5. A DNA sequence of "ACG" will code for the amino acid- (LS1-1)*Second MRNA baseGUUUPheUUCUCUUAUUGU .TyrCysUACSerUAA Stop UGA Stop AUGCUUAUCALeuUUGA EUCGUAG Stop UGG TrpGCGU 7HisCACCUUCCUCAUAGUCUCLeuCUACCCProCCACGCCye (C)AngCAACGAGinCUGCCGCAGCGGG.A CLaAUUAAUAshAACTheAAAAGUSerAGCACUAUC TeACCUGALIAACAAGAArgAGAUG MACGAAGGCUGCValGCAGAUAspGACAlaGAASluGAGGUUGGJGUCGlyGUAGUGGCGGGGCysThrTyrProOIt will not code for an amino acidFirst mRNA base (5' end of codon)Third MRNA base (3' end of codon)

Gene expression refers to the complex, highly-regulated biological process, which involves the…

5. A DNA sequence of "ACG" will code for the amino acid - (LS1- 1) * Second mRNA base C. UUU Phe UUC UCU UAU UGU Tyr Cys UCC Ser UCA UAC UGC UUA Leu UUG UAA Stop UGA Stop A A E UCG UAG Stop UIGG Trp AGU CUU CCU CAU CGU 7 His CAC CUC Leu CUA CCC Pro CCA CGC Ang Ca (C) CAA CGA Gin GU CUG CCG CAG CGG Trp(W G A C A AUU ACU AAU AGU Asn AUC le AAC The AGC ACC ACA AGA Arg AGG AIA AAA AUG Me ACG AAG GUU GCU GGJ GAU Asp GAC GGC Gly GGA GUC Val GUA GCC Ala GCA GAA Glu GAG GUG GCG GGG G. Cys O Thr O Tyr O Pro It will not code for an amino acid First mRNA base (5' end of codon) Third mRNA base (3' end of codon) 30

5. A DNA sequence of "ACG" will code for the amino acid - (LS1- 1) * Second mRNA base C. UUU Phe UUC UCU UAU UGU Tyr Cys UCC Ser UCA UAC UGC UUA Leu UUG UAA Stop UGA Stop A A E UCG UAG Stop UIGG Trp AGU CUU CCU CAU CGU 7 His CAC CUC Leu CUA CCC Pro CCA CGC Ang Ca (C) CAA CGA Gin GU CUG CCG CAG CGG Trp(W G A C A AUU ACU AAU AGU Asn AUC le AAC The AGC ACC ACA AGA Arg AGG AIA AAA AUG Me ACG AAG GUU GCU GGJ GAU Asp GAC GGC Gly GGA GUC Val GUA GCC Ala GCA GAA Glu GAG GUG GCG GGG G. Cys O Thr O Tyr O Pro It will not code for an amino acid First mRNA base (5' end of codon) Third mRNA base (3' end of codon) 30

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 17QP: Given the following mRNA, write the double-stranded DNA segment that served as the template....

