Bke2 Biochemistry Exercises

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Suggested answers to Group Excercise: Enzymes in glycolysis

  1. The enzyme hexokinase works along the principle of induced fit. Discuss the details of induced fit in this particular enzyme. Why is it important?

    Answer:
    Closing of active site creates non-polar environment which favours transfer of phosphate groups and provides effective discrimination against H2O as a substrate. See Horton p. 253 and 332 (or Stryer p. 499).


  2. How is glycolysis controlled and why is it necessary to control it? What enzyme steps and enzymes are controlled? Why these in particular?

    Answer:
    Glycolysis must be regulated to meet the different needs of the body/organism (e.g. long distance run versus digestion of a large meal). Regulation occurs at different levels, e.g.

    a) by regulation of the transport of metabolites over the cell membrane or

    b) regulation of enzymatic activity.

    In the latter case, the main points of control in glycolysis are the steps catalysed by hexokinase, phosphofructokinase (PFK) and pyruvate kinase. These enzymes all catalyse more or less metabolically irreversible reactions (Horton, section 11.5C, p 310) that are good targets for control. See further discussion in Horton, Chapter 12 (Stryer, chapter 19).


  3. Many intermediates in the pathway are phosphorylated. What is the advantage of having these phosphate groups? What are the enzymes called that put on or take off phosphate groups?

    Answer:
    (a) The phosphate groups are completely ionized at pH 7, which means that phosphorylated intermediates have a net negative charge. Since cell membranes are generally impermeable to charged molecules, phosphorylation of substrates helps to protect against loss by diffusion (out of the cell).

    (b) Phosphate groups often serve as recognition groups required for the proper binding of the intermediates to the active sites of their corresponding enzyme.

    (c) Phosphate groups are essential components in the enzymatic conservation of metabolic energy as they are ultimately transferred to ADP to yield ATP (see also section 11.7 about the metabolic roles of ATP).

    The enzymes are usually called kinases and phosphatases, respectively.


  4. Explain the metabolic role of alcohol dehydrogenase (ADH) in yeasts that are a) fermenting glucose to ethanol and b) aerobically oxidising ethanol.

    Answer:
    a) Yeast is facultative, i.e. it can live either aerobically or anaerobically (at least for some time). In fermentation, pyruvate is first decarboxylated to acetaldehyde by pyruvate decarboxylase. Alcohol dehydrogenase then reduces acetaldehyde to ethanol to regenerate NAD+ from the NADH produced in glycolysis. Under aerobic conditions, NAD+ is regenerated during oxidative phosphorylation, which requires oxygen. Reduction of acetaldehyde to ethanol (or reduction of pyrivate to lactate in mammals) allows glycolysis to proceed even in the absence of oxidative phosphorylation.

    b) Under aerobic conditions, alcohol dehydrogenase instead oxidises ethanol to acetaldehyde (the puruvate formed by glycolysis under aerobic conditions is oxidised via the citric acid cycle and oxidative phosphorylation to CO2 and H2O).


  5. Inspection of glycolysis reveals that it contains two steps where transfer of a phosphoryl group from ATP occurs (ATP is consumed) and two steps where ATP is formed, yet there is a net gain of two ATP in each cycle. Explain.

    Answer:
    Two triose phosphates are formed in the aldolase reaction. Hence the number of ATPs formed in the subsequent steps should be multiplied by 2.


  6. The TIM-barrel is a recurring motif in many proteins. What does it look like? It is named after a glycolytic enzyme. Which enzyme, and what reaction is this enzyme the catalyst of?

    Answer:
    The TIM-barrel motif is an a/b barrel consisting of eight alternating b-stands and eight a-helices forming a barrel-like structure with the b-strands in the middle surrounded by the helices on the outside. Triose phosphate isomerase (TIM) interconverts glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.


  7. Phosphofructokinase introduces a second phosphate group into fructose-6-phosphate. Why a second one?

    Answer:
    If fructose-6-phosphate were to be cleaved by aldolase first, then one of the products would not be phosphorylated and might have to be rapidly phosphorylated for protection against loss by diffusion (see Question 3 above). Phosphorylation before cleavage is more effective.


  8. How does a regulatory enzyme work? Use the enzyme phosphofructokinase as an example to explain (or any other example you know well). Do you know any other name for these proteins?

    Answer:
    Regulatory enzymes like PFK have a regulatory site in addition to the active site (see Horton section 5.10). Binding of an effector to the regulatory site causes a change in the enzyme's structure that affects the active site. These proteins are also referred to as allosteric proteins/enzymes.


  9. Phosphofructokinase is a multisubunit enzyme. Would it work in the same way if it were a monomeric enzyme?

    Answer:
    No. Phosphofructokinase is allosterically regulated. This implies the presence of a regulatory site in addition to the catalytic site. The presence of more than one subunit is the key to the cooperative behaviour of PFK as the allosteric signals are transmitted across the subunits by conformational changes.


  10. In order for ATP to serve as a regulator for phosphofructokinase the concentration of ATP has to reach a certain (rather high) level. Why is this? In mammals, the inhibition of phosphofructokinase by ATP is diminished when the AMP or ADP concentration is high. How can this observation be explained?

    Answer:
    ATP is a substrate of PFK as well as a regulator so there is competition between the active site and the allosteric site for ATP binding. ATP, ADP and AMP are in equilibrium with each other, and they all bind to the same regulatory site. In mammals, the effect on enzymatic activity is opposite for ATP (inhibitory) compared to ADP and AMP (activating).
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Exercise answers by Inger Andersson
Modified 97.09.19 by mowbray@xray.bmc.uu.se
Page updated 2000.08.18 by stefan@xray.bmc.uu.se
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