The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because<u> fructose 1,6-bisphosphate can undergo no other reactions than those of glycolysis.</u>
<h3>
What is phosphorylation?</h3>
- The crucial process of glycolysis involves the breakdown of glucose into two molecules of pyruvate. It involves a number of steps and many enzymes.
- It takes place over the course of ten phases, demonstrating how important and crucial phosphorylation is to the production of the final goods. Step 1 of the preliminary step (first half of glycolysis) and step 6 of the payout phase reactions are started by phosphorylation (second phase of glycolysis).
- Because fructose-6-phosphate cannot cross the cell membrane, it is forced to remain inside the cell. Step 3 involves phosphorylation, when fructose-6-phosphate is changed into fructose-1,6-bisphosphate.
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Yes, the taiga ecosystem is home to evergreens. The taiga system is a slightly warmer ecosystem then Tundra. You kind find a lot of taiga in Canada and Parts of Europe.
I hope this helps!
~kaikers
Answer: The correct answer is choice D.
Explanation: We can see on the cladogram that these two are in the same branch and lead to the same common ancestor. These two happen to share the most recent common ancestor, however.
Strong bases
Explanation:
KOH, CsOH and Ca(OH)₂ are all strong bases because they dissociate completely in aqueous solution.
A base is a substance that produces excess hydroxyl ions in aqueous solutions.
Strong bases would dissociate completely when in solution.
Bases of alkali and alkali earth metals are classified as strong because they exhibit full ionic characteristics.
They give full metals ions and hydroxide in solution.
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