CBSE Class 11 Biology Respiration In Plants Notes

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Respiration In Plants Class 11 Biology Revision Notes

Class 11 Biology students should refer to the following concepts and notes for Respiration In Plants in standard 11. These exam notes for Grade 11 Biology will be very useful for upcoming class tests and examinations and help you to score good marks

Respiration In Plants Notes Class 11 Biology

Introduction
Mechanism of breakdown of food materials within the cell to release energy, and the trapping of this energy for synthesis of ATP is called cellular respiration.
The complete combustion of glucose, which produces CO2 and H2O as end products, yields energy most of which is given out as heat.
The compounds that are oxidised during this process are known as respiratory substrates.
Usually carbohydrates are oxidised to release energy, but proteins, fats and even organic acids can be used as respiratory substrates, under certain conditions.

Calorific Value of protein, carbohydrate and fat :
The amount of heat liberated from complete combustion of 1 g food in a bomb calorimeter (a closed metal chamber filled with O2) is its gross calorific values.

Respiratory substrateGross Calorific ValuePhysiological Value
Carbohydrate4.1 kcal/g4.0 kcal/g
Protein5.65 kcal/g4.0 kcal/g
Fat9.45 kcal/g9.0 kcal/g

 

The actual amount of energy released by combustion of 1g of food is the physiological value of food.

Features of Cellular respiration :
(i) All the energy contained in respiratory substrates is not released free into the cells or in a single step.
(ii) Energy is released in a series of slow step wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP. (ATP acts as the energy currency of the cell)
(iii) Cellular respiration is an amphibolic process.
Reason : The carbon skeleton (Intermediates of respiration) produced during respiration is used as precursors for biosynthesis of other molecules in the cell.
(iv) Cellular respiration is an exergonic process.
Reason : The breaking of C–C bonds of complex compounds through oxidation within the cells, leading to release of considerable amount of energy.
(v) Cellular respiration is a downhill process.
Reason : Oxygen is a strong electron acceptor.

Do plants breathe ?

Yes, plants require O2 for respiration to occur and they also give out CO2. Hence, plants have systems in place that ensure the availability of O2. Plants, unlike animals, have no specialised organs for gaseous exchange but they have stomata and lenticels for this purpose.
There are several reasons why plants can get along without respiratory organs.
(1) Each plant part takes care of its own gas-exchange needs. There is very little transport of gases from one plant part to another. 
(2) Plants do not present great demands for gas exchange. Roots, stems and leaves respire at rates far lower than animals do. Only during photosynthesis are large volumes of gases exchanged and, each leaf is well adapted to take care of its own needs during these periods. When cells photosynthesise, availability of O2 is not a problem in these cells since O2 is released within the cell.
(3) The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.
Most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Types of respiration :
(A) On the basis of type of respiratory substrates :
(1) Floating respiration :
When carbohydrate or fats are oxidised inside the cell. Carbohydrates and fats are floating inclusions of cell thus, this is called floating respirations.
(2) Protoplasmic respiration :
When protein is oxidised inside the cell. This occurs in starved cell. Protein is constituent of protoplasm thus, this is called protoplasmic respiration.
(B) On the basis of presence or absence of O2 :
(1) Aerobic (2) Anaerobic/Fermentation

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Aerobic respiration : It is divided into following stages :
(1) Glycolysis
(2) Link reaction
(3) Krebs cycle
(4) Electron transport system and oxidative phosphorylation

(1) Glycolysis
(i) The term glycolysis has originated from the greek words, glycos for sugar and lysis for splitting.
(ii) The scheme of glycolysis was given by Gustav Embden, Otto Meyerhof and J.Parnas. Thus it is often referred to as the EMP pathway.
(iii) This process takes place inside cytoplasm of all living cells.
(iv) Glycolysis is called common pathway because it is a common step between aerobic and anaerobic respiration.
(v) Glycolysis is a chain process of ten chemical reactions, where 1, 3 and 10 reactions are irreversible.
(vi) In this process, glucose undergoes partial breakdown / oxidation to form two molecule of pyruvic acid. In plants this glucose is derived from sucrose (end product of photosynthesis) or from storage carbohydrates (starch).
(vii) Sucrose is converted into glucose and fructose by the enzyme, invertase and these two monosaccharides readily enter the glycolytic pathway. Glucose is the favoured substrate for respiration

Gross products of glycolysisNet products of glycolysis
2 molecules of pyruvic acid (CH3CO.COOH)molecules of pyruvic acid (CH3CO.COOH)
2 molecules of NADH2 molecules of NADH
4 molecules of ATP2 molecules of ATP

(2) Link reaction/Gateway step/Transition reaction
Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvate dehydrogenase.
This reaction require participation of several co-enzymes, including NAD+ and CoA.
Pyruvic acid + CoA + NAD+ (Mg2+/Pyruvate dehydrogenase) Acetyl CoA + CO2 + NADH(H+)
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis).

The acetyl CoA is called connecting link between glycolysis and Krebs cycle.

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(3) Krebs cycle
(i) Named after the scientist Hans Krebs who first elucidated it. It is also called TCA (tri carboxylic acid) cycle or CA (citric acid) cycle.
(ii) Krebs cycle occurs inside mitochondrial matrix of eukaryotic cells and cytoplasm of prokaryotic cells.
(iii) One turn of Krebs cycle involve four dehydrogenation , two decarboxylation and one substrate level phosphorylation.
(iv) OAA is considered as the first member of the cycle.
(v) All enzymes of Krebs cycle are located inside mitochondrial matrix except succinate dehydrogenase (Marker enzyme), which is located in inner membrane of mitochondria.

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(4) ETS and oxidative phosphorylation (Terminal oxidation of NADH and FADH2)
(i) It is associated with release and utilisation of the energy stored in NADH+H+ and FADH2.
(ii) NADH+H+ and FADH2 are oxidised through the electron transport system (ETS) and the electrons are passed on to O2 resulting in the formation of H2O.
(iii) ETS is present in the inner mitochondrial membrane of eukaryotes and plasma membrane of prokaryotes.

