If you conducted the analysis portion of Experiment 2, you calculated
1 answer:
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Answer: For three traits inherited in a dominant/recessive pattern, the probability of an offspring in a trihybrid cross between parents heterozygous for all three traits to be recessive at exactly two of the three loci is 23/64.
Explanation: Let A, B and C represent the dominant alleles and a, b and c represent the recessive alleles. The genotype of each of the parent will be AaBbCc since it has been said to be heterozygous for all the three traits. There are eight (8) possible different types of gametes from each of the genotype. This can be calculated using 2^n where n is the number of heterozygous loci.
The number of heterozygous loci in the genotype AaBbCc is three (3). 2^3 = 8. The gametes are;
ABC
ABc
AbC
Abc
aBC
aBc
abC and
abc.
Therefore, sixty-four possible offsprings are expected from the cross AaBbCc x AaBbCc.
Out of these offsprings, twenty-three (23) are expected to be heterozygous at two loci as boldly shown in the attached image.
