Maharashtra Board Class 10 Science Chapter 2 Life Processes in Living Organisms Part 1 Solutions

Std 10 Science Part 2 Chapter 2 Life Processes In Living Organisms Part - 1 Question Answer Maharashtra Board

Question 1. Fill in the blanks and explain the statements.
a. After complete oxidation of a glucose molecules, .......... number of ATP molecules are formed.
Answer: After complete oxidation of a glucose molecules, 38 number of ATP molecules are formed.
In simple words: When glucose is fully broken down in the presence of oxygen, it yields a total of 38 ATP molecules, which are the cell's energy currency.

🎯 Exam Tip: Remember the total ATP yield from complete glucose oxidation (38 ATP) as it's a fundamental concept in cellular respiration.

b. At the end of glycolysis, .......... molecules are obtained.
Answer: At the end of glycolysis, pyruvate molecules are obtained.
In simple words: Glycolysis, the initial stage of glucose breakdown, converts one glucose molecule into two molecules of pyruvate.

🎯 Exam Tip: Glycolysis is the first step in both aerobic and anaerobic respiration, producing pyruvate, ATP, and NADH2.

c. Genetic recombination occurs in .......... phase of prophuse of meiosis-l.
Answer: Genetic recombination occurs in pachytene phase of prophase of meiosis-l.
In simple words: During the pachytene stage of prophase-I in meiosis, homologous chromosomes exchange genetic material through a process called crossing over, leading to genetic recombination.

🎯 Exam Tip: Pachytene is crucial for genetic diversity due to crossing over, a key event in meiosis-I.

d. All chromosomes are arranged parallel to equatorial plane of cell in ............. phase of mitosis.
Answer: All chromosomes are arranged parallel to equutorial plane of cell in metaphase phase of mitosis.
In simple words: In mitosis, during the metaphase stage, all chromosomes line up precisely along the central equatorial plate of the cell, preparing for separation.

🎯 Exam Tip: Metaphase is characterized by the alignment of chromosomes at the metaphase plate, a visually distinct stage in cell division.

e. For formation of plasma membrane, phospholipid molecules are necessary.
Answer: For formation of plasma membrane, phospholipid molecules are necessary.
In simple words: The plasma membrane, which encloses cells, is primarily composed of a bilayer of phospholipid molecules that provide its structural integrity and fluidity.

🎯 Exam Tip: Phospholipids are fundamental to cell structure, forming the basic framework of all biological membranes.

f. Our muscle cells perform ............. type of respiration during exercise.
Answer: Our muscle cells perform anaerobic type of respiration during exercise.
In simple words: When muscle cells have insufficient oxygen during intense exercise, they switch to anaerobic respiration to produce energy, which also results in lactic acid accumulation.

🎯 Exam Tip: Anaerobic respiration in muscle cells during strenuous activity is a common example of how organisms adapt to oxygen scarcity.

Question 2. Write definitions.
a. Nutrition.
Answer: Nutrition: The process of taking nutrients in the body and utilizing them by an organism is known as nutrition.
In simple words: Nutrition is the biological process where an organism obtains and uses food for growth, metabolism, and repair.

🎯 Exam Tip: Focus on both the intake and utilization aspects when defining nutrition.

b. Nutrients.
Answer: Nutrients: The substances like carbohydrates, proteins, lipids, vitamins, minerals etc. which are components of the food are called nutrients.
In simple words: Nutrients are essential chemical substances found in food that living organisms need for energy, growth, and proper bodily functions.

🎯 Exam Tip: List a few key categories of nutrients (carbohydrates, proteins, fats, vitamins, minerals) to provide a comprehensive answer.

c. Proteins.
Answer: Proteins: Protein is a macromolecule which is formed by many amino acids which are joined by peptide bonds.
In simple words: Proteins are large, complex organic molecules made up of long chains of smaller units called amino acids, linked together by peptide bonds.

🎯 Exam Tip: Emphasize proteins as macromolecules formed from amino acids via peptide bonds.

d. Cellular respiration.
Answer: Cellular respiration: Oxidation of glucose and other food components which takes place inside the cell in presence or absence of oxygen, is known as cellular respiration.
In simple words: Cellular respiration is the process within cells that breaks down glucose to release energy, which can occur with or without oxygen.

🎯 Exam Tip: Highlight that cellular respiration involves glucose oxidation and can be either aerobic (with oxygen) or anaerobic (without oxygen).

e. Aerobic respiration.
Answer: Aerobic respiration: Cellular respiration taking place in presence of oxygen is known as aerobic respiration.
In simple words: Aerobic respiration is a type of cellular respiration that requires oxygen to efficiently break down glucose and generate a large amount of energy.

🎯 Exam Tip: The key differentiator for aerobic respiration is the requirement for oxygen and its high energy yield.

f. Glycolysis.
Answer: Glycolysis: The process occurring in the cell where a molecule of glucose is oxidized in step by step process forming two molecules of each of pyruvic acid, ATP, NADH2 and water, is called glycolysis.
In simple words: Glycolysis is the metabolic pathway that breaks down a glucose molecule into two pyruvate molecules, producing a net gain of ATP and NADH2, occurring in the cytoplasm.

🎯 Exam Tip: Remember the end products of glycolysis: two pyruvic acid molecules, two ATP, and two NADH2.

Question 3. Distinguish between
a. Glycolysis and TCA cycle
Answer:
Glycolysis:
• The process of glycolysis occurs in the cytoplasm of the cell.
• In glycolysis, one molecule of glucose is oxidized step-by-step to produce two molecules each of pyruvic acid, ATP, NADH2 and water.
• Glycolysis can take place in both aerobic and anaerobic respiration.
• The first step in cellular respiration is glycolysis where glucose is converted into pyruvate.
• Two molecules of pyruvate are obtained in glycolysis.
• Two molecules of ATP are used up in glycolysis.
• Four molecules of ATP are produced in glycolysis.
• CO2 is not produced during glycolysis.
TCA cycle:
• TCA cycle takes place in mitochondria.
• In TCA cycle, molecule of acetyl-co-A is completely oxidized and in the process CO2, H2O, NADH2, FADH2 and ATP is produced.
• TCA cycle takes place only during aerobic respiration.
• The second step in cellular respiration is TCA cycle.
• Pyruvate is converted into CO2 and H2O during TCA cycle.
• ATP molecules are not used up in TCA cycle.
• Two molecules of ATP are produced in TCA cycle.
• CO2 is produced in TCA cycle.
In simple words: Glycolysis is the initial breakdown of glucose in the cytoplasm, producing pyruvate and a small amount of ATP, and can occur with or without oxygen. The TCA cycle (Krebs cycle) is a subsequent aerobic process in the mitochondria, where acetyl-CoA is completely oxidized, generating significant amounts of electron carriers (NADH2, FADH2) and CO2.

🎯 Exam Tip: Focus on the location (cytoplasm vs. mitochondria), oxygen requirement (both vs. aerobic only), and main products (pyruvate vs. CO2, H2O, electron carriers) to differentiate effectively.

b. Mitosis and meiosis.
Answer:
Mitosis:
• In mitosis the chromosome number does not change. Diploid cells remain diploid, without change.
• One cell gives rise to two daughter cells in mitosis.
• Karyokinesis of mitosis has four stages, viz. prophase, metaphase, anaphase and telophase.
• Prophase of mitosis is not lengthy.
• Genetic recombination does not happen in mitosis as there is no crossing over.
• Mitosis is essential for growth and development.
• Mitosis takes place both in somatic cells and germinal cells.
Meiosis:
• In meiosis, the chromosome number is reduced to half. The diploid cells become haploid.
• One cell gives rise to four daughter cells in meiosis.
• Meiosis has two major stages, viz. meiosis-I and meiosis-II. Each is further subdivided into prophase, metaphase, anaphase and telophase.
• Prophase of meiosis-l is very lengthy.
• Genetic recombination takes place in homologous chromosomes as there is crossing over during prophase-l.
• Meiosis is essential for formation of gametes in sexual reproduction.
• Meiosis takes place in only germinal cells. It does not take place in somatic cells.
In simple words: Mitosis is a cell division that produces two genetically identical diploid daughter cells, essential for growth and repair. Meiosis, on the other hand, involves two divisions, resulting in four genetically diverse haploid daughter cells, crucial for sexual reproduction and gamete formation.

🎯 Exam Tip: Key distinctions include number of divisions, chromosome number in daughter cells (diploid vs. haploid), genetic identity (identical vs. diverse), purpose (growth/repair vs. reproduction), and occurrence (somatic/germinal vs. germinal only).

c. Aerobic and anaerobic respiration.
Answer:
Aerobic respiration:
• Oxygen is required for aerobic respiration.
• Aerobic respiration takes place in nucleus as well as in cytoplasm.
• At the end of aerobic respiration CO2 and H2O is formed.
• Energy is produced in large amount in aerobic respiration.
• Glucose is completely oxidized in aerobic respiration.
• 38 molecules of ATP are formed during aerobic respiration.
• Chemical reaction:
\( \text{C6H12O6} + 6\text{O2} \implies 6\text{H2O} + 6 \text{CO2} + 686 \text{ Kcal} \)
Anaerobic respiration:
• Oxygen is not required for anaerobic respiration.
• Anaerobic respiration occurs only in the cytoplasm.
• At the end of anaerobic respiration CO2 and C2H5OH are formed.
• Energy is produced in lesser amount in anaerobic respiration.
• Glucose is incompletely oxidized in anaerobic respiration.
• 2 molecules of ATP are formed during anaerobic respiration.
• Chemical reaction:
\( \text{C6H12O6} \implies 2 \text{ C2H5OH} + 2 \text{ CO2} + 50 \text{ Kcal} \)
In simple words: Aerobic respiration requires oxygen, produces a large amount of ATP (38 molecules), and completely oxidizes glucose to CO2 and H2O, primarily in the mitochondria. Anaerobic respiration occurs without oxygen, yields much less ATP (2 molecules), and incompletely breaks down glucose, often producing alcohol or lactic acid, exclusively in the cytoplasm.

🎯 Exam Tip: Compare and contrast the oxygen requirement, ATP yield, location within the cell, and end products for both types of respiration, ensuring correct chemical equations are included.

Question 4. Give scientific reasons.
a. Oxygen is necessary for complete oxidation of glucose.
Answer:
1. When glucose is completely oxidized in aerobic cellular respiration, it produces 38 molecules of ATP.
2. In cellular respiration, three processes take place one after the other, these are glycolysis, Krebs cycle and electron transport chain reactions.
3. In absence of oxygen only glycolysis can occur but further two reactions will not take place.
4. If glycolysis occurs in absence of oxygen, it produces alcohol.
5. By anaerobic glycolysis only two molecules of ATP are produced.
6. This results in less energy supply to the body. Therefore, oxygen is necessary for complete oxidation of glucose.
In simple words: Oxygen is vital for the Krebs cycle and electron transport chain, which follow glycolysis, enabling the full breakdown of glucose to generate 38 ATP molecules; without oxygen, only glycolysis occurs, yielding significantly less energy (2 ATP) and often producing byproducts like alcohol.

