Lab Test 2 Flashcards

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1
Q

Can you explain some purposes of generating genetically modified crops plants?

A
  • to encourage specific traits such as…
    high yield, undesirable to pests, delayed fruit ripening, reistant to drought
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2
Q

Can you briefly explain what is a GMO?

A

Farmers have been genetically modifying crops for centuries and crop breeding to encourage specific traits, such as high yield, is still an important part of agriculture today. However, there is now the option to place genes for selected traits directly into crop plants. These genes do not have to originate from the same plant species—in fact, they do not have to come from plants at all. GMO is the genetic modification of a plant in order for traits to be amplified or introduced into an organism.

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3
Q

What is the general purpose of PCR?

A

PCR is DNA replication in a test tube. PCR allows you to amplify specific sections of DNA and make millions of copies of the target sequence.

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4
Q

To prepare a DNA sample for PCR, do you know which reagents are required and what is their
respective function(s):

A

o Template DNA (what is this)?
This is the DNA molecule containing the segment of DNA you wish to amplify.
o How many primers/oligonucleotides per target DNA sequence to be amplified?
Small pieces of single stranded DNA (typically 18 to 25 bases long) that bind to complementary regions on opposite strands of the template. The primers serve as replication start points. Two primers are used to amplify one DNA segment.
o DNA polymerase
The enzyme used to replicate DNA. It adds bases onto the 3’ ends of the annealed primers. For use in PCR, the DNA polymerase should be stable at high temperatures. In PCR reactions, the enzyme Taq DNA polymerase isolated from bacteria that thrive in the hot springs is used.
o MgCl2
Provides magnesium ion (Mg2+) to the reaction. Mg2+ is a cofactor for DNA polymerase. Without Mg2+, DNA polymerase will not function.
o dNTPS
(deoxynucleoside triphosphates). dNTPs is the term used to refer to the four deoxyribonucleotides: dATP, dCTP, dGTP and dTTP needed to replicate the template DNA. The nucleotides are added by the Taq DNA polymerase enzyme onto the 3’ ends of the annealed primers.
-Buffers
Maintains the pH of the reaction at a level where the DNA polymerase is most active.

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5
Q

What is the sequence of the main steps in PCR in the thermal cycler? At what temperatures are each
of these steps performed? Why?

A

o Denaturation
The first level is at a high temperature and is used to break the hydrogen bonds between the complementary bases that hold the two strands of the template together.
-95°C water bath
o Annealing
The reaction is then taken down to a lower temperature at which point the two PCR primers can anneal to the template strands.
o Extension
the reaction is brought up to an intermediate temperature at which the DNA polymerase adds nucleotides onto the ends of the annealed primers.

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6
Q

What is the relationship between the number of cycles a PCR has and the amount of DNA produced?

A

The steps of denaturation, annealing, and extension, make up one complete cycle. Making enough copies of DNA for analysis by gel electrophoresis typically requires from 25 to 40 cycles.

During the reaction, each new DNA fragment made in one PCR cycle can serve as a template in the next. This results in a doubling of the amount of amplified PCR product with each cycle. Mathematicians call this exponential amplification and describe the rate at which new products accumulate by the expression 2^n, where n is the number of cycles. If a PCR amplification is carried through 40 cycles, for example, the DNA would be amplified by 2^40.

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7
Q

In experiment 5A, what two master mixes used?

A

PMM (455bp) and GMM (255bp)

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8
Q

What were the positive controls in experiment 5A? The negative control?

A

+ve GMM (A positive control is a group in an experiment that receives a treatment with a known result, and therefore should show a particular change during the experiment)

-ve: Non-GMO food (oats)

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9
Q

Why were the food samples mechanically ground as the first step of this experiment?

A

The food samples were ground up to release the DNA that we wanted to extract

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10
Q

Why was it important to keep all the DNA samples on ice during the last steps of DNA extraction
(before placing them into the thermal cycler for PCR)?

