Solve this :

$$\begin{gathered} 5. \mathrm{When} \mathrm{a} \mathrm{hybrid} \left(\mathrm{AaBbCcDd}\right) \mathrm{is} \mathrm{selfed} \mathrm{then} \mathrm{the} \mathrm{genotypes} \mathrm{AABbCCDd}, \mathrm{AaBBCcDd}, \mathrm{AaBbCcDd}, \mathrm{aabbcdd} \mathrm{would} \mathrm{be} \mathrm{in} \mathrm{a} \mathrm{proportion} \mathrm{of} : \\ \left(1\right) 2 : 4 : 8 : 21 \left(2\right) 4 : 8 : 16 : 1 \\ \left(3\right) 4 : 8 : 16 : 27 \left(4\right) 8 : 4 : 16 : 81 \end{gathered}$$
 

Dear student,
Please find below the solution to the asked query

​The correct option is 2.

The probability of a heterozygote ( say Zz) = 2/4 or 1/2 and the probability of a homozygote (say ZZ or zz) = 1/4
The total number of genotypes is 256. Therefore, the probability of given genotypes will be


a. AABbCCDd =  $\frac{1}{4} \mathrm{x} \frac{1}{2} \mathrm{x} \frac{1}{4} \mathrm{x}\frac{1}{2} = \frac{1}{64}$,    1 out of 64 means 4 out of 256 (as 256/64 = 4)
b. AaBBCcDd =  $\frac{1}{2} \mathrm{x} \frac{1}{4}\mathrm{x}\frac{1}{2}\mathrm{x}\frac{1}{2}=\frac{1}{32}$,       1 out of 32 means 8 out of 256 (as 256/32 = 8)
c. AaBbCcDd =  $\frac{1}{2}\mathrm{x}\frac{1}{2}\mathrm{x}\frac{1}{2}\mathrm{x}\frac{1}{2}=\frac{1}{16}$,        1 out of 16 means 16 out of 256 (as 256/16 = 16)
d. aabbccdd =  $\frac{1}{4}x\frac{1}{4}x\frac{1}{4}x\frac{1}{4}=\frac{1}{256}$,        1 out of 256 means same.


Therefore, the genotypes AABbCCDd, AaBBCcDd, AaBbCcDd, aabbcdd would be in a proportion of 4 : 8 : 16 : 1

Hope this information clears your doubt about the topic.

Regards