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The dyes that you separated using gel electrophoresis were: Orange G (yellow), Bromophenol blue (purple) and Xylene cyanole (blue). What electrical charge did these dyes carry?
They carried a negative electrical charge.
The dyes that you separated using gel electrophoresis were: Orange G (yellow), Bromophenol blue (purple) and Xylene cyanole (blue). What electrical charge did these dyes carry? [answer: They carried a negative electrical charge.] <---[part A] [actual question] -->What evidence allowed you to arrive at this conclusion
They dyes slowly parted from the black side [negative] and traveled to the positive side since the positive is attracted to the negative.
Molecular size can play a role in separation with small molecules moving through the gel matrix more rapidly than larger molecules. The formula (or molecular) weights for these dyes are Orange G (452.38), Bromophenol blue (669.98) and Xylene cyanole (538.62). From your results, did it appear that these molecules were separated clearly on the basis of size?
No, the dyes didn't separate according to size since Xylene cyanole [blue] migrated slower than bromophenol blue [purple].
Molecular size can play a role in separation with small molecules moving through the gel matrix more rapidly than larger molecules. The formula (or molecular) weights for these dyes are Orange G (452.38), Bromophenol blue (669.98) and Xylene cyanole (538.62). From your results, did it appear that these molecules were separated clearly on the basis of size? [answer: No, the dyes didn't separate according to size since Xylene cyanole [blue] migrated slower than bromophenol blue [purple].] <--[part A] [actual question] --> What other factors may have played a role in the separation of these dyes
molecular shape and degree of electronegativity
Which tube contained a single dye? A, B or C? [answer: tube C] <---[part A] [actual question] ---> Name this dye.
When aspirating a solution, why is it important to actually see the solution enter
the pipette tip
You may be aspirating air in the pipette tip instead.
Which antibiotic resistant gene, is found in pARA? In pKAN-R?
In pARA, there is amp[r].
In pKAN-R, there is kan[r].
One of these plasmids has been engineered to express a protein if the gene for that protein is inserted into a specific location. Which of the plasmids has been engineered for
Briefly outline, in your own words, the steps required to set-up the plasmid restriction digest.
If pKAN-R is digested with BamH I & Hind III, the rfp will be cut from plasmid. You will remove the rfp gene from pKAN-R and remove 40bp fragments from pARA. Then insert rfp gene in pARA, producing a recombinant DNA molecule.
Why do you suppose you are asked to set up two tubes without BamH I and Hind III?
to compare results; to see how BamH I & Hind III affects the solution; lets us know if enzyme is not working
In a 5'--->3' direction, what sequence of bases represents the "sticky-ends for each?
BamH I is GATC.
Hind III is AGCT.
Fill in the blanks: pARA digestion will yield ____ fragments and will be _____ base pairs in length.
2 fragments & 4018 bp in length
Fill in the blanks: pKAN-R digestion will yield _____ fragments and will be _____ bp and _____ bp in length.
2 fragments, 4706 bp, 702 bp in length
Assume you were given a culture of bacteria carrying one or both of these plasmids. Design a simple experiment that you could use to determine which of these plasmids, pARA or pKAN-R, the bacteria in the culture were carrying.
Grow bacterium & purify the plasmid from the cells.
Briefly define the term "recombinant DNA."
DNA molecule composed of ligated fragments taken from DNA molecules.
In order for two sticky ends to join together, what relationship needs to exist between them?
The nucleotide bases that compose the sticky ends must be complementary.
What properties of the restriction fragments produced in Lab 2 are needed to permit ligation of these fragments
Both plasmids were cut or digested with same restriction enzymes, BamH I and Hind III the resulting sticky ends must be complementary.
What is the importance step 1 in this protocol? [step 1: Obtain your A+ and K+ tubes from the rack at the front of the class. Place the two tubes in the 70°C water bath for 30 minutes. This heat exposure will denature (inactivate) any
BamH I and Hind III that might be active.]
to denature BamH I & Hind III. If the enzymes remain active, the restriction fragment can't be put together
Why was it important to place the A+ and K+ tubes in the 70°C water bath before
setting up the ligation reaction?