🎯 Exam Tip: Explain the stages of aerobic respiration and highlight how oxygen drives the more efficient energy-producing steps beyond glycolysis.

b. Fibres are one of the important nutrients. (Board's Model Activity Sheet)
Answer:
1. Fibres are indigestible substance.
2. They are thrown out along with other useless and undigested matter.
3. This aids in egestion. Some fibres also help in digestion of other substances.
4. Green leafy vegetables, fruits, cereals, etc. are considered as important in diet as they supply nutritious fibres.
5. Thus, fibres are considered as one of the important nutrients.
In simple words: Fibres, though indigestible, are crucial nutrients as they facilitate waste elimination (egestion), promote digestive health, and contribute to overall well-being by aiding in the absorption of other nutrients.

🎯 Exam Tip: Emphasize the role of fibres in digestion, egestion, and disease prevention, even though they don't provide direct energy.

c. Cell division is one of the important properties of cells and organisms.
Answer:
1. Cell division is very essential for all the living organisms.
2. The growth and development is possible only due to cell division.
3. The emaciated body can be restored only through the cell division which adds new cells.
4. Offspring is produced only through the cell division that take place in parents.
5. In asexual reproduction, mitosis helps to give rise to new generation.
6. In sexual reproduction, meiosis helps to form haploid gametes.
7. All such functions show that cell division is one of the important properties of cells and organisms.
In simple words: Cell division is fundamental for life, enabling growth, repair of damaged tissues, and reproduction (both asexual and sexual by forming gametes), thus ensuring the continuity and development of organisms.

🎯 Exam Tip: List the key roles of cell division: growth, repair/replacement, and reproduction (distinguishing between mitosis and meiosis functions).

d. Sometimes, higher plants and animals too perform anaerobic respiration.
Answer:
1. When there is deficiency of oxygen in the surrounding, the aerobic respiration is not possible.
2. In such case, to survive, higher plants switch over to anaerobic respiration.
3. In some animal tissues in case of oxygen deficiency cells perform anaerobic respiration.
In simple words: Higher plants and animals can temporarily resort to anaerobic respiration during oxygen scarcity to sustain energy production, although it is less efficient and typically a survival mechanism rather than a primary metabolic pathway.

🎯 Exam Tip: Explain that anaerobic respiration in higher organisms is a temporary adaptation to oxygen-deficient conditions, not their default mode of energy production.

e. Krebs cycle is also known as citric acid cycle.
Answer:
1. Sir Hans Kreb proposed this cycle and hence it is called Krebs cycle.
2. These are series of cyclic chain reactions which begins with acetyl-coenzyme-A molecules which act with molecules of oxaloacetic acid.
3. The reactions are catalysed with the help of specific enzymes.
4. The first molecule formed in this reaction is called citric acid. Therefore, Krebs cycle is also called citric acid cycle.
In simple words: The Krebs cycle is also known as the citric acid cycle because citric acid is the first molecule formed when acetyl-coenzyme-A combines with oxaloacetic acid at the start of this series of reactions.

🎯 Exam Tip: Mention both names (Krebs and Citric Acid Cycle) and explain the origin of the latter name by identifying citric acid as the initial product.

Question 5. Answer in detail.
a. Explain the glycolysis in detail.
Answer:
• Carbohydrates are converted to glucose after the process of digestion is completed. The oxidation of glucose for releasing energy is called glycolysis which takes place in cytoplasm.
• Glycolysis can occur in presence of oxygen or without oxygen too. The first type of glycolysis takes place in aerobic respiration and the second type is in anaerobic respiration.
• In aerobic respiration, there is step-wise oxidation of glucose molecule forming two molecules each of pyruvic acid, ATP, NADH2 and water.
• Later the pyruvic acid formed in this process is converted into molecules of Acetyl-Coenzyme-A along with two molecules of NADH2 and two molecules of CO2.
• During anaerobic respiration along with glycolysis there is fermentation too. This is incomplete oxidation of glucose and thus it results in formation of lesser energy.
• The process of glycolysis was discovered by Gustav Embden, Otto Meyerhof, and Jacob Parnas. Therefore, in their honour, glycolysis is also called as Embden-Meyerhof-Parnas pathway (EMP pathway). For the discovery they had performed experiments on muscles.
In simple words: Glycolysis is the initial metabolic pathway in the cytoplasm that breaks down glucose into two pyruvate molecules, producing ATP and NADH2. It can occur with or without oxygen and is also known as the EMP pathway, named after its discoverers.

🎯 Exam Tip: Include the definition, location, products (pyruvate, ATP, NADH2), oxygen dependency, and the alternative name (EMP pathway) with its origin.

b. With the help of suitable diagrams, explain the mitosis in detail.
Answer:
(1) There are two stages of mitosis. These are
(a) Karyokinesis or nuclear division and
(b) Cytokinesis or cytoplasmic division. Karyokinesis takes place in further four phases, viz prophase, metaphase, anaphase and telophase.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पशु कोशिका को प्रारंभिक प्रोफेज़ अवस्था में दिखाता है, जिसमें केंद्रक झिल्ली, केंद्रिका, सेंट्रियोल और पतले धागे जैसे गुणसूत्र स्पष्ट रूप से दिखाई दे रहे हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पशु कोशिका को देर से प्रोफेज़ अवस्था में दर्शाता है, जहाँ गुणसूत्र (दो क्रोमैटिड्स के साथ) संघनित हो गए हैं, सेंट्रियोल अलग हो गए हैं, और केंद्रक झिल्ली तथा केंद्रिका लुप्त हो रही हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पशु कोशिका को मेटाफेज़ अवस्था में प्रस्तुत करता है, जिसमें गुणसूत्र कोशिका के भूमध्यरेखीय तल पर व्यवस्थित हैं, स्पिंडल फाइबर सेंट्रोमियर से जुड़े हैं, और केंद्रक झिल्ली पूरी तरह से गायब है।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पशु कोशिका को एनाफेज़ अवस्था में दर्शाता है, जहाँ बहन क्रोमैटिड्स (जो अब बेटी गुणसूत्र बन गए हैं) स्पिंडल फाइबर द्वारा विपरीत ध्रुवों की ओर खींचे जा रहे हैं, जो केले के गुच्छे के समान दिखते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पशु कोशिका को टेलोफेज़ अवस्था में दर्शाता है, जिसमें गुणसूत्र विपरीत ध्रुवों पर विसंघनित हो रहे हैं, केंद्रक झिल्ली और केंद्रिका फिर से प्रकट हो रहे हैं, और कोशिका झिल्ली में एक संकुचन बन रहा है।
(a) Karyokinesis:
(i) Prophase: During prophase, condensation of chromosomes starts. The thin and thread like chromosomes start thickening. They are seen with their pair of sister chromatids. In animal cells the centrioles are seen to duplicate and move to opposite poles of the cell. Nuclear membrane and nucleolus disappear.
(ii) Metaphase: Chromosomes complete their condensation and each one is seen with its sister chromatids. The chromosomes are seen in equatorial plane of the cell. The spindle fibres are formed from polar region, where centrioles are present, and they attach themselves to the centromere of each chromosome. Nuclear membrane now disappears completely.
(iii) Anaphase: The centromeres of the chromosomes now divide forming two daughter chromosomes. The spindle fibres pull apart the chromosomes from equatorial region to the opposite poles. Chromosomes moving to the poles appear like bunch of bananas. One set of chromosomes reach each pole by the end of the anaphase.
(iv) Telophase: Telophase is reverse of events that occurred in prophase. The thickened chromosomes decondense. They again assume the thin and thread like appearance. Nuclear membrane and nucleolus appear again. The spindle fibres are completely lost. The cell looks as if it has two nuclei in one cytoplasm.
(b) Cytokinesis: In animal cells a notch develops in the middle of the cell. This notch goes on deepening down and later the cytoplasm divides into two. In plant cells, cell plate formation takes place and then cytokinesis takes place.
In simple words: Mitosis is a cell division process for growth and repair, involving two main stages: karyokinesis (nuclear division through prophase, metaphase, anaphase, telophase) and cytokinesis (cytoplasmic division), resulting in two identical daughter cells.

🎯 Exam Tip: Clearly describe the key events of each mitotic phase (chromosome condensation, alignment, separation, decondensation) and distinguish between animal and plant cytokinesis.

c. With the help of suitable diagrams, explain the five stages of prophase-I of meiosis.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोफेज़-I की लेप्टोटीन अवस्था में एक कोशिका को दर्शाता है, जिसमें संघनित गुणसूत्र, स्पष्ट सेंट्रियोल, केंद्रक झिल्ली और केंद्रिका दिखाई देते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोफेज़-I की जाइगोटीन अवस्था में एक कोशिका को दर्शाता है, जहाँ सजातीय गुणसूत्र युग्मित होकर (युग्मन) बाइवैलेंट बनाते हैं, जिसमें सेंट्रोमियर और केंद्रिका दिखाई देते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोफेज़-I की पैकीटीन अवस्था में एक कोशिका को दर्शाता है, जिसमें सजातीय गुणसूत्रों के गैर-बहन क्रोमैटिड्स के बीच क्रॉसिंग ओवर हो रहा है और केंद्रिका लुप्त हो रही है।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोफेज़-I की डिप्लोटीन अवस्था में एक कोशिका को दर्शाता है, जहाँ सजातीय गुणसूत्र अलग होना शुरू करते हैं लेकिन चियाज़माटा (X-आकार की संरचनाएँ) पर जुड़े रहते हैं, और प्रत्येक गुणसूत्र में दो क्रोमैटिड्स होते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोफेज़-I की डाइकाइनिसिस अवस्था में एक कोशिका को दर्शाता है, जहाँ चियाज़माटा समाप्त हो जाते हैं, केंद्रक झिल्ली टूट जाती है, और स्पिंडल फाइबर बनना शुरू हो जाते हैं, जो मेटाफेज़ की तैयारी को दर्शाता है।
Prophase-I: Prophase – I of meiosis is much longer phase of the meiosis.
It is subdivided into 5 substages, namely leptotene, zygotene, pachytene, diplotene, and diakinesis.
(1) Leptotene: Initially the chromosomes start condensation and they become compact during leptotene.
(2) Zygotene: In zygotene, homologous chromosomes start pairing. This pairing is called synapsis. The structure called synaptonemal complex develops to hold chromosomes in place during this pairing. Each chromosome's chromatid arm divides and forms structure called bivalent or tetrad.
(3) Pachytene: During pachytene stage, crossing over of non-sister chromatids of homologous chromosomes takes place. Genetic recombination is produced due to such exchange. The homologous chromosomes still remain paired together at the sites of crossing over.
(4) Diplotene: During diplotene, synaptonemal complex dissolves and the homologous chromosomes of the bivalents separate except at the point of crossing over. Thus, it looks like X-shaped structures called the chiasmata.
(5) Diakinesis: The last phase of prophase is for termination of chiasmata. The spindle fibres originate, and the nucleolus disappears, and the nuclear envelope breaks down.
In simple words: Prophase-I of meiosis is a prolonged stage divided into five sub-stages: leptotene (chromosome condensation), zygotene (homologous pairing and synapsis), pachytene (crossing over for genetic recombination), diplotene (separation with chiasmata remaining), and diakinesis (chiasmata termination and nuclear envelope breakdown).

