Answer:
Mother - bb XAXa
Father - Bb XAY
Son - Bb XaY
Daughter - bb XAXa
Explanation:
Given –
Mother has blue-eye and normal vision
Son has brown eye and is color blind
Let the allele for Blue eye be represented by “b” and the allele for brown eye be represented by “B”
Given brown (B) is dominant over blue (b), thus both BB and Bb represents brown color eyes
Color blindness is associated with X gene
Thus Xa represents X –linked recessive colorblindness
And XA represents normal vision
Genotype of blue-eye and normal vision mother is bb XAXa . Mother is the carrier of recessive allele Xa for color blindness
Since son is color blind , his genotype would be Bb XaY
So fathers genotype would be
Bb XAY
Daughters genotype bb XAXa
All of the above bc you can use all of those to transport
Answer:
I guess it is C.
Explanation:
Because a gene is the basic unit of heredity that determines a offspring's characteristic, my guess is that the answer is C
(Hope this helped!)
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