Lineweaver Burk plot to determine the Vmax and Km
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![1/Mean Reaction Rate (1/mm/min)
300
250
200
150
100
50
0
-0.05
0.00
◆no inhibitor
Inhibitor A
0.05
0.10
0.15
0.20
1/[ONPG] (1/mM)
y = 733.66x + 10.919
0.25
y = 390.8x + 10.77
0.30
0.35
Figure 4.4: Lineweaver-Burk plot showing the effect of an inhibitor on hydrolysis
of ONPG by b-galactosidase. Data points are the means of 36 replicates.
0.40](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e5f5ca1-9c38-413c-af6b-0878d611ceab%2F213f580d-3b63-4df7-9686-f8aa79a576b7%2Fn4qu26i_processed.jpeg&w=3840&q=75)
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- Below is a plot of Vo vs. [S] for a specific allosteric enzyme under different conditions. Which of the following best describes the graph? 4 (S] O Adding a positive modulator to #2 would result in curve 3. Curve 1 represents maximum inhibition. Line 4 is valid exclusively for curves 1 and 2. Adding a positive modulator to #1 would result in curve 2. O Curve 3 represents the effect of a negative modulator added to curve 2.A solution of [U 14C] glucose-1-phosphate (specific activity = 16,000 cpm/mmole) was incubated with glycogen and glycogen phosphorylase, an enzyme which adds glucose units on to glycogen. Radioactivity was incorporated into the glycogen primer at a rate = 2550 cpm/min. The rate of the enzymatic reaction in units of mmole glucose incorporated per minute is: (a) 0.016 mmol/min (b) 0.57 mmol/min (c) 0.16 mmol/min (d) 5.7 mmol/minCalculate the mass of invertase (in mg) and concentration of invertase (in mM) contained in a 25.0mL sample of yeast extract that has 3,000 total units of activity, assuming that pure invertase has a specific activity of 1,000 units/mg with a mass of 270kD.
- The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.Ibuprofen and indomethacin are clinically important inhibitors of prostaglandin H₂ synthase-1. Cells expressing this enzyme were incubated under four different conditions, after which the activity of the enzyme was measured by adding radiolabeled arachidonic acid and detecting newly-produced prostaglandin H₂: 1. 40 min without inhibitor (control) 2. 40 min with inhibitor 3. 40 min with inhibitor, after which the cells were resuspended in medium without inhibitor 4. 40 min with inhibitor, after which the cells were resuspended in medium without inhibitor and incubated for an additional 30 min The bar graphs show the results of the experiments with ibuprofen and indomethacin. The bars in each graph correspond to the four different incubation conditions. Ibuprofen Prostaglandin H₂ synthesized (relative to control) 100 80 60 40 20 I I I (1) (2) ll (3) (4) Prostaglandin H₂ synthesized (relative to control) 100k 80 60 40 20 I (1) Indomethacin (2) (3) (4)b. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Denote the point in the pathway in which each individual is blocked.
- a. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Draw the pathway leading to the production of protein E.Two co-op students at your start-up company have been asked to evaluate the rate of reaction occurring in a transparent gel particle containing immobilized mouse melanoma cells. The equation for the reaction rate ris is: Vmax T'As Km+CAs where R=3.2 mm, Vmax= 0.12 gmol s' m3, CAS = 41 gmol m3, and Km=0.8 gmol m3. One student reports a reaction rate of 1.6x10 gmol st; the other reports 1.6x10 1º gmol s1. You left your calculator on the bus this morning, but must know quickly which student is correct. Use an order-of- magnitude calculation to identify the right answer. Which student is correct?The activity of the enzyme β-galactosidase produced bywild-type cells grown in media supplemented with different carbon sources is measured. In relative units, thefollowing levels of activity are found:Glucose Lactose Lactose + glucose0 100 1Predict the relative levels of β-galactosidase activity incells grown under similar conditions when the cells arelacI−, lacIS, lacO+, and crp−.
- The protein fragments ABS1 and ABS2 of tropomodulin were produced as fusion proteins with chitin binding domain and purified by a chitin column. Explain the principles of this type of affinity chromatography and support it with a self‐drawn figure. This is with regards to Tropomodulin/ F-Actin complex.A number of auxotrophic mutant strains were isolated from wild-type haploid Neurospora crassa. These strains responded to the addition of certain nutritional supplements to minimal culture medium either by growth (+) or no growth (0) The data from this experiment are presented in the table below. Diagram a biochemical pathway, complete with positions of intermediates, that is consistent with the data. Indicate where in the pathway each mutant strain is blocked.If a particular cell type requires two passes through a Manton-Gaulin homogenizer to achieve a R of at least 0.90, what is the minimum number of passes required to achieve a disruption of at least 0.95 (if the operating pressure and temperature remain the same)? Explain what is meant by the extent of cell disruption is 0.9.