🎯 Exam Tip: For each sub-stage of Prophase-I, clearly state the key chromosomal event, especially synapsis in zygotene and crossing over in pachytene, as these are critical for genetic variation.

d. How do all the life processes contribute to the growth and development of the body?
Answer:
1. Different systems work in co-ordination with each other in the body of the living organisms. In human body the homoeostasis is very advanced.
2. Digestive system, respiratory system, circulatory system, excretory system, nervous system and all the external and internal organs in the bodywork independently but in coordination with each other.
3. The digested and absorbed nutrients of the food are transported to various cells with the help of circulatory system due to pumping of the heart.
Simultaneously, the oxygen absorbed in the blood by lungs is also transported to each cell by RBCs.
4. Mitochondria in every cell brings about oxidation of nutrients and produce energy required for all of these functions.
5. The control is exercised by the nervous system on all these actions. This keeps the organism alive and helps in growth and development of the same.
In simple words: Life processes like digestion, respiration, circulation, and excretion work in a coordinated manner, regulated by the nervous system, to supply nutrients and oxygen, produce energy, and remove waste, all of which are essential for the body's growth and development while maintaining homeostasis.

🎯 Exam Tip: Highlight the interdependence of different organ systems and their collective role in maintaining homeostasis, energy production, and nutrient supply for overall growth and development.

e. Explain the Krebs cycle with reaction.
Answer:
• Krebs cycle was proposed by Sir Hans Kreb. This cycle is named after him. It is also called tricarboxylic acid cycle or citric acid cycle.
• The acetyl-coenzyme-A molecules enter the mitochondria located in the cytoplasm.
• They participate in the chemical reactions taking place in Krebs cycle.
• In the cyclic chemical reactions, acetyl- coenzyme-A is completely oxidised
• It yields molecules of CO2, H2O, NADH2, FADH2 and ATP upon complete oxidation.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्रेब्स चक्र (सिट्रिक एसिड चक्र) को दर्शाता है, जो एसिटाइल-को-ए से शुरू होकर कार्बन डाइऑक्साइड, पानी, NADH2, FADH2 और ATP जैसे ऊर्जा-वाहक अणुओं का उत्पादन करता है। यह चक्र ऊर्जा उत्पादन में महत्वपूर्ण भूमिका निभाता है।
In simple words: The Krebs cycle, or citric acid cycle, is a mitochondrial process where acetyl-coenzyme-A is completely oxidized through a series of reactions, producing CO2, H2O, and high-energy carriers like NADH2 and FADH2, along with some ATP.

🎯 Exam Tip: Mention the discoverer (Krebs), alternative names (citric acid, tricarboxylic acid cycle), location (mitochondria), and the key products (CO2, H2O, NADH2, FADH2, ATP).

Question 5. How energy is formed from oxidation of carbohydrates, fats and proteins? Correct the dagram below.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कार्बोहाइड्रेट, वसा और प्रोटीन के ऑक्सीकरण से ऊर्जा उत्पादन की प्रक्रिया को दर्शाता है, जहाँ विभिन्न पोषक तत्व पाइरुविक एसिड और एसिटाइल को-एंजाइम-ए के माध्यम से क्रेब्स चक्र में प्रवेश करके CO2, H2O और ऊर्जा उत्पन्न करते हैं।
Answer:
(1) First of all the dietary carbohydrates are digested in the digestive system with the help of various enzymes and converted into glucose. Similarly, proteins are converted into amino acids and fats are broken down into fatty aid and glycerol (alcohol).
(2) Oxidation of carbohydrates takes place during cellular respiration. Glucose is oxidized by three steps during aerobic respiration, viz. glycolysis, tricarboxylic acid cycle or Krebs cycle and electron transfer chain.
(3) From one molecule of glucose two molecules of each pyruvic acid, ATP, NADH2 and water are formed during glycolysis. Pyruvic acid which is formed in this process is converted into Acetyl-Coenzyme-A along with release of two molecules each of NADH2 and CO2.
(4) In the next step, i.e. in TCA cycle, molecules of Acetyl-Co-A enter the mitochondria and a cyclic chain of reactions take place. Acetyl part of Acetyl- Co-A is completely oxidized through this cyclical process. The molecules CO2, H2O, NADH2, FADH2 are released in this process.
(5) In third step, i.e. in ETC reaction, NADH2 and FADH2 formed during first two steps are used for obtaining ATP molecules. 3 molecules of ATP are obtained from each NADH2 molecule and 2 molecules of ATP from each FADH2.
(6) Thus, one molecule of glucose upon complete oxidation in presence of oxygen yields 38 molecules of ATP. This is how from carbohydrates, energy is obtained.
(7) If carbohydrates are insufficient in diet, then proteins or lipids are used for energy production. Fatty acids derived from fats and amino acids derived from proteins are converted into Acetyl- Co-A. Acetyl-Co-A once again can yield energy through TCA cycle.
In simple words: Energy is generated by oxidizing carbohydrates (glucose), fats (fatty acids and glycerol), and proteins (amino acids), all of which are eventually converted into acetyl-CoA to enter the Krebs cycle and electron transport chain, yielding ATP.

🎯 Exam Tip: Explain how each macronutrient (carbohydrates, fats, proteins) is broken down into intermediate molecules that can feed into the common energy-producing pathways like the Krebs cycle.

Project

Project 1.
Use of ICT:
Collect videos and photographs of different life processes in living organisms. Prepare a presentation and present it on the occasion of science exhibition.

Project 2.
Books are my friend:
Read different Encyclopaedias of technical terms in biology and anatomy and other reference books.

Can You Recall?

Question 1. How are the food stuffs and their nutrient contents useful for body?
Answer: The food stuffs are digested and converted into soluble nutrients. These nutrients are carried by blood to every cell of the body. The oxygen inhaled at the time of respiration is also carried to every cell. In the body cells, this oxygen carries out oxidation of nutrients and thus energy is produced. The energy helps the body to carry out all its functions. The nutrients help in the growth and development of the body.
In simple words: Food is broken down into soluble nutrients and oxygen is transported to cells. Nutrients are oxidized to produce energy for body functions and support growth.

🎯 Exam Tip: Focus on the interconnected roles of digestion, respiration, circulation, and nutrient utilization for body functions and development.

Question 2. What is the importance of balanced diet for body?
Answer: Balanced diet has carbohydrates, proteins, fats, vitamins and minerals in the right proportion. Each nutrient carries a specific important function. In balanced diet all these nutrients are in right proportion. Since balanced diet is required for energy and nutrition, it is very important to maintain our health.
In simple words: A balanced diet provides all essential nutrients in correct proportions, which is vital for energy production, proper body function, and overall health maintenance.

🎯 Exam Tip: Emphasize the "right proportion" and "specific important function" of each nutrient in maintaining health.

Question 3. Which different functions are performed by muscles in body?
Answer: There are three types of muscles in our body. The voluntary muscles bring about all the movements according to our will. Involuntary muscles bring about all vital activities of the body. The visceral organs are under the control of involuntary muscles. The cardiac muscles control the movements of heart. Carbohydrates and proteins are stored in muscles.
In simple words: Muscles facilitate voluntary movements, control vital involuntary bodily functions like heartbeats and visceral organ actions, and also serve as storage sites for carbohydrates and proteins.

🎯 Exam Tip: Remember to differentiate between voluntary, involuntary, and cardiac muscles and their specific roles in the body.

Question 4. What is the importance of digestive juices in digestive system?
Answer: Digestive juice contains different enzymes. Enzymes act as catalysts and bring about the chemical reactions at faster pace. The digestive juices of stomach make pH of digestive tract acidic while that of intestinal juice make it alkaline.
In simple words: Digestive juices, containing enzymes, accelerate chemical reactions in digestion and regulate the pH levels in the digestive tract, which is crucial for nutrient breakdown and absorption.

🎯 Exam Tip: Highlight the catalytic role of enzymes and the pH regulation by different digestive juices as key functions.

Question 5. Which system is in action for removal of waste materials produced in human body?
Answer: Excretory system helps in the removal of nitrogenous waste materials produced in the human body.
In simple words: The excretory system is responsible for eliminating nitrogenous waste products from the human body, maintaining internal balance.

🎯 Exam Tip: Clearly state "Excretory system" and specify "nitrogenous waste materials" for accuracy.

Question 6. What is the role of circulatory system in energy production?
Answer: Due to circulatory system, glucose from digestive system and oxygen from respiratory system is transported to every cell. Red blood cells carry the oxygen as the blood is pumped by the heart. In every cell with the help of oxygen, glucose molecules yield the energy by the process of oxidation.
In simple words: The circulatory system transports glucose and oxygen to every cell, where oxygen helps oxidize glucose to produce energy, thus playing a direct role in energy production.

🎯 Exam Tip: Connect the circulatory system's function to the delivery of both glucose (from digestion) and oxygen (from respiration) for cellular energy release.

Question 7. How are the various processes occurring in the human body controlled? In how many ways?
Answer: The nervous system and the endocrine system brings about control by nervous and chemical coordination in the body. Due to such coordination different functions of the body are carried out in sequential and controlled manner.
In simple words: Processes in the human body are controlled in two primary ways: through rapid nervous coordination by the nervous system and slower chemical coordination by the endocrine system, ensuring sequential and regulated functions.

🎯 Exam Tip: Mention both the nervous and endocrine systems and their respective "nervous" and "chemical" coordination roles.

Use Your Brain Power:

Question 1. Many players are seen consuming some food stuffs during breaks of the game. Why may be the players consuming these food stuffs?
Answer:
1. Players require energy in greater amount.
2. They perspire heavily at the time of game or sport which results in the loss of water and electrolytes from their body.
3. This may affect their performance in sport. To prevent such unfavourable effect, they are given, juices or drinks.
4. This helps them to restore the balance of water and electrolytes in their body. It also gives enhanced energy required for the performance.
In simple words: Players consume food and drinks during breaks to quickly replenish lost energy and restore the balance of water and electrolytes due to heavy perspiration, which helps maintain their performance.

