Im going to say that its <span>d. The head of the phospholipid, which is hydrophilic, helps to control the movement of large hydrophobic molecules, and the tails of the phospholipid, which are hydrophobic, help to control the movement of large hydrophilic moleculeus. </span>
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.
To learn more about phosphorylation with the given link
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Catalase is an enzyme, O2 and H2O are substrates, and H2O2 is a reactant.
Catalase is a substrate, O2 and H2O are substrates, and H2O2 is an enzyme.
Catalase is a substrate, O2 and H2O are products, and H2O2 is an enzyme.
<span>Catalase is an enzyme, O2 and H2O are products, and H2O2 is a substrate.</span>
The answer is D, all of the above.
Answer:
i think distance and mass
Explanation:
