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BCHM218 > Module 3 Sections 1-5 > Flashcards

Module 3 Sections 1-5 Flashcards

Section 1: DNA Structure and Function Section 2: RNA Structure and Function Section 3: Quantification of Nucleic Acids Section 4: Amplification of DNA by PCR Section 5: Methods for DNA Sequencing

1
Q

nucleic acids

A

composed of chains of nucleotides

basic building blocks for DNA and RNA

Self-assemble into their 3D structure by weak forces and how atoms are arranged in space

3D helical structure of DNA is a result of base pairing and is the most energetically favourable conformation

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2
Q

oligonucleotide

A

a short nucleic acid containing 50 or fewer nucleotides

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3
Q

polynucleotide

A

longer nucleic acid

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4
Q

3 components of nucleotides

A
  1. A heterocyclic (cyclic compound with 1 or more ring structures that contain atoms of at least 2 different elements) base
  2. A 5-carbon sugar called pentose
  3. A phosphate group
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5
Q

chargaff’s rule

A

A + G = T + C

In DNA, there is always an equal percentage of purines and pyrimidines. This means for all DNA,

of A = # of T

of G = # of C

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6
Q

hydrogen bonding in DNA

A

2 strands of DNA molecules are held together via base pairing between the nitrogenous bases (hydrogen bonds)

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7
Q

G and C have

A

3 hydrogen bonds

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8
Q

A and T have

A

2 hydrogen bonds

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9
Q

DNA double helix

A

2 strands of DNA intertwine to form a right-handed double helix

Backbone of each strand is composed of alternating sugar and phosphate residues (sugar-phosphate backbone) and is negatively charged

Nitrogenous bases are positioned towards the center of the helix, letting them hydrogen bond with bases on the opposing strand

Has directionality each strand opposing each other (5 to 3 or 3 to 5)

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10
Q

why antiparallel

A

More energetically favourable than parallel because of the geometry of the component bases

Based on the linkages formed by carbons at the phosphate and OH groups on the pentose sugar

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11
Q

phosphodiester bonds

A

Link the nucleotide units in nucleic acids

5’-phosphate group of 1 nucleotide linked to the 3’-hydroxyl group of the next nucleotide

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12
Q

major groove of DNA purpose

A

Nucleotide sequence is primarily read by DNA binding proteins in the major grooves fund on the outside of the DNA strand for more accessibility

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13
Q

2 ways to stabilize the duplex

A
  1. hydrophobic stacking
  2. base pairing
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14
Q

hydrophobic stacking

A

Chemical properties of purines and pyrimidines

Bases are hydrophobic and are insoluble in water at near-neutral pH of the cell

Thus, bases align so that 2 or more are positioned with the planes of their rings in parallel like a stack of coins (looking down a barrel)

Stabilizes the helix by minimizing contact of the hydrophobic bases with water

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15
Q

base pairing

A

An extensive network of weak bonds within the double-stranded DNA structure that occurs between base pairs

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16
Q

functions of DNA

A
  1. Long-term storage of genetic information
  2. Acting as a template for DNA replication
  3. Coding for proteins
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17
Q

what does functions of DNA depend on

A

Highly dependent on its structure

Once disrupted, it can no longer carry out these critical functions

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18
Q

what makes DNA a good long-term storage of genetic information

A
  1. strand complementarity
  2. replication fidelity
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19
Q

strand complementarity

A

Hydrogen bonding

Most significant property of DNA that makes it a good information carrier

Specific base pairing within dsDNA allows exact copies to be made, allowing replication of genetic replication

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20
Q

replication fidelity

A

Structure of double helix can allow for the strands to be separated, and the original is used to synthesize a complementary strand

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21
Q

internal forces on DNA stability

A
  1. hydrophobic interactions
  2. van der waals interactions
  3. hydrogen bonding between paired bases
  4. ionic interactions
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22
Q

hydrophobic interactions

A

Stabilizes base pairing

Bases are hydrophobic and face the interior

Sugar-phosphate backbone is hydrophilic and faces the exterior, interacting with water

