SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Biology Respiration in Plants

12 questions · 12 still being checked

Exercises 12.1–12.12

  1. Exercise 12.1

    Differentiate between
    (a)
    Respiration and Combustion
    (b)
    Glycolysis and Krebs’ cycle
    (c)
    Aerobic respiration and Fermentation

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    (a) Respiration and Combustion
    Respiration oxidises the substrate in a series of slow, step-wise reactions controlled by enzymes; combustion oxidises it in one uncontrolled step.
    In respiration not all the liberated energy goes out as heat — it is trapped as chemical energy in ATP; in combustion most of the energy is given out as heat.
    Respiration is the breaking of the C–C bonds of complex compounds by oxidation within the cell.
    Both give the same end products from glucose, \(\displaystyle CO_2\) and \(\displaystyle H_2O\): \(\displaystyle C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}\). The difference lies in how the energy is released, not in what is formed.
    (b) Glycolysis and Krebs' cycle
    Glycolysis occurs in the cytoplasm; the Krebs' cycle occurs in the matrix of the mitochondria.
    Glycolysis is a linear chain of ten reactions from glucose to pyruvic acid; Krebs' cycle is a cyclic pathway that starts and ends with oxaloacetic acid (OAA).
    Glycolysis is only a partial oxidation of glucose and releases no \(\displaystyle CO_2\); in Krebs' cycle acetyl CoA is completely oxidised and \(\displaystyle CO_2\) is given off at two points per turn.
    Glycolysis needs no \(\displaystyle O_2\) and is present in all living organisms — in anaerobic organisms it is the only process in respiration; the Krebs' cycle runs only under aerobic conditions.
    Per glucose, glycolysis gives a net $\displaystyle 2$ ATP and $\displaystyle 2$ NADH + \(\displaystyle H^+\); each turn of the Krebs' cycle reduces \(\displaystyle NAD^+\) at three points and \(\displaystyle FAD^+\) at one point and makes one GTP (converted to ATP) by substrate-level phosphorylation.
    (c) Aerobic respiration and Fermentation
    Aerobic respiration degrades glucose completely to \(\displaystyle CO_2\) and \(\displaystyle H_2O\); fermentation accounts for only a partial breakdown of glucose, to ethanol and \(\displaystyle CO_2\) or to lactic acid.
    Aerobic respiration requires \(\displaystyle O_2\); fermentation takes place under anaerobic conditions.
    In eukaryotes aerobic respiration takes place in the mitochondria; fermentation is completed in the cytoplasm.
    Fermentation gives a net gain of only two ATP per glucose — less than seven per cent of the energy in glucose is released — whereas many more ATP (a net $\displaystyle 38$) are generated aerobically.
    NADH is oxidised to \(\displaystyle NAD^+\) rather slowly in fermentation, while the reaction is very vigorous in aerobic respiration.
    The products of fermentation are hazardous: either acid or alcohol accumulates, and yeasts poison themselves when the alcohol reaches about $\displaystyle 13$ per cent.

  2. Exercise 12.2

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

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Respiratory substrates are the compounds that are oxidised during respiration to release energy.
    The most common respiratory substrate is glucose — carbohydrates are usually the ones oxidised, and glucose is the favoured substrate.
    Proteins, fats and even organic acids can also be used as respiratory substrates in some plants under certain conditions.
    All other carbohydrates are usually first converted into glucose before they are respired.

