Obtaining Parental Lines - Autosomal Traits

Mutant Allele is Completely Recessive, Overdominant, Incompletely Dominant or Codominant.

 

Stock Centers:

The easiest way to get pure-breeding lines is to get them from a stock center.  However, this only works for a few, well-studied organisms.  In this case, some other geneticist has done the work for you.  In a problem set, students are given homozygous lines.  However, in experimental situations, these lines have to be constructed. The process may take several years depending on the organism and the particular genes involved.


How do Geneticists Get a New Parental Line?

The Scenario:

A new phenotype occurs in a population.  This new phenotype is due to a new mutant allele which we will call A1.  The wild-type allele we will call A2. At this point in time, nothing is known about the inheritance of the trait.  One cannot use F1 crosses to determine the inheritance patterns of A1 since F1 crosses rely on having two pure-breeding lines. Geneticists must rely on a series of matings. The genotypes of the parents may not be known.  Each mating will provide partial information about the inheritance of this new allele.

Preserve the Allele:  The first step is to preserve the new allele by making many copies of it.  This is done, of course, by breeding.  The type of mating will depend upon (1) the number of organisms with the new phenotype, (2) the gender of the organism with the new phenotype, (3) whether the allele is autosomal or sex-linked, (4) the dominance relations of the allele and the biology of the organism.

Number of Organisms With the New Allele: Autosomal Traits.

In the following discussion genotypes in black are associated with a wildtype phenotype. Genotypes in blue are associated with the mutant phenotype or, the case of incompletely dominant, overdominant or codominant traits, the genotype A1A1. Genotypes in red (A1A2) are associated with the incompletely dominant, overdominant or codominant mutant phenotype. 

Observation 1: In a wild-type population of lab organisms a new phenotype shows up in more than one individual and both males and females exhibit the new phenotype.  In this case, it is easiest to intermate the males and females with the mutant trait.  There are different outcomes depending on whether the heterozygote (A1A2)  has the wildtype phenotype, (the mutant allele is recessive), a unique phenotype (the mutant allele is incompletely dominant, overdominant or codominant) or has the mutant phenotype (the mutant allele is completely dominant).

In the mating shown below, the genotype of the mutant individuals is unknown.  They must have at least one A1 allele.  There are three basic outcomes.
 
Table 1:      
Cross Mutant X Mutant
  &   %
  (A1?)   (A1?)
       
       
Progeny

?????

   

A. If the mutant allele is recessive, all of the progeny will have the recessive phenotype.  That is, the mutant individuals have the genotype A1A1.  Their progeny with the recessive phenotype also have the genotype A1A1.  If these mutant progeny are allowed to intermate, they will establish a pure-breeding parental line. This situation is diagramed below

Table 2:      
Cross Mutant X Mutant
  &   %
  A1A1   A1A1
       
       
Progeny

A1A1

  A1A1

 

B. If the heterozygote has its own unique phenotype (incompletely dominant, overdominant or codominant), then the new mutants most like are heterozygous (A1A2)The progeny of a mutant x mutant cross (above) will show three phenotypes corresponding to the three genotypes A1A1:A1A2:A2A2 (wildtype) in a 1:2:1 ratio.  You merely need to pick all individuals with the correct phenotype corresponding to the A1A1 genotype and mass mate them to establish a parental line.

Table 3:      
Cross Mutant X Mutant
  &   %
  A1A2   A1A2
       
    Freq.  
Progeny

A1A1

1/4

A1A1

  A1A2 1/2 A1A2
  A2A2 1/4 A2A2

C. If the heterozygote trait is dominant, then the new mutants most likely are heterozygous (A1A2).  The progeny of a mutant x mutant cross (above) will show two phenotypes corresponding to the three genotypes A1A1:A1A2:A2A2 (wildtype) in a 3:1 phenotypic ratio.  Of those individual with the mutant phenotype only 1/3 of them will be homozygous. Fixing a dominant allele requires a great deal of effort.  This is discussed on another page.

