Question

Biochemistry Question

After visualizing VCTN amounts following question 4, and calculating the average VCTN production per timepoint, you obtain the following results [SPBRK) mM UM VCTN/min 2 6.00 - 4 8.19 6 9.01 11.68 10 11.79 13 12.12 15.06 19 14.62 23 14.65 15.41 Use a spreadsheet and graphing software to estimate the Km and the Vmax using a linear transformation. (Hint: You can use the Excel function "=slope(known_ys,known_xs)” to get a slope value, and "=intercept(known_ys,known_xs)" to get an intercept value to a specified number of sig figs. 16 27

Answer 1

[SPBRK] mM	VCTN umol/min	1/[SPBRK] 1/mM	1/VCTN (min/umol)
2	6.00	0.5	0.167
4	8.19	0.25	0.122
6	9.01	0.167	0.111
8	11.68	0.125	0.0856
10	11.79	0.1	0.0848
13	12.12	0.0769	0.0825
16	15.06	0.0625	0.0664
19	14.62	0.0526	0.0684
23	14.65	0.0435	0.0683
27	15.41	0.0370	0.0649

LINEVEAVER BURK PLOT iv 1/VCTN (min/umol) y = 0.225x + 0.060 R = 0.961 00 1/VCTN 1/[SPBRK (min/um ] 1/mM ol) 0.51 0.167 0.25 0.122 0.167 0.111 0.0856 0.0848 0.0769| 0.0825 0.0625 0.0664 0.0526 0.0684 0.0435 0.0683 0.037 0.0649 0.125 0.1 00 O 0.1 0.2 0.3 0.4 05 0,6 1/[SPBRK]

Lineweaver Burk equation : 1/Vo = (Km/Vmax) (1/[S] + 1/Vmax

the above equation is comparable to y = mx+c

slope, m = Km/Vmax and Y-intercept = 1/Vmax

the equation of above straight line graph : y = 0.225x+0.060

slope, m = Km/Vmax = 0.225

Y-intercept, c = 1/Vmax = 0.060

Vmax = 1/0.060 = 16.7 uM/min ( 3 significant figures)

Km/Vmax = 0.225

Km/16.67 = 0.225

Km = 0.225 x 16.67 = 3.8 mM ( 2 significant figures)