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4.Principles of Inheritance and Variation
medium
A girl of normal vision whose father was colourblind marries a man of normal vision whose father was also colourblind. Their sons would be (of total number of $sons$)
A
All colourblind
B
$50\%$ colourblind
C
All normal
D
$25\%$ colourblind
Solution
(b) Infact the girl with normal vision is carrier because her father is colourblind (daughter of colourblind father are either colourblind or carrier) and when she marries a normal man the possibility of their sons being colourblind is $50\%$ because the genotype of parents is $X^CX$ and $XY$,
so only half of the possible combinations of $XY$ have the $X$-linked recessive genes which exhibit the disease.
Standard 12
Biology
Similar Questions
:Match the columns :
Column $I$ |
Column $II$ |
$(1)$ Albinism | $(p)$ recessive gene represented by HbsHbs. |
$(2)$ Phenyl Ketonuria |
$(q)$ recessive gene represented by cc. |
$(3)$ Sickle cell anaemia |
$(r)$ recessive gene represented by aa. |
$4)$ Alkaptonuria | $(s)$ recessive gene represented by pp |
medium