E3 Flashcards

1
Q

You have measured the absorbance of a sample of DNA in solution and found that the ration A260nm/A280nm = 1.26. What can you conclude about this sample of DNA?

A

It is likely contaminated with protein.

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2
Q

Which is more stable, DNA or RNA?

A

DNA

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3
Q

What is the complementary sequence to CATTACGT?

A

ACGTAATG

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4
Q

Which piece of DNA will have the higher Tm, one with a cytosine plus guanine content of 30% or one with a cytosine plus guanine content of 50% if both are heated under the same experimental conditions?

A

50% cytosine plus guanine will have the higher Tm.

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5
Q

Chromatin interacts with the DNA backbone. Therefore, the primary structure of chromatin should have a high percentage of the following amino acids:

A

Lysine, Arginine

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6
Q

Which bases pair with one another in DNA? Classify each as a purine or a pyrimidine. How many hydrogen bonds form between each of these pairs?

A

Purine (A) bonds with Pyrimidine (T) w/ double H bond. Purine (G) bonds with Pyrimidine (C) w/ triple H bond.

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7
Q

Nucleobase

A
  • a nitrogen-containing compound. (C)ytosine, (G)uanine, (A)denine (found in both DNA and RNA) (T)hymine (found only in DNA), and (U)racil (found only in RNA) 5 in DNA & RNA,
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8
Q

Nucleoside

A

are components of nucleotides, have a nitrogenous base and a five-carbon carbohydrate group

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9
Q

Nucleotides

A

are simply a nucleoside with one or more phosphate groups attached

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10
Q

Nucleic acid

A

They are composed of nucleotides, monomer components: a 5-carbon sugar, a phosphate group and a nitrogenous base.

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11
Q

Describe and explain how the Griffith Experiment and the Avery-MacLeod-McCarty experiments demonstrated that DNA is the genetic material

A

showed transforming substance in bacteria could change harmless strains to virulent ones. Avery-MacLeod-McCarty Experiments (1944) identified this substance as DNA by destroying proteins, RNA, or DNA in the transforming mix, revealing the destruction of DNA stopped the transformation.

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12
Q

Chargaff observed that the total amount of A+G was equal to the total amount of C+T regardless of the source of DNA. Explain the structural basis for this observation

A

the structural stability and specificity of the DNA double helix. Chargaff’s rules of base pairing= transmission of genetic information.

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13
Q

Describe the major feature of the DNA double helix. What force is primarily responsible for the stability of the structure?

A

DNA made of two strands that wind around each other resemble a twisted ladder, helix-like shape. Each strand has backbone made of sugar (deoxyribose) and phosphate groups. base pairing between complementary strands and stacking between adjacent bases

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14
Q

Bacteria produce restriction endonucleases as a defense mechanism against phages. Explain why the enzymes do not cleave the bacteria’s genomic DNA?

A

restriction enzymes can’t cleave methylated DNA. bacterial genome is methylated, and resistant to enzymes. phage DNA is not methylated and so is cleaved by enzymes

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15
Q

LacZ

A

codes for beta-galactosidase

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16
Q

ApR

A

codes for beta-lactamase (ampicillin resistance)

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17
Q

Ori

A

origin of replication, so the bacteria can replicate the plasmid

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18
Q

Which restriction enzyme cut sites would you use to clone your blue pigment gene into pUC19 and why?

A

Any of the restriction cut sites located in the multiple cloning site (MCS), these sites only cut the plasmid once. Also the sites are in the lacZ gene and so facilitate blue/white screening.

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19
Q

You digest the plasmid and your blue pigment gene with the restriction enzymes chosen in step b. Explain how the two fragments will stick together, and explain how you will stitch the phosphate backbone back up

A

fragments stick by base pairing, bc of complementary sticky ends made by digesting gene and plasmid w/ same restriction enzymes. backbone would repair w/ adding DNA
ligase

20
Q

Following transformation of the plasmid, how would you ensure that the bacteria you isolate have successfully taken up the plasmid

A

By using antibiotic selection. Plate the bacteria on agar containing antibiotic (ampicillin), only bacteria that w/ the plasmid (and antibiotic resistance) will survive.

