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.
Answer:
atom is the answer I think
Answer: 3 Changes can occur to physical properties of a substance, but the chemical composition of the substance remains the same
Explanation:
Physical change mainly physical properties of a substance while the chemical changes remain unchanged. Once there is a change to the chemistry of the substance, this would be classified as a chemical change.
IT DIDN'T ROLL ROOSTERS CAN LAY EGGSS AHHH
Answer:
The final temperature was 612 °C
Explanation:
Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:
![\frac{V}{T}=k](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7BT%7D%3Dk)
When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:
![\frac{V1}{T1} =\frac{V2}{T2}](https://tex.z-dn.net/?f=%5Cfrac%7BV1%7D%7BT1%7D%20%3D%5Cfrac%7BV2%7D%7BT2%7D)
In this case:
- V1= 5.76 L
- T1= 22 °C= 295 °K (Being 0°C=273°K)
- V2=17.28 L
- T2=?
Replacing:
![\frac{5.76 L}{295 K} =\frac{17.28 L}{T2}](https://tex.z-dn.net/?f=%5Cfrac%7B5.76%20L%7D%7B295%20K%7D%20%3D%5Cfrac%7B17.28%20L%7D%7BT2%7D)
Solving:
![T2 =\frac{17.28 L}{\frac{5.76 L}{295 K}}=\frac{17.28 L*295 K}{5.76 L}](https://tex.z-dn.net/?f=T2%20%3D%5Cfrac%7B17.28%20L%7D%7B%5Cfrac%7B5.76%20L%7D%7B295%20K%7D%7D%3D%5Cfrac%7B17.28%20L%2A295%20K%7D%7B5.76%20L%7D)
T2= 885 °K = 612 °C
<u><em>The final temperature was 612 °C</em></u>