Question

How many types of zygotic combinations are possible between a cross $AaBBCcDd\times AAbbCcDD$?

• A. $32$
• B. $128$
• C. $64$
• D. $16$

Answer 1

The correct option is D $16$

Number of gamete possible in a parent is 2${}^n$ (where n- number of heterozygous allele in parent genotype)

number of zygotes possible are (gamete)X(gamete)

the number of gamete formation possible by parent 1 (AaBBCcDd) - 2${}^3$ = 8 (n=number of heterozygous allele = 3)

the number of gamete formation possible by parent 2 (AAbbCcDD) - 2${}^1$ = 2 (n=number of heterozygous allele = 1)

So, the number of zygotes possible are = gamete of parent 1 X gamete of parent 2

= 8 X 2

= 16

So, the correct answer to the question is '16'.