Exam 3 Flashcards

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1
Q

mutation

definition

A

an heritable change in genetic material

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2
Q

mutations positives and negatives

A
  • positive: give allelic variation, foundation for evoluntionary change
  • negatives: new mutations more likely to be harmful than beneficial, cause diseases
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3
Q

point mutation types

A

base subsitions two types:
* transition (change within pyrimidine and purine (C ⇔ T OR A ⇔ G)
* transversion (change b/w pyrimidine and purine (C,T ⇔ A,G)

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4
Q

transitions are _ common that transversions

A

more

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5
Q

transitions is a change

A

of a pyrimidine to another or a purine to another

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6
Q

transversion is a change of a

A

pyrimidine to a purine or vice versa

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7
Q

silent mutations

def

A

base subsitutions that do not alter the amino acid sequence of the polypeptide

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8
Q

silent mutations are possible because of

A

the degeneracy of the genetic code

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9
Q

missense mutations

def

A

base subsitutions where an amino acid sequence change occurs

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10
Q

nonsense mutations

A

base subsitutions that change a normal codon to a stop codon

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11
Q

frameshift mutations

def

A

addition or deletion that is not a factor of 3

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12
Q

silent mutations have _ effect on protein function

A

no

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13
Q

missense mutations have a _ effect on protein function

A

neutral or inhibitory

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14
Q

up promoter mutations…

A

increase transcription

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15
Q

a mutation in the promoter..

A

affects transcription

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16
Q

mutation in the regulatory element or operator site

A

disrupt the ability of the gene to be properly regulated

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17
Q

mutation in UTR

A

alter the ability of mRNA to be translated, or alter mRNA stability

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18
Q

mutation in the splice recognition sequence

effect

A

alter the ability of pre-mRNA to be properly spliced

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19
Q

forward mutation changes

A

the wild-type genotype into something

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20
Q

reverse mutation changes

A

a mutant allele back to the wild type

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21
Q

mutations are often characterized by

A

their differential ability to survive

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22
Q

suppressor mutations

def

A

reverse the phenotypic effect of another mutation

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23
Q

intragenic suppressors

def

A
  • the second mutation is within the same gene as the first mutation
  • typically the first mutations causes abnomality in protein and the second mutation restores normal protein structure
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24
Q

intergenic supressors

def

A
  • 2nd mutation is different gene than the first mutation
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25
Q

position effect

A

when a chromosome rearranges and a gene gets moved so the expression is altered

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26
Q

two common reasons for position effects

A
  1. movement to a position next to regulatory sequences
  2. movement to a hetrochromatic regions
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27
Q

animal cell two types

A
  • germ line cells
  • somatic cells
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28
Q

germ-line mutations are those that

A

occur directly in a sperm or egg cell, or in one of the precursor cells

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29
Q

somatic mutations are that occur

A

directly in a body cell that is not part of the germ-line

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30
Q

_ mutations are found in the whole body

A

germ-line mutations

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31
Q

somatic mutations result in _

A

patches of affected area
* earlier the mutation the larger the patch

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32
Q

_ mutations are passed onto gametes

A

germ line mutations

half the gametes

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33
Q

lamarck proposed that

A

physiological events determine whether traits are passed along to offspring

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34
Q

lederberg study objective was to

A

see if mutations are randomly occuring

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35
Q

in the lederberg experiment, if the physiological adaptation hypothesis is true,

A

then the number of tonr bacteria would be low unless there is selection for T1 resistance

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36
Q

lederberg experiment steps

A
  1. place bacteria on plate
  2. allow cells to divide
  3. transfer replicas from master plate to secondary plates containing t1 phages
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37
Q

results of lederberg experiment

A

mutations had occured randomly in absence of selection by T1
* no new colonies in the presence of T1

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38
Q

spotaneous mutations

A
  • result from abnomalities in the biological processes
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39
Q

induced mutations

def

A

caused by enviromental agents

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40
Q

mutagens

def

A

agents that alter DNA structure

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41
Q

spontaneous mutations arise from:

