Answer:
The correct answer is C fatty acid oxidation would stop when all of the CoA is bound as acetyl CoA.
Explanation:
Acetyl CoA is the principle end product of beta oxidation of even chain fatty acid such as palmitic acid.
When the cellelar label of actyl CoA increases at that time the excess acetyl CoA is converted to ketone bodies by the process called ketogenesis.
According to the question if the excess acetyl CoA is not converted to ketone bodies then it will interfere with the oxidation of fatty acid because fatty acid molecules will not get any CoA SH molecule to activate themselves to initiate a new round of beta oxidation.
As a result fatty acid oxidation will stop.
Option C is correct.
Lactic acid fermentation is a metabolic process in which a glucose molecule turns into lactate which again turn into lactic acid.
In this process a glucose molecule produces 2 molecule of lactate and releases 2 NADH. The NADH is produced from the reduction of NAD+ and ADP takes the phosphates to release 2 molecules of ATP.
Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.
“Aluminum” is most likely to
Answer:
For young organic materials, the carbon-14 (radiocarbon) method is used. The effective dating range of the carbon-14 method is between 100 and 50,000 years.
Explanation:
The answer is the explanation.
