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.
Diffusion is the process of a substance spreading out to evenly fill its container or environment. Rate of diffusion of a gas is inversely proportional to the molar mass of the gas.
Lighter(lower) the molar mass of the gas , faster will be its rate of diffusion and heavier (higher) the molar mass of the gas , slower will be its rate of diffusion.
We have to arrange the given gases from slowest rate of diffusion to fastest rate of diffusion that means we need to arrange gases from higher molar mass to lower molar mass.
Molar mass of given gases are :
Cl = 35.5 g/mol
Xe = 131.29 g/mol
He = 4.00 g/mol
N = 14.00 g/mol
So correct order for slowest rate of diffusion (highest molar mass) to fastest rate of diffusion (lowest molar mass) is :
Xe , Cl , N , He
Xe having the highest molar mass will have the slowest rate of diffusion and He with lowest molar mass will have the fastest rate of diffusion, so option 'c' is correct.
Note : Slowest rate of diffusion = High Molar Mass
Fastest rate of diffusion = Low Molar Mass