Electron carriers of ETS :
(i) Flavins (FMN)
(ii) FeS
(iii) Quinone (Ubiquinone or CoQ)
(iv) Cytochromes (Cyt b → Cyt c1 → Cyt c → Cyt a → Cyt a3)
ETS is consists of four complexes and fifth complex is ATP synthase which is associated with ATP synthesis.

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Special features of ETS :
(i) UQ (Co.Q) and Cyt c are mobile carrier of ETS.
(ii) Cytochrome c is a small protein attached to outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex-III (cytochrome bc1) and complex-IV (cytochrome c oxidase)

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(iv) The role of O2 is limited to the terminal stage of the process. The presence of oxygen is vital. Since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor.
Cyanide inhibits the activity of cytochrome c oxidase which catalyse the oxidation of cytochrome a3 and reduction of oxygen. In mitochondria of some plants alternative oxidase system is present in which ETS continues even in presence of cyanides. This type of respiration is known as cyanide resistance respiration or alternate electron pathway. eg. Spinach, Pisum

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Oxidative phosphorylation (Chemiosmotic theory / Coupling theory)
(i) During ETS of respiration CoQ (UQ) & FMN can releases H+ ions in perimitochondrial space and leads to differenctial H+ ion concetration across inner mitochondrial membrane. This differential H+ ion concentration across inner mitochondrial membrane leads to creation of proton gradiant (pH gradient) and Electrical potential (diffrence of charge). Both are collectively known as Proton motive force (PMF).
(ii) PMF do not allow stay of H+ ions in Perimitochondrial space (PMS) so they return towards the matrix through F0 part of ATPase selectively. Passage of 2H+ ions through F0 part or proton channel leads to synthesis of 1 ATP.

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(iii) Cytosolic or extra mitochondrial or glycolytic NADH transported to ETS by two type of shuttles (Only in eukaryotes) :
(a) Glycerol phosphate shuttle Common shuttle system eg.- all plants, nerves and muscles.
(b) Malate aspartate shuttle Heart, liver and kidney etc.
(iv) In prokaryotes, shuttle mechanism is absent. They always get 38 ATP from aerobic respiration of 1 glucose.
(v) Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH2 produces 2 molecules of ATP.

The respiratory balance sheet
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis,
TCA cycle and ETS pathway following one after another.
• The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesise any other compound.
• Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But this kind of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Theoretical energy calculation for complete oxidation of one glucose molecule :

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Amphibolic pathway
(i) Glucose is the favoured substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired, but then they do not enter the respiratory pathway at the first step.
(ii) Fats would need to be broken down into glycerol and fatty acids first. If fatty acids were to be respired they would first be degraded to acetyl CoA and enter the pathway. Glycerol would enter the pathway after being converted to PGAL.
(iii) The proteins would be degraded by proteases and the individual amino acids (after deamination) depending on their structure would enter the pathway at some stage within the Krebs cycle or even as pyruvate or acetyl CoA.
(iv) Since respiration involves breakdown of substrates, the respiratory process has traditionally been considered a catabolic process and the respiratory pathway as a catabolic pathway. Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of protein too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

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(v) Glycolysis is also known as oxidative anabolism or catabolic resynthesis, because it links with anabolism of fats and amino acids. An intermediate PGAL is used for the synthesis of glycerol later forms fats or lipid.
PGA is used for synthesis of Serine, Glycine, Cysteine. Alanine forms from pyruvate.
(vi) Acetyl Co-A is common meeting point (connecting link) between fat, carbohydrate and protein metabolism.

Amphibolism of Krebs cycle –
(1) Acetyl Co-A – Synthesis of fatty acids & GA (Gibberellic acid)
(2) Succinyl CoA – Synthesis of chlorophyll, Cytochromes, Phytochromes
(3) OAA & a-Ketoglutaric acid – Synthesis of Amino acids.
(4) OAA – Synthesis of Alkaloids.

Anaerobic respiration / Fermentation
In fermentation not much energy is released; less than seven percent of the energy in glucose is released and not all of it is trapped as high energy bonds of ATP.
The processes are hazardous, either acid or alcohol is produced. Yeasts poison themselves to death when the concentration of alcohol reaches about 13% percent.

Fermentation is of two types :
(A) Alcoholic fermentation ;
In this fermentation, say by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to CO2 and ethanol. The enzymes, pyruvate decarboxylase and alcohol dehydrogenase catalyse these reactions.

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(B) Lactic acid Fermentation :
Some bacteria produce lactic acid from pyruvic acid. In animal cells also like in muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid by lactate dehydrogenase.

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• The reducing agent is NADH + H+ which is reoxidised to NAD+ in both the processes.
During alcoholic fermentation triose phosphate (3PGAL) is the electron donor and acetaldehyde is acceptor, while during lactic acid fermentation although electron donor is triose phosphate but acceptor is pyruvic acid.

Pasteur effect :
It is an inhibitory effect of oxygen on the fermentation process.

Explanation :
The effect can be easily explained; as the yeast being facultative anaerobes can produce energy using two different metabolic pathways.
(i) While the oxygen concentration is low, the product of glycolysis, pyruvate is turned into ethanol and CO2 and the energy production efficiency is low (2 moles of ATP per mole of glucose).
(ii) If the oxygen concentration grows, pyruvate is converted into acetyl CoA that can be used in the citric acid cycle, which increases the efficiency to 36 moles of ATP per mole of glucose. 

Respiratory Quotient
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.

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RQ = Less than one
• During complete oxidation of protein and fat
• During protoplasmic respiration (In case of a starved cell)
• In case of mixed diet
• In case of germinating fatty seeds.

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Pure proteins or fats are never used as respiratory substrates because before entering the respiratory pathway they must be converted into such compounds which can enter into the glycolysis or link reaction or Krebs cycle at their respective stages.

RQ = More than one
• During complete oxidation of organic acids.
• In case of maturing fatty seeds.

