Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 29, Problem 10P
Interpretation Introduction
Interpretation:
To predict if the lap repressor interaction with inducer would be cooperative and, specify if it would be advantageous for the inducer to have cooperative binding to lac repressor.
Concept introduction:
DNA or Deoxyribonucleic acid is a molecule made of two chains which coil around one another. These form a double helix which carries instructions genetical in nature like related to reproduction, growth, development, functioning of the living organisms.
Lac operon repressor is homotetramer. The arrangement is such that this acts as a dimer’s dimer making the repressor bind to 2 separate operators.
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Direct mutagenesis of Ca2+ ATPase gene resulted in the replacement of two amino acid residues - Asn111 and Asn114 to Ala. These substitutions led to the reduction in Ca2+ transport activity by 10% and 50%, respectively. On the other hand, directed mutagenesis that resulted in the alteration of four Glu residues in the lumenal loop of this transport protein to Ala, did not affect the Ca2+ transport. Provide the possible explanation for the observed differences in the Ca2+ transport activity between the protein with Asn->Ala substitution and the protein with Glu->Ala substitution.
Now read this abstract from a 2013 journal article What is the authors' explanation of how Gal80p works?
Note UASG from the question above is the same as UASGAL
The DNA-binding transcriptional activator Gal4 and its regulators Gal80 and Gal3 constitute a galactose-responsive switch for
the GAL genes of Saccharomyces cerevisiae. Gal4 binds to GAL gene UASGAL. (upstream activation sequence in GAL gene pro-
moter) sites as a dimer via its N-terminal domain and activates transcription via a C-terminal transcription activation domain
(AD). In the absence of galactose, a Gal80 dimer binds to a dimer of Gal4, masking the Gal4AD. Galactose triggers Gal3-Gal80
interaction to rapidly initiate Gal4-mediated transcription activation. Just how Gal3 alters Gal80 to relieve Gals0 inhibition of
Gal4 has been unknown, but previous analyses of Gal80 mutants suggested a possible competition between Gal3-Gal80 and
Gal80 self-association interactions. Here we assayed Gal80-Gal80 interactions and tested for…
Chapter 29 Solutions
Biochemistry
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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- In terms of the efficient use of bacterial resources, explain why it is necessary for lac operon expression to be controlled both by the lac repressor and by a cyclic AMP receptor protein rather than just by the lac repressor alone.arrow_forwardThe diagram below shows a closeup of regulatory proteins binding to one of the UASG elements near the GAL7, GALI0, and GALI genes, which code for the protein products needed for yeast to use the sugar galactose. The red triangle symbolizes an "effector" molecule that binds to Gal80p. In this hypothesis (which has since been shown to be incorrect), what could be happening to Gal80p when it is bound to the effector molecule that causes it to change its position and uncover the Gal4p transcriptional activation domain. Hint: think about what effector molecules do upon binding to proteins such as the the Lac repressor protein or the CAP protein. Galactose absent, glucose absent Gal80p. _Activation domain Gal4p dimer -Binding domain UASG Galactose present, glucose absent Activation domain Gal80p- Binding domain UASG For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).arrow_forwardExplain why (a) inactivation of the O2 or O3 sequence of the lac operon causes only a twofold loss in repression, and (b) inactivation of both O2 and O3 reduces repression ∼70-fold.arrow_forward
- . An interesting mutation in lacI results in repressorswith 110-fold increased binding to both operator andnonoperator DNA. These repressors display a “reverse”induction curve, allowing β-galactosidase synthesis inthe absence of an inducer (IPTG) but partly repressingβ-galactosidase expression in the presence of IPTG. Howcan you explain this? (Note that, when IPTG binds a repressor, it does not completely destroy operator affinity,but rather it reduces affinity 110-fold. Additionally, ascells divide and new operators are generated by thesynthesis of daughter strands, the repressor must findthe new operators by searching along the DNA, rapidlybinding to nonoperator sequences and dissociating fromthem.)arrow_forwardSome elongation factors are evolutionarily related to the G-proteins involved in signal transduction. Provide a possible reason why this is the case.arrow_forwardINTERPRET DATA Develop a simple hypothesis that would explain the behavior of each of the following types of mutants in E. coli. Mutant a: The map position of this mutation is in the trp operon. The mutant cells are constitutive; that is, they produce all the enzymes coded for by the trp operon, even if large amounts of tryptophan are present in the growth medium. Mutant b: The map position of this mutation is in the trp operon. The mutant cells do not produce any enzymes coded for by the trp operon under any conditions. Mutant c: The map position of this mutation is some distance from the trp operon. The mutant cells are constitutive; that is, they produce all the enzymes coded for by the trp operon, even if the growth medium contains large amounts of tryptophan.arrow_forward
- Cell signaling The maximum secretion rates of proteins are one type of limiting constraint on signal-ing processes since a cell cannot send a signal faster than it can secrete it. These maximum signalingrates can be estimated with maximum rates of transcription and translation. Consequently, proteins withstrong promoters can be synthesized at much higher rates. Immunoglobulins, which are important signal-ing molecules, have very strong promoters. Their maximum secretion rate is on the order of 2000 to 8000antibody molecules/cell/second, which corresponds to approximately 1 pg/cell/hr. Part AWhat would be the maximum signal (in pg/cell/hr) that a cell could send for the following situations: (1)Signal molecule A has a MW = 100,000 daltons and is secreted at a rate of 1000 molecules/cell/second. (2)Signal molecule B has a MW = 50,000 daltons and is secreted at 5000 molecules/cell/second.Part BWhich molecule is secreted at a higher rate? By how much?Part CIf equivalent fluxes of both…arrow_forwardRepressors are inactivated either by interaction with a small-molecule inducer or by proteolytic cleavage. Why is it advantageous for a repressor like the lac repressor to be inactivated by binding to allolactose rather than by proteolytic cleavage?arrow_forwardneed helparrow_forward
- АСTIVITY For each of the 4 regulatory states of the lac shown in the following diagram, answer the following questions: оperon 1) Is glucose present? 2) Is lactose present? Synthesis of lac MRNA? Glucose present? Lactose present? NO Repressor y a NO CAMP-CRP complex y YES Transcription i p o y NOarrow_forwardExplain the role/importance of the localization of GTPase-activating protein (Ran-GAP) in BOTH nuclear export and import. What would occur if there was a loss of function mutation in Ran-GAP? What would occur if Ran-GAP was localized in the nucleus?arrow_forwardWhich of the proteins: the CAP or the lac repressor, does the eukaryotic nuclear receptor for the glucocorticoid dexamethasone primarily resemble?arrow_forward
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