**Problem Statement**Calculate the actual change in free energy at 37°C for the phosphoglycerate mutase reaction converting 3-PGA to 2-PGA in erythrocytes. Assume the following conditions after consuming a glass of orange juice:- Concentration of 3-PGA: 8 mM- Concentration of 2-PGA: 0.08 mM- Standard change in free energy (ΔG°') for the reaction: +4.5 kJ/mol**Options**:- ☐ +16.3 kJ/mol- ☐ -6.5 kJ/mol- ☐ -7.3 kJ/mol- ☐ -16.3 kJ/mol- ☐ +7.3 kJ/mol**Explanation**:To solve this problem, apply the formula for the actual change in free energy (ΔG) for a biochemical reaction:\[ \Delta G = \Delta G^\circ' + RT \ln\left(\frac{[ \text{Products} ]}{[ \text{Reactants} ]}\right) \]Where:- \( R \) is the universal gas constant- \( T \) is the temperature in Kelvin- \( \ln \) represents the natural logarithmPlug in the given concentrations and the standard change in free energy to find ΔG.

Phosphoglycerate Mutase: Phosphoglycerate Mutase is an enzyme involved in the glycolytic process.…

Answered: What is the actual change in free…

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

Problem 1P

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a newly identified bacterium called Nomore biochem is unable to synthesize ubiquinone. A mobile electron carrier called CXC3 is used as a substitute. From the information provided in the table, calculate delta G' and Keq value at 298K for the redox reaction that occurs in the Nomore Biochem electron transport chain. (constants: R=8.3 J/degree x mol, F= 96.1 kJ/v x mol Half Reactions E' (V) ubiquinone + 2e- + 2H+--> Ubiquinol + H2 0.045 NAD+ + 2e- + 2H+ --> NADH + H+ -0.320 CXC3 + 2e- + 2H+ --> CXC3H2 -0.450 explain the impact that using CXC3 instead of ubiquinone will have on ATP production in the cell. How might the cell adapt to this situation?
Given the initial rate data for the reaction A + B → C, determine the rate expression for the reaction. [A], M [B], M A[C]/At (initial) M/s 0.0418 0.122 2.73 x 10-4 0.0836 0.122 10.9 × 10 4 0.0418 0.488 2.73 x 104 O A[C] At ○ A[C] ΔΙ = 0.439 Ms[A][B]² = 1.27 M²s¹ [A][B] ● A[C] = 6.53 × 10³ s¹ [A] At O A[C] = 0.156 M's¹[A]² At ○ A[C] At = 5.35 x 102 M2s[A][B]
Could you explain step by step how to solve this? I'm not sure how to get all the numbers that are shown in the solved answer.
Certain bacteria can respire in anoxic environments using arsenic (V) as electron acceptor. The relevant unbalance half reactions are: H₂ AsO+H → H¸AsO +H₂O, logK = 10.84, AG = -14.5kca I m I ol-e CH₂O+H₂OH + CO2 (g), logK = 1.2, AG° = -1.63kca — ol-e 1) Balance the two half reactions 2) What is m the overall respiration reaction, standard free energy change \Delta GO 3) Is this process energetically more or less favorable than sulfate reduction? (\Delta GO for the reduction of sulfate to HS- is -5.78 kcal/mol - e-) 4) If [H2AsO4-] = [ H3ASO3] = 0.5 mM and pH = 7, estimate pe of the system
Zinc-air batteries are actively being researched because one reactant (oxygen) is provided in infinite supply for free. The standard cell potential for this battery is 1.65 V. If the cell potential is 1.61 V, what is the partial pressure of oxygen in the cell and a temperature is 25.0\deg C? The reaction in the cell converts Zn(s) and O2(g) into ZnO(s). 2Zn(s) 02 (g) -> 2ZnO(s)
On the right the Hill plot com- (b) pares the O2 binding properties of Hb Ya- kima with those of HbA in 0.1 M NaCl buff- Hb-Yakima ered to pH 7 with 0.01 M bis-Tris. Focus first on the line for "stripped Hb". This is the term for hemoglobin isolated from erythro- cytes with removal of all organic phosphate molecules that might bind to the protein in RBCS. You can see that 2,3-bisphospho- glycerate (BPG; labeled DPG according to old terminology) does not alter the O2 bind- ing affinity of Hb Yakima in contrast to HbA (although it was shown that BPG did bind to the deoxyHb Yakima molecule). Also, Hb Yakima is associated with markedly decreased allostery in the absence and presence of BPG, in comparison to HbA. IHP = inositol hexaphosphate, an artificial allosteric modifying ligand that binds more tightly than BPG. stripped Hb Hb-A •DPG +DPG n= 1.0 n = 2.3 n=2,5 +THP +IHP 0.5 0.5 1 5 10 50 po, ( mm Hg ) %3D On the right is a diagram copied from the lecture handout "Hemoglobin and Allo-…
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Triosephosphate isomerase catalyzes the conversion of glyceraldehyde 3-phosphate to dihydroxyacetone phosphate. This reaction has a Keq of 22.2. What is the standard free-energy change for this reaction? -7,680 kJ/mol -7.68 kJ/mol -3.34 kJ/mol -0.644 kJ/mol 7.68 kJ/mol
13. 0.9% (m/v) NaCI solution and 5% (m/v) glucose solution are both isotonic to red blood cells. SHOW your work and watch sig figs & units. c. convert the concentration from M to % (m/v) for a 0.342 M NaC solution. (HINT: convert to g/ml and then multiple by 100%)

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