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DYNAMIC STATE OF BODY CONSTITUENTS - CONCEPT OF METABOLISM
  All biomolecules have a turn over. This means that they are constantly being changed into some other biomolecules and also made from some other biomolecules. This breaking and making is through chemical reactions constantly occuring in living organisms. Together all these chemical reactions are called metabolism.
  Each of the metabolic reactions results in the transformation of biomolecules. A few examples for such metabolic transformations are; removal of CO2 from amino acids making an amino acid into an amine, removal of amino group in a nucleotide base; hydrolysis of a glycosidic bond in a disaccharide, etc.
  Majority of these metabolic reactions do not occur in isolation but are always linked to some other reactions. In other words, metabolites are converted into each other in a series of linked reactions called metabolic pathways.
  These metabolic pathways are similar to the automobile traffic in a city. These pathways are either linear or circular. These pathways criss-cross each other, i.e., there are traffic junctions. Flow of metabolites through metabolic pathway has a definite rate and direction like automobile traffic. This metabolic flow is called the dynamic state of body constituents.
  There is no uncatalysed metabolic conversion in living systems. Even CO2 dissolving in water, a physical process, is a catalysed reaction in living systems.

METABOLIC BASIS FOR LIVING
  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. 
  How do living organisms derive their energy? What strategies have they evolved? How do they store this energy and in what form? How do they convert this energy into work? All this study comes under a sub-discipline called 'Bioenergetics'.

THE LIVING STATE
  Chemical compounds in a living organism, otherwise called 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 4.5-5.0 mM, 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.
  One should remember from physics that systems at equilibrium cannot perform work. As living organisms work continously, 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.

ENZYMES

History
 Buchner discovered and isolated the enzyme zymase from yeast cells, while Kuhne coined the term enzyme.
 J.B. Sumner purified and crystalized urease enzyme from canavalia/Jack bean/Lobia plant.

Characteristics of enzymes
(1) Almost all enzymes are proteins. Though, there are some nucleic acids that behave like enzymes. These are called ribozymes.
 L19 RNAase (a ribozyme) was discovered by T. Cech from rRNA of a protozoan, Tetrahymena thermophila.
 RNAase P or Ribonuclease P (a ribozyme) was discovered by Altman from a prokaryotic cell.
(2) Enzymes are colloidal substances, which are macromolecules of amino acids and are synthesised by ribosomes under genetic control.
(3) Enzyme can be depicted by a line diagram.

An enzyme like any protein has -
(a) Primary structure : Amino acid sequence of the protein. Its lack active sites.
(b) Secondary structure : It is a helical structure which also lack active sites.
(c) Tertiary structure : In this structure backbone of the protein chain folds upon itself, the chain criss-crosses itself and hence many crevices or pockets are made such pockets represent active sites.
The catalytic structure of most of the enzymes are tertiary and globular.
(d) Quarternary structure : Represented by isoenzymes and active sites are present.
4) 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.
(5) Enzymes are very specific to their substrate or reactions. They are required in very small amount to catalyse a reaction. Catalytic power of an enzyme depends upon -
(a) Turn over number (b) Km constant
(a) Turn over number : It is the number of substrate molecules converted into products per unit time by a molecule of enzyme. Thus, catalytic power is directly proportional to turn over number. Carbonic anhydrase enzyme is considered as the fastest enzyme.

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(b) Km constant : This was coined by Michaelis and Menten. It is the concentration of substrate at which rate of reaction attains half of its maximum velocity.

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Catalytic power of an enzyme is inversely proportional to its Km value.
(6) Enzyme increase the rate of reaction several times by lowering down activation energy.
(7) Catalytic power of an enzyme remains same even outside the living system.
(8) Enzymes when not in use, represent inactive form, called zymogen or pro-enzyme. Pepsinogen is an inactive form of pepsin, similarly trypsinogen is an inactive form of trypsin.

Enzyme (Biocatalyst)Inorganic catalyst
Enzymes are thermo-sensitive and get damaged at high temperatures (say above 40°C)They work efficienty at high temperatures and high pressures.

However, enzymes isolated from organisms who normally live under extremely high temperatures (eg. hot vents and sulphur springs), are stable and retain their catalytic power even at high temperature. Thermal stability is thus an important quality of such enzymes isolated from thermophilic organisms. e.g. taq polymerase.

Uncatalysed reaction versus catalysed reaction

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In the absence of an enzyme this reaction is very slow, with about 200 molecules of H2CO3 being formed in an hour. However, in the presence of enzyme carbonic anhydrase inside cytoplasm the reaction speeds dramatically with about 600,000 molecules being formed every second. The enzyme has accelerated the reaction rate by about 10 million times.

How do enzymes bring about such-high rates of chemical conversions ?
The chemical or metabolic conversion refers to a reaction. The chemical which is converted into a product is called a substrate. Hence enzymes, i.e. proteins with three dimensional structures including an active site to convert a substrate (S) into a product (P). Symbolically, this can be depicted as:

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The y-axis represents the potential energy content. The x-axis represents the progression of the structural transformation or states through the "transition state". In above graph "P" is at a lower level than "S", thus this reaction is an exothermic reaction. (No need to supply energy in order to form the product.) However, whether it is an exothermic or spontaneous reaction or an endothermic or energy requiring reaction, the "S" has to go through a much higher energy state or transition state.

"The difference in average energy content of "S" from that of this transition state is called activation energy".
Enzymes eventually bring down this energy barrier making the transition of "S" to "P" more easy.
All changes (ES complex, EP complex) occured during transition state are transient and unstable.

Nature of enzyme action
Each enzyme (E) has a substrate (S) binding site in its molecular structure so that a highly reactive enzyme- substrate complex (ES) is produced. This complex is short-lived and dissociates into its product(s) P and the unchanged enzyme with an intermediate formation of the enzyme-product complex (EP). The formation of the ES complex is essential for catalysis.

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The catalytic cycle of an enzyme action can be described in the following steps:
1. First, the substrate binds to the active site of the enzyme, fitting into the active site.
2. The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
3. The active site of the enzyme, now in close proximity of the substrate breaks or form the chemical bonds of the substrate and the new enzyme- product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.

Cofactors
Enzymes are composed of one or several polypeptide chains. However, there are a number of cases in which non-protein constituents called cofactors are bound to the enzyme to make the enzyme catalytically active. In these instances, the protein portion of the enzymes is called the apoenzyme and non protein portion is called the cofactor.
Three kinds of cofactors may be identified: co-enzymes, prosthetic groups and metal ions.

