Biochem 451A Chapter 3

In these cases, ΔH must be negative for the process to become favorable as temperature decreases. In cases of cold denaturation, the sign on ΔS is dominated by the hydrophobic effect and is therefore negative. Clathrate cage formation would account for the sign of ΔH being negative (favorable) for cold denaturation.

Sometimes, however, proteins denature as temperature is decreased. How might this be explained? Check all that apply. In these cases, ΔH must be positive for the process to become favorable as temperature decreases. In cases of cold denaturation, the sign on ΔS is dominated by the hydrophobic effect and is therefore positive. In these cases, ΔH must be negative for the process to become favorable as temperature decreases. In cases of cold denaturation, the sign on ΔS is dominated by the hydrophobic effect and is therefore negative. Clathrate cage formation would account for the sign of ΔH being negative (favorable) for cold denaturation. Clathrate cage formation would account for the sign of ΔH being positive (favorable) for cold denaturation.

This glucose-6-phosphate concentration is never reached because glucose-6-phosphate is continuously consumed by other reactions, and so the reaction never reaches true thermodynamic equilibrium.

The answer you have obtained is an absurdly high value for the cell and in fact is never approached in reality. Explain why. This glucose-6-phosphate concentration is never reached because glucose-6-phosphate is unstable at these conditions, and so the reaction never reaches true thermodynamic equilibrium. This glucose-6-phosphate concentration is never reached because glucose-6-phosphate continuously dimerizes, and so the reaction never reaches true thermodynamic equilibrium. This glucose-6-phosphate concentration is never reached because glucose-6-phosphate is continuously consumed by other reactions, and so the reaction never reaches true thermodynamic equilibrium.

False

The change in enthalpy (ΔH) for the complete oxidation of a fatty acid is different depending on whether it occurs via a biochemical pathway or combustion to CO2 and H2O.

the same amount of energy

The combustion of palmitic acid in a bomb calorimeter yields energy in the form of heat released upon oxidation. From a thermodynamic perspective and with respect to the calorimeter experiment, what would you expect the combustion of the same amount of palmitic acid in our body to yield? the same amount of energy more energy less energy The calorimeter experiment is irrelevant to the combustion of palmitic acid in the human body.

The second reaction (C to D) must be more exergonic than the first reaction (A to B) is endergonic.

The conversion of A to B is coupled to the conversion of C to D. Which statement most accurately describes this sequence? The first reaction must be exergonic and the second must have a ΔG = 0. The second reaction (C to D) must be more exergonic than the first reaction (A to B) is endergonic. The first reaction (A to B) must be more exergonic than the second reaction (C to D) is endergonic. Both reactions are equally endergonic.

phosphoenolpyruvate to pyruvate

The conversion of ADP to ATP is endergonic by more than 30 kJ/mol. Which of the following hydrolyses could drive this conversion? Please consult Table 3.5. glucose-1-phosphate to glucose glucose-6-phosphate to glucose phosphoenolpyruvate to pyruvate PPi to 2 Pi

both gases more translational and rotational increases

The decomposition products are _______ and therefore have significantly ______ _____ freedom compared to the initial solid. Thus, entropy _____

is related to the change in free energy of the reaction.

The equilibrium constant of a reaction: is related to the change in free energy of the reaction. cannot be used to determine whether a reaction will proceed in the direction as written under non-standard conditions. is the same as the mass action ratio when the reaction is displaced from equilibrium. can change if the concentration of reactants and products are changed. is not related to the change in free energy of the reaction.

The value of Q must be less than that of K.

The terms Q and K describe mass action for reactions at non-equilibrium and equilibrium conditions, respectively. For a forward reaction to be favored in a living cell, which of the following statements must be true? The value of Q must be less than that of K. The value of K must be less than that of Q. No conclusion can be drawn about Q and K for a favorable process. None of the listed statements are true; cells are at equilibrium.

can predict whether a reaction will be thermodynamically favorable under standard conditions.

The ΔG o of a reaction: can predict whether a reaction will be thermodynamically favorable under standard conditions. will change if the temperature of the reaction is changed. is not related to the equilibrium constant. will change if the concentration of reactants and products are changed.

reduces more hydrophobic

This process _____ the entropy of the solvent water, which becomes _____ ordered in the clathrate structures. This is the _____ effect.

using the product immediately in the next step and maintaining a high starting-material concentration

Under standard conditions, a given reaction is endergonic (i.e., DG >0). Which of the following can render this reaction favorable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration? maintaining a high starting-material concentration using the product immediately in the next step keeping a high product concentration using the product immediately in the next step and maintaining a high starting-material concentration

Ionic.

What kind of bond is formed when lithium and fluorine combine to form lithium fluoride? Nonpolar covalent. Polar covalent. Redox. Ionic.

ΔS must be positive because the increase in isoenergetic conformations in the denatured state increases the entropy of the denatured state relative to the folded state.

