A graph of 1/vo vs 1/[S] is shown below. The y intercept is 6.6 and the slope is 133.3. Determine the value of Vmax and Km. 1/vo 120 100 80 40 20 0 0 Select one: O a. ● 0.1 Km=1.4 mM Vmax=0.01 mM/s O b. Km-0.15 mM Vmax=20 mM/s 0.2 c. Km 20 mM Vmax = 0.15 mM/s O d. Km= 0.0075 mM Vmax=0.01 mM/s O 0.3 0.4 1/[S] 0.5 0.6 0.7 0.8
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![A graph of 1/vo vs 1/[S] is shown below. The y intercept is 6.6 and the slope is 133.3.
Determine the value of Vmax and Km.
1/vo
120
100
80
40
20
0
0
Select one:
O a.
●
0.1
Km=1.4 mM
Vmax=0.01 mM/s
O b. Km-0.15 mM
Vmax=20 mM/s
0.2
c. Km=20 mM
Vmax = 0.15 mM/s
O d. Km= 0.0075 mM
Vmax=0.01 mM/s
O
0.3
0.4
1/[S]
0.5
0.6
0.7
0.8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17c1e71f-55d2-4ce6-93ef-b4a233349823%2Fd1b71c2a-674a-41e9-854c-59e6c9ea97e7%2Feaq5vof_processed.png&w=3840&q=75)
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- Determine the % inaccuracy by calculating the percent error of the micropipette at the set volume of 100 µL A. Standard deviation B. Covieffecrnt of varaiant C. % error (precision) Percent error = mean volume delivered -expected volume x 100 Expected volume RUN Weight Measurement (in g) 1 0.0985 2 0.0987 3 0.0985 4 0.0982 5 0.0984 6 0.0984 7 0.0981 8 0.0983 9 0.0982 10 0.0988 Total 0.9841 A. Standard deviation B. Covieffecrnt of varaiant C. % error (precision)I have a question, what are these ''a'' means in this plot? Normally its 1/vmax or -1/km. Please be specific. Thank you.Calculate the hemoglobin concentration of an unknown sample given the following info: concentration of known = 12.5 g/dl; absorbance of unknown = 0.48; absorbance of known = 0.65 a. 16.9 g/dl b. 9.2 g/dl c. 8.7 g/dl d. 11.3 g/dl e. 10.4 g/dl
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- In the experiment performed for the determination of cholesterol, the function of the calibration curve obtained using 25-400mg/dL standard cholesterol solutions was determined as y= 0.0009x. If the amount of cholesterol in the serum sample is determined as 3.11 g/L, what is the absorbance value of the sample? A. 0,280 B. 0.028 C. 0.208 D. 0.82A. Kinetics Data 1/[S], mM- 0.5714 0.4608 0.3333 0.1818 0.0952 1/Velocity (min/mM) Inhibited 0.7246 0.5988 0.4694 0.3367 0.26109 Uninhibited 0.5154 0.4424 0.3508 0.2816 0.2277 B. Lineweaver-Burk Plot and Analysis Paste your Microsoft Excel Graph of the kinetic analysis. Compute for Vmax and Km values and identify the mode of enzyme inhibition involved.DARK DARK CHOCOLATE A CHOCOLATE B Average systolic reading (mmHg) 117 110 68% of area Calculated value of 95% of area "1" Standard.Dev. + 3.38 + 7.97 (mmHg) 99.7% of area -3 -2 -1 For normally distributed data: 68% of the individual data values are within + "1" standard deviation of the mean. 95% of the individual data values are within + "2" standard deviations of the mean 99% of the individual data values are within + "3" standard deviations of the mean The values listed in the table above are for + "1" standard deviation. Add and subtract these 0thb values to and from the sample mean to gain a better appreciation for the blood pressures of the 09to975 individuals in each study. 811 values from the table above to Chocolate A: 68% of individuals had a blood pressure reading of (117 – 3.38) (117 + 3.38) 95% of individuals had a blood pressure reading of *+2 standard deviations, or twice the reported value to (117- 6.76) (117 + 6.76) use the values for tChocolate B: 68% of individuals…