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Home Q & A Board Biology-Related Homework Help Genetics and Developmental Biology
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happytime happytime
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9 years ago Edited: 9 years ago, happytime
The seeds in bush bean pods are each the product of an independent fertilization event. Green seed color is dominant to white seed color in bush beans.

If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny plant will consist of ...

4 green and 2 white seeds?

My thought process: I assume the order matters, so WGGGGW and GWGGGW are two of the many possibilities. I got a total of 12 possibilities. I know it's 0.75 chance for green and 0.25 chance for white. For the six seeds, I did (.75*.75*.75*.75*.25*.25) and got .01977. I then did .01977^12 due to 12 possibilities and got 3.57*10^-21.

Please help. ><
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rsb
wrote...
#1
9 years ago
Cross:

Gg x Gg

Phenotypes:

1/4 GG

1/2 Gg

1/4 gg


a)

Probability of 3 green seeds: 3 x 1/4 x 1/2 = 3/8

Probability of 3 white seeds: 3 x 1/4 = 3/4

b)

Probability of all green seeds: 6 x 1/4 x 1/2 = 6/8 OR 3/4

c)

Probability of 1 white seed: 1/4


Hope this helps! Slight Smile
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wizarddimwizarddim
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#2
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9 years ago
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happytime Author
wrote...
#3
9 years ago
Quote from: rsb (9 years ago)
Cross:

Gg x Gg

Phenotypes:

1/4 GG

1/2 Gg

1/4 gg


a)

Probability of 3 green seeds: 3 x 1/4 x 1/2 = 3/8

Probability of 3 white seeds: 3 x 1/4 = 3/4

b)

Probability of all green seeds: 6 x 1/4 x 1/2 = 6/8 OR 3/4

c)

Probability of 1 white seed: 1/4


Hope this helps! Slight Smile

How about 4 green and 2 white?

And I'm not so sure Pascal's triangle can be used since the probability of each successes are not the same.
wrote...
#4
3 years ago
Thank you!
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