You continue to study the expression of the hexose kinase gene and capture the following electronmicrograph of the gene being expressed.MRNA120ORI4060TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATCAAGCTCGAGAGCAСCAGCTСТАТGCGСТАСТАТААТGAССАКТАТАСССТАCGTGATAG5"3'RNApromoterpolymerase4a. Was this micrograph taken of a sample prepared from human cells or prokaryotic cells? How doyou know?4b. What is the sequence of the first 10 nucleotides of the transcript of this gene?5'3'---4c. What are the first 5 amino acids encoded by this gene?)Note - there is a codon tableavailable at the beginning of this exam.N'C'4d. Will translation stop at the UAA which begins at position 41? Explain your logic4e. You also study the expression of 3 different mutants for this gene. For each mutant answer thefollowing:Does this mutation change the sequence of the protein produced? Why or why not?If it does change the sequence of protein be sure to write out the new sequence.o If it does not change the protein sequence, what effect (if any) would you expect itto have on expression of the gene?120ORI40605'.TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTAT.3'3'...АAGCTCGAGAGCAGCAGCTCТАТСCGCTАСТАТААTGACCATTAТАССССТАСGTGATAG..5'promoteri. Mutant A has a single base pair substitution with the TIA being replaced with C/G base pair atposition 35 (position denoted by the * in the sequence above).ii. Mutant B has a 2 GIC pairs inserted between position 19 and 20 (position denoted by the ^ inthe sequence above).ii. Mutant C has had the first 5 base pairs deleted (position 1-5). Does this mutation change thesequence of the protein produced? Why or why not? If it does change the sequence of proteinbe sure to write out the new sequence. If it does not change the protein sequence, what effect(if any) would you expect it to have on expression of the gene?ribosome

Name: You continue to study the expression of the hexose kinase gene and ca
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Description: You continue to study the expression of the hexose kinase gene and capture the following electronmicrograph of the gene being expressed.MRNA120ORI4060TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATCAAGCTCGAGAGCAСCAGCTСТАТGCGСТАСТАТААТGAС

Gene expression includes transcription and translation. It leads to synthesize mRNA and protein…

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MRNA 20 ORI 40 - FTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGHAATATOGGGATGCACTATC AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCANTATACECCTACGTGATAG 60 promoter RNA polymerase 4a. Was this micrograph taken of a sample prepared from human cells or prokaryotic cells? How do you know? 4b. What is the sequence of the first 10 nucleotides of the transcript of this gene? 3' 4c. What are the first 5 amino acids encoded by this gene? available at the beginning of this exam. ) Note - there is a codon table N'. 4d. Will translation stop at the UAA which begins at position 41? Explain your logic ribosome

MRNA 20 ORI 40 - FTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGHAATATOGGGATGCACTATC AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCANTATACECCTACGTGATAG 60 promoter RNA polymerase 4a. Was this micrograph taken of a sample prepared from human cells or prokaryotic cells? How do you know? 4b. What is the sequence of the first 10 nucleotides of the transcript of this gene? 3' 4c. What are the first 5 amino acids encoded by this gene? available at the beginning of this exam. ) Note - there is a codon table N'. 4d. Will translation stop at the UAA which begins at position 41? Explain your logic ribosome