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23
Q

van der waals interactions

A

stacked bases interact through ring structures

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24
Q

hydrogen bonding between paired bases

A

GC is more stable than AT

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25
Q

ionic interactions

A

Negative charge of backbone phosphates are neutralized by interactions with cations

Na+ and Mg++ commonly interact with the backbone to neutralize the electrostatic repulsion between strands

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26
Q

external forces on DNA stability

A
  1. temperature
    above melting temp makes DNA unwind to single strand, destablizing it
  2. salt
    increase in salt conc = increase in duplex stability

sodium ions interact with the negatively charged DNA backbone and stabilizes it

  1. proteins
    DNA binding proteins are involved in the compaction of genomes and contribute to both the global and local structure of DNA
  2. organic solvents
    Carbon-based

Destabilize DNA helix by disrupting hydrogen bonds and solvating bases

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27
Q

electrostatic interactions using the sugar-phosphate backbone

A

Due to the negative charge of the sugar-phosphate backbones of both nucleic acids

Backbone of both DNA and RNA is hydrophilic, so the hydroxyl groups of the sugar residues form hydrogen bonds with water

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28
Q

phosphate groups at pH 7

A

have pKA near 2

completely ionized

negatively charged (neutralized by ionic interactions with positive charges on proteins, metal ions, or short linear organic molecules called polyamines)

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29
Q

polyamines

A

2 or more amine groups

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30
Q

types of coding RNA

A

mRNA

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31
Q

mRNA

A

Transient carriers of genetic information

Transcript copy of a gene that encodes a specific protein

Carries the encoded information from the nucleus to the ribosomes where the protein is produced

Coding sequence of mRNA determines the amino acid sequence of the protein

Different mRNA molecules adopt different 3D structures depending on what is most energetically favourable

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32
Q

non-coding RNA

A

Transfer RNA (tRNA)
- Present during translation

Ribosomal RNA (rRNA)
- Present during translation

Long non-coding RNAs (lncRNA)
- Can be important regulatory RNAs

Small nuclear RNAs (snRNAs)
- Play a role in gene regulations (splicing “snurps”)

MicroRNAs (miRNAs)
- Limit translation by binding to the 3’-end of target mRNAs

Small interfering RNAs (siRNAs)
- Can inhibit transcription of certain genes and viral DNA

Catalytic RNAs
- Ribozymes

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33
Q

RNA vs DNA similarities

A

both carriers of genetic information

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34
Q

DNA vs RNA

A

DNA - long term storage of genetic information
RNA - more transient (less permanent)

DNA - set of biological blueprints
RNA - helps carry out the guidelines found within these blueprints

DNA - double stranded
RNA - single stranded

DNA - not as versatile
RNA - structurally and functionally versatile

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35
Q

pentose sugar

A

called deoxyribose

In RNA, the deoxyribose sugar is replaced with ribose

Ribose is a 5-carbon sugar with a hydroxyl group at the 2’ carbon. This provides an additional site for hydrogen bonding, stabilizing the 3-dimensional folding of the polynucleotide

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36
Q

RNA base composition

A

RNA has uracil instead of thymine

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37
Q

non-canonical RNA base pairing

A

RNA can sometimes have A-A and G-U base pairing (helps stabilize the 3 dimensional folding of RNA, with surfaces capable of binding other molecules)

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38
Q

when is RNA less stable

A

under alkaline conditions (pH > 7) because of the additional hydroxyl group on the 2’ carbon of the pentose sugar (hydrolyzed rapidly, but DNA is not)

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39
Q

products of action of alkali on RNA

A

cyclic 2’,3’-monophosphates

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40
Q

cyclic 2’,3’-monophosphates yield the mixture

A

rapidly hydrolyzed further to yield a mixtuer of nucleoside 2’- and 3’-monophoshates