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Important examples of coenzymes and metal activators :

Examples of co-enzymeExamples of metal activator
Co. I (NAD/DPN) : Derivative of niacinFe++ : Cytochrome c oxidase, catalase, peroxidase, aconitase
Co. II (NADP/TPN) : Derivative of niacinCu++ : Cytochrome c oxidase, tyrosinase
FAD : Derivative of riboflavinZn++ : Carbonic anhydrase, alcohol dehydrogenase, carboxypeptidase
FMN : Derivative of riboflavinMg++ : Hexokinase, glucokinase, pyruvate kinase, PEPcase, RuBisCO
TPP : Derivative of thiamineCu++ : Cytochrome c oxidase, tyrosinase Pyruvate kinase
Co. Q : Derivative of ubiquinoneMn++ : arginase, ribonucleotide reductase, decarboxylase Co
Co. R : Derivative of biotinMo : Nitrogenase complex, nitrate reductase
Co. A : Derivative of panthothenic acidSe : Glutathione peroxidase

Classification and Nomenclature
Thousands of enzymes have been discovered, isolated and studied. Most of these enzymes have been classified into different groups based on the type of reactions they catalyse. Enzymes are divided into 6 classes each with 4-13 subclasses and named accordingly by a four-digit number.
(I) Oxidoreductases/dehydrogenases : Enzymes which catalyse oxidoreduction (oxidation-reduction) between
two substrates i.e. S and S' e.g. cytochrome c oxidase, dehydrogenase etc.
S reduced + S' oxidised ¾® S oxidised + S' reduced
(II) Transferases : Enzymes catalysing a transfer of a group, G (other than hydrogen) between a pair of substrate S and S'. e.g. transaminase, hexokinase etc.
S–G + S' → S + S–G
(III) Hydrolases : Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C–C, C–halide or P–N bonds. e.g. proteases, lipases, carbohydrases etc.
(IV) Lyases : Enzymes that catalyse removal of a group from substrate by mechanisms other than hydrolysis and leaving double bonds. e.g. aldolase.

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(V) Isomerases : Includes all enzymes catalysing inter-conversion of optical, geometric or positional isomers.
e.g. hexoisomerase, mutase etc.
(VI) Ligases/synthase : Enzymes catalysing the linking together of two compounds. Such enzymes catalyse joining of C–O, C–S, C–N, P–O etc. bonds. e.g. citrate synthase, DNA ligase etc.

Factors affecting enzyme activity
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 :
(1) Temperature
(2) pH
(3) Change in substrate concentration
(4) Inhibitor
(1) Temperature :
Enzymes generally function in a narrow range of temperature. Each enzyme shows its highest activity at a particular temperature called the optimum temperature. 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.
A general rule of thumb is that rate doubles or decreases by half for every 10°C change in either direction. Thus, value of Q10 for enzymatic activities is 2.

(2) pH :
Enzymes generally function in a very narrow range of pH. Each enzyme shows its highest activity at a particular pH called the optimum pH. Activity declines both below and above the optimum value.

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

(4) Inhibitors/Enzyme inhibition :
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 (inhibits) enzyme activity, the process is called inhibition and the
chemical is called an inhibitor.

It is of two types :
(1) Competitive inhibition :
 When the inhibitor closely resembles the substrate (substrate analogues) in its molecular structure and binds with active site of enzyme leads to inhibition of enzyme activity.
 It is competitive as well as reversible because inhibitor binds with active site (competitive) and can be deattached by increasing concentration of substrate. (reversible)

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eg. (a) Inhibition of succinate dehydrogenase by malonate. Malonate and succinate are structural analogues.
(b) Inhibition of folic acid synthesis by sulpha drugs.
p-amino benzoic acid (PABA) is a precursor of folic acid. Sulpha drugs are structural analogues of p-aminobenzoic acid, thus inhibits synthesis of folic acid. This competitive inhibition is often used in the controls of bacterial pathogens.

Reason : Bacteria requires folic acid for growth and multiplication.
In the presence of competitive inhibitor Km increases while Vmax. remains unchanged.
(2) Non competitive inhibition : It is again of two types :
(i) Non competitive irreversible :
• When the inhibitor binds with the site other than its active site (non competitive) and destroy the sulpha hydryl (S-H) group of enzyme.
• Here binding is irreversible because inhibitor can not be deattached by increasing substrate concentration

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eg. Cyanide binds with Cu center of Cyt a3 of cytochrome c oxidase and inhibits enzyme permanently.

(ii) Non competitive reversible : (Mostly)
• Some enzymes have allosteric site to control active site. This control is called modulation.
• Most of non competitive inhibitors bind reversibly with allosteric site and negatively change the configuration of active site. Such inhibition called non competitive reversible or negative allosteric modulation.

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eg. Phosphofructokinase/PFK (pacemaker enzyme) inhibited by excess of ATP (negative modulation).

• Some allosteric enzyme inhibited by the product of that biochemical reaction which is catalysed by them, such inhibition called feedback inhibition or retro inhibition or product inhibition.

""CBSE-Class-11-Biology-Respiration-In-Plants-25

In above reaction the enzyme hexokinase inhibited reversibly by excess of product (glucose-6-phosphate) "Therefore, all feedback inhibition are allosteric inhibition but all allosteric inhibiton are not feed back"
In the presence of non competitive inhibitor Km remains unchanged while Vmax. decreases.Km is not applicable for allosteric modulation.

""CBSE-Class-11-Biology-Respiration-In-Plants-26

Note : Allosteric site may also be use to increase enzymatic activity such phenomenon is called positive allosteric modulation. eg. Phosphofructokinase/PFK activated by excess of AMP (positive modulation).

Experiment - 4
Chlorophyll is essential for photosynthesis
(By experiement of varigated leaf)

Pick up a variegated leaf from coleus, pothos plant. Which has received light for a few hours. Take a piece of rice paper and place it over the surface of variegated leaf. Mark out the green and non green areas by means of a pencil. 'kill the leaf in boiling water and remove its chlorophyll by immersing it in warm spirit. Wash and test for starch. Only those parts of the leaf will give positive test which were originally green or with chlorophyll.

