A scientist finds an organism that cannot move. It has many cells, produces spores, and gets food from its environment. In which
1 answer:
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Answer:
C. Nernst equation
Explanation:
The Nernst Equation is used to find the equilibrium potential of an ion.
It tells you only what the equilibrium potential for an individual ion is, not what the summed effect of all ions is on the membrane potential.
The formula for Nernst Equation is:
=
In
where F is Faraday Constant, R = gas constant, 8.314 J/mol K
, T = temperature, in K, Z stoichiometric number of electrons in the reaction, is the Reduction Potential in V.
Kindly note that, if we have a squid giant axon at rest with normal intracellular and extracellular ion concentrations. If the membrane permeability to K+ ions is increased, the K+ equilibrium potential (Nernst potential) will stay the same.
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).
