Refer to the following illustration to answer the question. mobile genetic elements exon 1A insertion sequences O introns V element ends O inverted repeats O exons The element ends in the illustration above are: OUTRs exon 2A exon 2A exon 3A EXON 2A EXCISED BY TRANSOPSASE THAT RECOGNIZES THE ENDS OF TWO SEPARATE ELEMENTS
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- Given the following coding sequence for DNA, provide the sequence of the complementary(template) sequence. 5’ ATGCATAGATTAGGATATCCCAGATAG 3’If the sequence of a gene is: 3'TACATACCAACTGAGGATCGC5' 5'ATGTATGGTTGACTCCTAGCG3' And the RNA that is transcribed from this gene is: 5'AUGUAUGGUUGACUCCUAGCG3' Which strand of the DNA - top or bottom - is the template strand? Explain how you know.The following diagram represents a transcription unit in a hypothetical DNA molecule. 5′ … TTGACA … TATAAT … 3′ 3′ … AACTGT … ATATTA … 5′ Q. On the basis of the information given, is this DNA from a bacterium or from a eukaryotic organism?
- Based on one strand of a certain segment of DNA with the sequence below, answer the following question: 3’-ATGGATCCCATC-5’ What is a gene? What are some of the problems with this definition?If a restriction enzyme cuts between the G and the A whenever itencounters the sequence GAATTC, how many fragments will beproduced when the enzyme is digested with DNA with the followingsequence? TGAGAATTCAACTGAATTCAAATTCGAATTCTTAGCa. Two c. Fourb. Three d. FiveA small section of bacterial DNA template (antisense) strand has the following nucleotide sequence: CGA AAA GAG AAT A mutation in the above sequence involved a substitution of a single base, resulting in an incomplete protein due to a nonsense mutation. Which of the following gen sequences exemplifies the mutation described above? a. CGA UAA GAG AAT b. CGA AAA AAG AAT c. CGA AAA GAG ACT d. UGA AAA GAG AAT
- Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…Suppose the following point mutation occured in the DNA. Will this mutuation affect the protein resulting from this gene sequence? Original DNA sequence ATT TTA AAT GGC CCG CCC GGG Mutuated DNA sequence ATA TTA AAT GGC CAG CCC GGGWrite out the resulting DNA molecules after the following double stranded DNA molecule is digested with EcoRI: 5’-ATGTTAGAATTCTACGTCGAATTCAAATTT-3’ 3’-TACAATCTTAAGATGCAGCTTAAGTTTAAA-5’
- Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'The following fragment of DNA is from the template strand. First determine the amino acids of the protein encoded by this sequence by using the table at the end of this document. Then give the altered amino acid sequence of the proteins that will be found in each of the following mutations: 3’ – TAC AAG GCT CTA TTT GCC ACA ATC – 5’ The nucleotides are numbered 1-24 from left to right. Mutant 1: A transition at nucleotide 9 Mutant 2: A transition at nucleotide 11 Mutant 3: A T to A transversion at nucleotide 15 Mutant 4: A one-nucleotide deletion at nucleotide 7 Mutant 5: A transition at nucleotide 13 Mutant 6: An addition of GGA after nucleotide 6A segment of DNA has the following sequence:TTGGATGCTGAACCTACGACA. What would the sequence be immediately after reaction withnitrous acid? Let the letters H represent hypoxanthine and Urepresent uracil.B. Let’s suppose this DNA was reacted with nitrous acid. Thenitrous acid was then removed, and the DNA was replicated fortwo generations. What would be the sequences of the DNAproducts after the DNA had replicated two times? (Note:Hypoxanthinepairs with cytosine.) Your answer should containthe sequences of four double helices.