1.

A woman heterozygous for colour blindness marries a colorblind man. What be the ratios of carrier daughter

Answer»
Hi mate.



Thanks for asking this question.



Here is your answer,



We have,



A heterozygous woman


I. E the genotype is =


x \: {x}^{i}


And the colorblinded MAN.



I. e genotype is =

{x}^{i } y


By crossing these two PARENTS.



We GET,


(1)
carrier \: women \: (x \: {x}^{i} ) \:is \: 25\%

(2)

infected \: women \: ( \: {x}^{i} \: {x}^{i}) \: is \: 25\%

(3)

infected \: man \: ( {x}^{i}y) \: is \: 25\%

(4)

normal \: man \: (x \: y) \: is \: 25\%



The conclusion is given by the crossing which

is given in the image.


Please see the image and then see the

conclusion.




THANK you.




BE BRAINLY !!!!!!!!!!!!!


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