Campbell Biology: Concepts & Connections (8th Edition)
8th Edition
ISBN: 9780321885326
Author: Jane B. Reece, Martha R. Taylor, Eric J. Simon, Jean L. Dickey, Kelly A. Hogan
Publisher: PEARSON
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Chapter 5, Problem 15TYK
A biologist performed two series of experiments on lactase, the enzyme that hydrolyzes lactose to glucose and galactose. First, she made up 10% lactose solutions containing different concentrations of enzyme and measured the rate at which galactose was produced (grams of galactose per minute). Results of these experiments are shown in Table A below. In the second series of experiments (Table B), she prepared 2% enzyme solutions containing different concentrations of lactose and again measured the rate of galactose production.
Table A Reaction Rate and Enzyme Concentration
Table B Reaction Rate and Substrate Concentration
a. Graph and explain the relationship between the reaction rate and the enzyme concentration.
b. Graph and explain the relationship between the reaction rate and the substrate concentration. How and why did the results of the two experiments differ?
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Idris has successfully extracted enzymatic proteins from the fish viscera (intestines and stomach). After homogenization and centrifugation, he managed to pool the crude enzyme extract. He is characterizing the enzymes. Please help Idris by answering the followingquestions:(a). How do I determine the enzyme activity? Please give the unit. (b) How do I get the specific activity of this enzyme? Please give the unit.
Chapter 5 Solutions
Campbell Biology: Concepts & Connections (8th Edition)
Ch. 5 - Fill in the following concept map to review the...Ch. 5 - Label the parts of the following diagram...Ch. 5 - Which best describes the structure of a cell...Ch. 5 - Prob. 4TYKCh. 5 - The sodium concentration in a cell is 10 times...Ch. 5 - The synthesis of ATP from ADP and a. stores energy...Ch. 5 - Facilitated diffusion across a membrane requires...Ch. 5 - What are the main types of cellular work? How does...Ch. 5 - Why is the barrier of the activation energy...Ch. 5 - Relate the laws of thermodynamics to living...
Ch. 5 - How do the components and structure of cell...Ch. 5 - Sometimes inhibitors can be harmful to a cell;...Ch. 5 - Cells lining kidney tubules function in the...Ch. 5 - SCIENTIFIC THINKING Mercury is known to inhibit...Ch. 5 - A biologist performed two series of experiments on...Ch. 5 - Organophosphates (organic compounds containing...
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- A multi-enzyme complex Is made up of three polypeptide chains, A, B and C. A is associated with decarboxylase activity; B is a transacetylase, while C is a dehydrogenase. When the protein was placed in a nonpolar solvent, then run in PAGE, two protein bands were observed. Enzyme assays showed that one protein band exhibited decarboxylase activity while the other has both transacetylase and dehydrogenase activities. When the protein was also placed in an aqueous solvent at pH 5.0, then run in electrophoresis, two protein bands were also detected. Further enzyme assays also showed that one protein band exhibits transacetytase activity while the other has both decarboxylase and dehydrogenase activities. a. What types of non-covalent interactions are possible between A, B and C? b. Addition of urea, a reducing agent gave 4 bands in the PAGE profile with a subsequent loss of decarboxylase activity. What could be the reason for the observed result? Explain briefly in terms of the structure…arrow_forwardcreate a single illustration that will interrelate or link the two opposing pathways, the Glycogenesis and Glycogenolysis. I want you to include the enzyme in each step and include some important by-products as well. From the illustration, I want you to encircle the intermediate molecule to highlight the link between the two processes. Aside from the illustration, I want you to compare and contrast the two pathways in terms of function, number of reaction steps, and usage of UTP. You can tabulate this part to make it simpler.arrow_forwardLeia wants to determine the effect of enzyme concentration on enzyme activity using potato samples and hydrogen peroxide. However, her results do not follow the theoretical trend and thus she can't draw any conclusions from the data obtained. What might have caused errors in the experiment? I. Placing potato samples under room temperature II. Different treatments, with potato being proportional to the volume of hydrogen peroxide III. Placed the same amount of hydrogen peroxide for a different amount of potatoes IV. Placed the same amount of potatoes for a different volume of hydrogen peroxidearrow_forward
