BIOCHEMISTRY 2 TERM ACCESS
9th Edition
ISBN: 9781319402877
Author: BERG
Publisher: MAC HIGHER
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Question
Chapter 1, Problem 33P
Interpretation Introduction
Interpretation:
The average base differences between two human beings should be determined.
Concept introduction:
The building block of DNA and RNA is termed as the
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Compare and contrast primary and secondary active transport mechanisms in terms of energy utilisation and efficiency. Provide examples of each and discuss their physiological significance in maintaining ionic balance and nutrient uptake.
Rubric
Understanding the key concepts (clearly and accurately explains primary and secondary active transport mechanisms, showing a deep understanding of their roles)
Energy utilisation analysis ( thoroughly compares energy utilisation in primary and secondary transport with specific and relevant examples
Efficiency discussion
Use of examples (provides relevant and accurate examples (e.g sodium potassium pump, SGLT1) with clear links to physiological significance.
Clarity and structure (presents ideas logically and cohesively with clear organisation and smooth transition between sections)
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
Chapter 1 Solutions
BIOCHEMISTRY 2 TERM ACCESS
Ch. 1 - Prob. 1PCh. 1 - Prob. 2PCh. 1 - Prob. 3PCh. 1 - Prob. 4PCh. 1 - Prob. 5PCh. 1 - Prob. 6PCh. 1 - Prob. 7PCh. 1 - Prob. 8PCh. 1 - Prob. 9PCh. 1 - Prob. 10P
Ch. 1 - Prob. 11PCh. 1 - Prob. 12PCh. 1 - Prob. 13PCh. 1 - Prob. 14PCh. 1 - Prob. 15PCh. 1 - Prob. 16PCh. 1 - Prob. 17PCh. 1 - Prob. 18PCh. 1 - Prob. 19PCh. 1 - Prob. 20PCh. 1 - Prob. 21PCh. 1 - Prob. 22PCh. 1 - Prob. 23PCh. 1 - Prob. 24PCh. 1 - Prob. 25PCh. 1 - Prob. 26PCh. 1 - Prob. 27PCh. 1 - Prob. 28PCh. 1 - Prob. 29PCh. 1 - Prob. 30PCh. 1 - Prob. 31PCh. 1 - Prob. 32PCh. 1 - Prob. 33PCh. 1 - Prob. 34PCh. 1 - Prob. 35PCh. 1 - Prob. 36P
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- 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 0.61arrow_forwardConsider 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_forward
- Classify 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_forwardConsider 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_forward
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