Download Chapter 14 – From Gene to Phenoytpe

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Chapter 14 – From Gene to Phenoytpe
Questions to be addressed:
•
In typical crosses, what ratios are expected for genes with alleles
that do not show complete dominance/recessive relationships?
•
What ratios are expected from crosses if more than one gene
controls a particular characteristic?
•
What can these ratios tell us about the function of the genes?
Terminology (see also Glossary pages 607-634 and web site
http://helios.bto.ed.ac.uk/bto/glossary/)
pleiotrophic mutation: a mutation that affects several different phenotypic
characteristics
codominance: alleles which, when combined in the heterozygote show
aspects of both homozygotes
incomplete dominance: alleles which, when combined in the heterozygote
show a phenotype intermediate between the homozygotes
epistasis: the phenotype of a mutant allele at one gene overrides the
phenotype of a mutant allele at another gene, such that the double mutant
has the same phenotype as the first mutant
suppression: a mutation in one gene can cancel the effect of a mutation in
a second gene, resulting in a wild type phenotype
penetrance: the proportion of individuals with a specific genotype who show
that genotype phenotypically
expressivity: the degree to which a particular genotype is expressed in the
phenotype
Mendel’s findings:
•
one gene, two alleles
•
each gene controls a single different character
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•
one allele is dominant to the other
ratios =
=
3 : 1 = 1 : 2 : 1 (monohybrid)
heterozygotes
9 : 3 : 3 : 1 (dihybrid)
test cross of
1 : 1 (monohybrid)
heterozygotes
1 : 1 : 1 : 1 (dihybrid)
Complexity arises:
•
a gene may have multiple alleles: each gene consists of many
nucleotides, which when mutated result in different alleles
•
•
different alleles may have different phenotypes
a gene may be pleiotrophic: a single gene acts to control more than one
phenotypic characteristic
•
e.g. gene controlling pollen grain germination also controls root hair
growth
•
suggests a common underlying physiological or developmental
mechanism
•
more than one gene may control a single characterisic
•
since many biological pathways consist of a series of steps, each
controlled by a different protein (coded by a gene) , a single
characteristic may be the result of multiple gene activity
1. Each gene can have multiple alleles
•
How do you know if 2 mutant lines with the same phenotype are the
result of:
a) mutations in 2 genes in the same pathway
b) 2 alleles of the same gene
•
example: consider the trp biosynthetic pathway
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trpA+
X
trpB+
Y
trpC+
tryptophan
experiment: two new auxotrophic mutants for tryptophan are found in the
plant Arabidopsis thaliana
hypothesis 1: the two mutants are alleles of same gene, A (e.g. mutant 1 =
a1, mutant 2 = a2)
hypothesis 2: the two mutants are alleles of different genes A and B (e.g.
mutant 1 = a1, mutant 2 = b1)
experiment: cross two mutants together and the results will differ
depending on which hypothesis is correct
expected results for hypothesis 1
conclusion: F1 progeny is mutant, therefore mutations do not complement
one another, and the mutations are alleles of the same gene
expected results for hypothesis 2
conclusion: F1 progeny is wild type, therefore two mutants complement one
another, and must be mutations in different genes
RULES FOR A COMPLEMENTATION TEST
• can only be done with recessive mutations
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• If the mutations are in different genes, the two mutations will complement
one another (progeny will be wild type)
• If the mutations are alleles of the same gene, the two mutations will not
complement one another (progeny will be mutant)
VARIATIONS ON THE COMPLEMENTATION TEST – THE HETEROKARYON
(figure 6-2)
•
in organisms with a haploid life cycle, cannot do a complementation test
•
instead, fuse two cells
•
the resulting cells have 2 nuclei = heterokaryon in a common cytoplasm
•
gene products from both nuclei act within the common cytoplasm
Dominance relationships amongst alleles
Most frequently
•dominance = presence of enzyme function
A -----> functional enzyme
• recessiveness = absence of enzyme function
a -----> no functional enzyme
in Aa, presence of A (functional enzyme) will override presence of a (nonfunctional enzyme), therefore A will be dominant to a
Aa = A phenotype
Example 1
Phenotype depends on a threshold amount of protein.
eyeless + (ey+ )- required to form eyes in Drosophila
-each ey+ allele makes 50 units protein/cell
therefore, ey+ ey+ - makes 100 units of protein/cell
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Case I
-a cell needs 80 units of Ey functional protein to form an eye
-mutant allele ey-1 makes 0 units (a null allele, no protein is made)
-thus a homozygous mutant (ey-1ey-1) fly has no eye
What is the phenotype of an ey+ ey-1 fly?
