Lab Final Review Flashcards

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1
Q

Briefly explain the functions of the plasma membrane.

A

barrier, transport, structure, organization, recognition

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2
Q

How does a biological membrane differ from the dialysis tubing?

A

no proteins, phospholipid bilayer, no active transport

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3
Q

Explain the difference between simple diffusion and facilitated diffusion

A

facilitated diffusion requires a protein channel

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4
Q

What would happen if you ran the dialysis protocol, but failed to rinse off the sac after removing it from the large test tube? (How would the results differ from what you observed?)

A

Contamination, there would be outside solutes

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5
Q

Describe what you see under the high-power objective for sheep’s blood in 2M (2000mM) NaCl. Give an explanation.

A

This is a hypertonic solution, the cells would shrink

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6
Q

Why do you need the mortar and pestle for the wheat germ protocol but not the cheek cell protocol?

A

break cell wall

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7
Q

What is contained in the homogenate before the 1st centrifugation?

A

proteins and bits of cell wall, not needed

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8
Q

What is the purpose of Sodium Dodecyl Sulfate (SDS) in the experimental protocols?

A

it is a detergent that disrupted the bond between DNA and histones

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9
Q

Why do you use cold ethanol in this exercise?

A

DNA is insoluble with alcohol, this creates two distinct layers

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10
Q

Provide two reasons why you might not have isolated any DNA?

A

poured out precipitate instead of supernatant
denature in water bath if in there for too long

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11
Q

The restriction enzyme Alu I recognizes the sequence
5’….AGCT….3’ 3’….TCGA….5’ ↓
↑ How many times will Alu I cut the following sequence?
5’….GATACGTCTGAGCTCGTTAACACTCTGCATTGCTGACGTTAACTGAGCTCAGGGATAG….3’ 3’….CTATGCAGACTCGAGCAATTGTGAGACGTAACGACTGCAATTGACTCGAGTCCCTATC….5’

A

2X

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12
Q

What is different about the DNA fragments created by digestion with Alu I as com-pared to the DNA fragments created by digestion with EcoRI?

A

EcoRi creates sticky ends
Alu creates blunt ends

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13
Q

Would the EcoRI enzyme digest the DNA fragment above? Why or why not?

A

no, linear sequence not there

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14
Q

For circular DNA, the enzyme will

A

cut twice to create two fragments

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15
Q

Which of the DNA samples have the same number of restriction sites for the restriction endonuclease used? Write the lane numbers. (Assume all samples were circular plasmids before digestion.)

A

S2 because they have the same number of black boxes as CS

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16
Q

Which sample has the smallest DNA fragment? Which sample has the largest? Do you need to run a KB ladder to know which sample has the smallest DNA fragment? Why or why not?

A

Smallest: S4, lane 6 (fragments are further down)
Largest: S2, lane 4 (Fragments highest up)
No bc the fragments automatically order by size

17
Q

Assuming a circular piece of DNA (plasmid) was used as starting material, how many restriction sites were there in the DNA sample in lane 4? Now assume the plasmid used as starting material was linearized. How many restriction sites were there in the DNA sample in lane 4?

A

Restriction sites: 3 (plasmid)
Linear: 2

18
Q

Based on your analysis of the gel, what is your conclusion about the DNA samples in the picture? Do any of the samples seem to be from the same source? If so, which ones? Describe the evidence that supports your conclusion.

A

CS and S2 same source

19
Q

Refer to Table 7-1 to answer the following question. Are there any temperature values where the absorbance values for the “E” tube are quite close to the absorbance values for the “C” tube? If so, which temperature(s)? Provide a reasonable explanation for this outcome.

A

30 and 37C are similar likely because they are similar in temperature

20
Q

Explain how the crossing-over of homologous chromosomes leads to differences in gametes.

A

the alleles swap and lead to chromosomes with different genotypes

21
Q

Is it possible that one of the 2 cells that will result from completion of meiosis I contains both D alleles (DD)? Explain why or why not.

A

Yes

22
Q

How many T alleles will a gamete contain?

A

One

23
Q

Please draw a cell (2n = 6) in metaphase of mitosis.

A

Include spindle fibers and chromosomes at plate

24
Q

cell that is (2n = 4) undergoes meiosis. Please draw one of the four cells that result from completion of the second meiotic division.

A

nuclei with two haploid chromosomes chillin

25
Q

What advantage would there be for an organism to be able to turn on/off particular genes in response to certain conditions?

A

doesnt waste resources, adapt to environment

26
Q

Which plates could contain genetically transformed bacterial cells? Explain your answer

A

the +pGLO plates will most likely have transformed
the LB can’t differentiate

27
Q

Bacteria that had not taken up the pGLO plasmid could grow on which of the plates used in this experiment?

A

-pGLO/LB would grow E Coli because LB doesn’t contain ampicillin but does contain the nutrients that would allow E Coli to grow

28
Q

Which plates should be compared to determine if any genetic transformation has occurred? Why?

A

+pGLO LB/amp/ara and -pGLO/LB/amp
The -pGLO demonstrates that E Coli being used does not survive in the presence of ampicillin

29
Q

Which plate(s) should contain the greatest number of genetically modified bacteria?

A

+pGLO/LB/amp/ara plate

30
Q

Would it ever be possible for a bacterium that did not pick up the pGLO plasmid to live on the LB plates containing ampicillin?

A

Yes bc transfering plasmids can give it resistance

31
Q

How would you expect your transformation efficiency to be affected if you did not heat shock the bacteria? Why?

A

it would be lower because the heat shock opens up the pores, no bacteria would be present without it

32
Q

How would you be able to test if the colonies you observe on the LB-only plates are ampicillin resistant?

A

add ampicillin and add a bit of bacteria

33
Q

Calculate the transformation efficiency of the following experiment using the information and the results listed below and fill in the chart on the reverse side of this page.

DNA plasmid concentration: 0.5 μg/μl
250 μl CaCl2 transformation solution
10 μl pGLO plasmid solution
250 μl LB broth
100 μl spread on agar
10 colonies of transformants

A

0.5 ug/ul (DNA plasmid concentration) x 10uL (pGLO plasmid solution) = 5ug

250+10+250=510 total

100uL (spread on agar) /510uL (total)=0.196

5ugx0.196=0.98

10/0.98=10.2

34
Q

If a particular experiment were known to have a transformation efficiency of 3 × 103 bacteria/μg of DNA, how many colonies would be expected to grow on the LB/amp/ara plate? (Assume that the concentration ofDNA and fraction ofcells spread on the LB agar are 0.196)

A

3x10^3=x/0.196

x=588