Nascent form of the MRNA A. undergoes splicing only after capping 3. is also called hnRNA C. is polyadenylated after splicing D. participates in the spliceosome assembly for splicing E. joins snRNPs to form the spliceosome assembly (A) A, B, D, E only в) А, В, С only (С) А, В, С, D, E D B, C, D, E only
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Gene Interactions
When the expression of a single trait is influenced by two or more different non-allelic genes, it is termed as genetic interaction. According to Mendel's law of inheritance, each gene functions in its own way and does not depend on the function of another gene, i.e., a single gene controls each of seven characteristics considered, but the complex contribution of many different genes determine many traits of an organism.
Gene Expression
Gene expression is a process by which the instructions present in deoxyribonucleic acid (DNA) are converted into useful molecules such as proteins, and functional messenger ribonucleic (mRNA) molecules in the case of non-protein-coding genes.
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- Which of the following is false? Options: tRNAs are heavily processed but in a manner that is different than mRNAs mRNAs are protected from degradation by modifications at the 5' and 3' end rRNAs are all cleaved from a larger pre-rRNA precursor Group II splicing requires protein and RNA to excise intron sequences methylation of histone 3 at K4 results is mostly associated with activated transcriptionWhich of the following is the correct sequence for RNA splicing: 1. U2 binds to the branch site and ATP is hydrolyzed 2. U1 binds to the GU sequence at 5’ splice 3. U5/U4/U6 trimer binds 4. U2/U5/U6 are bound to lariat and 3’ site is cleaved and exons are ligated using ATP hydrolysis. 5. U1 is release, U5 shifts from exon to intron and U6 binds at the 5’ splice site 6. U4 is released, U6/U2 catalyzes transesterification, U5 binds exon at 3’ splice site, and the5’ site is cleaved. Group of answer choices 4,3,1,2,6,5 5,4,2,3,6,1 3,1,4,2,5,6 1,2,3,4,5,6 2,1,3,5,6,4Which statement is false: A) Each type of protein ( ex: hemoglobin vs trypsionngen) varies in the length and amino acid sequence of its peptide B) After the rpocess of transcription is complete, the mRNA that is produced will continue being tranlsated by ribosomes for the rest of the cells life. mRNA never breaks down C) A ribosome will bind to an mRNA and will translate the sequence by reading one codon at a time and adding one amino acid to the peptide chain. It will stop the translation once it encounters a stop codon D) The gene for a protein provides the information on the legth of the peptide, along w the amino acid sequence so the protein can be synthesized by a ribosome E) Once mRNA has left the nucleus, ribosomes will bind to it and will follow the instructions in its sequence to make the new protien
- The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post transcriptional processing which is essentially way too different from the original code in the DNA template. Given: Cytosine; a Protein with known amino acid sequence (amino acid sequence given below) MATIVNTKLGEHRGKKRVWLEGQKLLREGYYPGMKYDLELKDSQVVLRVKEEGKFTISKRERNGRVSPII DLTVQELATVFDGVEMLRVFIRNGAIVISAHHQQERVIERVNRLISKLENGESLSVCSLFHGGGVLDKAI HAGFHKAGIASAISVAVEMEGKYLDSSLANNPELWNEDSIVIESPIQAVNLSKRPPQVDVLMGGIPCTGA SKSGRSKNKLEFAESHEAAGAMFFNFLQFVEALNPAVVLIENVPEYQNTASMEVIRSVLSSLGYSLQERI LDGNEFGVIERRKRLCVVALSHGIDGFELEKVQPVRTKESRIQDILEPVPLDSERWKSFDYLAEKELRDK AAGKGFSRQLLTGDDEFCGTIGKDYAKCRSTEPFIVHPEQPELSRIFTPTEHCRVKGIPEELIQGLSDTI AHQILGQSVVFPAFEALALALGNSLWSWVGMMPIMVEVVDESQPVIGGEDFHWATALVDAKGTLKLSPAA KKQGMPFNIMDGQLAVYSPNGTKKSCGHEPCEYLPVMMSGDAIMVTSSLVH Requirement: Original DNA code (itemize the steps you would take to get to know the original DNA code of Cytosine in focus). Eucaryotic processing of hnRNA into mature mRNA includes all of the following steps except: ribosome attachment to the Shine-Dalgarno sequence 5’-addition of a 7-methylguanosine cap 3’-addition of a polyadenylated tail splicing together of exons excision of intronsFor the following characteristics, state whether they are true for the 5' CAP, the Poly-A tail, BOTH modifications, or NEITHER modification. Briefly (1 sentence) support your answer. Promotes translation of the mRNA at the 3' end. Acts to prevent degradation of the mRNA by exonucleases. Promotes splicing of the mRNA. Promotes binding of transcription factors to the Mediator complex.
- What would the sequence of the immature mRNA be? Place the sequence of ALL the transcript that would be synthesized from the gene. In each of the available line you should transcribe all nucleotides that form the immature mRNA. If the sequence indicated is not transcribed then write NA. WRITE THE NUCLEOTIDES IN CAPITAL LETTERS. Promoter: EXON 1: Intron 1: Exon 2: Intron 2: Exon 3 without terminator : Exon 3 terminator:Given the following DNA sequence: 3'-TACTTNGTNCTNTCN-5' where N stands for any nucleotide, give the complementary mRNA sequence. Indicate direction of strand as 3'--> 5' or 5'--> 3' as in the given sequence above. Give the amino acid sequence of your mRNA sequencelin No. 1. Indicate direction of strand as above. Use all lowercase letters, 3-letter name of amino acid separated by a hyphen (-), no spaces in-between.What potential polypeptides can be produced from the following mRNA sequence? There are more than one answer.. 5’ ...GGAGCUCGUUGUAUU... 3’ a. ser-ser-leu-tyr b. leu-cys-cys-ser-arg c. gly-ala-ser-trp-ile d. gly-ala-arg-cys-ile e. glu-leu-val-val f. You can't translate without a start codon. I know (d) is one of the answer but I'm stuck on how to find the rest. Please help.
- -Write down the complementary DNA sequence. TACCTAGCG CACATGTAGGTGGGCAAAGTT -Write down the complementary mRNA sequence for each of the following DNA sequence. A: TACCTAGCGCACATGTAGGTGGGCAAAGTT B: TAC ATG GTT ACA GTC TAT TAG ATG CTA TTT ACT TAG -If the first G changes to A what kind of mutation will happen? Show the change in amino acid sequence. This is base substitutions involve the replacement of one nucleotide with another. And it changes one amino acid coding, producing a missense mutation TAC CTA GCA CAC ATGTAGGTGGGCAAAGTT TAC CTA ACACACATGTAGGTGGGCAAAGTTThe diagram below shows an imaginary eukaryotic structural gene containing two exons. The exon nucleotides are numbered beginning at the transcription start site and a portion of the intron is not shown to save space: Help Center? transcription start site promoter U STACAGTATAAATGAATTAATTGACGTATGTCAATCGGTAAGT...TCAGGTACT U UUU} Phe UUG} Leu exon 1 3 ATGTCATATTTACTTAATTAACTGCATACAGTTAGCCATTCA...AGTCCATGAATGACTTATGTGCGGTTATTTACTGAT... Second letter C Predict the amino acid sequence of the polypeptide encoded by this structural gene. The genetic code is provided below.In eukaryotes there is not a consistent relationship between the length of the coding sequence of a gene and the length of the mature mRNA it encodes, even though one nucleotide in DNA = one nucleotide in pre-mRNA or primary transcript. Explain why this is so.