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2 years ago

In the reaction catalyzed by the pyruvate dehydrogenase complex, the two carbons constituting the acetyl group are

Chemistry

1 answer:

lara [203]2 years ago

4 0

Transferred to the lipoamide by an earlier intermediate in the process.

The pyruvate dehydrogenase complex (PDC) is a mitochondrial multienzyme complex composed of three different enzymes

<h3>What reaction is catalyzed by enzyme 2 of the pyruvate dehydrogenase complex ?</h3>

the pyruvate dehydrogenase complex is the bridge between glycolysis and the citric acid cycle

Five coenzymes are used in the pyruvate dehydrogenase complex reactions: thiamine pyrophosphate or TPP, flavin adenine dinucleotide or FAD, coenzyme A or CoA, nicotinamide adenine dinucleotide or NAD, and lipoic acid.
during the reactions catalyzed by the pyruvate dehydrogenase complex, it is first reduced to dihydrolipoamide, a dithiol or the reduced form of the prosthetic group, and then, reoxidized to the cyclic form.

Learn more about Pyruvate dehydrogenase complex here:

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Can anyone give me tips on how to present 4-6 mark questions in science assessments? I understand the logic but I find it hard t

Vlad1618 [11]

Answer:

- start by recycling the question

- explain the topic using key words related to that topic

- include a few examples/data if it's using a graph/photo

5 0

3 years ago

Some important points for hadpicking<br>

Svet_ta [14]

Answer:

Handpicking right?

It is a very simple and easy method of separation. It doesn't require any kind of equipment to proceed. It takes very less time when performed for small quantity of mixture. It doesn't require any kind of preparation

8 0

3 years ago

A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25Â°C. Calculate the p

natima [27]

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of $HC_2H_5CO_2$ = 0.1917 M

Volume of $HC_2H_5CO_2$ = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of $HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422$

As we known that at equivalent point, the moles of $HC_2H_5CO_2$ and KOH are equal.

So, Moles of KOH = Moles of $HC_2H_5CO_2$ = 0.0422 mol

Now we have to calculate the volume of KOH.

$\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}$

$\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}$

$\text{Volume of }KOH=0.00754$

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

$HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O$

Moles of $C_2H_5CO_2K$ = 0.0422 mol

$\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M$

At equivalent point,

$pH=\frac{1}{2}[pK_w+pK_a+\log C]$

Given:

$pK_w=14\\\\pK_a=4.89\\\\C=0.1855M$

Now put all the given values in the above expression, we get:

$pH=\frac{1}{2}[14+4.89+\log (0.1855)]$

$pH=9.08$

Therefore, the pH at equivalence is, 9.08

6 0

3 years ago

What volume of methanol should be used to make 1.25 L of a 21.0% by volume solution

Marat540 [252]

0.2625 litres is the volume of methanol should be used to make 1.25 L of a 21.0% by volume solution.

Explanation:

Given that:

Volume of solution = 1.25 litres

methanol to be present in the composition of 21% of the volume.

To make 21 % solution by volume of a substance it means that 21 L of the substance is present in 100 L of the solution which is made upto 100 L by adding the solvent.

So, 21 L in 100 L will make 21% solution

x litres of methanol in 1.25 L of solution

So, $\frac{21}{100}$ = $\frac{x}{1.25}$

x = 0.2625 litres

hence, 0.2625 litres of the methanol will be added and the volume will be made to 1.25 litres so that 21% solution by volume gets form.

4 0

3 years ago

If a gas at 4.5 atm expands to 3.0 L, what will the volume be at 13 atm?

lozanna [386]

Answer:

1.04 L

Explanation:

4.5 atm x 3 L/13 atm

7 0

3 years ago