A

to preserve the integrity of the samples and reagents and prevent thermal degradation of the proteins and enzymes being studied. Heat denatures the enzymes causing non-specific annealing and unwanted by-products.

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11
Q

Is gel electrophoresis used to separate only DNA bands? What other biomolecule have you used this technique to separate in our BZE labs?

A

We have used gel electrophoresis to separate proteins

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12
Q

How were the DNA bands in the lanes separated? By size? Shape? Structure?

A

Molecular weight

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13
Q

What is the unit for DNA size?

A

bp

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14
Q

Why was a DNA ladder loaded into a lane?

A

so we can determine around what size our DNA samples are when comparing it to known molecular weights

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15
Q

Why was loading dye (EZ-VisionTM) added to each DNA sample before it was loaded into a gel well?

A

So we can view the solution as it travels down the wells

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16
Q

Why does DNA migrate toward the positive electrode?

A

Due to its negative charge

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17
Q

Why was it possible to see the separated DNA bands in the gel?

A

The molecules can only pass through so far in the gel due to how it they arent all small enough to travel all the way towards the end

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18
Q

In the gel from this experiment, did a lane that didn’t have the GM promoter sequence
necessary mean that the food was not from a GMO?

A

No. It just means that it was not amplified

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19
Q

Are you able to label and give the function of cells and structures of different leaf cross sections;
cuticle, upper epidermis, lower epidermis, palisade mesophyll, spongy mesophyll, vascular tissue,
xylem, phloem, stomata, guard cells?

A

cuticle,
outside of leaf, usually waxy
upper epidermis,
lower epidermis,
palisade mesophyll,
the parenchyma (functional) tissue that lies between the upper and lower epidermis, excluding vascular tissues. the parenchyma (functional) tissue that lies between the upper and lower epidermis, excluding vascular tissues.
spongy mesophyll,
the parenchyma (functional) tissue that lies between the upper and lower epidermis, excluding vascular tissues. spongy mesophyll containing irregularly shaped cells beneath the palisade layer and having many intercellular spaces.
vascular tissue,
of vascular bundles, recognizable on the whole leaf as veins. (Made of xylem and phloem)
xylem, :
the xylem (upper portion, thick-walled cells, stained red)
phloem,
the phloem (lower portion, small, thin-walled cells, stained green).
stomata,
The stomata (singular stoma) are the systems made up of pores and guard cells through which gas exchange (CO2 in, O2 out) occurs and are usually located on the lower surface of the leaf.
guard cells,
); the guard cells, pairs of very small cells bordering stomata, can close or partially close to prevent excessive water loss

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20
Q

If you know the structural differences between different types of plant leaves, can you explain how
they reflect the adaptation of those plants to different environments with respect to: water conservation, gas exchanges & light absorption?

A

WATER
Water loss through stomata is replaced by upward diffusion of water from lower parts of the plant. This whole process is referred to as transpiration, a process of great advantage to plants since water moving up from the soil is rich in minerals. However, if the water lost in transpiration is greater than that replenished by roots (as may occur in dry conditions); the guard cells, pairs of very small cells bordering stomata, can close or partially close to prevent excessive water loss
AIR
stomata opening and closing depending on temperature and humidity
LIGHT ABSOPRTION
Different pigments to optimize light capture

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21
Q

Are you able to identify a mesophyte, xerophyte or hydrophyte leaf in a microscopic cross section by
characteristics and identify their components

A

hydrophyte: lots of air pockets in the spongy mesophyll
xerophyte: circles of space near lower epidermis
mesophyte: “uniform” look

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22
Q

Do you understand the structure and function of stomata:
o When are guard cells turgid? Flaccid? How are these states related to opening and closing
the pore?
o Why do guard cell have chloroplasts?
o How do the proton pumps in guard cells affect osmosis and pore-opening and closing?