3 types

A
  • depurination
  • deamination
  • tautomeric shift
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42
Q

depurination is the

A

removal of a purine (A G) from DNA

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43
Q

apurinic site is formed by

A

depurination

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44
Q

how does depurination cause mutation

A

sometimes repair system fails and the DNA polymerase adds a random base in the apurinic site (base missing there)

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45
Q

depurination has a _ chance of mutation in one or both daughter strands

A

75% of mutation in one of the daughter strand

wrong base can be filled in the a puranic site

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46
Q

deamination happens most with

A

cytosine

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47
Q

deamination of cytosine turns it into

A

uracil

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48
Q

_ is a hot spot for mutation

A

mythylated cytosine bases

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49
Q

tautomeric shift is when

A

thymine and guanine go from keto form to being in enol form
A and C convert from amino to imino form

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50
Q

to cause mutation, a tautomeric shift

A

must occur immediately prior to replication

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51
Q

oxidative stress is

A

an imbalance between the production of ROS and organism’s ability to break them down

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52
Q

ROS accumulation can cause

A

DNA damage and mutation

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53
Q

CAG repeat in coding sequence can cause

A

long tracks of glutamine that causes proteins to aggregate with each other

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54
Q

TNRE in non conding regions can

A
  • cause abnormal changes in RNA structure
  • methylated CpG islands (silence genes)
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55
Q

anticipation

A

TNRE occur more frequencly during gamete formation, so gets worse with each generation

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56
Q

mechanism of TNRE

A
  • TNRE contain C and G that create hairpin
  • polymerase slip off DNA, hairpin forms
  • DNA pol hops back on and beings synthesis from new location
  • hairpin spreads out, the gaps are filled by gap repair
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57
Q

chemical mutagen three types

A
  • base modifiers
  • intercalating agents (interfere with replication)
  • base analogues
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58
Q

base modifiers do what

A

covalently modify structure NT

59
Q

intercalating agents

A

contain flat planar structures that intercalate themselves in the double helix
* causes distortion of helix
* daughter strand may have additions or deletions
* frame shifts

60
Q

x ray and gamma rats have

A

shirt wavelngths and high energy

61
Q

ionizing radiation includes

A

x ray and gamma rays

62
Q

nonionizing radiation includes

A

UV light

63
Q

UV light causes the formation of

A

cross linked thymine dimers which can cause mutations when replicated

64
Q

mutation rate

A

likelihood that a gene will be altered by a new mutation

65
Q

mutation frequency

A

the number of mutant genes divided by the total number of genes in a population

66
Q

mutation frequency can get higher than mutation rates because of

A

natural selection and genetic drift

67
Q

how does the ames test work

A
  • a strain of bacteria that cannot make histindine is used bc of a point mutation
  • a second mutation may occur to reverse this
  • ames test monitors the rate at which the second mutation occurs
68
Q

_ can repair thymine dimers

A

photolyase

69
Q

alkyltransferase repairs

A

alkylated bases by transfering the methyl or ethyl group from the base to a cysteine side chain within the alkyltransferase

70
Q

base excision repair involves _ that can _

A

DNA N-glycosylases that can recognize an abnormal base and cleave the bond between it and the sugar in the DNA

71
Q

base excision repair steps

A
  1. N glycosylase recognize abnormal base and cleaves bond b/w base and sugar
  2. AP endonuclease cleaves DNA on 5’ end of missing base
  3. Creates nick in DNA
  4. ECOLI: DNA pol 1 removes damaged region and fills normal DNA, ligase seals
  5. EUKAR: DNA pol beta can fill base OR DNA pol delta/episilon do flap technique, ligase seals
72
Q

base excision repair is for

A

small repairs, one nucleotide

73
Q

nucleotide excision repair can repair

A

many types of DNA damage, more major problems

74
Q

nucleotide excision repair steps

A
  1. UvrA/B tracks along DNA in search
  2. After damage detected, UvrA released and UvrC binds
  3. UvrC makes cuts on both sides of the damage
  4. UvrD removes the damaged regions
  5. UvrB&C released
  6. DNA pol fills in the gaps, ligase seals
75
Q

mutations in genes involving nucleotide excision repair all cause diseases that

A

increase sensitivity to sunlight

76
Q

DNA pol have _ to recgonize _ and repair them

A

3’ to 5’ proofreading activity to detect base mismatches and fix them

77
Q

mis match repair system is specific

A

to the newly made daughter strand

78
Q

how does mismatch repair system know difference between parent and daughter strand