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41
Q

DNA backbone

A

serves a purpose in gene expression
DNA is maintained during cell division and during extended periods in nonreplication cells

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42
Q

DNA and RNA in alkaline conditions

A

DNA: resistant
RNA: degraded

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43
Q

RNA folding

A

secondary structure, can fold back on itself to form intramolecular base pairings

3D to fold into many different shapes

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44
Q

energetically favourable RNA structures

A

RNA base stacking hides the hydrophobic bases away from the hydrophilic surroundings

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45
Q

RNA base stacking in tRNA

A

all the bases are stacked.

base-triple interactions, helix-helix packing allows stable 3D folding

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46
Q

RNA secondary structures

A
  1. helical structures
  2. internal loops
  3. hairpin loops
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47
Q

helical structures

A

when the strand folds back onto itself, the paired strands are antiparallel to one another and form a right-handed helix

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48
Q

internal loops

A

separation of double-stranded DNA (single stranded RNA that folded back on itself) because lack of base pairing

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49
Q

hairpin loops

A

RNA folds back but there is an unpaired loop of bases at the end of a stem region

nucleotides within loops are arranged to maximize hydrogen bonding and base stacking, enhancing thermodynamic stability

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50
Q

most common type of RNA secondary structure

A

hairpin loops

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51
Q

stability of RNA structure

A

influenced by weak interactions - van der waals stabilize structures

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52
Q

metal ions in the stability of RNA structure

A

bind to specific sites to help shield the negative charge of thebackbone, allowing RNA to tightly pack together

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53
Q

other influences to the stability of RNA structure

A
  1. # of GC vs AU base pairs
  2. # of base pairs in a stem region
  3. # of base pairs in a hairpin loop (more than 10 or less than 5 bases requires more energy)
  4. # of unpaired bases - decrease stability
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54
Q

most accurate method of nucleic acid quantification

A

UV absorption - the light absorbed is directly proportional to the amount of proteinnucleic acid present in the sample

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55
Q

purines vs pyrimidines structure

A

purines = 2 rings
pyrimidines = 1 ring

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56
Q

ring structure

A

alternating single and double bonds that create resonance (partial double-bond character)

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57
Q

pyrimidine and purine structure

A

pyrimidine = planar
purine = nearly planar with a slight pucker

58
Q

spectrophotometry

A

measures the amount of light absorbed by a sample

59
Q

spectrophotometry steps

A
  1. Sample containing nucleic acids is placed inside a chamber and is exposed to ultraviolet light at a wavelength of 260 nm
  2. A photo-detector measures how much light passes through the sample – represented on an absorption spectrum
60
Q

UV light and nitrogenous bases

A

they absorb MORE light. the close interaction of stacked bases when bound within a nucleic acid decreases the absorption of UV light relative to that of a solution with the same conc. of free nucleotides

61
Q

beer’s law

A

calculating the concentration of nucleic acids in a solution - the darker a solution, the more concentrated it is as it is absorbing more light

62
Q

beer’s law equation

A

A260 = e260 * c * l

A260 = the absorbance of UV light as determined by the spectrophotometer

e260 = extinction coefficient

c = concentration

l = path length / distance that light travels through

63
Q

optical density

A

the amount of UV light able to pass through a solution

greatest at 260 nm - larger for DNA than for protein

64
Q

order of UV absorption

A

dsDNA < ssDNA &laquo_space;free nucleotides

65
Q

hypochromic effect

A

Large decrease in light absorption at 260 nm occurring as single strands of DNA anneal to form double-helical DNA

Forces that stabilize the DNA helix like hydrogen bonding and base stacking, limits the amount of resonance that can occur within the aromatic rings of the bases

Decrease in UV light absorbed as ssDNA anneals to form dsDNA

66
Q

hyperchromicity

A

Large increase in light absorption at 260 nm that occurs as DNA unwinds/melts

As DNA is denatured, the base pairs are disrupted and the 2 strands separate, forming ssDNA

