BIOL-UA 21 Lecture Notes - Lecture 18: Myod, Transcription Factor, Regulatory Sequence

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18 Mar 2016

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One genome encodes for many types of cells. We have a single genome which is the same in every cell however cells are all diferentiated by regulation in the transcription of the genome coding for speciic genes that code for speciic proteins. Transcription factors act to define cell identity. Transcription factors that regulate a number of target genes, example of target gene is myod. Transcription factor (red) binds to the dna and the binding event ensures that it binds to certain specific sequences and not other. Ebox dna sequence motifs = found upstream and are enhancers of genes that are myod targets. Myod coordinately regulates many genes that are expressed in muscle cells by binding to specific target sites. Always not a single transcription factor is responsible for one cell type. Not always on & of more of an attuning. Light switch that slides up or down.

Related Questions

QUESTION 1

How is a bone marrow transplant able to treat someone with leukemia?

		The donor blood contains antibodies that destroy cancer cells
		Donor blood contains stem cells that regenerate cells in the immune system
		Bone marrow contains induced pluripotent stem cells that turn into red blood cells
		Bone marrow contains tumor supressor genes that slow down leukemia growth
		Bone marrow contains pluripotent human embryonic stem cells

1.2 points

QUESTION 2

What would happen if siRNAs designed to target mRNA from an oncogene were introduced into cancer cells?

		Increased genomic instability would result from adding foreign nucleic acids
		New mutations would arise in the oncogene
		The oncogene would be converted back into a proto-oncogene
		The siRNAs would compete with the oncogene's active site
		Production of the oncogene protein would decrease

1.2 points

QUESTION 3

What is the purpose of chromatin-remodeling enzymes in human cells?

		The enzymes control the rate of translation of an mRNA transcript
		The enzymes add phosphate or methyl groups to histone tails
		The enzymes slide nucleosomes farther apart or closer to each other
		The enzymes send damaged proteins to the proteasome for degradation
		The enzymes cause DNA looping to occur

1.2 points

QUESTION 4

If you wanted to resurrect the mammoth, how would you do it?

		Use somatic cell nuclear transfer to insert the mammoth genome into the egg of an elephant
		Use transcription factors to re-program adult cells from elephants into mammoth cells
		Change the gene expression patterns of elephants to increase production of mammoth genes
		Use siRNAs to decrease expression of elephant-specific genes in an elephant embryo

1.2 points

QUESTION 5

How is eukaryotic transcription initiated?

		RNA polymerase is recruited to the start codon by transcription factors
		RNA polymerase begins transcription at the enhancer sequence
		RNA polymerase is activated by binding to GTP
		An initiator tRNA binds to the ribosome
		A complex of proteins forms at the promoter of a gene

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

		The mutation changes the number of domains in the enzyme, which makes it work more efficiently
		The mutation changes the amino acid sequence of the lactase protein
		The mutation increases the number of copies of the lactase gene that will be found in your genome
		The mutation changes whether the lactase sequence is found in an intron or exon
		The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

		The substrate will increase the reaction rate by binding to the allosteric site
		The substrate will increase the reaction rate by competing with the inhibitor for the active site
		The reaction rate will not change unless the inhibitor can be removed
		The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind
		The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

		Transcription regulators bind to the 5' UTR region of a gene
		Regulators bind via complementary base-pairing to certain DNA molecules
		Covalent bonds form between the transcription regulator and the atoms of the DNA backbone
		Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene
		Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

		DNA -> RNA -> protein
		All cells arise from pre-existing cells
		DNA provides the complete instructions to create a cell
		The identity of a cell is determined through gene expression patterns
		All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

		Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment
		Enzymes to digest lactose are only made when energy is low and lactose is available
		The activator can override the inhibition of the lac operon by the repressor
		The repressor can control the enhancer, while the activator can control the promoter
		When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

		Phosphorylation and acetylation of DNA affect its ability to be compacted
		Changes to the sequence of DNA change whether DNA will wrap around histone proteins
		Covalent modifications of histones affect the ability of the transcription initiation complex to form
		Histones provide the codon sequences needed for translation to occur
		The histone code affects which amino acids will get added to a polypeptide

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

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BIOL-UA 21 Lecture Notes - Lecture 18: Myod, Transcription Factor, Regulatory Sequence

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