Please draw the pre-mRNA that would be produced from this gene.Gray nucleotides indicate noncoding sequences. Label template strand: 5’-ACTGTGACCTTGACTGTCGATACGATG-3’ 3’-TGACACTGGAACTGACACGTATGCTAC-5’

Below is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'

The following DNA strand is a template strand (also called non-coding strand) of an E. coli gene. Arrow indicates the direction of transcription. The asterisk indicates t transcription initiation site GCAGTGACCGGATATAACGAAGAGGAATGCCGTACAAA 5' GCAGTGA 3' -10 Which of the following sequences accurately represents the RNA derived from this gene? C5' UCCUUAC 3' C3' UCCUUAC 5' C5' CGUCACU 3' +10 3' AGGAATG 5'

what is the anticodon sequence that would build this protein? AUGUUUGUACAUUUGUGUGGGAGUCACCUGGUUGAGGCGUUGUAUUUGGUUUGUGGCGAGCGCUUUUACCAGUUAGAGAAUUACUGA

Using the codon charts in your text (section 6.1), fill in the chart below. [  /8]   Original DNA sequence TAC  GGA  CAC  GTT  CGC  AAC mRNA sequence   tRNA anticodons   Amino acid sequence     Mutated DNA sequence TAC  GGA  CAC  ATT  CGC  AAC mRNA sequence   tRNA anticodons   Amino acid sequence   Type of mutation (highlight all that apply)             Frameshift         Nonsense         Missense     Silent                     Insertion     Deletion     Substitution

Below is the CODING sequence of a gene. An A is inserted at position 131. Insert the variant and write the mature mRNA sequence. Then write the amino acid sequence of the resulting transcript. AGAGCTACCAGCTTTAGCGGGATATAAACACACTTATCCTACTATCCCTA 51 TCTTCCCCCAAAGAGGGATCATGCCGTCGACGGCCAAACCGATCGAGCGG 101 AGGTTAAGGGCCATTGGAAAAGGACACTCTGTACGACTAAGAAAATAGC 151 AGCGACTGAGCTGAACGCTACGGGGGGGGGGGTATATACCCCCCCCCCCC 201 CCAGCTACGAGCGACTTTTTATTTTTTTTTTTTATTTTTTTTCTTTTTTA 251 TTTTTTTTGTTTTTTTTTTTCGCGACTACAGAGACTTCTAGCGATCGACG 301 AGCTAGCTAGCGCGGCATCGGACGCGATCGCGATCCTAGACTCTACGAT

The template strand of a gene has the sequence 5'-ATGCCTAGCCTAGGACT-3'. What will the sequence of the transcript from this gene be? 5' AGUCCUAGGCUAGGCAU 3 5 ATGCCTAGCCTAGGACT 3' 5 AUGCCUAGCCUAGGACU 3' 5' MAUGCCUAGCCUAGGACUAAAAAAAAAA 3" 5' AGTCCTAGGCTAGGCAT 3" V.

Please answer A, B, and C for codon sequence 3 only

The 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.

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…