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There are some mutations that alter the sequence of a bacterialpromoter. This can lead to higher levels of transcription than thewild type sequence. Explain how and why this canoccur in the bacteria as well as why similiarly positionedmutations that change eukaryotic promoter sequences are less likelyto up-regulate levels of transcription of the downstreamgene.

IF YOU DO NOT KNOW 100% FOR SURE PLEASE DO NOTANSWER.

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Jean Keeling
Jean KeelingLv2
28 Sep 2019

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Related questions

Multiple choice questions, I need the CORRECT ANSWERS, 3500points

1.The promoter is in region_____________.

a.1

b.2

c.3

d.4

e.5

2.The stop codon is identified in region __________

a.4

b.5

c.6

d.7

e.8

3.A mutation is region 8 would be expected to:

a.disrupt transcription

b.disrupt ribosome

c.disrupt DNA polymerase binding

d.disrupt splicing

e.none of the above

4.Which is responsible for the transcription of structural genesin eukaryotes?

a.RNA polymerase I

b.RNA polymerase II

c.RNA polymerase III

d.DNA polymerase

E.RNA Synthesis

5.The eukaryotic TATA box is directly bound by a basaltranscription factor known as:

a.TBF

b.TBP

c.TAF

D.TPP

E.TAT

6.mOST TRANSCRIPTIONAL ACTIVATOR proteins in eukarypotesfunction in the cell as:

a.monomers

b.dimers

c.trimers

d.tetramers

e.pentamers

7.The basal level of transcription of eukaryotic genes is quitehigh in vitro systems. This is because:

a.activation are present

b.formation of chromatin dos not occur in vitro

c.basal transcription factors are present

d.a and c

e.none of the above

8.Transcriptional repression by methylation of DNA involves:

a.methylation of adenine (A) bases of the sequence GATC

b.methylation of cytosine (C) bases of th sequence GATC

c.mthylation of cytosin (C) bases in CpG nuclotides

d.Methylation of guanosine (G) bases in CpG nucleotides

e.acetylation of hisone proteins

9.A eukaryotic DNA sequence that can increase the level oftranscription is called:

a. a transcription factor

b.a transcritpional activator

c.TATA box

d.an enhancer

e.a co-activator

10.Which is generally true of eukaryotic transcriptionalregulation?

a.activators and reprssors bind to the promoter while basalfactors bind to the enhances

b.basal factors bind to the promoter, while activators bind tothe enhancers

c.activators bind to the promoter while basal factors andrprssors bind to th enhancers

d.repressors bind to the promoter whil basal factors andactivators bind to the enhancers

e.activators and basal factors bind to the promtoer whilereprssors bind to the enhancers

11.You wish to sequence the promoters region of the mouse generab.To do this, you need to construct a library and then search forthe gene sequence.You have a DNA probe based on the sequence of theprotein coding region.Which of th follwoing strategies would youfollow in order to carry this out?

A.construct a genomic library and screen for the promoter usingPCR

b.construct a cDNA library and screen for the promotoer usingPCR

c.construct a genomic library and screen for the promoter usingPCR

d.Construct a cDNA library and screen for the promoter usingcolony hybridization

e.all of the above

12.if one were to examine th clones in a cDNA library derivedfrom th heart muscle cells and the clones in a cDNA library derivdfrom prancreatic cells, you would find that:

a.the two libraries would hav the same cloned sequences

b.the two libraries would have completely different clonesequences from one another

c.the two libraries would share some common clond sequences, butwould each have some sequences unique to that library, not to befound in the other

d.the sequences from both would also b the same as those foundin the genomic library

e.none of th above

13.In a particular population at Hardy-Weinberg equilibrium, thefrequency of recssive allele b for a particular gne is0.01.Assuming that there are only two alleles for the gene, whatwould b th expected frequency for dominant allele B in th nextgeneration?

a.0.01

b.0.02

c.0.98

d.0.99

e.none of the above

14.White wool in the sheep is controlled by a dominant allele R,and black wool by the recessive allele r.In an isolated popualtionof 6530 sheep, 514 are balck, 4981 are hterozygous, 1035 arhomozygous white. When will this population be expected to reachHardy-Weinberg equilibrium?

a.in the next generation

b.in two generations

c.it already is in equilibrium

d.impossible to tell

e.it will not reach equilibrium

15.A population has th following genotypic disturbutions:

