CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • question_answer
    The following is a simplified scheme showing the fate of glucose during aerobic and anaerobic respiration. Identify the end products that are formed at stagtes indicated as A, B, C, and D. Identify the correct option from those given below:

    A)  A = carbon dioxide and water, B = pyruvic acid, C = ethyl alcohol and carbon dioxide, D = lactic acid

    B)  A = pyruvic acid, B = carbon dioxide and water, C = lactic acid, D = ethyl alcohol and carbon dioxide

    C)  A = pyruvic acid, B = carbon dioxide and water, C = ethyl alcohol and carbon dioxide, D = lactic acid

    D)  A = pyruvic acid, B = ethyl alcohol and carbon dioxide, C = lacdc acid, D = carbon dioxide and water

    Correct Answer: B

    Solution :

    In the process of glycolysis one molecule of glucose breakdown into two molecules of pyruvic ocid. Krebs cycle (aerobic) leads to complete oxidation of pyruvic acid into \[C{{O}_{2}}\]and \[{{H}_{2}}O.\]Fermentation is an anaerobic process. Two most common types of fermentation are alcoholic fermentation (yeast) and lactic acid fermentation. Ethyl alcohol is end product of alcoholic fermentation whereas lactic acid, of lacetic acid fermentation.


You need to login to perform this action.
You will be redirected in 3 sec spinner