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Approximately 28 drops of liquid, from a medicine dropper or disposable pipette, equals 1 mL. Each drop represents what part of one mL? [Show your work.]

If 1mL/28 drops, then .036mL/drop=0.036mL

Approximately 28 drops of liquid, from a medicine dropper or disposable pipette, equals 1 mL. Each drop represents what part of one mL? [Show your work.] [answer: If 1mL/28 drops, then .036mL/drop=0.036mL] <---[part A][actual question]-->Approximately how many microliters are there in one drop of liquid?

.036mL X 1000 microliters/mL=36 microliters [per drop]

4 hairs, plucked from you head will yield about 1 microgram of DNA. How many micrograms of DNA can be extracted from a SINGLE hair root? [Show your work.]

If 1 microgram of DNA/4 hair roots, then .25 micrograms/root or 1 hair root=.25 micrograms of DNA.

4 hairs, plucked from you head will yield about 1 microgram of DNA. How many micrograms of DNA can be extracted from a SINGLE hair root? [Show your work.] [answer: If 1 microgram of DNA/4 hair roots, then .25 micrograms/root or 1 hair root=.25 micrograms of DNA.] <---[part A] [actual question]---->How many nanograms does this represent?

1 gram=1,000,000 micrograms
1 gram=1,000,000,000 ng [nanograms]
1,000,000 micrograms=1,000,000,000 ng
1 microgram=1000 nanograms
.25 micrograms=250 nanograms

ANSWER: 250 nanograms

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?

tube C

Which tube contained a single dye? A, B or C? [answer: tube C] <---[part A] [actual question] ---> Name this die.

Xylene cyanole

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.

After loading your gel, did any solution remain in tubes A, B or C

Yes

What could account for solution remaining in these tubes?

Air took up some space.

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
protein expression?

pARA

From what organism did the gene for mutant Fluorescent Protein come?

sea anenome

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. 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 [pARA & pKAN-R?]; lets us know if enzyme is not working

What is the RECOGNITION SEQUENCE for BamH I?

5'...GGATCC...3'[arrow pointing down on left 'G']
3'...CCTAGG...5'[arrow pointing up on left 'G']

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 & 40 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.

What does the term ligation indicate?

ligate-bind together

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 complete.

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 complete.

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.

DNA ligase is required to form which bond?

covalent bond

How did your actual gel results compare to your gel predictions?

They were poor compared to our actual gel predictions. Our bands did not spread as well. They were faint, disappearing halfway.

Are there any bands, appearing in your gel photo, that are not expected?

No, we were missing some bands that were supposed to appear.

What could explain the origin of these unexpected bands

could be result of dye splitting DNA.

Do you see evidence of the three plasmid forms in the uncut lanes?

No, don't see plasmid formations.

Is there evidence of more than one form of multimer?

NO, don't see any.

Why are the ligated plasmids so close to the well?

They're heaviest with the most base pairs.

Two of the 702 bp pKAN-R fragments, rfp gene fragments, may form a circularized fragment because each end of the fragments terminates in BamH I and Hind III sticky ends. Is there evidence
of a circularized 1404 bp fragment in the ligated lane?

No, our experiment went wrong due to wrong measurements.

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.

What is "transformation?"

process of taking up foreign pieces of DNA into bacteria cell

What does it mean if cells are "competent"?

ready to receive plasmids.

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?

4018 bp

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?

pARA

How is the P + culture treated differently from the P - culture?

P+ received recombinant plasmids.
P- didn't receive plasmids.

What is the purpose of the P- culture?

plasmid control

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?/or?] 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

If your actual results differed from your expected, propose some reasons that might explain these differences.

The reason why we messed up is because we added too much distilled water & the combo was too distilled.

How many red colonies were present on your LB/amp/ara plate?

None, the experiment went wrong.

How many red colonies were present on your LB/amp/ara plate?[answer: ] <--[previous question] [actual question] -->Why did the red colonies only appear on this plate and not the LB/amp plate?

Some genes did not ligate correctly.

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.

What is the purpose of boiling the cell sample?

lyse the cells & destroy some of the nucleases degrading DNA

What does PCR accomplish?

makes copies of selected gene segment or locus of 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.

How many base pairs is the Alu element?

around 300 bp

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.

With respect to the tPA gene, how many genotypes are possible?

3 ++, + -, - -

Calculate the frequency (percentage) of each genotype present

Alu+Alu+: 23.33%; Alu+Alu-: 20%; Alu-Alu-: 60%

Calculate the frequency of each allele present in your class

Alu+: 34.48%; Alu-: 65.52%

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