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Nady [450]
3 years ago
5

The oxidation of pyruvate is not the only process that generates acetyl-CoA. In particular, fatty acids are broken down through

oxidation, generating a great deal of acetyl-CoA, which can also enter the TCA cycle to generate high energy molecules. Is there a difference between between the TCA cycle using acetyl-CoA generated from fatty acid oxidation compared to that generated from carbohydrate oxidation?
A. The concentration of malate increases when reacting with acetyl-CoA derived from a fatty acid,
B. A molecule of acetyl-CoA derived from a fatty acid generates more FADH2 than acetyl-CoA from
C. A molecule of acetyl-CoA derived from a carbohydrate generates more NADH than acetyl-CoA but not when reacting with acetyl-CoA derived from a carbohydrate. WI derived from a carbohydrate derived from a fatty acid.
D. There is no difference in the TCA cycle, regardless of the source of acetyl-CoA.
Chemistry
1 answer:
Mnenie [13.5K]3 years ago
4 0

Answer:

The correct answer is D. There is no difference in the TCA cycle,regardless  of source of acetyl CoA.

Explanation:

Acetyl CoA is the central metabolite of aerobic respiration.Acetyl CoA is generated from both decarboxylation of pyruvate and from the beta oxidation of fatty acid in mitochondrial matrix.

            In both cases the generated Acetyl CoA molecule is same but it comes from different source.

  If the Acetyl CoA generated from beta oxidation of fatty acid enter into the TCA cycle it will generate same molecules of NADP and FADH2 as that of Acetyl CoA generated from carbohydrate metabolism

         

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How many moles are there in 990 grams of sugar, C12H22011? (Sig Figs, do not
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Number of moles = 2.89 mol

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Number of moles of sugar  = ?

Mass of sugar = 990 g

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Number of moles = mass/molar mass

Molar mass of C₁₂H₂₂O₁₁:

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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