A teenager finishes eating a hot fudge sundae and a milk shake. As her blood sugar rises, the liver recruits both Glucokinase (km = 10.0m M) and Hexokinase (km = 0.10 mM) to metabolize the glucose. Which of the following is a correct statement about the kinetics of these two enzymes?
A. Hexokinase will metabolize 100 times more glucose molecules
B. Hexokinase will lower the energy of activation more than glucokinase
C. Hexokinase has a much higher affinity to glucose than glucokinase
D. Hexokinase will have a higher Vmax than glucokinase
E. Hexokinase will be induced more than glucokinase by the rising concentration of glucose
Answer- C- Hexokinase has a much higher affinity for glucose than Glucokinase. It is not about number of Glucose molecules metabolized, or about energy of activation, Vmax or induction. Km signifies the affinity of the enzyme for its substrate.
Km and its significance
- The Michaelis constant Km is the substrate concentration at which Vi is half the maximal velocity (Vmax/2) attainable at a particular concentration of enzyme (Figure-1)
- It is specific and constant for a given enzyme under defined conditions of time, temperature and pH
- Km determines the affinity of an enzyme for its substrate, lesser the Km more is the affinity and vice versa
- Km value helps in determining the true substrate for the enzyme.
Figure-1- Km is the amount of substrate required to achieve the half maximum velocity (1/2 Vmax)
The statement “Glucokinase (km = 10.0m M) and Hexokinase (km = 0.10mM) to metabolize the glucose”, signifies that Glucokinase requires more substrate (glucose) concentration (10.0mM) to achieve the half maximum velocity (1/2 Vmax) in comparison to Hexokinase. At a very small glucose concentration of 0.10mM, half Vmax can be attained in case of hexokinase. Thus Hexokinase has higher affinity for glucose as compared to Glucokinase.Due to this reason only Hexokinase is the considered the enzyme to maintain the intracellular glucose concentration, since it can be active even in the presence of minute glucose concentration, whereas Glucokinase requires larger glucose concentration to be active. It is considered the enzyme to maintain blood glucose concentration. Hexokinase has a low Vmax, it gets saturated quickly by the rising glucose concentration, it is also product inhibited (feed back inhibition). It can not handle high glucose load. Glucokinase has a high Vmax, it can handle larger glucose load, it is not inhibited by the product.Glucokinase is present only in liver and pancreas whereas hexokinase is present in all the cells of the body
The relationship between substrate concentration , V max and Km can be understood by Michaelis-Menten Equation
The Michaelis-Menten equation is a quantitative description of the relationship among the rate of an enzyme catalyzed reaction [v1], the concentration of substrate [S] and two constants, Vmax and km (which are set by the particular equation).
The symbols used in the Michaelis-Menten equation refer to the reaction rate [v1], maximum reaction rate (V max), substrate concentration [S] and the Michaelis-Menten constant (km).
(1) When [S] is much less than km,
The term km + [S] is essentially equal to km. Since V max and km are both constants, their ratio is a constant (k). In other words, when [S] is considerably below km, V max is proportionate to k[S]. The initial reaction velocity therefore is directly proportionate to [S]- Figure-1. It is called first order reaction.
(2) When [S] is much greater than km,
The term km + [S] is essentially equal to [S]. Replacing km + [S] with [S] reduces the equation to
Thus, when [S] greatly exceeds km, the reaction velocity is maximal (V max) and unaffected by further increases in substrate concentration (figure-1). It is zero order reaction.
(3) When [S] = km
Equation states that when [S] equals km, the initial velocity is half-maximal. Equation also reveals that km is a constant and may be determined experimentally from—the substrate concentration at which the initial velocity is half-maximal. Thus km is the amount of substrate[S) required to achieve the half maximum velocity (Vmax/2) .
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