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Photosynthesis in Higher Plants Test - 71

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Photosynthesis in Higher Plants Test - 71
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  • Question 1
    1 / -0
    Petiole of Australian Acacia helps in?
    Solution
    A petiole joins the leaf to the stem and contains vascular tissue that gives an association from the stem to allow sap to enter the leaf and the results of photosynthesis (starches) to be transported from the leaf to whatever remains of the plant.
    So, the correct option is 'Photosynthesis'.
  • Question 2
    1 / -0
    Read the following statements : 
    a) $$F_0$$ part of ATPase is associated with breakdown of proton gradient.
    b) A H - carrier contributes in creation of proton gradient.
    c) Movement of electrons in ETS is coupled to pumping of protons into the lumen.
    d) Formation of $$NADPH + H^+$$ is related with the creation of proton gradient.
    How many of the above statements are correct ?
  • Question 3
    1 / -0
    In $$C_4$$ plant PEPcase enzyme involve in :- 
    Solution
    • Primary carboxylating enzyme, PEPcase in C4 plants is found in cytoplasm of mesophyll cells.
    • Phosphoenolpyruvate carboxylase (PEPcase) catalyzes the fixation of carbon dioxide with phosphoenolpyruvate to produce oxaloacetate and inorganic phosphate.
    So, the correct option is 'Carboxylation'.
  • Question 4
    1 / -0
    How many ATP - synthesised during complete oxidation of 3 molecule acetyl co A :
  • Question 5
    1 / -0
    Match the following and choose the correct combination from the option given 
    Sr.Column I SrColumn II
    A.Sulphur I.Chlorophyll
    B.ZincII.Nitrogenase 
    C.Magnesium III.Methionine
    D.MolybdenumIV.Auxin
    Solution
    Sulfur is essential for chlorophyll formation. Zinc is essentially important for the activity of nitrogenase. The presence of a magnesium deficiency, a low metabolic rate, or both conditions together, could prevent methionine.
    Auxin plays fundamental roles in plant growth and development AO family requires a molybdenum cofactor.
    So the correct option is 'A-I,B-II,C-III,D-IV'.
  • Question 6
    1 / -0
    In $$C_4$$ plant $$RuBisCO$$ activate during:-
    Solution
    RuBisCO catalyzes the key reaction in which RUBP acts as an acceptor for CO2 forming two molecules of PGA. The competing reaction with O2 leads to the formation of phosphoglycolate, which is recycled through photorespiration. Thus, the activity of Rubisco is majorly depended on light activation and dark inactivation. The light reactions of photosynthesis produce ATP and activation of RuBisCO is ATP-dependent, which furnishes a mechanism regarding the light-dependent activation of the enzyme. However, activation of RuBisco activase and transformation of RuBisCO in its inactive form need transport of electron and that can only occur in presence of light or during the day. 
    So, the correct answer is option A.
  • Question 7
    1 / -0
    In the pathway of dark reaction,which occur in all photosynthetic plants,primary $$CO_2$$ acceptor contains how many carbon(s)?
    Solution
    Carbon dioxide is captured by the chemical ribulose biphosphate (RuBP). RuBP is a 5-C chemical. Six molecules of carbon dioxide enter the Calvin Cycle, eventually producing one molecule of glucose.
    So, the correct option is '5'.
  • Question 8
    1 / -0
    The evolutionary strategies developed by $$C_4$$ plants are to?
  • Question 9
    1 / -0
    During photorespiration oxygen consuming reaction(s) occur in
    Solution
    The original oxygenation reaction by RUBISCO occurs in the chloroplast. In the peroxisome, glycolate from the chloroplast is converted to glyoxylate. This reaction consumes O2 and generates hydrogen peroxide. The O2 is released again, however, as the hydrogen peroxide is converted to water by the enzyme catalase. 
    So, the correct option is Stroma of chloroplasts & peroxisomes.
  • Question 10
    1 / -0
    How many Calvin cycle (with some modification) are required to produce 12 molecules of Fructose?
    Solution
    Calvin cycle fixes one molecule of carbon which can be used to make sugar. It takes three turns of the Calvin cycle to create one molecule of glyceraldehyde-3 phosphate. After six turns of the Calvin cycle, two molecules of glyceraldehyde-3 phosphate can be combined to make a fructose molecule. So, total 12 calvin cycle are required to produce 12 molecules of fructose.
    Thus, the correct answer is '12.'
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