BSC 114 Lecture Notes - Lecture 17: Acetyl Group, Oncogene, Chromatin

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plumtermite154

10 Nov 2016

School

University of Alabama

Department

Biological Sciences

Course

BSC 114

Professor

Daryl W.Lam

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Document Summary

Living organisms precisely regulate gene expression in response to environmental conditions. In multicellular eukaryotes, gene expression regulaates development and is responsible for differences in cell types: stem cells. Rna molecules play many roles in regulating gene expression in eukaryotes. One mechanism for control of gene expression in bacterial is the operon model. The operon can be switched off by a protein repressor. The repressor prevents gene transcription by binding to the operator and blocking rna polymerase. The repressor is the product of separate regulatory gene. Can be in an active or inactive form, depending on the presence of other molecules. A corepressor is a molecule that cooperates with a repressor protein to switch an operon off. E. coli can synthesize the amino acid tryptophan when it has insufficient tryptophan. By default, the trp operon is on and the genes for tryptophan synthesis are transcribed. When tryptophan is present, it binds to the trp repressor protien, which turns the operon off.

Related Questions

1. What is the term when the product of a biochemical pathway inhibits the activity of the first enzyme of a pathway?

		feedback inhibition.
		activator inhibition.
		diffusional regulation.
		enzyme activation.

2. Through control of gene expression, prokaryotes respond to changing __________ conditions.

		internal
		protein
		environmental
		genetic

3. A protein that negatively regulates transcription by binding to the operator is known as the

		operon
		repressor
		activator
		inducer

4. During attenutation, when tryptophan levels are high, the ________ stem-loop forms and transcription ________ the trpL gene.

		1-2, ends just past
		3-4, ends just past
		1-2, continues beyond
		3-4, continues beyond

5. The process of turning on the expression of a gene in response to a substance in the environment is called:

		constituation.
		repression.
		induction.
		catabolism.

6. What is a function of some specific transcription factors?

		to bind to the promoter region of the gene to facilitate the proper alignment of the RNA polymerase
		to bind to the centromere to induce meiosis
		to bind to enhancer regions to help regulate gene transcription
		to bind to the operator to recruit sigma factor

7. How does exposing an E.coli cell to glucose affect the regulation of the lac operon via CAP?

		cAMP binds to CAP and transcription is increased.
		cAMP binds to CAP and transcription is decreased.
		cAMP does not bind to CAP and transcription is increased.
		cAMP does not bind to CAP and transcription is decreased.

8. The lac repressor protein works when mutations in the repressor gene and structural genes are:

		in cis or in trans.
		in trans only.
		in cis only.
		in both prokaryotes and eukaryotes.

scarletwombat949

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

HELP ASAP;

Please help me understand my lab results, so be detailed to clearly understand . INclude answers to these questions what role does ampicilin (amp) and arabinose (ara) play in the experiment. Because GFP was isolated from jelly fisht, what modification to the GFP DNA had to be made to result in expression in E. coli?

BACKGROUND INFORMATION

Transformation of GFP and the pGLO Plasmid

GFP is a commonly used reporter protein in research labs, as the fluorescence creates a marker protein that can be used in many types of cell biology and biochemical studies. In basic research, GFP is often fused to a specific target protein of interest, creating a chimeric reporter protein. GFP has been used as a reporter protein to study blood vessel and tumor progression in mice, brain activity in mice, and malaria eradication in mosquitoes for instance (see website mentioned in notes).

In the first week you will use a procedure to genetically transform bacteria with the gene that codes for Green Fluorescent Protein. Following the transformation procedure, the bacteria express their newly acquired jellyfish gene and produce the fluorescent protein, which causes them to glow a brilliant green color under ultraviolet light. During the second week you will purify the protein away from all of the other proteins produced by the bacteria using chromatography, and during the third week you will evaluate the success of the purification procedure and estimate the size (molecular weight) of GFP by using sodium dodecylsulfate polyacrylamide gel electrophoresis (SDS-PAGE).

Fundamental to the process of genetic transformation of bacteria are plasmids. In addition to one large chromosome, bacteria naturally contain one or more small circular pieces of DNA called plasmids. Plasmid DNA usually contains genes for one or more traits that may be beneficial to bacterial survival. In nature, bacteria can transfer plasmids back and forth allowing them to share these beneficial genes. This natural mechanism allows bacteria to adapt to new environments. The occurrence of bacterial resistance to antibiotics is due in large part to the transmission of plasmids which contain antibiotic resistance genes.

The pGLO plasmid (Fig. 5, not a naturally occurring plasmid) contains the gene for GFP, and also contains the gene for the enzyme Î²-lactamase (bla), which provides resistance to the antibiotic ampicillin. The Î²-lactamase protein is produced and secreted by bacteria that contain the plasmid. Î²-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express Î²-lactamase can survive on plates that contain ampicillin. Only a very small percentage of the cells take up the plasmid DNA and are transformed. Untransformed cells cannot grow on the ampicillin selection plates. pGLO also incorporates a special gene regulation system, which can be used to control expression of the fluorescent protein in transformed cells. The gene for GFP can be switched on in transformed cells by adding the sugar arabinose to the cellsâ nutrient medium. Selection for cells that have been transformed with pGLO DNA is accomplished by growth on antibiotic plates. Transformed cells will appear white (wild-type phenotype) on plates not containing arabinose, and fluorescent green when arabinose is included in the nutrient agar medium.

for the first week section of lab

To each of two microfuge tubes, add 250ul of transformation buffer. Keep these tubes on ice.

Using a sterile loop, pick one colony of bacteria from the LB plate.

Transfer this colony to one of the microfuge tubes. Gently vortex the tube until the colony is dispersed. Repeat for the second tube.

Add 0.08ug of DNA to one of these tubes (+DNA tube).

Incubate both tubes on ice for 10 minutes.

Incubate both tubes at 42Â°C for 50 seconds.

Incubate both tubes on ice for 2 minutes.

Add 250ul of LB broth to each tube and incubate at room temperature for 10 minutes.

Flick each tube gently. Add 100ul of this transformation mix to the appropriate plates.

Results:

+DNA: LB/amp--> 1 colony, Glow: uncertain

+DNA: LB/amp/ara--> 9 colonies, Glow: yes

-DNA: LB/amp--> 7 colonies, GLow: no

-DNA: LB--> many colonies, Glow: no

BSC 114 Lecture Notes - Lecture 17: Acetyl Group, Oncogene, Chromatin

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