23 i am pretty sure i am studying that right now in science so
<span>mRNA: UACAUGGCCUUACGCUAA
tRNA: AUG UAC CGG AAU GCG AUU
a.a: Tyrosine, Methionine, Alanine, Leucine, and Arginine
DNA has 4 different bases, they are Adenine (A), cytosine (C), guanine (G), and Thymine (T). RNA also has 4 bases with three of them being identical to the DNA bases and Thymine being replaced with Uracil (U). These bases are generally represented by the 1st letter of their names. Each of the bases will join with a complementary base, so A always pairs with T or U, and C will pair with G. So to create the mRNA, simply replace every A with a U, every C with a G, every G with a C, and finally, every T with a A. So
mRNA: UACAUGGCCUUACGCUAA
Now for tRNA, there's a slight twist. It only comes in 3 base codons, You won't find a sequence of tRNA other than in 3 base codons. And each of those codons will be uniquely paired with an amino acid. In the ribosomes, the mRNA will be sequentially scanned 3 bases at a time allowing for a matching tRNA sequence to bind to the exposed 3 bases, this will cause the next amino acid to be bound into the protein being constructed. So split the mRNA into 3 base sequences and calculate the complement to get the tRNA. A simple shortcut is to look at the original DNA sequence and simply replace a T bases with U. So
tRNA: AUG UAC CGG AAU GCG AUU
Notice the spaces every 3rd base. THIS IS REQUIRED. These is no continuous length of tRNA. You'll only find it in 3 base lengths and each of them will be bound with an amino acid.
For the amino acid that's coded to the RNA, you'll need to use a lookup table in your text book, or one you can find online. Then it's a simple matter of matching each 3 base sequence to the amino acid. For the sequence given we have:
AUG - Tyrosine
UAC - Methionine
CGG - Alanine
AAU - Leucine
GCG - Arginine
AUU - STOP
Notice the AUU doesn't decode to a specific amino acid. It instead indicates to the ribosome to stop the production of the protein. So the amino acid sequence for the originally given DNA sequence is:
Tyrosine, Methionine, Alanine, Leucine, and Arginine.</span>
Answer:
the relative frequency of an allele at a particular locus in population
Explanation:
What I know is that amino acid has side chain, amine groupe ,and groupe of carboxyl so what make them difference is side group where the is 20 different R-group which give each individual characteristics
