Question

When a hybrid (AaBbCeDd) is selfed then the
genotypes AABBCCDd, AaBBCcDd,
AaBbCeDd, aabbccdd would be in a proportion
of
(1) 2:4:8:21 (2) 4:2:1:16
(3) 4:8:16:27 (4) 8:4:16:81​

Answer 1

Answer:

1.... 2:4:8:21......

Answer 2

The given question have a mistake.

When a hybrid (AaBbCcDd) is selfed then the  genotypes AABBCCDd, AaBBCcDd, AaBbCcDd, aabbccdd would be in a proportion  of 2:8:16:1

• Aa x Aa = AA(1/4) : Aa(1/2) : aa(1/4)

• Bb x Bb = BB(1/4) : Bb(1/2) : bb(1/4)

• Cc x Cc = CC(1/4) : Cc(1/2) : cc(1/4)

• Dd x Dd = DD(1/4) : Dd(1/2) : dd(1/4)
• The numbers in the bracket shows the probability of getting the phenotype
• Outcome of every gene is independent of each other so here product rule will be applied.
• The probability of outcome of

1. AABBCCDd =1/4*1/4*1/4*1/2 = 1/128
2. AaBBCcDd =1/2*1/4*1/2*1/2 = 1/32
3. AaBbCcDd =1/2*1/2*1/2*1/2 = 1/16
4. aabbccdd = 1/4*1/4*1/4*1/4 = 1/256

• The ratio will be 2:8:16:1