.) Give the name of chemical bond in which energy liberated during degradation of metabolites, is stored.

5.) Define anabolic pathways and catabolic pathways.

3.) Glucose becomes lactic acid in our skeletal muscle is the catabolic pathway reaction, which constitute degradation of biomolecule and release energy.

4.) In Living organism energy liberated during degradation of metabolites stored in the form of chemical bonds i.e. ATP. Adenosine triphosphate (ATP) is the most important form of energy currency in living systems.

5) Anabolic pathway – Metabolic pathways which leads to a more complex structure from a simpler structure are termed as anabolic pathways or biosynthetic pathways.

Catabolic pathway – Metabolic pathways which leads to a simpler structure from a complex structure are termed as catabolic pathways.

Proteins are heteropolymers containing strings of amino acids. Structure of molecules means different things in different contexts. In inorganic chemistry, the structure invariably refers to the molecular formulae (e.g., NaCl, MgCl2, etc.). Organic chemists always write a two dimensional view of the molecules while representing the structure of the molecules (e.g., benzene, naphthalene, etc.). Physicists conjure up the three dimensional views of molecular structures while biologists describe the protein structure at four levels. The sequence of amino acids i.e., the positional information in a protein – which is the first amino acid, which is second, and so on – is called the primary structure of a protein. A protein is imagined as a line, the left end represented by the first amino acid and the right end represented by the last amino acid. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid. A protein thread does not exist throughout as an extended rigid rod. The thread is folded in the form of a helix, only some portions of the protein thread are arranged in the form of a helix. In proteins, only right handed helices are observed. Other regions of the protein thread are folded into other forms in what is called the secondary structure. In addition, the long protein chain is also folded upon itself like a hollow woollen ball, giving rise to the tertiary structure. This gives us a 3-dimensional view of a protein. Tertiary structure is absolutely necessary for the many biological activities of proteins.

Some proteins are an assembly of more than one polypeptide or subunits. The manner in which these individual folded polypeptides or subunits are arranged with respect to each other (e.g. linear string of spheres, spheres arranged one upon each other in the form of a cube or plate etc.) is the architecture of a protein otherwise called the quaternary structure of a protein (Fig. 9.4 d). Adult human haemoglobin consists of 4 subunits. Two of these are identical to each other. Hence, two subunits of α type and two subunits of β type together constitute the human haemoglobin (Hb).

In a polypeptide or a protein, amino acids are linked by a peptide bond which is formed when the carboxyl (-COOH) group of one amino acid reacts with the amino (-NH2 ) group of the next amino acid with the elimination of a water moiety (the process is called dehydration). In a polysaccharide the individual monosaccharides are linked by a Glycosidic bond. This bond is also formed by dehydration. This bond is formed between two carbon atoms of two adjacent monosaccharides. In a nucleic acid a phosphate moiety links the 3’-carbon of one sugar of one nucleotide to the 5’-carbon of the sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group of sugar is an ester bond. As there is one such ester bond on either side, it is called phosphodiester bond. Nucleic acids exhibit a wide variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson – Crick Model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel i.e., run in the opposite direction. The backbone is formed by the sugar-phosphate-sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand.There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase.

1.) To form polypeptide molecules, number of amino acids joined together by _______________ bond.

a.) Covalent bond

b) Glycosidic bond

c) Peptide bond

d) Phosphodiester bond

2.) Number of monosaccharides are joined together by _____________ to form polysaccharide.

a.) Phosphodiester bond

c) Hydrogen bond

d) Ester bond

3.) Define N-terminal amino acid and c-terminal amino acid.

4.) Explain how amino acid chain formed in the formation of polypeptide molecule.

5.) Name the bond present between nitrogen bases ( A and G / T and C ) of nucleic acid.

3) The first amino acid present in amino acid chain is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid.

4) When the carboxyl (-COOH) group of one amino acid reacts with the amino (-NH2) group of the next amino acid, they form peptide bond between them. This way formation of amino acid chain continuous which leads to the polypeptide.

5) The nitrogen bases A and G of one strand compulsorily base pairs with T and C, respectively, there are two hydrogen bonds between A and T and three hydrogen bonds between G and C.

