Upload
derek-dalton
View
220
Download
0
Tags:
Embed Size (px)
Citation preview
1
Lecture 9 One Gene One enzyme
2
Genes
---TTGACAT------TATAAT-------AT-/-AGGAGGT-/-ATG CCC CTT TTG TGA---AACTGTA------ATATTA-------TA-/-TCCTCCA-/-TAC GGG GAA AAC ATT
(-10)(-35)
PROMOTER
5’3’
3’ 5’
antisense
sense
RIBOSOMEBINDING
SITE
U-/-AGGAGGU-/-AUG CCC CUU UUG UGA
5’ 3’
Met Pro leu leu stp
Prokaryotic Genes
When ALL OF THESE RULES ARE SATISFIED THEN AND ONLY THEN WILL A PIECE OF DNA GENERATE A PROTEIN.EUKARYOTES ARE EVEN MORE COMPLICATED.
Is there a ribosome binding site upstream of the ATG
Is there a promoter upstream of the ribosome binding site
3
One gene One enzyme hypothesis
In the next few lectures, the following questions will be Addressed:
What is the structure of a gene?
How does a gene function?
How is information stored on the gene?
What is the relationship between genotype and phenotype?
4
The work of biochemists showed that chemical compounds in the cell are synthesized through a series of intermediates-a biochemical pathway
Ornithine Citruline Arginine
Enzyme1 Enzyme2
Glutamic acid-
How do you link genes to enzymes
5
How does a gene generate a phenotype?
The experiments of Beadle and Tatum in the 1940’s provided the first insight into gene function.
They developed the one gene/one enzyme hypothesis
This hypothesis has three tenets:
1 Products are synthesized as a series of steps2 Each step is catalyzed by an unique enzyme3 Each enzyme is specified by a unique gene
The logic:
Precursor Int1 Int2 Product
EnzA EnzB EnzC
GeneA GeneB GeneC
6
Consequences of mutations
Precursor Int1 Int2 Product
EnzA EnzB EnzC
GeneA GeneB GeneC
Lets say we know the biochemical pathway.
With this pathway, what are the consequences of a mutation in geneB?
Would the final product be produced?
Would intermediate2 be produced?
Would intermediate1 be produced?
What happens if we add intermediate1 to the media?
What happens if we add intermediate2 to the media?
7
Neurospora
Beadle and Tatum analyzed biosynthetic mutations in the haploid fungus Neurospora(Red bread mold)
It had the advantage in that it could be grown on a defined growth medium.
Given salts like Na3 citrate,KH2PO4,NH4NO3,MgSO4,CaCl2
and sugars like sucrose
Neurospora can synthesize the amino acids, vitamins etc required and grow to form colonies on agar plates.
8
Prototroph: a strain that
utilizes sugar, salt and water to grow.
Auxotroph: Mutant strain that
needs a specific amino acid or vitamin along with sugar, salt and water to grow.
9
Arginine biosynthetic mutants
Beadle and Tatum set out to identify genes involved in the biosynthetic pathway that led to the production of the amino acid arginine.
Neurospora has approximately 15,000 genes and only 4-5 of these genes are involved in synthesizing arginine.
How do you identify five genes from 15,000?
The POWER OF GENETICS!!!!!!
Typically the organism is exposed to a strong mutagen.
This randomly mutagenizes genes.
Then you look for a mutant in the pathway of interest
10
Logic of experiment
ARGININE BIOSYNTHESIS PATHWAY
Irradiate (mutagenize) spores.
Grow on medium containing arginine
Transfer to medium lacking arginine
DO THEY GROW OR NOT?
If the cells cannot grow on medium lacking arg, then they must have a mutation in a gene required for making ARGININE
Mutant needed arginine to grow.
Conclusion: Enzyme for making arginine was missing
11
The method
To identify mutants
Transfer mutants to minimal media (water, sugar, salts)
Strain1 and 7 can grow on complete media but not minimal media.They have a mutation in a gene required for growth on minimal media!!! Vitamins and Amino acids are missing- CONCLUSION?
Complete
All mutants grow
minimal
1 2 3 4 5 6 7 8 9 10
Irradiate spores.
Take mutant spores. Plate individual spores on complete media(sugar, salts, water, AND vitamins AND all 20 amino acids).
12
Analogy
In the class:
There are two kinds of students:
Students who can climb treesStudents that cannot climb trees.