A

o When are guard cells turgid? Flaccid? How are these states related to opening and closing
the pore?
When guard cells are turgid, they become kidney-shaped (in dicotyledon plants) causing the stoma to open; when guard cells are flaccid, they take on a semi-circular shape causing the stoma to close. A guard cell’s shape is
determined by its water content which, in turn, is controlled by its intracellular osmolarity. When guard cell osmolarity becomes greater than the extracellular fluid, water enters the cell via osmosis causing turgidity and stomatal opening. Conversely, when intracellular osmolarity becomes lesser than the extracellular fluid, water leaves the cell causing flaccidity and stomatal closure.
o Why do guard cell have chloroplasts?
They control the stomatal pore, which serves as a channel for exchange of gas by balancing between CO2 uptake for photosynthesis and water loss through transpiration. As a result, chloroplasts in the guard cells have become potential tool for manipulation toward improvement of plant productivity through photosynthesis.
o How do the proton pumps in guard cells affect osmosis and pore-opening and closing?
The absorption of water is inhibited when guard cells block the stomata’s opening

23
Q
  • Do you understand the difference between absorption & reflection of light, & that the light that is
    absorbed “powers” photosynthesis
A

light from e- excitation (caused by the absorption of photons) is harvested for photosytensis via PSII and PSI. Additionally, all light that is not absorbed is reflected, emitting a plant’s green hue.

24
Q

Do you understand how paper chromatography separates the pigments (what other molecules can
chromatography separate?

A

the pigments are seperated based on the molelcules’ polarity. double bonds and oxygen molecules impact the polairty of the lipid.
Chromatography can be used to separate the components in a mixture of biomolecules (such as amino acids or lipids). The molecules are separated based on their solubility in a particular solvent vs. their attraction to a particular solid material.

25
Q

What is the stationary phase and mobile phase of chromatography? Why is selecting the
appropriate eluent so important to chromatography? What eluent was used in this experiment?

A

The mixture is loaded onto a sheet (either paper or a plastic sheet that has been coated with a thin layer of solid material) called the adsorbent. Then the sheet is placed in a chamber containing a shallow volume of solvent called the eluent. Upon adding the sheet to the chamber, the eluent is pulled through the adsorbent by capillary action .

The adsorbent makes up the stationary phase of the system while the moving eluent creates the mobile phase of the system. Molecules that spend more time in the mobile phase will travel further up the sheet . Molecules that spend more time in the stationary phase will travel the least

26
Q

Are you able to explain the experiment in this lab conducted in simple terms using proper
vocabulary and ingredients (spinach, chlorophyll, xanthophyll, carotenes, chromatography,
absorbance)

A
  1. Apply spinach pigment concentrate onto the the given paper sheet several times until a opaque green line appears on the paper.
  2. Place the paper pigment closest to the ethanol (eluent) and observe as the pigment begin to move up the paper.
  3. remove the paper and allow it to dry
  4. Observe the separation of the following pigments (chlorophyll a, b, and carotenoids)
    5.measure the distances of how far each pigment travelled, thereby determining r factor
27
Q

Why doesn’t chlorophyll a fluoresce in an intact plant?

A

Because the energy has an outlet (energy been harvested for photosynthesis), instead of having no outlet. Since the conc has no outlet to send the energy from the photons to, it fluoresces.

28
Q

What does a larger
Rf value indicate for a pigment? A smaller Rf value?

A

small: most polar
big : least polar

29
Q

Understand how and why the different pigments were separated using the structural
difference between chlorophyll a and b as an example.

A

B: more polar, has more double bond and oxygen rendering it more -ve
A: made only of hydrocarbons: less oxygens

30
Q

Describe each pigments’s light absortion

A

A: High @ 400nm, low from 450-650nm, high again
B: low from 400-450, low from 500-650, slight high after 650
C: high from 400-550, low for anything after 550

31
Q

Are you able to identify a plant pigment based on observing its absorption spectrum?
o Given a data set, be able to draw a graph depicting absorbance at different wavelength

A

based on how it can absorb different wavelengths will allow us to imagine the absorption graph

32
Q

What is a transgenic organism? What was the transgenic organism you created in this lab?

