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Sagot :
We have then the following:
Parent 1 (XX)
Parent 2 (XY)
Offspring 50 % XX (red-green colour defficiency)
50 % XY
Assumming that this condition is linked to the sexual chromosome of the parent (XX), since the probability of the offspring is 50 % XX, we have:
RR: expresed gene
Rr: " " "
rr: non-expresed gene
We say that it's a gene linked to the sexual chromosome because is a gene present in those ones. In other words, males who inherit only one X chromosome from their mother, and a Y chromosome from their father, and females who inherit both X chromosome from their both parents, have differents rates to inherit the pathology, since there must be at least two chromosomes with the gene of that trait in women to develop that condition, while in men if the inherited chromosome carries the gene bearing the patalogy, they would have that condition.
So the genotypes of the parent 1 (XX, female) would be RG or rg, and the parent 2 (XY) being rg
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