Experts please tell how to solve this

Dear students.

Number of gametes = 2n.  n = number of heterozygous pair.
in this number of  heterozygous pair. are 3 ( Aa , Bb ,Dd)

Therefore, gamete will be 23 = 8. aBCDE, abCDE, abCdE, aBCdE , ABCDE, AbCDE , AbCdE and ABCdE.

8 kinds of gametes will be formed, if A and B are linked then gametes have AB and ab ,no combination such aB and Ab are not formed . 
so number of gametes are  4 , ABCDE, ABCdE , abCDE and abCdE
or 2n-m
 where n is 
number of  heterozygous pair
m is number of gene linked .

Regards.

 

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