BIOCHEMISTRY
9th Edition
ISBN: 2818440090622
Author: BERG
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 16, Problem 55P
Interpretation Introduction
Interpretation:
Function of fructose-2,6-bisphosphate in the presence of allosteric enzymes should be determined.
Concept introduction:
Phosphofructokinase enzyme is an allosteric inhibitor that regulates the glycolysis process. It increases or decreases the rate of glycolysis by analyzing the energy requirements in the cell. For instance, high ATP to ADP value decreases the PFK activity, leading to a decrease in glycolysis rate.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Examine the metabolic pathway. The enzymes that catalyze each step are identified as "e" with a numeric subscript.
e₁
e3
e4
A
B
с
1°
B'
02
e5
e6
e7
E
F
Which enzymes catalyze irreversible reactions?
ப
e
ez
☐ ez
e4
☐
ப
es
26
5
e7
Which of the enzymes is likely to be the allosteric
enzyme that controls the synthesis of G?
€2
ез
e4
es
26
5
e7
An allosteric enzyme that follows the concerted model has an allosteric coefficient (T/R) of 300 in the absence of substrate.
Suppose that a mutation reversed the ratio.
Select the effects this mutation will have on the relationship between the rate of the reaction (V) and substrate
concentration, [S].
ㅁㅁㅁ
The enzyme would likely follow Michaelis-Menten kinetics.
The plot of V versus [S] would be sigmoidal.
The enzyme would mostly be in the T form.
The plot of V versus [S] would be hyperbolic.
The enzyme would be more active.
Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also
known as ẞ-lactamase), an enzyme present in some penicillin-resistant
bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kDa.
The amount of penicillin hydrolyzed in 1 minute in a 10.0 mL. solution
containing 1.00 x 10 g of purified penicillinase was measured as a
function of the concentration of penicillin. Assume that the concentration of
penicillin does not change appreciably during the assay.
Plots of V versus [S] and 1/V versus 1/[S] for these data are shown.
Vo (* 10 M minute"¹)
7.0
6.0
5.0
4.0
3.0
20
1.0
0.0
о
10
20
30
1/Vo (* 10 M1 minute)
20
103
90
BO
70
50
[S] (* 100 M)
40
50
60
y=762x+1.46 × 10"
[Penicillin] (M)
Amount hydrolyzed (uM)
1
0.11
3
0.25
5
0.34
10
0.45
30
0.58
50
0.61
Chapter 16 Solutions
BIOCHEMISTRY
Ch. 16 - Prob. 1PCh. 16 - Prob. 2PCh. 16 - Prob. 3PCh. 16 - Prob. 4PCh. 16 - Prob. 5PCh. 16 - Prob. 6PCh. 16 - Prob. 7PCh. 16 - Prob. 8PCh. 16 - Prob. 9PCh. 16 - Prob. 10P
Ch. 16 - Prob. 11PCh. 16 - Prob. 12PCh. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Prob. 15PCh. 16 - Prob. 16PCh. 16 - Prob. 17PCh. 16 - Prob. 18PCh. 16 - Prob. 19PCh. 16 - Prob. 20PCh. 16 - Prob. 21PCh. 16 - Prob. 22PCh. 16 - Prob. 23PCh. 16 - Prob. 24PCh. 16 - Prob. 25PCh. 16 - Prob. 26PCh. 16 - Prob. 27PCh. 16 - Prob. 28PCh. 16 - Prob. 29PCh. 16 - Prob. 30PCh. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Prob. 33PCh. 16 - Prob. 34PCh. 16 - Prob. 35PCh. 16 - Prob. 36PCh. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - Prob. 43PCh. 16 - Prob. 44PCh. 16 - Prob. 45PCh. 16 - Prob. 46PCh. 16 - Prob. 47PCh. 16 - Prob. 48PCh. 16 - Prob. 49PCh. 16 - Prob. 50PCh. 16 - Prob. 51PCh. 16 - Prob. 52PCh. 16 - Prob. 53PCh. 16 - Prob. 54PCh. 16 - Prob. 55PCh. 16 - Prob. 56PCh. 16 - Prob. 57PCh. 16 - Prob. 58PCh. 16 - Prob. 59PCh. 16 - Prob. 60PCh. 16 - Prob. 61P
Knowledge Booster
Similar questions
- Consider the four graphs shown. In each graph, the solid blue curve represents the unmodified allosteric enzyme and the dashed green curve represents the enzyme in the presence of the effector. Identify which graphs correctly illustrate the effect of a negative modifier (allosteric inhibitor) and a positive modifier (allosteric activator) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes for each type of modifier. Negative modifier Reaction velocity - Positive modifier Substrate concentration - Reaction velocity →→→→ Substrate concentration Answer Bankarrow_forwardConsider the reaction: phosphoglucoisomerase Glucose 6-phosphate: glucose 1-phosphate After reactant and product were mixed and allowed to reach at 25 °C, the concentration of each compound at equilibrium was measured: [Glucose 1-phosphate] = 0.01 M [Glucose 6-phosphate] = 0.19 M Calculate Keq and AG°'. Код .0526 Incorrect Answer 7.30 AG°' kJ mol-1 Incorrect Answerarrow_forwardClassify each phrase as describing kinases, phosphatases, neither, or both. Kinases Phosphatases Neither Both Answer Bank transfer phosphoryl groups to acidic amino acids in eukaryotes may use ATP as a phosphoryl group donor remove phosphoryl groups from proteins catalyze reactions that are the reverse of dephosphorylation reactions regulate the activity of other proteins catalyze phosphorylation reactions PKA as an example turn off signaling pathways triggered by kinasesarrow_forward
