Obtaining Parental Lines - Sex-Linked ZZ-ZW

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


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 I will call A1.  The wildtype allele I 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 since such formal crosses rely on having two pure-breeding lines. Geneticist must rely on a series of crosses.  Each cross will provide partial information about the inheritance of this new allele.

Preserve the Allele:  The first step is to preserve the new allele by breeding by making many copies of it.  This is done, of course, by breeding.  The type of breeding 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:  Sex-Linked Traits ZZ-ZW system.

General Principles.  In general one can make a few guesses about the dominance of a new mutation in a sex-linked system.  In ZZ-ZW sex determination system.  Females only need one copy of A1 to show a mutant phenotype no matter what the dominance.  Males need to have two copies of A1 to show a mutant phenotype if the trait is recessive. They need only one copy of A1 if the trait is completely dominant or the heterozygote has it own unique phenotype (incompletely dominant, overdominant or codominant).  Although it is not infallible, if a new mutation occurs first in males or in both males and females, it is probably not a recessive trait.

Females have two possible genotypes A1W and A2W and can have only two phenotypes.  Males have three possible genotypes A1A1, A1A2 and A2A2 and may have two or three phenotypes.

In the following examples, genotypes in black are wildtype (A2A2;A2W).  Genotypes in blue represent the mutant phenotype or, when the heterozygote has its own phenotype, the phenotype corresponding to the genotypes A1W and A1A1.  Genotypes in red represent the mutant phenotype corresponding to the genotype A1A2. when the heterozygote has its own phenotype. Genotypes in black are wildtype.

 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 mate the males and females with the trait.  There are different outcomes depending on whether the heterozygote (A1A2) has the wildtype phenotype, (mutant is recessive), a unique phenotype (mutant is incompletely dominant, overdominant or codominant) or has the mutant phenotype (mutant is dominant).

Table 1:      
Cross Mutant X Mutant
  &   %
  A1W   (A1?)
       
       
Progeny

?????

   
The initial cross is between two mutants.  We know that the mutant males must have one A1 allele but the other allele is unknown.  The best guess is that the male has the genotype A1A2 (see above).  The genotype of the females, however, is known.  Females with the mutant phenotype have one A1 allele and a W-chromosome.

Possible Outcomes:

a.  If the mutant allele is recessive, all of the progeny will have the recessive phenotype.  That is, the mutant males have the genotype A1A1.  Their sons with recessive phenotype have the genotype A1A1 and their daughters have the genotype A1W.  If these mutant progeny are allowed to intermate, they will establish a pure-breeding parental line. 

b.   If the heterozygote has it own unique phenotype (incompletely dominant, overdominant or codominant), then the mutant males most like are heterozygous (A1A2).  In this case, the phenotypes of the sons should be different from the phenotypes of the daughters. 

Table 2:      
Cross Mutant x Mutant
  &   %
  A1W x A1A2
       
Progeny A1W  

A1A1

  A2W  

A1A2

The progeny of a mutant x mutant cross (above) should show three different phenotypes - two mutant phenotypes and a wildtype phenotype.  If the females have a mutant phenotype that differs from one of the mutant phenotypes of the males, the inheritance must be incompletely dominant, overdominance or codominant. 

c.  If the mutant trait is dominant, then the mutant males most like are heterozygous (A1A2).  The females will have the same mutant phenotype as the mutant males. The mutant females will have the genotype A1W.

Table 3:      
       
Cross Mutant x Mutant
  &   %
  A1W x A1A2
       
Progeny A1W  

A1A1

  A2W  

A1A2

The progeny of a mutant x mutant cross (above) will show either mutant or wildtype phenotypes. 

Hedge Your Bets: At the same time as you carry out the mutant x mutant cross, it is desirable to mate mutant males to several wild-type females (wildtype & x  mutant % - preferably from a stock not known to be segregating the new mutation) to preserve the allele.  If the trait is recessive, 1/2 of the daughters and none of the sons will 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 will have a mutant phenotype. The sons will have a different mutant phenotype than the daughters.  If the trait is dominant, 1/2 of the progeny should be wildtype and 1/2 of the progeny should be mutant.