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Class 11th Biology - Excretory Products and Their Elimination Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Biology Subject - Excretory Products and Their Elimination, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Excretory products and their elimination case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

Animals accumulate nitrogenous wastes, carbon dioxide, water and some inorganic ions as metabolic wastes, which have to be eliminated totally or partially. (a) Name the most toxic and the least toxic nitrogenous wastes. (b) Kidneys do not play any significant role in the elimination of Ammonia. How is ammonia eliminated by animals? (c) What term is given to the animals which excrete urea as their major nitrogenous waste? (d) How is carbon dioxide eliminated by human beings?

A survey of the animal kingdom shows that different groups of animals have different types of excretory structures. They depend on their habitats and the nature of the excretory products. (a) Human kidneys perform not only excretion of nitrogenous wastes, but also another function. Name that function. (b) Name the excretory structures of (i) Amphioxus (ii) Crustaceans (iii) Insects (iv) Platyhelminths (v) Annelids.

Nephrons are the structural and functional of human kidneys; they are the complex tubular structures. (a) What are the two major parts of a nephron? (b) What is renal corpuscle? (c) Name the two types of nephrons in human kidneys.

The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such mechanism carried out by the juxta glomerular apparatus is shown in the following flow chart. Fill in the blanks A, B. C, D, E and F in the flow chart. The (A) cells release Renin when there is a fall in (B) \(\downarrow\)   Renin converts (C) in the in blood into (D) \(\downarrow\)   D activates the adrenal cortex to release (E) \(\downarrow\) E raises the reabsorption of Na + and water and raises the (F)

The renal tubule consists of Bowman's capsule, proximal convoluted tubule, Henle's loop, distal convoluted tubule and the collecting tubule. 70-80o/o of the glomerular filtrate is reabsorbed in the PCT. (a) How is the structure of PCT suited for this absorption? (b) Why is glomerular filtration considered as ultra filtration? (c) How does the descending limb of Henle's loop differ from the ascending limb in its permeability to substances?

Mammals/humans can excrete hyperosmotic urine or hypo-osmotic urine depending on the availability of water in the body. The regulation of kidney functioning is under both neural and hormonal control. (a) Where are the osmoreceptors located in our body? (b) Mention any two factors that activate the osmoreceptors (c) Name the hormone secreted under water scarcity. Mention the two ways in which it acts.

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Excretory products and their elimination case study questions with answer key answer keys.

(a) (i) Ammonia is the most toxic. (ii) Uric acid is the least toxic (b) Ammonia is eliminated (i) by diffusion through body surface. (ii) through surface of gills (in flshes). (c) Ureoteiic animals. (d) Carbon dioxide is removed by the lungs

(a) Osmoregulation (b) (i) Protonephridia (ii) Green glands (iii) Malpighian tubules (iv) Flame cells (v) Nephridia.

(a) Glomerulus and renal tubule are the two parts of a nephron. (b) The Bowman's capsule along with the glomerulus in its cavity, is called renal cotpuscle. (c) Cortical nephrons and juxtamedullary nephrons.

(a) A - Efferent arteriole B - Afferent arteriole C - Glomerulus D - Bowman's capsule (b) It continues as the proximal convoluted tubule (PCT). (c) The fine capillary network formed by the efferent arteriole around the renal tubule, is called peritubular capillaries.

A - Juxtaglomerular B - Glomerular filtration rate C -  Angiotensinogen  D - Angiotensin II E - Aldosterone F - Blood pressure and glomerular filtration rate.

(a) PCT is lined by simple cuboidal brush-bordered epithelium that increases the surface area for absorption. (b) Since almost all the constituents of plasma except proteins, arefilterg4 it is called ultrafiltration. (c) Ascending limb'is permeable to electrolytes, but impermeable to water. Descending limb is permeable to water, but impermeable to electrolytes.

(a) Osmoreceptors are located in the hypothalamus. (b) They, are activated by changes in the (i) blood volume (ii) body fluid volume and (iii) ionic concentration. (c)  Antidiuretiq hormone/Vasopressin (i) It induces water reabsorption by DCT. (ii) It constricts the blood vessels, increases the blood pressure and thereby the GFR.

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