The resonance within the bases are no longer constrained, so the UV light absorption of a single-stranded DNA is higher than double-stranded at the same concentration

67
Q

dsDNA viscosity

A

highly viscious at pH 7 and 25 degrees cels

exposed to extreme pH or above 80 degrees
viscosity decreases, indicating denaturation

68
Q

dsDNA denaturation

A

melting and 2 strands break apart - hydrogen bonds and base stacking interactions are disrupted

69
Q

does DNA absorb more light in single or double stranded form

A

single-stranded form - the resonance within nitrogenous bases are no longer constrained by the forces that stabilize the DNA double helix

70
Q

how to determine the melting point of DNA

A
  1. track the absorbance of an aqueous solution across a range of temperatures
  2. the temp at which half the DNA in the sample is denatured (50% dsDNA and 50% ssDNA)
71
Q

denaturation/melting point uses

A
  1. determining base composition
  2. analyzing structure and function of DNA (interactions holding the strands together)
  3. estimating the amount of G and C bases
72
Q

GC base pairs vs AT base pairs in melting points

A

higher content of GC = higher melting point (3 hydrogen bonds instead of 2)

73
Q

melting temperature equation

A

T = 0.41 (%G+C) + 69.3 degrees

under standard conditions and a salt conc. of 0.2 M

74
Q

renaturation

A

reconstruction of a nucleic acid (or protein) to its original form after denaturation

75
Q

reannealing

A
  1. 2 strands must not be completely separated (dozen residues must remain intact)
  2. when pH/temp returns to normal, the separated segments of the double helix spontaneously rewind to form an intact duplex
76
Q

renaturing completely separated DNA

A
  1. Complementary sequences in the 2 strands find each other by random collisions and form a short segment of double helix
  2. The remaining unpaired bases come together as base pairs and the 2 strands zip themselves together to form the double helix
77
Q

how are RNA-DNA hybrids made

A

2 samples of isolated DNA or RNA are heated to cause denaturation and then are mixed. the sample is left to cool, letting the complementary strands to form a duplex whether it is a DNA-DNA, RNA-DNA, or RNA-RNA hybrid

78
Q

factors that influence DNA hybridization

A
  1. DNA concentration
    More concentrated sample of DNA has more frequent intermolecular collisions = faster hybridization rate
  2. base pair mismatch
    When hybridizing 2 nucleic acid segments that are completely complementary, mismatched nucleotides can interrupt base pairing, stacking interactions, and DNA stability
  3. pH
    In a range of 5-9, there is only a minor effect on DNA hybridization

pH < 5 causes the liberation of all bases from the helix

pH > 9 causes the conversion of dsDNA to ssDNA

79
Q

stringency of DNA hybridization

A

The extent that hybridization can occur between 2 non-complementary strands of nucleic acid - i.e the degree of complementarity required between 2 strands in order for them to hybridize

80
Q

high stringency

A

Hybridization occurs only when the 2 strands are highly compatible

81
Q

low stringency

A

Hybridization occurs even in the presence of some base mismatches

82
Q

increased stringency is used when

A

Duplex formation is not favoured

High temperatures – close to melting point

Low-salt concentrations

Presence of organic solvents

83
Q

decreased stringency is used when

A

Hybrid formation is favoured

Low temperature – below melting point

High-salt concentrations

Absence of organic solvents

84
Q

what does PCR rely on

A

DNA polymerases (enzymes able to synthesize DNA strands using a pre-existing DNA template and free deoxyribonucleotides. They add nucleotides (primers) to pre-existing strands)

85
Q

primers used in PCR

A

2 synthetic primers are prepared, complementary to sequences on opposite strands of the target DNA at positions defining the ends of the segment to the amplified

extended by a DNA polymerase

86
Q

simple PCR reaction mixture contains:

A

DNA sample containing the segment to be amplified

A pair of synthetic oligonucleotide primers

Deoxynucleoside triphosphates (dNTPS)

