Glycolysis
A series of chemical reactions in the cytosol of a cell in which the six-carbon glucose molecule is split into 2 three-carbon pyruvate molecules. 2 ATPs are required at the beginning of glycolysis during during the energy-investment phase. 4 ATPs are produced later in glycolysis. leading to a net gain of 2 ATPs. In the process of glycolysis, 2 molecules of NAD+ are reduced to 2 molecules of NADH. All Living organisms can perform glycolysis
Oxidation of Pyruvate
The process in the mitochondria by which the three-carbon molecule pyruvate is oxidized to a two-carbon acetyl group and a molecule of carbon dioxide, and a molecule of NAD+ is reduced by NADH by the enzyme pyruvate dehydrogenase
Citric Acid Cycle
Series of chemical reactions in the matrix of the mitochondria in which 3 molecules of NADH, one molecule of FADH2, one molecule of ATP, and 2 molcules of CO2 are produced. Also known as the Krebs cycle
Oxidative Phosphorylation
The process during cellular respiration in which the electron transport chain creates a proton gradient is used by ATP synthase to produce ATP. Oxygen serves as the final or terminal electron acceptor in the electron transport chain. Oxidative phosphorylation occurs on the inner membrane of the mitochondria and yields more ATP than substrate-level phosphorylation
Substrate-Level Phosphorylation
The enzyme-catalyzed formation of ATP by the transfer of a phosphate group to ADP
Aerobic
Any process that requires oxygen
Anaerobic
Any process that does not require oxygen
Fermentation
The anaerobic process that oxidizes NADH to NAD+
NAD+
A coenzyme that functions as an electron carrier in cellular respiration. NAD+ is reduced to NADH during glycolysis, the oxidation of pyruvate, and the Krebs cycle
FAD
A coenzyme that functions as an electron carrier in cellular respiration. FAD is reduced to FADH2, during the Krebs cycle
Coenzyme A
A coenzyme that helps oxidize pyruvate during cellular respiration and delivers a two-carbon acetyl group to the Krebs cycle
Electron Transport Chain
A series of protein complexes that transfers electrons using oxidation and reduction reactions. The free energy released during transfer of electrons is used to pump protons across a membrane. Electron transport chains are found on the thylakoid membrane of the chloroplast and on the inner membrane of the mitochondria
Alcohol Fermentation
A fermentation process that oxidizes NADH to NAD+ and splits pyruvate into ethanol and carbon dioxide
Lactic Acid Fermentation
A fermentation process that oxidizes NADH to NAD+ and converts pyruvate into lactic acid
Students investigated the amount of substrate consumed in an enzyme-catalyzed reaction during a 10-minute period under different temperature conditions. Their results are shown in the dad table below:
Temperature in Degrees Celsius → Number of Moles of Substrate Consumed in 10 Minutes
0 → 0
10 → 2.1
20 → 8.2
30 → 15.3
40 → 11.1
50 → 0
Which of the following is the best explanation for the number of moles of substrate consumed at 50 degrees Celsius?
(A) An inhibitor molecule was present that slowed the rate of the reaction
(B) At 50 degrees, the enzyme became denatured and no longer functioned normally
(C) The optimum temperature of the enzyme-catalyzed reaction was 50 degrees Celsius
(D) At 50 degrees Celsius, the number of collisions between the enzyme and substrate were greatly reduced
(B) High temperatures denature most enzymes, changing an enzyme’s shape so that it is no longer effective. Choice (A) is incorrect because an inhibitor would slow the rate of reaction at all temperatures, not just at 50 degrees Celsius. Choice (C) is incorrect because, if 50 degrees Celsius was the optimum temperature of the enzyme, more substrate would be consumed, not less. At higher temperatures, the number of collisions between an enzyme and its substrate is increased, not reduced, so choice (D) is incorrect
Students investigated the amount of substrate consumed in an enzyme-catalyzed reaction during a 10-minute period under different temperature conditions. Their results are shown in the dad table below:
Temperature in Degrees Celsius → Number of Moles of Substrate Consumed in 10 Minutes
0 → 0
10 → 2.1
20 → 8.2
30 → 15.3
40 → 11.1
50 → 0
Which of the following is the best explanation for the number of moles of substrate consumed at 10 degrees Celsius?
(A) 10 degrees Celsius is the optimum temperature for the enzyme
(B) The enzyme is denatured at 10 degrees Celsius
(C) The number of collisions between the substrate molecules and the enzyme at 10 degrees Celsius is greatly reduced
(D) An inhibitor molecule was present that slowed the rate of reaction
(C) At lower temperatures, the kinetic energy of all molecules involved in the reaction is reduced and fewer collisions will occur between the enzyme and the substrate. Fewer collisions will lead to fewer reactions. (A) is incorrect because, of 10 degrees Celsius was the optimum temperature for the enzyme, more substrate would be consumed than at another temperature, not less. (B) is incorrect because, if the enzyme was denatured, no reaction would occur, and no molecules of substrate would be consumed. There is no indication that an inhibitor is present, and an inhibitor would slow the reaction at all temperatures, so choice (D) is incorrect
Students investigated the amount of substrate consumed in an enzyme-catalyzed reaction during a 10-minute period under different temperature conditions. Their results are shown in the dad table below:
Temperature in Degrees Celsius → Number of Moles of Substrate Consumed in 10 Minutes
0 → 0
10 → 2.1
20 → 8.2
30 → 15.3
40 → 11.1
50 → 0
Based on the experimental data, what is the optimum temperature for the enzyme-catalyzed reaction?
(A) 10
(B) 20
(C) 30
(D) 40
(C) The greatest number of moles of substrate are consumed at 30 degrees Celsius , so that is the most likely optimum temperature of the enzyme. Choices (A), (B), and (D) are incorrect because fewer moles of substrate are consumed at all three of those temperatures than at 30 degrees Celsius
The inhibition of an enzyme-catalyzed reaction by a competitive inhibitor will be likely overcome by
(A) the addition of a noncompetitive inhibitor
(B) the addition of more substrate
(C) the addition of an acid
(D) the addition of heat
(B) Competitive inhibitors bind at the enzyme’s active site, so adding more substrate would likely allow the substrate to outcompete the competitive inhibitor for binding to the active site. Choice (A) is incorrect because a noncompetitive inhibitor binds to the enzyme’s allosteric site and would less likely mitigate the effects of a competitive inhibitor. The addition of an acid or of heat would likely denature the enzyme, lowering its efficiency, so choices (C) and (D) are incorrect