21
Q

How would you know which bacteria have a plasmid with your gene inserted?

A

Blue/white screening. Place X-gal in agar, if bacteria produce lacZ colonies will be blue, if not the colonies will be white. Since the gene in the middle of lacZ, that will make the protein inactive, so the white colonies will have the correct gene insert.

22
Q

You are now interested in inserting your blue pigment gene into a tomatoes plant. You assume that people would love to eat blue tomatoes. Explain how you might go about inserting this gene into the plant

A

You would use the Ti plasmid from Agrobacterium tumefaciens. remove tumor causing genes from plasmid and replace with blue pigment gene. plasmid would be reintroduced back into the bacteria, that would be used to infect the plants. plasmid would go into the plant’s chromosome changing it with the gene.

23
Q

What is Taq polymerase? Where was it isolated and from which organism? Why is it crucial for the polymerase chain reaction?

A

Taq polymerase is a heat stable DNA polymerase. If was isolated from the bacteria Thermus aquaticus in a hotspring in Yellowstone National park. The enzyme is important in PCR because it does not denature at high temperature, so new enzyme does not have to be added at each cycle of the reaction

24
Q

Describe using a diagram how the polymerase chain reaction works and results in exponential amplification of DNA. What steps are involved, what temperatures are required, and what components are needed for the reaction to occur.

A
  1. Denaturation: 94 °C, DNA strands separate
  2. Annealing: Cool to ~55 °C, synthtetic DNA primers form duplex with the separated DNA strands.
  3. Extension: 72 °C, stable DNA polymerase copies the DNA chain Repeat steps 1-3 25-30 times

To perform the reaction you will need -heat stable DNA polymerse (Taq or equivalent)
-dNTP (deoxy nucleotide triphosphates)
-template
-primers
-Mg2+

25
Q

You wish to make a single nucleotide change in your gene for the blue pigment. Describe how you could accomplish this

A

use site-directed mutagenesis. Perform a PCR using primers that have desired mutation. The two primers must anneal to the same place on the gene, but to both Strands of DNA. You will end up with a mixture of parental DNA and new mutated DNA. The parental DNA is methylated while the new mutated DNA is not. Use DpnI to digest the parental (non-mutated) DNA. Transform bacteria

26
Q

Explain how the CRISPR-Cas system works as a bacterial immune system

A

short palindromic repeats of DNA coded in genome. Invading DNA from viruses incorporated in-between CRISPR sequences.

27
Q

Briefly outline one of two strategies for recreating extinct species.

A

CRISPR - Sequence the passenger pigeon genome
See how it is different from existing pigeon species

Use CRISPR technology to change existing pigeon DNA to be passenger pigeon DNA

The other approach would be somatic nuclear transfer (see you lecture notes)

28
Q

You grow a culture of E. coli in media containing 15NH4Cl, you then switch the bacteria to media containing 14NH4Cl for three generations (an eightfold increase in population). What would be the ratio of hybrid DNA (15N-14N) to light DNA (14N-14N) at this point?

A

After 3 generation there would be a ratio of 2 to 6

29
Q

On the diagram draw arrows indicating the electron flow that occurs in the formation of the new phosphodiester bond. In words describe what is happening. What are the products of the reaction?

A

DNA polymerase catalyzes the formation of a new phosphodiester bond. The 3’OH group on the end of the growing chain is deprotonated to become a nucleophile (O-). It attacks the 5’- phosphate on the incoming deoxy nucleotide triphosphate forming a new phosphodiester bond, releasing pyrophosphate

30
Q

Why is DNA synthesis discontinuous on the lagging strand?

A

The DNA polymerase has only 1 catalytic activity, the formation of a new DNA chain in the 5’ to 3’ direction. The enzyme can only attach nucleotides to a free 3’OH group. Since the parental
DNA is anti-parallel, the two strands go in opposite directions (one is 5’ to 3’, the other is 3’ to 5’ This means when the two strands are being copies, only the 3’ to 5’ strand will be continuously copied, as its new complementary strand would be the correct 5’ to 3’ direction.