A

. In transgenesis, the goal is to add a foreign gene to an organism, express the gene product, and alter a phenotype (trait) in that organism. Once added to its genome, the organism will also be able to pass on this gene to its offspring. Transgenesis creates transgenic organisms and is considered a component of genetic engineering.

WE USED E.COLI

33
Q

Do you know what a horizontal gene transfer event is, its relationship to bacteria transformation
and how it is different from a vertical gene transfer?

A

Although bacteria reproduce asexually, these organisms often transfer DNA from one cell to another within a population (and sometimes between species) by a process called horizontal gene transfer (HGT).

This differs from vertical gene transfer, which is the passage of a plasmid from mother to daughter cells during division.

34
Q

Be able to define what bacterial plasmids are and what their purpose is in bacteria.

A

In addition to chromosomal DNA, bacteria (also archaea and yeast) contain small circular DNA molecules called plasmids. These extrachromosomal DNA molecules contain one or more genes for traits that are beneficial to cell survival in new environments (e.g. resource use, antibiotic resistance). These plasmids can be transferred by HGT to other bacterial cells, which will spread these beneficial traits and adapt a population to a new environment.

35
Q

Do you understand the function and the relationship between the GFP, beta lactamase (bla) and ara
C genes on the pGLO plasmid?

A

GFP (C)
GFP Gene
(Gene of Interest)
Encodes the amino acid sequence of the GFP protein. Expression produces the GFP protein.

bla (C)
Beta-Lactamase Gene
(Selectable Marker Gene)
Encodes the amino acid sequence of the beta-lactamase enzyme that confers resistance to certain antibiotics. Expression of the enzyme acts as a selectable marker. A selectable marker allows you to select for cells that took in the plasmid and are expressing its genes.

araC (C)
Ara C Protein Gene
(Regulatory Protein Gene)
Encodes the amino acid sequence of the Ara C protein that acts as a gene expression regulatory protein. The protein can repress expression from the ParaBAD promoter in the absence of the sugar arabinose but activate expression from the same promoter when arabinose is present.

P (B)
Bacterial Promoter (Regulatory Element) Recruits RNA polymerase and directs transcription of a linked transcription unit in bacteria. ParaBAD is a regulated promoter and can be controlled by the AraC protein. ParaC and Pbla are constitutive promoters in bacteria (always on).

36
Q

What phenotype does the bla gene give to transformed bacteria? Is this gene constitutively
expressed or regulated in the pGLO plasmid?

A

resistance to certain antibiotics. resistance to amipicllin. It is constitutively expressed

37
Q

How does the ara C gene regulate the GFP gene? Briefly describe this arabinose operon.

A

Encodes the amino acid sequence of the Ara C protein that acts as a gene expression regulatory protein. The protein can repress expression from the ParaBAD promoter in the absence of the sugar arabinose but activate expression from the same promoter when arabinose is present.

38
Q

What is the difference between a bacterial lawn
and a bacterial colony? What could alter a bacteria’s morphology?

A
  • lawn: indistinguishable colonies versus distinguishable, quantifiable colonies
  • morphology can change based on food sources and changes in genes
39
Q

Do you understand the importance of the control treatments that were used in the experiment (uncut pGLO)

A

(uncut pGLO)
This was important because it allows for the determining the original structures before the plasmids
were cut. Comparing the uncut plasmids to the cut ones ensure that there can be an understanding of
how the plasmid was cut by the enzymes involved. the plasmid is circular and not cut at random
places. Eliminates strange unaccounted bands
in the DNA samples.

40
Q

a) Did the number of colonies observed for the –pGLO LB plate match your prediction? What
information about the experiment does this control give you?

b) Did the number of colonies observed for the –pGLO LB/amp plate match your prediction? What
information about the experiment does this control give you?

