Concept explainers
To analyze:
Consider a single base substitution that creates α-globin gene mutant Hb Constant Spring (HbCS), whose product contains
Introduction:
The α-globin gene mutant Hb Constant Spring (HbCS) is created by a single base substitution mutation. Single base substitution is a type of mutation in which one base is substituted by another base (for example - switching of base U to A). Such mutation can produce a different type of peptide chain or can result in premature termination or increase in the length of the peptide chain. In case of Hb Constant Spring (HbCS), the protein chain produced by the mutant gene contains
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GENETIC ANALYSIS: AN INTEG. APP. W/MAS
- A mutant E. coli strain is found with a mutation affecting some of its tRNA(Cys). The wild type normally produces a tRNA that recognizes the codon 5’ UGC 3’, and is charged with the amino acid Cysteine (Cys) (its notation is tRNA(Cys)). The mutant tRNA is still charged with Cysteine, but the mutation is in its anticodon that now has the sequence 5’- UCA-3’. How will some of the proteins produced in these E. coli cells be different from the proteins produced in the wild type cellsarrow_forwardThree E. coli tRNA molecules with the anticodon sequences CGG, OGG , and UGG are charged with the same amino acid. What is the amino acid ?arrow_forwardThe following four mutations have been discovered in a gene that has more than 60 exons and encodes a very large protein of 2532 amino acids. Indicate which mutation would likely cause a detectable change in the size of the mRNA and/or the size of the protein product. Consider a detectable change to be >10% of the wild-type size. A table of the genetic code is shown below. First letter 0 00 U O A บบบ UUC UUA UUG U CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu >Leu AUU AUC lle AUA AUG Met >Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Second letter C Ser Pro Thr Ala CAU CAC CAA CAG UAU UGU Tyr UAC UGC UAA Stop UGA UAG Stop UGG AAU AAC AAA AAG A GAU GAC GAA GAG His Gin Asn Lys Asp G Glu CGU CGC CGA CGGJ AGU AGC AGA AGG GGU GGC GGA GGG O AAG576UAG (changes codon 576 from AAG to UAG) Cys Stop Trp O GUG326AUG (changes codon 326 from GUG to AUG) Arg Ser Arg Gly DUAG DUA G DCAG DO AG deletion of codon 779 insertion of 1000 base pairs into the sixth intron (this particular…arrow_forward
- Consider the tryptophan codon 5′ - UGG - 3′ in the standard genetic code . Can a single base change in this codon create a synonymous mutation? Can a single base change in this codon create a nonsense codon?arrow_forwardA wildtype gene produces the polypeptide sequence: Wildtype: Met-Ser-Pro-Arg-Leu-Glu-Gly Each of the following polypeptide sequences is the result of a single mutation. Identify the most likely type of mutation causing each, be as specific as possible. M1:Met-Ser-Ser-Arg-Leu-Glu-Gly missense mutation M2:Met-Ser-Pro M3:Met-Ser-Pro-Asp-Trp-Arg-Asp-Lys M4:Met-Ser-Pro-Glu-Gly nonsense mutation frameshift insertion in frame deletion M5:Met-Ser-Pro-Arg-Leu-Glu-Gly in frame insertionarrow_forwardGiven the following mRNA and amino acids, construct a polypeptide from this tRNA strand. tRNA UAA CCA UUA UAA mRNA Amino Acids AUU = isoleucine AAU = asparginine GGU = glycine GUC = valine GAG = glutamic acidarrow_forward
- A mutation is found in a tRNA-encoding gene. The wild type (non-mutant) allele (version) produces a tRNA that recognizes the codon GAA, and is charged with the amino acid glutamic acid (Glu). The mutant tRNA is still charged with Glu, but it recognizes the codon UAA. What effect will this have on translation in these cells? How will the proteins produced be different? Speculate: is this mutation more likely to be beneficial or harmful?arrow_forwardConsider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?arrow_forwardAnother thalassemic patient had a mutation leading to the production of an mRNA for the β chain of hemoglobin that was 900 nucleotides longer than the normal one. The poly(A) tail of this mutant mRNA was located a few nucleotides after the only AAUAAA sequence in the additional sequence. Propose a mutation that would lead to the production of this altered mRNA.arrow_forward
- As described earlier, DNA damage can cause deletion or insertion of base pairs. If a nucleotide base sequence of a coding region changes by any number of bases other than three base pairs, or multiples of 3, a frameshift mutation occurs. Depending on the location of the sequence change, such mutations can have serious effects. The following synthetic mRNA sequence codes for the beginning of a polypeptide: 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCUUAUC-AUGUUU-3′ First, determine the amino acid sequence of the polypeptide. Then determine the types of mutation that have occurred in the following altered mRNA segments. What effect do these mutations have on the polypeptide products? a. 5′-AUGUCUCCUACUUGCUGACGAGGGAAGGAGGUGGCUUAUCA-UGUUU-3′ b. 5′-AUGUCUCCUACUGCUGACGAGGGAGGAGGUGGCUUAUCAU-GUUU-3′ c. 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCCCUUAUC-AUGUUU-3′ d. 5′-AUGUCUCCUACUGCUGACGGAAGGAGGUGGCUUAUCAU-GUUU-3′arrow_forwardIn HbS, the human hemoglobin found in individuals with sickle-cell anemia, glutamic acid at position 6 in the beta chain is replaced by valine. Q.) Show that one of the glutamic acid codons can be converted to a valine codon by a single substitution mutation (i.e., by changing one letter in one codon).arrow_forwardSeveral experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that encodes methionine initiator tRNA (tRNAiMet) was located and changed; specifically, the nucleotides that specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was 5′ –CCA–3′ instead of 5′ CAU–3′. When this mutated gene was placed in a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of these proteins contained extra aminoacids, and others contained fewer amino acids than normal. a. What do these results indicate about how the ribosome recognizes the starting point for translation in eukaryotic cells? Explain your reasoning. b. If the same experiment had been conducted on bacterial cells, what results would you expect? c. Explain why some of the proteins produced contained extra amino acids while others contained fewer amino acids than normal.arrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning