A man who has the trait for colour blindness marries a normal woman. What percentage of their children would be sufferers, carriers and normal respectively
25%, 25% and 50%
25%, 50% and 25%
50%, 25% and 25%
25%, 37.5% and 37%
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Discussions (37)
no option is correct... it should be 0% suffer, 50% Carrier and 50% normal....
let the trait for color blindness be Tt
normal to be TT
Tt × TT
TT, TT, Tt and Tt
no option is right
it should be ,50% sufferer, 50% carrie and 0% normal
if X⁵ is the trait for color blind..
X⁵X⁵. XY
X⁵X, X⁵Y. X⁵X,X⁵Y
all the male will be carrier since it affect only one X, All the female will be sufferer sine Y chromosomes is not s*x link (female will always be either sufferer or normal,.
so its 50,50 and 0........ no noormal
no amount of crossing can give you the answer to this question
because the question itself is wrong
males cannot be carriers for s*x linked traits
it's either they have the disease or they don't because they have only one X chromosome which expresses itself fully
in my own understanding, s*x linked characters are located on the X chromosome. a man has XY chromosome, so its either his a suffer or his normal
i dont think the question is correct.
the answer s supposed to be
Normal:50%
carriers:50%
sufferers:0
because the males are not havin s*x linked characters on their X chromosome in this situations...hence it would have been fully expressed..and the females are the carriers and nobody is fully suffering it because the mom is normal
The answer is wrong, it's B.
Hereditary.
Let's use a trait for Colour blindness as Aa
Then a Normal as Aa because u don't know if the woman is homozygous normal or heterozygous normal, so the best to use is Heterozygous normal
Aa X Aa
The products are AA, Aa, Aa, aa
Therefore aa is the sufferer and it is just one i.e 25%
Aa is the carrier and it is two i.e 50%
AA is the Normal and it is one i.e 25%.
