I. Overview - genes and gene products

• One gene - one polypeptide: ________________

II. Allelic variation

A. Many alleles are possible in a population, but in a diploid individual, there are only two alleles....The Vrs1 example, again

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

 

III. Relationships between alleles at a locus

A. 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....

B. Incomplete (partial) dominance __________________

Red X ivory gives a pink F1. The F2 phenotypic ratio is 1 red: 2 pink: 1 ivory. 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. Note: 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.

C. Codominance __________________

Many biochemical and molecular markers show codominant inheritance. In the case of codominance, both of the alleles that are present in a heterozygote can be detected. A way of visualizing codominance is through electrophoresis. (Electrophoresis animation) 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.

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 automated 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.  (Data courtesy of I Vales, OSU)

 

	F1 ¯
A ®  B ®

 

 

D. Overdominance __________________



E. Illustration of consequences of different combinations of alleles at a locus: Human hemoglobin alleles and malaria (Details)

 

IV. 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. Details in Olsen et al. 2007.

Parental, F1, and F2 phenotypes: Parent 1            X           Parent 2 (low cyanide)            (low cyanide) ¯ F1 (high cyanide) ¯  F2(9 high cyanide: 7 low cyanide)

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

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>>>>>>>>>>>>>

 

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)

A visually stimulating tour of epistasis in cucurbits: epistasis is only as far away as the garden, farm, store, or dinner table!
A genetically stimulating tour of vernalization sensitivity in barley : something as simple as growth habit can involve all types of allelic variants and interactions

A reminder 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. 
• 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).
  • The degree of homozygosity of the parents will likely be a function of their mating biology, e.g. cross vs. self-pollinated.
• Mendelian analysis is straightforward when one or two genes determine the trait.
• Expected and observed ratios in cross progeny will be a function of
  • the degree of homozygosity of the parents
  • the generation studied
  • the degree of dominance
  • the degree of interaction between genes
  • the number of genes determining the trait

IV. Applications of Mendelian analysis to practical questions: Are mutations in the same or in different genes?

A. Complementation: Determining if two mutations (each conferring the same phenotype) are

• in the same gene (alleles at the same locus)
• in different genes (alleles at different loci) and the two genes are part of a pathway
  
  
• Assumptions: the mutation(s) must be recessive and each of the parents must be homozygous for the recessive mutation.
• Interpreting the results: If the F1 progeny are all wild type, the mutations must be in different genes.
  

  Phenotype: Flower color. 

A. Both parents white; F1 white 

• Assume:  1 locus, complete dominance: W_ = red; ww = white
• Infer: 
  • Parents: ww X  ww   
  • F1:  ww = white 
• Interpretation: The two parents have the same (or functionally equivalent) alleles for flower color
• Further proof: F2 or DH all white 

B: Both parents white; F1 red 

• Assume:  2 loci, where at least one dominant allele at each locus necessary for red phenotype. 
• Infer: 

• Parents: W1W1w2w2 X  w1w1W2W2
• F1: W1w1W2w2 = red

• Interpretation: The two parents have the different alleles at 2 loci determining the phenotype and these alleles complement one another in the F1 to give flower color.
• Further proof: 
  • DH: expect 1 red: 3 white (W1W1W2W2 vs: W1W1w2w2, w1w1W2W2,w1w1w2w2)
  • F2: expect 9 red: 7 white (figure it out!)

B. Allelism tests: Applying the concept to a series of crosses between accessions with the same phenotype:

Determining if accessions with the same phenotype have

• functionally equivalent alleles at the same locus
• alleles at different loci that lead to the same phenotype
  

 

• Interpreting the results: If the F1 and all subsequent progeny generations are like the two parents, the parents have the same (or functionally equivalent) alleles at a locus. If generations after the F1 segregate for the traits, the parents must have different alleles at different loci

• Phenotype: Disease resistance
• Assume:  2 loci, where one dominant allele at either locus is sufficient to confer resistance  

	Accession 1	Accession 2	Accession 3
Phenotype	Resistant	Resistant	Resistant
Genotype	R1R1r2r2	R1R1r2r2	r1r1R2R2

A. Accession 1 x Accession 2  

• Observe: 
  • F1:  resistant 
  • F2 or DH all resistant 
• Infer: Parents: R1R1r2r2  
• Interpretation: The two parents have the same (or functionally equivalent) alleles for resistance at the same locus/ loci 

B. Accession 1 x 3  (or 2 x 3)

• Observe: 
  • F1:  resistant 
  • DH: 3 resistant: 1 susceptible 
  • F2: 15 resistant: 1 susceptible
• Infer: Parents: R1R1r2r2  and r1r1R2R2
  • DH: R1R1R2R2, R1R1r2r2, r1r1R2R2 vs r1r1r2r2
  • F2: figure it out!
• Interpretation: The two parents have the resistance alleles at different loci

If mapping is cheap and accurate, you could simply map all the alleles and determine if the cosegregate. Of course, cosegregation could be due to tight linkage....and so you are drawn into sequencing and functional assays....

 

V. Related concepts 

Pleiotropy: __________________

Penetrance: __________________

Key concepts:

• Alleles at a locus can show a range of relationships including codominance, partial dominance, dominance, and overdominance.
• Alleles at many biochemical and DNA loci show codominant inheritance, and this allelic relationship can be visualized via electrophoresis.
• The same phenotype in different individuals can be due to identical alleles at the same locus, an allelic series at the same locus, or alleles at different loci. The possibilities can be explored using complementation and allelism tests.
• Pleiotropy is the situation where a single gene can affect multiple phenotypes.
• Penetrance describes the proportion of individuals with the same genotype that show the same phenotype and expressivity describes the degree of commonality of expression across a sample of individuals with the same genotype.
• Epistasis and suppression are examples of interaction of alleles at two or more loci.
• The phenotypic manifestations of epistasis and suppression are altered phenotypic ratios.
• Epistasis is probably the rule rather than the exception. 

Text: Chapter 6

Useful links:

Nicely illustrated examples of Epistasis from NDSU