Problem 5. Suppose a cell expresses HK-I and 5 mM is the concentration of glucose in the blood. Compare the cellular concentration of glucose (relative to 5 mM) based on the G6P concentration in the cell being low or high.
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HK-1 is a type of hexokinase encoded by the HK1 gene. This enzyme produces glucose-6-phosphate by phosphorylating glucose. It can phosphorylate other hexoses too. Conversion of glucose to glucose-6-phosphate initiates the process of glycolysis which is associated with the production of energy-yielding molecules. Hexokinase is present on the mitochondrial outer membrane.
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- Work 3. The formation and secretion of hydrochloric acid Complete the scheme, specify the starting substances for the formation of HCI, the enzyme, the reaction product, the secreted substances and the substances entering the cell, specify the energy-consuming carrier. Substrates Enzyme Hydrochloric acid secretion is stimulated by: receptors - receptors - (ATP) energy-consuming transport O Carrier Channels receptors Prostaglandins E2 (PGE2) Img. 16. Secretory processes in the parietal cell (increase or inhibit?) gastric secretion. 2 Stomach cavity Interstitial fluidNeed help with the problem 1Critical thinking question PROBLEM: After nursing, Baby A developed abdominal pain and diarrhea. Upon admission to the hospital, tests were performed to identify the problem. Lactose was administered on an empty stomach, and the blood concentration of glucose was determined over time, as shown in the following graph. The baby's results were compared to a control curve generated on tests from children who did not suffer after nursing. Suggest a possible cause of the baby's discomfort. Blood glucose concentration (mm) 10 O 1 Control results Baby A results 2 Time (hrs) 3
- Problem 1. Greco et al. have developed a regression curve fitting package in 1982 and tested it on Vp vs. [S] profiles of Hexosaminidase enzyme (see the figure below). The reactions rates were experimentally measured at various substrate concentrations in the presence or absence of an inhibitor. P-nitrophenyl-N-acetyl-ß-D- glucosaminide (NAG) is the substrate. 2-acetamido-2,4-dideoxy-4-fluoro-D-galactono-1,5-lactone (ADFGL) is the inhibitor. 1 0.9 0.8 -- No Inhibitor 0.7 - 0.2 mM Inhibitor 0.6 0.5 0.4 0.3 0.2 0.1 2 [S] (mM) Using the data given in the Figure, answer the following questions. A) Determine Vmax and Km of the reaction in the absence of inhibitor. B) Determine the inhibition type? C) Determine the equilibrium constant (K;) associated with the enzyme-inhibitor complex. V, (mM/s)Activity 1: Practice Problems 1. In humans, phenylketonuria (PKU) is a disease caused by an enzyme deficiency at step A in the following simplified reaction sequence, and alkaptonuria (AKU) is due to an enzyme deficiency in one of the steps summarized as step B: A B phenylalanine tyrosine CO₂ + H₂O A person with PKU marries a person with AKU. What phenotypes do you expect for their children? wildtype O PKU O AKU O Both PKU and AKUQuestion: Determine the Km and Vmax for this enzyme/substrate combination. [Substrate] (mM) V0 (mM/min) 0.25 0.183 0.50 0.356 1.00 0.665 2.50 1.45 5.00 2.35 What is the concentration of substrate necessary to achieve a turnover rate of 1.00 mM/min?
- Homework 1 (a) The kinetic data given below are for the reaction catalyzed by prostaglandin endoperoxide synthase. Focusing here on the first two columns, determine the Vmax and Km of the enzyme. (b) Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the synthesis of prostaglandins, ibuprofen reduces inflammation and pain. Us ing the data in the first and third columns of the table, determine the type of inhibition that ibuprofen exerts on prostaglandin endoperoxi' Rate of formation [Arachidonic acid] (mm) Rate of formation of PGG2 (mm/min) of PGG, with 10 mg/mL ibuprofen (mM/min) 0.5 23.5 16.67 1.0 32.2 25.25 1.5 36.9 30.49 41.8 44.0 2.5 37.04 3.5 38.91PROBLEMS 8.1 An enzyme-catalysed reaction was found to be affected by two inhibitors A and B. The following results were obtained at fixed total enzyme čoncentration: Substrate conc" Initial velocity (absorbance units per minute) (mmol l-) With 1 mmol I-IB Uninhibited With 1 mmol I-1À 50 20 0.684 1.08 0.653 1.01 0.649 0.476 0.374 0.311 10 1.43 0.468 1.02 0.363 0.296 3.3 0.798 2.5 2.0 0.657 0.549 0.250 Comment on these results.1N(10 M min'y1 3. To the right is a Lineweaver-Burk plot for enzyme X along with two additional (A) 501 lines from testing a low and high concentration of a new inhibitor. a. Mark which line is the enzyme with no inhibitor, low concentration of inhibitor and high concentration of inhibitor. b. What is the KM for this enzyme? -50 50 150 What is the Vmax for this enzyme? 1/S (M-1) С. d. What type of inhibition occurs in the presence of the new inhibitor? -25
- PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?PROBLEMS 8.1 An enzyme-catalysed reaction was found to be affected by two inhibitors A and B. The following results were obtained at fixed total enzyme čoncentration: Substrate conc" Initial velocity (absorbance units per minute) (mmol l-) With 1 mmol I-B Uninhibited With 1 mmol l- A 0.684 50 20 1.08 0.653 0.468 10 1.43 1.01 0.649 0.476 0.374 0.311 5 1.02 0.363 0.798 0.657 3.3 0.296 2.5 0.250 2.0 0.549 Comment on these results. 8.2 The system investigated in problem 7.1 was investigated again under identical conditions but in the presence of an inhibitor, giving the following data: 40.0 6.67 10.0 156 20.0 Substrate conc" (mmol 1-1) Initial velocity (umol 1- min-1) 5.0 100 122 222 278 Determine the type of inhibition. If K, for this system is 2.9 mmol 1-', calculate the inhibitor concenträtion present.Computation: Ratio Strength, PPM, mg%Show your complete solution.1. Diabetes is diagnosed by any of the following: two consecutive fasting blood glucose tests that are equal to or greater than 125 mg/dL, any random blood glucose that is greater than 200 mg/dL, and a two hour oral glucose tolerance test with any value over 200 mg/dL.A.) what is yje equivalent value express in terms of milligrams percent?Equivalent value of 126 mg/dL in mg% = ?Equivalent value of 200 mg/dL in mg% = ?B.) how many milligrams of glucose would be present in 10-mL sample serum (use the FASTING BLOOD SUGAR LEVEL)?