Difference between revisions of "Complementation"
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If there are mutations at two different genes then crossing true-breeding (homozygous) lines together generates double heterozygote offspring with a wild-type phenotype. This is complementation and indicates the two mutants occur at different genes. | If there are mutations at two different genes then crossing true-breeding (homozygous) lines together generates double heterozygote offspring with a wild-type phenotype. This is complementation and indicates the two mutants occur at different genes. | ||
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| + | =Examples= | ||
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| + | ==Abstract example== | ||
Say we have three mutations, a[-], b[-], and c[-], that result in mutant wing phenotypes in Drosophila. These mutant phenotypes are recessive to wild-type. First we cross a[-] and b[-] together. | Say we have three mutations, a[-], b[-], and c[-], that result in mutant wing phenotypes in Drosophila. These mutant phenotypes are recessive to wild-type. First we cross a[-] and b[-] together. | ||
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This double heterozygote has a wild-type phenotype because the phenotype corresponding to a[1] and the phenotype corresponding to c- are both recessive to wild-type. We can infer that these are alleles of two different genes because they complemented each other (restored a wild-type phenotype). We rename c- the c[1] allele now that we know it is an allele of a different gene. | This double heterozygote has a wild-type phenotype because the phenotype corresponding to a[1] and the phenotype corresponding to c- are both recessive to wild-type. We can infer that these are alleles of two different genes because they complemented each other (restored a wild-type phenotype). We rename c- the c[1] allele now that we know it is an allele of a different gene. | ||
| − | Links | + | ==Actual example== |
| + | |||
| + | =Exceptions= | ||
| + | |||
| + | Exceptions (except for dominant mutant phenotypes) are rare but they can give insights into gene products with functions that are closely associated. | ||
| + | |||
| + | ==Dominance== | ||
| + | |||
| + | Cases where the mutant phenotype is dominant to wild-type, such as gain-of-function hypermorphs violate the assumptions used in complementation testing. | ||
| + | |||
| + | ===Example=== | ||
| + | |||
| + | (agouti vy and - amorph) | ||
| + | |||
| + | ==Intragenic complementation== | ||
| + | |||
| + | If mutations that disrupt alternative promoters or alternative splicing variants are responsible for the altered phenotype then having two mutant alleles that are disrupted in different ways can restore a wild-type phenotype. | ||
| + | |||
| + | ===Example=== | ||
| + | |||
| + | ==Unlinked noncomplementation== | ||
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| + | If gene products from two different genes directly interact, such as subunits in a complex, or their functions are closely associated, such as steps in the same biochemical pathway, they can fail to complement in a double heterozygote. | ||
| + | |||
| + | ===Example=== | ||
| + | |||
| + | =Links= | ||
http://www.wormbook.org/chapters/www_complementation/complementation.html | http://www.wormbook.org/chapters/www_complementation/complementation.html | ||
Revision as of 12:12, 13 October 2017
Complementation tests if two different alleles that result in observable phenotypes are part of the same gene or are alleles of different genes. If you have generated a range of mutants phenotypes for a trait it allows these mutations to be quickly grouped by genes (rather than, for example, conducting fine scaled recombinant mapping). Complementation assumes simple dominance where the mutant phenotype is recessive to wild-type.
If there are mutations in the same gene then when homozygous lines are crossed together there is not a wild-type allele present in the heterozygote. Thus, the heterozygote has a mutant phenotype. This is non-complementation and indicates that the two mutants occur at the same gene.
If there are mutations at two different genes then crossing true-breeding (homozygous) lines together generates double heterozygote offspring with a wild-type phenotype. This is complementation and indicates the two mutants occur at different genes.
Contents
Examples
Abstract example
Say we have three mutations, a[-], b[-], and c[-], that result in mutant wing phenotypes in Drosophila. These mutant phenotypes are recessive to wild-type. First we cross a[-] and b[-] together.
a-/a-, c+/c+ x b-/b-, c+/c+
↓
a-/b-, c+/c+
a[-] and b[-] are two different alleles of the same gene and the wild-type phenotype is not restored (they failed to compliment). We can rename these as the a[1] and a[2] alleles.
Now we cross a[1] and c[-] together.
a[1]/a[1], c+/c+ x a+/a+, c-/c-
↓
a[1]/a+, c+/c-
This double heterozygote has a wild-type phenotype because the phenotype corresponding to a[1] and the phenotype corresponding to c- are both recessive to wild-type. We can infer that these are alleles of two different genes because they complemented each other (restored a wild-type phenotype). We rename c- the c[1] allele now that we know it is an allele of a different gene.
Actual example
Exceptions
Exceptions (except for dominant mutant phenotypes) are rare but they can give insights into gene products with functions that are closely associated.
Dominance
Cases where the mutant phenotype is dominant to wild-type, such as gain-of-function hypermorphs violate the assumptions used in complementation testing.
Example
(agouti vy and - amorph)
Intragenic complementation
If mutations that disrupt alternative promoters or alternative splicing variants are responsible for the altered phenotype then having two mutant alleles that are disrupted in different ways can restore a wild-type phenotype.
Example
Unlinked noncomplementation
If gene products from two different genes directly interact, such as subunits in a complex, or their functions are closely associated, such as steps in the same biochemical pathway, they can fail to complement in a double heterozygote.
Example
Links
http://www.wormbook.org/chapters/www_complementation/complementation.html
