The fraction of the offspring that will show the phenotype produced by the recessive c allele is 1024/4096 = 0.25 A-BbccDdE-Ff = 25%
Explanation:
You can calculate the fraction of the offspring that will show the phenotype produced by the recessive c allele by making the punnet square for each gene and then multipling the phenotypic proportions, like this:
Cross) AAbbCcDDEeff x AaBBCcddEEFF
Cross For each gene by separately:
Parental ) AA x Aa
Gametes) A A A a
F1) 2/4 AA
2/4 Aa
Parental ) bb x BB
Gametes) b b B b
F1) 4/4 Bb
Parental ) Cc x Cc
Gametes) C c C c
F1) 1/4 CC
2/4 Cc
1/4 cc
Parental ) DD x dd
Gametes) D D d d
F1) 4/4 Dd
Parental ) Ee x EE
Gametes) E e E E
F1) 2/4 Ee
2/4 EE
Parental ) ff x FF
Gametes) f f F F
F1) 4/4 Ff
So, fraction of the offspring that will show the phenotype produced by the recessive c allele is:
4/4 A- x 4/4 Bb x 1/4 cc x 4/4 Dd x 4/4 E- x 4/4 Ff =
The fraction of its offspring will show the phenotype produced by the c allele is 1/4
Explanation:
Although there are six genes shown in the question, only one of them, the "C" gene is our concern.
Because different genes assort independently, we can ignore all the others and just think about this one gene. Then, our cross is really just Cc × Cc. From there, it should be easy to figure out that ¾ of the offspring will be either CC or Cc and show the dominant "C" phenotype, while ¼ will be cc and show the recessive "c" phenotype.