  3. Exercise 12.3

    Give the schematic representation of glycolysis?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Draw glycolysis as a single vertical flow chart, one arrow per step, with the co-substrates written on small side arrows to the right (the book's Figure $\displaystyle 12.1$).
    Head the chart Glucose (6C) at the top.
    Arrow down to Glucose-$\displaystyle 6$-phosphate (6C); side arrow ADP ← ATP (enzyme hexokinase). Label this an ATP-utilising step.
    Arrow down to Fructose-$\displaystyle 6$-phosphate (6C) — an isomerisation, no side arrow.
    Arrow down to Fructose $\displaystyle 1,6$-bisphosphate (6C); side arrow ADP ← ATP. Second ATP-utilising step.
    From this box draw a fork into two boxes side by side: Triose phosphate (dihydroxyacetone phosphate) (3C) and Triose phosphate (glyceraldehyde-$\displaystyle 3$-phosphate, PGAL) (3C), with a double-headed arrow between them to show they are interconvertible.
    Arrow down from PGAL to $\displaystyle 2$ × Triose bisphosphate ($\displaystyle 1,3$-bisphosphoglyceric acid, BPGA) (3C); side arrows \(\displaystyle NAD^+\) → NADH + \(\displaystyle H^+\), and inorganic phosphate entering.
    Arrow down to $\displaystyle 2$ × Triose phosphate ($\displaystyle 3$-phosphoglyceric acid, PGA) (3C); side arrow ADP → ATP. First ATP-synthesising step.
    Arrow down to $\displaystyle 2$ × $\displaystyle 2$-phosphoglycerate.
    Arrow down to $\displaystyle 2$ × phosphoenolpyruvate (PEP); side arrow showing \(\displaystyle H_2O\) leaving.
    Arrow down to $\displaystyle 2$ × Pyruvic acid (3C) at the foot; side arrow ADP → ATP. Second ATP-synthesising step.
    Write below the chart: $\displaystyle 2$ ATP used, $\displaystyle 4$ ATP formed, so the net gain is $\displaystyle 2$ ATP and $\displaystyle 2$ NADH + \(\displaystyle H^+\) per glucose; all of it in the cytoplasm.
    NCERT_Solution_Class11_Biology_Ch12_Q12-3

  4. Exercise 12.4

    What are the main steps in aerobic respiration? Where does it take place?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    The two crucial events of aerobic respiration, and where each happens:
    Complete oxidation of pyruvate by the step-wise removal of all the hydrogen atoms, leaving three molecules of \(\displaystyle CO_2\) — this takes place in the matrix of the mitochondria.
    Passing on of the electrons removed as part of the hydrogen atoms to molecular \(\displaystyle O_2\), with simultaneous synthesis of ATP — this is located on the inner membrane of the mitochondria.
    The main steps, in order:
    Pyruvate, the final product of glycolysis (which happens in the cytoplasm), is transported from the cytoplasm into the mitochondrial matrix.
    Oxidative decarboxylation: pyruvic acid is converted to acetyl CoA by pyruvic dehydrogenase, which needs \(\displaystyle NAD^+\) and Coenzyme A — \(\displaystyle CO_2\) and NADH + \(\displaystyle H^+\) are released. Two NADH are produced from the two pyruvic acid molecules of one glucose.
    Krebs' (tricarboxylic acid) cycle: acetyl CoA is completely oxidised in the matrix, giving \(\displaystyle CO_2\), NADH + \(\displaystyle H^+\), \(\displaystyle FADH_2\) and GTP.
    Electron transport system and oxidative phosphorylation: NADH + \(\displaystyle H^+\) and \(\displaystyle FADH_2\) are oxidised on the inner mitochondrial membrane, the electrons pass to \(\displaystyle O_2\) forming \(\displaystyle H_2O\), and ATP is made by ATP synthase.
    The whole of aerobic respiration therefore takes place inside the mitochondria, and it requires \(\displaystyle O_2\).

  5. Exercise 12.5

    Give the schematic representation of an overall view of Krebs’ cycle.

