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Which of the following statements are FALSE? Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation... SHOW MORE V When S is converted to P by an enzyme the reaction equilibrium is shifted to the right. Enzymes are distinguished from inorganic catalysts because enzymes display specificity toward their reactants i.e substrates and plain catalysts do not. Vmax for an enzyme-catalyzed reaction is twice the rate observed when the concentration of substrate is equal to the Km: The Lineweaver-Burk plot is used to solve graphically the ratio of products to reactants for any starting substrate d concentration. An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1000 times greater than the Km for that substrate. After 9 minutes 1 % of the substrate had been converted to product and the amount of product formed e in the reaction mixture was 12 micromoles. In a separate experiment one-third as much enzyme and twice as much substrate was used. The amount of time required to make 12 micromoles of product is 27 minutes. f In competitive inhibition an inhibitor binds only to the ES complex. For many enzymes the slowest (rate-limiting) step is the reaction that actually releases the product. Under these conditions k2 can be ignored and Km becomes equivalent to the dissociation constant for the ES complex. The difference in free energy content (AG) between substrate (or reactant) and product for a reaction reflects the relative amounts of each compound present at equilibrium. i The rate of conversion from substrate to product depends on the free-energy (G) difference between them.

Which of the following statements are FALSE? Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation... SHOW MORE V When S is converted to P by an enzyme the reaction equilibrium is shifted to the right. Enzymes are distinguished from inorganic catalysts because enzymes display specificity toward their reactants i.e substrates and plain catalysts do not. Vmax for an enzyme-catalyzed reaction is twice the rate observed when the concentration of substrate is equal to the Km: The Lineweaver-Burk plot is used to solve graphically the ratio of products to reactants for any starting substrate d concentration. An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1000 times greater than the Km for that substrate. After 9 minutes 1 % of the substrate had been converted to product and the amount of product formed e in the reaction mixture was 12 micromoles. In a separate experiment one-third as much enzyme and twice as much substrate was used. The amount of time required to make 12 micromoles of product is 27 minutes. f In competitive inhibition an inhibitor binds only to the ES complex. For many enzymes the slowest (rate-limiting) step is the reaction that actually releases the product. Under these conditions k2 can be ignored and Km becomes equivalent to the dissociation constant for the ES complex. The difference in free energy content (AG) between substrate (or reactant) and product for a reaction reflects the relative amounts of each compound present at equilibrium. i The rate of conversion from substrate to product depends on the free-energy (G) difference between them.

Biochemistry
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
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Which of the following statements are FALSE? Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation... SHOW MORE V When S is converted to P by an enzyme the reaction equilibrium is shifted to the right. Enzymes are distinguished from inorganic catalysts because enzymes display specificity toward their reactants i.e substrates and plain catalysts do not. Vmax for an enzyme-catalyzed reaction is twice the rate observed when the concentration of substrate is equal to the Km: The Lineweaver-Burk plot is used to solve graphically the ratio of products to reactants for any starting substrate d concentration. An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1000 times greater than the Km for that substrate. After 9 minutes 1 % of the substrate had been converted to product and the amount of product formed e in the reaction mixture was 12 micromoles. In a separate experiment one-third as much enzyme and twice as much substrate was used. The amount of time required to make 12 micromoles of product is 27 minutes. f In competitive inhibition an inhibitor binds only to the ES complex. For many enzymes the slowest (rate-limiting) step is the reaction that actually releases the product. Under these conditions k2 can be ignored and Km becomes equivalent to the dissociation constant for the ES complex. The difference in free energy content (AG) between substrate (or reactant) and product for a reaction reflects the relative amounts of each compound present at equilibrium. i The rate of conversion from substrate to product depends on the free-energy (G) difference between them.
Transcribed Image Text:Which of the following statements are FALSE? Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation... SHOW MORE V When S is converted to P by an enzyme the reaction equilibrium is shifted to the right. Enzymes are distinguished from inorganic catalysts because enzymes display specificity toward their reactants i.e substrates and plain catalysts do not. Vmax for an enzyme-catalyzed reaction is twice the rate observed when the concentration of substrate is equal to the Km: The Lineweaver-Burk plot is used to solve graphically the ratio of products to reactants for any starting substrate d concentration. An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1000 times greater than the Km for that substrate. After 9 minutes 1 % of the substrate had been converted to product and the amount of product formed e in the reaction mixture was 12 micromoles. In a separate experiment one-third as much enzyme and twice as much substrate was used. The amount of time required to make 12 micromoles of product is 27 minutes. f In competitive inhibition an inhibitor binds only to the ES complex. For many enzymes the slowest (rate-limiting) step is the reaction that actually releases the product. Under these conditions k2 can be ignored and Km becomes equivalent to the dissociation constant for the ES complex. The difference in free energy content (AG) between substrate (or reactant) and product for a reaction reflects the relative amounts of each compound present at equilibrium. i The rate of conversion from substrate to product depends on the free-energy (G) difference between them.
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