Answer to the Tips Questions of the Final Exam

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  • 8/6/2019 Answer to the Tips Questions of the Final Exam

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    Answer to the Tips Questions of the Final Exam

    ~!!!(>w< )/ D

    For Franki, ~!!! XD

    9.(a) The higher the % of oxygen , the higher amount the phosphate is absorbed(1)

    But when the % increased from 2.1 to 21.0 by ten times. The phosphate intake

    increased only by 0.01, represent that other factors such as the no of protein carrier for

    active uptake of minerals are limiting.

    (b) From 0 to 5, the rate decrease lower then more rapid from 2 to 5. (1)

    Beyond 5, the rate of phosphate uptake decrease in a lesser extend(1)

    Due to the electron transport chain is inhibited by DNP(1),

    Fewer ATP is produced(1)

    Since active trasport need ATP(1)

    Its rate decrease(1)

    (c) (i)Decrease(1)

    (ii) The Krebs cycle is inhibited,(1) therefore,

    Since the cycle generate 3 NADH, 1 FADH2, that they go to the electron transport

    chain to produce ATP, (1)

    The above are not produced, (1)

    Fewer ATP produced in the electron transport chain (1)

    Fwer active uptake of phosphate.(1)

    (d) make DNA, ATP, cell wall

    10(a) In RESTING stage (1) of an individual

    (b) (i) In the age of growth, the growth rate is the faster for a new born baby.(1)

    ALSO baby, have a large surface area to volume ratio(1)

    Therefore heat loss is large(1)

    High basal metabolic rate to recover the large amount of heat lost(1)

    (ii) Male is higher than female(1)

    Since male have a higher work load than female(1), they need to work(1)

    Higher basal metabolic rate to support high energy demands work(1)

    (iii) the BMR of BOTH female and male decrese , due to aging (1)

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    And generally decrease of capacity of metabolic rate of an individual(1)

    The BMR decrease with increasing age(1)

    ~!!!!!!!!!!!!!!

    11.(a) the body is flattened, therfore the surface area to volume ratio is incresed(1)

    And the distance for the diffusion of the food molecules to reach the body cell is

    decreased(1)

    BOTH lead to a higher diffusion rate(1)

    Help the parasite to get more oxygen and nutrient from host for growth(1)

    (b) Glucose is water soluble, therfore polar molecules pass through the cell membrane

    by carrier protein and channel protein.(1)

    If the acidity of the intestine is high, the SPECIFIC carrier and channel protein is

    DENATURED, (1)

    and they can no longer bind with glucose (1)and

    cannot transport them into the intestine.(1)

    (c)Fewer glucose absorbed by the host cell. SO more glucose left in the intestine (1)

    The higher the concentration gradient between the parasite and the intestine

    glucose(1)

    The parasite can get more glucose by diffusion.(1)

    (d) The carbon dioxide is used to synthsis the Krebs cycles intermediate,

    And NADH is used to provide the energy.(1)

    SO, the H atom carrier/energy carrier NAD+ is REGENERATED(1)

    Then the regenerated NAD+ move out the mitochondria (1)

    Being the H carrier of the process turing triose phosphate to pyruvate

    CONTIUOUSLY (1)

    Allowing glycolysis to continue(1)

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    12.

    (c) To ensure that BOTH the 2 factors are not limiting(1)

    To obtain the MAXIMUM photosynthetic rate measured(1)

    (d)The duration of light flash is the extend of photochemical reaction(1)

    The measure of duration of dark period is the carbon fixation(1)

    The duration of light flash is CONSTANT in all experiment(1),

    So the amount of ATP and NADPH produced are controlled/constant each time(1)

    ONLY the duration of carbon fixation varie(1)

    When duration of dark period incresed from 3 to 17, the EXTEND of increse of

    photosynthesis is only 1 unit.

    SHOwing the carbon fixation is much more rate limiting (1)

    (e) ALL ATP and NADPH are used up in 17 ms already.(1)

    13.

    b(i) Nuclei, Ribosome

    (ii) Pyruvate is permeable through the mitochondrial membrane, BUT NOT

    glucose(1)

    Pyruvate is converted to acetyl CoA, relesing 1 CO2 (1)

    The 6-C compound turn to 5-C then 4-C generate 2 CO2 TOTALLY (1)

    (iii) Cyanide only inhibit the ELECTRON TRANSPORT CHAIN (1)

    So the NADH produced cant be oxidized / donate their H to the electron transport

    chain (1)

    NADH thus ACCUMMULATE in the cell, (1)

    To allow glycolysis to continue (1)

    The NADH react with the pyruvate , forming lactate/lactic acid(1)

    (c) Ethanol (1) and CO2 (1)

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