A

a)The colonies observed for the -PGLO LB formed a lawn, meaning that it is as predicted and the
cultures did not grow. This due to how it was given broth and was not in the presence of ampicillin.
b)Yes, it matches our prediction; no colonies are observed. This tells us that the bacteria do not have
resistance to ampicillin; the ampicillin prevented the bacteria from splitting and this plate acts as a -ve
control.

41
Q

Why was the –pGLO in LB plate a positive control for this experiment?
Why was the –pGLO in LB + ampicillin plate a negative control experiment?

A

LB : because we knew that something would grow with no ampicillin present
LB+ a: Because we expected nothing to grow in the presence of ampicillin

42
Q

Can you explain from the results of the four bacterial plates why the pGLO plasmid was successfully
transformed in this experiment?

A

It was sucessful because not only did the +pGLO araC glow, but there was ampicillin resistance

43
Q

What does transformation efficiency measure? How would you determine this efficiency measure
from the four plates in this experiment?

A
  • How many colonies had reproduced with altered genes
44
Q

What was the purpose of the transformation solution, CaCl2?

A

? It is postulated that the Ca2+ cations of the transformation solution (50 mM
CaCl2, pH 6.1) neutralizes the repulsive negative charges of the phosphate backbone of the
DNA and the phospholipids of the cell membrane to allow the DNA to enter the cells during the
heat shock procedure coming up in the next part of the protocol (heat shock in procedure P2).
This is described as making the bacterial cells artificially competent.

45
Q

What was the purpose of the heat shock procedure (ice-heat-ice)? Why were the restriction enzyme(s) + pGLO plasmid samples placed in a 37oC bath and then heatshocked at 65oC?

A

? The heat shock procedure (ice-heat-ice) increased the permeability of the
cell membrane to DNA. While the mechanism is not known, the duration of the heat shock is
critical and has been optimized for the type of bacteria used and the transformation conditions
employed. The pGLO plasmid DNA entered the permeable cell membrane which the help of the
CaCl2 transformation solution.

46
Q

Why was conducting the experiment with asceptic technique (as much as possible) so important?

A

To avoid other bacteria growing on our plates along with the e.coli

47
Q

Can you explain the use of restriction endonucleases by bacteria to destroy viral DNA and how they
protect their own DNA by methylation?

A

Bacterial restriction endonucleases target viral DNA,
not the bacterial cells’ own DNA, because bacteria
modify their own DNA, using enzymes known as
methylases that add methyl (CH3) groups to some
of the nucleotides in the bacterial DNA.When nucleotides within a restriction
endonuclease’s recognition sequence have been
methylated, the endonuclease cannot bind to that
sequence. Consequently, the bacterial DNA is
protected from being degraded at that site.Viral DNA, on the other hand, has not been
methylated and therefore is not protected from
enzymatic cleavage.

48
Q

Be able to explain the pGLO restriction digest experiment conducted in simple terms using proper
vocabulary and ingredients (ie. DNA extraction, restriction endonucleases, restriction sites,
electrophoresis, agar, EcoRV, HindIII, loading dye).

A

watch a vid queen

49
Q

Can you describe what a recognition site is and restriction fragments?

A

In biology recognition site is the site recognized by a restriction enzyme to cleave DNA is called a recognition site.
These sites are located on a DNA molecule containing specific sequences of nucleotide (4-8 base pair)

A restriction fragment is a DNA fragment resulting from the cutting of a DNA strand by a restriction enzyme (restriction endonucleases), a process called restriction.

50
Q

Can you explain what are “sticky ends” and “blunt ends” when a restriction site is cut by its enzyme? Why does this matter?

A

The fragments generated by a restriction
enzyme can be either sticky or blunt. Sticky
ends are created when the restriction enzyme cuts each strand at different points within the restriction site. The ends created have overhangs, a single stranded region.

Some enzymes cut both stands at the same point within the restriction site and generate blunt ends with no overhangs.