A^M A^M=1787, A^M A^O=2011, A^OA^O=2331.Is this population atH-W equilibrium?

a.yes, with allele frequencies of p=0.45562 and q=0.54437

b.yes, with frequencies of p=0.54437 and q=0.37862

c.no, with allele frequencies of p=0.45562 and q=0.37862

d.no, with xpectd genotypic ratios of p^2=1272, 2pq=3040, andq^2=1826

e.yes, with expctd genotypic frequencies of p^2=1778, 2pq=2015,and q^2=2336

16.The result of the Luria-Delbruck fluctuation exprimentindicatd that:

a. most bacteria are naturally resistant to phage

b.a low level of bacteria become resistant to phage by mutationupon exposure to phage

c.most bacteria become resistant to phage by mutation uponexposure to phag

d.bacteria become resistant to phag by random spontaneousmutation

e.bacteria become resistant to phag by acquiring resistanttraists upon repeated exposure

17.DNA probes are used for:

a.restriction ndonuclease analysis

b.sequencing

c.PCR

d.DNA library analysis

e.all of th above

18.Colony hybridization is used to:

a.isloate a gne of interest froma DNA library

b.analyze the presence or number of copies of a particular DNAsegment in a chromosome

c.determine the size of a gene or sequencing of interest

d.replace the more tediuos method of library screening.

e.all of the above

19.For the colony hybridization procedure, which of thefollowing is true:

a.individual colonies are isolated and analyzed by DNAsequencing

b.the DNA fragments are separated by size prior to transfer ontonitrocellulose membrane

c.the plasmid DNA must be denatured before hybridization

d.the relative positions of DNA fragments on the agarose gel arepreserved after transfer to the nitrocllulose membrane

e.the cloned target sequence is detected by staining withthidium bromide

20.The following strand of DNA is to be used for sequencing3'CGACTGACTCGACGGCCTAGCTATAG.The primer is GCTGACTG.The number ofDNA fragments resulting from the ddCTP tube is expectd to be:

a.7

b.5

c.8

d.4

e.6

21.Which of the following is not true of heterozygosity for agene deletion?

a.it is often detrimental if there is a gene dosage effect.

b.as long as one wildtype copy of any deleted gene is presnt onthe homologous chromosome, no phenotypic consequnces willresult.

c.it could be detected by Southern blot analysis

d.it could be detected by PCR analysis

e.none of the above

22.Robertsonian translocation arise from:

a.breaks at or near the chromosome of two acrocentricchromosomes

b.breaks at or near th centromeres of two metacentricchromosomes

c.breaks at or near th centromeres of one acrocentric and onemetacentric chromosome

d.breaks in a large and a small chromosomes

e.transposable elements

If you are able to answer any or all of these that would begreat!!!

25. Talked about the warburg effect ( aerobic glycolysis) inwhich cancer cells preferentially use fermentation instead of goingthrough oxidative phosphorylation. Answer the followingquestions?

A) If cells are going to preferentially use fermentation, theyneed to transport higher levels of glucose into the cell. Explainwhy this is so?

B) Cells that preferentially use fermentation had higher levelsof the GLUT1 carrier protein at the membrane. explain the mechanismGLUT1 uses to bring glucose into the cancer cells?

C) If the GLUT1 transporter is composed of two main domains,what are their likely functions? what is the nature of the aminoacids within each domain?

D) What are two ways cells could increase the amount of GLUT1transporter on the cell membrane? Briefly explain each.

E) What is the one way the rate of movement of glucose into thecell through GLUT1 carriers could be altered ( either increase ordecreasd)? Describe a change you could make and be sure to indicatehow it would alter the rate of glucose uptake( either increase ordecrease)?

F) A drug called fasentin inhibits GLUT1 carriers. However, whenyou treat FTC133 cells with fasentin, they do not die. What is onehypothesis for how they are surviving in the absence ofglucose?

G What is a prediction that you could make based on yourhypothesis?

H) In the paper, the authors conclude that adding the PTEN backinto the cancer cells via transfection decreases the amount ofglucose that FTC133 cells take up. You hypothesize that HeLa cellsmight benefit from having additional PTEN protein. YOu transfectDNA coding for the PTEN protein into the HeLa cells and perform awestern blot to confirm PTEN expression. Unfortunately, you do notsee any expression of PTEN. What is one possible reason you dontget PTEN expression in your HeLa cells? ( HINT: you know yourantibodies are working and that transfection was successful, sothink about factors controlling gene expression)

I, Since you couldnt get PTEN expressions in the HeLa cells, youdecide to return to the FTC133 cells in order to test if the lipidphosphatase activity of PTEN is required for its ability todecrease glucose consumption. What type of mutation ( silent,missense, nonsense? frameshift or not?) would you introduce intothe PTEN gene in order to destroy the catalytic ability of PTENprotein while retaining the rest of the protein's structure andfunction? Explain your answer.