- Consider the reaction. kp S P kg What effects are produced by an enzyme on the general reaction? AG for the reaction increases. The rate constant for the reverse reaction (kr) increases. The reaction equilibrium is shifted toward the products. The concentration of the reactants is increased. The activation energy for the reaction is lowered. The formation of the transition state is promoted.arrow_forwardThe graph displays the activities of wild-type and several mutated forms of subtilisin on a logarithmic scale. The mutations are identified as: • The first letter is the one-letter abbreviation for the amino acid being altered. • The number identifies the position of the residue in the primary structure. ⚫ The second letter is the one-letter abbreviation for the amino acid replacing the original one. • Uncat. refers to the estimated rate for the uncatalyzed reaction. Log₁(S-1) Wild type S221A H64A -5 D32A S221A H64A D32A -10 Uncat. How would the activity of a reaction catalyzed by a version of subtilisin with all three residues in the catalytic triad mutated compare to the activity of the uncatalyzed reaction? It would have more activity, because the reaction catalyzed by the triple mutant is approximately three-fold faster than the uncatalyzed reaction. It would have less activity, because the reaction catalyzed by the triple mutant is approximately 1000-fold slower than the…arrow_forwardB Substrate Product AL Product Substrate Reaction progress- Reaction progress- omplete the passage describing the two reactions. In reaction A, the stability of the substrate is (AG) of the reaction is positive, Incorrect Answer greater than the stability of the product. The free-energy change Incorrect Answer so the reaction is considered In reaction B, the stability of the substrate is (AG) of the reaction is less than Incorrect Answer endergonic and Incorrect Answer not spontaneous. Incorrect Answer the stability of the product. The free-energy change negative, so the reaction is considered Incorrect Answer exergonic and spontaneous. Incorrect Answer Incorrect Answerarrow_forward
- Match the descriptions and compounds with the terms competitive, uncompetitive, and noncompetitive inhibition. Competitive inhibition Uncompetitive inhibition Noncompetitive inhibition Answer Bank inhibitor binds to the enzyme-substrate complex only Vmax remains the same but Ky increases inhibitor and substrate can bind simultaneously lowers Vmax and KM doxycycline sulfanilamide KM remains unchanged, but Vis lower prevents substrate from binding to the active site Rounduparrow_forwardDraw a pentapeptide made of the following amino acids: glycine, tyrosine, lysine, glutamate, and leucine at pH 1, pH 7, and pH 12. Draw the correct stereochemistry for each pentapeptide. Calculate the charge of the three compounds you've drawn and the PI.arrow_forward7. a. Complete the following redox reaction with the missing products NADH H b. Provide an arrow-pushing mechanism for the oxidation reaction. Use the explicit structure of NADH or NAD' as needed. For simplicity, use only the nicotinamide portion of the moleculearrow_forward
- Discuss the differences between eukaryotic and prokaryotic cells with regards to their genetic materials. Including in your discussion the structure and organisation of genetic material, as well as any implications these differences may have on cellular functions and evolution. Rubric Understanding of genetic material differences (provides a comprehensive and accurate explanation of the differences between eukaryotic and prokaryotic cells in terms of their genetic material, including the structure, organisation and function of genetic material in each cell type. Demonstrates a thorough understanding of the topic) Analysis of implications (analyze the differences between eukaryotic and prokaryotic cells in terms of their genetic material with depth and insight, discussing the implications of these differences on cellular functions and evolution. Provides specific examples and explanations.)arrow_forwardWhen was the dihydropyridine calcium channel blocker isradipine first patented and by whom? Please provide information on the origin and history of isradipine and who owns it/manufactures it.arrow_forward9) Below, there is a representation of an SDS-PAGE gel. Assuming the samples in the MW standard have masses of: 66 kDa, 45 kDa, 36 kDa, 29 kDa, 24 kDa, 20.1 kDa, and 14.2 kDa, a) Figure 4: indicate where each of the measurements were taken and label as in II.6. figure 2 above. b) As in II.7. Table 1 above create Table 2 using the data below. Determine the r.f. values for the MW standards, plot the relative mobility versus the log of the mass for the standards, and use the best fit straight line to determine the molecular weights of the proteins in the whey, peak 1, and peak 2 lanes. (5 pts—this will be scaled up appropriately if your gel did not develop properly) dye MW Whey Peak 1 Peak 2arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning
BiochemistryBiochemistryISBN:9781305961135Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougalPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Biochemistry
Biochemistry
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Cengage Learning
Biochemistry
Biochemistry
ISBN:9781305961135
Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal
Publisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax