answer

watching

views

3 Apr 2018

You plate a sample containing two different bacterial species onto MacConkey agar and Mannitol Salt Agar plates. Nothing grows on the Mannitol Salt Agar. One colony type grows up on the MacConkey plate. It is pink in color. What can you conclude about your two mystery bacteria (explain how you arrived at your conclusions)? What media would you use next to isolate the second species and why did you pick that media?

answer

watching

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Sixta KovacekLv2

5 Apr 2018

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Related questions

This experiment is about Affinity Purification of Green Fluorescent Protein (GFP)

please answer the below questions

1. What is a bacterial colony?

2. Why did you pick one green colony and one white colony from your agar plate(s)? Why do you think you picked one of each color? What could this prove?

3. How are these items helpful in this cloning experiment? a. ultraviolet (UV) light - b. incubator - c. shaking incubator

4. Explain how placing cloned cells in nutrient broth to multiply relates to your overall goal of purifying the fluorescent protein.

5. You have used a bacterium to propagate a gene that produces a green fluorescent protein. Identify the function of these items you need in this procedure. a. Centrifuge â b. Lysozyme â c. Freezer â

6. Can you explain why both liquid cultures fluoresce green?

7. Why did you discard the supernatant in this part of the protein purification procedure?

8. Can you explain why the bacterial cellsâ outer membrane ruptures when the cells are frozen. What happens to an unopened soft drink when it freezes?

9. What was the purpose of rupturing or lysing the bacteria?

10. Briefly describe hydrophobic interaction chromatography and identify its purpose in this lab.

This lab is about Crystal Growth of Hen Egg White Lysozyme so please answer the following question

1. What condition yielded the "best" crystals of lysozyme?

2. What changes did your group try to make, and what effect did they have on the crystals?

3. How many of your conditions yielded crystals? How many did the entire class get? Give the percentage (number of conditions that gave crystals/number of different conditions that were set up) for your attempts as well as for the entire class. What does this tell you about protein crystallization?

4. What methods can you think of to determine if the crystals are lysozyme and not salt (many salt crystals are colorless as well)? (Hint: think of protein properties and ways to assay and look at proteins).

5. Many protein crystals are formed at or near the pI of the protein being studied. Can you rationalize any reason why this observation might be made (in other words, why is the pI so good at helping form protein crystals).

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

In a genetic experiment:

The use and limits of RNA interference as a reverse genetics technique for knocking down gene expression were tested. The wild-type and mutants worms for the strains were selected. The RNAi feeding strains and control bacterial were used for the the RNAi experiment. The eri-1 C. elegan strains (wild-type) were used for RNAi treatment and mutant strains that were used ( unc-22 and dpy-13) C. elegans. Both carbencillin and IPTG wered added to the RNAi feeding plates and E.coli containing the RNAi transgenes were spread on the plates. The antibodic carbacillin only allows growth of bacteria which contains the RNAI construct. IPTG induces the expression of the polymerase which then transcribes the siRNA in the bacteria. 6 NGM plates without bacteria were used and 20ul of 12.5 mg/ml carbencillin concentration were spread on the 6 of the plates. The carbencillin dried and then IPTG was spread (25ug/ml) on each plate. Then 6 plates with the RNAi bacterial strains were used. Theree of plates for strain one and the other three for strain the second strain. When IPTG dried , 150 ul of the appropraite bacteria was added onto each plate. 2NGM plates (without carbenicillin or IPTG) was spread with 150 ul of OP50 bacterial culture. OP50 is the regular E. coli strain for culturing C. elegans and does not contain the RNAi constructs. These plate were used as the control plates6. Then the plates were incubated for two days. Then the adult hermaphrodites from the RNAi feeding plates and the op50 plates were removed. Then from the eri-1 wildtype stock, 2 or 3 L4 hermaphrodites or
young adults were transferred onto each RNAi feeding plate. Then L4 worms were transferred to an intermediate plate first. After they have a minute to move away from each other and the eggs, and transfered to the RNAi plates. On the two control plates seeded with OP50, 2 L4 or young adult hermaphrodites were placed there. Plates were obtain as the reference mutants (unc-22, bli-1 or unc-13). Then the plates were incubated to allow the worm to develop and lay eggs in two days. Then the adult hermaphrodites from your RNAi feeding plates and the op50 plates were removed and then placed back into the incubator.

The op500 was the negative control plate wtih 80 and 300 progeny and unc-22 had around 50. (reference plates)

Mating in the RNAi plates

Dpy (60) x wildtype (15) and unc( 70) x wildtype ( 8) progeny

the phenotype was movement and it seemed similar phenotype within the progeny.

Can you help me understand the results and what is happening genetically in the experiment ? Are the for dyp-15 x unc-22 mating cross and unc-22 x eri-1 wildtype C. elegans. Based on the data:

Briefly discuss the results of the RNAi experiment. How closely did the RNAi-treated worms phenocopy the mutant worms?

Was the gene knockdown 100% effective?

Was the RNAi treatmentequally effective for the two genes? If not, discuss this result and why that could occured.

Please be specific and detailed

Genetics HELP ASAP

Weekly leaderboard

Home

Homework Help 3,900,000

Biology 460,000

Unlock all answers

Related textbook solutions

Molecular Cell Biology

Biology: Science for Life with Physiology

Concepts of Biology

Biology

Essentials of Biology

Related questions

Related Documents

BIO 226 Study Guide - Midterm Guide: Exponential Growth, Eosin, Hemolysis

BIO 2200 Lecture Notes - Lecture 3: Polymerization, Motility, Numerical Aperture

BIOS 111 Lecture Notes - Lecture 12: Crystal Violet, Peptide, Meat Extract

Weekly leaderboard