🎯 Exam Tip: Focus on the twin needs of energy replenishment and electrolyte balance for athletes during intense activity.

Question 2. Many times, we experience dryness in mouth.
Answer:
1. In our body there is 65-70% water. This proportion is always maintained.
2. Sometimes we lose lots of water either through perspiration or due to unavailability of water for a long time. In such situations, we experience dryness in our mouth.
3. Dryness is a natural feeling which creates urge in us to drink water, thereby the proportion of water in the body is brought back to its normal levels.
In simple words: Dryness in the mouth signals water loss from the body, often due to perspiration or insufficient intake, prompting us to drink and restore the body's essential water balance.

🎯 Exam Tip: Explain the body's homeostatic mechanism for water balance and how mouth dryness is a signal for dehydration.

Question 3. Oral rehydration solution (Salt-sugar- water) is frequently given to persons experiencing loose motions.
Answer:
1. Loose motions cause lot of loss of water from the body.
2. This may result in dehydration. This can be lethal if ignored.
3. Especially in case of young children this is a very serious fatal problem.
4. Thus, to bring back the normal proportion of water and electrolytes, oral rehydration solution or ORS is given to the patient who suffers from loose motions.
In simple words: ORS is given during loose motions to combat severe dehydration and electrolyte loss, which can be life-threatening, especially in children, by restoring the body's fluid and salt balance.

🎯 Exam Tip: Highlight the risk of dehydration and electrolyte imbalance due to loose motions and how ORS effectively counteracts these.

Question 4. We sweat during summer and heavy exercise.
Answer:
1. During summer, the environmental temperatures are high.
2. This causes rise in our body temperature. Exercising also cause rise in the temperature. But since we can regulate our body temperature to a constant level, the sweat, glands get automatically stimulated.
3. This induces perspiration.
4. The sweat evaporates and causes fall in the body temperature. Thus, for regulation of body temperature, we sweat during summer or even after heavy exercise.
In simple words: We sweat during summer and heavy exercise to regulate body temperature. Increased external temperature or physical activity raises body heat, stimulating sweat glands; the evaporation of sweat then cools the body down.

🎯 Exam Tip: Explain that sweating is a thermoregulatory mechanism to maintain a constant body temperature when faced with external heat or internal heat generation.

Question 5. What do you mean by diploid (2n) cell?
Answer:
• The cells in which chromosome number is double are known as diploid cells.
• Male and female gametes unite together in the process of fertilization. Their chromosomes mix together in the zygote, therefore, the chromosome number is always diploid.
• E.g. Diploid chromosome no. in human beings is 46. We have 46 chromosomes in each of our body cells.
In simple words: A diploid (2n) cell contains two complete sets of chromosomes, one from each parent, as seen in most somatic cells after fertilization, like human cells having 46 chromosomes.

🎯 Exam Tip: Define diploid by stating "double set of chromosomes" (2n) and provide a clear example like human somatic cells.

Question 6. What do you mean by haploid (n) cell?
Answer:
• The cells with only one set of chromosomes is known as haploid cell.
• At the time of sexual reproduction, there is meiosis. In meiosis chromosome number of the parental germ cells are reduced to half. Therefore, gametes are haploid.
• The haploid chromosome number (n) in human beings is 23.
• Sperm and ovum both are haploid carrying 23 chromosomes each.
In simple words: A haploid (n) cell contains a single set of chromosomes, typically found in gametes (sperm and ovum) formed through meiosis, ensuring that the offspring receives a correct chromosome number after fertilization.

🎯 Exam Tip: Clearly state that haploid cells have "one set of chromosomes" (n) and are characteristic of gametes, along with an example.

Question 7. What do you mean by homologous chromosomes?
Answer:
• Every species has definite number of chromosome pairs in their diploid cells.
• In every pair, the two chromosomes are alike in shape, type and genes located over them.
• Such chromosomes are called homologous chromosomes.
• E.g. In human diploid cell, pair of chromosome no. 1 shows chromosome no. 1 from mother and chromosome no. 1 from father. These two chromosomes are homologous to each other.
In simple words: Homologous chromosomes are a pair of chromosomes (one from each parent) in a diploid cell that are similar in size, shape, and gene sequence, ensuring proper genetic inheritance.

🎯 Exam Tip: Emphasize that homologous chromosomes are a pair, one from each parent, similar in characteristics and gene positions.

Question 8. Whether the gametes are diploid or haploid? Why?
Answer: The cells that give rise to gametes are diploid (2n). But by meiosis they give rise to gametes which are haploid (n). Two haploid gametes undergo fertilization and the zygote formed becomes once again diploid (2n).
In simple words: Gametes are haploid (n) because they are formed by meiosis, a reduction division, ensuring that when two gametes fuse during fertilization, the resulting zygote restores the diploid (2n) chromosome number of the species.

🎯 Exam Tip: Explain that gametes are haploid due to meiosis, and this is crucial for maintaining a constant chromosome number across generations after fertilization.

Question 9. How are the haploid cells formed?
Answer: Diploid cells undergo meiosis, which is a reduction division. In this way haploid cells are formed.
In simple words: Haploid cells are formed when diploid cells undergo meiosis, a specialized type of cell division that reduces the chromosome number by half.

🎯 Exam Tip: State "meiosis" and "reduction division" as the core concepts for haploid cell formation.

Question 10. What is the importance of haploid cells?
Answer:
1. The gametes that take part in the sexual reproduction should be haploid.
2. Otherwise the chromosome number will not be maintained at constancy. E.g. Parents have 2n = 46 chromosomes in their cells.
3. If meiosis does not take place in them, the gametes formed will also contain 46 chromosomes.
4. The resultant offspring will have 46 + 46 = 92 chromosomes.
5. Such skewed number will produce large scale abnormalities.
6. But due to meiosis, the gametes formed are haploid and thus the chromosome number is maintained constant for every species. Gametes are haploid cells, this is the most important fact.
In simple words: Haploid gametes are essential for sexual reproduction because they maintain a constant chromosome number across generations; without them, fertilization would lead to a doubling of chromosomes in each generation, causing severe abnormalities.

🎯 Exam Tip: Focus on the role of haploid cells in maintaining species' chromosome number constancy across generations during sexual reproduction.

Internet Is My Friend

Question. Collect information.
(a) What are symptoms of diseases like night blindness, rickets, beriberi, neuritis, pellagra, anaemia, scurvy?
Answer:

🎯 Exam Tip: For such questions, creating a mental table or mnemonic for each disease and its most prominent symptom can aid recall.

Question. (b) What do you mean by coenzymes?
Answer: Co-enzyme is a non-protein compound that is necessary for the functioning of an enzyme. It is bound to the enzyme as a catalyst. This increases the rate of reaction. Co-enzymes always act along the enzymes. They cannot work independently. But the same molecule of coenzyme can be used again and again. Many co-enzymes are vitamins or derived from vitamins. When vitamin intake is too low, then an organism also lacks the co-enzymes that catalyse reactions. Water-soluble vitamins, which include all B complex vitamins and vitamin C, lead to the production of co-enzymes. Two of the most important and widespread vitamin-derived coenzymes are Nicotinamide Adenine Dinucleotide (NAD) and co-enzyme A.
In simple words: Coenzymes are non-protein molecules, often derived from vitamins, that assist enzymes in accelerating biochemical reactions by binding to them, and they can be reused multiple times.

🎯 Exam Tip: Define coenzymes as non-protein helpers of enzymes, emphasizing their role as catalysts and their frequent origin from vitamins.

Question. (c) Find the full forms of FAD, FMN, NAD, NADP.
Answer:

🎯 Exam Tip: Memorize the full names for these common coenzymes, as they are frequently tested for their biochemical relevance.

Question. (d) How much quantity of each vitamin is required every day?
Answer:

🎯 Exam Tip: While exact values can vary, understanding the typical units (μg vs mg) and relative amounts for each vitamin is useful.

Choose The Correct Alternative And Write Its Alphabet Against The Sub-Question Number:

Question 1. The process of glycolysis occurs in ..........
(a) cytoplasm
(b) mitochondria
(c) nucleus
(d) cell membrane
Answer: (a) cytoplasm
In simple words: Glycolysis, the initial stage of glucose breakdown, exclusively takes place in the cytoplasm of the cell.

🎯 Exam Tip: Remember the location of glycolysis (cytoplasm) as a fundamental fact in cellular respiration.

Question 2. ATP is called .......... of the cell.
(a) energy currency
(b) combustion fuel
(c) storage of glucose
(d) protein depot
Answer: (a) energy currency
In simple words: ATP (Adenosine Triphosphate) is termed the "energy currency" because it is the primary molecule cells use to store and transfer energy for various metabolic activities.

🎯 Exam Tip: ATP's role as the universal energy donor in cells is a key concept; associate it directly with "energy currency."

Question 3. Excess of carbohydrates are stored in liver and muscles in the form of .............
(a) sugar
(b) glucose
(c) glycogen
(d) protein
Answer: (c) glycogen
In simple words: When there's an excess of carbohydrates, they are converted into glycogen and stored in the liver and muscles for later use.

🎯 Exam Tip: Distinguish between glucose (circulating sugar) and glycogen (storage form of glucose) and their respective locations.

Question 4. Chemically vitamin B2 is ...........
(a) Riboflavin
(b) Nicotinamide
(c) Cyanacobalomine
(d) Pantothetic acid
Answer: (a) Riboflavin
In simple words: Vitamin B2 is chemically known as Riboflavin, a vital nutrient involved in energy metabolism.

🎯 Exam Tip: Learn the common names and chemical names for important vitamins, especially the B-complex group.

Question 5. Somatic and stem cells undergo type of ............ division.
(a) meiosis
(b) mitosis
(c) budding
(d) cloning
Answer: (b) mitosis
In simple words: Somatic cells and stem cells undergo mitosis, a type of cell division that produces two genetically identical daughter cells, crucial for growth and repair.

🎯 Exam Tip: Remember that mitosis is characteristic of somatic cells for growth and repair, while meiosis is for gamete formation.

Question 6. We get ........... energy from carbohydrates.
(a) 9 kcal/gm
(b) 9 cal/gm
(c) 4 cal/gm
(d) 4 kcal/gm
Answer: (d) 4 kcal/gm
In simple words: Carbohydrates provide approximately 4 kilocalories of energy per gram when metabolized by the body.

🎯 Exam Tip: Know the energy yield for major macronutrients: 4 kcal/gm for carbohydrates and proteins, and 9 kcal/gm for fats.

Question 7. Which of the following vitamins is necessary for synthesis of NADH2?
(a) Vitamin B2
(b) Vitamin B3
(c) Vitamin
(d) Vitamin K
Answer: (b) Vitamin B3
In simple words: Vitamin B3, also known as Niacin, is essential for the synthesis of NADH2, a coenzyme crucial for various metabolic reactions.

🎯 Exam Tip: Connect specific B vitamins to their roles in coenzyme synthesis, such as B3 (Niacin) for NAD/NADH2.

Write Whether The Following Statements Are True Or False:

Question 1. Glucose is oxidized step by step in the cells during the process of respiration at the body level.
Answer: False. (Glucose is oxidized step by step in the cells during the process of cellular respiration.)
In simple words: The breakdown of glucose in cells occurs during cellular respiration, not respiration at the body level, which refers to gas exchange.

🎯 Exam Tip: Differentiate between "body level respiration" (gas exchange) and "cellular respiration" (biochemical oxidation of glucose).

Question 2. In aerobic respiration, glucose is oxidized in three steps.
Answer: True
In simple words: Aerobic respiration involves three main steps-glycolysis, Krebs cycle, and electron transport chain-to fully oxidize glucose.

🎯 Exam Tip: Remember the three stages of aerobic respiration: glycolysis, Krebs cycle (TCA cycle), and electron transport chain.

Question 3. Glycolysis is also called Embden-Meyerhof-Paarnas pathway.
Answer: True
In simple words: Glycolysis is indeed known as the Embden-Meyerhof-Parnas (EMP) pathway, named after its discoverers.

🎯 Exam Tip: Knowing alternative names for metabolic pathways like glycolysis (EMP pathway) can be helpful in exams.

Question 4. Molecules of pyruvic acid formed in this glycolysis are converted into molecules of acetyl-co-enzyme A.
Answer: True
In simple words: After glycolysis, pyruvic acid is converted into acetyl-coenzyme A, which then enters the Krebs cycle.

🎯 Exam Tip: Understand the metabolic link between glycolysis and the Krebs cycle, where pyruvic acid is transformed into Acetyl-CoA.

Question 5. Excess of ATP molecules obtained from proteins are not stored in the body.
Answer: False. (Excess of amino acids obtained from proteins are not stored in the body.)
In simple words: The statement is false; while ATP isn't stored in excess from proteins, excess amino acids derived from proteins are not stored directly but are either converted or excreted.

🎯 Exam Tip: Clarify that while ATP is energy currency, the body doesn't store excess ATP from any source; instead, excess amino acids (protein breakdown products) are converted or eliminated, not stored as protein itself.

Question 6. Proteins of animal origin are called 'first class' proteins.
Answer: True
In simple words: Animal-derived proteins are termed 'first-class' because they contain all essential amino acids in proportions suitable for human needs.

🎯 Exam Tip: Understand that 'first class proteins' refer to those providing all essential amino acids, predominantly found in animal sources.

Question 7. The disease related with the deficient synthesis of insulin is heart disease.
Answer: False. (The disease related with the deficient synthesis of insulin is diabetes.)
In simple words: The disease caused by deficient insulin synthesis is diabetes, not heart disease, as insulin regulates blood sugar levels.

🎯 Exam Tip: Correctly associate insulin deficiency with diabetes and not directly with heart disease, although diabetes can increase heart disease risk.

Match The Columns:

Question. Protein | Part of the body
(1) Haemoglobin | (a) muscles
(2) Ossein | (b) skin
| (c) bones
| (d) blood
Answer: (1) Haemoglobin - blood
(2) Ossein - bones.
In simple words: Haemoglobin is a protein found in the blood, responsible for oxygen transport, while Ossein is a protein component of bones, providing structural integrity.

🎯 Exam Tip: Focus on linking specific proteins to their primary location or function in the body.

Question. Protein | Part of the body
(1) Keratin | (a) muscles
(2) Myosin | (b) skin
| (c) bones
| (d) blood
Answer: (1) Keratin - skin
(2) Myosin - muscles.
In simple words: Keratin is a structural protein prominent in skin, hair, and nails, while Myosin is a motor protein primarily found in muscles, crucial for contraction.

🎯 Exam Tip: Understand the function and location of key structural and contractile proteins like keratin and myosin.

Find The Odd One Out:

Question 1. Progesterone, Estrogen, Testosterone, Insulin
Answer: Insulin. (All the others are hormones produced with the help of fatty acids.)
In simple words: Insulin is the odd one out because it's a protein hormone regulating blood sugar, while progesterone, estrogen, and testosterone are steroid hormones derived from fatty acids.

🎯 Exam Tip: Categorize hormones based on their chemical structure (steroid vs. protein) and primary function to identify the odd one out.

Question 2. Actin, Ossein, Myosin, Melanin
Answer: Melanin. (All the others are proteins concerned with locomotion of the body.)
In simple words: Melanin is the odd one out as it's a pigment for skin and hair color, whereas Actin, Ossein, and Myosin are proteins primarily involved in structural support and muscle movement.

🎯 Exam Tip: Group proteins by their functional roles (e.g., structural, contractile, pigmentary) to identify the anomaly.

Question 3. Lipids, Carbohydrates, Fatty acids, Proteins
Answer: Fatty acids. (All the others are food constituents; fatty acid is soluble nutrient.)
In simple words: Fatty acids are the odd one out because they are a component or product of digestion (a soluble nutrient), while lipids, carbohydrates, and proteins are major food constituents (macronutrients).

🎯 Exam Tip: Differentiate between primary food groups (macronutrients) and their digested components (soluble nutrients).

Question 4. Alcohol, Vinegar, Pyruvic acid, Lactic acid.
Answer: Pyruvic acid. (All the others are chemical substances formed by the process of fermentation.)
In simple words: Pyruvic acid is the odd one out as it's an intermediate in glycolysis, while alcohol, vinegar, and lactic acid are products typically formed during various fermentation processes.

🎯 Exam Tip: Identify pyruvic acid as a key intermediate in respiration, distinct from end products of fermentation like alcohol or lactic acid.

Question 5. Tricarboxylic acid cycle, Citric acid cycle, Krebs cycle, EMP pathway.
Answer: EMP pathway. (All the other terms are synonymous to each other.)
In simple words: The EMP pathway (glycolysis) is the odd one out because Tricarboxylic acid cycle, Citric acid cycle, and Krebs cycle are all synonymous terms referring to the same metabolic pathway.

🎯 Exam Tip: Recognize that Tricarboxylic acid cycle, Citric acid cycle, and Krebs cycle are different names for the same metabolic pathway, while the EMP pathway refers to glycolysis.

Considering The Relationship In The First Pair, Complete The Second Pair By Using A Word Or Group Of Words:

Considering The Relationship In The First Pair, Complete The Second Pair By Using A Word Or Group Of Words:

Question 1. Process that occurs in the cytoplasm : Glycolysis :: Process that occurs in the mitochondria .........
Answer: Krebs cycle
In simple words: Glycolysis happens in the cytoplasm, while the Krebs cycle (TCA cycle) takes place in the mitochondria as part of cellular respiration.

🎯 Exam Tip: Remember the cellular locations for key metabolic pathways, as this is a common point of distinction in exams.

Question 2. Skin: Keratin :: Blood : .......
Answer: Haemoglobin
In simple words: Keratin is a primary protein in the skin, just as hemoglobin is the main protein found in blood.

🎯 Exam Tip: Familiarize yourself with major proteins and their primary locations/functions in the body.

Question 3. Energy obtained from protein : 4 kcal :: Energy obtained from fats / lipids : ........
Answer: 9 Kcal
In simple words: Proteins provide 4 kcal of energy per gram, while fats (lipids) provide more than double, specifically 9 kcal per gram.

🎯 Exam Tip: Know the energy yield per gram for carbohydrates, proteins, and fats, as this is a fundamental concept in nutrition.

Question 4. Breakdown of glucose molecule : Glycolysis :: Formation of glucose from proteins : ...............
Answer: Gluconeogenesis
In simple words: Glycolysis is the breakdown of glucose, whereas gluconeogenesis is the process where glucose is synthesized from non-carbohydrate sources like proteins.

🎯 Exam Tip: Understand the differences between glucose breakdown (catabolism) and glucose formation (anabolism) pathways, and their respective names.

Question 5. Condensation of chromosomes : Prophase :: Formation of spindle fibres : .........
Answer: Metaphase
In simple words: Chromosomes condense during prophase, and the spindle fibers are fully formed and attached to chromosomes during metaphase.

🎯 Exam Tip: Be able to identify the key events and structures associated with each stage of mitosis and meiosis.

Question 6. Division of nucleus : Karyokinesis :: Division of cytoplasm :: ..........
Answer: Cytokinesis.
In simple words: Karyokinesis refers to the division of the cell nucleus, while cytokinesis is the process of cytoplasmic division following nuclear division.

🎯 Exam Tip: Clearly differentiate between karyokinesis and cytokinesis as distinct, though often sequential, parts of cell division.

Write Definitions:

Question 1. Gluconeogenesis.
Answer: Gluconeogenesis: Formation of glucose through non-carbohydrate sources such a protein is called gluconeogenesis.
In simple words: Gluconeogenesis is the body's way of making new glucose from non-sugar sources, like proteins, when carbohydrate stores are low.

🎯 Exam Tip: Focus on the "new formation" aspect and the non-carbohydrate precursors to distinguish gluconeogenesis from other glucose pathways.

Question 2. Fermentation.
Answer: Fermentation: Conversion of pyruvic acid produced in the process of glycolysis into other organic acids or alcohol with the help of some enzymes is called fermentation.
In simple words: Fermentation is an anaerobic process where pyruvate is converted into products like alcohol or lactic acid, regenerating NAD+ for glycolysis.

🎯 Exam Tip: Emphasize the anaerobic nature and the regeneration of NAD+ as critical aspects of fermentation.

Name The Following:

Question 1. Products formed after complete oxidation of acetyl part present in the molecule of acetyl-coenzyme-A.
Answer: Molecules of \( \text{CO}_2 \), \( \text{H}_2\text{O} \), \( \text{NADH}_2 \), \( \text{FADH}_2 \) and ATP.
In simple words: When acetyl-CoA is fully oxidized in the Krebs cycle, it produces carbon dioxide, water, and energy carriers like NADH2, FADH2, and ATP.

🎯 Exam Tip: List all end products of aerobic respiration, including the key energy carriers and waste products.

Question 2. Place where electron transfer chain reaction take place.
Answer: Mitochondria present in the cytoplasm of the cell.
In simple words: The electron transport chain, which generates most of the cell's ATP, occurs within the mitochondria.

🎯 Exam Tip: Cellular location is crucial; specify the inner mitochondrial membrane for the electron transport chain.

