In reaction 1 of the Krebs cycle, acetyl‑CoA formed in the pyruvate dehydrogenase reaction condenses with the four‑carbon compound to form <em>citrate </em>with the elimination of coenzyme A. Since the product has three carboxyl groups, this pathway is referred to as the cycle. In reaction 2 of the Krebs cycle, this product then undergoes to form<em> isocitrate. </em>The enzyme is called aconitase because the compound cis‑aconitate is the <em>intermediate product</em> of the reaction. Reaction 3 eliminates CO2 to form the five‑carbon dicarboxylic acid <em>α-cetoglutarate. </em>Oxidation also occurs, with electrons transferred from the substrate to <em>COO-</em> . Consequently, this reaction is an oxidative decarboxylation.
In the image, you can see the reaction 2 in Krebs cycle is a two steps reaction with an intermediate cis-aconitase and a product called isocitrate.
Answer:
n = 0.0022 mol
Explanation:
Moles is denoted by given mass divided by the molar mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molar mass .
From the information of the question ,
w = 0.108 g
As we known ,
The molar mass of titanium = 47.867 g / mol
The mole of titanium can be caused by using the above relation , i.e. ,
n = w / m
n = 0.108 g / 47.867 g / mol
n = 0.0022 mol
Answer:1.
Explanation: This reaction is catalyzes by pyruvate dehydrogenase. Pyruvate being the end product of glycolysis has many fates after glycolysis,one of which is to enter the TCA(Tricarboxylic acid cycle) cycle. It is first converted to actetate by the action of pyruvate dehydrogenase. This enzyme converts pyruvate to acetate releasing CO2 and NADH because this oxidative decarboxylation of pyruvate is coupled with reduction of NAD+ which can feed into the electron transport chain.
Answer: The percent yield of the reaction is 77.0 %
Explanation:
According to stoichiometry:
2 moles of 
produces = 2 moles of 
2.18 moles of is produced by=
of
Mass of =
percent yield =
Answer:
i think its A
Explanation:
bc the nucleus is the very center of whatever the organism is.
