We can't answer as we do not know what book you are referring to.
The newly spliced mRNA binds to a ribosome. tRNA molecules migrate towards the ribosome, these tRNA molecules carries a specific amino acid. The ribosome allows two tRNA molecules into the ribosome at a time. The tRNA molecules have complementary anti-codons to the codons present on the mRNA strand. Two tRNA move into the ribosome and their anti-codons join to complementary codons on the mRNA strand. As one molecule leaves the ribosome, its amino acid forms a peptide bond with an amino acid on the adjacent tRNA molecule, with the help of ATP and an enzyme. As the ribosome moves along the the mRNA strand, a polypeptide chain is created. The ribosome stops reading the mRNA strand when it reaches a stop codon.
AAUUCCGG
Proteins are formed through transcription, translation, and protein synthesis. In transcription, where a strand of DNA is read by mRNA polymerase, which copies the DNA codes (adenine, thymine, guanine, and cytosine) by making their opposites (ex. DNA strand has thymine, so mRNA strand has adenine, but thymine is NOT on mRNA but is instead called uracil). Next in translation, the mRNA leaves the cell's nucleus. The codons on the strand (set of 3 nucleobases, ex. UUU or AAA, etc.) call for an anticodon (a set of codons carrying a protein) to attach to it. For instance if a codon is read UUU, then the anticodon carried the AAA protein. Another example, if the codons read AUG, the anticodon reads UAC. The anticodons bring in proteins until the mRNA has no more codons to copy. And thats how proteins are made in organisms.
Because of the asymmetry in pyrimidine and purine use in coding sequences, the strand with the greater coding content will tend to have the greater number of purine bases (Szybalski's rule). Because the number of purine bases will to a very good approximation equal the number of their complementary pyrimidines within the same strand and because the coding sequences occupy 80-90% of the strand, there appears to be a selective pressure on the third base to minimize the number of purine bases in the strand with the greater coding content and that this pressure is proportional to the mismatch in the length of the coding sequences between the two strands. The origin of the deviation from Chargaff's rule in the organelles has been suggested to be a consequence of the mechanism of replication. During replication the DNA strands separate. In single stranded DNA, cytosine spontaneously slowly deaminates to adenosine (a C to A transversion). The longer the strands are separated the greater the quantity of deamination. For reasons that are not yet clear the strands tend to exist longer in single form in mitochondria than in chromsomal DNA. This process tends to yield one strand that is enriched in guanine (G) and thymine (T) with its complement enriched in cytosine (C) and adenosine (A), and this process may have given rise to the deviations found in the mitochondria. Chargaff's second rule appears to be the consequence of a more complex parity rule: within a single strand of DNA any oligonucleotide is present in equal numbers to its reverse complementary nucleotide. Because of the computational requirements this has not been verified in all genomes for all oligonucleotides. It has been verified for triplet oligonucleotides for a large data set. Albrecht-Buehler has suggested that this rule is the consequence of genomes evolving by a process of inversion and transposition. This process does not appear to have acted on the mitochondrial genomes. Chargaff's second parity rule appears to be extended from the nucleotide-level to populations of codon triplets, in the case of whole single-stranded Human genome DNA A kind of "codon-level second Chargaff's parity rule" is proposed as follows: Codon populations where 1st base position is T are identical to codon populations where 3rd base position is A: « % codons Twx ~ % codons yzA » (where Twx and yzA are mirror codons i.e TCG and CGA). Codon populations where 1st base position is C are identical to codon populations where 3rd base position is G: « % codons Cwx ~ % codons yzG » (where Cwx and yzG are mirror codons i.e CTA and TAG). Codon populations where 2nd base position is T are identical to codon populations where 2nd base position is A: « % codons wTx ~ % codons yAz » (where wTx and yAz are mirror codons i.e CTG and CAG). Codon populations where 2nd base position is C are identical to codon populations where 2nd base position is G: « % codons wCx ~ % codons yGz » (where wCx and yGz are mirror codons i.e TCT and AGA). Codon populations where 3rd base position is T are identical to codon populations where 1st base position is A: « % codons wxT ~ % codons Ayz » (where wxT and Ayz are mirror codons i.e CTT and AAG). Codon populations where 3rd base position is C are identical to codon populations where 1st base position is G: « % codons wxC ~ % codons Gyz » (where wxC and Gyz are mirror codons i.e GGC and GCC).
Translation ends when a stop codon is reached. The stop codons are: * UAA * UAG * UGA
We can't answer as we do not know what book you are referring to.
Codons are found on messenger RNA, while anticodons are found on transfer RNA
I think codons are found on dna. Anticodons are found only on trna.
The newly spliced mRNA binds to a ribosome. tRNA molecules migrate towards the ribosome, these tRNA molecules carries a specific amino acid. The ribosome allows two tRNA molecules into the ribosome at a time. The tRNA molecules have complementary anti-codons to the codons present on the mRNA strand. Two tRNA move into the ribosome and their anti-codons join to complementary codons on the mRNA strand. As one molecule leaves the ribosome, its amino acid forms a peptide bond with an amino acid on the adjacent tRNA molecule, with the help of ATP and an enzyme. As the ribosome moves along the the mRNA strand, a polypeptide chain is created. The ribosome stops reading the mRNA strand when it reaches a stop codon.
termination codons
termination codons
A codon consists out of a 3 nucleotid long DNA piece. That 3 nucleotids code for an amino acid.
Assuming it's 5' to 3', The complementary strand would be 3' G-A-A-T-C-C-G-A-A-T-G-G-T 5'
The DNA strand contains the genetic information needed for cellular function.
The complimentary strand of MRNA would be AAUUCCGG.
The complementary (partner) strand to the segment ACTGT would be TGACA. This is because in DNA, A binds to T and C binds to G.
Eukaryotes are cells in which DNA is contained in a nucleus. Codons describe sections of 3 base pairs in DNA which code for an amino acid. So, anything with DNA has codons, therefore eukaryotes have codons.