Question

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

Answer 1

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 !!!!!!!!!!!!!