Genetics question: determine the wild type mRNA?

A fragment of a wild-type polypeptide is sequenced for sevenamino acids. The same polypeptide region is sequenced in fourmutants.
Wild-type N . . . Thr-His-Ser-Gly-Leu-Lys-Ala . . . Cpolypeptide
Mutant 1 N . . . Thr-His-Ser-Val-Leu-Lys-Ala . . . C
Mutant 2 N. . . Thr-His-Ser-C
Mutant 3 N . . . Thr-Thr-Leu-Asp-C
Mutant 4 N . . . Thr-Gln-Leu-Trp-Ile-Glu-Gly . . . C

Determine the wild-type mRNA sequence.

I have already tried:
5'-A-C-U-C-A-U-A-G-U-G-G-U-C-U-U-A-A-A-G-C-U-3'

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Jean KeelingLv2

28 Sep 2019

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Related questions

Consider the following peptides:

#1 gly-lys-phe-met-asp-asp-leu-ile-glu-val-trp-ala-gly

#2 lys-lys-phe-gly-arg-his-val

#3 gly-lys-ala-phe-glu-asp-arg-arg-gln-ala-val

#4 val-ala-ile-trp-phe

a) Predict the order of elution of these peptides from a gel filtration column.

b) Which of these peptides would bind the most tightly to a hydrophobic (reverse phase) column at pH = 7?

c) Which of these peptides would bind the most tightly to a cation exchange column at pH = 6.5? Which would bind the least tightly?

d) Draw a simple sketch to show what a separation of these peptides would look like on a SDS gel.

bluehamster938

Consider the peptide:
NH₂-GAPAGPAGTGKTETTKDLAKSMALLCWFNCS-COOH

Table I. pK values of some amino acids

amino acid	pK of side chain or terminal group
Alanine (Ala or A)	uncharged
Arginine (Arg or R)	12.5
Asparagine (Asn or N)	uncharged
Aspartic acid (Asp or D)	3.9
Cysteine (Cys or C)	8.3
Glutamic acid (Glu or E)	4.3
Glycine (Gly or G)	uncharged
Histidine (His or H)	6.0
Leucine (Leu or L)	uncharged
Lysine (Lys or K)	10.8
Phenylalanine (Phe or F)	uncharged
Serine (Ser or S)	uncharged
Tryptophan (Trp or W)	uncharged
Tyrosine (Tyr or Y)	10.9
Valine (Val or V)	uncharged
peptide N terminus	7.8
peptide C terminus	3.6

Using Table I above, determine the net charge of the peptide at pH9.

(the answer is not -1 or 1)

A few amino acids arepost-translationally removed from the C-terminal end of theb-lactamase enzyme fromB. imaginarium (i.e. - after translation and releasefrom the ribosome, a protease chews off a few amino acids).The purified wildtype enzyme is 327 amino acids inlength and the sequence of its C-terminal amino acids is shownbelow aligned with the C-terminal amino acids found in a mutantstrain of B.imaginarium, which containsof a frameshift mutation that abolishes processingif its C-terminus. This frameshift mutant was caused by insertionor deletion of a single nucleotide. Usethis information to answer the questions below.

wild-type: Ser Asn Ala LysAsp Trp

mutant: Ser Asn Ala Ile Arg Leu Asp Trp Gln MetLys

A. What was the mutational event that gave rise to the frameshiftmutant? Determine precisely where the single nucleotide insertionor deletion occurred in the b-lactamase codingsequences.

B. What is the length of the wildtype enzyme prior to processing(how many amino acids)?

C. Deduce, as far as possible, the actual sequence of theC-terminus of the wild-type enzyme prior to processing.