Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
7th Edition
ISBN: 9781305081086
Author: STOKER, H. Stephen
Publisher: Brooks Cole
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Chapter 24, Problem 24.43EP
Interpretation Introduction
Interpretation: To determine the overall reaction equation for the conversion of pyruvate to
Concept introduction: Pyruvate is the end product in the glycolysis. The production of the fate of pyruvate varies with the nature of the organism and the cellular conditions. The three common fates of pyruvate are as follows:
Aerobic reactions need oxygen while anaerobic reactions don’t need oxygen. Pyruvate forms
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Chapter 24 Solutions
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
Ch. 24.1 - The primary site within the human body where...Ch. 24.1 - What is the first site within the human body where...Ch. 24.1 - What effect does gastric juice in the stomach have...Ch. 24.1 - Prob. 4QQCh. 24.1 - Which of the following substances is needed for...Ch. 24.2 - Prob. 1QQCh. 24.2 - Prob. 2QQCh. 24.2 - Prob. 3QQCh. 24.2 - Prob. 4QQCh. 24.2 - Prob. 5QQ
Ch. 24.2 - Prob. 6QQCh. 24.2 - Prob. 7QQCh. 24.3 - Prob. 1QQCh. 24.3 - Prob. 2QQCh. 24.3 - Prob. 3QQCh. 24.3 - Prob. 4QQCh. 24.3 - Accumulation of which of the following substances...Ch. 24.4 - Prob. 1QQCh. 24.4 - The net yield of ATP for the complete oxidation of...Ch. 24.4 - Prob. 3QQCh. 24.5 - Prob. 1QQCh. 24.5 - Prob. 2QQCh. 24.5 - Prob. 3QQCh. 24.6 - Prob. 1QQCh. 24.6 - Prob. 2QQCh. 24.6 - Prob. 3QQCh. 24.6 - Which of the following statements about ATP...Ch. 24.6 - Prob. 5QQCh. 24.7 - Prob. 1QQCh. 24.7 - Prob. 2QQCh. 24.8 - Prob. 1QQCh. 24.8 - Prob. 2QQCh. 24.8 - Prob. 3QQCh. 24.9 - Which of the following hormones promotes the...Ch. 24.9 - Which of the following pairs of hormones increases...Ch. 24.10 - Prob. 1QQCh. 24.10 - Prob. 2QQCh. 24.10 - Prob. 3QQCh. 24 - Where does carbohydrate digestion begin in the...Ch. 24 - Very little digestion of carbohydrates occurs in...Ch. 24 - Prob. 24.3EPCh. 24 - Prob. 24.4EPCh. 24 - Prob. 24.5EPCh. 24 - Prob. 24.6EPCh. 24 - Prob. 24.7EPCh. 24 - Prob. 24.8EPCh. 24 - Prob. 24.9EPCh. 24 - Prob. 24.10EPCh. 24 - Prob. 24.11EPCh. 24 - Prob. 24.12EPCh. 24 - Prob. 24.13EPCh. 24 - Prob. 24.14EPCh. 24 - Prob. 24.15EPCh. 24 - Prob. 24.16EPCh. 24 - Prob. 24.17EPCh. 24 - Prob. 24.18EPCh. 24 - Prob. 24.19EPCh. 24 - Prob. 24.20EPCh. 24 - Prob. 24.21EPCh. 24 - Prob. 24.22EPCh. 24 - Prob. 24.23EPCh. 24 - Prob. 24.24EPCh. 24 - Prob. 24.25EPCh. 24 - Prob. 24.26EPCh. 24 - Prob. 24.27EPCh. 24 - Prob. 24.28EPCh. 24 - Prob. 24.29EPCh. 24 - Prob. 24.30EPCh. 24 - Prob. 24.31EPCh. 24 - Prob. 24.32EPCh. 24 - Prob. 24.33EPCh. 24 - Prob. 24.34EPCh. 24 - Prob. 24.35EPCh. 24 - Prob. 24.36EPCh. 24 - Prob. 24.37EPCh. 24 - Prob. 24.38EPCh. 24 - Prob. 24.39EPCh. 24 - Prob. 24.40EPCh. 24 - Prob. 24.41EPCh. 24 - Prob. 24.42EPCh. 24 - Prob. 24.43EPCh. 24 - Prob. 24.44EPCh. 24 - Prob. 24.45EPCh. 24 - Prob. 24.46EPCh. 24 - Prob. 24.47EPCh. 24 - Prob. 24.48EPCh. 24 - Prob. 24.49EPCh. 24 - Prob. 24.50EPCh. 24 - Prob. 24.51EPCh. 24 - Prob. 24.52EPCh. 24 - Prob. 24.53EPCh. 24 - Prob. 24.54EPCh. 24 - Prob. 24.55EPCh. 24 - Prob. 24.56EPCh. 24 - Prob. 24.57EPCh. 24 - Prob. 24.58EPCh. 24 - Prob. 24.59EPCh. 24 - Prob. 24.60EPCh. 24 - Prob. 24.61EPCh. 24 - Prob. 24.62EPCh. 24 - Prob. 24.63EPCh. 24 - Prob. 24.64EPCh. 24 - Prob. 24.65EPCh. 24 - The liver, but not the brain or muscle cells, has...Ch. 24 - Prob. 24.67EPCh. 24 - Prob. 24.68EPCh. 24 - Prob. 24.69EPCh. 24 - Prob. 24.70EPCh. 