Post on 13-Mar-2020
transcript
Genes to Phenotypes
"A set of genes represents the individual components of the
biological system under scrutiny"
Modifications of the "3:1 F2 monohybrid ratio" and gene
interactions are the rules rather than the exceptions"
One gene - one polypeptide: an oversimplification
Allelic variation
A. Many alleles are possible in a population, but in a diploid
individual, there are only two alleles
B. Mutation is the source of new alleles
C. There are many levels of allelic variation, ranging from
DNA sequence changes with no change in phenotype to
large differences in phenotype due to effects at the
transcriptional, translational, and/or post-translational levels
Vrs1 again
Relationships between alleles at a locus Complete dominance, partial dominance, codominance
Complete dominance: Deletion, altered transcription,
alternative translation. The interesting case of aroma in
rice: a loss of function makes rice smell great, and patent
attorneys salivate....
Incomplete (partial) dominance
Example: Red X ivory gives a pink F1. The F2 phenotypic
ratio is 1 red: 2 pink: 1 ivory.
Explanation: Red pigment is formed by a complex series of
enzymatic reactions. Plants with the dominant allele at the
I locus produce an enzyme critical for pigment formation.
Individuals that are ii produce an inactive enzyme and thus
no pigment. In this case, II individuals produce twice as
much pigment as Ii individuals and ii individuals produce
none. The amount of pigment produced determines the
intensity of flower color.
Incomplete (partial) dominance
Example: Red X ivory gives a pink F1. The F2 phenotypic
ratio is 1 red: 2 pink: 1 ivory.
Perspectives: Enzymes are catalytic and heterozygotes
usually produce enough enzyme to give normal
phenotypes. This is the basis for complete dominance.
However, upon closer examination, there are often
measurable differences between homozygous dominant
and heterozygous individuals. Thus, the level of
dominance applies only to a specified phenotype.
Codominance
Many molecular markers show codominant inheritance.
Both of the alleles that are present in a heterozygote can
be detected.
A way of visualizing codominance is through
electrophoresis.
www.mun.ca
Codominance
An application of electrophoresis is to separate proteins or DNA
extracted from tissues or whole organisms. An electric charge is run
through the supporting media (gel) in which extracts, containing
proteins or DNA for separation, are placed. Proteins or DNA fragments
are allowed to migrate across the gel for a specified time and then
stained with specific chemicals or visualized via isotope or fluorescent
tags. Banding patterns are then interpreted with reference to
appropriate standards. The mobility of the protein or DNA is a function
of size, charge and shape.
Codominance The following illustration shows codominant alleles at a simple
sequence repeat (SSR) locus (HvHVA1) in the F1 and doubled-haploid
lines of barley. Lanes 1-17; 19-21 are doubled haploids and lane 18 is
the F1. The forward primer was unlabeled and the reverse primer was
labeled at its 5’ end with a fluorochrome. The PCR products were
analyzed using an ABI PRISM 377 DNA sequencer. The doubled
haploids are homozygous for one of the two parental alleles (BCD47
allele = 138 bp = A; Baronesse allele = 122 bp = B). The F1 shows
both alleles.
Overdominance
Heterozygote advantage
Aa >AA, aa
Epistasis: Interaction between alleles at different
loci
Example: Duplicate recessive epistasis (Cyanide
production in clover). Identical phenotypes are produced
when either locus is homozygous recessive, or when both
loci are homozygous recessive.
Parental, F1, and F2 phenotypes:
Parent 1 X Parent 2
(low cyanide) (low cyanide)
F1
(high cyanide)
F2 (9 high cyanide: 7 low cyanide)
Parent 1 X Parent 2 (AAbb low cyanide) (aaBB low cyanide) F1 (high cyanide AaBb) F2 …………………9……………….. ……..3……… ……….3……… ..1…. AABB AABb AaBB AaBb AAbb Aabb aaBB aaBb aabb High cyanide>>>>>>>>>>> Low cyanide>>>>>>>>>>>>>>>>>
Cyanide in clover: The mechanism
Precursor Enzyme 1 (AA; Aa) Glucoside Enzyme 2 (BB; Bb) Cyanide
If Enzyme 1 = aa; end pathway and accumulate precursor; if Enzyme 2 = bb; end
pathway and accumulate glucoside
Assigning parental genotypes
More phenotypic ratios with two-locus epistasis
Key concept is modification of expected 2 locus ratios
•In F2 see variations on 9:3:3:1 (e.g. 9:3:4 or 15:1)
•In DH see variations on 1:1:1:1 (e.g. 3:1)
Examples of 2-locus epistasis
Recessive epistasis: Complete dominance at both loci, but homozygous
recessive condition at one of the two loci is epistatic to the other.
Example: coat color in mouse.
Locus 1: A color; a albino. Locus 2:B agouti; b black. Locus 1, when
homozygous recessive, epistatic to Locus 2.
