A Species of rodent has either white or tan fur (W, w) where tan is recessive. Monohybrid crosses result repeatedly in litters with twice as many white as tan offspring. Consider the lethal alleles and explain the 2:1 phenotype ratio.
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Although the question does not mention lethal alleles, it appears that the dominant homozygote genotype is lethal. so the original cross is Ww x Ww:
______W______w___
W____WW____Ww__
w____Ww_____ww___
Since the WW genotype is not viable, there are 2 Ww = white offspring for each 1 ww = tan offspring.
______W______w___
W____WW____Ww__
w____Ww_____ww___
Since the WW genotype is not viable, there are 2 Ww = white offspring for each 1 ww = tan offspring.
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So normally in a cross where 2 alleles are segregating independently, you would get 3 possible genotypes. You would get 1 WW, 2 Ww, and 1 ww
ww is recessive and those are your tan rodents.
Ww are your heterozygotes. They are the white rodents
For some reason, WW is lethal, normally these rodents would be white as well, but since they're dead, you don't see them. If they weren't lethal, you would expect to see 3 white for every 1 tan. Since you only see 2 white for every 1 tan, that's why you suspect a lethal allele.
ww is recessive and those are your tan rodents.
Ww are your heterozygotes. They are the white rodents
For some reason, WW is lethal, normally these rodents would be white as well, but since they're dead, you don't see them. If they weren't lethal, you would expect to see 3 white for every 1 tan. Since you only see 2 white for every 1 tan, that's why you suspect a lethal allele.