Lab 1 Flashcards

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1
Q

Lab

approaches used to identify genes in bacteria

A
  • mutagenesis
    • relies on loss of phenotype
    • causes changes in DNA structure that in turn results in phenotype changes
    • Allows rapid identification of genetics associated with a phenotype change
    • assesses role for a given gene in a particular function
    • Two common methods
      • use of transposons (mobile genetic elements)
      • use of chemicals
    • disadvantages
      • organism may have several related structural genes expressing the same phenotype
      • loss of a particular phenotype may be the result of mutations in a gene that is important in the expression of a phenotype, but is not the actual structural gene for that phenotype
        • i.e. regulatory genes involved in the expression of the target gene
  • cloning for function
    • relies on gain of phenotype (gene function)
    • offers a direct method to isolate and identify a gene by expression of its phenotype in another organism
  • Reverse genetics
    • opposite of forward genetics (mutations and direct cloning)
    • working with genome sequences to identify gene function through mutagenesis or other forms of altering gene expression
    • becoming the new standard
    • easier to sequence genome and decipher function
  • Biochemical approach
    • isolate protein
    • determine amino acid sequence
    • design degenerate primers for PCR
    • identify gene
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2
Q

Lab

Transposon mutagenesis

A
  • highly preferred
  • uses transposons (mobile genetic elements)
  • advantage
    • transposon insertions automatically result in a physical “tagging” of the gene location
    • the mutated gene can be easily identified by physically locating the presence of the transposon in genome
      • for example
        • cloning transposon or portion of it, w/DNA region where transposon inserted
        • DNA sequences flanking transposons can be used as probes to identify homologous sequences from wildtype DNA
  • Successful use contingent on contingent upon the tractability (susceptibility) of the host organism
  • disadvantages
    • Lack of host susceptibility
    • potential for less random insertions throughout the genome
      • may insert in regions that are “hot spots” (specific nucleotide sequences preferred for insertion), as opposed to “cold spots” (specific nucleotide sequences where they don’t insert).
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3
Q

Lab

Chemical mutagenesis

A
  • DNA changes are not readily detected at the physical level by molecular methods that are less sensitive than nucleotide sequencing
  • recovery of the corresponding wildtype gene is based on restoration (complementation) of the mutated phenotype
    • requires transferring an entire library of clones that collectively make up the genome of the wildtype organism into the mutant strain
  • advantage
    • most organisms are susceptible to mutagens
    • several chemicals that differ in mutation modes of action are available for use
    • less likely to induce mutations in “hot spots”
    • Manipulation of conditions influencing mutation rate can generate single- or multiple mutations per cell
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4
Q

Lab

Cloning for function (direct cloning)

A
  • relies on gain of phenotype (gene function)
  • expression of a gene/group of genes originating from one bacterium in a different bacterium that lacks the phenotype (heterologous expression)
  • conducted by constructing a genetic “library” of clones consisting of a number of DNA fragments each contained in a plasmid vector
    • can consist of fragments of an organism’s entire genome, or a subset
    • library is transferred to a second organism
    • individual clones are maintained separately in individual cells
    • library then screened for in a second, heterologous organism for phenotype acquisition
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5
Q

Lecture

Genotype

A

genetic information encoded within the genome of an organism

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6
Q

Lecture

Phenotype

A
  • observable characteristic
  • based on genotype and environment
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7
Q

Lecture

mutation

A
  • A heritable change in an organism’s genome (genotype)
  • caused by
    • alteration of single base units in DNA
    • deletion
    • insertion
    • rearrangement of larger sections of DNA in genes or chromosomes
  • can occur
    • Naturally. Spontaneous mutations
      • during DNA replication
      • rate of 10-4 to 10-10 per generation: 1 in 10,000 to 1 in 10 billion
      • between 10-1 and 10-2 per individual
      • cellular factors maintain replication fidelity such as proofreading and mismatch repair
    • mutagens
      • agents that speed up the natural rate of mutation
    • Directed mutagenesis/site specific mutations
      • through genetic manipulation
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8
Q

Lecture

Effects of Mutagens

A
  • Physical manifestation
    • structural change of DNA molecule
    • Results in genetic change during replication
  • Internally coded
    • inheritable information
    • change in genetic information – change in base sequence
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9
Q

Lecture

Structural changes

What chemical alterations result in mutations?