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Draw the citric acid cycle as a closed loop, running clockwise, with the pyruvate step entering it from outside (the book's Figure $\displaystyle 12.3$). Give the carbon number in brackets after every acid.
    Start outside the loop with Pyruvate (3C). Arrow to Acetyl coenzyme A (2C), with side arrows: CoA in, \(\displaystyle NAD^+\) → NADH + \(\displaystyle H^+\), and one \(\displaystyle CO_2\) out. This is the link reaction, not part of the cycle itself.
    Acetyl CoA feeds into the loop, condensing with Oxaloacetic acid (OAA, 4C) and water to give Citric acid (6C); mark the enzyme citrate synthase and show CoA being released.
    Citric acid → (isomerised to isocitrate) → α-ketoglutaric acid (5C); side arrows \(\displaystyle NAD^+\) → NADH + \(\displaystyle H^+\) and one \(\displaystyle CO_2\) out.
    α-ketoglutaric acid → succinyl-CoA → Succinic acid (4C); side arrows \(\displaystyle NAD^+\) → NADH + \(\displaystyle H^+\), one \(\displaystyle CO_2\) out, and GDP → GTP (substrate level phosphorylation; the GTP then makes one ATP from ADP).
    Succinic acid → Malic acid (4C); side arrow \(\displaystyle FAD^+\) → \(\displaystyle FADH_2\).
    Malic acid → back to Oxaloacetic acid (4C); side arrow \(\displaystyle NAD^+\) → NADH + \(\displaystyle H^+\). The loop closes here.
    Write CITRIC ACID CYCLE in the middle of the loop, and mark the whole figure as taking place in the mitochondrial matrix.
    Write the summary equation under the diagram:
    \(\displaystyle \text{Pyruvic acid} + 4NAD^+ + FAD^+ + 2H_2O + ADP + Pi \rightarrow 3CO_2 + 4NADH + 4H^+ + FADH_2 + ATP\)
    Per turn the cycle itself gives $\displaystyle 3$ NADH + \(\displaystyle H^+\), $\displaystyle 1$ \(\displaystyle FADH_2\), $\displaystyle 1$ ATP (via GTP) and $\displaystyle 2$ \(\displaystyle CO_2\); OAA must be continuously replenished for the cycle to keep running.
    NCERT_Solution_Class11_Biology_Ch12_Q12-5

  6. Exercise 12.6

    Explain ETS.

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    The electron transport system (ETS) is the metabolic pathway through which electrons pass from one carrier to another; it is present in the inner mitochondrial membrane.
    Its job is to release and use the energy stored in NADH + \(\displaystyle H^+\) and \(\displaystyle FADH_2\) by oxidising them and passing the electrons on to \(\displaystyle O_2\), which results in the formation of \(\displaystyle H_2O\).
    The carriers, in order:
    Complex I (NADH dehydrogenase) oxidises the NADH produced in the matrix during the citric acid cycle and transfers the electrons to ubiquinone, located within the inner membrane.
    Complex II feeds ubiquinone with reducing equivalents from the \(\displaystyle FADH_2\) generated during the oxidation of succinate in the citric acid cycle.
    Complex III (cytochrome \(\displaystyle bc_1\) complex) oxidises the reduced ubiquinone (ubiquinol) and transfers the electrons to cytochrome c.
    Cytochrome c is a small protein attached to the outer surface of the inner membrane; it acts as a mobile carrier between complex III and complex IV.
    Complex IV (cytochrome c oxidase complex) contains cytochromes a and \(\displaystyle a_3\) and two copper centres.
    Complex V (ATP synthase) is coupled to this flow: as electrons pass from complex I to IV, ATP is produced from ADP and inorganic phosphate.
    The yield depends on the electron donor: oxidation of one molecule of NADH gives $\displaystyle 3$ ATP, and one molecule of \(\displaystyle FADH_2\) gives $\displaystyle 2$ ATP.
    \(\displaystyle O_2\) acts as the final hydrogen acceptor. Its role is limited to the terminal stage, but it is vital because it drives the whole process by removing hydrogen from the system.

  7. Exercise 12.7

    Distinguish between the following:
    (a)
    Aerobic respiration and Anaerobic respiration
    (b)
    Glycolysis and Fermentation
    (c)
    Glycolysis and Citric acid Cycle