DNA polymerase

87
Q

PCR procedure steps

A
  1. denaturation
    Briefly heated to separate the strands

Enables the components of the reaction to interact with the single-stranded DNA template

  1. annealing
    Cooled so the synthetic primers can anneal to the DNA template

High concentration of primers increases the likelihood that they will anneal to each strand before they can reanneal too one another

  1. elongation
    Temperature is slightly increased for synthesis of a complementary DNA strand

DNA polymerase recognizes the 3’ ends and extends the strands along the targeted segment

  1. amplification
    Steps 1-3 process is repeated for a second time
  2. repeat
    Steps 1-3 are repeated 25-30 times, amplifying the DNA segment between the primers until it can be analyzed or cloned
88
Q

the design of primers

A

comprised of short sequences of oligonucleotides that are synthesized commerically

89
Q

parameters for designing a good PCR primer set

A
  1. 18-25 nucleotides in length
  2. 40-60% GC content
  3. Annealing temperature in the range of 50-60 degrees Celsius
  4. 1 or 2 GC residues at the 3’ end of the DNA strands
  5. Minimal secondary structure and base repeating
  6. Complementary to sequence chosen for amplification
90
Q

melting temp formula for oligonucleotides

A

Tm = 2 degrees (A+T) + 4 degrees (G+C)

91
Q

annealing temp formula

A

Ta = Tm - 5 degrees

92
Q

what is gel electrophoresis used for

A

determine the size of the PCR product

93
Q

gel electrophoresis

A

Technique for separating mixtures of large charged molecules like proteins or nucleic acids by causing them to move through a gel matrix when an electric field is applied

94
Q

gel electrophoresis steps

A
  1. The sample of interest is added to a slot at one end of the gel. The gel matrix is composed of agarose (a kelp-derived material that does not disrupt nucleic acid base pairing)
  2. Voltage is applied to the gel. Since the backbone of nucleic acids are negatively charged, both DNA and RNA will migrate towards the positive end of the gel in the electric field
  3. Larger molecules tend to move more slowly than smaller ones, so the molecules are separated based on their size – larger molecules are retained closer to the top and smaller molecules migrate towards the bottom of the cell
95
Q

gel electrophoresis: ethidium bromide

A

used to visualized the PCR product loaded into each well - fluoresces when exposed to UV light, used to detect nucleic acids in a gel

96
Q

gel electrophoresis ethidium bromide intercalating agent

A

planar molecules = inserts itself in between nucleic acid strands in a nonspecific manner

97
Q

reverse transcriptase PCR

A

used to amplify segments of RNA like mRNA gene expression products

98
Q

RT PCR characteristics

A

RNA is not amplified by PCR, instead it is converted to cDNA, amplified to PCR

can amplify and sequence gene segments without introns

can quantify mRNA levels as a measure of gene expression using quantitative PCR

99
Q

what does quantitative PCR use

A

SYBR green

100
Q

SYBR green

A

a fluorescent label that fluoresces substantially brighter when bound to dsDNA

101
Q

SYBR purpose

A

allows the quantification of dsDNA product inthe reaction after the extension stage of every PCR cycle

measures the amount of PCR product present - the more double stranded products produced with each cycle, the more fluorescence detected

102
Q

quantitiative PCR steps

A
  1. DNA is denatured at 95 degrees in a solution containing SYBR green
  2. Primer annealing (55-65 degrees)
  3. Beginning of synthesis
  4. Extension (60-72 degrees)
  5. Repeated
103
Q

cycle threshold

A

cycle number at which the threshold is first surpassed - should not be reached without target DNA or cDNA

104
Q

no template control (NTC)

A

added in quantitative PCR to control the amplification of non-specific DNA products

105
Q

problems with qPCR

A

SYBR green may bind to non-desired amplification product and produce a florescent signal

Non-specific products will melt at a different temperature than the desired PCR product, giving a melt curve with sharp decreases at more than one temperature

