BIOL 112 Lecture Notes - Lecture 10: Chromosome, Transfer Rna, Consensus Sequence

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28 Jan 2015

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BIOL 112 Full Course Notes

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Cells can sense and respond to their environment by changing their gene expression. Two mechanisms for gene regulation in which a small molecule or signal enters the cell and changes gene expression are called: Additional bacterial responses within the human environment may also include synthesis and secretion of harmful toxins. These responses are due to the ability of bacteria to regulate their gene expression. The regulation of gene expression is also important for multicellular eukaryotic organisms. Abnormal regulation of gene expression can lead to developmental defects as well as diseased conditions such as cancer. 1. 1 environmentally regulated gene expression. Genes that are expressed all the time are called constitutively expressed genes. Rna polymerase subunits, ribosome components, and amino acyl trna synthetases are all examples of constitutively expressed genes. The level of expression of these genes varies. Constitutively expressed genes are always being transcribed to some extent, though.

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True or False????? HELP

1. In the field of epidemiology, increased prevalence of a disease in a population is the universal signal that disease control and treatment are not effective.

2. Cytoplasmic membranes of prokaryotes typically contain more proteins than cytoplasmic membranes of eukaryotes microorganisms.
3. Sanger (or dideoxy) DNA sequencing produces relatively short DNA sequence data, but is by far the least expensive DNA sequencing method currently available.
4. The (italicizes) abbreviation lac Z refers explicitly to the beta galactosidase protein/ enzyme.
5. Archaeal pathogens have been linked to some of the most virulent infectious diseases. 6. A bacterial mutant that is unable to make an amino acid required for growth (relative to its wild-type strain is termed an auxotroph.
7. When counting plaques to enumerate phage particles (as in the BIOL2026 Laboratory), the plaque represents lacK, not presence, of bacterial growth.
8. Because of competition with plant roots for nutrients, microbial numbers in the rhizosphere are typically much lower than in surrounding bulk soil.
9. Chemolithoautotrophy is the strategy that facilitates the existence of biological communities deep beneath the earthâs surface and near hydrothermal vents.

10. In a primary treatment (1 Â°) system of a municipal sewage/ wastewater treatment plant, the main goal is to remove ammonium (NH4+) using specific bacterial enrichment cultures.
11. The small intestine in human is the main site of microbial breakdown of partially digested food.
12. Root nodules of legumes contain nitrogen fixing bacteria.
13. Anoxygenic photosynthesis involves using/ removing electrons from H2O.
14. Algae and Cyanobacteria utilize generally the same two-photosystem biochemical mechanisms in the light- reaction of photosynthesis.
15. In bacteria, the only known reason proton motive force is generated is to power ATP synthase in cellular respiration/ oxidative phosphorylation.
16. Anaerobic respiration is only carried out by prokaryotes, however fermentation can be carried out by both prokaryote and eukaryal microorganism.

17. The general structure and biochemical âmotorsâ of flagella are similar between Bacteria and eukaryotic cells, but fundamentally different in the Archaea.
18. Antibiotic resistance genes appeared in microbes (for the first time ever) only after human .. producing and releasing massive amounts of antibiotics to the environment.
19. Experiment by Esther Lederberg, and then later by Luria and Delbruck (that won them a Nobel prize), illustrated that genetic mutations in bacteria arose in the absence of selective pressures, rather than in response to selection.
20. Quaternary sensing is a mechanism used by bacteria to sense other bacteria and regulate gene expression accordingly (e.g. In the case of expressing the liac operon or initiating EPS production in biofilms).

Please I need help answering those questions i always lose bunch of points i'm not doing it correctly.

Use information from Nature Reviews Microbiology to answer the following question.What is the role of EF-G and EF4 in protein synthesis? Under what circumstances does EF4 participate in translation and what is its role?

from this reading The Highly Conserved LepA Is a Ribosomal Elongation Factor that Back-Translocates the Ribosome Qin Y,Polacek N, Vesper O, Staub E, Einfeldt E, Wilson D.N, and Nierhaus N.H. Cell 127: 721-733 (2006).

Q2. In which species is Lep A found and in which ones is it absent? Is the result expected based on your knowledge of the endosimbiosis theory?

