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lesya [120]
4 years ago
7

Given the following genotypes for two parents, AABBCc × AabbCc, assume that all traits exhibit simple dominance and independent

assortment. What proportion of the progeny of this cross will be expected to phenotypically resemble the first parent with the genotype AABBCc?
Biology
1 answer:
LuckyWell [14K]4 years ago
5 0

Answer:

When all the conditions cit e in the question is applicable, which are the independent assortment and simple dominance, a cross between these two parents will lead to an expectation of 75% of the progeny to resemble the parent with a genotype AABBCc.

Explanation:

Independent assortment is a phenomenon in which the alleles particular for a trait in an organism can separate into two independently to each other during the process of meiosis, while simple dominance is a term that help to express the occurrence of both dominant and recessive allele for a particular trait, where the presence of at least one of the dominant allele will lead to the expression of the trait. Having known the meaning of these conditions, we then apply it to solving the question.

Parent 1's genotype=AABBCc

Assorted alleles to be produced=ABC and ABc.

Parent 2's genotype=AabbCc

Assorted alleles to be produced=AbC, Abc, abC, abc.

Crossing these alleles using the punnets square will lead to: AABbCC, AABbCc, AaBbCC, AaBbCc, AABbCc, AABbcc, AaBbCc, AaBbcc. Which have a genotypic ratio of=1AABbCC: 2AABbCc: 1AABbcc: 1AaBbCC: 2AaBbCc: 1AaBbcc. What is expected from the question is phenotypic ratio which is 6ABC, and 2ABc. Therefore, the percentage of those that will resemble the first parent will be 6/8 X 100 = 75%.

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