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Lecture 9
One Gene One enzyme
1
Genes DNA sequences that code for RNA- prm, RBS,
ORF, Term
Is there a ribosome binding site upstream of the ATG
Is there a promoter upstream of the ribosome binding site
Prokaryotic Genes
PROMOTER
3’
5’
antisense
---TTGACAT------TATAAT-------AT-/-AGGAGGT-/-ATG CCC CTT TTG TGA
---AACTGTA------ATATTA-------TA-/-TCCTCCA-/-TAC GGG GAA AAC ATT
sense
3’
(-35)
(-10)
RIBOSOME
BINDING
SITE
5’
5’
3’
U-/-AGGAGGU-/-AUG CCC CUU UUG UGA
Met Pro leu leu stp
When ALL OF THESE RULES ARE SATISFIED THEN AND ONLY
THEN WILL A PIECE OF DNA GENERATE A PROTEIN.
2
EUKARYOTES ARE EVEN MORE COMPLICATED.
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?
3
Huntington's Disease
Huntington's disease (HD) results from degeneration of
neurons, in certain areas of the brain. This degeneration
causes uncontrolled movements, loss of intellectual faculties,
emotional disturbance and early death
This disease is caused by a single dominant mutation on the
forth chromosome.
Each child of an HD parent has a 50-50 chance of inheriting
the mutation.
A person who inherits the mutation will sooner or later develop
the disease!
What is the normal function of the Huntington gene?
What happens in the mutant?
Can it be blocked?
To understand this disease we need an interdisciplinary
approach.
DNA
Geneticist
Bioinformatics
Protein complex
Biochemist
RNA
Molecular biologist
Cellular phenotype
Cell biologists
Protein
Biochemist
Organism phenotype
Physicians
4
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
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
2
3
Products are synthesized as a series of steps
Each step is catalyzed by an unique enzyme
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
Irradiate spores.
Take mutant spores. Plate individual
spores on complete media
(sugar, salts and water, AND
vitamins AND all 20 amino acids).
Complete
1
2
3
4
5
6
7
8
9 10
All
mutants
grow
To identify mutants
Transfer mutants to minimal media
(water, sugar, salts)
minimal
Strain1 and 7 can grow on complete media but not minimal 12
media.
They have a mutation in a gene required for growth on minimal
media!!!
Analogy
In the class:
There are two kinds of students:
Students who can climb trees
Students 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.
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Conclusion- strain1
Strain1 and 7 are defective in either amino acid
production or Vitamin production
Take Strain 1
Complete media
(salt+sugar+
Vitamin + amino acids)
Minimal media
(salt+sugar)
Minimal media
(salt+sugar)
+ 20 amino acids
Minimal media
(salt+sugar)
+ vitamins
Conclusion: strain1 is defective in the production of
Vitamins and the mutant is rescued by adding back
vitamins
14
Conclusion- strain7
Strain1 and 7 are defective in either amino acid
production or Vitamin production
Take strain 7
Complete media
(salt+sugar+
Vitamin + amino acids)
Minimal media Minimal media Minimal media
complete media
(salt+sugar)
(salt+sugar)
(salt+sugar)
(salt+sugar)
+ 20 amino acids + vitamins
Vitamin + amino acids
Conclusion:
15
Conclusion- strain7
Strain1 and 7 are defective in either amino acid
production or Vitamin production
Complete media
(salt+sugar+
Vitamin + amino acids)
Minimal media Minimal media Minimal media
complete media
(salt+sugar)
(salt+sugar)
(salt+sugar)
(salt+sugar)
+ 20 amino acids + vitamins
Vitamin + amino acids
Conclusion: strain7 is defective in the production of
Amino acids and the mutant is rescued by adding back
amino acids
Which of the 20 amino acids does strain7 fail to produce
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Which amino acid
Minimal
Minimal
Minimal
-
Minimal
media + vitamin + all 20 amino acid
media + vitamin + lysine
media + vitamin + glutamine
Growth
No growth
No growth
media + vitamin + arginine
Growth
Mutant7 is in a gene required for the production of Arginine.