A
  • Tautomeric shifts
    • transient rearrangement of bonding to form a structural isomer
    • results in altered base pairing during replication
  • deamination
    • loss of exocyclic amino group
    • changes base, and thus base pairing capabilities
      • i.e. Cytosine to Uracil
  • loss of bases by depurination and depyrimidination
  • physical changes to DNA (eg. Cross linking or thymine/pyrimidine dimers)
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10
Q

Lecture

Structural changes

Tautomeric shifts

A
  • transient rearrangement of bonding to form a structural isomer
  • results in altered base pairing during replication
  • i.e.
    • in guanine or thymine
      • C=O (keto) converted to C-OH (enol)
      • double bond moves from exocyclic to endocyclic
    • in adenine and cytosine
      • NH2 (amino) to NH (imino)
      • double bond moves from endocyclic to exocyclic
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11
Q

Lecture

Structural changes

Tautomeric shifts: Altered Pairing

A
  • in guanine or thymine
    • C=O (keto) converted to C-OH (enol)
  • in adenine and cytosine
    • NH2 (amino) to NH (imino)
  • T’ (enol) pairs with G
  • C’ (imino) pairs with A
  • G’ (enol) pairs with T (keto)
  • A’ (imino) pairs with C (amino)
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12
Q

Lecture

Structural changes

Deamination

A
  • loss of exocyclic amino group
  • changes base, and thus base pairing capabilities
  • i.e. Cytosine to Uracil
    • Loss of an amine (NH2) group and is replaced by an Oxygen
    • changes C to U
    • Uracil is equivalent to T
    • Cytosine should base pair with G, but now base pairs with A
    • 5-Methylcytosine deamination occurs naturally but less than Cytosine deamination
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13
Q

Lecture

Structural changes

A
  • depurination and depyrimidination
  • loss of a base from the phosphate sugar backbone of a DNA polymer
  • results in random insertion of base during replication, subsequently changing the G to T
  • in image
    • 1 = shows GC pairing
    • 2 = depurination of G (missing)
    • 3 = G is randomly replaced with A by DNA polymerase
      • 1 in 4 chances it will pick the right one
      • better than terminating replication early because of missing base
    • 4 = A is then base paired with T
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14
Q

Lecture

Structural changes

physical changes to DNA

A
  • Cross linking or thymine/pyrimidine dimers
  • Pyrimidine dimers (typically TT) form when DNA is exposed to Ultraviolet light
    • results in covalent bond
    • on the same DNA strand
    • physical linkage does not allow dna polymerase to read the strand correctly
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15
Q

Lecture

Types of Mutagens

Environmental

A
  • UV light
  • Ionizing Radiation: structural damage or breakage
  • damage is done by oxidative damage
    • i.e. conversion of Guanine to 8-oxo-7-hydro-guanine (GO)
  • See Scenario 1 in image
    • unrepaired GO causes a transversion, a base change from pyrimidine to purine or vice versa
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16
Q

Lecture

Types of Mutagens

Chemical - Alkylating agents

A
  • broad group of compounds that function in base modification by adding an alkyl (methyl or ethyl) group to DNA base
  • Prevents correct base reading during replication
  • Can lead to
    • depurination/depyrimidination
    • dimers/covalent linkages that lead to DNA strand breakage
  • examples
    • Aflatoxins
      • naturally occurring toxins produced by fungi
      • involved in rot fruits, vegetable and grains
    • Nitrosoguanidine
      • aka N-methyl-N-nitro-N-nitrosoguanidine, NTG or MNNG
      • used in experiments conducted in this class
      • can perform alkylation or depurination
      • functions primarily at replication
        • need replicating bacteria, in exponential phase
        • can cause multiple mutations/clustering
  • Most susceptible site for alkylation w/in DNA
    • the 6’-O position on guanine bases
    • Oxygen is attacked by an electrophilic mutagen
    • An alkyl is added converting G → O-6-Ethylguanine
    • This leaves only two N sites for H-bonds in O-6-Ethylguanine
    • O-6-eG pairs with T instead of C
    • transition conversion: purine to purine change (or pyrimidine to pyrimidine)
17
Q

Lecture

Types of Mutagens

Chemical - Cross-linking agents

A
  • between strands
  • e.g. eg mustard or mitomycin
  • Prevents strand separation, blocking replication
  • categorized as alkylating agents
  • covalent bond formation results in strand breakage
  • If DNA repair occurs, it typically results in regions of DNA deletion
18
Q

Lecture

Types of Mutagens

Chemical - Base analogs

A
  • molecules which have a very similar structure to one of the four nitrogenous bases used in DNA: adenine, guanine, cytosine or thymine
  • They form a structure similar to one of the DNA nucleotides
  • can be used to form the new strand in semi conservative replication
  • causes mispairings
  • e.g.
    • 5 bromouracil
      • has two forms, a keto and enol
      • keto pairs with A, and enol paris with G
    • 5-methylcytosine
      • similar in structure to cytosine and thymine
      • continues to pair with G
19
Q

Lecture

Types of Mutagens

Chemical - Intercalating agents

A
  • hydrophobic heterocyclic ring molecules
  • resemble the planar ring structure of base pairs
  • Insertion of these agents between adjacent base pairs, distorts the DNA double helix and interferes w/DNA replication, transcription, and repair
  • e.g.
    • ethidium bromide
      • widely used in molecular biology as a stain for DNA
      • intercalated molecule fluoresces on exposure to ultraviolet light.
    • acridine orange
      • histological stains
      • intercalation may cause DNA Pol to “stutter” and copy the molecule as an extra base pair, introducing a frameshift mutation.
    • actinomycin D
20
Q

equation to make a specific volume of a dilute solution from a stock solution

A

C₁V₁ = C₂V₂

  • C₁ - [] of stock solution
  • V₁ - volume of stock solution needed to make dilute solution
  • C₂ - [] of dilute solution
  • V₂ - final volume of dilute solution
21
Q

dilution factor

A
  • factor by which [dilute solution] is reduced compared to [stock solution]
  • DF = C₁ / C₂ or
  • DF = V₂ / V₁ or
22
Q

