Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 30, Problem 9P
Interpretation Introduction
Interpretation:
The reason for 100-fold larger rate for a protein identifying a specific position along a DNA molecule should be determined.
Concept introduction:
DNA synthesis can be defined as the process by which the deoxynucleic acids such as adenine, thymine, cytosine, and guanine are associated together to form DNA.
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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 30 Solutions
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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
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