Ends with overhangs are considered “sticky”
because the overhangs are complementary
and can hybridize by base pairing. 5’ vs. 3’
overhang refers to whether the singlestranded region of a sticky end represents the 5’ end or 3’ end of the DNA molecule of that strand.

If a restriction enzyme
generates sticky ends, then any fragment
created with this enzyme will have the same
sticky ends. This property can be used to join DNA fragments together with the help of the enzyme ligase to create recombinant DNA

51
Q

Are you able to describe how restriction endonucleases work and be able to predict how often DNA
will be cut by different restriction endonucleases based on the restriction site?

A

Restriction enzymes are important tools to analyze DNA and create recombinant DNA for genetic
engineering. These enzymes are endonucleases found in bacteria as part of a primitive protective
system called restriction modification that defends these organisms from infection by viruses called
bacteriophages (Latin/Greek roots for bacteria eaters).

52
Q

When preparing the pGLO plasmid for digestion with various restriction enzymes, why was a buffer
added?

A

to create optimal reaction conditions for specific restriction enzymes, or facilitate multiple enzymatic reactions.

53
Q

Describe gel electrophoresis and what it was used for in this experiment:
o What did this procedure separate?
o What is the unit of size for DNA?
o What were the two purposes of adding the 6X loading dye to each digested sample before
loading them onto a gel? (How do you know when to stop
o What is the purpose of loading a lane in the gel with DNA ladder?
o What was the purpose of loading a lane in the gel with uncut plasmid? Why were there two
bands in this lane?
o What is the overall charge of DNA (positive or negative)?
o Why would DNA bands move towards the positively charged electrode in a gel when there is
a current running through the gel?
o Why are smaller DNA bands migrating further down the gel than larger DNA bands?
o What happens if you don’t pay attention to the gel and leave it running for hours? Where do
the DNA bands go? Do they stay in the gel?

A

o What did this procedure separate?
- nucleic acids
o What is the unit of size for DNA?
-bp (base pairs)
o What were the two purposes of adding the 6X loading dye to each digested sample before
loading them onto a gel? (How do you know when to stop)
- we can see how far the solutions have traveled so they dont run off the gel
o What is the purpose of loading a lane in the gel with DNA ladder?
-The DNA ladder is a set of standard DNA
fragments of known sizes. It can be used as a tool (type of positive control) to estimate the size of
an unknown fragment in an experimental sample.
o What was the purpose of loading a lane in the gel with uncut plasmid? Why were there two
bands in this lane?
Molecules in gel electrophoresis are separated in the gel based on molecular size. This means
that supercoiled uncut plasmids can migrate faster than circular open uncut plasmids.
However, when the plasmid is cut, they end up larger and take longer to migrate to the positive
node compared to the supercoiled uncut plasmid. But they’re both the same size! Shape and structure is what is causing plasmid
the difference. Supercoiled uncut plasmid takes up less space and travels
more easily through the pores in the agarose compared to relaxed uncut plasmid
o What is the overall charge of DNA (positive or negative)?
-ve
o Why would DNA bands move towards the positively charged electrode in a gel when there is
a current running through the gel?
DNA is negatively charged, therefore, when an electric current is applied to the gel, DNA will migrate towards the positively charged electrode.
o Why are smaller DNA bands migrating further down the gel than larger DNA bands?
They are smaller therefore they move faster
o What happens if you don’t pay attention to the gel and leave it running for hours? Where do
the DNA bands go? Do they stay in the gel?
THEY RUN OFF THE GEL BC THEY DIFFUSE (?)

54
Q

Are you able to determine which restriction sites belong to which restriction enzyme if you
are given a completed gel and a plasmid map?

A

-U pGLO (uncut) 2 Can’t use the DNA ladder to estimate size of circular
DNA forms. Standards are linear DNA fragments.
-E pGLO / EcoRV, 1 5000-6000bp
-H pGLO / HindIII, 2 4000-4500bp and 750bp
-D pGLO / EcoRV +, 3, 3000-4000bp, 1000-1050bp, and 750bp