What must be the sign of ΔS for the change: native → denatured? ΔS must be positive because the increase in conformations with different energy in the denatured state decreases the entropy of the denatured state relative to the folded state. ΔS must be negative because the increase in isoenergetic conformations in the denatured state decreases the entropy of the denatured state relative to the folded state. ΔS must be positive because the increase in isoenergetic conformations in the denatured state increases the entropy of the denatured state relative to the folded state. ΔS must be negative because the increase in conformations with different energy in the denatured state increases the entropy of the denatured state relative to the folded state.

Biological systems are highly ordered so entropy changes are not relevant.

Which of the following statements is FALSE? Biological systems expend energy to overcome entropy. The entropy of a biological system can decrease. Biological systems are highly ordered so entropy changes are not relevant. The entropy of an isolated system will tend to increase to a maximum value. Entropy is a measure of disorder.

Organisms are open systems as they can create energy from their environments.

Which of the following statements is FALSE? Organisms are open systems as they can exchange both energy and materials with their environments. Energy can be transferred between a system and the surroundings. In an open system energy can be converted from one form into another. Organisms are open systems as they can create energy from their environments. In biochemical processes, energy can neither be created or destroyed.

A hydrogen atom is transferred to the atom that loses an electron.

Which of the following statements is not true of most cellular redox reactions? A hydrogen atom is transferred to the atom that loses an electron. Changes in potential energy can be released as heat. The electron acceptor is reduced. The reactant that is oxidized loses electrons.

Oxygen holds electrons more tightly than hydrogen does, and the net charge is zero.

Which of the following statements is true of the bonds in a water molecule? Oxygen acts as the electron acceptor and is oxidized. The electron in each hydrogen atom is completely transferred to the oxygen atom, and each hydrogen atom has a net charge of +1. There is equal sharing of the electrons between the oxygen atom and the two hydrogen atoms, and the net charge is zero. Oxygen holds electrons more tightly than hydrogen does, and the net charge is zero.

Electronegativity

Which term describes the degree to which an element attracts electrons? Reduction. Oxidation. Electronegativity. Polarity.

Anion and cation

Which terms describe two atoms when they form a bond in which electrons are completely transferred from one atom to the other? Polar and nonpolar. Ionic and covalent. Proton and electron. Anion and cation.

ΔG < 0.

A biochemical reaction will proceed in the direction as written if: ΔH < 0. ΔG < 0. ΔH > 0. ΔG > 0. ΔG = zero.

concentration

A reaction at equilibrium can be driven in one direction or the other by changing the ________ of reactants or products.

equilibrium

A reaction at its lowest energy state for the system and with equal rates in the forward and reverse directions is said to be at ________.

NaCl will migrate (diffuse) across the membrane until there is an equal concentration on both sides.

A solution of 2M NaCl in water is separated from pure water by a semipermeable membrane. Which of the following is true? The crossing of NaCl is an endergonic process. NaCl will migrate (diffuse) across the membrane until there is an equal concentration on both sides. Water will move from the 2M NaCl solution to the pure water compartment. Nothing will happen because the system is at ΔG = 0.

False

A thermodynamically unfavorable reaction can become favorable when coupled to a highly endergonic reaction.

The value of TΔS must equal the value of ΔH.

At the melting temperature of water, which of the following statements is true? The value of TΔS must be greater than the ΔH component. The reaction is spontaneous because ΔH is greater than -TΔS. The value of TΔS must equal the value of ΔH. There is not enough information to answer this question.

Yes, regardless of the sign of ΔH, if ΔS is positive, a reaction can be rendered favorable by increasing the temperature.

Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive ΔG at RT be made favorable by increasing the reaction temperature? only if both ΔH and ΔS are negative No, an unfavorable reaction cannot be rendered favorable. yes, but only if ΔH is positive and ΔS is negative Yes, regardless of the sign of ΔH, if ΔS is positive, a reaction can be rendered favorable by increasing the temperature.

This reaction is favorable.

Consider the following reaction: NADH + FAD + H+ → NAD+ + FADH2. Which of the following statements is correct? The ΔE° for this reaction is negative. This reaction is favorable. FAD is the reductant. NADH is the oxidant.

Water boiling and mixing cola and root beer sodas are favorable processes, but assembling a house of cards is not.

From the standpoint of entropy, consider the following processes: water boiling, mixing cola and root beer sodas, and assembling a house of cards. Based on your understanding, which of the following statements is true? Water boiling is a favorable process, but mixing cola and root beer sodas and assembling a house of cards are not. Water boiling and mixing cola and root beer sodas are favorable processes, but assembling a house of cards is not. Water boiling and mixing cola and root beer sodas are not favorable processes, but assembling a house of cards is.

Hydrogen, polar.

Gaseous hydrogen burns in the presence of oxygen to form water: 2H2 + O2 → 2H2 O + energy Which molecule is oxidized and what kind of bond is formed? Hydrogen, polar. Oxygen, polar. Hydrogen, nonpolar. Oxygen, nonpolar.

