how thymidine relieve the inhibition of the TTP
biosynthetic pathway?
and why FUdR inhibitor will be detected in S-phase?
thank you
Thymidine relieves the inhibition of the TTP (Thymidine triphosphate) biosynthetic pathway. Thymidine salvage by Thymidine kinase 2 enzyme is necessary for the TTP biosynthetic pathway. TTP is generated from thymidine by the action of thymidine kinase, thymidylate kinase, and nucleoside diphosphate kinase. Thymidine reacts with ATP to give thymidine monophosphate and ADP by help of enzyme Thymidine kinase. Thymidine monophosphate is phosphorylated to thymidine diphosphate by the help of enzyme thymidylate kinase. Thymidine diphosphate is further phosphorylated to thymidine triphosphate by the help of enzyme nucleoside diphosphate kinase. This is how thymidine relieves the inhibition of the TTP biosynthetic pathway.
FUdR inhibitor will be detected in S-phase because FUdR results in the cell cycle arrest at the S-phase. FUdR acts as an inhibitor of DNA replication by binding to thymidylate synthase during the S-phase.
how thymidine relieve the inhibition of the TTP biosynthetic pathway? and why FUdR inhibitor will be...
In feedback inhibition, the inhibitor of the biochemical pathway is Multiple Choice Ο the final product of the biochemical pathway. Ο a substance that is produced towards the middle of the biochemical pathway. Ο the product of the enzyme inhibited. Ο the substrate of the enzyme inhibited. Ο O a product of another biochemical pathway.
In feedback inhibition of a metabolic pathway, where does the inhibitor bind? Once the substrate moves into active site of the enzyme, What process typically regulates the enzymes involved in metabolic reactions?
7. Propionate is commonly used as an inhibitor of microbial growth. This growth inhibition comes from propionate being converted into propionyl-CoA (shown below) and then entering the TCA cycle. Explain why this results in growth inhibition. (9 points) S-CoA
Biochemistry help 1) Write the names of the degradative and the biosynthetic enzymes of glycogen metabolism that are targeted to regulate glycogen metabolism. 2) ) Discuss, by drawing structures, how hormones regulate the enzymes you name in Question 1. 3) Discuss, with structures, the reactions of the non-oxidative phase of the pentose phosphate pathway and show how the non-oxidative phase is used to synthesize all the pentoses in nature.
Microbiology In the competitive inhibition of enzyme activity, which statement is correct? Inhibitor directly competes with the substrate. Less substrates must be added in order to reach Vmax. The number of active sites is unlimited. None of the above. In the noncompetitive inhibition of enzyme activity, which statement is correct?Inhibitors will cause a conformational change in the enzyme, more substrate must be added to reach Vmax, Enzymes bound to the inhibitor can still bind substrate, none of the above. ....
BIOCHEMISTRY Propose a structure for a suicide inhibitor of the enzyme Enolase. Fully explain how and why the structure is likely to operate as asuicide inhibitor. (Think about what molecular characteristics are required for a suicide inhibitor). Draw a plot that illustrate the kinetics of the suicide inhibition and explain the curves/lines on the plot.
a. Identify the type of inhibition occurring. Explain how you know. b. Describe the difference or similarity between the Vmax and Km values for the different conditions C. Having identified the type of inhibitor indole is, explain what this means for where indole binds to chymotrypsin. + Inhibitor 1/VO Control (no inhibitor present) 1/[S] MacBook At 0 BASE 7
Please show how to calculate Km and Vmax for no inhibitor/low
inhibitor given graph. Show how to solve for a.,a/a etc.
Lineweaver Burk #4: No inhibitor, Low inhibitor 0.2 ▲ No inhibitor Low inhibitor 0.15 0.05 0.2 0.15 0.1 0.05 0.05 0.1 0.15 02 5 1/IS] in units of 1/mM Fill in the blanks. Show your work. No inhibitor Kmno Vmax,o- Vmax,w = ๙ Vmax,o Solve for ๙ inhibitor Krmkw =픕Km,o Solve for 픕 Hint treat, as a single number....
QUESTION 2 (A) The pentose phosphate pathway is comprised of two phases. Why are both phases necessary for the survival of the cell? (2pts) (B) In the non-oxidative phase of the pentose phosphate pathway ribose 5-phosphate is converted to fructose 6-phosphate. Why does this pathway require so many enzymatic steps? What benefit does each step provide? (4 pts) (C) Many cancer drugs are effective because they reduce cell proliferation by reducing the amount of base material needed for cell division...
1. Graphic depictions of enzyme kinetic data can aid in evaluation of reversible enzyme inhibition. Please use the data below to determine type of inhibition for using two different concentrations (in blue and purple). Be sure to include graphical representation (Michaelis-Menton, Lineweaver-Burk plots), type of inhibition, appropriate kinetic schemes and rate equations. The enzyme concentration at its initial state E, was kept constant. All data points are given in arbitrary units. Completely justify your answer. Use Excel or graphing paper...