- Idris has successfully extracted enzymatic proteins from the fish viscera (intestines and stomach). After homogenization and centrifugation, he managed to pool the crude enzyme extract. He is characterizing the enzymes. Please help Idris by answering the following questions:(a) How do I determine the protein/enzyme concentration? Please give the unit.arrow_forwardA mutation has occurred that results in phosphofructokinase not being able to bind ATP in its allosteric site. What impact will this mutation have on the production of ATP in the cell? Select one: a. If no ATP can bind to the allosteric site, then phosphofructokinase will not be able to add the phosphate to fructose-6-phosphate to make fructose-1,6-bisphosphate and glycolysis will not work. So no pyruvate, not cellular respiration. b. If ATP cannot bind to the allosteric site, phosphofructokinase will not be activated to make more ATP by substrate-level phosphorylation. c. No impact on the production but will not be able to effectively shut off over production of ATP with feedback inhibition.arrow_forwardPyridoxal phosphate (PLP) is a coenzyme for the enzyme ornithine aminotransferase. The enzyme was purified from cells grow in PLP = deficient media as well as from cells grown in media that contained pyridoxal phosphate. The stability of the two different enzyme preparations was then measured by incubating the enzyme at 37°C for different lengths of time and then assaying for the amount of enzyme activity remaining. The following results were obtained. (a) Why does the amount of active enzyme decrease with the time of incubation? (b) Why does the amount of enzyme from the PLP deficient cells decline more rapidly?arrow_forward
- You design an experiment to explore the effects of enzyme concentration on the rate of glucose (product) production. In the experiment, you prepare 5 tubes with equal amounts of water and substrate but altered the amount of enzyme. Tube #1 did not have enzyme, Tube #2 had 100 mg of enzyme, Tube #3 had 200 mg of enzyme, Tube #4 had 300 mg of enzyme, Tube #5 had 400 mg of enzyme. The results are summarized in the graph. Select all that apply. Given enough time all tubes (#1-5) would reach 1100 mg/dL. The substrate in tube #1 was denatured at 40 minutes. The fastest rate of reaction was in Tube #4. Adding more enzyme to Tube #5 at 40 minutes would produce more product. Tube #4 and #5 have run out of substrate at 40 minutes.arrow_forwardPlease asnwer the following question: Suppose you specifically want to measure the amount of beta-glucose present in a sample. You have an enzyme that reacts specifically with the beta form of the sugar,and produces a signal that is easy to measure. However, every time you try to measure the amount of beta-glucose, mutarotation occurs and the alpha form of the sugar is converted into beta!! Can you suggest a simple way of overcoming this problem so you could measure only the beta form of the sugar? HINT: Is there a way you can prevent mutarotation?arrow_forwardA laboratory in Japan performed a similar experiment on earthworms from different localities of the Japanese archipelago, but instead of linking trypsin activity within worms to anatomical features, they were interested in geographical differences in the ability of the worms to breakdown environmental pollutants using the enzyme polyphenol oxidase. The protocol was similar to the one you used, except that enzyme activity was measured from 100 µl of worm extract. Data obtained are displayed below: Locality Replicate Amount of product Concentration of total protein in worm extract (µg/ml) liberated over 10 minutes (pmol) Hokkaido 1 101 4 87 3 3 107 4 4 88 3 5 89 Hon-shu 1 126 2 2 120 3 3 156 4 4 153 5 5 119 4 Куu-shu 1 103 4 2 99 5 3 94 4 108 3 5 110 5 Okinawa 1 88 3 2 90 4 3 96 4 100 112 5 5 3 From the data in the table, calculate the specific activity of polyphenol oxidase (in units of umol/min/mg protein) from each locality, presenting the data as mean + standard error (n=5).arrow_forward
- What are the correct names of the items, I, II and III in the reaction given below? a. Fructose-6P / Phosphofructokinase / Ribose-1,6P b. Glucose-6P / Hexokinase / Glucose-1,6 P c. Fructose-6P / Glucokinase / Glucose-1,6 P d. Fructose-6P/ Phosphofructokinase / Fructose-1,6 P e. Fructose-6P / Phosphoglucose isomerase / Fructose-1,6 Parrow_forwardIdris has successfully extracted enzymatic proteins from the fish viscera (intestines and stomach). After homogenization and centrifugation, he managed to pool the crude enzyme extract. He is characterizing the enzymes. Please help Idris by answering the followingquestions: (a) How do I run the experiment to find the kinetic properties of the enzyme, KMand Vmax? (b) The enzyme decreased in activity in the presence of NaCl. How do I find out if NaCl is a competitive or non-competitive inhibitor? Explain.arrow_forwardCompounds I, II, and III are in the following biochemical pathway: precursor compound I enzyme A enzyme B compound II ► compound III enzyme Mutation a inactivates enzyme A, mutation b inactivates enzyme B, and mutation c inactivates enzyme C. Mutants, each having one of these defects, were tested on minimal medium to which compound I, II, or III was added. Fill in the results expected of these tests by placing a plus sign (+) for growth or a minus sign (-) for no growth in the table below. Minimal medium to which is added Strain with mutation Compound I Compound II Compound IIIarrow_forward
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Enzyme Kinetics; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=FXWZr3mscUo;License: Standard Youtube License