DNA sense or template DNA sense or template mRNA tRNA protein 5', 3' amino or carboxyl 5' T U A A RG T T U G C G A G T/C 5', 3' amino or carboxyl 3'
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- What is thee tRNA anticodon for the first 5’-ACGAUC-3’?TAC-ATA-ACG-CGA-CAA-CTA-AAA-ACT Write the amino acid sequence of the protein that would be formed by translating this plece of DNA. You can use the three letter abbreviations for the amino acid in each box, do not use the name of the full amino acids, Use the abbrevlation of the amino adid exactly as Itswritten in the table (including the appropriate capltalizations For example it your answer for amino acid 1 is"Methionine" you would write MET in the box (not Met or met) Second Later UUU UUUC Phe UCU UAU Ser UAC UAA UAG Tyr UGU Cys U Leu UCA ucG UGC Stop uGA Beop UUG Step UGG Trp G Cuu C cuC CUA CUG CU Leu Coc CCA CG CAU Pro cGu CAC CAA CAG His CGC Arg CGA CGG 1st 3rd letter AUU A AUG AUA AUG ACU le AAU AAC AGU ACC Asn Ser U leBer ACC ACA Mct ACG The AAA AAG ACA Lys AGG Arg acu Val GCC GCA GCG GAU GAC GAA GAG GOU GGC GGA GGG Arp G GUC GUA GUG Ala Gly Glu Seovence of the protein, 1st amino acid. 2nd arnino acid: 3rd amino acid: 4th amino acid Sth amino acid: 6th amino acid: 7th amino…With this DNA sequenec: - 5'-GCAATGGAGAGAATCTGCGCG-3'- - 3'-CGTTACCTCTGTTAGACGCGC-5' - -Identify the sequencce of RNA in which the protein product will be detrnemined. -What will be the protein product sqeuence? -What will be the pl of the protein product?
- This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.5' UGG CAA UCC UAC GAU 3' - 1. Here is the MRNA sequence from a section of a gene (it is the middle of the sequence, so it has no AUG). What is the template sequence of this gene? - 2. Are any of these codons in the MRNA non-degenerate? If so, indicate which one. e 3. 4 a) Translate this mRNA section. Give the 3 letter codes for the amino acids. b) Indicate on the peptide which is the C terminus and which is the N terminus. e 4. Is it possible for a single base pair substitution to cause a truncation in this peptide? If so, e explain how. e 5. Write out the sequence of the anticodon in the tRNA that would bind to the fourth codon in the e MRNA. e 6. Write out a possible miRNA that could regulate the expression of this gene
- 5'- ATTGTGATATGGCCACCTGCCACCTGGAGAGCAGCT GATTAG-3' What oligopeptide is encoded by the above sequence? Please use the one-letter code for your answers. (You will need to consult the Table of the Genetic Code for this question) 1st letter UUU Phe UCU UCC Leu UCA UCG U UUC UUA UUG CUU C CUC CUA CUG AUU A AUC AUA AUG U GUU G GUC GUA GUG CCU Leu CCC CCA CCG ACU lle ACC ACA Met ACG GCU Val GCC GCA GCG с Second Letter Ser Pro Thr Ala UAU UAC A | AAU AAC AAA AAG GAU GAC GAA GAG Tyr CAU CAC CAA Gin CAG 1 UAA Stop UGA Stop A UAG Stop UGG Trp G His Lys UGU UGC Asp Glu G 3rd Asn AGU Ser U letter AGC AGA AGG Cys U CGU CGC Arg CGA CGG GGU GGC GGA GGG DCAG DUAG DUCAG Arg Gly UCAGBM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple48 Second letter If any single nucleotide is deleted from the DNA sequence shown below, what type of mutation is this? UUU U UUC UUA UCU Phe UCC UAU UGU Cys ANTISENSE 5' GGACCCTAT3' UAC Tyr UGC UAA Stop UGA Stop UAG Stop UGG Trp Ser UCA Leu UUGL" UCG CUU CU CAU) His CAC) CGU CGC CGA Arg CUC C Leu Pro CAA1 Gin CAG) CUA CCA CUG CG CGG AAU AAC Asn AGC AAA AAG Lys AGG Arg AUU ACU AGU Ser AUC lle ACC Thr AUA ACA AGA AUG Met ACG GCU GCC GUU GAU] GGU GUC Val GAC Asp GGC Ala Gly GUA GCA GAA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a FRAMESHIFT SILENT NONSENSE MISSENSE Third letter First letter
- Translate the following DNA sequence into amino acids 5'ATAGTACCGCAAATTTATCGCT3 O met-ala-phe-lys-stop O met-ala-phe-lys- met-tyr-his-gly-val-stop-met-gly O met-ala-ser-gly-thr-stop O tyr-his gly-val-stop-met-ly O ala-phe-lys stopDetermine the sequence of a polypeptide treated with trypsin and chimotripsine. Below are the fragments generated with each treatment. Determine the original sequence for both fragmentations (reduerde that they must be equal in the order of amino acids) Quimotripsina 1. Leu-His-Lys-Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly-Pro-Ser 1. Gln-Gln-Ala-Gln-His-Leu-Arg-Ala-Cys-Gln-Gln-Trp 2. Arg-lle-Pro-Lys-Cys-Arg-Lys-Phe Trypsin 1. Arg 2. Ala-Cys-Gln-GIn-Trp-Leu-His-Lys 3. Cys-Arg 4. Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly- Pro-Ser 5. lle-Pro-Lys 6. Light 7. Phe-Gin-Gln-Ala-Gln-His-Leu-ArgSecond letter A UUU UCU) UC UCA UCG UAU UUC Phe UUA Tyr UGU UGCJ Ser UAA Stop UGA Stop A UAG Stop UGG Trp UUG Leu CAUHIS CUU CUC CUA CUG CCU* C ССА CCG CGU His САС Leu CGC Arg CGA Pro CAA Gin CGGJ Gln Which amino acid is carried by the TRNA with the anticodon 5'-UCA-3? ACU ACC ACA AAU AAC. AGU AGC AGA AUU Ser Asn AUC Ile A AUA Thr AAA Lys AAG Lys AGG Arg AUG Met ACG GAU GGU] GUU GUC GUA GUGJ GCU GCC GCA GCG GAC Asp Ala GAA GGC Gly GGA Val GAG Glu GGGJ Isoleucine. None-this is a stop codon. Aspartic acid. Histidine. IV. Leucine O V. Third letter UCAG UCAG UCAG First letter