Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 10, Problem 7P
Interpretation Introduction
Interpretation:
The effect of a mutation in an allosteric enzyme that resulted in a
Concept introduction:
Enzymes being the catalysts increase the
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I. A protein, X, was Isolated from a pathogenlc mlcroorganism. The proteln Is a vlrulence
factor whose path0genlclty lies In a heptapeptide of unknown sequence. After trypsin
cleavage of the heptapeptide from protein X, the peptlde's compOsition and sequence
was determined. The fOllowing were the results of the sequenclng process:
1. When the peptide was treated with dinitrofluorobenzene (DNFB), DNP-asp
and a mixture of amino acids were produced.
2. When the same Intact peptide was treated with streptococcal protease, a
pentapeptide of composition asp, asN, cys, gly and ser and 2 amlno acids
were released.
3. When the heptapeptlde was also treated with hydrOxylamine HCI, a
tripeptide and a tetrapeptide were obtained. The C-terminal amino acid of
the tripeptide was asN.
1) What is the sequence of the heptapeptide if it is composed of cys, asp, lys, asN, gly
and ser only?
2) What is the pl of the heptapeptide?
GTTTTCACTGGCGAGCGTCATCTTCCTACT
8. What is the function (e.g. transcriptional regulation, transmembrane signaling, kinase, protease, etc.) of the protein(s) encoded by the gene.
Homozygosity for extremely rare mutations in a humangene called SCN9A cause complete insensitivity topain (congenital pain insensitivity or CPA) and a totallack of the sense of smell (anosmia). The SCN9A geneencodes a sodium channel protein required for transmission of electrical signals from particular nerves inthe body to the brain. The failure to feel pain is a dangerous condition as people cannot sense injuries.The SCN9A gene has 26 exons and encodes a1977-amino acid polypeptide. Consanguineous matings in three different families have resulted in individuals with CPA/anosmia. In Family 1, a G-to-Atransition in exon 15 results in a truncated protein that is898 amino acids long; in Family 2, deletion of a singlebase results in a 766-amino acid polypeptide; and inFamily 3, a C-to-G transversion in exon 10 yields a458-amino acid protein.a. Hypothesize as to how each of the three SCN9Amutations affects gene structure: Why are truncatedproteins made in each case? b. How would you…
Chapter 10 Solutions
Biochemistry (Looseleaf)
Ch. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3PCh. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - Prob. 11PCh. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Prob. 15PCh. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43P
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- AAAGAGAAAAGAAUA to AAAGAGAAAUGAAUA. Suppose the codon sequence has a single base pair mutation If the old protein sequence was Lys-Glu-Lys-Arg-Ile, what will be the new sequence encoded by the mutant gene? (Use the 3-letter amino acid abbreviations with hyphens and no spaces in between, i.e. Ser-Asn-Tyr-Leu-Pro.) Submit Answer Retry Entire Group No more group attempts remainarrow_forwardCytosine deaminationoccurs ~100 cytosinesper genome per day in a human genome. Eukaryotic cells also contain residues of 5’-methylcytosine, which is involved in regulating gene transcription rates. Mutation of 5’-methylcytosine by deaminationconverts it to thymine. This presents the cell with a much more severe problem than normal cytosine deaminationof cytosine to uracil. Why?arrow_forwardSickle cell hemoglobin DNA CACGTAGACTGAGG ACAC.. Sickle cell hemoglobin MRNA Sickle cell hemoglobin AA sequence 4. What type of mutation is this? Please explain why.arrow_forward
- . The double-stranded circular DNA molecule thatforms the genome of the SV40 tumor virus can be denatured into single-stranded DNA molecules. Becausethe base composition of the two strands differs, thestrands can be separated on the basis of their densityinto two strands designated W(atson) and C(rick). When each of the purified preparations of the single strands was mixed with mRNA from cells infectedwith the virus, hybrids were formed between the RNAand DNA. Closer analysis of these hybridizationsshowed that RNAs that hybridized with the W preparation were different from RNAs that hybridized withthe C preparation. What does this tell you about thetranscription templates for the different classes ofRNAs?arrow_forwardHN, NH. NH H2N° А. В. Consider the 3 structures shown. Which of these is (normally) NOT present in either DNA or RNA? (select all that apply) А В none of the above ZI C.arrow_forward. Explain why DNA is stable in the presence of alkali (0.3 M KOH), while RNA is quantitatively degraded to 2'- and 3'-nucleoside monophosphates under these conditions.arrow_forward
- On average, how many phosphoanhydride bonds (P;-P; bonds) are directly hydrolyzed in thecourse of synthesizing a 200 amino acid protein? Assume that you begin with the mature mRNA,ribosomal subunits, tRNAs, free amino acids, and all necessary factors.arrow_forward. A geneticist examined the amino acid sequence of aparticular protein in a variety of E. coli mutants. Theamino acid in position 40 in the normal enzyme isglycine. The following table shows the substitutionsthe geneticist found at amino acid position 40 in sixmutant forms of the enzyme.mutant 1 cysteinemutant 2 valinemutant 3 serinemutant 4 aspartic acidmutant 5 argininemutant 6 alanineDetermine the nature of the base substitution thatmust have occurred in the DNA in each case. Whichof these mutants would be capable of recombinationwith mutant 1 to form a wild-type gene?arrow_forwardThe 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 +arrow_forward
- b. Which one of the following a cell mutants will be able to switch at least once? [Select]arrow_forwardConsider the following two nonhomologous wildtype chromosomes, where letters or numbers represent genes, the "-" represents the centromere of each chromosome, and chromosomes are shown on separate lines. ABCDE-FGHIJK 123-45678 Identify the type of rearrangement shown in each of the following (A-C) and then identify whether it is balanced or unbalanced. Assume that the individual is diploid and heterozygous for the rearrangement. A. • ABCDE-FGHIJKGH 123-45678 Rearrangement: [Select] • Balanced or Unbalanced: [Select] B. • ABCDGF-EHIJK 123-45678 Rearrangement ✓ [Select] • Balanced or Un pericentric inversion Robertsonian translocation deletion tandem duplicate paracentric inversion C. reciprocal translocation ABCDE dispersed duplicate nonreciprocal translocation 123-45arrow_forward8a. Given the following mutated sequence (with respect to the normal sequence), what TYPE of mutation occurred: AAACGTTAC 8b. Where did the mutation take place?arrow_forward
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