How many exons does tra-RA contain? Q7. How many introns does tra-RA contain?
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Q6. How many exons does tra-RA contain?
Q7. How many introns does tra-RA contain?
Gene is a portion of the genome that can be transcribed or a functional unit of the genome that can be transcribed. The genome size of a human is 3200MB categorized into 2 parts repetitive sequences and genes or gene-related sequences. Genes are made up of exons (coding sequences) that code for a particular protein or enzyme while introns are non-coding sequences. In eukaryotes, introns and exons are present on hnRNA. During RNA processing, splicing of introns occur. It required 2 esterification reactions that will remove introns from mRNA. mRNA contains only coding sequences (exon) which further do the translation to form the particular protein.
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- summarize these results using concise language in a neat table; Control : 5’ ATGTACGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ This is the coding strand of DNA and hence this DNA sequence is similar to mRNA sequence. So the mRNA sequence is : 5’ AUGUACGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ Mutant 1: 5’ ATGTACGAGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence 5’ AUGUACGAGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ The bold Adenine is the mutated base which is substituted in place of Cytosine. So the codon change from GCG to GAG. GCG codes for Alanine but GAG codes for Glutamic acid. So the amino acid sequence changes. Hence this mutation is missense mutation where a base substitution results in change in amino acid sequence. Mutant 2: 5’ ATGTATGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence: 5’ AUGUAUGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ In this mutation, Cytosine is replace by Thymine and hence the codon…The human RefSeq of the entire first exon of a geneinvolved in Brugada syndrome (a cardiac disordercharacterized by an abnormal electrocardiogram andan increased risk of sudden heart failure) is:5′ CAACGCTTAGGATGTGCGGAGCCT 3′The genomic DNA of four people (1–4), three ofwhom have the disorder, was subjected to singlemolecule sequencing. The following sequences represent all those obtained from each person. Nucleotidesdifferent from the RefSeq are underlined. Individual 1:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGCGGAGACT 3′Individual 2:5′ CAACGCTTAGGATGTGAGGAGCCT 3′Individual 3:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGGCGGAGCCT 3′Individual 4:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGTGGAGCCT 3′a. The first exon of the RefSeq copy of this gene includes the start codon. Write as much of the aminoacid sequence of the encoded protein as possible,indicating the N-to-C polarity.b. Are any of these individuals homozygotes? If so,which person and what allele?c. Is…Utilizing Hind III and EcoR V Restriction Enzyme with Pet41 and the following gene of interest... a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg aactgcaatt ggacggtttc 781…
- To determine the reproducibility of mutation fre-quency measurements, you do the following experiment.You inoculate each of 10 cultures with a single E. coli bac-terium, allow the cultures to grow until each contains 106cells, and then measure the number of cells in each culturethat carry a mutation in your gene of interest. You were sosurprised by the initial results that you repeated the experi-ment to confirm them. Both sets of results display the sameextreme variability, as shown in Table Q5–1. Assuming thatthe rate of mutation is constant, why do you suppose thereis so much variation in the frequencies of mutant cells indifferent cultures?Which of the following set(s) of primers a–d couldyou use to amplify the following target DNA sequence, which is part of the last protein-coding exonof the CFTR gene?5′ GGCTAAGATCTGAATTTTCCGAG ... TTGGGCAATAATGTAGCGCCTT 3′3′ CCGATTCTAGACTTAAAAGGCTC ... AACCCGTTATTACATCGCGGAA 5′a. 5′ GGAAAATTCAGATCTTAG 3′;5′ TGGGCAATAATGTAGCGC 3′b. 5′ GCTAAGATCTGAATTTTC 3′;3′ ACCCGTTATTACATCGCG 5′c. 3′ GATTCTAGACTTAAAGGC 5′;3′ ACCCGTTATTACATCGCG 5′d. 5′ GCTAAGATCTGAATTTTC 3′;5′ TGGGCAATAATGTAGCGC 3′⦁ Original: ATTTGAGCCMutated: ATTGAGCC. This is an example of what kind of mutation?
- #2 EcoRI --- 5’ G ↓AATTC 3’ 5’ ACG ACGTATTAGAATTCTTA TCCGCCGCCGGAATTCT CATCA 3’ 3’ TGC TGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:The sequence of a region of interest in a DNA template strand is3′–ATACGACTAGTCGGGACCATATC–5′. If the primer in a dideoxysequencing experiment anneals just to the left of this sequence, drawthe sequencing ladder that will be obtained.1. You are required to use the pET 22b expression vector in a cloning experiment. If thevector is 5.5 kilobasepairs in length and the mass of the DNA that you have addedrepresents 8.26 x 109 molecules. Determine the mass of vector DNA (in nanogram) thatyou have added to your reaction [1basepair ~ 660 Da or g/mol, Avogadro’s constant =6.023 x 1023 molecules/mol] 2. You have added 4 μl of a 5 μM stock primer to a 25 μl PCR reaction. What is the amountof primer in nanomol that you have added to the reaction?
- The amino acid sequence of part of a protein has beendetermined:N . . . Gly Ala Pro Arg Lys . . . CA mutation has been induced in the gene encodingthis protein using the mutagen proflavin. The resultingutant protein can be purified and its amino acidsequence determined. The amino acid sequence of themutant protein is exactly the same as the amino acidsequence of the wild-type protein from the N terminus of the protein to the glycine in the preceding sequence. Starting with this glycine, the sequence ofamino acids is changed to the following:N . . . Gly His Gln Gly Lys . . . CUsing the amino acid sequences, one can determinethe sequence of 14 nucleotides from the wild-typegene encoding this protein. What is this sequence?e. four-base, not overlapping4. An example of a portion of the T4 rIIB gene in whichCrick and Brenner had recombined one + and one −mutation is shown here. (The RNA-like strand of theDNA is shown.)wild type 5′ AAA AGT CCA TCA CTT AAT GCC 3′mutant 5′ AAA GTC CAT CAC TTA ATG GCC 3′a. Where are the + and − mutations in the mutant DNA?b. The double mutant produces wild-type plaques.What alterations in amino acids occurred in thisdouble mutant?c. How can you explain the fact that amino acids aredifferent in the double mutant than in the wild-typesequence, yet the phage has a wild-type phenotype?1.Your goal is to design a strategy to create thisfinal pDHFR plasmid for fusion protein expressionfrom the materials available:You have an empty pET21a expression vector and another vectorthat contains the fusion protein nucleotide sequence (pBluescript + GST-DHFR-His). (#) indicatesbp location ofenzyme cut: a.EcoRIonly, HindIII only, NotIonly, EcoRI and HindIII, EcoRI and NotI, or HindIII and NotI? b.Briefly explain the rationale for your selection. Please connect your rationale to the need for compatible ends without any further manipulation and required directionality of the DHFR fusion protein coding sequence relative to the T7 promoter that will drive its expression. c.What is the final size(kb)of your desired pDHFR plasmid ligation product based onyour design? d. Complete the chart providedto indicate thesize of all fragments(in bp)resultingfrom complete digestion of each plasmid givenyour strategy, andthe fragment(in kb)from each you would isolate for ligation.Please mind…