Failure of anti-oxidant function results in the hydroxylation of an aromatic acid of Enzyme Z andits activation, so that it degrades protoporphyrin to porphyrin, an unstable product. When hit bylight, this product further degrades to form a compound responsible for the lesions andexcruciating pain the man suffers. The mutation also affected an amino acid at the N-terminal ofEnzyme X. Sequencing of the first seven (7) amino acids at the N-terminal of the normal enzymegave the following sequence:trp-arg-asp-leu-ser-gly-hisWhen the cDNA was sequenced by the Sanger method utilizing ddCTP, the following productswere obtained:TetranucleotideHexanucleotideNonanucleotideDecanucleotideDodenucleotideOctadecanucleotideNonadecanucleotide21-nucleotideWhat is the sequence of the bases in the mRNA coding for the peptide above?

Sanger DNA sequencing is carried out using ddCCTP for the chain termination. The given peptide is…

Answered: Failure of anti-oxidant function…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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The intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.
The intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +
Most of the mutations that Yanofsky recovered were missense mutations. However, Yanofsky also recovered a nonsense mutation that changed amino acid number 15 into a stop codon. This codon normally encodes Lysine. Does the recovery of this mutation support the hypothesis that this Lysine residue is critical in the function of the tryptophan synthetase protein? Why or why not?
If the following nucleotide sequence represents the active domain of the COVID19’s M-protein 5’ ---- 5’ GGGUACAUGGUAGCCCCCGUCGAGAAAACACCC …. 3’ a) describe a potential mutation that may occur and the mechanism that could fix it b) if the repair mechanism is faulty, explain the consequences for COVID19 & that of the infected individual
Shown below are two homologous lengths of the alpha and betachains of human hemoglobin. Consult a genetic code dictionary and determine how many amino acid substitutionsmay have occurred as a result of a single nucleotidesubstitution. For any that cannot occur as a result of a singlechange, determine the minimal mutational distance. Alpha: ala val ala his val asp asp met proBeta: gly leu ala his leu asp asn leu lys
A mutant has no activity for the enzyme isocitrate lyase.Does this result prove that the mutation is in the geneencoding isocitrate lyase?
A normal polypeptide and a mutant of the polypeptide were hydrolyzed by an endopeptidase under the same conditions. The normal and mutantpolypeptide differ by one amino acid. The fingerprints of the peptides obtained from the two polypeptides are shown below. What kind of amino acid substitution occurred as a result of the mutation? (That is, is the substituted amino acid more or less polar than the original amino acid? Is its pI lower or higher?) (Hint: Photocopy the fingerprints, cut them out, and overlay them.)
. Seven E. coli mutants were isolated. The activity ofthe enzyme β-galactosidase produced by cells containing each mutation alone or in combination with othermutations was measured when the cells were grown inmedium with different carbon sources.Lactose +Glycerol Lactose GlucoseWild type 0 1000 10Mutant 1 0 10 10Mutant 2 0 10 10Mutant 3 0 0 0Mutant 4 0 0 0Mutant 5 1000 1000 10Mutant 6 1000 1000 10Mutant 7 0 1000 10F′ lac from mutant 0 1000 101/ mutant 3F′ lac from mutant 0 10 102/ mutant 3Mutants 3 + 7 0 1000 10Mutants 4 + 7 0 0 0Mutants 5 + 7 0 1000 10Mutants 6 + 7 1000 1000 10Assume that each of the seven mutations is one andonly one of the genetic lesions in the following list.Identify the type of alteration each mutation represents.a. superrepressorb. operator deletionc. nonsense (amber) suppressor tRNA gene (assumethat the suppressor tRNA is 100% efficient in suppressing amber mutations)d. defective CRP–cAMP binding sitee. nonsense (amber) mutation in the β-galactosidase genef.…
In studies of the amino acid sequence of wild-type and mutant forms of tryptophan synthetase in E. coli, the following changes have been observed: Determine a set of triplet codes in which only a single-nucleotide change produces each amino acid change.
The protein encoded by the cystic fibrosis gene is 1480amino acids long, yet the gene spans 250 kb. How is thisdifference possible?
An individual carries a somatic mutation that changes a lysinecodon into a glutamic acid codon. Prior to acquiring this mutation,the individual had been exposed to UV light, proflavin, and5-bromouracil. Which of these three agents would be the mostlikely to have caused this somatic mutation? Explain your answer.
A man in his early 30s suddenly developed weakness in hishands and neck, followed weeks later by burning musclepain—all symptoms of late-onset muscular dystrophy. Hisinternist ordered genetic tests to determine whether he had oneof the most common adult-onset muscular dystrophies—myotonicdystrophy type 1 (DM1) or myotonic dystrophy type 2 (DM2). Thetests detect mutations in the DMPK and CNBP genes, the onlygenes known to be associated with DM1 and DM2. While awaitingthe results of the gene tests, the internist explained that thedisease-causing mutations in these genes do not result in changesto the coding sequence. Rather, myotonic dystrophies resultfrom increased, or expanded, numbers of tri- and tetranucleotiderepeats in the 3 untranslated region of the DMPK or CNBP genes.The doctor went on to explain that the presence of RNAs withexpanded numbers of repeats leads to aberrant alternative splicingof other mRNAs, causing widespread disruption of cellular pathways.This discussion…

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