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Respiration in Plants Test - 44

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Respiration in Plants Test - 44
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  • Question 1
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    Which of the following is the connecting link between glycolysis and Krebs cycle?
    Solution
    Glycolysis is the 1st step of cellular respiration and is common to both aerobic and anaerobic organisms because it occurs in the absence of oxygen. The product of glycolysis is pyruvate which is oxidised to Acetyl CoA and it is acetyl CoA that enters mitochondria and takes part in Krebs cycle. This process is known as acetylation. Therefore acetyl CoA is a connecting link between glycolysis and Krebs cycle. 
    Hence, the correct answer is option A.
  • Question 2
    1 / -0
    Enzyme encolase catalyses the conversion of $$2$$PGA to phosphoenol pyruvic acid in presence of ____________, which is the cofactor.
    Solution
    Enolase is a metalloenzyme responsible for the catalysis of the conversion of 2-phosphoglycerate (2-PG) to phosphoenolpyruvate (PEP), the ninth and penultimate step of glycolysis. In this reaction magnesium ion acts as a cofactor.
  • Question 3
    1 / -0
    During anaerobic respiration the conversion of pyruvate into acetaldehyde, along with co-enzyme TPP, the cofactor required is
    Solution
    When Pyruvate undergoes decarboxylation by removal of carbon dioxide to form Acetaldehyde co enzyme TPP and cofactor Zn is required. Thus the correct answer is option D.
  • Question 4
    1 / -0
    Anaerobic respiration, that occurs after glycolysis is also called as
    Solution
    Glycolysis is a common step for both aerobic and anaerobic respiration. Next step after glycolysis is fermentation in anaerobic respiration. Fermentation starts with pyruvate (the end product of glycolysis). Depending on the organism, pyruvate can either be fermented into ethanol or lactate. 
    Hence, the correct answer is option A.
  • Question 5
    1 / -0
    The final electron acceptor during ETS in respiration is
    Solution
    The final stage of cellular respiration is the electron transport chain (ETS). The ETS is a series of molecules embedded in the mitochondrial membrane. The first molecule accepts protons and electrons from the products of the Krebs cycle. The electrons are passed from molecule to molecule and finally react with oxygen and protons to form water. Without oxygen the final reaction cannot occur. In the absence of oxygen the ETS stops working and no ATP is generated.
  • Question 6
    1 / -0
    Pyruvate kinase enzyme catalyses
    Solution
    Pyruvate kinase enzyme catalyses third irreversible step of glycolysis. It is the the final step of glycolysis in which phosphoenolpyruvate converts into pyruvate with the help of the enzyme pyruvate kinase. This reaction involves the transfer of a phosphate group. The phosphate group attached to the 2′ carbon of the PEP is transferred to a molecule of ADP, yielding ATP. Hence option C is correct.
  • Question 7
    1 / -0
    In the electron transport chain during terminal oxidation, the cytochrome, which donates electrons to O$${_2}$$ is 
    Solution
    The electron transport system occurs in the inner membrane of the mitochondria. The electron transport system contains mainly six components arranged in the sequence of NAD, FMN, cytochrome b, cytochrome c, cytochrome a and cytochrome a$${_3}$$.
    Electrons from NADH produced during citric acid are oxidized by an NADH dehydrogenase (complex I) and electrons are then transferred to ubiquinone located within the inner member. Ubiquinone also receives reducing equivalents via FADH$${_2}$$ , through the activity of the enzyme, succinate dehydrogenase (complex II). The reduced ubquinone called ubquinol is then oxidised with the transfer of electrons to cytochrome c, (complex III) cytochrome c is a small protein attached to outer surface of the inner membrane and is the mobile carrier for transfer of electrons between complex between III and IV. Complex IV is cytochrome c oxidase complex containing cytochromes a and a$${_3}$$ and 2 copper centers.
    When electrons pass from one carrier to another via complex I to IV, in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP.
    At each step of electron transport, the electron acceptor has a higher electron affinity than the electron donor. The energy from such electron transport is utilised in transporting protons (H+) from the matrix, across the inner membrane to its outer chamber.
    This creates a higher proton concentration outside the inner membrane than in the matrix. This difference in proton concentration on the outer and inner sides of the inner mitochondrial membrane is known as proton gradient.
    At the end of the chain the electrons and the accompanying protons are combined with oxygen to form water. Oxygen is thus, the terminal electron acceptor of the mitochondrial respiratory chain.

  • Question 8
    1 / -0
    Who contributed to the research process on fermentation?
    Solution

    Fermentation is the process pyruvate into lactic acid or alcohol in absence of oxygen. Schwann Theodor is considered as the founder of modern histology, the study of the structure of plant and animal tissues. Schwann discovered pepsin, the digestive enzyme, in the stomach epithelium, or membrane tissues, of animals. He also conducted valuable research on the processes of fermentation, purefaction, and muscular and arterial contraction. Therefore, the correct answer is option D.
  • Question 9
    1 / -0
    Citric acid cycle is the alternate name of which of the following?
    Solution
    Citric acid cycle or Krebs cycle or tricarboxylic acid cycle are alternative names. Krebs cycle is also known as citric acid cycle, because the first stable compound in this cycle is citric acid and since chemically, citric acid is tricarboxylic acid in nature, it is also known as tricarboxylic acid cycle (TCA cycle). 
  • Question 10
    1 / -0
    Which of the following substances yield less than $$4 {Kcal}/{mol}$$, when its phosphate bond is hydrolysed?
    Solution

    The reaction in which the hydrolysis of the phosphate bond in ATP takes place gives rise to free energy range of 7.3 Kcal/mol, when ADP is hydrolysed, the reaction also gives free energy in the range of 7.3 Kcal/mol. 
    Creatine phosphate is also a high energy phosphate compound. The reason for high free energy is due to difference of free energy contents in reactants and products. The process of glycolysis involves the conversion of glucose to glucose-6-phosphate by phosphorylation process. Hydrolysis of glucose 6-phosphate gives an yield of -3.3 Kcal/mol, which is less than 4 Kcal/ mol. Thus, the correct answer is option C.
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