106
Q

what is sanger sequencing also known as

A

dideoxy chain-termination method

107
Q

sanger method purpose

A

method for determining the nucleotide sequence of DNA

108
Q

what does sanger sequencing require

A

DNA polymerases (to mock the mechanism of DNA synthesis - the 3’-hydroxyl group of the primer reacts with an incoming nucleotide to form a new phosphodiester bond )

labelled primer and dideoxynucleotides (for enzymatic synthesis of a DNA strand complementary to the strand under analysis)

109
Q

dideoxynucleotides

A

chain-elongating inhibitors of DNA polymerase, used in the Sanger method for DNA sequencing

110
Q

why are dideoxynucleotides used in sanger method?

A

ddNTPS (have H at both 2’ and 3’ positions) interrupt DNA synthesis because they lack the 3’-hydroxyl group needed for the next step

111
Q

steps of sanger sequencing

A
  1. DNA Denaturation
    - Apply heat to DNA sample causing the dsDNA to ssDNA, forming template and complementary strands
  2. Primer
    - DNA primer is annealed to the template strand, allowing nucleotides to be added later
    - Primer is added because DNA polymerase requires a free 3’ hydroxyl group to add free nucleotides
    - Primer was radiolabeled in original Sanger sequencing, which allowed the products of the DNA synthesis to be detected by an autoradiogram
  3. Free Nucleotides (dNTPS)
    - 4 reaction mixtures are set up and the template strand (with primer) is added to each one along with DNA polymerase and free nucleotides
  4. Modified Nucleotides (ddNTPS)
    - Used to terminate the synthesis reaction (no 3’-hydroxyl group, has only –H)
    - ddNTPS are added to each reaction mixture
    - Only 1 type of ddNTP is added to each reaction mixture, ddATP (green), ddTTP (red), ddCTP (blue), or ddGTP (yellow)
    - Added at much lower concentration than dNTPS, allowing for the extension of the synthesized strand with unmodified dNTPS before a ddNTP is added to terminate the extension
  5. Chain Termination
    - ddNTPS lack a 3’-OH group required for the formation of a phosphodiester bond between 2 nucleotides
    - Causes DNA polymerase to cease extension of DNA
  6. Gel Electrophoresis
    - Sample is collected from reaction mixtures
    - DNA is separated by size using gel electrophoresis
    - Each reaction mixture is added to a separate lane of the gel; the radiolabeled primer is detected by autoradiography
    - All possible chain lengths produced are separated by 1 nucleotide
    - Shorter fragments run further on the gel than the longer ones
112
Q

dye-termiantor sanger sequencing vs. sanger sequencing

A
  1. All ddNTPS are added to the same reactin (blue, green, yellow and red)
  2. Products of different size are separated by size using capillary electrophoresis

The fluorescently labeled segments are excited by a laser and the wavelength of the fluorescent emission (red, green, yellow or blue) is detected, one nucleotide at a time

113
Q

how is dye-termiantor sanger sequencing different from sanger sequencing

A
  1. all ddNTPS are added to the same reactin (blue, green, yellow, red)
  2. Products of different size are separated by size using capillary electrophoresis. The fluorescently labeled segments are excited by a laser and the wavelength of the fluorescent emission (red, green, yellow or blue) is detected, one nucleotide at a time
114
Q

in vitro PCR amplification

A

For DNA segments flanked by regions with known sequences, PCR primers can be designed to amplify the region of interest

If the sequence is unknown, synthetic adapters with known sequences can be ligated (enzyme joining 2 nucleic acid fragments) to the ends to serve as primer binding regions, enabling PCR amplification

115
Q

in vivo DNA replication

A

Segment is incorporated into a vector by a process called molecular cloning

Once inserted into bacteria, the vector can be replicated as the bacteria proliferate, enabling amplification of the DNA segment of interest

116
Q

molecular cloning

A

method to provide large quantities of purified DNA for sequencing

117
Q

general idea of molecular cloning

A

Isolation and generation of recombinant DNA molecules that are placed in organisms for replication and study

