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klio [65]
3 years ago
11

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

h the OH on it in DHAP is labeled B) the carbon with the =O on it in GAP is labeled C) both the carbon with the =O on it in DHAP and the carbon with the OH on it in GAP are labeled D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled E) the label is lost before pyruvate is formed
Chemistry
1 answer:
Oliga [24]3 years ago
3 0

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone  phosphate (DHAP).  In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.  

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

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4.71

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3 years ago
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Explanation:

To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.

Based on this premise, the number of moles of species on the reactant and product side must be the same;

 

      Li   +   Br₂  →    LiBr

Put a,b and c as the coefficient of each species

   aLi  + bBr₂ → cLiBr

balancing Li;

       a  = c

 balancing Br;

        2b  = c

 let a  = 1;

       c  = 1

        b = \frac{1}{2}  

or a = 2, b = 1 , c = 2

        2Li   +   Br₂  →    2LiBr

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Using the same method;

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balancing P;

          a  = c

balancing Cl;

          2b  = 3c

let a = 1;

     c  = 1

      b  = \frac{3}{2}

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  a  = 2, b  = 3, c  = 2

    2P   +   3Cl₂  →    2PCl₃

iii,

   H₂   +   SO₂  →   H₂S   +   H₂O

use coefficients a,b,c and d;

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balancing H;

    2a = 2c + 2d

balancing S;

     b  = c

balancing O

    2b  = d

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      c = 1

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    a  = 3

    3H₂   +   SO₂  →   H₂S   +   2H₂O

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3 years ago
How many grams are in 8.3 moles of CaCl2?
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3 years ago
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What's the balanced equation for cellular respiration?
Alik [6]
<span>Correct answer is:

C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy

But how to get there?

Let's start with simple explanation of what exactly is cellular respiration. 

Cellular respiration is a multistage biochemical oxidation process of organic substances when prime product is energy (ATP - adenosine triphosphate) and other are released waste products. Cellular respiration takes place even if other metabolic processes are stopped, but cellular respiration may differ in particular organism groups.Some reactions during whole process of cellular respiration are similar in all types of living organisms.

Cellular respiration is prime indication of declining living processes.Only viruses which are on the edge of living organism and chemical particle are not performing cellular respiration.But to the point :P

In cellular respiration all substrates which are in the cell might be organic, but mostly we are using sugar oxidation - glucose in the presence of oxygen. Chemical formula of sugar looks like this:

C_{6}H_{12}O_6

Oxygen is just

O_2

so for now we have just part of the equation:

C_{6}H_{12}O_6+O_{2}\Rightarrow

But what would be on the right hand side?

It's quite simple, remember equation of full combustion? If we want to burn something we need oxygen like in the equation, so the product of this equation would be carbon dioxide, water and of course energy (ATP).Carbon dioxide formula looks like this:

CO_{2}

As a reminder water formula:

H_{2}O

Full formula would look like that:

C_{6}H_{12}O_6+O_{2}\Rightarrow\ CO_{2}+H_{2}O+energy

But still as you see this equation is unbalanced, after balancing it would like that:

C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy

At the end I would like to explain one more thing. Energy which has been released during this process is part of high-energy connection which might be used to perform chemical reactions in the cell or to move organism for example in muscles. We need to remember that production of ATP is not happening with 100% efficiency and part of this energy is released as heat.</span>
5 0
3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
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