Drosophila SAC

[ggxoxo]

Hi

could someone please explain to me how to prove one allele is dominant over the other while only using ratios as evidence?

Thanks

[Russ]

If it's dominant, it'll always be expressed if present right? So if you cross two heterozygotes (Aa with Aa) then you'd expect a 3:1 ratio of the dominant phenotype. AA, Aa, aA will be the "A" phenotype and only aa will be the "a" phenotype.

If it's not dominant, you'd expect something different - AA will still be the "A" phenotype and aa will still be the "a" phenotype, but the other two will be different (what they are exactly will depend on the relationship). So if you see a 1:2:1 ratio, there's no complete dominance

[ggxoxo]

Does this hold true for sex-linked inheritance?

[scar]

Kinda.   Females can still have the Hetrozygous form  (ie. they can be 'carriers' of a recessive trait and not show it themselves) since they have 2 X chromosomes to play with.   Males only have 1 X so they can't be carriers.  The term for only 1 chromosome is Hemizygous - a term that often crops up in exams!!

[simpak]

An easier way is to just cross purebreeding mutants with purebreeding wild types and if 100% of the offspring are wild type then the mutation is recessive (the heterozygote shows the wild type phenotype).

For sex linked inheritance, you would look at the phenotype of the heterozygous female (if X linked, so most common).  So you would cross a purebreeding wild type female with a recessive male and if 100% of /females/ show the wild type phenotype then the gene is X linked recessive.  Here it wouldn't be useful to use the male because, as stated above, he is hemizygous so he will show a phenotype resulting from whatever allele he has regardless of whether it is 'dominant' or 'recessive'.