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2 years ago

Which enzyme is not part of gluconeogenesis?

Chemistry

1 answer:

lisov135 [29]2 years ago

6 0

Phosphoenol pyruvate enzyme is not part of gluconeogenesis.

<h3>Phosphoenol pyruvate</h3>

The ester formed when pyruvate and phosphate are combined to form an enol results in phosphoenol pyruvate (2-phosphoenolpyruvate, or PEP). As an anion, it exists. In biochemistry, PEP is a crucial intermediary. Involved in glycolysis and gluconeogenesis, it boasts the highest-energy phosphate bond yet discovered in an organism (61.9 kJ/mol). It also plays a role in carbon fixation and the manufacture of a number of aromatic chemicals in plants. In bacteria, it provides energy for the phosphotransferase system. Enolase reacts with 2-phosphoglyceric acid to produce PEP as a result. Pyruvate kinase (PK) converts PEP to pyruvic acid, and this process produces adenosine triphosphate (ATP) via substrate-level phosphorylation. One of the main units of currency for chemical energy in cells is ATP.

Learn more about Phosphoenol pyruvate here:

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Iridium has only two naturally occurring isotopes. Ir-191 with arelative mass of 190.96058 and Ir-193 with a relative mass of192

OLEGan [10]

<u>Answer:</u> The fractional abundance of Ir-191 is 0.372

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Ir-191 be x and that of Ir-193 isotope be (1-x)

<u>For isotope 1 (Ir-191):</u>

Mass of isotope 1 = 190.96058 amu

Fractional abundance of Ir-191 = x

<u>For isotope 1 (Ir-193):</u>

Mass of isotope 1 = 192.96292 amu

Fractional abundance of Ir-193 = (1 - x)

Average atomic mass of iridium = 192.217 amu

Putting values in equation 1, we get:

$192.217=[(190.96058\times x)+(192.96292\times (1-x))]\\\\x=0.372$

Hence, the fractional abundance of Ir-191 is 0.372

6 0

3 years ago

Help me with my work please

blsea [12.9K]

Answer:

i am pretty sure it is compound

8 0

3 years ago

2 Na + 1 Cl2 → 2 Naci<br> How many grams of NaCl are created from 23,2 grams Cl2?

evablogger [386]

37 grams of NaCl (when I mean equivalent I mean the ratio of the equation is 1:2 for moles or Cl2 and NaCl

3 0

2 years ago

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

schepotkina [342]

Answer:

The right solution is " $8.5\times 10^{-19} \ joule$ ".

Explanation:

As we know,

1 mole electron = $6.023\times 10^{23} \ no. \ of \ electrons$

Total energy = $513 \ KJ$

= $513\times 1000 \ joule$

For single electron,

The amount of energy will be:

= $\frac{513\times 1000}{(6.023\times 10^{23})}$

= $8.5\times 10^{-19} \ joule$

8 0

3 years ago

How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat

stira [4]

Answer:

- 278.85 J

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J

8 0

3 years ago