NEED HELP BY TONIGHT PLEASE!!! Please help me, I am deeply confused in this clas
ID: 134481 • Letter: N
Question
NEED HELP BY TONIGHT PLEASE!!!
Please help me, I am deeply confused in this class and trying so hard to understand. Please give detailed answers. I know some of these might be confusing but your help will really help me learn. The teacher requires very very detailed answers and he gives nearly no (and I mean literally nearly zero) partial credit so I'm dying here. Thank you sooooooo much, you have no idea.
Also, I am nearly leaglly blind and I am dyslexic so if you could please write clear it would help so much. I just want to learn like everyone else. SUPER SUPER THANK YOU
An expedition to Zambia has discovered a new species of bacteria that seems to rely on the catabolism of this hexose: 2. CH2OH ?? ?? ?? ?? You discover that this bacteria has evolved a metabolic pathway that is analogous to glycolysis in eukaryotic cells, which you have named allolysis. The allolytic pathway seems to feature similar enzymes as glycolysis, but ones that have evolved to use this stereoisomer of glucose, and its metabolic products, as substrates. [7 pts] Indicate the chiral carbon centers in the diagram above [1 pt] a. b. Based on this information, do you suspect that the products of allolysis could be used as a substrate for steps of aerobic metabolism (IE. pyruvate oxidation, citric acid cycle, and oxidative phosphorylation)? Why or why not? 12 pts]Explanation / Answer
Answer-a: This is Allose sugar. C3 is chiral centre.
Answer-b: Yes, products of allolysis can be used as a substrate for the aerobic pathways becase this pathway is analogous to glycolysis. This means that, even though the starting substrate for both the pathways can be different but both pathways will converge at a point where they will continue their respective pathway having similar products. SO, according to the question, both pathways must produce pyruvate as their end product and it can further go on to carry out aerobic metabolic pathways.
Answer-c: There will be net gain of 20 ATP molecules.
2 ATP gained from glycolysis
6 NADH gives 15 ATP [6 x 2.5]
2 FADH2 gives 3 ATP [2 x 1.5]
Answer-d: Due to presence of isomerase, aldose sugar gets converted into ketose sugar, allose into allulose. Keto sugar allulose-6-phosphate will be converted into fructose-6-phosphate by epimerase enzyme which will continue the pathway to yied pyruvate as end product.
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