A

parent strand is methylated while daughter is not

79
Q

mismatch repair system steps

A
  1. MutS finds mismatch
  2. MutS and L bind to MutH which is already bound to a sequence
  3. MutH cuts nonmethylated strand
  4. MutU seperates DNA
  5. Endonuclease digest DNA to point where MutS is binded
  6. Base mitchmatach is yeeted
  7. DNA pol fill and ligase seals
80
Q

homologous recombination is only available during _ because _

A

during S and G2 pahses because sister chromatid is needed

81
Q

homologous recombination is error_ because

A

error free because sister chromatids are genetically identical

82
Q

homologous recombination steps

A
  1. end processing of broken strand (ends are chewed in)
  2. strand exchange
  3. non-broken strands act as a template for fixing thhe break
  4. resolution by cutting strands and ligase reattching, part of unbroken strand now on fixed strand
83
Q

non homologous end joining is error_ because

A

error prone because processing may result in a deletion

84
Q

nonhomologous end joining steps

A
  1. end binding proteins bind to each broken end
  2. other proteins form a cross bridge
  3. gap is repaired and ligase seal
85
Q

DNA pol 3 in e coli is unable to

A

replicate lesioned DNA

86
Q

translesion synthesis is possibel because

A

translesion-replicating pol contain an active site with a loose pocket so can accomodate weird structures

87
Q

a negative to the translesion synthesis process

A

translesion -replicating pol have low fidelity, high mutation rate

88
Q

homologous recombination occurs in

A

meiosis 1

89
Q

homologous recombination involves

A

alignment of pair of chromosomes, breakage at analogous locations and exchange of segments

90
Q

homologous recombination steps

holiday model

A
  1. both chromosomes nicked at identical locations
  2. DNA strands to the left of the nicks invade and link to the strands to the right of the nicks
  3. creates a holiday junction
  4. holiday junction migrates (brand migration)
  5. creates two hetroduplex regions
  6. hetroduplex regions have some base mismatches
91
Q

recent models of recombination steps

A
  • no nick in the same location
    1. DNA helix incur break in both strands of one chromatid
    2. Ends processed
    2. strand invasion fors D-loop
    3. gap repair synthesis fills in
    4. branch migration
92
Q

gene conversion

A

two different alleles become one

93
Q

gene conversion occurs

A
  1. dna mismatch repair
  2. DNA gap repair synthesis: double stranded break area loses a small part
94
Q

A mutation changes a codon to another codon that does not change the original amino acid to a different amino acid. This is a

A

silent mutation

95
Q

What is a key characteristic(s) of the repeat sequences that promote trinucleotide repeat expansion?

A

contain a C and G

96
Q

During the Ames test, if a substance is a mutagen, there will be_________ colonies on the plates in which the cells had been exposed to the mutagen, because the mutagen converts some of the cells from ____________.

A

more, his- to his+

97
Q

During nucleotide excision repair, which of the following does not happen?

  • detection of a DNA lesion
  • making cuts on either side of the lesion in the same strand
  • removing the damaged strand
  • making a double stranded cut in the DNA
A
  • making a double stranded cut in the DNA
98
Q

During nucleotide excision repair, which of the following does not happen?

  • detection of a DNA lesion
  • making cuts on either side of the lesion in the same strand
  • removing the damaged strand
  • making a double stranded cut in the DNA
A
  • making a double stranded cut in the DNA
99
Q

A gene exists in two alleles, which we will call B and b. The gene is 1123 bp in length, and the B and b alleles exhibit single base pair differences at six different sites. If gene conversion changed the b allele into the B allele, which mechanism would you favor to explain the conversion?