Separates a specific gene or DNA segment from a larger chromosome and incorporating it into a small molecule of carrier DNA

modified DNA is introduced into a host cell and replicated

118
Q

steps of molecular cloning

A
  1. Obtaining the DNA segment to be cloned
    - Enzymes called restriction endonucleases act as precise molecular scissors, recognizing specific sequences in DNA and cleaving genomic DNA into smaller fragments suitable for cloning
    - Alternatively, genomic DNA can be sheared randomly into fragments of a desired size
    - If the sequence of targeted genomic regions is known, some DNA segments that will be cloned are simply synthesized
  2. Selecting an appropriate carrier molecule of DNA capable of self-replication
    - Called cloning vectors and act as carriers of new DNA
    - Ex: plasmid vectors and bacterial artificial chromosomes
  3. Joining 2 DNA fragments covalently
    - The enzyme DNA ligase links the cloning vector to the DNA fragment to be cloned (inserted)
    - Composite DNA molecules of this type, comprising covalently linked segments from 2 or more sources are called recombinant DNA
  4. Moving recombinant DNA from the test tube to a host organism
    - The host organism provides the enzymatic machinery for DNA replication
    - Bacteria are often used for this purpose
  5. Selecting or identifying host cells that contain recombinant DNA
    - Cloning vectors have features that allow the host cells to survive in an environment where cells lacking the vector would die (ex: antibiotic resistance)
    - Cells containing the vector are selectable in that environment
119
Q

cloning vector

A

DNA molecule known to replicate autonomously (self-replicate) in a host

120
Q

recombinant DNA

A

a segment of DNA is placed within a cloning vector, allowing for its replication

121
Q

engineered plasmid DNA

A

circular DNA molecule found in bacteria that replicates separately from the bacterial chromosome

122
Q

when are plasmids useful

A

when cloning fragments are less than 15,000 bp in length

123
Q

plasmids purpose

A

Plasmids have specialized sequences that enable them to use the cell’s resources for their own replication and gene expression to survive

Naturally occurring plasmids have a symbiotic role in the cells (ex: performing new functions for the cell, resistance to antibiotics, etc)

124
Q

components of plasmid DNA

A
  1. Ori (Origin of replication)
    - Sequence where replication is initiated by cellular enzymes
    - Required to propagate the plasmid
  2. Restriction Sequences
    - Several unique restriction sequences are targets for restriction endonucleases
    - Providing sites where the plasmid can be cut to insert foreign DNA
  3. Number of Base Pairs
    - Facilitates both its entry into cells and the biochemical manipulation of the DNA
    - Generated by trimming away many DNA segments from a larger parent plasmid (sequences that the molecular biologist does not need)
  4. Antibiotic Resistance
    - Have genes that confer resistance to antibiotics tetracycline and ampicillin
    - Allows the selection of cells that contain the intact plasmid or a recombinant version of the plasmid using these antibiotics
125
Q

stages of molecular cloning experiment

A
  1. plasmid generation
  2. transformation and antibiotic selection
126
Q

plasmid generation steps

A
  1. The pBR322 plasmid DNA is cleaved at the restriction site by a restriction endonuclease called Pstl. The foreign DNA contains the Pstl complementary ends
  2. Foreign DNA fragments are ligated into the plasmid. Successful integration disrupts the AmpR gene, making the plasmids no longer resistance to ampicillin
127
Q

transformation and antibiotic selection

A
  1. Plasmid DNA is introduced into the bacterial cells by transformation, whereby the cells and plasmids are incubated at 0 degrees in a calcium chloride solution and then heat shocked by raising the temperature to 43 degrees (chemical transformation). Alternatively, the cells can be subjected to a high-voltage pulse to allow the plasmid DNA to enter (electroporation transformation)
  2. The cells are grown on agar plates with tetracycline to select only for those that have taken up the plasmid. Individual colonies are transferred to matching positions on additional plates. 1 plate has tetracycline and the other has tetracycline and ampicillin
  3. Cells that grow only on the tetracycline agar plates contain recombinant plasmids with disrupted ampicillin resistance, hence the foreign DNA. Cells with pBR322 without foreign DNA retain ampicillin resistance and grow on both plates mean that cloning was unsuccessful
128
Q

bacterial artifical chromosomes (BACs)