""CBSE-Class-11-Biology-Respiration-In-Plants-27

Experiment - 5
Measurment of photosynthesis by the rate of evolution of oxygen
(By Wilmott Bubbler)

Take a fresh shoot of Hydrilla plant remove its apical 10cm long portion by giving an oblique cut while keeping it submerged in water. Place it in a test tube 3/4th full of water with the cut end upwards as such or gently tied to a glass rod for support. Add some sodium bicarbonate in the water of test tube. Give a fresh oblique cut to the shoot end & place the test tube in sunlight. Bubbles of oxygen gas are seen coming out of the cut end of hydrilla plant count the no. of bubbles per minute.Rate of reaction increases by adding NaHCO3- in H2O.

""CBSE-Class-11-Biology-Respiration-In-Plants-28

Experiment - 6
Measurement of R.Q. by means of Ganongs respirometer.

Ganong's respirometer has a large stoppered bulb,a graduated side tube and a levelling tube. Both the stopper and neck of the bulb have a hole which can be used to equalise the atmospheric pressure inside the apparatus.
Place the respiring material in the bulb of Ganong's respirometer. Pour saturated common salt solution (CO2 is sparingly soluble in it) in the levelling tube. Adjust the level of solution in the levelling tube and note the reading. After an hour note the reading. Calculate the rise (e.g. when fats are respiratory substrate) or fall (e.g., when organic acids are substrate) of salt solution level as +ve, or – ve, cc. Insert-KOH crystal over the salt solution level in the graduated tube through the levelling tube. KOH shall absorb CO2. Take a new reading. Let it be V2CC.V2CC is the volume of CO2 librated during respiration. The O2 used is V2 plus V1. R.Q. will therefore be R.Q = V/ V2+(-ve or + ve V1)

""CBSE-Class-11-Biology-Respiration-In-Plants-29
If v1 is zero, positive or negative, R.Q. will be accordingly unity, less than unity or more than one respectively.

Experiment - 7
Bell jar experiments for demonstration of transpiration

Take a small well watered potted plant. Cover the external surface of the pot and its soil throughly with polythene bag. Place the potted plant on a glass slab in a cool place invert a dry bell-jar over it. Seal the edges of bell-jar with vaseline. After some time drops of water will appear on the inner side of the bell-jar. The water drops have been formed by the condensation of water va- pours given out by shoot of the plant in transpiration.

""CBSE-Class-11-Biology-Respiration-In-Plants-30

Experiment - 8
Compare the loss of water from the two sides of a leaf
By four leaf experiment

Remove four leaves of Pipal. They should be almost equal size. Seal the cut end of petioles with vaseline. Tie a thread to the petiole of each in the fashion of a loop. Apply vaseline to the upper surface of A, lower surface of B and both the surfaces of C. Keep the leaf D as such to function as control.
Weigh the leaves one by one. For weighing each leaf is picked up through the thread attached to its petiole.
After initial weighing keeping the leaves hanging from stands by means of their threads for a few hours. Examine them closely. Leaves A & D wilt while B and C remain turgid. Weigh them again. It will be seen that max loss of weigh occurs in D followed by A. Loss of weight is negligible in B. It is nill in case of C. In D leaf both surfaces loss water, while in A only the lower surface is exposed to transpire. The negligible loss of water in B shows that upper surface transpires little.

""CBSE-Class-11-Biology-Respiration-In-Plants-31

Experiment - 9
Measurement of Transpiration
By Ganong's potometer

Fill Ganong's Patometer completely with water. Close the hole of the capillary tube with your finger. Insert a fresh leafy shoot, cut under water, into the mouth of the wide tube by means of cork. Make the connections air tight by applying melted wax or vaseline to the joint. Place the apparatus in dry place or sunlight. Remove your finger from the free end. An air bubble will enter the graduated capillary tube dip the free end in a beaker containing water. The air bubble will be seen moving forwards. Note the time taken for the air bubble to travel from one end to other end in tube. As soon as the air bubble reaches the other end open the stop cock of the reservior till it is pushed back to the beginning of the horizontal tube. Repeat the experiment. Note the amount of water absorption in different conditions.
V = p r2 l V = volume of water which absorbed/transpired water
p = 3.14
l = distance of movement by air bubble

""CBSE-Class-11-Biology-Respiration-In-Plants-32

Experiment - 10
Demonstration of Geotropism by Clinostat

Geotropic curvature results is due to unilateral gravity stimulus can be demonstrated by using a Clinostat. Clinostat consist of a disc attached to the tip of an horizontal rod, which can be rotated with the help of a clockwork. 
Fix a small young potted plant on a clinostat in horizontal position and rotated, neither the root will bend down nor the stem will curve upward (Nullify the geotropism). It is because in such case the effect of gravity will be uniform all.

""CBSE-Class-11-Biology-Respiration-In-Plants-33
Round the stem and root. If however, the plant lies in horizontal position & not rotated, the roots will receive gravity stimulus only on their lower sides or the effect of gravity will be unilateral. Unilateral stimulus of gravity causes unequal distribution of auxin in root tip. (more auxin on upper side than on lower side). This in turn results in more growth on the upperside and less growth on lowerside and ultimately a +ve geotropism is observed. However in case of stem the more conc. of Auxin on lower side promotes more growth on that side so stem is negative geotropic. (Recent findings have implicated ABA in geotropism)

Experiment - 11
Demonstration of osmosis by Potato Osmoscope

Take a large sized potato tuber. Cut one side flat. Bore a cavity from the other side in such a way that only a thin base is left 
intact on the flat side. Also remove the skin near the edges of the flat end because the skin of the tuber is impermeable to water. Now place the tuber on its flat cut end in a petri dish half full of water. Fill half of the cavity of the potato tuber by 10-20% sugar solution. Mark the level of sugar solution in the cavity with the help of pin. After some time the level of sugar solution rises in the cavity.

""CBSE-Class-11-Biology-Respiration-In-Plants-34

The rise in the level of sugar solution in the cavity of the potato tuber indicate that it has absorbed water from the petri dish. The two are separated from each other by a large number of cells.
The entry of water into sugar solution proves that (i) Sugar solution is an osmotically active solution. (ii) the cytoplasm of all the cells of the tuber that lie between the sugar solution and the water of the petri dish act as a single semi-permeable membrane (iii) Water enters the sugar solution when it is separated from it by a semi-permeable membrane, the phenomenon being known as osmosis.

