Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Where would you expect to find the 14C label when alpha-ketoglutarate is converted to succinate?

Answers

Answer 1
Answer:

Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive ^(14)C while none of succinate would contain ^(14)C, and all of carbon dioxide released would contain ^(14)C.

Explanation:

When oxaloacetate in the citric acid cycle is labeled with ^(14)C in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the ^(14)C) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. CO_2 released had all the radioactivity.


Related Questions

A generic solid, X, has a molar mass of 78.2 g/mol. In a constant-pressure calorimeter, 12.6 g of X is dissolved in 337 g of water at 23.00 °C.X(s) yeilds X(aq)The temperature of the resulting solution rises to 24.40 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there\'s negligible heat loss to the surroundings. How much heat was absorbed by the solution?What is the enthalpy of the reaction?
A weather balloon has a volume of 200.0 L at a pressure of 760 mm Hg. As it rises, the pressure decreases to 282 mm Hg. What is the new volume of the balloon? (Assume constant temperature)
g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN
What is the question mark
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.3 atm of sulfur dioxide gas and 0.79 atm of oxygen gas at 31.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.47 atm Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Consider the following reaction: Br2(g) + Cl2(g) ⇌ 2BrCl(g), Kp=1.112 at 150 K.
A reaction mixture initially contains a Br2 partial pressure of 751 torr and a Cl2 partial pressure of 737 torr at 150 K.
Calculate the equilibrium partial pressure of BrCl.

Answers

Answer:

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Final answer:

To calculate the equilibrium partial pressure of BrCl, use the equilibrium constant expression and substitute the given partial pressures of Br2 and Cl2. The equilibrium partial pressure of BrCl is approximately 0.0375 atm.

Explanation:

To calculate the equilibrium partial pressure of BrCl, we need to use the equilibrium constant expression:



Kp = ([BrCl]^2) / ([Br2] * [Cl2])



Given that the equilibrium partial pressures of Br2 and Cl2 are 0.450 atm and 0.115 atm, respectively, we can substitute these values into the expression:



1.112 = ([BrCl]^2) / (0.450 * 0.115)



Simplifying the expression, we find that the equilibrium partial pressure of BrCl is approximately 0.0375 atm.

Learn more about Equilibrium partial pressure of BrCl here:

brainly.com/question/35557199

#SPJ12

Why is it important to know what temperature scale is being used in a given situation?

Answers

It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.

Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius 

What volume of a solution is needed to dissolve 1.0 mol of KOH to make a solution whose pH is 13.55? A. 2.82 L
B. 12.43 L
C. 4.77 L
D. 3.35 L

Answers

Answer:

Option A. 2.82 L

Explanation:

Step 1:

Data obtained from the question.

pH = 13.55

Step 2:

Determination of the pOH of the solution.. This is illustrated below:

pH + pOH = 14

pH = 13.55

13.55 + pOH = 14

Collect like terms

pOH = 14 - 13.55

pOH = 0.45

Step 3:

Determination of the concentration of the OH ion.

This is illustrated below:

pOH = - Log [OH-]

pOH = 0.45

0.45 = - Log [OH-]

- 0.45 = Log [OH-]

[OH-] = antilog (- 0.45)

[OH-] = 0.355 M

Step 4:

Determination of the molarity of KOH. This is illustrated below:

First, we'll write the dissociation equation of KOH as follow:

KOH —> K+(aq) + OH-(aq)

From the balanced equation above,

1 mole of KOH produced 1 mole of OH-.

Therefore, 0.355 M of KOH will definitely produce 0.355 M of OH-.

The molarity of KOH is 0.355 M

Step 5:

Determination of the volume of the solution needed to dissolve 1 mole of KOH. This is illustrated below:

Mole of KOH = 1 mole

Molarity of KOH = 0.355 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 1/0.355 M

Volume = 2.82L

On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.

Answers

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

Which two structures will provide a positive identification of a plant cell under a microscope? A.) Lysosomes, cell wall. B.) large central vacuole, cell wall. C.) large central vacuole ribosomes. D.)nucleoid, chloroplasts.

Answers

The Right Answer Is D.) Nucleoid chloroplasts. 
the answer is B large central vacuole and cell wall. they are the easiest/biggest things to see under a microscope to identify a plant cell

Consider the following reaction at equilibrium. What effect will adding 1.4 mole of He to the reaction mixture have on the system? 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(g) + 2 SO2(g) Consider the following reaction at equilibrium. What effect will adding 1.4 mole of He to the reaction mixture have on the system? 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(g) + 2 SO2(g) The reaction will shift to the left in the direction of reactants. No effect will be observed. The reaction will shift to the right in the direction of products. The equilibrium constant will increase. The equilibrium constant will decrease.

Answers

Answer:

The reaction will shift to the left in the direction of reactants.

Explanation:

According to Le Chatelier's principle, when an external constraint is applied to a chemical system in equilibrium, the system adjust in order to annul the effect impose on it by the external system.

Also, from the principle, the addition of an inert gas can affect the equilbrium of a gaseous system, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

1. Addition of an inert gas at constant volume:

When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.  Hence, there will be no effect on the equilibrium.  

2. Addition of an inert gas at constant pressure:

When an inert gas is added to a system in equilibrium at constant pressure, then the total volume will increase(i.e. the number of moles per unit volume of various reactants and products will decrease). Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.  

Considering the given reaction in equilibrium:

2H₂S(g) + 3O₂(g) ⇌ 2H₂O(g) + 2SO₂(g)

The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the backward direction because the number of moles of reactants is more than the number of moles of the products.

Final answer:

Adding 1.4 moles of He to the reaction mixture will have no effect on the equilibrium of the system.

Explanation:

Adding 1.4 moles of He to the reaction mixture will have no effect on the system. The equilibrium of the reaction will not shift to the left or right, and there will be no change in the equilibrium constant. This is because He is considered an inert gas and does not participate in the reaction.

Learn more about Effect of adding He to a reaction mixture here:

brainly.com/question/16201332

#SPJ12