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Strategy for Figuring Out the Underlying Genetic System from Data |
The starting point of most of genetics is the single gene trait. This is, after all, where Mendel started. Unfortunately, single gene Mendelian traits are actually relatively rare. Complex phenotypes usually involve two or more genes.
| Example: An investigator crossed two pure-breeding stocks of wheat hooded x short-awned. The F1 progeny were all hooded. The F1 were intercrossed to provide an F2 . The F2 progeny had 450 hooded, 150 long-awned, and 200 short-awned phenotypes. The reciprocal F1 and F2 cfosses gave nearly identicl results (not shown). Explain and diagram the genetic system that would give these results. |
Treat this problem as laboratory data.
(1) What do we actually already know about the system?
We know that some someone has already verified the purity of the parental lines ("two pure-breeding stocks of wheat"). This means that the Parental lines are homozygous and always breed true. Most likely, the homozygous nature of the lines has been further verified by test crosses.
(2) What can we infer from the fact that the lines are pure-breeding?
We can infer that neither
hooded nor short-awned is a
recessive lethal. Recessive lethals must, by definition, be heterozygous.
We can also
infer that neither hooded or short-awned are dominant lethals.
We can also know that if hooded and short-awned are conditional lethals,
that the
conditions are not met that would cause death.
We can infer there
is 100% penetrance in the parental lines because each parental line has only one phenotype.
We can infer that there is 100% penetrance in the heterozygote since all F1 all of the progeny have one phenotype (hooded)
One gene, two genes or more
Once we have eliminated penetrance and lethality as confounding factors, there are only a few patterns that you might see in a monohybrid cross. Other forms of gene action such as pleiotropy and expressivity, however, do not change expected Mendelian ratios.
In a monohybrid cross (AA x aa) we expect to see three genotypes and either 2 (complete dominance) or 3 (incomplete dominance, codominance, overdominance) phenotypes.
For a single gene, once we cross two pure-breeding parental lines and look at the phenotype of the F1 and F1' progeny, we know the dominance pattern and whether the gene is autosomal or sex-linked (review F1). With that information, we can estimate the distribution of phenotypes in the F2 generation.
For single autosomal trait, the phenotypic ratio in the F2 generation is:
1/4 P1 phenotype; 1/2 F1 Phenotype: 1/4 P2 phenotype
---- with complete dominance this becomes the familiar 3:1 ratio
---- with incomplete dominance, codominance, or overdominance, this becomes the familiar 1:2:1 ratio.
| Rule of Thumb:
Once you have eliminated all possible confounding factors described above. For an autosomal trait, if a new phenotype shows up in the F2 generation or you show deviation from the 1:2:1 ratio or the 3:1 ratio you are not dealing with a single autosomal gene!! |
Referring to the above example:
A new phenotype, long-awned, showed up in the F2 generation. Therefore we can reject the idea that we are dealing with a single autosomal gene. We need to look for a more complex model.
The obvious place to start is a two gene model. We have to build and test a new model based on two interacting genes.
| 1. | Since the phenotype in the F1 generation was hooded, we know that the phenotype associated with the genotype AaBb is hooded (for a two-gene model). |
| 2. | With two genes, complete dominance and independent assortment, we expect to see a 9:3:3:1 phenotypic ratio in the F2 generation. However, we observe a 9:3:4 ratio. In order to get the "4"in this ratio we hypothesize that the genotype aabb is included in the short-awned phneotype. |
| 3. | From the first two assumptions, we assume that the pure-breeding hooded strain has the genotype AABB. Thus, the pure-breeding short-awned strain would be aabb. |
| 4. | We must assume that recessives at one of the genes, for example, aa, give short-awned phenotypes no matter what the genotype is at the B locus. |
| 5. | We assume that recessives at the other gene (bb) give the long-awned phenotype if combined with dominants at the A locus (AA or Aa) |
| 6. | Since one parental strain and the F1 were hooded, we can assume that any A_B_ genotypes are also hooded. |
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Tentative Model |
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| Partial genotype | Phenotype |
Genotypes |
| A _ B_ | hooded | AABB, AABb, AaBB, AaBb |
| aa _ _ | short-awned | aaBB, aaBb, aabb |
| A_ bb | long-awned | AAbb, AaBb |
Check on the system (the crosses under the tentative model).
| Parental | Hooded (AABB) | x | short-awned (aabb) |
| F1 | Hooded (AaBb) | x | Hooded (AaBb) |
| F2 |
9 Hooded (A_B_); 3 long-awned (A_bb): 4 short-awned (aa___) |
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My model is consistent with the observed data. Proving that the model is correct, however, would involve doing additional experimental crosses.
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