THCA 100 This reaction is Entropy is THC ( Leafly Mur AD RNA polymerase ATGACGOATCAGCCOCAAG GOAATTGOOGACATAA VACUOCCUAQUCOGCQUU RNA Transorpt TAUTOCOTAGTCOD 99 ATTEOR TATT (In this picture, RNA is being made by complementary base pairing with DNA.) This reaction is Entropy is
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- 5'-TAGCTGATCGAATATGCGGTCTCTATCTTCGTAGACGA-3' 3'-ATCGACTAGCTTATACGCCAGAGATAGAAGCATCTGCT -5' Determine the amino acids that will be encoded by this sequence Second letter First letter U C A G U UUU Phe UUC UUA UUG Leu CUU CUC CUA CUG Leu GUU GUC GUA GUG Val UCU UCC UCA UCGJ AUU AUC lle AUA ACA AUG Met ACG CCU CCC C CCA CCG ACU ACC GCU GCC GCA GCG Ser - Pro Thr Ala A UGU UACTyr Cys UGC. UAA Stop UGA Stop A UAG Stop UGG Trp G CAC His CAA Gin CAG GAUT GAC Asp GAA AAU Asn ACC Ser AGU AAG LYS AA Glu GAGJ Oa. N-Met-Arg - Ser-Leu-Ser - Ser-C Ob. N-Met-Pro-Arg - Asn-Asp - Ser-C d. N-Met-Lys - Val-Glu-Ala-C Oc. N-Asp-Pro-Lys - Ser - Val-Ile-C Oe. N- Met-Ala-Asp-Pro-Lys - Ser-C G CGU CGC CGA CGG AGA AGG. GGU GGC GGA GGG Arg SCAO Gly U UCAG UUA DUAG Arg G Third letter 133’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.Based on sequences A,B,C. Provide an anticodon sequence that would build this protein. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
- EcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .COMPLEMENTARY DNA SEQUENCE OF GACGGCTTAAGATGCName: Date: 2. The sequence of a fragment of one strand of DNA is AATTGCATATACGGGAAATACGACCGG. Transcribe this s sequence into MRNA. er bns eldst eboo oi ebitqeqylog erlt to noihiog eri qu elsm bluow tsri abios onime Jlaw as ye s 1ot noitem atelomet AHG 3. The following MRNA ştrand is being used to asemble a polyp
- BM4_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 letterThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: bottom strand is the noncoding strand). 5'-ААCGCATGAGAAAGCCCCCCGGAAGATCACСТТСCGGGGGCТТТАТАТААТТАGC-3' 3'-тTGCGTACтстттCGGGGGGCCTTCTAGTGGAAGGCCCCCGАААТАТАТТААТтCG-5' (i) Draw the structure of hairpin loop that will be formed during transcription. (ii) Illustrate how the hairpin loop structure initiates the termination of transcription.
- 5-ccuaaucg-34 3'-acctgcctataccggattagetetgatectaagcatgtc-5 The diagram above shows an RNA primer hydrogen-bonded to a DNA template. Which letter indicates the site where DNA polymerase would add nucleotides to this structure? OA OB D Any of these is possible C5'- 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 UCAG5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC Find start codon and stop codon UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’