 

14.Respiration in Plants

Points to Remember :

• The breaking of C-C bonds of complex compounds through oxidation within the cells, leading to release of considerable amount of energy is called respiration.

• The compound that oxidized during this process is known asrespiratory substrates.

• In the process of respiration the energy is released in a series of slow step-wise reactions controlled by enzymes and is trapped in the form of ATP.

• ATP acts as the energy currency of the cell.

Glycolysis :

• The term has originated from the Greek word, glycos =glucose, lysis = splitting or breakdown means breakdown of glucose molecule.

• It is also called Embeden-Meyerhof-Paranus pathway. (EMP pathway)

• It is common in both aerobic and anaerobic respiration.

• It takes place outside the mitochondria, in the cytoplasm.

• One molecule of glucose (Hexose sugar) ultimately produces two molecules of pyruvic acid through glycolysis.

• Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate, catalyzed by hexokinase.

• This phosphorylated form of glucose is then isomerizes to produce fructose-6-phosphate.

• ATP utilized at two steps:

o First in the conversion of glucose into glucose-6-phosphate

o Second in fructose-6-phosphate→fructose 1, 6-diphosphate.

• The fructose-1, 6-diphosphate is split into dihydroxyacetone phosphate and 3-phosphoglyceraldehyde (DPGA).

• In one step where NADH + H+ is formed form NAD+; this is when 3-phosphogleceraldehyde (PGAL) is converted into 1, 3-bisphophoglyceric acid (DPGA).

• The conversion of 1, 3-bisphophoglyceric acid into 3-phosphoglyceric acid is also an energy yielding process; this energy is trapped by the formation of ATP.

• Another ATP synthesized when phosphoenolpyruvate is converted into pyruvic acid.

• During this process 4 molecules of ATP are produced while 2 molecules of ATP are utilized. Thus net gain of ATP is of 2 molecules.

Fermentation :

• There are three major ways in which different cells handle pyruvic acid produced by glycolysis:

o Lactic acid fermentation.

o Alcoholic fermentation.

o Aerobic respiration.

• Alcoholic fermentation :

o The incomplete oxidation of glucose to achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted into CO2 and ethanol.

o The enzyme pyruvic acid decarboxylase and alcohol dehydrogenase catalyze these reactions.

o NADH + H+ is reoxidised into NAD+.

• Lactic acid fermentation:

o Pyruvic acid converted into lactic acid.

o It takes place in the muscle in anaerobic conditions.

o The reaction catalysed by lactate dehydrogenase.

o NADH + H+ is reoxidised into NAD+.

• Aerobic respiration:

o Pyruvic acid enters into the mitochondria.

o Complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO2.

o The passing on the electrons removed as part of the hydrogen atoms to molecular oxygen (O2) with simultaneous synthesis of ATP.

Aerobic Respiration:

• The overall mechanism of aerobic respiration can be studied under the following steps :

• Glycolysis (EMP pathway)

• Oxidative Decarboxylation

• Krebs’s cycle (TCA-cycle)

• Oxidative phosphorylation

Oxidative decarboxylation:

• Pyruvic acid formed in the cytoplasm enters into mitochondria.

• Pyruvic acid is converted into Acetyl CoA in presence of pyruvate dehydrogenase complex.

• The pyruvate dehydrogenase catalyses the reaction require several coenzymes, including NAD+ and Coenzyme A.

• During this process two molecules of NADH are produced from metabolism of two molecules of pyruvic acids (produced from one glucose molecule during glycolysis).

• The Acetyl CoA (2c) enters into a cyclic pathway, tricarboxylic acid cycle.

Tri Carboxylic Acid Cycle (Krebs cycle) or Citric acid Cycle :

• This cycle starts with condensation of acetyl group with oxaloacetic acid and water to yield citric acid. This reaction is catalysed by citrate synthase.

• Citrate is isomerised to form isocitrate.

• It is followed by two successive steps of decarboxylation, leading to formation of α-ketoglutaric acid and then succinyl-CoA.

• In the remaining steps the succinyl CoA oxidized into oxaloacetic acid.

• During conversion of succinyl CoA to succinic acid there is synthesis of one GTP molecule.

• In a coupled reaction GTP converted to GDP with simultaneous synthesis of ATP from ADP.

 

MCQ Questions for NCERT Class 11 Biology Respiration in Plants

 Question. In respiration,

(a) 2 PGA are formed in glycolysis and none in Krebs’ cycle
(b) 6 PGA in glycolysis, 3 PGA in Krebs’ cycle
(c) 8 PGA in glycolysis, 3 PGA in Krebs’ cycle
(d) PGA formation does not occur in respiration.

Answer : A

Question. If R.Q. is less than 1.0 in a respiratory metabolism, it would mean that
(a) carbohydrates are used as respiratory substrate
(b) organic acids are used as respiratory substrate
(c) the oxidation of the respiratory substrate consumed more oxygen than the amount of CO2 released 
(d) the oxidation of the respiratory substrate consumed less oxygen than the amount of CO2 released.

Answer : C

Question. Dough kept overnight in warm weather becomes soft and spongy due to
(a) absorption of CO2 from atmosphere
(b) imbibition
(c) fermentation
(d) all of these.

Answer : C

Question. The electron transport chain helps to
(a) cycle NADH+H+ back to NAD+
(b) maintain surplus of hydrogen ions in the intermembrane space of mitochondria
(c) synthesise ATP through ATP synthase
(d) all of these.

Answer : D

Question. Select the wrong statement with respect to glycolysis.
(a) It occurs outside mitochondria.
(b) It is an anaerobic phase.
(c) Glucose undergoes partial oxidation to form two molecules of pyruvic acid.
(d) Glucose is phosphorylated to glucose-6-phosphate by isomerase enzyme.

Answer : D

Question. In aerobic respiration, one glucose molecule produces
(a) 10NADH + 4FADH2 + 4 ATP
(b) 12NADH + 4FADH2 + 4ATP + GTP
(c) 12NADH + 4GTP + 2FADH2
(d) 10NADH +2FADH2 + 4ATP.