Hedge Your Bets: At the same time as you carry out the mutant x mutant cross, it is desirable to mate many mutant individuals to wild-type individuals (preferably from a stock not known to be segregating the new mutation) to preserve the allele.  If the trait is recessive, none of the progeny should show the new phenotype.  If the heterozygote has its own phenotype (and the mutant individual is heterozygous ), then 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant.  If the trait is dominant (and the mutant individual is heterozygous), 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant.

Example:

As a graduate student at the University of Illinois I noticed a number of orange-eyed blow flies (Phormia regina) in one of my stocks.  Wildtype Phormia all have red eyes.  I mated several pairs of orange-eyed flies in an attempt to get a pure breeding line. At the same time, I mated several orange-eyed males to many red-eyed females in order to preserve the mutation.

The cross of orange x orange produced only orange offspring. The cross of orange male to wildtype red female ( or wildtype females to orange male) produced only red-eyed progeny.  The results of the cross were consistent with orange eyes being a recessive trait. 

 In this case, the orange x orange cross produced several hundred progeny.  After one generation, I had a pure-breeding line of orange-eyed flies.  However, the progeny of this stock continued to be observed each generation to ensure that only orange-eyed flies were produced.

 

Observation 2:  A single individual with a unique phenotype shows up in the population and it is a male.

Mutant males are great since, in general, they can be mated to multiple wildtype females.  If the trait is recessive, none of the progeny should show the new phenotype.  If the heterozygote has its own phenotype (and the mutant individual is heterozygous ) , then 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant.  If the trait is dominant (and the mutant individual is heterozygous), 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant (slightly different rules apply if the trait is sex-linked either XX-XY system or ZZ-ZW system).  At this point, it is impossible to tell the difference between a completely dominant trait and one where the heterozygote has its own phenotype.

 2A. If the mutation is recessive, all of the progeny of a mutant x wildtype cross will have a wildtype phenotype and the genotype A1A2 The F1 progeny are allowed to interbreed.  There should be two phenotypes among the F2 progeny: wildtype to mutant in a 3:1 ratio. The 1/4 of the progeny with the mutant phenotype have the genotype A1A1.  These A1A1 individuals can be mated to establish a pure-breeding parental line.

Table 4: Single male with Genotype A1A1
       
Cross Wildtype X Mutant
  &   %
  A2A2   A1A1
       
       
Progeny Gen 1

A1A2

X A1A2
       
Progeny Gen 2 A1A1 1/4 A1A1
 

A1A2

1/2

A1A2

  A2A2 1/4 A2A2

2B. If the heterozygote has its own phenotype (and the original mutant individual was heterozygous ), the mutant progeny from the first mating will have the genotype A1A2 and they will have the same phenotype as their mutant father.  These A1A2 individuals are allowed to interbreed. A new phenotype should be seen among the progeny corresponding to the genotype A1A1 The A1A1 individuals can be intermated to establish a pure-breeding parental line.

Table 5: Single male with Genotype A1A2 and heterozygote has its own phenotype
       
Cross Wildtype X Mutant
  &   %
  A2A2   A1A2
       
Progeny Gen 1

A1A2

X

A1A2

       
Progeny Gen 2 A1A1 (1/4)   A1A1 (1/4)
  A1A2 (1/2)   A1A2 (1/2)
  A2A2(1/4)   A2A2(1/4)

 

2C.  If the mutation is dominant (and the original mutant individual was heterozygous ), 1/2 of the progeny will have a mutant phenotype from the first mating and will have the genotype A1A2.   These individual are allowed to interbreed.  There should be two phenotypes among the F2 progeny - mutant:wildtype in a 3:1 ratio.  The individuals with the dominant phenotype have either the A1A1 or the A1A2 genotypeFixing a dominant allele requires a great deal of effort.  This is discussed on another page.

Table 6: Single male with Genotype A1A2 and heterozygote has the single mutant phenotype.
       
