It's Mutualism, since they share a mutual relationship
Answer:
(E) Phosphoglucose isomerase (phosphohexose isomerase or glucose-6-phosphate isomerase)
Explanation:
Glyocolysis is break down of glucose into pyruvate.
Isomerization:The second step of glycolysis involves the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P). This reaction occurs with the help of the enzyme phosphoglucose isomerase (PI). As the name of the enzyme suggests, this reaction involves an isomerization reaction.
Yeast mutant lacking phopshoglucose isomerase unable to performed this step and that's why it unable to grow in media containing glucose while in fructose break down this step is not required so it is able to grow in media containing fructose.
Deletion of phosphoglucose isomerase, prevents growth on glucose; therefore, phosphoglucose isomerase mutant is able to grown in a fructose medium.
Answer: the steel wool in the 3.0 M HCl reacts fastest.
Explanation:
Answer:
Here are the answers:
a. 4 Cell determination as an issue in the *rest is missing*
b. 4 They assumed that different ways of separating an embryo into two parts would be equivalent as far as the fate of the two parts was concerned.
c. 4 I and III only
Explanation:
The passage demonstrates the importance of two factors in the development of an embryo: cleavage planes of division of embryonic cells and cell differentiation.
Cleavage Planes:
Cleavage basically refers to the division of the zygote into a large number of cells called blastomeres. Cleavage planes are geometrical lines or orientations along which cleavage takes place. Since, all embryonic cells are the precursors of some type of body cells, the cleavage planes determine if the cells are adequate for growth and development.
Cell Differentiation:
Cell differentiation is the transition of an undifferentiated cell into a specialized one. For example, stem cells are undifferentiated cells that develop into progenitor cells that mature into a specific cell lineage. For an embryo to regenerate, the presence of adequate embryonic stem cells is crucial. Embryonic stem cells are present in abundance before the gastrulation phase of embryonic development, after which they rapidly start differentiating.
Answer:
The correct option is 3. "At pH 6.5 the enyzme is 50% active"
Explanation:
For the titratable group to be protonated and cause the enzyme to be in the active state, it needs to have gained a hydrogen cation (H+). In order for that to happen, there must be enough hydrogen cations in the environment of the enzyme, and hence, an acidic pH is required in this case.
