Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
6th Edition
ISBN: 9798214167343
Author: GARRETT
Publisher: Cengage Learning US
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Chapter 29, Problem 6P
Interpretation Introduction
Interpretation:
The length of the promoter in nm, the number of turns of B-DNA in it and the number of nucleosomes connected to it need to be determined.
Concept introduction:
In genetic science, a promoter could be a region of
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9. Which one of the compounds below is the major organic product obtained from
the following reaction sequence, starting with ethyl acetoacetate?
요요.
1. NaOCH2CH3
CH3CH2OH
1. NaOH, H₂O
2. H3O+
3. A
OCH2CH3
2.
ethyl acetoacetate
ii
A
3. H3O+
OH
B
C
D
E
7. Only one of the following ketones cannot be made via an acetoacetic ester
synthesis. Which one is it?
Ph
کہ
A
B
C
D
E
2. Which one is the
major organic
product obtained
from the following
reaction sequence?
HO
A
OH
1. NaOEt, EtOH
1. LiAlH4
EtO
OEt
2. H3O+
2. H3O+
OH
B
OH
OH
C
-OH
HO
-OH
OH
D
E
.CO₂Et
Chapter 29 Solutions
Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
Ch. 29 - Prob. 1PCh. 29 - The Events in Transcription Initiation Describe...Ch. 29 - Substrate Binding by RNA Polymerase RNA polymerase...Ch. 29 - Comparison of Prokaryotic and Eukaryotic...Ch. 29 - Prob. 5PCh. 29 - Prob. 6PCh. 29 - Prob. 7PCh. 29 - Alternative Splicing Possibilities Suppose exon 17...Ch. 29 - Prob. 9PCh. 29 - Prob. 10P
Ch. 29 - Post-transcriptional Modification of Eukaryotic...Ch. 29 - Prob. 12PCh. 29 - Prob. 13PCh. 29 - The Lariat Intermediate in RNA Splicing Draw the...Ch. 29 - Prob. 15PCh. 29 - Prob. 16PCh. 29 - Prob. 17PCh. 29 - Prob. 18PCh. 29 - Figure 29.15 highlights in red the DNA phosphate...Ch. 29 - Chromatin decompaction is a preliminary step in...
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- 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 e7arrow_forwardAn 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.arrow_forwardPenicillin 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 0.61arrow_forward
- 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
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