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.

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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