Answer:
1/64
Explanation:
Parent 1: AaBbCc
Parent 2 : AaBbCc
When the cross is checked for individual gene:
Aa X Aa =
A a
A AA Aa
a Aa aa
Probability of producing an AA organism is 1/4
Similarly, probability of producing a BB from Bb X Bb would be 1/4
Probability of producing a CC from Cc X Cc would be 1/4
Hence, probability of producing an AABBCC would be:
1/4 * 1/4 * 1/4 = 1/64
The forked line method is an efficient way of getting the genotypic and phenotypic ratios when making crosses that involve many genes. In the exposed example, the probability of getting the genotype AABBCC is 1/64.
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We will assume these are autosomal genes that assort independently and express complete dominance.
We need to get the genotype AABBCC.
To avoid making a Punnett square, we can just use the the forked line method.
We will perform crosses separately and then multiply the combined proportions to get the possible genotypes of the progeny.
Lets us see,
Gene A
Parental) Aa x Aa
F1) 1/4 AA, 1/2 Aa, 14 aa
Gene B
Parental) Bb x Bb
F1) 1/4 BB, 1/2 Bb, 14 bb
Gene C
Parental) Cc x Cc
F1) 1/4 CC, 1/2 Cc, 14 cc
Possible genotypes
.
.
.
.
And so on, until you get the genotypes of the whole progeny.
Note: In the attached files you will find a better image.
To the goal of this problem, we just need to get the probability of having individuals with the genotype AABBCC. ⇒⇒ 1/64
The probability of producing the genotype AABBCC is 1/64.
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