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andreyandreev [35.5K]
1 year ago
9

How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos

e in the preparatory phase?
Chemistry
1 answer:
svet-max [94.6K]1 year ago
7 0

In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place.  During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).

The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

For more information on phosphorylation click on the link below:

brainly.com/question/7465103

#SPJ4

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How many significant figures are in the measurement, 0.005890 g?
pantera1 [17]

4 significant figures. Use a rule called the Atlantic-Pacific rule where if the period is absent (Atlantic), you would start counting the numbers from right to left. If the period is present (Pacific), then start counting the numbers from left to right. Since the period is present, start from 5 and there are 4 digits including 5.

6 0
3 years ago
A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267
kirill115 [55]
Ok so, remember that t<span>he average atomic mass is what is seen on the periodic table. It is the average mass of all of the isotopes with their frequency taken into account. What you need to do is add the products of the masses and frequencies Just like this:</span>

<span>0.903*267.8 + 0.097*270.9
When you add it the result is what you are looking for</span>
4 0
3 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
A fuel cell designed to react grain alcohol with oxygen has the following net reaction:
sertanlavr [38]

Answer:

b. E = 2,28V

Explanation:

The maximum work is the same than ΔG. As ΔG could be written as:

ΔG = nFE <em>(1)</em>

Where n is moles of electrons transferred, F is faraday constant (96485 J/Vmol) and E is the voltage of the cell.

For the reaction:

CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l)

The oxidation state of C in CH₃OH is -2 but in CO₂ is +4, that means transferred electrons are +4 - -2 = <em>6e⁻</em>

Replacing in (1):

1320x10³ J = 6mol e⁻×96485J/Vmol×E

<em>E = 2,28V</em>

<em></em>

I hope it helps!

8 0
3 years ago
Use the normal boiling points: propane, C3H8, –42.1˚C; butane, C4H10, –0.5˚C; pentane, C5H12, 36.1˚C; hexane, C6H14, 68.7˚C; hep
svlad2 [7]

Answer:

The answer to your question is 126.1°C

Explanation:

                  Boiling point         Difference of boiling points  

C₃H₈             - 42.1°C

C₄H₁₀             -   0.5°C                 41.6 °C

C₅H₁₂               36.1°C                  36.6°C                       41.6 - 36.6 = 5°C          

C₆H₁₄               68.7°C                 32.6°C                        36.6 - 32.6 =4°C

C₇H₁₆               98.4°C                 29.7°C                        32.6 - 29.7 = 2.9°C

We can observe on the table that the difference of boiling points diminishes 1°C when the hydrocarbon has one more carbon, then the difference of temperature between the hydrocarbon of 8 carbons and the hydrocarbon of 7 carbons must be 2°C.

So, this difference is 29.7°C - 2°C = 27.7°C.

And the boiling point of octane is approximately 98.4 + 27.7°C = 126.1°C

7 0
3 years ago
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