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Pavel
5 days ago
9

An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs

trate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.
Chemistry
1 answer:
eduard [2.6K]5 days ago
6 0

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

Explanation:

Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

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This equation allows for computation of P at 10,000 m beneath the ocean's surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density found at a depth of 10,000 m in the ocean is

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For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

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Regarding the specified chemical reaction:

DHAP\rightleftharpoons G_3P

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The relationship between standard Gibbs free energy and equilibrium constant is as follows:

\Delta G^o=-RT\ln K_{eq}

The free energy change can be calculated using the following equation:

\Delta G=\Delta G^o+RT\ln Q

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\Delta G=-RT^o\ln K_{eq}+RT\ln Q

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T = temperature of the cell = 37^oC=[273+37]K=310K

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Substituting the values into the equation yields:

\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol

Thus, the change in free energy for the reaction is -5339.76 J/mol

  • For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, -5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol

Consequently, the free energy change's units are kJ/mol

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For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

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