The heat exposure will denature [inactivate] any BamH I & Hind III that might be active.
What do you think might have happened if this step was omitted? [step: place the A+ and K+ tubes in the 70°C water bath before setting up the ligation reaction]
The ligase wouldn't be able to connect because restriction enzymes don't allow, keeps cutting it.
Could two rfp fragments join together and circularize in the Ligase tube?
Yes, the plasmids can make new parts & form recombinant DNA.
In the DNA molecule, there are two kinds of chemical bonds: covalent chemical bonds and hydrogen bonds. Brieﬂy describe how these bonds differ in strength and where, in the DNA molecule, you would ﬁnd them.
Hydrogen bonds are weaker. They are found between nucleotide bases. Covalent bonds are stronger & are found between sugar & phosphate groups of DNA.
During ligation, which of the bonds (hydrogen or covalent) form ﬁrst? Where do they form? Which bonds form next and where do they form?
Hydrogen bonds form first. Hydrogen bonds form between complementary bases. Covalent bonds come next & they form between sugars & phosphate groups.
Beginning with the restriction digest of pARA and pKAN-R, briefly describe the steps required for constructing your recombinant plasmid from pARA and pKAN-R restriction fragments. Explain your answer including the manipulations of these two plasmids and their restriction fragments.
1. Combine 2.5x buffer with pARA & enzymes. Use pKAN-R, not pARA. When enzymes incubated, then will split to 2 parts each.
2. Heat tubes will denature enzymes with 5x buffer, ligase, & distilled water.
3. Control tubes come together & hydrogen bonds form at sticky ends of fragments. Ligase makes covalent bond stronger. The plasmid with rfp and ara genes will be used.
How would you know if a bacterium gets transformed with a plasmid containing the amp[r] gene?
It will make a protein allowing it to live in areas with ampicillin.
How would you know if a bacterium gets transformed with a plasmid containing the amp[r] gene? [answer: It will make a protein allowing it to live in areas with ampicillin.] <--[previous question] [actual question] --> What would be the size of the fragment carrying this gene?
How would you know if a bacterium gets transformed with a plasmid containing the amp[r] gene? [answer: It will make a protein allowing it to live in areas with ampicillin.] What would be the size of the fragment carrying this gene? [answer: 4018 bp]<--[previous questions][actual question]--> From what plasmid did this fragment originate?
How is the P + culture treated differently from the P - culture?
P+ received recombinant plasmids.
P- didn't receive plasmids.
What do all of the cells growing on the LB/amp and LB/amp/ara plates have in common?
resist to ampicillin
What do all of the cells growing on the LB/amp and LB/amp/ara plates have in common? [answer: resist to ampicillin] <--[previous question] [actual question] -->What single RESTRICTION FRAGMENT must they all contain to grow on plates with ampicillin?
They must all have the amp[r] gene located in 4018 fragment.
Would you expect that all of the cells growing on the LB/amp/ara plate were transformed with the same plasmid? Explain.
No. Many possibilities. All need amp[r] on pARA fragment.
How might you determine which of the cells on the LB/amp/ara plate contain pARA-R, the recombinant plasmid that you've made by ligating the rfp gene with the large pARA restriction fragment?
red will have pARA-R
Would you expect that some of the bacteria on the LB/amp plate were transformed with pARA-R? Brieﬂy describe how you might test your answer
There is some bacteria on the P+ side of the plate since ampicillin kills bacteria cell walls, & pARA-R resists to it, maybe we can inject arabinose into the cell & see if they grow because arabinose is needed to express the rfp gene.
Why is it important that the cells you collect from your mouth are mixed with Chelex?
Chelex beads will bind divalent magnesium ions; reducing, degrading of DNA so we are able to lyse the cell & liberate the DNA.
The "primers" that are used in this PCR are unique to what locus (location or DNA segment)?
tPA gene; location is in intron-Alu element
What is an "Alu element?"
part of DNA coding for RNA molecule that aids secretion of newly formed polypeptides from the cell.
Does everyone carry an Alu element in the region that we are amplifying by PCR?
No, b/c the locus is dimorphic; it has 2 forms, some forms may have Alu element.
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