The glucose uniporter GLUT1 has a Km of 1.5 mM, and GLUT2 has a Km of...
The glucose uniporter GLUT1 has a Km of 1.5 mM, and GLUT2 has a Km of 20mM for D-glucose. At a concentration of 5 mM, what is the rate of GLUT1 transport relative to GLUT2 transport? The Vmax may be assumed to be the same for both.
a. 3.8-fold slower
b. equal
c. 3.8-fold faster
Please show the steps I would take to find the answer
Homework Answers
Answer: Option A is
correct
Explanation:
Km = Substrate concentration at which the reaction rate is half the
Vmax.
Km is inversely proportional to the enzyme affinity.
Greater the Km, lower is the affinity.
Lesser the Km, greater is the affinity.
In the given case, GLUT-1 has less Km compared to GLUT-2.
So, GLUT-2 will have less reaction rate at the given [S] as
compared to GLUT-1.
The rate of transport for each uniporter can be calculated using the Michaelis-Menten equation:
v = (Vmax*[S]) / (Km + [S])
Where v is the rate of transport, [S] is the substrate concentration, Vmax is the maximum rate of transport, and Km is the Michaelis constant.
For GLUT1:
v1 = (Vmax*[S]) / (Km1 + [S])
For GLUT2:
v2 = (Vmax*[S]) / (Km2 + [S])
We are given that [S] = 5 mM, Km1 = 1.5 mM, and Km2 = 20 mM. We are also told that Vmax is the same for both transporters. Therefore, the relative rate of transport for GLUT1 compared to GLUT2 can be calculated as:
v1 / v2 = [(Vmax*[S]) / (Km1 + [S])] / [(Vmax*[S]) / (Km2 + [S])]
v1 / v2 = (Km2 + [S]) / (Km1 + [S])
v1 / v2 = (20 + 5) / (1.5 + 5)
v1 / v2 = 25 / 6.5
v1 / v2 = 3.8
Therefore, the answer is (a) 3.8-fold slower.
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2. Does the data presented in the paper support the hypothesis
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3.According to Chaudhuri, what is the potential role of thew
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