Under normal growth conditions when supermarkets are open, both kinds of students live happily
When supermarkets are closed
Students who can climb trees grow happily because they can climb trees and eat fruit
Students who cannot climb trees do not grow. They cannot climb trees, and go hungry.
13
Conclusion- strain1
Strain1 and 7 are defective in either amino acid production or Vitamin production
Complete media(salt+sugar+
Vitamin + amino acids)
Minimal media(salt+sugar)
+ 20 amino acids
Minimal media(salt+sugar)+ vitamins
Minimal media(salt+sugar)
Conclusion: strain1 is defective in the production ofVitamins and the mutant is rescued by adding backvitamins
Take Strain 1
14
Conclusion- strain7
Strain1 and 7 are defective in either amino acid production or Vitamin production
Complete media(salt+sugar+
Vitamin + amino acids)
Minimal media(salt+sugar)
+ 20 amino acids
Minimal media(salt+sugar)+ vitamins
Minimal media(salt+sugar)
Conclusion: strain7 is defective in the production ofAmino acids and the mutant is rescued by adding backamino acids
Which of the 20 amino acids does strain7 fail to produce
complete media(salt+sugar)
Vitamin + amino acids
Take strain 7
15
Which amino acid
Minimal media + vitamin + all 20 amino acidGrowthMinimal media + vitamin + lysineNo growthMinimal media + vitamin + glutamineNo growthMinimal media + vitamin + arginineGrowthMutant7 is in a gene required for the production of Arginine.
Beadle and Tatum found that three mutants could not produce arginine
Arg1 Arg2 Arg3
Precursor -----> ornithine -----> citrulline -----> arginine
enz1 enz2 enz3
The biochemical pathway for arginine synthesis was kind of known. Ornithine and citrulline are closely related to arginine and were thought to be precursorsThe pathway for arginine biosynthesis is :
16
Add back
Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not
Ornithine CitrullineArginine
Mutant1
Mutant2
Mutant3
Precursor -----> ornithine -----> citrulline -----> arginine
enz1 enz2 enz3
There are three different enzymes required for arginine synthesis
Enz1, enz2 and enz3
Beadle and Tatum isolated three different mutations in genes (three genes)
Arg1 Arg2 Arg3
?????Which mutant gene codes for which enzyme????
17
Add back
Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not
Ornithine CitrullineArginine
Mutant1 + ++
Mutant2 - ++
Mutant3 - -+
Precursor -----> ornithine -----> citrulline -----> arginine
enz1 enz2 enz3
Arg1 Arg2 Arg3
18
Mutant in Arg1- only precursor made
Add ornithine or citrulline to media, downstream enzymes are functional and pathway continues---> arginine synthesized
Mutant in Arg2- You need to supplement media with citrulline for the pathway to continue. Adding the precursor or ornithine does not help.
Mutant in Arg3-You need to supplement media with arginine. Adding the precursor, ornithine or citrulline does not help.
These experiments demonstrated that a single gene (mutation) coded for a single enzyme.
In addition, the combination of appropriate mutations and intermediates enabled Beadle and Tatum to define the biochemical pathway leading to Arginine synthesis.
The Results Also show that THREE different Genes/enzymes are necessary for ONE phenotype- synthesis of ARG!
This would affect phenotype ratios in a cross
19
Analogy
UCSC
Walnut Laurel Bay
20
Another example
I get three mutants for a particular pathway
I add back various intermediates in this pathway and determine the results
CompoundE B N A
Mut1 - - + +
Mut2 - - + -
Mut3 + - + +
What is the order of the compounds and mutations in the pathway?
21
CompoundE B N A
Mut3 + - + +
Mut1 - - + +
Mut2 - - + -
B----> E----> A----> N
mut3 mut1 mut2
CompoundB E A N
Mut3 - + + +
Mut1 - - + +
Mut2 - - - +
CompoundE B N A
Mut1 - - + +
Mut2 - - + -
Mut3 + - + +
Rearrange the mutants
Rearrange the compounds
Another example
22
The steps in a biochemical pathway identified by this procedure are dependent on the available intermediates and mutations.
This procedure does not identify every step in the pathway
This process does not identify every step in the pathway!
B----> E----> A----> N
B----> E----> S-----> A----> N
This process might also identify multiple mutants for the steps in the pathway!
