Question

Some amino acids are post-translationally removed from the C-terminal end of the beta-lactamase enzyme from B. imaginarium (i.e. - after it is translated and released from the ribosome, a protease chews off a some amino acids).  The wild-type enzyme, which has had the amino acids removed from the C’-terminus, is 246 amino acids in length and the C-terminal amino acids are shown below aligned with the C-terminal amino acids of a frameshift mutant, which – due to a frameshift mutation - is not processed at its C-terminus. The wildtype and frameshift mutant have identical coding sequences except for the insertion or deletion of a single nucleotide in the DNA encoding the C-terminal region of the protein.  Use this information to answer the questions below.                   

codon#: 243 244 245 246

wt:     gly leu asn ser-COOH

mutant: gly his gln gln tyr trp met asn his glu gly-COOH

1. What was the mutational event that gave rise to the frameshift mutant?  (identify which codon contains the inserted or deleted nucleotide and the position within the codon where the mutation occurred.
2. Deduce the number of amino acids in the wild-type enzyme prior to processing of the C-terminus.
3. Deduce, as far as possible, the sequence of the C-terminus of the wild-type enzyme prior to processing (you will not get a unique answer due to the degeneracy of the genetic code).

Note:  I have included a photo of the  Excel spreadsheet my professor provided to help solve this problem. I need help ASAP please!

F8 x V fx CA(A/G) K L M NO possible codons amino acid deduced codons GGN gly GGN CUN leu WT F Wt C-terminus AAC/U) AG(U/C) asn ser Mutant deduced codons amino acid possible codons gly GGN his gln gin tyr trp CAU/C) CAA/G) CAA/G) UA(U/C) UGG met AUG asn his glu AA(U/C) CA(U/C) GAA/G) gly CCN stop UA(A/G), UGA amino acid # 243 244 245 246 247 248 249 250 251 252 253 2nd Base Ser CA CGU Leu CCA RNA Amino RNA Amino RNA Amino RNA Amino codon acid codon acid codon acid codon acid UUU Phe UCU UAU Tyr UGU Cys U U 0 C Pha UAC Tyr UGC Cys UUA Leu UAA Stop UGA Stop UUG UAG Stop UGG CUU CA 0 His CUC CCC CAC Hig C GC CU A Lau CAA Gin CGA CUG CAG Gin CGG AUU lle AAU As AGU AUC lle ACC The AAC Asn AGC AUA lle ACA The AAA Lys AGA Arg AUG Met ACG The AAG Lys AGG Arg GUU Val GCU GGU GUC Val GCC Ala GAC Asp GGC GUA Val GCA Ala GAA GluGGA GlyA GUG Val GCG Ala GAG Giu GGG Gly Lou 1st Base Base ACU GAU

Answer 1

The C terminus sequence of enzymes is given. It is processed and only a few amino acids are left. I the case of mutant, frameshift mutation abolished the processing of the C-terminus sequence.

{A} The mutational event that gave rise to a frameshift mutation, is involved with the protease enzyme which will remove the amino acid in the wild type. It got effected in the mutant, resulting in the absence of processing. A single nucleotide deletion or insertion is not possible.

The number of amino acids that are present before the processing is 11 amino acids which are seen in the mutant due to the absence of processing.