Which of the following stretches of RNA is most likely to form a hairpin with 4 bases making up the turn? OUUGCGCAUAACG GCUAUUCCGCUA CACGAUUCGCAC OGCUCUAAGGAGC
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- For the messenger RNA sequence below, find the beginning of the amino acid coding sequence and translate the sequence using the genetic code provided below. 5' - AAUUAUGGGCAAUAUGCCGGGCcGGUUAAGCG - 3' Second Letter A UGU cys u UGC Phe UCU UU U UUC UUA UAU Tyr Ser UAC UAA UAG Leu UCA Stop UGA Stop UUG UCG Stop UGG Trp CUU CU CAU His CGU c cuc Leu ccc ССА CCG Pro CÁC CAA CAG CGC CGA CGG Arg CUA CUG Gin 1st 3rd letter Ser u letter AUU ACU AAU AAC AAA AAG Asn A AUC AUA AUG lle ACC ACA AGU AGC AGA AGG Thr Lys Arg Met ACG GUU G GUC GUA GUG GCU GAU GAC GAA Asp GGU Val GCC Ala GGC Gly GCA GGA Glu GCG GAG GGG GUse the codon chart to determine the following RNA strand in amino acids (Remember to write it the same way the strand is): ACA-AGG-UUA-UGA second letter C A UAU Tyr U UUU UCU UGU Phe Cys UUC UCC UAC UGC C Ser UAA stop | UGA stop| A UAG stop UGG Trp UUA UCA UUG Le UCG CUU CCU CAU CGU His CUC ССС CAC CGC Leu Pro Arg CUA ССА САА CGA Gln СCG CAG CGG CUG AGU AAU Asn AUU ACU Ser AGC S AGA Arg AAC AUC } lle A AUA АСC Thr AAA Lys АСА AUG Met ACG AAG AGG GUU GCU GAU GGU Asp GAC GGC Gly GGA GCC GUC Val Ala GAA GAG } GUA GCA Glu GGG GUG GCG Your answer first letter ACUCAGUCAGUCAG third letterA segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)
- Please by using the first base of each as the first triplet in a condo, how do I translate two almost identical RNA strand into sequences? You will notice that the second strand has a point deletion (the u in bold) with respect to the first strand – comment on how this has affected the resulting peptide chain. aguuguuaucgaaaacugcgaguaaauauccugagggcgcgaagcaacc aguuguaucgaaaacugcgaguaaauauccugagggcgcgaagcaaccPart of a sequence of DNA from a person without this genetic disease is: TAG TAA AAA CCA CCC AGG Part of a sequence of DNA from a person with a genetic disease is: TAG TAA CCA CCC AGG The possible codons for some amino acids are shown in the table. Amino acid Codons glycine GGU GGC GGA GGG isoleucine AUU AUC phenylalanine UUU UUC serine UCU UCC UCA UCG Which amino acid is missing from a person with this genetic disease? serine glycine O phenylalanine isoleucineThe following segment of DNA codes a protein. The uppercase letters represent Exons, the lowercase letters introns. Draw the pre- mRNA, the mature mRNA and translate the codons using the genetic code to form the protein. Identify the 5’UTR and 3UTR 5’- AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3’
- Which of these single strand RNA sequences could form a hairpin secondary structure? 5' AAAAAAAAAAAAAAAAAAA 3' 5' ACACACACACACACACAC 3' 5' CCCGGGGUUUUCCCCGGG 3' 5' UUUUUUUUUCCCCCCCCC 3' 5’ UUUGGGUUUGGGUUUGGG 3’Which of the following RNA sequences is an inverted repeat and can form a stem-loop structure? What is the correct option from the choices below? GGUAGGCAUUACGGAUGG GGUAGGCAUUAGCCUACC GAGAGAGAUUAGAGAGAG GGGGGGGAUUAGGGGGGG GAGAGUGUAAGUGUGAGAThe DNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U A G U UUU1 UUC UUA LOU Leu CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala CAU His CAC CAA CAG Gin AAU Asn AAC AAA 1 Lys AAG LYS G {}a UAA Stop UGA Stop A UAG Stop UGG Trp GAU 1 GAC Asp GAA GIU Glu GAGJ UGU UGC CGU CGC CGA CGG AGU AGC AGA AGG Cys GGU GGC GGA GGG Arg Ser Arg DOA DOA DOA DUTO Third letter Gly
- In the copies of each sequence below, divide the sequences into codons (triplets) by putting a slash between each group of three bases. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGPlace sequences a, b and c into the correct order based on start and stop codons. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence B TCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTC AATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTC GCGCCGAAAAAGATATGGTranslate the following mRNA nucleotide sequence into an amino acid sequence, starting at the first base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’