Answer : D

Question. The number of molecules of pyruvic acid formed from one molecule of glucose at the end of glycolysis is
(a) 1
(b) 2
(c) 3
(d) 4.

Answer : B

Question. Succinate is oxidised to fumarate in Krebs’ cycle by
(a) removal of hydrogen
(b) loss of electrons
(c) addition of oxygen
(d) removal of oxygen.

Answer : A

Question. Net gain of ATP in glycolysis is
(a) 6
(b) 2
(c) 4
(d) 8.

Answer : D

Question. Enzyme helping in oxidative decarboxylation of pyruvic acid is
(a) pyruvate kinase
(b) pyruvate dehydrogenase
(c) malate dehydrogenase
(d) succinate dehydrogenase.

Answer : B

Ques. How many ATP molecules are produced by aerobic oxidation of one molecule of glucose?
(a) 2
(b) 4
(c) 38
(d) 34 

Answer: C

Ques. Net gain of ATP molecules, during aerobic respiration, is
(a) 40 molecules
(b) 48 molecules
(c) 36 molecules
(d) 38 molecules. 

Answer: C

Ques. When one molecule of ATP is disintegrated, what amount of energy is liberated?
(a) 8 kcal
(b) 38 kcal
(c) 7 kcal
(d) 4.5 kcal 

Answer: C

Ques. The ultimate respiratory substrate, yielding maximum number of ATP molecules, is
(a) glycogen
(b) ketogenic amino acid
(c) glucose
(d) amylose. 

Answer: C

Ques. Out of 38 ATP molecules produced per glucose, 32 ATP molecules are formed from NADH/FADH2 in
(a) respiratory chain
(b) Krebs’ cycle
(c) oxidative decarboxylation
(d) EMP. 

Answer: A

Ques. When one glucose molecule is completely oxidised, it changes
(a) 36 ADP molecules into 36 ATP molecules
(b) 38 ADP molecules into 38 ATP molecules
(c) 30 ADP molecules into 30 ATP molecules
(d) 32 ADP molecules into 32 ATP molecules.

Answer: B

Ques. Out of 36 ATP molecules produced per glucose molecule during respiration
(a) 2 are produced outside glycolysis and 34 during respiratory chain
(b) 2 are produced outside mitochondria and 34 inside mitochondria
(c) 2 during glycolysis and 34 during Krebs’ cycle
(d) all are formed inside mitochondria. 

Answer: B

Ques. EMP can produce a total of
(a) 6 ATP
(b) 8 ATP
(c) 24 ATP
(d) 38 ATP. 

Answer: B

Ques. Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins?
(a) Glucose-6-phosphate
(b) Fructose 1, 6-bisphosphate
(c) Pyruvic acid
(d) Acetyl CoA

Answer: D

Ques. Which of the metabolites is common to respirationmediated breakdown of fats, carbohydrates and proteins?
(a) Pyruvic acid
(b) Acetyl CoA
(c) Glucose - 6 - phosphate
(d) Fructose 1, 6 - bisphosphate 

Answer: B

Ques. Aerobic respiratory pathway is appropriately termed
(a) parabolic
(b) amphibolic
(c) anabolic
(d) catabolic. 

Answer: B

Ques. Plants, but not animals, can convert fatty acids to sugars by a series of reactions called
(a) photosynthesis
(b) Kreb’s cycle
(c) glycolysis
(d) glyoxylate cycle. 

Answer: D

Ques. Link between glycolysis, Krebs’ cycle and β-oxidation of fatty acid or carbohydrate and fat metabolism is
(a) oxaloacetic acid
(b) succinic acid
(c) citric acid
(d) acetyl CoA. 

Answer: D

Ques. NADP+ is reduced to NADPH in
(a) HMP
(b) Calvin Cycle
(c) glycolysis
(d) EMP. 

Answer: A

Ques. Respiratory Quotient (RQ) value of tripalmitin is
(a) 0.09
(b) 0.9
(c) 0.7
(d) 0.07. 

Answer: C

Ques. Apparatus to measure rate of respiration and R.Q. is
(a) auxanometer
(b) potometer
(c) respirometer
(d) manometer. 

Answer: C

Ques. R.Q. is ratio of
(a) CO2 produced to substrate consumed
(b) CO2 produced to O2 consumed
(c) oxygen consumed to water produced
(d) oxygen consumed to CO2 produced. 

Answer: B

Ques. R.Q. is
(a) C/N
(b) N/C
(c) CO2/O2
(d) O2/CO2.

Answer: C

Important Questions for NCERT Class 11 Biology Respiration in Plants

Short Answer Type Questions

Question. Write the overall equation of respiration.

Answer. C6H12O6 + 6O2 → 6CO2 + 6H2O + Energy

Question. Describe the three phases of respiration.

Answer. The three phases of respiration are : (i) External respiration : It is a physical process that involves exchange of oxygen of environment (air or water) and carbondioxide of the blood at the respiratory organ. (ii) Internal (tissue) respiration : It is also a physical process that involves exchange or oxygen of blood and carbon-dioxide of the cell at the cellular level. Both the exchange of gases occur on the principle of diffusion. (iii) Cell respiration : It is an enzymatically- controlled stepped chemical process in which glucose is oxidised inside the mitochondria to produce energy-rich ATP molecules with high-energy bonds. So respiration is a biochemical process.

Question. What is meant by the statement ‘aerobic respiration is more efficient’?

Answer. The aerobic respiration is a high energy yielding process. During the process of aerobic respiration as many as 36 molecules of ATP are produced for every molecule of glucose that is utilized. This shows that aerobic respiration produces much more energy than anaerobic respiration, which produces only 2 ATP molecules.

Question. Give an account of various intermediates in Krebs’ cycle.