Q3.

3a What is the percentage of identity between LepA and its ortholog genes?

3b. (5pts) Indicate what domains are conserved between Ecoli EFG and LepA, and which ones are not.

3c. What do grey and black boxes in the sequence alignment indicate?

Q4. Analysis of in vivo experiments shown in Figure 3. Only analyze graph on the left side of the figure.

4a. What is the difference between the bacteria whose growth are shown by the blue and the green curves? What can be concluded about the effect of EFG on bacterial growth?

4b. Which protein is expressed by bacteria whose growth is shown by the yellow curve? Does the expression of this protein affect bacterial growth? Compare the yellow and blue curves. How can this result be explained?

4c. Explain why the addition of IPTG results in lower bacterial growth. Compare the yellow and red curves?

4d. Why is the green curve an important control of the experiment? The green curve.

The next questions are based on the reading of the article Elongation factor 4 (EF4/LepA) accelerates protein synthesis at increased Mg2+ concentrations.

Elongation factor 4 (EF4/LepA) accelerates protein synthesis at increased Mg2+ concentrations Pech M, Karim Z, Yamamoto H, Kitakawa H, Qin Y, and Nierhaus K. PNAS 108: 3199-3203(2011)

Studies done in E. coli using a knockout mutant of the lepA gene (ÎlepA , lepA gene has been deleted) did not show any phenotype when grown in rich LB medium. This result was unexpected due to the high level of conservation of EF-4 among species.

Q5. Equal amounts of wild type and mutant cells (lacking the lepA gene) were mixed and co-cultured for 24h. The population distribution was determined every 4 hours. Figure 1 indicates changes in the % of mutant cells over several generations.

Q5a. What is the question that researchers addressed in the experiments shown in Figure 1?

Q5b. Which curve represents the control of the experiment? Do the two strains grow at similar rate? How do you know?

Q5c. Do high salt concentration and/or low temperature have any effect on the growth of mutant cells?

Q5d. What can you conclude about the role of LepA on bacteria growth? When is this protein important?

Q6.

6a. (6pts) Where in the cell is EF4 located? Did you expect that this would be the EF4 location? Why or why not?

What is the question that the researchers want to address in the experiment shown in Figure 2?

6c. Analyze the green and red bars graph shown in Figure 2. What technique was used to detect EF4? Is there any change in EF4 localization?

Q7. Summarize the main points of the model shown in Figure 4

ochregerbil875

HELP ASAP; Please help me understand my lab results, so be detailed to clearly understand

Then , Please include answers to the question below to include in your response

1. what role does ampicilin (amp) and arabinose (ara) play in the experiment.

2) 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 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.

The transformation procedure involves three main steps. These steps are intended to introduce the plasmid DNA into the E. coli cells and provide an environment for the cells to express their newly acquired genes. 1) Cells are first treated with a solution of CaCl2, which is thought to neutralize the repulsive negative charges of the phosphate backbone of DNA and the phospholipids of the cell membrane, allowing the DNA to adhere to the cells. This is referred to making the cells âcompetentâ (i.e. the cells are capable, or competent, to take-in exogenous DNA). 2) Cells are then subjected to a heat shock, which is thought to increase the permeability of the cell membrane, allowing the cells to take-in the plasmid DNA. 3) Cells are allowed to grow in a ârecoveryâ phase in the absence of ampicillin. During this time the cells begin to produce the Î²-lactamase protein, which will allow them to survive when they are subsequently placed on agar plates containing ampicillin.

For the first week section of lab

To eac

h 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 t

his transformation mix to the appropriate plates.

Results For plates:

+DNA: Has the E. Coli pGlo plasmid and - DNA plates do no have EColi pGLo plasmid .

Amp means ampiclin and Ara means arabinose on agar plates

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

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

--DNA LB/AMP- 1 colony so no colony growth and no glovw

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

Please explain significance of result or thand the reason for the results in scientific terms based on the background information provided and your own knowledge to be detailed. Would these results be expected as wll.

BIOL 112 Lecture Notes - Lecture 10: Chromosome, Transfer Rna, Consensus Sequence

BIOL 112 Full Course Notes

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