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klio [65]
2 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]2 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) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.

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The cytochromes are <u>proteins that contain heme prosthetic groups</u>. Cytochromes <u>undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome</u>:

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The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes <em>b</em> and <em>c₁</em>, and a Fe-S center. This second complex reduces cytochrome <em>c</em>, a water-soluble membrane peripheral protein. Cytochrome <em>c</em>, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes <em>a</em>, <em>a₃</em> and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.

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7 0
3 years ago
The limestone used in building the Egyptian pyramids was used because it is _____________
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2 years ago
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The right line is a 90° clockwise rotation of the left line about the origin. Click the 90° clockwise button. Are these lines th
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Answer:

Switch the coordinates and change the sign of the second one by multiplying it by negative 1.

Explanation:

Here are some examples and a more general way to understand the problem.

Consider the point (1,1), a 90 degree rotation clockwise about the origin would move it into the 4th quadrant.

The new point is (1,-1) , similarly (-4,2)-> (2,4), (-4,3)-> (3,4)

We take a point p= (x,y) the the result of rotation p 90 clockwise about the orgin is a new point p'=(x',y')= (-y, x). .

In the case of p=(1,0) the new point is p'= (0, -1)

One can use a matrix where the first row is cos(a), sin(a) and the second row is

-sin(a) cos(a) for any clockwise rotation of a degrees about the origin.

If we let a=90 degrees we have

[0 1] as the first row and [-1 0] as the second row. So the matrix is:

|0 1|

|-1 0|

Call that matrix M

So a point p= (x,y) can be multiplied by M as follows Mp=p' where p' is the rotated point.

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x'=1*0+0*1=0

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How about the point on the y axis (0,1), it should go to the point (1,0)

0*1+1*1=1 and -1*0+0*1 gives you the pont (1,0) ( we don't see the negative sign because -0 is just 0)

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3 years ago
In the following reaction, how many grams of silver can be produced from 49.1 g copper?
notsponge [240]

166.4 g Ag grams of silver can be produced from 49.1 g of copper.

<h3>What is a mole?</h3>

A mole is a very important unit of measurement that chemists use. A mole of something means you have 602,214,076,000,000,000,000,000 of that thing, like how having a dozen eggs means you have twelve eggs.

1 mol Cu + 2 mol AgNO_3 → 2 mol Ag + 1 mol Cu(NO3)2

63.55 g Cu —> 2 x 107.688 g Ag

63.55 g Cu gives 215.376 g of Ag

So, 49.1 g Cu —> \frac{215.376 g X 49.1 g}{63.55 g}

= 166.4 g Ag

Hence, 166.4 g Ag grams of silver can be produced from 49.1 g of copper.

Learn more about moles here:

brainly.com/question/26416088

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