In Drosophila, vestigial (partial formed) wings (vg) are recessive to normal long wings (vg+), and the gene for this trait is autosomal. The gene for the white eye trait is on the X chromosome. Suppose a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigial-winged male.
A. What will be the genotypes and phenotypes of F1?
B. What will be the genotypes and phenotypes of F2?
C. What will be the genotypes and phenotypes of a cross of F1 back to each parent?
F1 back to each parent?
wvg+ w+vg Yvg ♀ wvg, wvg+ , w+vg+, w+vg
w+wvg+vg ♀ w/Yvg+vg ♂ ♂wvg+, wvg, Yvg+, Yvg
3.A Color blindness is a recessive X-linked genetic disease and blood type is determine by autosomal gene. A man who had color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman's father had color blindness.
(a) What are the genotypes of the man and woman?
(b) What proportion of their children will have color blindness and type B blood?
(c) What proportion of their children will have color blindness and type A blood?
(d) What proportion of their children will have color blindness and type AB blood?
ww vg+vg+ X w+/Y vgvg
You are correct about the F1:
w+w vg+vg (all females are wild type) and w/Y vg+vg (all males are white eyed)
But now, what about the F2? We have to set up the Punnett Square. I've put the gametes for the female on topic and the gametes for the male down the left side. This is a very complicated Punnett Square and it requires a lot of detailed concentration to get it right. One slip and you mess up.
w+vg+ w+vg wvg+ wvg
wvg+ w+w vg+vg+ w+w vg+vg ww vg+vg+ ww vg+vg
wvg w+w vg+vg w+w vgvg ww vg+vg ww vgvg
Y vg+ w+Y vg+vg+ w+Y vg+vg wY vg+vg+ wY vg+vg
Y vg w+Y vg+vg w+Y vgvg wY vgvg+ wY vgvg
So, what do we have? The top two rows inside the square are the females, right? And the two bottom rows are the males.
You should be able to assign the phenotypes. I'll give you a few examples and you can do the rest:
w+w vg+vg wild type female
ww vg+vg ...