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%
They are both made up of organisms
Answer:
start codon - AUG
in-sequence amino acid - AAA, UGC, AUC, CAC, GCA, ACU
stop codon - UAA, UAG, UGA
Explanation:
From the amino acid chart, the start codon is represented by the codon AUG which also code for the amino acid methionine. There are three stop codons including UAA, UAG, and UGA. The stop codons signal the end of the translation process of the mRNA. Every other codon within the mRNA chain codes for different amino acids except methionine.
Hence, using the attached amino acid chart:
start codon - AUG
in-sequence amino acid - AAA, UGC, AUC, CAC, GCA, ACU
stop codon - UAA, UAG, UGA
<em>C) Winter.</em>
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<em>I used someone else's answer and it was incorrect so I did it myself</em>
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