Question 3. Two co-enzymes involved in cellular respiration.
Answer: NAD - Nicotinamide Adenine Dinucleotide and FAD Flavin Adenine Dinucleotide.
In simple words: NAD and FAD are two vital coenzymes that carry electrons during cellular respiration, contributing to ATP production.

🎯 Exam Tip: Remember both the abbreviations (NAD, FAD) and their full names, along with their role as electron carriers.

Question 4. Scientist who discovered the TCA cycle.
Answer: Sir Hans Krebs.
In simple words: The TCA cycle, also known as the Krebs cycle, was discovered by Sir Hans Krebs.

🎯 Exam Tip: Associating names with discoveries helps in remembering scientific concepts and can be a direct question in exams.

Question 5. Steps of anaerobic respiration.
Answer: Glycolysis and fermentation.
In simple words: Anaerobic respiration typically involves two main stages: glycolysis, which breaks down glucose, and fermentation, which processes the resulting pyruvate.

🎯 Exam Tip: Understand that anaerobic respiration does not use oxygen and comprises glycolysis followed by a fermentation step.

Question 6. Most abundantly found protein nature.
Answer: An enzyme RUBISCO present in plant chloroplasts.
In simple words: RUBISCO is the most abundant protein on Earth, crucial for photosynthesis in plants.

🎯 Exam Tip: Know the significance of RUBISCO in carbon fixation and its high abundance in the plant kingdom.

Give Scientific Reasons:

Question 1. We feel exhausted after exercising.
Answer:
• When we undertake constant exercises, there may be shortage of oxygen for the cells.
• Therefore, our muscles and other tissues perform anaerobic respiration in such condition.
• In this process, lactic acid is formed.
• Molecules of ATP produced in oxidation of food are also much less.
• Thus, there is less energy in the body and accumulation of lactic acid too. All this brings about a feeling of exhaustion.
In simple words: Intense exercise can lead to oxygen debt in muscles, causing them to switch to anaerobic respiration, which produces less ATP and builds up lactic acid, leading to fatigue and exhaustion.

🎯 Exam Tip: Explain the physiological processes (oxygen shortage, anaerobic respiration, lactic acid production, reduced ATP) that contribute to exhaustion after exercise.

Answer The Following Questions In Detail:

Question 1. Write the forms to which the following food materials are converted after digestion:
(a) Milk
(b) Potato
(c) Oil
(d) Chapati.
Answer:
(a) Milk: Proteins (casein) are converted into amino acids. Lactose sugar is converted into glucose. Lipids are converted into fatty acids and glycerol.
(b) Potato: Carbohydrates (starch) are converted into glucose.
(c) Oil: Lipids are converted into fatty acids and glycerol.
(d) Chapati: Carbohydrates (starch) are converted into glucose.
In simple words: During digestion, complex food items are broken down into simpler absorbable forms: milk components (proteins, lactose, lipids) become amino acids, glucose, fatty acids, and glycerol; potato (starch) and chapati (starch) become glucose; and oil (lipids) becomes fatty acids and glycerol.

🎯 Exam Tip: Clearly state the original nutrient and its final, absorbable digested form for each food item mentioned.

Question 2. On which two levels does respiration take place in living organisms?
Answer:
1. In organism respiration takes place at two levels, viz. Body level and Cellular level.
2. Respiration at body level: The exchange of respiratory gases such as oxygen and carbon dioxide between body and surrounding is called respiration at body level.
3. Cellular respiration: Oxidation of nutrients inside the cell with or without oxygen is called cellular respiration.
In simple words: Respiration occurs at two levels: body level, involving gas exchange with the environment (breathing), and cellular level, where nutrients are oxidized within cells to produce energy.

🎯 Exam Tip: Distinguish between the macro-level process of breathing (body respiration) and the micro-level energy production (cellular respiration).

Question 3. Answer the following questions:
(a) Write main types of vitamins.
(b) Name water soluble vitamins.
(c) Name fat soluble vitamins.
Answer:
(a) Write main types of vitamins.
Answer: A, B, C, D, E and K are main types of vitamins.
(b) Name water soluble vitamins.
Answer: Water soluble vitamins are B and C.
(c) Name fat soluble vitamins.
Answer: Fat soluble vitamins are A, D, E and K.
In simple words: Vitamins are essential nutrients, broadly categorized into water-soluble (B and C) and fat-soluble (A, D, E, K) types, each playing distinct roles in bodily functions.

🎯 Exam Tip: Memorize the two categories of vitamins (water-soluble vs. fat-soluble) and the specific vitamins belonging to each group.

Question 4. Answer the following questions:
(a) Why some living organisms have to perform anaerobic respiration?
(b) Give two examples of such living organisms.
Answer:
(a) Why some living organisms have to perform anaerobic respiration?
Answer: Some bacteria and lower organisms do not live in the presence of oxygen. In order to survive, they have to perform anaerobic respiration. Sometimes, muscle cells and erythrocytes also perform anaerobic respiration when there is lack of enough oxygen.
(b) Give two examples of such living organisms.
Answer: Yeast and bacteria.
In simple words: Some organisms perform anaerobic respiration because they either cannot tolerate oxygen or face oxygen-deficient environments; examples include yeast and many bacteria, and even human muscle cells under strenuous conditions.

🎯 Exam Tip: Provide reasons for anaerobic respiration (obligate anaerobes, facultative anaerobes, oxygen scarcity) and specific examples of organisms.

Question 5. Which is the energy currency of the cell? Explain it in detail.
Answer:
• ATP or Adenosine triphosphate is the 'energy currency' of the cell.
• Chemical composition of ATP is as follows: it is a triphosphate molecule having adenosine ribonucleoside. The nitrogenous compound-adenine, pentose sugar-ribose and three phosphate groups are present in ATP.
• In this energy-rich molecule the energy remains trapped in the bonds by which phosphate groups are attached to each other.
• ATP molecules are stored in the cells. As per the need, energy is derived by breaking the phosphate bond of ATP.
• During cellular respiration, the oxidation of glucose yields 38 molecules of ATP. Whenever required they are consumed to liberate energy.
In simple words: ATP (Adenosine triphosphate) is the cell's energy currency, a molecule with high-energy phosphate bonds that release energy when broken, powering cellular activities.

🎯 Exam Tip: Define ATP, describe its chemical structure (adenine, ribose, three phosphates), and explain how it stores and releases energy.

Question 6. How is energy obtained during starvation or hunger?
Answer:
• Due to starvation or hunger, there is less supply of nutrients and energy to the body. In such condition, the stored carbohydrates in the body also deplete.
• In such condition, fats and proteins present in the body are utilized.
• Fats or lipids are converted into fatty acids and proteins are broken down to amino acids.
• Fatty acids and amino acids both are converted to acetyl-coenzyme-A.
• Acetyl-coenzyme-A can undergo series of cyclic reactions and oxidised to liberate energy in the form of ATP molecules.
In simple words: During starvation, the body first uses stored carbohydrates, then breaks down fats into fatty acids and glycerol, and finally proteins into amino acids, converting them to acetyl-CoA to generate energy through the Krebs cycle.

🎯 Exam Tip: Explain the sequence of nutrient utilization (carbohydrates -> fats -> proteins) and the common intermediate (acetyl-CoA) for energy production during starvation.

Question 7. Why glycolysis is also called EMP pathway?
Answer: Process of glycolysis was discovered by Gustav Embden, Otto Meyerhof, and Jacob Parnas along with their colleagues. They performed experiments on muscles to understand glycolysis. Hence, in their honour, glycolysis is also edited Embden-Meyerhof-Parnas pathway or EMP pathway.
In simple words: Glycolysis is named the EMP pathway in honor of Gustav Embden, Otto Meyerhof, and Jacob Parnas, who extensively researched and elucidated its biochemical steps.

🎯 Exam Tip: Acknowledge the scientists responsible for identifying the glycolysis pathway, linking their names to the EMP designation.

Question 8. How are proteins obtained? What are the components of the proteins?
Answer:
• Protein, is a macromolecule which is formed by amino acids.
• When digestion of protein takes place, it forms different amino acids. These amino acids are transported to each cell by blood circulation.
• By protein synthesis, these amino acids are again used to make different kinds of proteins which our body needs.
• Animal proteins are said to be 'first class proteins' as they contain good quality amino acids.
• 4 Kcal/gm energy is obtained from the proteins.
In simple words: Proteins are large molecules made of amino acids; they are obtained from diet, digested into amino acids, and then resynthesized into various functional proteins our body needs.

🎯 Exam Tip: Define proteins as polymers of amino acids, describe their dietary source, and explain their role in protein synthesis and energy supply.

Question 9. Where and in which forms the amino acids formed after digestion of food are used in the body?
Answer:
(1) After digestion of proteins, amino acids are formed. These amino acids are used to synthesise proteins in different forms. e.g.
• In blood-Haemoglobin and antibodies are formed.
• In skin - Melanin and keratin are formed.
• In bones – Ossein is formed.
• In pancreas-Insulin and trypsin are synthesized.
• Pituitary and all other glands produce hormones by utilising amino acids.
• In muscles – Actin and myosin are formed.
• In all the cells, plasma membrane is formed by proteins. All enzymes are also synthesised using the amino acids.
In simple words: Digested amino acids are transported to cells throughout the body and used to synthesize various essential proteins, including hemoglobin for blood, melanin and keratin for skin, ossein for bones, hormones (insulin, pituitary hormones), muscle proteins (actin, myosin), enzymes, and plasma membrane components.

🎯 Exam Tip: Provide diverse examples of where amino acids are used to build specific proteins in different body tissues and their functions.

Question 10. What are fatty acids? What are the different uses of fatty acids ?
Answer:
(1) The fatty acids are components of the lipids. When lipids are digested, it forms fatty acids and alcohol (glycerol).
(2) There are certain chemical bonds between fatty acids and alcohol.
(3) Fatty acids are very essential for the health.
(4) After digestion, fatty acids are absorbed into the blood and transported to the cells.
(5) Different types of cells produce their own substances from these fatty acids.
E.g. (a) Plasma membrane is produced from phospholipids.
(b) Hormones like testosterone, progesterone, estrogen, aldosterone are produced from fatty acids.
(c) The axonal coverings around the neurons are also made from fatty acids.
In simple words: Fatty acids are building blocks of lipids, formed during fat digestion, and are vital for health, used by cells to build structures like plasma membranes (as phospholipids), hormones (like testosterone, estrogen), and protective coverings around nerve cells.

🎯 Exam Tip: Explain that fatty acids are lipid monomers and list their diverse physiological roles, from structural components to hormone synthesis.

