For AaBBCcDd x AaBbCcdd what is the probability that any ind

For AaBBCcDd x AaBbCcdd,

what is the probability that any individual offspring will the dominant phenotype for all traits?

I understand the branched method of calculating probabilities,

but feel I can speed up the process instead of caluclating each individual type and then adding them,

my question is do I add up the probabilities for AA and Aa, so 1/4 + 1/2, or multiply them,

and then do I multiply or add those to the probability of phenotypically dominant B, which is 1/2 BB and 1/2 Bb, so 1,

and so on until I get to D.

EDIT: okay, so I saw another similar question answered where I just use the punnett square and multiply each individual probablilities for the Aa, Bb, and then multiply those together, so my question is, instead of having to do those for the first aaBbCcDdee and then the aabbCCddEe, is there a way to find out both in one fell swoop?

Thanks for clearing that up,

please explain any logic used,

Max

Solution

the probaility of AA and Aa are to be added as both are possible , which then has to be multiplied to BB + Bb(the probality of gene B) as that has to be incorporated.