Aconitase catalyzes the conversion of Citrate to Isocitrate in the TCA cycle. The standard free energy change (AG°) for this reactionis +6.7 kJ/mol. The observed free energy change (AG) for the reaction in pig heart mitochondria is +0.8 kJ/mol. The ratio of[isocitrate]/[citrate] in these mitochondria is:Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.a10b.11000.01ef0.330.10h0.25i2

The citric acid cycle, which produces ATP, GTP, NADH, and FADH2 molecules, is one of the most…

Answered: Aconitase catalyzes the conversion of…

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?
Calcium is an important regulator of the citric acid cycle. Calcium is transported across the mitochondrial inner membrane by a Ca²- 2+ uniporter that is driven by the negative potential inside the matrix. Part A Assuming a membrane potential across the inner membrane of 172 mV (inside negative), calculate the ratio of the [Ca2+] in the matrix to that in the cytoplasm ([Ca2+]m/[Ca²+]c) that would exist at equilibrium (i.e., AG = 0). Express your answer using two significant figures. [Ca2+]m [Ca2+] = ΤΟ ΑΣΦ Submit Request Answer Part B ? Cytoplasmic [Ca2+] is on the order of 10-7 M in a healthy cell. Based on your answer in Part A, calculate the [Ca2+] that would exist in the matrix at equilibrium. Express your answer to two significant figures and include the appropriate units. ☐ μÅ ? Value Units
If the enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM, what is its Km? 0.125 mM 0.25 mM 0.50 mM 1.25 mM 5.0 mM
You are isolating mitochondria from insect cells and incubating in a test tube with 0.005 M FADH2, 0.05 M ADP and 0.05 M Pi. Assuming, these 3 can enter mitochondria at no cost,no glucose/products of glucose metabolism remain in the isolated mitochondria, and oxygen is present. Part 1) If all expected reactions go to completion, how much ATP is expected formed? (0 M)(0.02 M) (0.03 M) (0.05 M) (0.1 M) (0.005 M) (0.01 M) (0.015 M) Part 2) The ratio of FADH2/FAD at completion of all expected reactions would be? (2) (>>2) (0) (1) Part 3) After completion of all expected reactions, ratio of H ion concentration inside vs. outside the mitochondrial inner membrane should be? (<1) (~1) (>1)(0) Please provide brief explanation
In active muscle cells, the pO2 is about 10 torr at the cell surface and 1 torr at the mitochondria(the organelles where oxidative metabolism occurs). Calculate the percentage of bound oxygentransported to the mitochondria of muscle cells by myoglobin (KD = 2 torr).
The cytochromes are heme-containing proteins that function as electron carriers in the mitochondria. Calculate the difference in the reduction potential (AE°') and the change in the standard free energy (AG°) when the electron flow is from the carrier with the lower reduction potential to the higher. cytochrome c₁ (Fe³+) + e¯ = cytochrome c₁ (Fe2+) E°' = 0.22 V cytochrome c (Fe³+) + e¯ = cytochrome c (Fe²+) E°' = 0.254 V Calculate AE°' and AG°'. AE°' = AG°' = V kJ/mol
It has been shown recently that the African naked GLUT5-O Fructose mole-rat can withstand up to 18 min of complete anoxia (ab- sence of O2) in the presence of high levels of CO2 without ap- parent brain damage. The basis of its capacity to tolerate such unusual conditions is that the brain, heart, and liver in the mole- rat switch to fructose-driven glycolysis. In contrast to normal mice, these tissues in the naked mole-rat have high levels of GLUT5 specific for cellular uptake of fructose. The available pathways for breakdown of fruct ose for metabolic energy production are shown on the right. KHK stands for ketohexosekinase, an enzyme that is markedly upregulated in the mole-rat compared to normal mice. Conse- quently mice must metabolize fructose largely via hexokinase (HK). In the diagram ALDOB, ALDOA, etc. represent aldolase isozymes. KHK HK Fructose-6-P PFK Fructose-1-P Fructose-1,6-BP ALDOB ALDOC ALDOA ALDOC Glyceraldehyde DHAP GAЗP Glyc-3-P 3-PGA Lactate Pyruvate mouse naked…
Calculate the ΔG for Malate dehydrogenase reaction of CAC if the concentration of oxaloacetate is 1x 10-8 M, malate is 0.2 mM, NAD+ is 10 mM and NADH is 0.1 mM in rat liver mitochondria. ΔGo’ for this reaction is 30 kJ/mole. Please write the units of the final answer and the formula you are using to solve. R = 0.008314kJ/K*mol, T = 298K

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