how many different types of gametes can be formed by f1 progeny resulting from the following cross AABBCC ×aabbcc also state the reason a) 3b)8 c)27d)64
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A cross b/w AABBCC and aabbcc gives the result :
AaBbCc
F1 :AaBbCc
We know, types of gametes is given by 2^n
Where n= no of heterozygous
.
Here n=3
So no of gametes: 2^3=8 ie. option(b)
AaBbCc
F1 :AaBbCc
We know, types of gametes is given by 2^n
Where n= no of heterozygous
.
Here n=3
So no of gametes: 2^3=8 ie. option(b)
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