Give Explanations For The Following Statements:

Question 1. After complete oxidation of a glucose molecules, 38 number of ATP molecules are formed.
Answer:
I. Glycolysis: No. of ATP molecules formed = 4
No. of ATP molecules used = 2
II. Krebs cycle : No. of ATP molecules formed = 2
III. ETC Reaction :
\( \text{NADH}_2 \): 10 \( \text{NAD}_2 \) x 3 ATP = 30 ATP
\( \text{FADH}_2 \): 2 \( \text{FADH}_2 \) x 2 ATP = 4 ATP
Total ATP molecules produced = (4+2+34) = 40 ATP
ATP molecules used = 2 ATP
Therefore, total ATP molecules = 38 ATP
In simple words: The complete oxidation of one glucose molecule during aerobic respiration yields 38 ATP molecules, calculated by summing ATP directly produced in glycolysis and the Krebs cycle, and the ATP generated from NADH2 and FADH2 via the Electron Transport Chain, minus the ATP consumed.

🎯 Exam Tip: Break down the ATP yield calculation by each stage of aerobic respiration (glycolysis, Krebs cycle, ETC) to demonstrate the total net production.

Question 2. At the end of glycolysis, pyruvate molecules are obtained.
Answer: The process of glycolysis takes place m the cytoplasm of the cell. One molecule of glucose is gradually oxidized step by step forming two molecules of each pyruvic acid, ATP, \( \text{NADH}_2 \) and water. Of these, pyruvate or pyruvic acid takes part in the further reactions.
In simple words: Glycolysis is the initial breakdown of glucose in the cytoplasm, resulting in two molecules of pyruvate, along with ATP and NADH2, which then proceed to subsequent stages of cellular respiration.

🎯 Exam Tip: Emphasize that glycolysis is the first stage, occurs in the cytoplasm, and produces two pyruvate molecules from one glucose molecule.

Question 3. Genetic recombination occurs in pachytene phase of prophase of meiosis-l.
Answer: In prophase of meiosis I there are total 5 stages. Of these in pachytene the process of crossing over takes place between homologous chromosomes as chromosomes come near each other forming synapsis.
In simple words: Genetic recombination, or crossing over, happens during the pachytene stage of Prophase I in meiosis when homologous chromosomes exchange genetic material, leading to increased genetic diversity.

🎯 Exam Tip: Highlight the pachytene stage as the specific point for crossing over and its importance for genetic variation.

Question 4. All chromosomes are arranged parallel to equatorial plane of cell in metaphase of mitosis.
Answer: In mitosis, the metaphase is the stage when dividing chromosomes lie on the equatorial plane of the cell. They are later pulled by the spindle fibres to the opposite poles.
In simple words: During metaphase of mitosis, all chromosomes align along the cell's equatorial plane, forming the metaphase plate, which ensures equal distribution to daughter cells.

🎯 Exam Tip: The alignment of chromosomes at the metaphase plate is the hallmark of metaphase, ensuring accurate segregation during cell division.

Question 5. For formation of plasma membrane, phospholipid molecules are necessary.
Answer: Upon the digestion of fats, fatty acids and glycerol are formed. The fatty acids can be converted into phospholipid which are essential molecules for development of plasma membrane.
In simple words: Phospholipid molecules, derived from digested fatty acids, are crucial for building the plasma membrane, which forms the outer boundary of cells.

🎯 Exam Tip: Connect the digestion of fats to the formation of phospholipids, emphasizing their structural role in the cell membrane.

Question 6. Our muscle cells perform anaerobic type of respiration during exercise.
Answer: When the proportion of oxygen is less, then the cells switch over to anaerobic respiration. When we are exercising there is increased demand of oxygen for muscle cells. If this is not fulfilled, they perform anaerobic respiration during exercise.
In simple words: During intense exercise, when oxygen supply to muscle cells is insufficient, they switch to anaerobic respiration to produce energy, resulting in lactic acid accumulation.

🎯 Exam Tip: Explain the direct link between high oxygen demand during exercise and the muscle cells' compensatory shift to anaerobic respiration.

Question 7. Excess of carbohydrates are stored in liver and muscles in the form of glycogen.
Answer: The carbohydrates which are not used to produce energy cannot be stored in the body in the form of glucose. This glucose is therefore converted into complex compound called glycogen. Glycogen is stored in muscles and liver.
In simple words: When the body has excess carbohydrates, they are converted into glycogen, a complex storage form, which is then stored primarily in the liver and muscles for later energy use.

🎯 Exam Tip: Identify glycogen as the storage form of glucose and specify the liver and muscles as its primary storage sites.

Complete The Paragraph By Choosing The Appropriate Words Given In The Brackets:

Question 1. (gamete, crossing over, haploid, Meiosis-II, meiosis-l, diploid)
_________ is just like mitosis. In this stage, the two haploid daughter cells formed in _________ undergo division by separation of recombined sister chromatids and four _________ daughter cells are formed. Process of _________ production and spore formation occurs by meiosis. In this type of cell division, four haploid (n) daughter cells are formed from one _________ cell. During this cell division, _________ occurs between, the homologous chromosomes.
Answer: Meiosis-ll is just like mitosis. In this stage, the two haploid daughter cells formed in meiosis-l undergo division by separation of recombined sister chromatids and four haploid daughter cells are formed. Process of gamete production and spore formation occurs by meiosis. In this type of cell division, four haploid (n) daughter cells are formed from one diploid cell. During this cell division, crossing over occurs between the homologous chromosomes.
In simple words: Meiosis-II mirrors mitosis, where haploid cells from Meiosis-I divide, separating sister chromatids to produce four haploid daughter cells. This process, crucial for gamete and spore formation, involves crossing over between homologous chromosomes during Meiosis-I.

🎯 Exam Tip: Understand the sequence of meiosis, distinguishing events of Meiosis-I (crossing over, homologous chromosome separation) from Meiosis-II (sister chromatid separation, like mitosis).

Question 2. (external, inhalation, alveolar, breathing, respiration, exhalation)
Release of energy from the assimilated food is called _________. Inhalation and exhalation is called ........ When ....... is done, air enters the lungs. The oxygen from this air enters the blood while carbon dioxide from the blood exits from the blood. Through exhalation, \( \text{CO}_2 \) is given out. This gaseous exchange occurs through _________ membrane. This is called _________ respiration. The RBCs carry oxygen to every cell.
Answer: Release of energy from the assimilated food is called respiration. Inhalation and exhalation is called breathing. When inhalation is done, air enters the lungs. The oxygen from this air enters the blood while carbon dioxide from the blood exits from the blood. Through exhalation, \( \text{CO}_2 \) is given out. This gaseous exchange occurs through alveolar membrane. This is called external respiration. The RBCs carry oxygen to every cell.
In simple words: Respiration is the energy release from food. Breathing, encompassing inhalation and exhalation, moves air into and out of the lungs. Gaseous exchange across the alveolar membrane, where oxygen enters the blood and carbon dioxide exits, is termed external respiration, with RBCs transporting oxygen.

🎯 Exam Tip: Clearly differentiate between the terms respiration (energy release) and breathing (gas exchange), and recognize the role of the alveolar membrane in external respiration.

Read The Paragraph And Answer The Questions Given Below:

1. Dietary fibre - found mainly in fruits, vegetables, whole grains and legumes - is probably best known for its ability to prevent or relieve constipation. But foods containing fibre can provide other health benefits as well, such as helping to maintain a healthy weight and lowering your risk of diabetes, heart disease and some types of cancer. Dietary fibre, also known as roughage or bulk, includes the parts of plant foods your body can't digest or absorb. Unlike other food components, such as fats, proteins or carbohydrates - which your body breaks down and absorbs - fibre isn't digested by your body. Instead, it passes relatively intact through your stomach, small intestine and colon and out of your body.

Questions And Answers :

Question 1. Which food items provide rich fibre content?
Answer: Fruits, vegetables, whole grains and legumes give rich amount of dietary fibre.
In simple words: Foods like fruits, vegetables, whole grains, and legumes are excellent sources of dietary fiber.

🎯 Exam Tip: List common food sources rich in dietary fiber.

Question 2. Enlist the advantages of fibres in diet.
Answer: Fibres help to relieve constipation and help in maintaining a healthy weight and lowering risk of diabetes, heart disease and some types of cancer.
In simple words: Dietary fibers aid in preventing constipation, managing weight, and reducing the risk of conditions like diabetes, heart disease, and certain cancers.

🎯 Exam Tip: Focus on the main health benefits of fiber, particularly digestive health and disease prevention.

Question 3. Are fibres digested in the body?
Answer: No, fibres are not digested in the body but are passed on without any alteration.
In simple words: No, fibers are not digested by the human body; they pass through the digestive system largely unchanged.

🎯 Exam Tip: Clearly state that fiber is indigestible, distinguishing it from other macronutrients.

Question 4. Which is the path through which fibres pass in the digestive tract?
Answer: Fibres pass through stomach, small intestine and colon.
In simple words: Fibers travel through the stomach, small intestine, and then the colon before being eliminated from the body.

🎯 Exam Tip: Outline the correct sequence of organs that fiber traverses in the digestive system.

Question 5. What is a roughage?
Answer: Roughage is the fibre content of the food which consists of plant matter which cannot be digested by the human enzymes, hence form undigested bulk matter in the faeces.
In simple words: Roughage, also known as dietary fiber, is the indigestible plant material in food that adds bulk to stool and aids in bowel movements.

🎯 Exam Tip: Define roughage by its indigestible nature and its role in forming fecal bulk.

2. The substances formed by specific chemical bond between fatty acids and alcohol are called lipids. Digestion of lipids consumed by us is nothing but their conversion into fatty acids and alcohol. Fatty acids are absorbed and distributed everywhere within the body. From those fatty acids, different cells produce various substances necessary to themselves. Ex. the molecules called phospholipids which are essential for producing plasma membrane are formed from fatty acids. Besides, fatty acids are used for producing hormones like progesterone, estrogen, testosterone, aldosterone, etc. and the covering around the axons of nerve cells. We get 9 Kcal of energy per gram of lipids. Excess of lipids are stored in adipose connective tissue in the body.

Questions And Answers:

Question 1. Define lipids.
Answer: Lipids are molecules formed of fatty acids and glycerol (alcohol) which have specific bonds between them.
In simple words: Lipids are biomolecules composed of fatty acids and glycerol, characterized by their insolubility in water and diverse biological functions.

🎯 Exam Tip: Key elements of a lipid definition include its components (fatty acids, glycerol) and its general property of being water-insoluble.

Question 2. What happens to fats that are eaten in excess?
Answer: When excess of fats are eaten, they are stored in adipose connective tissue.
In simple words: Excess fats consumed are stored in specialized adipose (fat) connective tissue in the body.

🎯 Exam Tip: Connect excess fat intake directly to storage in adipose tissue, emphasizing the body's energy reserve mechanism.