31
Q

How are the discontinuous fragments joined to
make a continuous strand of DNA?

A
32
Q

the structure of the anti-retroviral drug AZT. What nucleotide does the drug resemble?

A

nucleotide Thymidine

33
Q

Describe a plausible mechanism for how this drug AZT would inhibit the growth of HIV.

A

The drug lacks a 3’OH group (azide group in its place), which means it cannot have additional nucleotides, attaches to it. If the drug becomes phosphorylated, and incorporated into the virus’ growing DNA chain, the reverse transcriptase will be unable to add new nucleotides, thus terminated DNA replication.

34
Q

Given what you know about DNA replication, why does the drug kill the virus but not
people?

A

The reverse transcriptase has no proofreading ability so cannot remove the drug form its growing DNA chain. DNA pol III has proofreading ability so it can remove the drug from the growing DNA chain, making it less toxic to humans than to the virus

35
Q

You have discovered a new class of DNA polymerase in bacteria from Mars. The polymerase lacks 3’ to 5’ exonuclease activity, would you expect it to be more or less accurate that E. coli DNA polymerase III? Explain your answer

A

You would expect this new polymerase to be less accurate than DNA polymerase III. The 3’ to
5’ exonuclease activity is the proofreading activity of the enzymes that allows them to remove nucleotides that have been incorrectly added to the chain

36
Q

Turnover number

A

is the speed that the polymerase can copy DNA

37
Q

The Processivity

A

is how many nucleotides can be added to the new chain of DNA before the
polymerase dissociates from the chain

38
Q

5’ to 3’ polymerization activity

A

is the ability to make new DNA

39
Q

5’ to 3’ exonuclease activity

A

is the repair activity

39
Q

What would you expect the major function(s) of each of the DNA polymerases to be
in DNA replication and why?

A

Pol A = major polymerase involved in DNA replication. high turnover number = rapid copying of genome. high processivity to copy a large portion of the genome before falling off. has polymerization activity and proofreading activity.

Pol B = proofreading and repair. small turnover number and processivity unlikely to copy whole genome. has both proofreading and repair activity.

40
Q

the 3’ to 5’ exonuclease activity

A

is the proofreading activity.

41
Q

Consider the following DNA duplex:
5’-ATGCCAGGTCAAGTTAGTAA-3’
3’TACGGTCCAGTTCAATCATT-5’

What is the sequence of the template strand (5’ to 3’)?

A

5’-TTACTTAACTTGACCTGGCAT-3’

42
Q

Consider the following DNA duplex: 5’-ATGCCAGGTCAAGTTAGTAA-3’
3’-TACGGTCCAGTTCAATCATT-5’

What is the sequence of the coding strand (5’ to 3’)?

A

5’-ATGCCAGGTCAAGTTAGTAA-3’

43
Q

Consider the following DNA duplex: 5’-ATGCCAGGTCAAGTTAGTAA-3’
3’-TACGGTCCAGTTCAATCATT-5’

What is the sequence of the mRNA produced from the DNA (5’ to 3”)?

A

5’-AUGCCAGGUCAAGUUAGUAA-3’

44
Q

Provide a definition for a promoter. What are the roles of Pribnow box, the -35 region and
the TSS?

A

The promoter region of DNA that is not transcribed into mRNA, serves a role in gene regulation. The Pribnow box is a AT rich sequence where the DNA duplex unwinds and the transcription bubble forms. The -35 region is where the sigma factor bind and initiate transcription. The TSS is the transcription start site, it is the first nucleotide from the DNA that gets transcribed.

45
Q

What are sigma factors? Provide an example of how different sigma factors can be used to control gene expression.

A

Sigma factors are a subunit of RNA polymerase that bind the promoter region to initiate transcription. The sigma factors only bind specific sequences in the promoter, thus provide a way of controlling gene expression by producing different sigma factors at different times. An example is the control of early, middle and late genes by different sigma factors. The early genes are transcribed by the early sigma factor, which also produces the middle genes sigma factor.

This will then turn on the middle genes, which produces the late genes sigma factor. This mechanism ensures sequential control of gene expression and different periods of the bacterial
life cycle

46
Q
A