A

gap repair synthesis

100
Q

According to the double-strand break model for homologous recombination, what happens right after DNA strand degradation at the double-stranded break site?

A

formation of a D loop

101
Q

According to the Holliday model for homologous recombination, what is/are the possible end result(s) of the resolution step?

A

nonrecombinant or recombinant chromosomes with a heteroduplex region

102
Q

5-bromouracil would cause what type of mutation

A

transition mutation

thymine analoug

problem set 10

103
Q

what type of mutation would nitrous acid make

A

transition mutation

PS 10

104
Q

what type of mutation proflavin cause

A

frameshift mutation

intercalating agent

105
Q

A mutagen changes cytosine to uracil. Chemically, what is the mutagen doing to cytosine to convert it to uracil? Give an example of a mutagen that can do this. Which DNA repair system(s) would be able to repair this defect?

A

This spontaneous mutation is called a deamination of cytosine which changes a cytosine to a uracil. An amino group is removed from the cytosine which makes it a uracil and the DNA repair enzyme removes it because uracil is not found in the DNA sequence. NH2 in the base is switched into an O.

106
Q

recombinant DNA technology is

A

the use of in vitro molecular techniques to isolate and manipulate fragments of DNA

107
Q

process of cloning genes into vectors

A
  1. endonucleases cut genes into sticky ends that can base pair to another DNA sequence with complementary sequence
  2. other DNA is cut with the same enzyme
  3. incubate the DNA together
  4. sticky ends hydrogen bond
  5. add DNA ligase
  6. result can be a recirculized vector, recombinant with target or recombinant without target gene
  7. test to see which vector has target gene
  8. if recombinant with right target, gene will be white and beta galac inactive
108
Q

when insert gene, it has ampr and lacz why

A
  • amp r shows that cells have taken up the vector
  • lax z shows the vector is the targetted gene one
109
Q

As described in your textbook, one way to determine if a segment of chromosomal DNA has been inserted into a vector is to put XGal in the growth media. Following overnight growth…

A

white colonies contain a vector with an inserted fragment of chromosomal DNA; the lacZ gene has been inactivated

110
Q

sticky ends are most stable when they

A

have more C and Gs

111
Q

how is functionality of the Lac Z gene determined in recombinant vectors?

A

X-gal is cleaved by beta galactosidase into a blue dye
* so white colonies have a inactive lax z due to insertation of a DNA fragment

112
Q

amplification of a clones genes occurs in two ways

A
  • vector gets replicated by host many times
  • bacterial cell divides approx every 20 mins
113
Q

complementary DNA

A

DNA made from RNA

114
Q

cDNA has

A

only the coding parts of the gene
* allows expression of the encoded protein

115
Q

how is cDNA made

A
  1. add poly-dT primer to mRNA that binds poly A tail of mRNA
  2. add reverse transcriptase and dNTPs
  3. add RNaseH to cut up the mRNA template and make RNA primers
  4. add DNA pol 1 and DNA ligase to synthesize the second DNA strand
116
Q

genomic library

A

contains whole genomes, starting material is chromosomal DNA
* contains colonies of dna that carry seperate peices of chromosomal DNA

117
Q

cDNA library

A

starting material is cDNA

118
Q

how are genomic libraries made

A
  • create hybrid vectors (technique of recombi vectors)
  • transform plasmid vectors into DNA
  • each bacterial colony holds a different peice of chromosomal DNA
119
Q

making cDNA library

A
  1. isolate mRNA
  2. make cDNA
  3. add linker DNA that is a sequence recognized by restriction enzymes
  4. cut cDNA and plasmid DNA with RE and ligate cDNA into vectors
  5. transform bacteria
  6. each colony has cDNA peice
120
Q

PCR requires enough

A

knowledge of the gene to have short primers

121
Q

starting material for PCR

A
  • template DNA
  • oligonucleotide primers
  • dNTPs
  • Taq polymerase
122
Q

Why is Taq polymerase used in PCR as opposed to DNA polymerase from another source such as E. coli?

A

Taq polymerase is resistant to heat denaturation.