Prepare 100 ml of 10mM Tris buffer

from a 1 M Tris stock and determine DF

A
  1. determine variables
    • C₁ - 1M = 1000 mM
    • V₁ - unknown
    • C₂ - 10 mM
    • V₂ - 100 ml
  2. solve for unknown
    • C₁V₁ = C₂V₂
    • V₁ = C₂V₂ / C₁
    • V₁ = (10 mM × 100 ml) / 1000 mM
    • V₁ = 1 ml
    • 1 ml of 1 M Tris stock needed
  3. determine amt of water needed
    • amt of water = V₂ - V₁
    • amt of water = 100 ml - 1 ml = 99 ml
    • 99 ml of water + 1 ml of 1 M Tris stock = 100 ml solution
  4. determine dilution factor
    • DF = C₁/C₂ or V₂/V₁
    • C₁/C₂: DF = 1000 mM / 10 mM = 100
    • V₂/V₁: DF = 100 ml / 1 ml = 100
23
Q

optical density (OD)

A
  • based on absorbance detection mode
  • determines which portion of light passes through a sample/suspension of microorganisms
  • Particles in solution scatter light
  • the more particles (microorganisms) can be found in a solution, the more light is scattered by them
  • a replicating population of bacteria or yeast increases light scattering and measured absorbance values
  • measures light scattering and not absorbance by the absorbing molecules
  • OD600
    • measurement of the optical density at 600 nm
      *
24
Q

SERIAL DILUTION

  • We are starting with an OD of 0.6 = 1.8 × 10⁸ c/ml​
  • we want to get a reasonable # of 30 - 300 colonies per plate
  • final volume is 10 mL
  • use five tubes
  1. Calculate the range of c/ml that will fall between 30 - 300 colonies
    • compare 1.8 to range
    • we need to move the decimal in 1.8 _____ places to the ____, converting 1.8 to _____, a number that falls w/in 30-300
  2. Calculate the final c/ml qty needed to fall w/in 30 - 300 range
    • we moved X decimal places to the right to get 180
    • so 10⁸ - 10X decimal places = 10Y
    • so 10⁸ - 10Y = _____
    • so _____ c/ml is the final qty we need to get to fall w/in range of 30-300 range
  3. You want an over and under estimate of ____ so include the following [] as part of serial dilution
  4. Using the 5 tubes w/total volume of 10mL, calculate the 5 concentrations for serial dilutions:
  5. Get the OD, dilution factor, volume to transfer and buffer volume for each dilution series (tube) in step four
A
  1. Calculate the range of c/ml that will fall between 30 - 300 colonies
    • two
    • right
    • 180
  2. Calculate the final c/ml qty needed to fall w/in 30 - 300 range
    • two
    • 10⁶
    • 10<span>2</span>
    • 1.8 × 10²
  3. Overestimate / underestimate
    • 1.8 × 10³ and 1.8 × 10¹
  4. You’ve got the starting concentration and calculated the last 3, so pick any numbers that decrease for the 2 and 3rd concentrations
    • 1.8 × 10⁸ c/ml
    • 1.8 × 10⁶ c/ml
    • 1.8 × 10⁴ c/ml
    • 1.8 × 10³ c/ml
    • 1.8 × 10² c/ml
    • 1.8 × 10¹ c/ml
  5. Get the dilution factor, volume to transfer, and buffer amount for each tube in step four
    • OD
      • OD = [Initial/First] - 1.8 × 10⁶
      • OD = 10⁸ - 10⁶ = 10²
      • OD for all tubes
        • OD = 10⁸ - 10⁶ = 10²
        • OD = 10⁸ - 10⁴ = 10⁴
        • OD = 10⁸ - 10³ = 10⁵
        • OD = 10⁸ - 10² = 10⁶
        • OD = 10⁸ - 10¹ = 10⁷
    • Dilution factor
      • 10⁸ - ⁶ = 10² = 100
      • DF = 1 : 100
      • DF for all tubes
        • DF = 10⁸ - 10⁶ = 10²
        • DF = 10⁶ - 10⁴ = 10²
        • DF = 10⁴ - 10³ = 10¹
        • DF = 10³ - 10² = 10¹
        • DF = 10² - 10¹ = 10¹
    • calculate volume to transfer
      • ttl vol. / DF = 10ml / 100 = 0.1 mL
    • calculate buffer volume
      • ttl vol - txfr volume = buffer volume
        10 mL - 0.1mL = 9.9 mL buffer​