Answer:
c. parasite
Explanation:
the barnacle benefits from the host while the host is harmed from the barnacle
Sports drinks are the best choice.
Answer:
Dissolve 47.025 grams of sucrose in enough water to produce 250 ml of solution.
Explanation:
You need to prepare 250. mL of a 0.550 M aqueous solution of sucrose, C12H22O11 (aq),
which is used frequently in biological experiments.
Based on your answer above, what is the value of x?
Solution:
A 0.550 M aqueous solution contains 0.550 mole of sucrose per liter of solution.
250 ml = 0.250 liter
Thus a 0.250 liter of a 0.550 M aqueous solution of sucrose contains 0.250 * 0.550 = 0.1375 mole of sucrose
Recall:
Mass = number of moles * mass of 1 mole
mass of 1 mole of C12H22O11
= 12(12) + 1(22) + 16(11)
= 144 + 22 + 176
= 342 grams
Mass = 0.1375 * 342 = 47.025 grams of sucrose.
Dissolve 47.025 grams of sucrose in enough water to produce 250 ml of solution.
Answer:
C. THE CONVERSION OF FRUCTOSE 1,6-BISPHOSPHATE to fructose- 6- phosphate is not catalyzed by phosphofructokinase -1, the enzyme involved in glycolysis.
Explanation:
This statement is true as the enzyme involved in this step is FRUCTOSE-1,6-BISPHOSPHATASE.
Gluconeogenesis is the coversion of non-carbohydrate molecules (lactic acid, amino acids, glycerol) through the pyruvic acid into glucose in the cells.
This process takes place mainly in the liver and occurs during periods of fasting, starvation, low carbohydrate diets.
The pathway of gluconeogenesis involves eleven steps of enzymatic catalyzed reactions.
In the conversion of fructose 1,6- bisphosphate to fructose-6-phosphate is catalyzed by fructose 1,6-bisphosphatase and not by phosphofructokinase -1 which is involved in glycolysis. This step is a rate-limiting step of the pathway.
The conversion of glucose-6-phospahte to glucose is not catalyzes by hexokinase but glucose -6- phosphatase.
