A. The type of nucleotide mutation that occurred in the mutant sequence is a [Select] B. The type of amino acid mutation that occurred in the mutant sequenc is a [Select] mutation. C. This mutation resulted from a misincorporation that went unnoticed b DNA polymerase during DNA replication. Which DNA repair system was tasked with recognizing and correcting this error? [Select]

A. The type of nucleotide mutation that occurred in the mutant sequence is a [Select] B. The type of amino acid mutation that occurred in the mutant sequenc is a [Select] mutation. C. This mutation resulted from a misincorporation that went unnoticed b DNA polymerase during DNA replication. Which DNA repair system was tasked with recognizing and correcting this error? [Select]

Name: Please solve and explain! Thank you. The wildtype version of a very sh
Uploaded: 2024-03-20T06:42:41.000Z
Channel: Bartleby
Description: Please solve and explain! Thank you. The wildtype version of a very short gene is:DNA:5'-GTACGGATGCTTTACGTAGGGTGA-3'3'-CATGCCTACGAAATGCATCCCACT-5'and is translated to:N-Met-Leu-Tyr-Val-Gly-CResearchers discover a mutant version of this gene with the

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 18QP: The following is a portion of a protein: met-trp-tyr-arg-gly-pro-thr-Various mutant forms of this...

See similar textbooks

Similar questions

A 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 insertion
MCAD deficiency is an inborn error of metabolism. The coding strand is shown for the wild-type gene. The TATA box and kozak sequences are shown in parenthesis. Wild-type: 5’-ATGGCC[TATAT]ATGTCACTTGACTACGCAGCC[GCCACCATGG]ATATAGATAATGCGCGCATAGCATACTGAGGGTAGTAG-3’ What is the resulting polypeptide from the wild-type protein?
A gene encodes a protein with the following amino acid sequence: Met-Trp-His-Arg-Ala-Ser-Phe A mutation occurs in the gene. The mutant protein has the following amino acid sequence: Met-Trp-His-Ser-Ala-Ser-Phe An intragenic suppressor mutation restores the amino acid sequence to that of the original protein: Met-Trp-His-Arg-Ala-Ser-Phe Give at least one example of base changes that could produce the original mutation and the intragenic suppressor.
A gene encodes a protein with the following amino acid sequence: Met-Trp-His-Arg-Ala-Ser-Phe A mutation occurs in the gene. The mutant protein has the following amino acid sequence: Met-Trp-His-Ser-Ala-Ser-Phe An intragenic suppressor mutation restores the amino acid sequence to that of the original protein: Met-Trp-His-Arg-Ala-Ser-Phe Give at least one example of base changes that could produce the original mutation and the intragenic suppressor. (Consult the genetic code in Figure 15.10.)
With this DNA sequenec: - 5'-GCAATGGAGAGAATCTGCGCG-3'- - 3'-CGTTACCTCTGTTAGACGCGC-5' - -Identify the sequencce of RNA in which the protein product will be detrnemined. -What will be the protein product sqeuence? -What will be the pl of the protein product?
The BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINY
Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'
The most common MCAD mutation is shown below. The coding strand is shown for both the WT and mutant. The TATA box and kozak sequences are in parenthesis. What type of mutation is present? Wild-type:5’-ATGGCC[TATAT]ATGTCACTTGACTACGCAGCC[GCCACCATGG]ATATAGATAATGCGCGCATAGCATACTGAGGGTAGTAG-3’ Mutant:5’-ATGGCC[TATAT]ATGTCACTTGACTACGCAGCC[GCCACCATGG]ATATAGATAATGCGCGC AGAGCATACTGAGGGTAGTAG-3’ Answer: Is this a transition mutation? because there is an exchange of G instead of A? It kind of confuses me a little. help
5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. Write the resulting amino acid sequence using the 3 letter code. Write the answer in a all capital letters. Leave a space between the amino acids. Do not write 5' and 3'. 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a T, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutation 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a A, then the result will be A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutation
BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple
A portion of the sequence from the DNA coding strand of the chick ovalbumin gene is shown. Determine the partial amino acid sequence of the encoded protein. CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA-(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA Enter the 3-letter abbreviation for each amino acid in sequence, separated with dashes, and no spaces (example: xxx-xxx-XXX-XXX...) The amino acid sequence is .1104bp..…........
A double mutant produced by uneven crossing-over contains two single nucleotide mutations that result in frame shifts and are separated by about 20 base pairs. The first is an insertion, while the second is a deletion. The amino acid sequences of the wildtype and mutant polypeptide in this region of the protein are as follows: Wildtype: Lys – Lys – Tyr – His – Gln – Trp – Thr – Cys – AsnDouble Mutant: Lys – Gln – Ile – Pro – Pro – Val – Asp – Met – Asn a) What are the original and double mutant mRNA sequences. You may find it useful to use the conventional symbols Y for pyrimidine, R for purine, N for any nucleotide, and H for A,C, or T. 2. b) Which nucleotide was inserted? 3. c) Which nucleotide was deleted?