Hello Mahmood; I haven't done a Mendelian genetics problem for many years! Here's a try. Grey is dominant and black is recessive,

Normal wing is dominant and small wing is recessive.

To get 3:1 rations in the F1s both parents must be heterozygous for both traits...something like GgNn x GgNn. Make a punnett square out of that and I think it works. What do you think? Best regards, Jim Des Lauriers

thanks alot for your answer James Des Lauriers , i have answer what do you think about it?

1. For grey colour it is a case of incomplete dominance , the grey colour (AB) results from combination of black colour (BB) and white colour (AA), For the wings it is a case of complete dominance , as the normal wing is completely dominant to the undeveloped wing (recessive gene) 2. consider that dominant gene for normal wings (NN) and the recessive gene of undeveloped wing (nn), grey colour (AB) , black colour (BB) , white colour (AA), So the genotype of parents is (ABNn) for both parents .

Mahmood; I agree with the way you did the wing trait. However, if you do the color trait as incompletely dominant, then you should get some white F1s and the question doesn't mention anything about white. So, I'd guess that they wouldn't accept that answer.

My answer assumes that the traits are autosomal dominant ones and that the parents are both heterozygous for both traits. I also assumed that the remark about females is true but is a trick and that about 25% of the males would also have small wings. If that isn't true, then there isn't enough information to solve the problem…I think.

P1 GgNn x GgNn Their gametes would look like…

GN Gn gN gn

GN GGNN GGNn GgNN GgNn

Gn GGNn GGnn GgNn Ggnn

gN GgNN GgNn ggNN ggNn

gn GgNn Ggnn ggNn ggnn

That arrangement gives you the 3:1 ratio that the question wants.

Incidentally, what stimulated you to ask this question?

Best regards, Jim Des Lauriers

Nice Dear James Des Lauriers