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mylen [45]
3 years ago
7

(Mendelian Genetics) Consider three gene pairs Aa, Bb, and Cc. The uppercase letter is the dominant allele and each gene pair as

sorts independently of each other. What is the probability of obtaining an offspring displaying the A B C phenotype (that is having the dominant phenotypes for each of the three genes) from a cross of Aa Bb CC X Aa Bb cc parents?
Biology
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

9/16

Explanation:

The cross was Aa Bb CC X Aa Bb cc

Because genes assort independently, we can analyze each of them separately in the cross, calculate the probability of getting offpsring with a dominant allele, and then multiply each of those probabilities to obtain the overall probability of having A_B_C_ progeny.

<u>A/a</u>

Aa x Aa ---> 3/4 A_, 1/4 aa

<u>B/b</u>

Bb x Bb ---> 3/4 B_, 1/4 bb

<u>C/c</u>

CC x cc ---> 1 C_

3/4 × 3/4 × 1 = 9/16 A_B_C_ offspring.

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Answer:

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Explanation:

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The genotype of a blue-eyed, dimpled chin child would be either of ppDd or ppDD, but in this case, the mother is ppdd and as such, the child cannot have two copies of D allele. Hence, the genotype of the child can only be ppDd.

In order for the child to have blue eyes (pp), it means that the father has to have the non-pigmented allele (p). This also means that the father is heterozygous for eye pigmentation (Pp).

<em>Hence, the genotype of the father is now limited to </em><em>PpDD</em> <em>and </em><em>PpDd</em>.

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2 years ago
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vaieri [72.5K]

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