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KATRIN_1 [288]
2 years ago
14

What is the fate of glucose 6‑phosphate, glycolytic intermediates, and pentose phosphate pathway intermediates in this cell? Gly

colytic intermediates can only enter the pentose phosphate pathway through conversion to pyruvate and subsequent gluconeogenesis. The oxidative pentose phosphate pathway reaction catalyzed by glucose 6‑phosphate dehydrogenase is slowed down. Most of the glucose 6‑phosphate enters the pentose phosphate pathway. One molecule of glyceraldehyde 3‑phosphate and two molecules of fructose 6‑phosphate are used to generate three molecules of ribose 5‑phosphate. Most of the glucose 6‑phosphate enters the glycolytic pathway and is converted to fructose 6‑phosphate and glyceraldehyde 3‑phosphate. Under the given conditions, all triose phosphates are converted to pyruvate by the glycolytic pathway.
Chemistry
1 answer:
madreJ [45]2 years ago
3 0

Answer:

The Phosphorylated  glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And  represented the fate of  glucose -6-phosphate.

The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .

These intermediates are converted to ribose 5-phosphates in the presence of transketolase  and transaldolase enzymes.And they are finally   converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.

Basically the phosphate pathway reaction is very slow due to enzyme catalysis.

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A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the person’s weight, in pounds, when expressed to
Sauron [17]

Answer:

87 kilograms =191.8 pounds

4 0
2 years ago
Read 2 more answers
How many grams of CaF2 are present in 1.25 L of a 0.15 M solution of CaF2? How do I find the grams I am confused on that part?
dybincka [34]

Answer:

Mass = 14.64 g

Explanation:

Given data:

Volume of solution = 1.25 L

Molarity of Solution = 0.15 M

Mass of CaF₂ = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

We will calculate the number of moles of CaF₂ and then determine the mass by using number of moles.

0.15 M =  number of moles of solute / 1.25 L

number of moles of solute = 0.15 M ×  1.25 L

number of moles of solute = 0.1875 mol/L × L

number of moles of solute = 0.1875 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.1875 mol ×78.07 g/mol

Mass = 14.64 g

7 0
2 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is
GenaCL600 [577]
1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V  = 7.9 atm
3 0
2 years ago
Read 2 more answers
Identify structure(s) for all constitutional isomers with the molecular formula C4H8 that have one double bond. Select all that
Alchen [17]

Answer:

a) But-1-ene

b) E-But-2-ene

c) Z-But-2-ene

d) 2-Methylpropene

Explanation:

In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u>C_4_H8 in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>

<u />

For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>

<u />

Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>

<u />

See figure 1 to further explanations.

I hope it helps!

<u />

<u />

8 0
2 years ago
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