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.
Using the given formula, the density of the material is 2.015 g/mL
<h3>Calculating Density </h3>
From the question, we are to determine the density of the material
From the given formula
Density = Mass / Volume
And from the given information,
Mass = 65.5 g
and volume = 32.5 mL
Putting the parameters into the equation,
Density = 65.5/32.5
Density = 2.015 g/mL
Hence, the density of the material is 2.015 g/mL.
Learn more on Calculating density here: brainly.com/question/24772401
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The answer is A. planning a hiking trip
A topographical map would not help studying plant growth over time unless you are looking for a better altitude to plant said plants.
A topographical map would not help studying rainfall for one year unless your location was <em>really</em> so high or low that it affected your weather
And most of all, a topographical map would not be useful for planning a cuise across the Atlantic Ocean because the elevation of the sea is zero!
A topographical map <em>would</em> be useful for planning a hiking trip because there are many factors and details that a hike should have. Which includes height, distance, paths, and elevation.
An intensive property is a property that does not change depending on how much mass of it you are considered. An example of an intensive property is density. No matter how much water you examine, the density of the sample will be 1g/cm³.