J) Next, you decide that you want to purify large amounts ofyour wild type and mutant PTEN proteins to test their catalyticabilities in in vitro. Your labmate suggests that you transform (insert) your DNA into bacterial cells because they are rapidlygrowing and are know to express large amounts of protein under theright conditions. You consider this but are unsure as to how wellyour human PTEN protein might be expressed in these cells. Based onwhat you know about the genetic code and about gene expression inprokaryotic and eukaryotic cells describe (i) one reason why yourgene might be able to be expressed in bacterial cells under theright conditions and (ii) two reasons why your PTEN gene might notbe expressed properly. (iii) Then, decribe one modification thatcould be made to your PTEN DNA to increase the likelihood that itwill be expresssed in the bacteria.

The bacterium E. coli, which is a major model system for the study of genes and DNA, has a singular circular chromosome of about 4x106 bp. In contrast, human diploid cells have __1__ chromosomes, of __2__ different types (don’t forget that the X ¹ Y!). The total amount of DNA found in a haploid cell such as a sperm or egg is called the organism’s “C-value”; the human C-value is __3__.

Also in contrast to the E. coli circular chromosome is the fact that eukaryotic chromosomes are not circular, but instead are __4__. Eukaryotic chromosomes are complexed heavily with protein to form a DNA-protein complex generically called __5__. The proteins found can be considered to fall into two general classes: __6__ and __7__. The first class is made up of 5 major proteins called __8__, __9__, __10__, __11__, and __12__. On a weight to weight basis, these proteins make up about __13__% of the total mass of a chromosome. Since these proteins must associate relatively non-specifically with DNA, they are rich in the basic amino acids __13__ and __14__, which gives the proteins a net __15__ charge.

One goal of packaging the DNA into chromatin is to compact the genome so it will fit into a nucleus. The basic unit of chromatin structure is the __16__, which has __17__ base pairs of DNA wrapped around a “core particle” composed of __18__ copies each of the core histones. Each nucleosome is separated by a stretch of “linker” DNA so that the repeating unit is approximately 200 bp. Lets do a little arithmetic: If we consider that the nucleosome is approximately __19__ nm in diameter, that there is about 200 bp in this structure, and that 200 bp of DNA would normally occupy __20__ nm if stretched out straight (recall that each base is spaced about 0.34 nm from the next), we can see that the nucleosome alone allows for a __21__-fold compaction ratio.

The nucleosome structure is further compacted by histone __22__, which is not a part of the core particle. Instead, this histone __23__ (does what?) to cause the nucleosomal 10 nm fiber to condense into a higher-order structure referred to as the __24__ nm fiber. In the nucleus, chromosomes do not just float around freely: they are tethered to a proteinaceous structure called the __25__ at specific DNA regions called “SARs”, which stands for __26__. SARs are rich in topoisomerases, to allow the winding and unwinding of the DNA that accompanies replication and transcription to occur, despite this constraint.

The 30 nm fiber discussed above is the form in which transcriptionally active DNA is believed to exist in the nucleus; however, further condensation leads to a highly condensed inactive form called __27__; in this terminology, the active form is referred to as __28__. The inactive condensed form can be divided into two types: one type is essentially invariant between cell types, and since it is “on all t time”, it is referred to as __29__. The other type comes and goes, depending on the expression of the genes present in that region of the chromosome; it is referred to as __30__.

Eukaryotic chromosomes carry the genes, of course, but also have other functional regions. One such region is the place where the spindle apparatus attaches during mitosis and meiosis; this region of the chromosome is called the __31__. Another important part of the linear eukaryotic chromosome is the ends; these are called __32__. In humans, the end of each chromosome is composed of many repeats of the short sequence __33__. One more type of functional region on each chromosome is the place where DNA replication starts; in humans, each chromosome has several of these __34__. Thus, eukaryotic chromosomes carry several different types of functional sequences, which together allow for their __35__ during S phase, __36__ during mitosis and meiosis, and the __37__ of their genes

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