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

This is the most likely case if the mutation is recessive.  This single mutant female should be mated to multiple wildtype males (to mitigate potential male infertility).

Table 4:      
       
Cross Wildtype x Mutant
  &   %
  A1W x A2A2
       
  A2W  

A1A2

A.   Fixing a Recessive Allele: If the mutation is recessive, all of the sons will have the genotype A1A2.  The only female with an A1 allele, however is the mother. If the mother is still alive and still fertile, she should be mated back to her sons (see Table 5 below).  Half of the progeny from this mother-son cross will be mutant.  These mutants can be intermated to produce a parental line.

Table 5:      
       
Cross Wildtype x Mutant
  &   %
  A1W x A1A2
       
Progeny A1W  

A1A1

  A2W   A1A2

If the mother is not alive, it takes longer to fix a recessive line. Heterozygous males (A1A2) are mated to wildtype females (A2W - the only females available).  This produces mutant and wildtype females (see Table 6).  The wildtype females are discarded.  The males are all wildtype but one half of the males are heterozygous A1A2 and the other half of the males are homozygous A2A2.

Table 6:      
       
Cross 1 Wildtype x Mutant
  &   %
  A2W x A1A2
       
Progeny A1W  

A1A2

  A2W   A2A2

 

The mutant A1W females are individually mated to their brothers.  These matings are of two different types.

Cross 2a produces both wildtype and mutant males and females.  The mutant progeny can be intermated to produce the homozygous parental line. 

 Cross 2b produces only wildtype progeny.  The heterozygous females might be kept for additional breeding (always preserve the allele) but the males would be discarded.
 

Table 7
 
Cross 2a Wildtype x Mutant
  &   %
  A1W x A1A2
       
Progeny A1W  

A1A1

  A2W   A1A2

 

Cross 2b Wildtype x Mutant
  &   %
  A1W x A2A2
       
Progeny A2W  

A1A2

       

 

B.  Fixing an incompletely dominant, overdominant or codominant allele.

All of the sons will have the genotype A1A2.  The only female with an A1 allele, however, is the mother. As before, if the mother is still alive and still fertile, she should be mated back to her sons.  Half of the progeny from this cross will be mutant corresponding to the A1A1 or A1W genotypes.  These mutants can be intermated to produce a parental line.

If the mother is not alive, heterozygous males are mated to wildtype females (the only females available). This produces mutant and wildtype females. The wildtype females are discarded.  Half of the males are all homozygous wildtype (A2A2) and half of the males are heterozygous (A1A2).  Heterozygous males can be recognized by their phenotype. The wildtype males can be discarded.

Table 8      
       
Cross 1 Wildtype x Mutant
  &   %
  A2W x A1A2
       

Progeny

A1W  

A1A2

  A2W   A2A2

 

The red mutant males are mated to their blue mutant sisters.  The second cross produces hemizygous mutant females and homozygous mutant males.  These "blue" mutants can be intermated to produce the homozygous parental line. 
 
Table 9:      
       
Cross 2 Wildtype x Mutant
  &   %
  A1W x A1A2
       

Progeny

A1W  

A1A1

  A2W   A1A2

 

C. Fixing a dominant allele is discussed on another page.

The interpretation of the outcome of the above crosses will depend on the fecundity of the female.  If females have only one or two progeny, it is possible that the allele will be lost.

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

 It is extremely unlikely that the new mutation is recessive (see above) so the male will be heterozygous.  This single male should be mated to numerous wildtype females.  The phenotype of the daughters of this mating will give the genotype of their father.  If 1/2 of the females have a wildtype phenotype and the other 1/2 of the females have a mutant phenotype, then the father was heterozygous. 

 a.  If the allele is dominant, 1/2 of the males will have a wildtype phenotype and the other 1/2 of the females have a blue mutant phenotype. 

b.  If the heterozygote has its own phenotype, then the mutant phenotype in the females will differ from that of their father and their heterozygous brothers.  In the latter case, the males can be mated to their sisters to establish a pure breeding line (as above). 

c.  If the allele is dominant, see the next page .

Parental Lines -
Parental Lines - Autosomal
Parental Lines - Sex-Linked XX-XY
Parental Lines - Sex-Linked ZZ-ZW

Mendelian Genetics Index