Cross Wildtype X Mutant
  &   %
  A2A2   A1A2
       
Progeny Gen 1

A1A2

X

A1A2

       
Progeny Gen 2 A1A1 (1/4)   A1A1 (1/4)
  A1A2 (1/2)   A1A2 (1/2)
  A2A2(1/4)   A2A2(1/4)

 

Observation 3: A single individual with a unique phenotype shows up in the population and it is a female.

This is the least desirable situation since the number of progeny is limited.  The single female should be mated to several wildtype males.  This is a safety measure since one or more of the males might be infertile. The expected results are no different from those for a single mutant male (see above).   If the trait is recessive, none of the progeny should show the new phenotype.  If the heterozygote has its own phenotype (and the mutant individual is heterozygous), then 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant.  If the trait is dominant (and the mutant individual is heterozygous), 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant (slightly different rules apply if the trait is sex-linked either XX-XY system or ZZ-ZW system).  At this point, it is impossible to tell the difference between a completely dominant trait and a trait where the heterozygote as its own phenotype.

The interpretation of the outcome will depend on the fecundity of the female.  A certain number of progeny must be produced to interpret these data.

Table 7:  Probability of observing a mutant progeny depends on the total number of progeny.

Female is heterozygous and 1/2 of the progeny should have the new phenotype (Mutant is dominant or heterozygote has its own phenotype)
 

Progeny Obtained Wild-type Progeny Only Probability
1 1 .5
2 2 .25
3 3 .125
4 4 .0625
5 5 .0313
6 6 .0156
7 7 .0078
8 8 .0039
9 9 .0020
10 10 .0010

By chance, it is possible to obtain only wildtype A2A2 individuals in a cross of a heterozygote mutant individual (A1A2) with a homozygous wildtype individual (A2A2).  From the table above, we need to obtain at least 7 progeny to have better than a 99% of detecting a non-recessive allele.  Obviously 10 or more progeny would be better.  This is fairly easy to obtain with a single mutant male since he can be mated to multiple females.  A single mutant female, however, may produce a very small litter.  On the other hand, even a single offspring is informative if the mutant phenotype shows up.

Summary: Except for the number of progeny produced, there is no difference whether you initially have a single mutant male or a single mutant female.

3A. If the mutation is recessive all of the F1 progeny will have the genotype A1A2 These F1 individuals are allowed to interbreed.  There should be two phenotypes among the F2 progeny, wildtype:mutant in a 3:1 ratio. The 1/4 of the progeny with the mutant phenotype will have the genotype A1A1.  These A1A1 individuals can be mated to establish a pure-breeding parental line. Again, it is easy to be defeated by numbers.  Only 1/4 individuals will have the mutant phenotype and the homozygous genotype A1A1.   In organisms of limited fecundity,  you may obtain no homozygous individuals or all of your mutant individuals may be of one sex. A geneticist must be flexible and plan additional crosses based on the genotypes and sexes of the available individuals.

3B.  If the heterozygote has its own phenotype (and the original mutant individual was heterozygous), the mutant progeny from the first mating will have the genotype A1A2 These individual are allowed to interbreed. A new phenotype should be seen among the progeny corresponding to the genotype A1A1 The A1A1 individuals can be mated to establish a pure-breeding parental line. As before, it is easy to be defeated by numbers.  Only 1/4 individuals will have the mutant phenotype and the homozygous genotype A1A1.  However, you can also distinguish heterozygotes with the genotype A1A2 Again, a geneticist must be flexible and plan additional crosses based on the genotypes and sexes of the available individuals.

3C. If the mutation is dominant (and the original mutant individual was heterozygous), the mutant progeny from the first mating will have the genotype A1A2 These individual are allowed to interbreed.  There should be two phenotypes among the F2 progeny mutant:wildtype in a 3:1 ratio.  The individuals with the dominant phenotype have either the A1A1 or the A1A2 genotypeFixing a dominant allele requires a great deal of effort.  This is discussed on another page.

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