B----> E----> A----> N
Mut3mut5
Mut1mut4
Mut2
23
This rationale currently is being used in many laboratories to elucidate more complex pathways in multicellular organisms
キ Development- formation of the body axis
キ Behavior- courtship and mating
キ Biological clocks
キ Aging
キ Cell cycle and Cancer
Review
キ Biochemical processes occur as a series of discrete stepwise reactions
キ Each reaction is catalyzed by a single enzyme
キ Each enzyme is specified by a unique gene
Solving biochemical pathways:
The more mutations that a compound rescues, the later in the pathway the compound is located
Conversely, the later a mutation is in a pathway, the fewer compounds will rescue it:
xxxxxxxxxx
24
25
Temperature-sensitive mutations
The one gene/one enzyme concept explains a number of genetic phenomena
Temperature-sensitive mutations
Some mutations exhibit a phenotype at one temperatures (the restrictive temperature) but function normally at another temperature (permissive temperature).
Reasons: Slight destabilization/alteration of the 3D conformation of the enzyme or its ability to interact with other proteins
Low temp- structure of enzyme- normal- activity normal
High temp- structure of enzyme-altered- No activity
These kinds of conditional mutants allow you to turn on and off a function of a protein.
Temperature sensitive mutants
26
Cold sensitive
Protein is functional at high temp and inactive at low tempActive at 30C but inactive at 15C
Heat sensitive
Protein is functional at low temperature but inactive at high temperatureActive at 23C but inactive at 32C
K253E
Cs
Interacts very stably with RFC
C752T
Ts
Mis folding
PCNA
27
Dogs and cats that are white with black feet or vice versa
The genes for coat color are normal at one temperatures but are inactive at another temperatures
One of the genes for coat color is Albino - in cats
This gene affects melanin production. The normal or dominant form, C, is 'full color'. Various mutant alleles. These mutants are temperature sensitive -
Ts mutant and Cat coat color
28
In order of decreasing dominance we have C, Cb, Cs and c.
C is wild-type or full color. It is dominant to all other alleles.
Cb- 'Burmese' factor- it causes a slight lightening of color and is slightly temperature sensitive.
Cs- 'Siamese' factor; it has a much greater lightening effect and is very temperature sensitive.
c is the most recessive form, also known as albino. In the homozygote cc this causes complete absence of any pigment and white fur.
Cb is incompletely dominant over Cs; the heterozygote (Cb/Cs) gives a phenotype intermediate between Burmese and Siamese, known as Tonkinese.
xxxxxxxxx
29
30
Biosynthetic pathways at the grocery store
Most of the red and blue colors found in higher plants are a result of pigments synthesized from one of two metabolic pathways, the carotenoid or the anthocyanin pathway.
The biosynthetic pathway for corn kernel color is as follows:
Precursor-----> Chalcone ----> Flavanone ---->Anthocyanins
(white) (yellow) (white) (blue)
Grocery store corn is usually yellow. Which step in the pathway must be mutated to produce yellow corn?
Beadle/Tatum Results Also show that THREE different Genes/enzymes are necessary for ONE phenotype- synthesis of ARG!
Similarly for blue corn multiple genes/enzymes are required.
Let’s see how this would affect phenotype ratios in a cross
31
Precursor -----> ornithine -----> citrulline -----> arginine
enz1 enz2 enz3
Arg1 Arg2 Arg3
Arg1 mutant -- phenotype is unable to make arginineArg2 mutant-- phenotype is unable to make arginineArg3 mutant-- phenotype is unable to make arginine
What would happen if you crossed different Arg mutants with one another such as arg1 with arg2?
32
Multiple Genes affect single phenotype - 9:7
Precursor----> intermediate----> productwhite white blue
EnzA EnzB
AB Ab aB ab
AB
Ab
aB
ab
AABB AABb AaBB AaBb
AAbB AabB
aABB
aAbB
aABb
9 A-B- blue3A-bb white3aaB- white1aabb white
AAbb Aabb
aaBB aaBb
aAbb aabBaabb
AaBb x AaBb
Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.
33
Mutants and Genetic pathways
Altered PHENOTYPE RATIOS!