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    (a) Aerobic respiration and Anaerobic respiration
    Aerobic respiration takes place in the presence of \(\displaystyle O_2\); anaerobic respiration (fermentation) takes place under anaerobic conditions.
    Aerobic respiration is a complete oxidation of the substrate to \(\displaystyle CO_2\) and \(\displaystyle H_2O\); anaerobic respiration is an incomplete oxidation, ending in ethanol and \(\displaystyle CO_2\) or in lactic acid.
    In eukaryotes the aerobic steps occur inside the mitochondria (glycolysis first in the cytoplasm); anaerobic respiration is completed entirely in the cytoplasm.
    Aerobic respiration gives a net gain of $\displaystyle 38$ ATP per glucose; anaerobic respiration gives a net gain of only $\displaystyle 2$ ATP, less than seven per cent of the energy in glucose being released.
    In aerobic respiration \(\displaystyle O_2\) is the final hydrogen acceptor and NADH is reoxidised vigorously; in fermentation pyruvic acid or its derivative is the hydrogen acceptor and NADH is oxidised to \(\displaystyle NAD^+\) rather slowly.
    Aerobic respiration is common in higher organisms; fermentation occurs in many prokaryotes, unicellular eukaryotes, germinating seeds, and in muscle when oxygen is inadequate.
    (b) Glycolysis and Fermentation
    Glycolysis is the breakdown of glucose to pyruvic acid; fermentation begins where glycolysis ends and converts that pyruvic acid to \(\displaystyle CO_2\) and ethanol, or to lactic acid.
    Glycolysis is a chain of ten enzyme-controlled reactions; fermentation takes only one or two — pyruvic acid decarboxylase and alcohol dehydrogenase for alcohol, lactate dehydrogenase for lactic acid.
    Glycolysis occurs in all living organisms and under both aerobic and anaerobic conditions; fermentation takes place only under anaerobic conditions.
    Glycolysis produces NADH + \(\displaystyle H^+\); fermentation consumes it — NADH + \(\displaystyle H^+\) is the reducing agent and is reoxidised to \(\displaystyle NAD^+\), which lets glycolysis continue.
    Glycolysis yields a net $\displaystyle 2$ ATP per glucose; the fermentation steps themselves yield no ATP.
    Both occur in the cytoplasm.
    (c) Glycolysis and Citric acid cycle
    Glycolysis happens in the cytoplasm; the citric acid cycle happens in the matrix of the mitochondria.
    Glycolysis is a linear pathway of ten steps; the citric acid cycle is a closed cycle beginning and ending with oxaloacetic acid.
    The substrate of glycolysis is glucose (6C), which is only partially oxidised to two molecules of pyruvic acid (3C); the substrate of the citric acid cycle is acetyl CoA (2C), which is completely oxidised.
    No \(\displaystyle CO_2\) is released in glycolysis; two molecules of \(\displaystyle CO_2\) are released per turn of the citric acid cycle.
    Glycolysis does not need \(\displaystyle O_2\) and is present in all living organisms; the citric acid cycle operates only under aerobic conditions.
    Glycolysis yields a net $\displaystyle 2$ ATP and $\displaystyle 2$ NADH + \(\displaystyle H^+\) per glucose; one turn of the citric acid cycle yields $\displaystyle 3$ NADH + \(\displaystyle H^+\), $\displaystyle 1$ \(\displaystyle FADH_2\) and $\displaystyle 1$ GTP (converted to ATP).

  8. Exercise 12.8

    What are the assumptions made during the calculation of net gain of ATP?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Four assumptions are made, and they make the balance sheet a theoretical exercise only:
    There is a sequential, orderly pathway functioning, with one substrate forming the next, and with glycolysis, the TCA cycle and the 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 the pathway at any of the intermediary stages.
    These assumptions are not really valid in a living system: all pathways work simultaneously rather than one after another, substrates enter and are withdrawn as and when necessary, ATP is used as and when needed, and enzymatic rates are controlled by multiple means.
    On these assumptions there can be a net gain of $\displaystyle 38$ ATP for one molecule of glucose respired aerobically.

  9. Exercise 12.9

    Discuss “The respiratory pathway is an amphibolic pathway.”