A

Have origin of replication, antibiotic resistance, restriction sites and a reporter gene (enables the detection or measurement of gene expression)

BACs have stable origins of replication to support very long segments of cloned DNA

129
Q

limitations of sanger sequencing

A

Slow and expensive

Read lengths are only up to 1000-1500 bases

Sequences for a large segment of DNA need to be broken down, analyzed one at a time and compiled together

130
Q

next-generation sequencing purpose

A

rapid sequencing of large DNA segments

131
Q

next gen sequencing steps

A
  1. These large DNA segments (could be an entire human genome) are fragmented to smaller segments (300-400 bp) and are sequenced simultaneously
  2. The sequences for overlapping fragments are then aligned to generate a consensus sequence for the entire DNA segment
132
Q

what is used in reversible terminator sequencing

A

modified nucleotides bearing a reversible terminator (RT)

133
Q

RT sequencing vs sanger sequencing

A

sanger = labeled with fluorescent tags
RTS = the fluorescent tags are bound to the modified nucleotides using a cleavable linker region

134
Q

steps of RTS

A
  1. library preparation
  2. cluster generation
  3. sequencing
  4. data analysis
135
Q

RTS library preparation

A

DNA fragmentation:
- Large DNA segments are cleaved into 300-400 base pair fragments
- Fragmented ends are repaired and a single adenosine nucleotide is added to the 3’ ends of the fragments to prevent them from ligating to each other

adaptor ligation:
- Ligated to the 5’ and 3’ ends of the fragmented DNA segment
- Terminal sequences = essential to the next stage of sequencing, known as Cluster Generation
- Index sequences = allow DNA libraries from different samples to be processed and separately and pooled together in the same run
- Each index is like a unique barcode for DNA fragments from a specific sample. After the run, the sequences from individual samples can be separated from each other based on this indexing barcode
- Primer binding sequences = serve as binding regions for sequencing primers
- Allows for paired sequencing from both ends of the DNA fragment

136
Q

RTS cluster generation

A

Involves the amplification of individual sequences from the DNA library to form clusters of clonal (all from the same gene sequence) DNA segments

Takes place in the flow cell, a piece of acrylamide-coated glass that is coated with oligonucleotides

137
Q

RTS cluster generation steps

A
  1. DNA library is added to the flow cell. terminal sequences in the adapters allow single DNA segments to hybridize with the oligonucleotides bound to the surface of the flow cell. a DNA polymerase is used to extend an oligonucleotide that is complementary to the bound DNA molecule
  2. the original template is washed away, leaving only the newly synthesized strand that is covalently bound to the flow cell
  3. the adapter sequence at the 3’ end of the bound DNA molecules hybridizes with a nearby oligonucleotide. the bridge is extended and the 2 strands are denatured
  4. process is repeated, forming a cluster of forward and reverse strands - reverse strands are hydrolyzed and washed away, leaving a cluster of unidirectional clonal strands
138
Q

RTS sequencing

A

3 components added to the flow cell:

  1. 4 fluorescently labeled reversible terminator nucleotides (RT-dATP, RT-dCTP, RT-dGTP, RT-dTTP)
  2. a sequencing primer that can hybridize with the 3’ adapter region of the bound DNA segment
  3. DNA polymerase
139
Q

results of RTS sequencing

A

graph with many coloured dots - each dot represents a stage of amplification and the colour determines the next nucleotide (e.g, green cluster = next nucleotide is A)

140
Q

RTS data analysis

A

aligned to a reference genome once all the reads are generated

“depth” of coverage = # of times that a specific base pair appears in a sequence read

BCHM218 (4 decks)

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