24 - Prob. 24.71EPCh. 24 - Prob. 24.72EPCh. 24 - Prob. 24.73EPCh. 24 - Prob. 24.74EPCh. 24 - Prob. 24.75EPCh. 24 - Prob. 24.76EPCh. 24 - Prob. 24.77EPCh. 24 - Prob. 24.78EPCh. 24 - Prob. 24.79EPCh. 24 - Prob. 24.80EPCh. 24 - Prob. 24.81EPCh. 24 - Prob. 24.82EPCh. 24 - Prob. 24.83EPCh. 24 - Prob. 24.84EPCh. 24 - Prob. 24.85EPCh. 24 - Prob. 24.86EPCh. 24 - Prob. 24.87EPCh. 24 - Prob. 24.88EPCh. 24 - Prob. 24.89EPCh. 24 - Prob. 24.90EPCh. 24 - Prob. 24.91EPCh. 24 - Prob. 24.92EPCh. 24 - Prob. 24.93EPCh. 24 - Prob. 24.94EPCh. 24 - Prob. 24.95EPCh. 24 - Prob. 24.96EPCh. 24 - Prob. 24.97EPCh. 24 - Prob. 24.98EPCh. 24 - Prob. 24.99EPCh. 24 - Prob. 24.100EPCh. 24 - Prob. 24.101EPCh. 24 - Prob. 24.102EPCh. 24 - Prob. 24.103EPCh. 24 - Prob. 24.104EPCh. 24 - Prob. 24.105EPCh. 24 - Prob. 24.106EPCh. 24 - Prob. 24.107EPCh. 24 - Prob. 24.108EPCh. 24 - Prob. 24.109EPCh. 24 - Prob. 24.110EPCh. 24 - Prob. 24.111EPCh. 24 - Prob. 24.112EPCh. 24 - Prob. 24.113EPCh. 24 - Prob. 24.114EPCh. 24 - Prob. 24.115EPCh. 24 - Compare the biological functions of glucagon and...Ch. 24 - Prob. 24.117EPCh. 24 - Prob. 24.118EP
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- Write a balanced equation for the conversion in the glyoxylate cycle of two acetyl units, as acetyl-CoA, to oxaloacetate.arrow_forwardFatty acids and triglycerides are an important source of nutrition and a dense form of stored energy. Digestion of fats yields more energy per gram than digestion of carbohydrates. Each turn on the B-oxidation spiral results in the formation of a new acetyl COA molecule. This molecule can then be used to generate more energy in A) fatty acid synthesis B) ketolysis C) glycolysis D) tricarboxylic acid cycle E) oxidative phosphorylationarrow_forwardWhat compound is produced during strenuous exercise, when the muscles are depleted of oxygen? pyruvate carbon dioxide lactate ethanol NADPHarrow_forward
- How many molecules of ATP are obtained from the complete (including the fourth stage of catabolism) metabolism of one molecule of a 16-carbon saturated fatty acyl-CoA?arrow_forwardDetermine the number of ATP molecules that can be formed from the complete oxidation of 10 molecules of acetyl CoA. The overall net equation for the complete oxidation of acetyl CoA is provided below. Acetyl CoA + 3NAD+ + FAD + GDP + Pi + 2H₂O → 2CO2 + HS-CoA + 3NADH + 3H+ + FADH2 + GTP (Given: The oxidation of one NADH yields 2.5 ATP; the oxidation of one FADH2 yields 1.5 ATP; and one GDP yields 1 ATP.) O 10 ATP O 7,5 ATP O 100 ATP O 75 ATParrow_forwardGlycolysis is the first step in the oxidation of fatty acids to produce energy. True or Falsearrow_forward
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- Which of the following substances can be considered as metabolic intermediates? Water ATP Pyruvic acid Glucosearrow_forwardBelow is the overall net equation for the complete oxidation of pyruvate. Calculate the number of ATP molecules that can be produced from the complete oxidation of 6 molecules of pyruvate. Pyruvate + 4NAD+ + FAD + GDP + Pi + 2H₂O → 3CO2 + 4NADH + 4H+ + FADH2 + GTP (Given: The oxidation of one NADH yields 2.5 ATP; the oxidation of one FADH2 yields 1.5 ATP; and one GDP yields 1 ATP.) O 75 ATP O 60 ATP O 12.5 ATP O 32 ATParrow_forwardDuring the overall process of glycolysis, in which of the following occurs for each glucose molecule processed. net loss of two ATP molecules net loss of four ATP molecules net gain of two ATP molecules net gain of four ATP moleculesarrow_forward
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