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- 1. Describe which enzymes are required for lactose and tryptophan metabolism inbacteria when lactose and tryptophan, respectively, are (a) present and (b)absent. 2. Contrast positive versus negative regulation of gene expression. Describe therole of the repressor in an inducible system and in a repressible system.The a/a operon is an inducible operon that controls the production of the sugar arabinose. When arabinose is present in a bacterium it binds to the protein AraC, and the complex binds to the initiator site to promote transcription. In this scenario, AraC is a(n). activator inducer repressor operatorFor the E. coli lac operon, when lactose is present: a. and glucose is absent, cAMP binds and activates catabolicactivator protein (CAP). b. and glucose is absent, the level of cAMP decreases. c. activated CAP binds the repressor protein to remove it from the operator gene. d. the cell prefers lactose over glucose. e. RNA polymerase cannot bind to the promoter.
- INTERPRET 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.The yeast gene SER3, whose product has a role in serine biosynthesis, is repressed during growth in nutrient-rich medium, so little transcription takes place, and little SER3 enzyme is produced, under these conditions. In an investigation of the repression of the SER3 gene, a region of DNA upstream of SER3 was found to be heavily transcribed when SER3 is repressed (J. A. Martens, L. Laprade, and F. Winston. 2004. Nature 429:571–574). Within this upstream region is a promoter that stimulates the transcription of an RNA molecule called SRG1 RNA (for SER3 regulatory gene 1). This RNA molecule has none of the sequences necessary for translation. Mutations in the promoter for SRG1 result in the disappearance of SRG1 RNA, and these mutations remove the repression of SER3. When RNA polymerase binds to the SRG1 promoter, the polymerase travels downstream, transcribing the SGR1 RNA, and passes through and transcribes the promoter for SER3. This activity leads to the repression of SER3. Propose…Recall that the nuclear membrane disintegrates late in prophase of mitosis in most eukarvotic cells. Once the membrane is reformed in telophase in a daughter cell, several components of gene expression mignt therefore be "caught" out in the cytoplasm when they would otherwise onlv ever be found inside the nucleus. Consider where the following components of gene expression are made and where they runction. Which component is normally never found in the cytoplasm outside the nucleus? O A. Spliced intron • B. RNA polymerase O C. Histones • D. DNA polymerase
- i)Describe attenuation control and how it is used to regulate gene expression. ii)Give a specific example of how this works? iii)Could this be used in eukaryotes? why ?or why not?4. Which or which (can select more than one alternative) of the control methods of genetic expression in eukaryotes NOT USED in prokaryotic cells?to. control how often a gene is expressedb. control how an mRNA is processedc. control which mRNA is exported from the nucleus to the cytosold. control which mRNA is translated to proteins in the ribosomeand. control how quickly proteins are destroyed once synthesized 5. Unlike DNA that usually forms a helical structure, different RNA molecules assume bends in a wide variety of 3D shapes. This is due to :to. RNA contains uracil and ribose as sugarb. the bases in the RNA cannot form hydrogen bonds with each otherc. nucleotides in RNA have a different chemical structure than they have in DNAd. Although RNA is single-stranded, complementary bases can pair with each other. 6. Viruses consist of a layer of protein that surrounds your genetic material. the genetic material in the HIV virus is:to. Double-stranded DNA (dsDNA)b. Single-stranded DNA…Below are data from Jimenez et al (2022) depicting cellular expression patterns of the tumor suppressor transcription factor p53 (a gene commonly mutated in cancers). They explored oscillating expression patterns found in a mutated human cell line. Panel D shows measurements of changes in p53 mRNA levels measured using two different approaches; figure E shows measurements of protein levels from the same samples. The numbers on either side indicate the observed fold change (FC) in relative abundance of these molecules (note that any difference in values obtained between the two methods are not relevant for this question). Answer part b as pictured in the question
- For each of the E. coli strains that follow, indicate theeffect of the genotype on the expression of the trpEand trpC genes in the presence or absence of tryptophan. [In the wild type (R+ P+ o+ att+ trpE+ trpC+),trpC and trpE are fully repressed in the presence oftryptophan and are fully expressed in the absence oftryptophan.]R = repressor gene; Rnproduct cannot bind tryptophan; R− product cannot bind operatoro = operator for the trp operon; o− cannot bind repressoratt = attenuator; att− is a deletion of the attenuatorP = promoter; P− is a deletion of the trp operonpromotertrpE− and trpC− are null (loss-of-function) mutationsa. R+ P− o+ att+ trpE+ trpC+b. R− P+ o+ att+ trpE+ trpC+c. RnP+ o+ att+ trpE+ trpC+d. R− P+ o+ att− trpE+ trpC+e. R+ P+ o− att+ trpE+ trpC−/R− P+ o+ att+trpE− trpC+f. R+ P− o+ att+ trpE+ trpC−/R− P+ o+ att+trpE− trpC+g. R+ P+ o− att− trpE+ trpC−/R− P+ o− att+trpE− trpC+Brenner’s m mutant phages (m1–m6) described inFig. 8.8 were suppressed when grown in suppressor(su−) mutant bacteria; they produced full-length Mproteins that functioned like wild-type M protein.a. What gene do you think was mutant in the su−bacteria?b. When the m− phages were propagated in the su−bacterial strain, not all of the proteins made by themutant m alleles were identical to wild-type Mprotein. How did some of them differ?4.) Fully explain the role Gal3 has in the expression of galactose inducible (gal) genes.