123
Q

Why is Taq polymerase used in PCR as opposed to DNA polymerase from another source such as E. coli?

A

Taq polymerase is resistant to heat denaturation.

124
Q

PCR cycle steps

A
  1. Denaturation- DNA strands are seperated with high temp
  2. Primer annealing- Primers bind to DNA strands *lower temp *
  3. Primer extension nucleotides are added to the primers, synthesis of complementary strands slightly higher temp
125
Q

after many cycles of PCR

A

there will be fragments that only contain the region of interest and many copies of it

126
Q

_ can be used to amplify small samples

A

PCR

127
Q

reverse transcriptase PCR

A

used to amplify RNA

128
Q

reverse transcriptase PCR steps

A
  1. add reverse transcriptase to RNA along with primer at 3’ end and dNTPs
  2. make cDNA
  3. do normal PCR with the cDNA
129
Q

reverse transcriptase PCR is _ specific

A

highly

130
Q

RT PCR is used to

A

quantitate the amount of specfic gene or mRNA in a sample

131
Q

how does RT PCR work

A
  1. Taqman Probe is complementary to PCR product, has reporter fluorenscent molecule on one side and quencher that absorbs the fluorenscents
  2. during primer annealing, both primer and TaqMan bind to DNA
  3. Taw polymerase digests the TawMan probe which seperates the reporter and quencher
  4. reporter fluorescence is now detected
  5. see fluorescence = strand has been made
132
Q

During real-time PCR, why does the level of fluorescence given off by the TaqMan probe increase over time?

A

Taq polymerase cleaves the TaqMan probe, which separates the quencher and the reporter.

133
Q

phases of fluorescence in RT PCR

A

first linear, then pass the cycle threshold and linear growth then plateau

134
Q

chain termination

A

ddNTPs added to growing DNA strand so the strand can no longer grow

135
Q

dideoxy sequencing

A
  1. make single stranded recombinant vector
  2. mix recombinant vector and primer
  3. add normal dNTPs alot
  4. add fluoresc ddNTPs a little
  5. add DNA pol
  6. strands made with ddNTPs chain terminating
  7. seperate sequences by length
  8. use colors to derieve sequence
136
Q

site directed mutagenesis allows researchers to see how

A

mutations affect:
* expression
* function of a protein
* phenotype of an organism

137
Q

Cas-CRSPR explained

A
  1. sgRNA has the crRNA with the spacer and repeat, linker, tracrRNA
  2. spacer of sgRNA binds to complementary region of target gene
  3. cas9 cleaves the target gene in both strands
  4. nonhomologous end joinin has no donor so creates small deletion in gene
  5. homologous recombination repair uses donor and creates point mutation/point fixing
138
Q

Which component of the CRISP-Cas system is needed if the goal is to create a point mutation (such as changing one codon into a different codon) in a target gene, but it is not needed to create a deletion in a target gene?

A

donor DNA

139
Q

northern blotting is used to identify

A

a specific RNA within a mixture of many RNA molecules

140
Q

northern blotting procedure

A
  1. a mizture containing many different RNAs is extracted from cells
  2. the RNAs are seperated from each other by gel electrophoresis
  3. the RNAs in the gel are then blotted onto a nitrocellulose or nylon filter
  4. the filter is placed into a soliution containing a labeled probe
  5. the RNA that is complementary to the labeled probe is the one detected as a dark band
141
Q

interpreting northern blots

A

how far up the line shows how much expressed/made in cell
* thickness of band shows molecular weights
* three different molecular weights means the mRNA is alternatively spliced

142
Q

western blotting is used to

A

detect a specific protein

143
Q

procedure of western blotting

A

mizture of many different proteins are extracted from cells and seperated by SDS Page
* protein bands within the gel are blotted onto a nitrocellulose or nylon filter
* filter placed in a primary antibody that recognizes protein of interest
* secondary antibody conjugated to alkaline phosphatase recignizes the constant region of the primary antibody
* colorless dye added
* alkaline phospha make dye black
* protein of interst revealed as a dark band