Given the mRNA sequence ,5’-AUGUACAAGGUCGGAUGA-3’ which of the following amino acid sequence would result from translation? Tyr-met-val-lys-gly Met-val-lys-tyr-gly Met-tyr-lys-val-gly Ser-arg-leu-glu-his-val Gly-met-val-lys-tyr
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Given the mRNA sequence ,5’-AUGUACAAGGUCGGAUGA-3’ which of the following amino acid sequence would result from translation?
Tyr-met-val-lys-gly
Met-val-lys-tyr-gly
Met-tyr-lys-val-gly
Ser-arg-leu-glu-his-val
Gly-met-val-lys-tyr
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- A certain mRNA strand has the following nucleotide sequence: 5AUGACGUAUAACUUU3 What is the anticodon for each codon? What is the amino acid sequence of the polypeptide? (Use Figure 13-5 to help answer this question.) Figure 13-5 The genetic code The genetic code specifies all possible combinations of the three bases that compose codons in mRNA. Of the 64 possible codons, 61 specify amino acids (see Figure 3-17 for an explanation of abbreviations). The codon AUG specifies the amino acid methionine and also signals the ribosome to initiate translation (start). Three codonsUAA, UGA, and UAGdo not specify amino acids; they terminate polypeptide synthesis (stop).Given the following mRNA, write the double-stranded DNA segment that served as the template. Indicate both the 5 and the 3 ends of both DNA strands. Also write out the tRNA anticodons and the amino acid sequence of the protein encoded by the mRNA message. DNA: mRNA: 5-CCGCAUGUUCAGUGGGCGUAAACACUGA-3 protein: tRNA:Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?
- Which amino acid sequence will be generated during translation from the following small mRNA: …CCC-AUG-UCU- UCG-UUA-UGA-UUG…? (Hint: Remember where translation starts and stops.) (a) Met-Glu-Arg-Arg-Glu-Leu (b) Met-Ser-Ser-Leu-Leu (c) Pro-Met-Ser-Ser-Leu-Leu (d) Pro-Met-Ser-Ser-Leu (e) Met-Ser-Ser-LeuTranslate the mRNA into a protein using the genetic code. (Amino acid chain.) AUG( met)-UUU( Phe)-GUA( Val) - CAU (his)-UUG(leu) -UGU(cys)- GGG(gly)- AGU(ser)- CAC (his) - CUG(leu) GUU(Val)- GAG (Glu)-GCG(ala)- UUG(Leu)- UAU(tyr)- UUG (leu)-GUU(Val)- UGU(cys)- GGC(gly)-GAG( Glu)-CGC(arg) - GGC(gly) - UUU(phe) - UUC(phe).During the termination of translation, what is the correct polypeptide sequence which will be released by the ribosome? 5' - AUG - UAU - CUC - UUU - 3' (mRNA codon sequence) 3' - UAC - AUA - GAG - AAA - 5' (tRNA anticodon sequence) A.START - Tyr - Leu - Phe B. Tyr - Ile - Glu - Lys C. START - Ile - Glu - Lys D. Ser - Tyr - Gly - Cys
- UAA is a stop codon. Why does the UAA sequence in the segment of mRNA 5′-G-C-A-U-G-G-A-C-C-C-C-G-U-U-A-U-U-A-A-A-C-A-C-3′not cause protein synthesis to stop?For the following mRNA sequence (reading from left to right) what will be the amino acid sequence following translation? (Use chart provided) AUGCCAGUUGAAUAA A. Pro-Val-Met-Leu-His B. Met-Val-Tyr-Pro C. Met-Pro-Val-Glu D. Correct answer not given E. Met-His-Phe-Ala-ArgTranslate the following mRNA: 5-A U G A A A U U U C U U U A G G U C G A A -3 NH3+- Met-Leu-Phe-Val- COO- NH3+- Met-Thr-Val-Ser- COO- NH3+- Met-Glu-Gln-Ser- COO- NH3+- Met-Asp-Ser-Pro- COO- NH3+- Met-Lys-Phe-Leu- COO-
- During transcription, a portion of mRNA is synthesized with the following base sequence. 5'-AUG-GUA-CCA-CAU-UUG-UGA-3' Q.) Write the primary structure of the polypeptide coded for by this section of mRNA.For the codon sequence : 5’ GGA – AUA – UGG – UUC – CUA – 3’ Write the Amino acid sequence produced in each of the following ways: Translation proceeds in a normal manner A mutation changes GGA to GGG A mutation changes GGA to CGAThe template strand of a double helical segment of DNA consists of the following sequence: 5’-GTAGCCTTAAGCGATCACCGTCCGTATTACTAGTGGCCAGACTCTTTTCACTCTCATGTATAGTTG-3’ What is the nucleotide order in the complementary mRNA that can be transcribed from the given DNA strand?
