In solution, 80% of the fructose 6-phosphate is in the beta-anomeric form and 20% is in the alpha-anomeric form...
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It should be 2,5 -Anhydro...(in beta configuration).© BrainMass Inc. brainmass.com October 24, 2018, 6:23 pm ad1c9bdddf
Both derivatives can bind to PFK about the same -- indicative from the Km. However, only the mannitol derivative can be phosphorylated. The glucitol derivative can't be phosphorylated at all. On the contrary, it is an ...
The solution is a succinct explanation.
Specific rotation of glucose molecule (anomers)
Q1. ?-D-Galactopyranose has [?]D = +150.7o, and ?-D-galactopyranose has [?]D = +52.8o. When either of them is dissolved in water and allowed to reach equilibrium, the specific rotation of the solution will be +80.2o. What are the percentages of each anomer at equilibrium? (Original formatted question attached.)
I really have no ideal how to attempt this question but I will provide a solution:
150.7 + 80.2 = 230.9
150.7/ 230.9 * 100 % = 65.3%
52.8 + 80.2 = 133
52.8/133 * 100% = 40%
Right away, I know this answer is wrong because the percentages of the two anomers do not add up to 100%.View Full Posting Details