If ΔH is positive (energy is required to break the noncovalent interactions in the folded state), the +,+ situation pertains. We expect ΔS for denaturation to be positive due to the increase in conformational entropy. Under these conditions, the denaturation process goes from being unfavorable at lower temperature to being favorable at higher temperature.

Given what you have learned so far about ΔH and ΔS, explain why this is reasonable. Check all that apply. If ΔH is positive (energy is required to break the noncovalent interactions in the folded state), the +,- situation pertains. If ΔH is positive (energy is required to break the noncovalent interactions in the folded state), the +,+ situation pertains. If ΔH is negative (energy is released to break the noncovalent interactions in the folded state), the -,- situation pertains. We expect ΔS for denaturation to be positive due to the increase in conformational entropy. If ΔH is negative (energy is released to break the noncovalent interactions in the folded state), the -,+ situation pertains. We expect ΔS for denaturation to be negative due to the increase in conformational entropy. Under these conditions, the denaturation process goes from being unfavorable at lower temperature to being favorable at higher temperature. Under these conditions, the denaturation process goes from being unfavorable at higher temperature to being favorable at lower temperature.

ΔG=ΔH−TΔS positive negative positive positive positive hydrophobic negative positive reduces

Since _____, a _____ value forΔS results in a _____ contribution to ΔG (T is always _____ in the Kelvin scale). Thus, for proteins to be stably folded (which requires ΔG to be _____ for the process described in the question), it would appear that ΔH must be large and _____. This is generally true; however, this simple analysis does not consider the _____ effect, which results in a more _____ overall ΔS. Thus, the requirement that ΔH is large and _____ is not absolute because the hydrophobic effect _____ the magnitude of ΔS.

If ΔH∘ and ΔS∘ are independent of temperature, a graph of lnK versus 1/T should be a straight line with slope −ΔH∘/R. These values can also be determined from direct fits to the K versus T data using nonlinear curve-fitting software.

How could you use values of K determined at different temperatures to determine ΔH∘ for the reaction? (Hint: Recall that you are treating ΔH∘ and ΔS∘ as constants.) If ΔH∘ and ΔS∘ are independent of temperature, a graph of lnK versus 1/T should be a straight line with slope −ΔH∘/R. These values can also be determined from direct fits to the K versus T data using nonlinear curve-fitting software. If ΔH∘ and ΔS∘ are dependent of temperature, a graph of lnK versus 1/T should be a straight line with slope ΔS∘/R. These values can also be determined from direct fits to the T versus K data using nonlinear curve-fitting software. If ΔH∘ and ΔS∘ are dependent of temperature, a graph of K versus T should be a straight line with slope −ΔS∘/R. These values can also be determined from direct fits to the lnK versus 1/T data using nonlinear curve-fitting software. If ΔH∘ and ΔS∘ are independent of temperature, a graph of K versus T should be a straight line with slope ΔH∘/R. These values can also be determined from direct fits to the lnK versus 1/T data using nonlinear curve-fitting software.

The reverse reaction is favored.

If a mixture was prepared containing 1 M glucose-6-phosphate and 1×10−3M glucose-1-phosphate, what would be the thermodynamically favored direction for the reaction? The direct reaction is favored. The reverse reaction is favored.

True

In a general redox reaction, the reductant becomes oxidized and the oxidant becomes reduced.

True

Life is an irreversible process, such that it never comes to equilibrium.

1. First we should write equation for ΔG∘: ΔG∘=ΔH∘−TΔS∘ and ΔG∘=−RTlnK 2. Then combine equations from the first step: ΔH∘−TΔS∘=−RTlnK 3. Divide both sides by −RT. 4. We obtain the following equation: −(ΔH∘RT)+ΔS∘R=lnK 5. Rearranging the previous equation we get the final equation: lnK=−ΔH∘RT+ΔS∘R

Show from this, and equations given in the chapter in the textbook, that lnK=−ΔH∘RT+ΔS∘R where K is the equilibrium constant. 1. First we should write equation for ΔG∘: ΔG∘=ΔH∘−TΔS∘ and ΔG∘=−RTlnK 1. First we should write equation for ΔG∘: ΔG∘=ΔH∘+TΔS∘ andΔG∘=RTlnK 2. Then combine equations from the first step: ΔH∘−TΔS∘=−RTlnK 2. Then combine equations from the first step: ΔH∘+TΔS∘=−RTlnK 3. Divide both sides by −RT. 3. Divide both sides by RT. 4. We obtain the following equation: −(ΔH∘RT)−ΔS∘R=lnK 4. We obtain the following equation: (ΔH∘RT)+ΔS∘R=lnK 4. We obtain the following equation: −(ΔH∘RT)+ΔS∘R=lnK 5. Rearranging the previous equation we get the final equation: lnK=−ΔH∘RT+ΔS∘R