Case 2
a cell requires 80 units of Ey protein to form an eye
allele ey-2 makes 35 units of protein (a partial loss of function)
-cells in a homozygous mutant (ey-2 ey-2) fly make 70 units, therefore this fly
has no eye
Example 2
The phenotype varies with the concentration of protein.
allele ELF+ - necessary to delay flowering (ELF = early flowering)
- makes 50 units protein/cell
ELF+ ELF+ - produces 100 units protein cell
(makes flowers in 24 days)
allele elf - makes 0 units protein/cell
elf elf - produces 0 units protein/cell
ELF+ elf - produces 50 units/cell
(flowers in 0 days)
(flowers in 12 days)
-the heterozygote is intermediate between two homozygous individuals
= INCOMPLETE DOMINANCE
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Question: if heterozygotes are crossed, what ratio is expected in the
progeny?
Conclusion: If the heterozygote has a phenotype distinct from either
homozygote, the phenotypic ratio will be the same as the genotypic ratio.
Dihybrid crosses involving altered dominance/recessive relationship
Trait 1
Trait 2
ELF+ ELF+ = flowers in 24 days
ELF+ elf = flowers in 12 days
elf elf = flowers in 0 days
RR = red
Rr = red
rr = white
CROSS = ELF+ elf Rr
X
progeny? (3:1 x 3:1 = 9:3:3:1?)
ELF+ elf Rr
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3/4 R_
3/16 ELF+ELF+R_
1/4 rr
1/16 ELF+ELF+rr
3/4 R_
6/16 ELF+elf R_
1/4 rr
2/16 ELF+elf rr
3/4 R_
3/16 elf elf R_
1/4 rr
1/16 elf elf rr
1/4 ELF+ ELF+
1/2 ELF+ elf
1/4 elf elf
ratio = 3 : 1: 6 : 2 : 3 : 1
(= 1:2:1 x 3:1)
Example 3
What if alleles produce proteins having different and independent functions?
Blood groups
•
blood is categorized into different types based on how blood reacts to a set
of antibodies
•
each antibody interacts with a particular antigen (protein) present on the
cell surface
•
the presence of such an interaction is recognized by blood clumping
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one antigen (protein) is produced by gene I
•
-3 alleles (IA, IB, i) which produce different antigens recognized by different
antibodies
IA reacts with A antibody
I B reacts with B antibody
i - reacts with no antibody (null allele)
genotype
antigen
blood group
IA IA
IB IB
A
B
A
B
ii
none
O
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What if we consider heterozygous combination?
genotype
antigen
blood group
IA I B
IB i
A and B
B
AB
B
IA i
A
A
CODOMINANT - heterozygote has phenotypic characteristics of both
homozygotes
therefore,
IA and IB are codominant to one another
IA is dominant to i
I B is dominant to I
Question?
A baby has blood type O. Two couples claim to be the baby’s parents:
• couple 1 has blood groups A and B
• couple 2 has blood groups AB and O
Which are the parents, couple 1 or couple 2?
conclusion:
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Ambiguity in terminology
e.g. sickle cell anemia, a tropical disease caused by a mutation in the gene
coding for hemoglobin
HbA HbA -wild type, blood cells round, NO anemia
HbS HbS - blood cells sickle shaped, ANEMIC
HbA HbS - blood cells normal, except under low O2, NO anemia
extract hemoglobin and electrophorese
Hemoglobin types
present
S and A
S
A
Genotype
Hb S Hb A
Hb S Hb S
Hb A Hb A
Phenotype
Sickle-cell /
Normal cell
Sickle-cell
Normal shape
Origin
Positions
to which
hemoglobins
have migrated
(see Figure 14-8)
If we consider different phenotypic characteristics, the dominance/recessive
descriptions vary:
phenotypic characteristic = disease
phenotypic characteristic = migration
phenotypic characteristic = cell shape under low O2
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Lethal Alleles
e.g. manx cat X manx cat
(manx = talless due to abnormal spinal development, Figure 14-11)
results: progeny = 2 manx cat and 1 tailed cat
Conclusion: manx cats are not true breeding, moreover, 2:1 does not look
like a standard Mendelian ratio
Simplest hypothesis: this is a Mendelian ratio, but one class is combined
with another or missing (a modified Mendelian ratio)
- what standard ratios might we consider for such a modification?
hypothesis:
wild type (tailed) =
manx (tailless) =
experiment: cross 2 manx cats and look for lethality during embryogenesis
results: 1/4 of embryos have arrested development (lethal)
conclusion: M+ is necessary for spinal cord development
M M -> die early in embryonic development due to spinal cord malformation
M+ M - have sufficient M+ protein to make most of the spinal cord, but the
tail is absent
Part 4: Genes controlling same characteristic
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example 1: Recessive epistasis
-the dark red colour of wild type Drosophila eyes results from a biochemical
pathway with two steps, each controlled by a different gene (v+ and s+)
v+
vermillion
s+
scarlet
dark red
recessive allele v
(vv flies = vermilion)
recessive allele s
(ss flies = scarlet)
CROSS:
vermillion fly X scarlet fly
PARENTAL:
vvs+s+ X v+v+ss
v+vs+s
(wildtype)
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v+_ s+_
3
v+ _ ss
3
vv s+_
1
vvss
F1:
F2:
(wild type)
(scarlet)
(vermillion)
(?)