Beadle and Tatum found that three mutants could not produce
arginine
Arg1
Arg2
Arg3
The pathway for arginine biosynthesis is :
Precursor -----> ornithine -----> citrulline -----> arginine
enz1
enz2
enz3
17
Beadle and Tatum found that three mutants could not produce
arginine
Arg1
Arg2
Arg3
The biochemical pathway for arginine synthesis was kind of
known. Ornithine and citrulline are closely related to arginine
and were thought to be precursors
Precursor -----> ornithine -----> citrulline -----> arginine
enz1
enz2
enz3
18
Add back
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????
Instead of arginine, if they added ornithine or citrulline to the
media, some mutants were rescued and others were not
Arginine
Ornithine
Citrulline
Mutant1
Mutant2
Mutant3
19
Add back
Precursor -----> ornithine -----> citrulline -----> arginine
enz1
enz2
enz3
Arg1
Arg2
Arg3
Instead of arginine, if they added ornithine or citrulline to the
media, some mutants were rescued and others were not
Mutant1
Arginine
Ornithine
Citrulline
+
+
+
Mutant2
-
+
+
Mutant3
-
-
+
20
Mutant in Arg1- only precursor made
Add ornithine or citrulline to media, downstream enzymes are
functional and pathway continues---> arginine synthesized
Mutant in Arg2You need to supplement media with citrulline for the pathway
to continue. Adding the precursor or ornithine does not help.
Mutant in Arg3You 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!
Let’s see how this would affect phenotype ratios in a cross
21
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.
22
Heat sensitive mutants
Cold sensitive
Protein is functional at high temp and inactive at low temp
Active at 30C but inactive at 15C
Temperature sensitive
Protein is functional at low temperature but inactive at high
temperature
Active at 23C but inactive at 32C
PCNA
Interacts very
stably with RFC
Mis folding
Cs
Ts
K253E
C752T
23
An example of a Ts mutation:
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 -
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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 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.
25
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----->
(white)
Chalcone ---->
Flavanone ---->
Anthocyanins
(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
26
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---->
yellow
intermediate---->
white
EnzA
Parental cross
product
blue
EnzB
white
x
yellow
27
Multiple genes affecting a single phenotype
Precursor---->
yellow
EnzA
intermediate---->
white
EnzB
ｷ
A, B = normal alleles
ｷ
a, b= nonfunctional mutant alleles
Parental cross:
AAbb
x
white
F1
product
blue
aaBB
yellow
AaBb (blue)
x
AaBb (blue)
F2
28
Multiple genes affecting a single phenotype
Precursor---->
yellow
intermediate---->
white
EnzA
product
blue
EnzB
F2
AB
Ab
aB
ab
AB
Ab
aB
ab
9 A-B3A-bb
3aaB1aabb
blue
white
yellow
yellow
29
Multiple genes affecting a single phenotype
Precursor---->
yellow
intermediate---->
white
EnzA
product
blue
EnzB
F2
AB
Ab
aB
ab
AaBB
AaBb
AabB
Aabb
AB
AABB
AABb
Ab
AAbB
AAbb
aB
ab
aABB
aAbB
aABb
aAbb
aaBB
aabB
4:3:9
Y:W:B
aaBb
aabb
9 A-B3A-bb
3aaB1aabb
blue
white
yellow
yellow
30
Labradors
Parental Cross:
black
x
yellow
BBEE
bbee
BbEe (black)
x
BbEe (black)
Yellow------->
E
brown--------> black
B
Given the pathway show above, what phenotypic ratios would be
produced in progeny from the dihybrid cross: BbEe x BbEe
EB
Eb
eB
eb
EB
9:3:4
Eb
eB
eb
31
Labradors recessive Epistasis give 9:4:3 ratio
Parental Cross:
black
x
