Self Studies

Respiration in Plants Test - 84

Result Self Studies

Respiration in Plants Test - 84
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    In a germinating seed, when protein is aerobically oxidized, the R.Q. value will be
    Solution
    The respiratory quotient (RQ) is the ratio of 
    RQ = CO$$_{2}$$ eliminated / O$$_{2}$$ consumed.
    When carbohydrates are being oxidized in the organism and the requisite oxygen is available, the RQ is 1. In the oxidation of fats, the RQ is 0.7 and in the oxidation of proteins, 0.8. Whenever the substrate is protein, RQ is always less than one (0.5-0.9). So, in germinating seeds, when protein is aerobically oxidized, the R.Q value will be less than one. Thus, option A is correct and other options are incorrect.
  • Question 2
    1 / -0
    In TCA cycle, when succinate is converted to fumarate
    Solution
    Succinate and fumarate are the intermediates of TCA cycle. When succinate is converted to fumarate, 2 protons are removed which are taken by FAD and convert into FADH$$_{2}$$. This reaction is catalysed by succinate dehydrogenase. 

  • Question 3
    1 / -0
    The wrong statement amongst the following is 
    Solution
    Anaerobic oxidation reactions cannot be increased by aeration. Under anaerobic conditions, the rate of glucose metabolism is faster, but the amount of ATP produced is smaller. When exposed to aerobic conditions, the ATP production increases and the rate of glycolysis slows, because the ATP produced acts as an allosteric inhibitor for phospho-fructokinase 1, the third enzyme in the glycolysis pathway. 
  • Question 4
    1 / -0
    Which of these steps in Krebs cycle indicates substrate level phosphorylation? Conversion of 
    Solution
    In Krebs cycle, during the conversion of Succinyl-CoA to succinic acid, GTP is produced from GDP which being unstable immediately changes to ATP. Since ATP synthesis takes place without entering ETC directly by the substrate, it is known as substrate-level phosphorylation. 
    So, the correct answer is option C.
  • Question 5
    1 / -0
    How many ATP molecules are formed in ETS from the reduced nicotinamide adenine dinucleotide generated in two rounds of Krebs cycle?
    Solution
    One Krebs cycle results in the production of 3 $$NADH_2$$. So, at the end of two cycles, 6 $$NADH_2$$ are produced. One molecule of $$NADH_2$$ produces 3 ATP which means 18 ATP are produced from 6 $$NADH_2$$ molecules.
  • Question 6
    1 / -0
    How many times $$CO_2$$ is released in aerobic respiration?
    Solution
    Respiration proceeds in two stages. First stage is glycolysis, during which one molecule of hexose glucose is converted to two molecules of pyruvic acid. No oxygen is consumed nor any carbon dioxide is released during glycolysis. Glycolysis is followed by oxidative decarboxylation of pyruvic acid, in which each molecule of three carbon atom containing pyruvic acid is converted to two carbon atom containing acetyl CoA and a molecule of carbon dioxide is released. Next stage is the citric acid cycle, followed by the last stage the electron transport chain and oxidative phosphorylation. In the course of the citric acid cycle, two molecules of $$CO_2$$ and the equivalent of eight hydrogen atoms (8 protons and 8 electrons) are removed, forming three NADH and one $$FADH_2$$. The $$CO_2$$ produced accounts for the two carbon atoms of the acetyl group, that entered the citric acid cycle. At the end of each cycle, the four carbon oxaloacetate (which condenses with acetyl CoA in the first reaction of citric acid cycle) has been regenerated, and the cycle continues. Thus three times carbon dioxide is released during the course of aerobic respiration- once during oxidative decarboxylation of pyruvate and two during the Krebs or citric acid cycle.

  • Question 7
    1 / -0
    Number of oxygen molecules required for aerobic oxidation of one pyruvate is
    Solution
    A- Aerobic respiration occurs in two steps-1. Glycolysis which does not require oxygen and 2. the krebs cycle that requires oxygen to breakdown the pyruvate molecules. In krebs cycle oxidation of three carbon compound pyruvate into ATP takes place. For every molecule of Pyruvate to be broken down, 6 molecules of oxygen is required. 
    So, the correct answer is '6'
  • Question 8
    1 / -0
    The respiratory quotient (R.Q) of some of the compounds are 4, 1 and 0.7. These compounds are identified respectively as 
    Solution
    The respiratory quotient (RQ) is the ratio of $$CO_2$$ produced to $$O_2$$ consumed, while food is being metabolized.
    RQ = $$CO_2$$ eliminated / $$O_2$$ consumed
    The respiratory quotient for carbohydrate metabolism can be demonstrated by the chemical equation for oxidation of glucose:
    $$C_6H_{12}O_6$$ + 6 $$O_2$$ ---> 6 $$CO_2$$+ 6 $$H_2O$$
    RQ = 6 $$CO_2$$ / 6 $$O_2$$ = 1
    The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen. So, during fat oxidation $$CO_2$$ produced is less than $$O_2$$ used, giving R.Q less than one.
    When organic acids are broken down as respiratory substrates under aerobic conditions the RQ is more than one. Organic acids contain more oxygen than carbohydrates and therefore require less oxygen for their oxidation. R.Q of oxalic acid is 4.
  • Question 9
    1 / -0
    If all the lenticels of stem are blocked, the first to die will be
    Solution
    A lenticel is the opening present on the periderm of the secondarily thickened organs and the bark of woody stems and roots of dicotyledonous flowering plants. It is a porous tissue consisting of cells with large intercellular spaces. It helps in the direct exchange of gasses between the internal tissues and atmosphere through the bark, which is otherwise impermeable to gases. Lenticels are found in most of the woody trees but absent is  woody climbers. If lenticels are blocked then root will die first due to lack of gaseous exchange. 
    Thus, option C is correct.
  • Question 10
    1 / -0
    In which of the following steps of Krebs cycle, $$CO_{2}$$ is evolved?
    Solution
    Isocitrate dehydrogenase catalyzes oxidative decarboxylation of isocitrate to form alpha-ketoglutarate. $$Mn^{2+}$$ in the active site interacts with the carbonyl group of the intermediate oxalosuccinate, which is formed transiently but does not leave the binding site until decarboxylation converts it to alpha-ketoglutarate. $$Mn^{2+}$$ also stabilizes the enol formed transiently by decarboxylation. There are two different forms of isocitrate dehydrogenase in all cells, one requiring NAD as an electron acceptor and the other requiring NADP. 
    So, the correct answer is option B.
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now