albinism in humans is controlled by a recessive gene 'a' if both paren...
parent A........... parent BAa. Aaoffsprings ~= AA, Aa, Aa, aa1/4 of total offspring would be expected to show the recessive trait ofalbinism..3/4 of total would be normal skin are...normal skin= 4×3/4×(1/4)^3=4×3/4×1/64=12/256 or 3/64..samje ho!!
albinism in humans is controlled by a recessive gene 'a' if both paren...
mujhe question aur options dono me doubt hi. kuch to correction hi.well maine jo logic lgaya hi wo bta rha hu.Albinism Autosomal Recessive disorder hi, means Albinism ka phenotype hme "aa" me dekhne ko milega.Parents "Aa" genotype ke hi, i.e. both parents are Carrier for Albinism.ab agr iska normal monohybrid cross krate hi to hme Genotype 1:2:1 ke ratio me milega, but Phenotype 3:1 me milega, i.e.Unaffected Individuals - 3/4Affected Individuals - 1/4ab question me 1 normal and 3 Albinism ka case puch rha hi, to single mating me to ye impossible hi. as we know ki 1 cross me more than 1 affected Individual ho hi nahi sakta.means ye to confirm hi ki total 4 cross krane padenge. to ab 1 affected individual ke chances 1/4 hi, to 3 affected individual ke liye chance 3/4 ho jaynge.similarly, 1 unaffected individual ke chance 3/4 hi.means 4 cross me 3 Affected and 1 non affected hone ke chance 3/4 hone chahiye.please let me know if there is any correction.
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