A meadow ecosystem includes many species of grasses and small shrubs, several herbivores, and a few carnivores. A fungus coloniz
2 answers:
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The correct answer is "transmission electron microscope".
The transmission electron microscope can magnify the organelles of the cell clearly; as in light microscopes wherein most organelles cannot be visualized. TEM works by using a beam of electrons to pass through a specimen to make a clear image. Because of its superior magnification, mitochondrial defects can be visualized easily.
Attached is a sample picture of a transmission electron microscope image of a mitochondria.
The transmission electron microscope can magnify the organelles of the cell clearly; as in light microscopes wherein most organelles cannot be visualized. TEM works by using a beam of electrons to pass through a specimen to make a clear image. Because of its superior magnification, mitochondrial defects can be visualized easily.
Attached is a sample picture of a transmission electron microscope image of a mitochondria.
Answer:
B. Glucokinase acts on glucose only at high glucose concentrations.
Explanation:
To understand the answer, first, we have to know that the Michaelis Menten kinetics requires to have a constant concentration of enzyme quantity to see what happens with the rate.
To determine that we can evaluate how much time takes the total consumption of substrate or the time taken to obtain a product.
We know that when an enzymatic reaction occurs, the enzyme accelerates the reaction by minimizing the activation energy.
Suppose we have the enzyme on this case glucokinase and hexokinase, and we going to make them react with the substrate, in this case, the glucose.
Any of both enzymes will be abbreviated as E, and glucose as a substrate will be abbreviated as S. And we have the following reaction:
E + S = ES = E + P
When ES is the enzyme-substrate complex, and P is the product of the reaction, in these case, when the hexokinase or glucokinase takes the glucose the product is the phosphorylation of the glucose.
We can see this reaction, for both enzymes or any enzyme as a graph (watch image attached).
This image explains that when you have a constant concentration of enzyme and you add more substrate, the velocity will rise but, when the reaction reaches the maxim velocity (Vmax), this value will be constant, and it is because all the enzyme active site are full with substrate already, so even if you add, the reaction will not accelerate.
Take a look at the graph and note the Km value related to the
1/2 Vmax. And the Km is finally just the substrate concentration when the half value of velocity is reached.
So, for any enzyme reaction, the Km value is the concentration of the substrate when the half higher value of velocity will be reached.
So for glucokinase, who has a Km value of 10 mM, and hexokinase with a Km = 0.1 mM we can say that glucokinase requires more quantity of substrate (glucose) to reach the maximum velocity, but hexokinase reaches the maximum velocity a very small quantity of glucose concentration. And that behavior has to be related to the higher affinity of glucose with the hexokinase instead of glucokinase.
For that reason the answer must be :
B. Glucokinase acts on glucose only at high glucose concentrations.
As you must know, glucokinase is an enzyme produced on the liver and has a higher specify for D-glucose instead for other hexoses, the glucokinase has a poor activity at lower blood concentration of glucose ( fast periods, before breakfast).
