Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 4, Problem 16P
Interpretation Introduction
Interpretation:
The template and primer related to DNA synthesis should be defined.
Concept introduction:
DNA synthesis by Watson-Crick methods explains how genetic information is transferred to succeeding generations. Template and primer are the basic components that initiate and elongate the DNA synthesis.
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Students have asked these similar questions
The pKa values for glutamic acid are 2.19, 9.67, 4.25. Sketch out the titration curve for this amino
acid and include all of the pKa values and the pl.
Calculate the isoelectronic point, pl, from the pKa values for histidine, arginine and asparagine.
The free energy released by the hydrolysis of ATP under standard conditions is -30.5 kJ/mol.
If ATP is hydrolyzed under standard conditions except at pH 5.0, is more or less free energy released? Why?
More free energy is released because the increased [H+] stabilizes the negative charge on the ADP molecule.
Less free energy is released because an acidic environment depletes cellular ATP levels.
Less free energy is released because the reaction favors ATP production over hydrolysis due to the higher [H+]
in solution.
More free energy is released because the total cellular concentrations of ATP, ADP, and P; are greater at the lower
pH.
Correct Answer
Chapter 4 Solutions
Biochemistry (Looseleaf)
Ch. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10P
Ch. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52P
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- Consider a system consisting of an egg in an incubator. The white and yolk of the egg contain proteins, carbohydrates, and lipids. If fertilized, the egg transforms from a single cell to a complex organism. How does the entropy change in both the system (developing chick) and suroundings (the egg environment) drive the irreversible process of chick development? ☐ The release of glucose from sucrose, which produces energy needed for chick development, decreases entropy in the surroundings. Chick development increases entropy in the system, which causes a concominant decrease in entropy in the surroundings. Carbohydrates, proteins, and lipids within the egg break down into CO2 and H2O, which increases entropy in the surroundings. Chick development decreases entropy in the system, but this is smaller than the concominant increase in entropy in the surroundings.arrow_forwardThe amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form -NH or as the free base -NH2, because of the reversible equilibrium R-NH =R-NH₂ + H+ In what pH range can glycine be used as an effective buffer due to its amino group? pH 8.6 to pH 10.6 In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the -NH form? Correct Answer Correct Answer 45 How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH to 10.0? 10 mL When 99% of the glycine is in its -NH form, what is the numerical relation between the pH of the solution and the pKa of the amino group? pH = pKa - 2 Correct Answer Correct Answerarrow_forwardThe glycolytic enzyme Phosphofructokinase (PFK) catalyzes the following reaction: Fructose-6-phosphate (F6P) + ATP → Fructose-1,6-bisphosphate (F1,6BP) + ADP AG"=-14.2 kJ/mol This is considered the enzymatic step that commits a sugar substrate to glycolysis. a) Calculate the standard free energy of hydrolysis of fructose-1,6-bisphosphate. b) What is the equilibrium constant for this coupled reaction? c) ATP is a known inhibitor of PFK. If the cellular concentrations of ATP and ADP are 5 mM and 1.0mM respectively, and the concentrations of F6P and F1,6BP are 2mM, what is the free energy change of the system?arrow_forward
- 2) Consider the following reaction: A + 2B 3C + D At equilibrium the concentration of the reactants and products are: [A] = 20.0 mM [C] = 3.0 mM [B] = 4.0 mM [D] = 50.0 mM Calculate (a) the equilibrium constant and (b) AG". Comment on which side of this reaction is more likely to occur.arrow_forwardGlycine is a diprotic acid, which can potentially undergo two dissociation reactions, one for the a-amino group (NH), and the other for the carboxyl (-COOH) group. Therefore, it has two pK₁ values. The carboxyl group has a pK₁ of 2.34 and the α-amino group has a pK2 of 9.60. Glycine can exist in fully deprotonated (NH2-CH2-COO¯), fully protonated (NH3-CH2-COOH), or zwitterionic form (NH3-CH2-COO¯). Match the pH values with the corresponding form of glycine that would be present in the highest concentration in a solution of that pH. fully deprotonated form NH2-CH2-COO- fully protonated form NH–CH,–COOH zwitterionic form NH–CH,−COO Answer Bank pH 7.0 pH 11.9 pH 6.0 pH 8.0 pH 1.0arrow_forwardThe AG of hydrolysis of a sugar phosphate (S-O-P) to the free sugar (S-OH) is -26.6 kJ/mol in a hypothetical cell in which the steady-state concentrations of sugar phosphate, free sugar, and inorganic phosphate are 1.0 mM, 0.20 mM, and 50.0 mM, respectively. S-O-P + H2O S-OH + Pi (a) What is the AG°' for this reaction? (b) In the cell, S-O-P is formed by the transfer of a phosphate group from ATP. What would the AG be for the transfer of the g-phosphate from ATP to this sugar (S-OH)? [AG for ATP hydrolysis is -31 kJ/mol.]arrow_forward
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