Answer. The various intermediates of Krebs’ cycle are utilised to synthesise other compounds through anabolic reactions / pathway such as : (i) Acetyl CoA provides 2-carbon compounds for the synthesis of fatty acids, cutin, aromatic compounds and isoprenoids for forming phytol chain of chlorophyll, carotenoids, steroids, terpenes, gibberellins, etc. (ii) a-ketoglutarate of Krebs’ cycle produces an important amino acid called glutamate. (iii) Succinyl CoA takes part in synthesis of pyrrole compounds of chlorophyll, cytochrome and phytochrome. (iv) Oxaloacetate produces another important amino acid called aspartate. It also forms pyrimidines and alkaloids. (v) It forms GTP which is an important component of signaltransduction system.

Question. What are respiratory substrates? Name the most common respiratory substrate.

Answer. The organic substances, which are catabolised in the living cells to release energy are called as respiratory substrates. Though any food stuff-carbohydrate, fat or protein may act as a respiratory substrate, the most common respiratory substrate is glucose.

Question. Write two energy yielding reactions of glycolysis.

Answer. The conversion of BPGA to 3-phosphoglyceric acid (PGA), is an energy yielding process; this energy is trapped by the formation of an ATP. Another ATP is synthesised during the conversion of PEP to pyruvic acid.

Question. Give, three aerobic conditions where RQ is more than unity.
Answer. Three aerobic conditions where RQ is more than unity is as follows: (i) 2(COOH)2 + O2 → 4CO2 + 2H2O RQ = 4 CO2/1 O2 = 4.0 Oxalic acid (ii) C4H6O5 + 3 O2 → 4CO2 + 3H2O RQ = 4CO2/3O2 or 1.3 Malic acid (iii) 2C4H6O4 + 7O2 → 8CO2 + 6H2O RQ = 8CO2/7O2 or 1.14 Succinic acid.

Long Answer Type Questions 

Question. Where does glycolysis occur in cell? Describe the sequence of reactions in it. Mention the end product.

Answer. Glycolysis : Glycolysis means “the splitting of sugar”.    
A glucose molecule is converted into 2 molecules of pyruvic acid during glycolysis. It occurs in cytosol or cytoplasm of the cell.
Sequence of reactions in glycolysis :
The series of reactions occurring in glycolysis is as follows:

Question. Elucidate the formation or production of 36 or 38 molecules per glucose molecule oxidised in an aerobic cell. Also mention why there is difference in net gain of ATP even in aerobic respiration.

Answer. The total ATP production from the complete oxidation of a glucose molecule to CO2 and H2O under aerobic conditions are.
(i) Glycolysis provides 2 ATP molecules and 2NADH + 2H+.
(ii) Pyruvate oxidation yields 2 NADH + 2H+ only.
(iii) Krebs’ cycle gives 2 GTP molecules,
6 NADH + 6H+ and 2FADH2. Generally no distinction is made between ATP and GTP because GTP is changed into ATP in the cytoplasm by an enzyme nucleoside diphosphate
kinase. Therefore, GTP is regarded ATP in the concerned calculations.
(iv) ETS produces 32 or 34 ATP molecules, and is the major source of energy for a cell. Its yield is as under :
– 2 NADH molecules from glycolysis give 4 ATP molecules if their electrons are introduced into route 2 of ETC by the less efficient shuttle, or 6 ATP molecules if their electrons are passed by the more efficient shuttle into route 1 to ETC.
– 2 NADH molecules from pyruvate oxidation yield 6 ATP molecules in route 1 of ETC.
– 6 NADH molecules from Krebs’ cycle yield 18 ATP molecules in route 1 of ETC.
– 2 FADH2 molecules from Krebs’ cycle yield 4 ATP molecules in route 2 of ETC.
32 or 34 ATP from electron transfers, when added to 4 ATP from glycolysis and Krebs’ cycle, give a grand total of 36 or 38 ATP for each glucose molecule fully oxidised to CO2 and H2O.
The shuttle system seems to vary with the species. Thus, a glucose molecule on complete oxidation produces 36 ATP in most eukaryotic cells, but also forms 38 ATP in some
species. In aerobic prokaryotes, heart, liver and kidneys, 38 ATP molecules are produced per glucose molecule oxidised.
Glucose + 36/38 Pi + 6O2 → 6CO2 + 42/44 H2O
+ 36/38 ATP + Heat
The 42/44 H2O molecules shown in the products above include 6H2O molecules produced in the last step of ETSby combination of hydrogen and oxygen, and 36/38 H2O molecules released as by products of the synthesis of ATP.
36/38 ADP + 36/38 Pi 36/38 ATP + 36/38 H2O In the prokaryotic cells, aerobic cell respiration of a glucose molecule always provides 38 ATP molecules because NADH molecules formed during glycolysis are not to enter the
mitochondria.

 

Please click the link below to download pdf file for CBSE Class 11 Biology Respiration in Plants Notes.

Chapter 10 Cell Cycle and Cell Division
CBSE Class 11 Biology Cell Cycle And Cell Division Notes
Chapter 13 Photosynthesis in Higher Plants
CBSE Class 11 Biology Photosynthesis In Higher Plants Notes
Chapter 14 Respiration in Plants
CBSE Class 11 Biology Respiration In Plants Notes
Chapter 15 Plant Growth and Development
CBSE Class 11 Biology Plant Growth And Development Notes
Chapter 16 Digestion and Absorption
CBSE Class 11 Biology Digestion and Absorption Notes
Chapter 17 Breathing and Exchange of Gases
CBSE Class 11 Biology Breathing And Exchange Of Gases Notes
Chapter 18 Body Fluids and Circulation
CBSE Class 11 Biology Body Fluids And Circulation Notes
Chapter 19 Excretory Products and their Elimination
CBSE Class 11 Biology Excretory Products And Their Elimination Notes
Chapter 2 Biological Classification
CBSE Class 11 Biology Biological Classification Notes
Chapter 20 Locomotion and Movement
CBSE Class 11 Biology Locomotion And Movement Notes
Chapter 21 Neural Control and Coordination
CBSE Class 11 Biology Neural Control And Coordination Notes
Chapter 5 Morphology of Flowering Plants
CBSE Class 11 Biology Morphology Of Flowering Plants Notes
Chapter 6 Anatomy of Flowering Plants
CBSE Class 11 Biology Anatomy Of Flowering Plants Notes
Chapter 7 Structural Organisation in Animals
CBSE Class 11 Biology Structural Organisation In Animals Notes

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