The one gene/one enzyme helps explain altered phenotype ratios observed in a standard dihybrid cross: (2 genes segregating independently)
If the Two genes being analyzed affect the same genetic pathway
Precursor----> intermediate----> productyellow white blue
EnzA EnzB
Parental cross white x yellow
34
Multiple genes affecting a single phenotype
AB Ab aB ab
AB
Ab
aB
ab
AABB AABb AaBB AaBb
AAbB AabB
aABB
aAbB
aABb
9 A-B- blue3A-bb white3aaB- yellow1aabb yellow
Precursor----> intermediate----> productyellow white blueEnzA EnzB
F2
AAbb Aabb
aaBB aaBb
aAbb aabBaabb
4:3:9Y:W:B
Parental cross: AAbb x aaBB
white yellow
F1 AaBb (blue) x AaBb (blue)
35
Labradors recessive Epistasis give 9:4:3 ratio
Parental Cross: black x yellow
BBEE bbee
BbEe (black)
x
BbEe (black)
Given the pathway show above, what phenotypic ratios would be
produced in progeny from the dihybrid cross: BbEe x BbEe
Yellow-------> brown--------> blackE B
EB Eb eB eb
EB
Eb
eB
eb
EEBB EEBb EeBB EeBb
EEBb EeBb
EeBB
EeBb
EeBb
EEbb Eebb
Eebb eebb
eeBB eeBb
eeBb
4:3:9Y:Br:Bl
Recessive epistasisHomozygous ee gene alleles mask effect of B gene allelese is epistatic to BE works upstream of B
Epistasis= When the Allele of One Gene Mask the Expression of Allele of Second Gene
Genetic pathway
36
Enz V+
Precursor -----Brown pigment \(white) \ transporter W+
--------- Red
/Precursor ----- Vermilion pigment /(white) Enz B+
WT -- Brown
WT -- Vermilion
WT -- White
37
Summer squash color 12:3:1 (Dominant Epistasis)
GeneW determines pigment production (repressor of GeneY)W No pigmentw Pigment (color? Depends on Y gene)
GeneY determines color of pigmentY yellowy green
Cross a heterozygous white squash to itself
WwYy x WwYy
9 W-Y- white3 W-yy white3 wwY- yellow1 wwyy green
Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.
Dominant allele of one gene hides effects of both alleles of second gene= 12:3:1 ratioDominant allele of one gene hides effect of dominant allele of second gene = 13:3 ratio
Dominant Epistasis 12:3:1
38
In Leghorn chickens
Colored feathers are due to a dominant gene, C;
White feathers are due to its recessive allele, c.
CC= color
Cc= color
cc= white
Another dominant gene, I, inhibits expression of color while ii allows expression of color In birds with genotypes CC or Cc or cc if there is II or Ii the birds are white!
Therefore both CCII, CcII, CCIi, CcIi and cc– are ALL white.
Only birds that are colored are C-ii
Dominant Epistasis 13:3
In white leghorn and white wyandotte chickens, a dominant B allele masks color production associated with the dominant A allele of a second gene.
Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.
Dominant allele of one gene hides effects of both alleles of second gene= 12:3:1 ratioDominant allele of one gene hides effect of dominant allele of second gene = 13:3 ratio
B/b is the epistatic gene. Any chicken with a dominant B in their genome will have white feathers. Being homozygous recessive bb at this locus enables the expression of genes coded for at the hypostatic locus (A).At the hypostatic locus A/a the dominant allele A codes for colored feathers while the recessive a codes for no color. Hence, a chicken that is homozygous recessive aa will also be white giving you a 13:3 ratio
Dominant Epistasis give 13:3 ratio
40
41
The 9:3:3:1 ratio in the F2 suggests two genes control coat color
You have to know the rules for each color
Gene Interaction: Range of Phenotypes From Combined Action of Alleles of Two Genes
Multiple genes regulate a single phenotype
42
Pepper Color
Gene 1: R=red r=yellow
Gene 2: Y=absence of chlorophyll (no green)
y=presence of chlorophyll (green)
Possible genotypes:
R-/Y- : red (red/white no chlorophyll) R-/yy : browny orange (red/green chlorophyll) rr/Y- : yellow (yellow/white no chlorophyll) rr/yy : green (yellow/green chlorophyll)
Two genes affect Chicken Combs
43
4 different chicken comb phenotypes result:
Rose Combs (R-pp) Walnut Combs (R-P-) Pea Combs (rrP-) Single Combs (rrpp)
Multiple gene inheritance
a. so far been discussing traits that are governed only by one gene
b. far from the truth. Many human traits, such as height, skin color etc are determined by multiple genes. Multigenic ("many gene") traits exhibit a mode of inheritance that would have surprised Gregor Mendel.
