Answer:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.
Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin. Palmitate oxidation however, does not involve carboxylation.
Explanation:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.
Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme. The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.
Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.
Answer:
Explanation:
-mass of solute, density of solute, and volume of solution
-volume of solute, mass of solution, and density of solution
Answer:
Its B
Explanation:
Its pretty easy. Im three days late my bad my guy
Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
![V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]](https://tex.z-dn.net/?f=V%28t%29%3DV_0%2B%28f_i-f_o%29%2At%5C%5C%5C%5CV%28t%29%20%3D%202%20%2B%286.33%2F1000-3.25%2F1000%29%2At%3D2%2B0.00308%2At%20%5C%2C%20%5C%2C%20%5Bm3%5D)
Try to sound out the words ! :)