Question 3. Which hormones regulating reproductive functions are produced from fatty acids?
Answer: Progesterone, estrogen and testosterone are the reproductive hormones produced from fatty acids.
In simple words: Reproductive hormones like progesterone, estrogen, and testosterone are synthesized from fatty acids.

🎯 Exam Tip: Remember that steroid hormones, including sex hormones, are derived from lipids (fatty acids).

Question 4. How is plasma membrane of the cells formed?
Answer: The digested fats are absorbed in the form of fatty acids. These are converted back to phospholipids from which plasma membrane of cells is formed.
In simple words: The plasma membrane is primarily formed from phospholipids, which are synthesized from absorbed fatty acids after fat digestion.

🎯 Exam Tip: Link fatty acids to phospholipid synthesis, which in turn forms the fundamental structure of the cell's plasma membrane.

Question 5. What happens to lipids when their digestion is completed? How much energy do they provide?
Answer: After complete digestion of lipids they are converted to fatty acids and glycerol. 1 gm of lipid provides 9 kcal of energy.
In simple words: Upon complete digestion, lipids break down into fatty acids and glycerol, providing 9 kcal of energy per gram.

🎯 Exam Tip: State both the end products of lipid digestion and the energy yield per gram accurately.

Diagram Based Questions:

Question 1. Draw a neat diagram of the structure of chromosome and label the parts: (a) Centromere (b) p-arm
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक क्रोमोसोम की संरचना का आरेख है, जिसमें स्पष्ट रूप से इसके भुजाएँ (p Arm, q Arm), सेंट्रोमीयर और DNA को दर्शाया गया है। यह आरेख छात्रों को क्रोमोसोम के विभिन्न भागों की पहचान करने और उनके कार्यात्मक महत्व को समझने में मदद करता है।

🎯 Exam Tip: Practice drawing and labeling the key components of a chromosome, including p-arm, q-arm, and centromere, as it's a fundamental concept in genetics.

Question 2. Sketch and label the diagram to show ATP - the energy currency of the cell.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एडेनोसिन ट्राइफॉस्फेट (ATP) और एडेनोसिन डाइफॉस्फेट (ADP) की संरचना और उनके ऊर्जा संबंधी संबंध को दर्शाता है। इसमें एडेनाइन, राइबोस शुगर, और फॉस्फेट बॉन्ड्स को दर्शाया गया है, साथ ही यह भी दिखाया गया है कि कैसे एक फॉस्फेट बॉन्ड के टूटने से ऊर्जा निकलती है और ADP बनता है।

🎯 Exam Tip: Be able to draw and label the structure of ATP, showing its adenine, ribose, and three phosphate groups, and how breaking a phosphate bond releases energy.

Question 3. Mitochondria and Krebs cycle:
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख माइटोकॉन्ड्रिया की आंतरिक संरचना और क्रैब्स चक्र के विभिन्न घटकों को दर्शाता है। इसमें प्यूरुविक एसिड, फैटी एसिड और एसिटाइल-कोएन्जाइम-ए के माइटोकॉन्ड्रिया में प्रवेश को दिखाया गया है, जहां क्रैब्स चक्र होता है, जिससे \( \text{NADH}_2 \), \( \text{FADH}_2 \), और \( \text{CO}_2 \) जैसे उत्पाद बनते हैं।

🎯 Exam Tip: Understand the structure of the mitochondrion and how its compartments are involved in the Krebs cycle, linking the inputs and outputs of the cycle.

Question 4. Observe the diagrams 2.8 and 2.9 given on the Textbook page no. 19 and answer the following questions.
ℹ️ चित्र व्याख्या (Diagram Explanation): आरेख 2.8 और 2.9 अर्धसूत्रीविभाजन (Meiosis) के विभिन्न चरणों को दर्शाते हैं। आरेख 2.8 अर्धसूत्रीविभाजन-I के चरणों को दिखाता है, जिसमें प्रोफेज़-I (लेप्टोटीन से पैकिटीन तक समरूप गुणसूत्रों का युग्मन और क्रॉसिंग ओवर), मेटाफेज़-I (भूमध्यरेखीय तल पर गुणसूत्रों का संरेखण), एनाफेज़-I (समरूप गुणसूत्रों का पृथक्करण), और टेलोफेज़-I (दो पुत्री कोशिकाओं का निर्माण) शामिल हैं। आरेख 2.9 अर्धसूत्रीविभाजन-II के चरणों को दिखाता है, जिसमें प्रोफेज़-II, मेटाफेज़-II, एनाफेज़-II और टेलोफेज़-II शामिल हैं, जो अनिवार्य रूप से माइटोसिस के समान होते हैं, जिससे अंततः चार अगुणित पुत्री कोशिकाएँ बनती हैं।

(a) Which peculiarity do you observe in the figure of Metaphase-I of meiosis ?
Answer: The chromosomes are seen lying on the equatorial plane in the metaphase-I of meiosis.
In simple words: In Metaphase-I of meiosis, homologous chromosome pairs align centrally along the equatorial plate of the cell.

🎯 Exam Tip: Identify the key event in Metaphase-I: alignment of homologous pairs, not individual chromosomes, at the metaphase plate.

(b) What is the important difference between Telophase-I and Telophase-III of meiosis?
Answer: In figure of Telophase-I the diploid chromosomes are seen in two daughter cells. In Telophase-Il four daughter cells are seen with haploid chromosomes in them.
In simple words: Telophase-I results in two haploid cells, each with replicated chromosomes, while Telophase-II concludes with four haploid cells, each containing unreplicated chromosomes.

🎯 Exam Tip: Focus on the number of cells and the ploidy level (diploid vs. haploid) and chromosome state (replicated vs. unreplicated) to differentiate between Telophase-I and Telophase-II.

(c) Which figure shows phenomena of crossing over?
Answer: The third figure of Prophase-I shows phenomena of crossing over.
In simple words: The third figure depicting Prophase-I shows the event of crossing over, where genetic material is exchanged between homologous chromosomes.

🎯 Exam Tip: Recognize the visual representation of crossing over (chiasmata formation) in diagrams of Prophase-I.

Question 5. Label the diagram below? Which phase of cell division is seen in the above diagram?
Answer: The above figure shows Telophase-II of Meiosis-II.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख अर्धसूत्रीविभाजन-II के टेलोफेज़-II चरण को दर्शाता है, जिसमें क्रोमोसोम ध्रुवों पर एकत्रित हो रहे हैं, नाभिकीय झिल्ली फिर से बन रही है और कोशिका द्रव्य विभाजित होने लगता है, जिससे चार अगुणित पुत्री कोशिकाओं का निर्माण होता है। इसमें गुणसूत्र, सेंट्रियोल, नाभिकीय झिल्ली और न्यूक्लियोलस को दिखाया गया है।

In simple words: The diagram illustrates Telophase-II of Meiosis-II, characterized by chromosomes reaching the poles, nuclear envelopes reforming around them, and the cell preparing for cytokinesis, leading to four haploid daughter cells.

🎯 Exam Tip: Identify the distinct features of Telophase-II, such as chromosomes at poles, nuclear envelope reformation, and the onset of cytokinesis, differentiating it from other stages of cell division.

Question 5. Label the diagram below? Which phase of cell division is seen in the above diagram?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख अर्धसूत्रीविभाजन-II के टेलोफेज-II चरण को दर्शाता है। इसमें चार पुत्री कोशिकाएँ दिखाई गई हैं, प्रत्येक में गुणसूत्र और नए नाभिकीय झिल्ली बन रही हैं, और सेंट्रियोल, पुत्री कोशिका के गुणसूत्र, नाभिकीय झिल्ली और न्यूक्लियोलस जैसे घटक स्पष्ट रूप से लेबल किए गए हैं। The phase of cell division seen in the above diagram is **Telophase-II of Meiosis-II**.
In simple words: This diagram illustrates the final stage of meiosis II, where the cell divides into four haploid daughter cells, each forming new nuclei around its chromosomes.

🎯 Exam Tip: Understanding the visual characteristics of each meiotic phase, particularly the number of cells and chromosome arrangement, is crucial for identification in exams.

Question 6. Observe and label the diagram: (Text Book Page No. 13)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख मानव श्वसन प्रणाली को दर्शाता है, जिसमें नाक, मुँह, स्वरयंत्र, श्वासनली, ब्रोन्की, फेफड़े, डायाफ्राम और एल्वियोलस जैसे प्रमुख अंग स्पष्ट रूप से लेबल किए गए हैं। यह श्वसन मार्ग और गैस विनिमय से संबंधित संरचनाओं का एक व्यापक दृश्य प्रस्तुत करता है। The diagram shows the **Human respiratory system** with the following labeled parts: Nose, Mouth, Larynx, Trachea, Bronchi, Lungs, Diaphragm, and Alveolus.
In simple words: The diagram displays the human respiratory system, detailing the various organs involved in breathing, from the entry points of air to the gas exchange units in the lungs.

🎯 Exam Tip: Accurately labeling diagrams of human organ systems, such as the respiratory system, is a common exam question that tests anatomical knowledge.

Activity based questions:

Question 1. Complete the following chart and state which process of energy production it represents:
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चार्ट ऊर्जा उत्पादन की प्रक्रिया को दर्शाता है, जिसमें लिपिड, कार्बोहाइड्रेट और प्रोटीन जैसे विभिन्न खाद्य पदार्थ ऊर्जा में परिवर्तित होते हैं। इसमें दर्शाया गया है कि लिपिड से फैटी एसिड, कार्बोहाइड्रेट से ग्लाइकोलाइसिस के माध्यम से पाइरुविक एसिड, और प्रोटीन से अमीनो एसिड बनते हैं। ये सभी अंततः एसिटाइल को-एंजाइम-ए में परिवर्तित होकर क्रेब्स चक्र में प्रवेश करते हैं, जहाँ कार्बन डाइऑक्साइड, पानी और ऊर्जा का उत्पादन होता है। The chart shows process of energy production through aerobic respiration of carbohydrates, proteins and fats. This process outlines how: (1) Lipids are converted into Fatty acids. (2) Carbohydrates undergo Glycolysis to form Pyruvic acid. (3) Proteins are broken down into Amino acids. (4) Fatty acids, Pyruvic acid, and Amino acids are further converted into Acetyl Co-enzyme A. (5) Acetyl Co-enzyme A then enters the Krebs Cycle, which ultimately yields CO2, H2O, and Energy.
In simple words: This chart illustrates how the body derives energy from different food sources like carbohydrates, fats, and proteins by breaking them down into simpler molecules that feed into the Krebs cycle for energy generation.

🎯 Exam Tip: Understanding this metabolic flowchart is critical for explaining how the body produces energy from various macronutrients during aerobic respiration.