c. most phenotypes that we are aware of are distributed in a bell-shaped curve like human height
d. often multiple genes affect such traits
e. height in plants might be affected by three genes each possessing two alleles
f. the dominant allele of each gene might add 1 cm to basic height of plant
g. the recessive allele of each gene would not affect the basic 10 cm height
h. aabbcc X AABBCC
i. F1 generation selfs itself
j. 1/64 6/64 15/64 20/64 15/64 6/64 1/64
k. the more genes affecting a trait, the smoother is the bell curve
l. the environment also affects phenotype smoothing off the curve even more
45
The height of plants is controlled by 4 genes. Alleles A, B, and C contribute 3 cm to the plant's height. Alleles that are recessive do not contribute to the height.
In addition Gene L is always found in a dominant
condition and always contributes 40 cm to the height.
a) What would be the height of a plant with the genotype AABBCCLL? 3+3+3+3+3+3+40
b) What would be the height of a plant with a genotype aabbccLL? 0+0+0+0+0+0+40
c) What would be the height of the offspring produced from a cross between the plants in a) and b)? AaBbCcLL 3+0+3+0+3+0+40
d) What would be the heights of the offspring produced from a cross between AaBbCcLL and AaBbCcLL?
Genes for hair color
46
Hair Color Hair color is controlled by multiple genes on
chromosomes 3, 6, 10, and 18. The more dominant alleles that appear in the
genotype, the darker the hair!
Continuously varying traits are also called quantitative traits.
Additive Gene Interaction Model for Continuous
Variation
Skin Pigmentation
ABC, AbC, aBC, ABc, abC, Abc, aBc, abc
Continuously varying traits are also called quantitative traits.The width/height of each sub-phenotype class indicates the number of variable Genes/alleles
Additive Gene Interaction Model for Continuous Variation
Multi-gene Disease
50
Multigenic diseases result from less severe mutations in more than one gene. Any of these mutations alone might not affect a trait, but together, they can lead to significant phenotypic differences.
Complex interactions between genes.
Number of traits result from mutations in single genes- MONOgenic trait.
Interaction among the phenotypic effects of different genes- epistasis, adds a layer of complexity to the study of genetic disease.
Genes don't function alone; rather, they constantly interact with one another.
These gene-gene interactions result in an output phenotype. Certain genes are known to modify the phenotype of other genes.
This implies that multiple genes may interact to increase or decrease disease susceptibility.
If the effect of the disease-bearing gene is masked or altered by the effects of a second gene (by say altering expression level of the disease bearing gene), then identifying the first gene can be complicated.
In addition, if more than one genetic interaction occurs to cause a disease, then identifying the multiple genes involved and defining their relationships becomes even more difficult.
Alzheimers
51
Alzheimer's disease, is a progressive neurodegenerative disorder that causes memory loss and dementia.A mutation in a gene called apolipoprotein E4 was associated with a higher risk of developing Alzheimer's. While having one or two copies of mutant apolipoprotein E4 increase one's risk of Alzheimer's, not all carriers of apolipoprotein E4 develop the disease. This suggested that gene-gene interactions were involved.They confirmed 27 different genetic interactions in 4 different biochemical pathways: cholesterol metabolism, beta-amyloid production, inflammation, oxidative stress. Some interactions were synergistic, while others were antagonistic. The synergistic interactions indicate that the pair of involved genes together increase the risk of Alzheimer’s. The strongest synergistic interactions involved the pairing of apolipoprotein E4 mutation with mutations in three different genes: alpha(1)-antichymotrypsin, β-secretase, and butyrylcholinesterase K. These genes are not acting alone, but in a pathway that affect one another.