yellow
BBEE
bbee
BbEe (black)
x
BbEe (black)
Yellow------->
E
brown--------> black
B
Given the pathway show above, what phenotypic ratios would be
produced in progeny from the dihybrid cross:
BbEe x BbEe
EB
EB
Eb
eB
eb
EEBB
EEBb
EeBB
EeBb
Eb
eB
eb
EEBb
EeBB
EeBb
EEbb
EeBb
Eebb
EeBb
eeBB
eeBb
4:3:9
Y:Br:Bl
Eebb
eeBb
Recessive epistasis
Homozygous ee gene
alleles mask effect of
B gene alleles
e is epistatic to B
eebb
Epistasis= When the Alleles of One Gene Mask the Expression of Alleles of a Second Gene
32
Gene interactions give 9:7
Precursor---->
white
intermediate---->
white
EnzA
AB
product
blue
EnzB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AAbB
AAbb
AabB
Aabb
aB
ab
aABB
aAbB
aABb
aAbb
aaBB
aabB
aaBb
aabb
9 A-B3A-bb
3aaB1aabb
blue
white
white
white
33
WT -- Brown
WT -- Vermilion
WT -- White
Enz V+
Precursor -----Brown pigment
(white)
\
\
 Red
Precursor ----- Vermilion pigment /
(white)
Enz B+
transporter W+
---------
/
34
Gene Interaction: A range of Phenotypes Arise From Combined
Action of Alleles of Two Genes
The 9:3:3:1 ratio in the F2 suggests two genes control
coat color.
35
Multiple genes regulate a single phenotype
 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 : brown/orange (red/green chlorophyll)
rr/Y- : yellow (yellow/white no chlorophyll)
rr/yy : green (yellow/green chlorophyll)
36
Two genes affect Chicken Combs
 4 different chicken comb phenotypes result:




Rose Combs (R-pp)
Walnut Combs (R-P-)
Pea Combs (rrP-)
Single Combs (rrpp)
37
Multiple genes for hair color
 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!
38
Multiple genes affect a single phenotype- additive
effects
39
Height
Additive Gene Interaction for Continuous Variation
Continuously varying traits are also called quantitative traits.
Additive Gene Interaction Model for Continuous Variation
Continuously varying traits are also called quantitative traits.
 The height of plants is controlled by 4 pairs of alleles. 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 homozygous 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?
42
Biochemical Pathways and Linked Genes
Precursor---->
yellow
intermediate---->
white
EnzC
GeneC
Parental C-D
C-D
product
blue
EnzD
GeneD
x
c-d
c-d
F1
The F1 is testcrossed
The following F2 progeny are produced:
50 yellow
40 blue
10 white
43
Biochemical Pathways and Linked Genes
Precursor---->
yellow
intermediate---->
white
EnzC
product
blue
EnzD
GeneC
GeneD
Parental C-D
C-D
F1
x
C-D
c-d
x
c-d
c-d
c-d
c-d
The following F2 progeny are produced:
50 yellow 40 blue 10 white
Parental
Recomb
C-D
c-d
blue
40
c-d
c-d
yellow
40
C-d
c-d
white
10
c-D
c-d
yellow
10
What is the map distance between these two genes?
Map Distance+#Recombinants/Total Progeny x 100%
2(10)/100= 20 Map Units
44
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.
45
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.
46
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.
47
48
Another example
I get three mutants for a particular pathway
I add back various intermediates in this pathway and determine
the results
Mut1
Compound
E
B
-
N
+
A
+
Mut2
-
-
+
-
Mut3
+
-
+
+
What is the order of the compounds and mutations in the
pathway?
49
Another example
Mut1
Compound
E
B
-
N
+
A
+
Mut2
-
-
+
-
Mut3
+
-
+
+
Rearrange the mutants
Mut3
Compound
E
B
+
-
N
+
A
+
Mut1
-
-
+
+
Mut2
-
-
+
-
Rearrange the compounds
Mut3
Compound
B
E
+
A
+
N
+
Mut1
-
-
+
+
Mut2
-
-
-
+
B----> E----> A----> N
mut3
mut1
mut2
50
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
51