answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pavel
1 month 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 [2.7K]1 month 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.

You might be interested in
You want to test the effect of pH on the distribution of Artemia in your 35cm long testing chamber. If you measure a pH = 1 at o
Tems11 [2777]

Answer:The pH measured 10 cm from the most acidic end is 3.42.

Explanation:

The pH at one end = 1The pH at the other end = 13

The chamber length = 13 cm

The change in pH concerning the chamber's length from the acidic end is

Thus, the pH at a distance of 10 cm from the most acidic end is 3.42.

x=\frac{13-1}{35 cm}=\frac{12}{35} pH/cm

7 0
2 months ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
Tems11 [2777]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

Where P represents the pressure of each component at equilibrium.

Starting with an initial pressure of H2S at 3.00 atm, the equilibrium concentrations are:

H2S = 3.00 atm - 2X

H2 = 2X

S2 = X

Substituting these values into the equation gives:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Calculating X yields:

X = 0.008945 atm

In equilibrium, the pressure of S2 is X, so the pressure stands at 0.008945 atm

7 0
1 month ago
It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it
Anarel [2989]
3,048 minutes. Explanation: 762 divided by 5, then multiply that number by 20.
5 0
1 month ago
Other questions:
  • (g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra
    14·1 answer
  • Which of the following phrases describes valence electrons?
    15·1 answer
  • What evidence is there from your results that the characteristic color observed for each compound is due to the metal ion in eac
    14·1 answer
  • 1 point There are 118 known elements. How many elements have atoms, in their ground-state, with core electrons whose quantum num
    8·1 answer
  • At a certain temperature, the reaction 2NO + Cl2 ⇌ 2NOCl has an equilibrium constant Kc of 45.0. A chemist creates a mixture wit
    5·1 answer
  • 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu
    9·1 answer
  • Molecular bromine is 24 percent dissociated at 1600 k and 1.00 bar in the equilibrium br2 (
    6·1 answer
  • Element X reacts with element Y to give a product containing X3+ ions and Y2− ions.
    10·1 answer
  • Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy
    14·1 answer
  • The molecules that make up soap have a region that contains polar covalent bonds and a region that contains nonpolar covalent bo
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!