The APP gene produces a transmembrane protein that is modified, cleaved by secretase and then inserted into membranes. Function of APP is not fully known- could be membrane or Ca+ trafficking in neurons. APP interacts with chymotrypisn and Apolipoprotein. Alterations in these interactions lead to formation of APP plaques (amyloid fibers) leading to Alzheimers. ApoE4 mutation leads to alterations in cholesterol in membranes. The ApoE4 allele causes greater aggregation of APP and greater amyloid deposition.
xxxxx
52
53
Biochemical Pathways and Linked Genes
The F1 is testcrossed
The following F2 progeny are produced:
50 yellow
40 blue
10 white
Precursor----> intermediate----> productyellow white blue
EnzC EnzD
GeneC GeneD
Parental C-D x c-dC-D c-d
F1
54
Biochemical Pathways and Linked Genes
The following F2 progeny are produced:
50 yellow 40 blue 10 white
Precursor----> intermediate----> productyellow white blue
EnzC EnzD
GeneC GeneD
Parental C-D x c-dC-D c-d
F1 C-D c-dc-d x c-d
Parental C-D blue 40c-d
c-d yellow 40c-d
Recomb C-d white 10c-d
c-D yellow 10c-d
What is the map distance between these two genes?
Map Distance+#Recombinants/Total Progeny x 100%
2(10)/100= 20 Map Units
55
One gene: one polypeptide
The concept of 1 gene/enzyme was modified to the concept of: 1 gene/ 1 protein
Almost all enzymes are proteins but not all proteins are enzymes. Many proteins provide structural rather than enzymatic roles.
For example polymers of the protein actin provide structural integrity to the eukaryotic cell.
Perhaps the most notable example of this comes from studies of Hemoglobin.
Hemoglobin is an iron carrying protein found in the red blood cells and is responsible for transporting oxygen from the lungs to the cells of the body.
56
Hb
Hemoglobin consists of four polypeptides (proteins) each associated with a specific Heme group (Heme is a small iron containing molecule to which oxygen can attach) Adults contain 2 alpha polypeptides and 2 beta polypeptides
Alpha polypeptide = 141 amino acids
Beta polypeptide= 146 amino acids
Over 300 known hemoglobin variants are known and each is the result of a specific mutation
Most of these are the result of a single amino acid substitution
キ Hb A:
キ Hb S:
キ Hb C:
These results demonstrate that:
1. Genes specify proteins that are not enzymes
2. Mutations can disrupt a single amino acid out of the many that make up the protein.
57
Hb
Hemoglobin consists of four polypeptides (proteins) each associated with a specific Heme group (Heme is a small iron containing molecule to which oxygen can attach) Adults contain 2 alpha polypeptides and 2 beta polypeptides
Alpha polypeptide = 141 amino acids
Beta polypeptide= 146 amino acids
Over 300 known hemoglobin variants are known and each is the result of a specific mutation
Most of these are the result of a single amino acid substitution
キ Hb A: val his leu thr pro *glu *glu
キ Hb S: val his leu thr pro *val* glu
キ Hb C: val his leu thr pro *lys* glu
These results demonstrate that:
1. Genes specify proteins that are not enzymes.
2. Mutations can disrupt a single amino acid out of the many that make up the protein.
58
Muscular Dystrophy
Through these techniques, it was found that DMD patients have a mutation in a single gene.
The normal function of the gene is to enable muscle fibers to make a protein called dystrophin.
Dystrophin localizes to the plasma membrane in muscle cells.The normal dystrophin protein stabilizes the muscles during muscle contractions.
Muscle fibers in people affected with DMD are extremely deficient in dystrophin.
Without this protein, the plasma membrane ruptures during muscle contraction and degeneration of the muscle tissue occurs.
59
Alkaptonuria
Degenerative disease. Darkening of connective tissue, arthritis
Darkening of urine
1902 Garrod characterized the disorder- using Mendels rules- Autosomal recessive. Affected individuals had normal parents and normal offspring.
1909 Garrod termed the defect- inborn error (genetic) of metabolism. Homogentisic acid is secreted in urine of these patients. This is an aromatic compound and so Garrod suggested that it was an intermediate that was accumulating in mutant individuals and was caused by lack of enzyme that splits aromatic rings of amino acids.
Garrods results and his explanation were ignored
1958 La Du showed that accumulation of homogentistic acid is due to absence of enzyme in liver extracts
1994 Seidman mapped gene to chromosome 3 in human
1996 Gene cloned and mutant identified P230S &V300G
2000 Enzyme principally expressed in liver and kidneys
60
Pathways
Biologists and clinicians want to address the question of how altering a particular set of base pairs that make up the 3 billion base pairs in the human genome led to this phenotype.
Formal genetics provides little information about these intermediary steps.
Since the 1970’s, a set of techniques have been developed that enable us to elucidate each step in a pathway
These techniques are generally placed under the rubric of Molecular Biology or Molecular Genetics
xxxxxxxx
61