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    The respiratory pathway is called amphibolic because it is involved in both catabolism (breaking down) and anabolism (synthesis) — so it cannot be called a purely catabolic pathway.
    As a catabolic pathway — where substrates enter:
    Glucose is the favoured substrate; all carbohydrates are usually first converted into glucose before being respired.
    Fats are first broken into glycerol and fatty acids. Fatty acids are degraded to acetyl CoA and enter there; glycerol enters after being converted to PGAL.
    Proteins are degraded by proteases, and the individual amino acids — after deamination — enter at some stage within the Krebs' cycle, or as pyruvate or acetyl CoA, depending on their structure.
    As an anabolic pathway — where intermediates are withdrawn:
    The very same compounds are withdrawn from the respiratory pathway when the cell has to synthesise those substrates.
    When the organism needs to synthesise fatty acids, acetyl CoA is withdrawn from the respiratory pathway for it.
    In the same way, respiratory intermediates form the link during the synthesis of proteins as well as their breakdown.
    Since respiratory intermediates are both produced by breakdown and drawn off for synthesis, it is better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

  10. Exercise 12.10

    Define RQ. What is its value for fats?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    RQ (respiratory quotient), also called the respiratory ratio, is the ratio of the volume of \(\displaystyle CO_2\) evolved to the volume of \(\displaystyle O_2\) consumed in respiration.
    \(\displaystyle RQ = \dfrac{\text{volume of } CO_2 \text{ evolved}}{\text{volume of } O_2 \text{ consumed}}\)
    For fats the RQ is less than 1. For the fat tripalmitin it works out to about $\displaystyle 0.7$:
    \(\displaystyle 2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O + \text{energy}\), so \(\displaystyle RQ = \dfrac{102}{145} = 0.7\)
    For comparison, RQ is $\displaystyle 1.0$ for carbohydrates completely oxidised, and about $\displaystyle 0.9$ for proteins.
    RQ depends on the type of respiratory substrate used; in living organisms the substrates are usually more than one, as pure fats or pure proteins are never used alone.

  11. Exercise 12.11

    What is oxidative phosphorylation?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Oxidative phosphorylation is the synthesis of ATP using the energy of oxidation-reduction, as electrons pass along the electron transport system to \(\displaystyle O_2\).
    It is named so to contrast it with photophosphorylation: there it is light energy that is used to produce the proton gradient required for phosphorylation, whereas in respiration it is the energy of oxidation-reduction that is used for the same purpose.
    The electrons moving from complex I to complex IV are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate.
    ATP synthase has two major components: the \(\displaystyle F_1\) headpiece, a peripheral membrane protein complex carrying the site for ATP synthesis from ADP and inorganic phosphate, and \(\displaystyle F_0\), an integral membrane protein complex forming the channel through which protons cross the inner membrane.
    The passage of protons through the \(\displaystyle F_0\) channel is coupled to the catalytic site of \(\displaystyle F_1\); for each ATP produced, \(\displaystyle 4H^+\) pass through \(\displaystyle F_0\) from the intermembrane space into the matrix, down the electrochemical proton gradient.
    Yield: oxidation of one NADH gives $\displaystyle 3$ ATP and one \(\displaystyle FADH_2\) gives $\displaystyle 2$ ATP; \(\displaystyle O_2\) is the ultimate acceptor of electrons and gets reduced to water.

  12. Exercise 12.12

    What is the significance of step-wise release of energy in respiration?

    The book prints no answer for this

    NCERT publishes no answers for this textbook, so there is nothing in the book to check this working against. It has also not yet been read through by hand.

    Step-wise release lets the cell trap the energy as ATP instead of losing it as heat. If glucose were burnt in one step, as in combustion, most of the energy would be given out as heat and would be of no use to the cell.
    Oxidising glucose in several small steps makes some steps just large enough for the released energy to be coupled to ATP synthesis — energy released in one big burst could not be coupled at all.
    Each step is controlled by an enzyme, so the rate of energy release can be regulated by the cell according to its needs.
    The energy is stored in ATP, the energy currency of the cell, which is broken down whenever and wherever energy is needed, rather than being released free into the cell all at once.
    The many intermediate steps also yield carbon skeletons that serve as precursors for the biosynthesis of other molecules — this is what makes the pathway amphibolic.