Lab 1

Question 1

Lab

approaches used to identify genes in bacteria

Answer 1

• 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

Question 2

Lab

Transposon mutagenesis

Answer 2

• 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).

Question 3

Lab

Chemical mutagenesis

Answer 3

• 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

Question 4

Lab

Cloning for function (direct cloning)

Answer 4

• 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

Question 5

Lecture

Genotype

Answer 5

genetic information encoded within the genome of an organism

Question 6

Lecture

Phenotype

Answer 6

• observable characteristic
• based on genotype and environment

Question 7

Lecture

mutation

Answer 7

• 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

Question 8

Lecture

Effects of Mutagens

Answer 8

• 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

Question 9

Lecture

Structural changes

What chemical alterations result in mutations?

Answer 9

• 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)

Question 10

Lecture

Structural changes

Tautomeric shifts

Answer 10

• 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

Question 11

Lecture

Structural changes

Tautomeric shifts: Altered Pairing

Answer 11

• 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)

Question 12

Lecture

Structural changes

Deamination

Answer 12

• 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

Question 13

Lecture

Structural changes

Answer 13

• 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

Question 14

Lecture

Structural changes

physical changes to DNA

Answer 14

• 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

Question 15

Lecture

Types of Mutagens

Environmental

Answer 15

• 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

Question 16

Lecture

Types of Mutagens

Chemical - Alkylating agents

Answer 16

• 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)

Question 17

Lecture

Types of Mutagens

Chemical - Cross-linking agents

Answer 17

• 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

Question 18

Lecture

Types of Mutagens

Chemical - Base analogs

Answer 18

• 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

Question 19

Lecture

Types of Mutagens

Chemical - Intercalating agents

Answer 19

• 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

Question 20

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

Answer 20

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

Question 21

dilution factor

Answer 21

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

Question 22

Prepare 100 ml of 10mM Tris buffer

from a 1 M Tris stock and determine DF

Answer 22

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

Question 23

optical density (OD)

Answer 23

• 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
    *

Question 24

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 - 10^X decimal places = 10^Y
   • so 10⁸ - 10^Y = _____
   • 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

Answer 24

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​