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A part of an mRNA molecule with the sequence 5-UGC GCA-3 is being translated by a ribosome. The following activated tRNA molecules are available. tRNA Anticodon Amino Acid 3-GGC-5 Proline 3-CGU-5 Alanine 3-UGC-5 Treonine 3-CCG-5 Glycine 3-ACG-5 Cysteine 3-CGC-5 Alanine Which two of them can bind correctly to the mRNA so that a dipeptide can form? a. cysteinealanine b. prolinecysteine c. glycinecysteine d. alaninealanine e. threonineglycine
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:
3. Using your Genetic Code, translate and group the following mRNA bases. Example: UAC CAG UGG UUU CUA AUC Your Answer: AUG GUC ACC AAA GAU UAG Met – Val - Thr- Lys – Asp - Stop 5' CUCGGUACGUAAUAGAUCUAUCAUCC GAGACUCUUAAAU 3 b. 5' CUAAUAUUGUGAGAU 3'
(ol soupe bannos96 SAM 3 A small strand of DNA has this sequence: 0pxbeter owl nade hoparg TACCGGAAACTG ATGGCCTTTGAC a. If the TOP strand is the template strand, what will be the mRNA made from this DNA read from left to right? What process allows for the mRNA to be made? b. If this is a eukaryotic mRNA, where does transcription occur? What would happen if the mRNA was not processed? c. What is the sequence of the protein made (use the genetic code below)? What process allows for the protein to be made?
6. How many amino acids will the mRNA sequence "AUG GAC CUG UCG UGA" produce? (LS1-1) * Second MRNA base C. UUU Phe UUC UCU UAU UGU Cps UGC U Tyr UAC UCC Ser UCA UUA Leu UUG UAA Stop UGA Stop A Gu (G) 19) UCG UAG Stop UGG Trp G. Tye AGU A CUU CCU CAU His CAC CGU CUC Leu CUA CC CGC Cys (C) Pro CAA Gin CAG Arg CGA CCA CUG. CCG CGG Arg R AU ACU A C AAU Asn AAC Thr Le AGU Ser AGC AUC e ACC C AUA ACA Lys IK AGA Lys AGG AAA AUG M ACG AAG The GUU GCU GAU GGU Asa GAC GCC Val GCA GGC Ala Gly GUA GAA GGA Glu GSG GUG GCG GAG First mRNA base (5' end of codon) Third mRNA base (3' end of codon)
w/opCulGACU GAC UC 4C According to the Genetic Code Sheet below, which of the following amino acid sequences corresponds to this MRNA strand? CỤC AAG UGC UUC PHE GLU ASP SER ALA TYR U A STOP A GU VAL U CIS U U G A STOP IG TRP ARG AC U LEU SER UG PRO ASN HIS THE GLN MET ILE ARG O a lys-leu-cys-phe O b glu-cys-pro-phe leu-lys-cys-phe O d leu-glu-leu-val U...pdf
6. How many amino acids will the mRNA sequence "AUG GÁC CUG UCG UGA" produce? (LS1-1) * Second mRNA base G UUU Phe UUC UCU UAU UGU Tyr UAC Cys UCC Ser UCA UGC C Phe Gly (G) Leu UAA Stop UGA Stop A Glu (F) (L) UUA Leu UUG Ser Asp (E) (D) (S) Tyr (Y) UCG UAG Stop UGG Trp GUC Ala CUU (A) A CCU CAU CGU His CUC Leu CUA CC Pro ССА CAC CGC Arg CGA Cys (C) Val CAA Gln CAG G U (V) G Trp (W) CUG CCG CGG Arg (R) G U A C A C Leu AUU ACU AAU Asn AGU Ser AGC (L) Ser (S) C AUC Ile ACC AAC Thr Lys (K) UG A AUA Pro AGA Arg AGG ACA AAA Lys AAG Asn (N) (P) AUG Met or start ACG His Thr (H) (T) Gln GUU GCU GAU Asp GAÇ GGU (Q) lle Arg (R) (1) GUC Val GUA GCC Ala GCA GGC Gly GGA GAA Glu OGAG GUG GCG GGG O 1 3 4 First mRNA base (5' end of codon) - U A GUca Gbc AG Third mRNA base (3' end of codon)
2. Given a DNA template molecule of 5' - ATGGCTCCTACCTACTAGTTAACATATGG-3', generate the mRNA sequence and then the polypeptide sequence, starting at the first start codon. mRNA: 5'- CCAU AUG UUA ACU AGU AGG UAG GAG CCA U - 3' Polypeptide: (N) Met-Leu - Thr - Ser - Arg (C)
1. Using the genetic code, determine the sequence of amino acids encoded by the mRNA codocn sequence 5' GCC-AUG-GUA-AAA-UGC-GAC-CCC 3' 2. Using the genetic code, determine the sequence of amino acids encoded by the mRNA codocn sequence 5' CAU-CCU-CAC-ACU-GUU-UGU-UGG 3'
3’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.
2. Transcribe the following gene sequence to the appropriate MRNA codons: TCA TTT CCA CAG ACA * AGU AAA GGU GUc UGU 3. Translate the mRNA codons (i.e. your answer to question 2) into the appropriate amino acids using the following Universal Genetic Code Chart. * Universal Genetic Code Chart Messenger RNA Codons and the Amino Acids for Which They Code SECOND BASE A UCU UAU UGU UUU UUC UUA PHE TYR CYS UCC UCA UAC UGC SER UAA UGA STOP LEU STOP UUG UGG} TRP UCG UAG CUU CCU CAU CGU HIS CUC CCC CAC CGC LEU PRO ARG CCA CCG CUA CAA CGA GLN CUG CAG CG AUU AUC ACU AAU AGU AGC AGA ASN SER ILE ACC AAC THR AAA AAGS ACA AUA AUG MET or START LYS ARG ACG AGG S GCU GCC GUU GAU GGU ASP GUC GAC GGC VAL ALA GLY GUA GCA GCG GAA GGA GGG GLU GUG GAG トエーCロ ASE A, L-RST BASE
RNA Write the mRNA and the polypeptide made from the RNA. Translation DNA Template strand Transcription TTT TT TACGGCGTTAGACAAGTGCGTGAGTACACA ATGCCGCAATCTGTTCACGCACTCATGTGT ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬||