4) The active site of an enzyme a) remains rigid and does not change shape. b) is found at the center of globular enzymes. c) is complementary to the rest of the molecule. d) contains amino acids without sidechains. e) None of the above choices are correct. 1) What Is the name glven to the unligue combination of amino beld residues lit in entryme molecule that ensures complementary Interaction with the robstrate and direct particlpation in the act of catalysis: a) active site. b) allosteric site: c) molecular site d) cofactor: e) coenzyme 2) What is the name of the polypeptide part of the enzyme: a) apoenzyme; b) Isoenzyme, c) coenzyme, d) holoentyme; e) prosthetic group 3) What is the name of the section of the enryme molecule to which effectors bind causing a decres or Increase in entymatic activity: a) active site, b) allosteric site c) catalytic site; d) cofactor; e) anchor area? 5) The Michaelis-Menten equation. The Km for the Enzyme "A" for the substrate "B" was determined be 300mu M. When the substrate "B"concentration was set to 160 uM, the initial rate of the reaction w found to be 65.0mu mol/(mLast s) What is Vmax for Enzyme "A" under these conditions?

Question

4) The active site of an enzyme a) remains rigid and does not change shape. b) is found at the center of globular enzymes. c) is complementary to the rest of the molecule. d) contains amino acids without sidechains. e) None of the above choices are correct. 1) What Is the name glven to the unligue combination of amino beld residues lit in entryme molecule that ensures complementary Interaction with the robstrate and direct particlpation in the act of catalysis: a) active site. b) allosteric site: c) molecular site d) cofactor: e) coenzyme 2) What is the name of the polypeptide part of the enzyme: a) apoenzyme; b) Isoenzyme, c) coenzyme, d) holoentyme; e) prosthetic group 3) What is the name of the section of the enryme molecule to which effectors bind causing a decres or Increase in entymatic activity: a) active site, b) allosteric site c) catalytic site; d) cofactor; e) anchor area? 5) The Michaelis-Menten equation. The Km for the Enzyme "A" for the substrate "B" was determined be 300mu M. When the substrate "B"concentration was set to 160 uM, the initial rate of the reaction w found to be 65.0mu mol/(mLast s) What is Vmax for Enzyme "A" under these conditions?

Answer 1

4) The correct answer is e) None of the above choices are correct.The active site of an enzyme is the region where the substrate binds and where the catalytic reaction takes place. It is not rigid and can change shape depending on the substrate and other factors. The active site is not found at the center of globular enzymes, nor is it complementary to the rest of the molecule. It does not contain amino acids without sidechains.1) The correct answer is a) active site.The unique combination of amino acid residues in an enzyme molecule that ensures complementary interaction with the substrate and directs participation in the act of catalysis is called the active site.2) The correct answer is a) apoenzyme.The polypeptide part of the enzyme is called the apoenzyme. It is the protein component of the enzyme, and it requires a cofactor or coenzyme to become an active enzyme (holoenzyme).3) The correct answer is b) allosteric site.The section of the enzyme molecule to which effectors bind, causing a decrease or increase in enzymatic activity, is called the allosteric site. Effectors are molecules that bind to the allosteric site and induce a conformational change in the enzyme, affecting its activity.5) To find the Vmax for Enzyme "A" under these conditions, we can use the Michaelis-Menten equation:V = (Vmax \* [S]) / (Km + [S])Where:V is the initial rate of the reactionVmax is the maximum velocity of the reaction[S] is the substrate concentrationGiven:V = 65.0 μmol/(mL·s)[S] = 160 μMKm = 300 μMWe can rearrange the equation to solve for Vmax:Vmax = (V \* (Km + [S])) / [S]Substituting the given values:Vmax = (65.0 μmol/(mL·s) \* (300 μM + 160 μM)) / 160 μMVmax = (65.0 μmol/(mL·s) \* 460 μM) / 160 μMVmax = 149.375 μmol/(mL·s)Therefore, the Vmax for Enzyme "A" under these conditions is 149.375 μmol/(mL·s).

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