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 13QP

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In the topic about prokaryotic gene expression regulation, we learned about the LacI-, LacIS, and LacOc mutations. Based on your knowledge of allosteric regulation, invent and describe a hypothetical mutation that would produce constitutive expression of the Lac operon but is not a mutation that occurs in the Lac operator (that is, it must be a different mutation than a LacOc type mutation) and is not a mutation to the DNA binding domain of LacI (that is, it must also be a different mutation than a LacI- type mutation). Clearly mention where in the prokaryotic genome this mutation would occur, and explain the logic behind how that mutation would produce constitutive Lac operon expression.
Arginine is one of the amino acids that serves as a building block for protein synthesis. Arginine biosynthesis depends on the expression of the arg operon, which encodes ArgC (a reductase) and ArgB (acetylglutamate kinase). Which of the following statements regarding mutation of the arginine biosynthesis repressor ArgR is true? O A 1-bp insertion in the argR gene would result in a mutant able to grow in the absence of arginine, O A 1-bp deletion in the argB gene would result in a mutant unable to transcriptionally regulate the arg operon encoding argB. O A 1bp insertion in the argC gene would result in a mutant able to grow in the absence of arginine. O A 1-bp deletion in the argR gene would result in a mutant that represses the arg operon when grown in the presence of arginine. None of the above.
You have isolated two different mutants (reg1 and reg2) causing constitutive expression of the emu operon (emu1 emu2). One mutant contains a defect in a DNA-binding site, and the other has a loss-of-function defect in the gene encoding a protein that binds to the site. Is the DNA-binding protein a positive or negative regulator of gene expression? Explain. To determine which mutant has a defect in the site and which one has a mutation in the binding protein, you decide to do an analysis using F′ plasmids. Assuming you can assay levels of the Emu1 and Emu2 proteins, what results do you predict for the two strains (i and ii; see descriptions below) if reg2 encodes the regulatory protein and reg1 is the regulatory site? Explain. F′ (reg1− reg2+ emu1− emu2+)/reg1+ reg2+ emu1+ emu2− F′ (reg1+ reg2− emu1− emu2+)/reg1+ reg2+ emu1+ emu2−
If you were given a tube containing only a purified protein of interest but had unlimited time and resources, how would you determine the in vivo expression levels of that protein? Is there another level of regulation beyond the protein expression level which might not be observable with the method you described?
Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.
You are teaching a class on the regulation of eukaryotic gene expression. In order to demonstrate this complex process, you decide to draw for the class a typical eukaryotic gene/transcription unit with its major regions, such as the promoter regions, where the RNA polymerase II and transcription factors would bind From the list given - choose all components that you think are part of a typical eukaryotic gene From the list given - choose all the regulatory sequences that you think would control the expression of this eukaryotic gene From the list given - choose all of the regulatory proteins that would bind the eukaryotic gene to control its expression
Searching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4. GAL4 activates the transcription of galactose genes. Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism. The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T. In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added. These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to. The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced. Results showed that only 10 different DNA sequences had GAL4 bound. Since the…
Consider 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?
You are interested in finding out in which organ GTF2H5 is highly expressed experimentally. Describe a method in recombinant DNA technology to measure gene expression, explain your rationale, and step by step, what are needed for this method to work
A strain of bacteria possesses a temperature-sensitive mutation in the gene that encodes the rho subunit. At high temperatures, rho is not functional. When these bacteria are raised at elevated temperatures, which of the following effects would you expect to see? Explain your reasoning for accepting or rejecting each of these five options. a. Transcription does not take place. b. All RNA molecules are shorter than normal. c. All RNA molecules are longer than normal. d. Some RNA molecules are longer than normal. e. RNA is copied from both DNA strands.
Consider the following DNA sequence, which codes for a short polypeptide: 5'-ATGGGCTTAGCGTAGGTTAGT-3' Determine the mRNA transcript of this sequence. You have to write these sequences from the 5' end to the 3' end and indicate those ends as shown in the original sequence in order to get the full mark. How many amino acids will make up this polypeptide? Determine the first four anticodons that will be used in order to translate this sequence.
You then make a screen to identify potential mutants (shown as * in the diagram) that are able to constitutively activate Up Late operon in the absence of Red Bull and those that are not able to facilitate E. Coli growth even when fed Red Bull. You find that each class of mutations localize separately to two separate regions. For those mutations that prevent growth even when fed Red Bull are all clustered upstream of the core promoter around -50 bp. For those mutations that are able to constitutively activate the operon in the absence of Red Bull are all located between the coding region of sleep and wings. Further analysis of each DNA sequence shows that the sequence upstream of the promoter binds the protein wings and the region between the coding sequence of sleep and wings binds the protein sleep. When the DNA sequence of each is mutated, the ability to bind DNA is lost. Propose a final method of gene regulation of the Up Lateoperon using an updated drawn figure of the Up Late…