recall the pathway
v+
vermillion
s+
scarlet
dark red
What colour will flies nonfunctional at both v+ and s+ be?
ratio =
•
we say that phenotype v (vermilion) is epistatic to phenotype s (scarlet)
(epistatic = stands over)
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NOTE: if one gene is epistatic to another, it usually means that the
epistatic gene acts first (is upstream) in the biochemical (or
developmental) pathway, and that the two genes act independently (on
different steps within the same pathway
example 2: Complementary gene action
•
mutations in either gene gives the same phenotype
trpA+
X
trpB+
Y
trpC+
tryptophan
trpC+ mutant allele c - auxotrophic for trp
trpB+ mutant allele b - auxotrophic for trp
ratio =
Example 3: Suppression
•
a mutation in one gene compensates for a mutation in a second gene
•
good evidence that two gene products interact
A_B_ =
aaB_ =
aabb =
A_bb =
b suppresses (compensates for) a
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Example:
protein A
+
protein B
functional protein (dimer)
-a protein containing 2 different subunits = heterodimer
-a protein containing 2 identical subunits = homodimer
clavata mutants = clv1 and clv3
mutation in CLV1 (clv1) causes the division of cells in apical meristem to become
uncontrolled
----> enlarged and club-shaped apex
mutation in CLV3 (clv3) also causes the division of cells in apical meristem to
become uncontrolled
-----> enlarged and club-shaped apex
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WILD TYPE
response
membrane
cytoplasm
CLV3
protein
CLV1
protein
response
signal
MUTANT
responses
membrane
membrane
cytoplasm
cytoplasm
CLV3
protein
clv1
protein
NO
response
NO
response
signal
WILD TYPE
response
membrane
cytoplasm
clv3
protein
clv1
protein
response
signal
clv3
protein
CLV1
protein
NO
signal
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phenotypic ratio =
What if there was an allele of CLV3 (clv3*) which could bind with both wild
type CLV1 and mutant clv1?
CLV1CLV1CLV3CLV3 =
CLV1CLV1clv3*clv3* =
clv1clv1 CLV3CLV3 =
clv1clv1clv3*clv3* =
phenotypic ratio =
Example 4: Genetic Redundancy (duplicate genes)
(two genes doing the same thing – loss of one gene can be compensated for
by presence of other gene)
chemotaxis in Caenorhabditis elegans (a nematode)
WAN1 and WAN2 - both expressed in sensory neurons
wan1 - mutation causing lack of WAN1 protein
wan2 - mutation causing lack of WAN2 protein
WAN1WAN1 WAN2WAN2 -moves to chemical (wild type response)
wan1wan1 WAN2WAN2 - moves to chemical
WAN1WAN1 wan2wan2 - moves to chemical
wan1wan1wan2wan2 - wanders
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-wild type copy of one WAN gene can compensate for loss of the other WAN
gene
-only see a phenotype if both functions are lost
RATIO =
MODIFIED DIHYBRID RATIOS RESULTING FROM GENE INTERACTIONS
Type of gene interaction
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A_B_
3
A_bb
1
aabb
Phenotypic
ratio
3
1
9:3:3:1
None (four discrete phenotypes)
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Complementary gene action
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Supression
9
3
4
13:3
9
3
4
9:3:4
Recessive epistasis of
aa action on B and b alleles
Dominant expistasis of
3
3
aaB_
9:7
7
12
3
1
12:3:1
1
15:1
A acting on B and b alleles
Duplicate genes
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Haploid Organisms
1) interaction between alleles
(e.g. dominance/recessive relationships)
-can only define if phenotype is expressed in the diploid stage
2) interaction between genes
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(epistasis, suppression, redundancy)
example: genes controlling spore (haploid) pigmentation in Neurospora
(fungus)
al = albino
al+ = orange
ylo = yellow
ylo+ = orange
conclusion: al is epistatic to ylo, suggesting the following biosynthetic
pathway
Penetrance and Expressivity (Figure 14-22)
PENETRANCE - the percentage of individuals with a given genotype who
exhibit the phenotype associate with that genotype
B_ = normal vision
bb = blind
but, of 100 people of genotype bb, only 20 are blind
therefore, allele b has a penetrance of 20%
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EXPRESSIVITY - the extent to which a given phenotype is expressed in an
individual
A_ = normal pigmentation
aa = albino
-one individual has genotype aa, but has 30% of normal skin pigmentation