Hexokinase can use glucose or fructose as a substrate. the km forglucose is 0.15 mm and for fructose km = 1.5 mm.
which of the following statement is true?
a. for which substrate does hexokinase have the highest affinity? why?
b. calculate v as a percentage of vmax for each substate when [s]equals 0.15mm.

a) glucose because Km is less for glucose.

b) For glucose, V as a percentage of Vmax = 50 %

   For fructose, V as a percentage of Vmax = 9 %

Explanation:

a) Hexokinase has highest affinity for glucose because it's Km for glucose is low.

When enzyme [E] binds a substrate [S] an [ES] complex is formed. It is represent as under:

                  [E]  +   [S]    ⇌    [ES]   ⇌   [E] +  [P]

As soon as  [ES] is formed it may either form [E] & [P] in a forward reaction or dissociate back into [E] & [S]. Km is known as dissociation constant which represents the extent to which [E] & [S] are being formed. A higher value of Km represents formation of more [E] & [S] back from [ES]. It also means that binding affinity of enzyme with substrate is less or we can say that Km is inversely proportional to binding affinity.

                    Km  ∝  1 / binding affinity.

So, it simply means that lower the Km more will be the affinity of the enzyme for the substrate. In the given question, Km for glucose is 0.15 mM which is less than fructose (Km for fructose = 1.5 mM) that is why glucose is the answer.  

b) The formula for Michaelis–Menten equation is as under:

                    V = Vmax × [S] / ([S] + Km)

                    V / Vmax =  [S] / ([S] + Km)

Here, for glucose Km = 0.15 mM and [S] = 0.15mM.

So V as a percentage of Vmax for glucose will be as under:

                    V / Vmax =  0.15/0.15 + 0.15

              →    V / Vmax =  0.15/0.30

              →    V / Vmax =  0.5 = 1/2 = 50 %

Here, for fructose Km = 1.5 mM and [S] = 0.15mM.

So V as a percentage of Vmax for fructose will be as under:

                    V / Vmax =  0.15/0.15 + 1.5

              →    V / Vmax =  0.15/1.65

              →    V / Vmax =  0.09 = 9/100 = 9 %

interphase   is the cell cycle stage in which the cell acquires nutrients, creates and uses proteins and other molecules and starts the process of cell division by replicating the dna. it is the longest stage in the eukaryote cell cycle.

the following   can go wrong during interphase:   a base pair could be mismatched during the duplication process causing a mutation.   correct answer: c

explanation:

sheep is a herbivore animal whereas grasshopper belongs to the class chordates but both of them feed on greens.  

if grasshopper and sheep feeding on a pasture land, i do not think both of them will have any impact on each other. however, the number of the species can make difference. since sheep feed on greens, there is a possibility there left no pasture land for the grasshoppers or they may get extinct.