Phenotypic ratio: 9:3:4
Dominant epistasis: Presence of dominant allele at one locus always
gives the same phenotype, regardless of the allelic constitution at the
other locus. Only when the dominant epistatic locus is homozygous
recessive can the allelic constitution of the second locus be expressed.
Example: fruit color in summer squash.
Locus 1: A white; a color. Locus 2: B yellow; b green.
Phenotypic ratio: 12:3:1
Duplicate genes with cumulative effects: Complete dominance at
both loci. Interaction between dominants at both loci gives a new
phenotype.
Example: fruit shape in summer squash.
Locus 1: A sphere; a long. Locus 2: B sphere; b long. A_B_ gives new
phenotype:disc.
Phenotypic ratio: 9:6:1
Duplicate genes with cumulative effects: Complete dominance at
both loci. Interaction between dominants at both loci gives a new
phenotype.
Example: fruit shape in summer squash.
Locus 1: A sphere; a long. Locus 2: B sphere; b long. A_B_ gives new
phenotype:disc.
Phenotypic ratio: 9:6:1
Duplicate dominant genes: Dominant allele at each locus produces
same effect, but not a new phenotype when there are dominant alleles at
both loci, as in the previous example.
Example: seed capsule shape in Shepherd's purse.
Locus 1: A triangular; a ovoid. Locus 2: B triangular; b ovoid.
Phenotypic ratio: 15:1
Dominant and recessive interaction: Complete dominance at each locus.
Locus 1, when dominant allele present, is epistatic to locus 2. Locus 2,
when homozygous recessive, is epistatic to Locus 1, if there is no dominant
allele present at Locus 1.
Example: feather color in the chicken.
Locus 1: A color inhibition; a color appearance. Locus 2: B color; b white.
Phenotypic ratio: 13:3
A visually stimulating tour of epistasis in cucurbits
prepared by Dr. Paul Kusolwa
Fruit shapes in squash
Di locus
Dominant to spherical or pyriform
Duplicate genes with cumulative effects
When Di is present together with Spherical S
locus Di is dominant = Disc fruits
When Di present with recessive s = spherical fruits
When didi/ss = long or pyriform fruits
Modified Ratio = 9: 6: 1
6 Spherical = Di/s & di/S_
1 Pyriform di/s
9 Discs
Di_S_
Complementary interaction
Pathway involving two genes
Wt = warty fruits Dominant to
non warty wt
Hr = hard rind
hr = intermediate texture
9:7
B_
Hr_
Wt_
Y_
Bicolor fruits = locus B pleiotropic for fruits
and leaves
For yellow and green color patches BB or
Bb
Extent of yellow or green
dosage of another incompletely
dominant, additive to modifier
genes. Ep1 and Ep 2
Ep1 and Ep2 alleles are involved
in enlarging the yellow patches
Genotypes
Bb with a dosage of 0 to 1 Ep
alleles = bicolor green and yellow
fruits
Dosage of 2-4 dominant alleles
Ep extends the yellow coloration
Genotypes
9 B_Ep_ Extended yellow
3 B_epep Yellow narrow
3 bbEp_ Green extended
1bbepep green
12:4
Three or more genes interactions in ornamental gourds
Gb = green bands; gb for no bands
Gr/G = green rind (gr/g buff skin) L-2 = color intensity (yellow /orange)
Gr
Gb
L-2
Ep-1/2
B_ bicolor
L-2 intensity
Pro ovary
momo-1
MoMo -1
Complementary recessive
for loss of green fruit color
before maturity
mo-1 Mature orange color
mo-2 Mature orange
A genetically stimulating tour of vernalization sensitivity in barley : Something as simple as growth habit can involve all types of allelic variants and interactions
Reminders from the first lecture......
Mendelian genetic analysis: the "classical" approach to
understanding the genetic basis of a difference in phenotype is
to use progeny to understand the parents.
• If you use progeny to understand parents, then you make
crosses between parents to generate progeny populations
of different filial (F) generations: e.g. F1, F2, F3;
backcross; doubled haploid; recombinant inbred, etc.
Reminders from the first lecture......
Mendelian genetic analysis: the "classical" approach to
understanding the genetic basis of a difference in phenotype is
to use progeny to understand the parents.
• The genetic status (degree of homozygosity) of the
parents will determine which generation is appropriate for
genetic analysis and the interpretation of the data (e.g.
comparison of observed vs. expected phenotypes or
genotypes).
o The degree of homozygosity of the parents will
likely be a function of their mating biology, e.g.
cross vs. self-pollinated.
Reminders from the first lecture......
Mendelian genetic analysis: the "classical" approach to
understanding the genetic basis of a difference in phenotype is
to use progeny to understand the parents.
• Mendelian analysis is straightforward when one or two
genes determine the trait.
• Expected and observed ratios in cross progeny will be a
function of
o the degree of homozygosity of the parents
o the generation studied
o the degree of dominance
o the degree of interaction between genes
o the number of genes determining the trait