Exactly 989527/1048576, or approximately 94.37%
Since each trait is carried on a different chromosome, the two traits are independent of each other. Since both parents are heterozygous for the trait, each parent can contribute 1 of a possible 4 combinations of the alleles. So there are 16 possible offspring. I'll use "a", "A", "b", "B" to represent each allele and the possible children are aabb, aabB, aaBb, aaBB, aAbb, aAbB, aABb, aABB, Aabb, AabB, AaBb, AaBB, AAbb, AAbB, AABb, and AABB
Of the above 16 possibilities, there are 7 that are homozygous in an undesired traint and 9 that don't exhibit the undesired trait. So let's first calculate the probability of "what are the chances that all 5 children not exhibiting an undesired trait?" and then subtract that result from 1. So
1-(9/16)^5 = 1 - 59049/1048576 = 989527/1048576 which is approximately 0.943686485 = 94.3686485%
So the answer is exactly 989527/1048576, or approximately 94.37%
<span>b. faults
Thank me if I was right!
Hoped it helped!
</span><span>
</span>
Answer:
Answer b is the right one
Answer:
B
Explanation:
ribosomes allow the Rough ER to make protiens
