<span>A Simple Compound Microscope</span>
DNA<span> contains codes for proteins, which are necessary for the growth and functioning of an </span>organism<span>. </span>DNA<span> separates into long single strands that make up each part of an</span>organism<span>. </span>DNA<span> produces the energy an </span>organism<span> needs in order to grow. </span>DNA<span> folds into the nucleus of each of the </span>cells<span> of an </span>organism<span>.</span>
The answer is A because it talks abt the environment and stuff
The Translation initiated is <u>Option D.All of the listed answers are correct. </u>
At the initiation of translation ribosomes and tRNA bind to the mRNA. tRNA is located at the first docking site of the ribosome. The anticodon of this tRNA is complementary to the start codon of the mRNA where translation begins. After binding to the mRNA, the ribosome initiates translation at the start codon AUG and moves the mRNA transcript one codon at a time until it reaches the stop codon.
When tRNA recognizes and binds to the corresponding codon in the ribosome, it transfers the corresponding amino acid to the end of the growing amino acid chain. tRNA and ribosomes then continue to decode the mRNA molecule until the entire sequence is translated into protein. tRNA acts as an adapter molecule during the translation process. Formerly known as soluble RNA or sRNA. As an adapter, it connects amino acids to nucleic acids.
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Answer:
B) 1/2 X 1/2 = 1/4
Explanation:
It happens when both the parents are heterozygous for the trait. The cross related to this result is shown as under:
Father Mother
Bb x Bb
/ \ / \
Gametes: B b B b
| | | |
Probability: 1/2 1/2 1/2 1/2
The probability of formation of 'b' gamete from father is 1/2 because there are only two gametes 'B' and 'b' and out of these two one will be assorted as 'B' and another one as 'b' and from mother also the probability of formation of 'b' gamete is 1/2.
Now the assortment of gametes with each other is an independent event i.e. any gamete from father can fuse with any gamete of mother so the overall probability of formation of 'bb' genotype will be 1/2 x 1/2 = 1/4.
