Cross Turquoise,5

Write the genotypes of all parents and progeny as completely as possible?

Beetles of a certain species may have green (wg), blue(wb), or turquoise(wt) wing covers. Virgin beetles were selected from a polymorphic laboratory population and mated to determine the inheritance of wing-cover color. The crosses and results were as follows:
Cross Parents Progeny
B=Blue G= Green T= Turquoises

1) shows that blue is dominant over green, no other way to get 100% offspring with two differently phenotyped parents.
2)shows there are some turquiose alleles there as turquoise was produced from two blue parents. This means blue is dominant over turquoise as the alleles weren’t expressed in the parents.
3)Same logic as 2) above, shows green is dominant over turquoise.

So these show order of dominance: B>G>T so we can work out genotypes.

1) all blue means no recessive alleles in blue parent, so they are BB. The green one could be either GG (So all progeny BG) or GT (half progeny BG and half BT).
2)the proportions show that each parent has a T allele as there was a chance to produce turqusoise offspring. So both are BT. (1/4 BB, 1/2 BT and 1/4 TT).
3)same as above, each must have a T allele as there’s Turquoise offspring so both are GT. (1/4GG, 1/2GT, 1/4 TT).
4)turquoise offspring produced means the blue parent must have a T allele. Turquoise parent must have two T alleles as that’s the only way to be turqoiuse. So BT and TT. (1/2BT and 1/2 TT)
5)both must have a green allele each to make green offspring. So both BG. (same as (2) but G rather than T)
6)blue parent must have g allele as there are g offspring. The green parent may or may not have a T allele as it is still recessive to both the B and G allele of the blue parent. So BG and GG(same as (4) but G rather than T) or GT (1/2 BG, 1/4 GG and 1/4 GT)
7)both parents must have T allele to produce turquoise offspring. so BT and GT (1/4 BG, 1/4BT, 1/4 GT and 1/4 TT)
8)TT and TT. (progeny TT too)

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