6. Give two separate definitions of the Michaelis constant. Discuss what each te

Question

6. Give two separate definitions of the Michaelis constant. Discuss what each tells you about the reaction it describes K the Michaelis constant, is the ratio (k kz)/k It is also equal to the substrate concentration at which the rate of the enzyme-catalyzed reaction is half the maximum. biology: Km "usually also gives an indication of the intracellular concentration of the substrate. 7. Discuss the advantages and disadvantages of plots of [S] vs Vo and 1/S] vs 1Ne Do these change if you are using modern computational tools or paper and pencil? Linear plots make for easier calculation in the absence of computational tools The following data represents an enzymatic reaction in the absence of inhibitor (1) and in the presence of two different inhibitors (2) and (3). Use the data to determine Km and Vmax for each set of conditions and determine the mechanism of each inhibitor 8. SI (mM) (2)VoHM-- 1.17 2.10 4.0 5.7 7.2 7.5 3)Vo(HM- 0.77 1.25 2.00 2.50 2.86 3.3 3.3 2.5 6.3 7.6 9.0 3.1 10 20 Km 10 9. Show algebraically that the substrate concentration is equal to the K, when initial reaction velocity equals exactly one half of the maximum reaction velocity Divide by Vmax Substitute ½ for Vow- 2S/ IS) Multiply by 2 Multiply by KmIS Subtract [S) mIS 21S mIs) 10-Question 1 proves that a substrate concentration equal to 1 times the Km results in a reaction velocity of 50% of

Explanation / Answer

Ques 8)

we will plot graph to calculate the asked quantities in the question

(i) take inverse of [S] and Vo

(ii) plot 1/ [S] on x-axis and 1/ Vo on y axis

(iii) calculate regression equation , Vmax = 1/ intercept

Km / Vmax = slope

Let us see the graphs

a) a/[S] v/s Vo without inhbitor

y = 0.301x + 0.098

Vmax = 1/ intercept = 1/ 0.098 = 10.20

Km / Vmax = slope

Km = slope X vmax = 0.301 X 10.20 = 3.07 = 3.1

2) inhibitor 1

y = 0.754x + 0.1

Vmax = 1/ intercept = 1/ 0.1= 10

Km / Vmax = slope

Km = slope X vmax = 0.754X 10 = 7.54 = 7.5

As Vmax is almost same as without ihibitor so must be competitive inhibitor

3) inhibitor 2:

y = 0.998x + 0.300

Vmax = 1/ intercept = 1/ 0.3= 3.33

Km / Vmax = slope

Km = slope X vmax = 0.998 X 3.33 = 3.32

Vmax less than Vmax , mixed inhibition

without inhibitor [S] vo 1/[S] 1/V0 1 2.5 1 0.40 2 4 0.5 0.25 5 6.3 0.2 0.16 10 7.6 0.1 0.13 20 9 0.05 0.11

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