Obtaining Parental Lines - Sex-Linked XX-XY

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 wild-type 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, and (4) the dominance relations of the allele and the biology of the organism.

Number of Organisms With the New Allele:  Sex-Linked Traits XX-XY system.

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

In the XX-XY sex-determination system, males have two possible genotypes A₁Y and A₂Y and can have only two phenotypes.  Females have three possible genotypes A₁A₁, A₁A₂ and A₂A₂ and may have two or three phenotypes. In the XX-XO sex determination system, males have two possible genotypes A₁O and A₂O and can have only two phenotypes.  Females have three possible genotypes A₁A₁, A₁A₂ and A₂A₂ and may have two or three phenotypes. Both systems show the same inheritance patterns at the phenotypic level. 

 Observation 1:  In a wild-type population  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).

If the mutant allele is recessive, all of the progeny will have the recessive phenotype.  That is, the mutant females have the genotype A₁A₁.  Their daughters with recessive phenotype have the genotype A₁A₁  and their sons have the genotype A₁Y.  If these mutant progeny are allowed to intermate, they will establish a pure-breeding parental line. 

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

The progeny of a mutant x mutant cross (above) should show three different phenotypes. (Genotypes with the same color font have the same phenotypes).  If the females have a mutant phenotype that differs from the mutant phenotype of the males, the inheritance must be incompletely dominant, overdominance or codominant. 

• The daughters from this mating should show two different mutant phenotypes corresponding to the homozygous and heterozygous genotypes.  These will occur in equal frequencies (1/2) .

• The sons  from this mating should show a  mutant and a wildtype phenotype.  These will occur in equal frequencies (1/2) .

• You merely need to pick all individuals with the correct mutant phenotype (the blue one) corresponding to the A₁A₁ and A₁Y genotypes and mass mate them to establish a parental line.

If the heterozygote trait is dominant, then the mutant females most like are heterozygous (A₁A₂).  The females will have the same mutant phenotype as the mutant males. The mutant males will have the genotype A₁Y.

The progeny of a mutant x mutant cross (above) will show two the same two phenotypes in both males and females.

• The daughters from this mating should show a mutant and a wildtype phenotype corresponding to the homozygous (A₁A₁) and heterozygous (A₁A₂) genotypes.  These phenotypes will occur in equal frequencies (1/2) .

• The sons  from this mating should show a  mutant and a wildtype phenotype corresponding to the A₁Y and A₂Y genotypes.  These will occur in equal frequencies (1/2) .

• 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 ) , 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 male.

This is the most likely case if the mutation is recessive.  This single mutant male should be mated to  multiple wildtype females. 

•  If the trait is recessive, none of the progeny (male of female) should show the new phenotype. 
•  If the heterozygous females have a mutant phenotype but that phenotype is different from that of their father, then the trait is incompletely dominant, overdominant or codominant.  The males will be wildtype.
• If the heterozygous females have a mutant phenotype and that phenotype is the same as that of their father of their father, then the trait is completely dominant.   The males will be wildtype.

Fixing a Recessive Allele:  If the mutation is recessive all of the daughters will have the genotype A₁A₂.  The only male with an A₁ allele, however is the father. If the father is still alive and still fertile, he should be mated back to his daughters (see below).  Half of the progeny from this cross will be mutant.  These mutants can be mass mated to produce a parental line.

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

Fixing an incompletely dominant, overdominant or codominant allele.

All of the daughters will have the genotype A₁A₂.  The only male with an A₁ allele, however,  is the father. As before, if the father is still alive and still fertile, he should be mated back to his daughters.  Half of the progeny from this cross will be mutant) corresponding to the A₁A₁ or A₁Y genotypes.  These mutants can be mass mated to produce a parental line.

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

Fixing a dominant allele requires a great deal of effort.  This is discussed on another page.

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.   It  is extremely unlikely that the new mutation is recessive (see above) so the female will be heterozygous.  This single female should be mated to several wildtype males.  This is a safety measure since one of the males might be infertile.  The phenotype of the males will give the female genotype.  If 1/2 of the males have a wildtype phenotype and the other 1/2 of the males have a mutant phenotype, then the female was heterozygous.  If the allele is dominant, 1/2 of the females will have a wildtype phenotype and the other 1/2 of the females have a mutant phenotype. If the heterozygote has its own phenotype, then the mutant phenotype in the males will differ from that of their mother and their heterozygous sisters.  In the latter case, the males can be mated to their sisters to establish a pure breeding  line (as above).  If the allele is dominant see the next page .

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

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