C?
Electrons are found orbiting around the nucleus of an atom in defined energy levels.
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.
ΔG > 0
is always true for the freezing of water.
Explanation:
- The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
- The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
- Entropy will decline elsewhere, provided it increases by at least as much elsewhere.
Pool smell is due, not to chlorine, but to chloramines, chemical compounds that build up in pool water when it is improperly treated. Chloramines result from the combination of two ingredients: (a) chlorine disinfectants and (b) perspiration, oils and urine that enter pools on the bodies of swimmers.
In the equation given above, there is conservation of MASS, CHARGE AND ENERGY.
These three parameters are usually conserved during the course of chemical reactions. When any of these parameter experience a reduction during the course of chemical reaction, such loss is always gained by other elements involved in the same reaction, so that at the end of the day, they are not considered as lost.
