20 ORF for gene X e for gene X above has a hypomorphic mutation (which you name Mutation #1) that is on the diagram. Based on the experimental data below, what specific sequence e mutation most likely in? pe Суре direction of transcription 70 3' +1 in
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![5'
3'
ORF for gene X
The sequence for gene X above has a hypomorphic mutation (which you name Mutation #1) that is
not indicated on the diagram. Based on the experimental data below, what specific sequence
element is the mutation most likely in?
Wild Type
Northern
Gene X
O -10 region
O 3' of the ORF
ribosome binding site
O -35 region
Mutant
O within the ORF
Wild Type
Western
Gene X
direction of transcription
Mutant
3'
+1
5'](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c5565a-787b-43d3-b23e-780453faf6da%2F67562c1a-8e16-4124-8f01-62ccc3c9bd02%2Frga1k3_processed.png&w=3840&q=75)
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- 5' 3' ORF for gene X +1 You find a mutation elsewhere in the genome (not within the sequence above) that you decide to call Mutation #2. When Mutation #1 in gene X (from the above question) is combined with Mutation #2 in the same organism, you get the phenotype in the blots below. This second mutation is most likely in what specific factor? Wild Type Northern Gene X O ribosome O sigma factor O DNA polymerase RNA polymerase Double mutant Wild Type Western Gene X direction of transcription Double mutant 3' 5'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 LeadpleCynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X G
- Fig 2 shows gastric organoids cultured from a TffCre; XYZI/fl mouse. Ki67 has been used to identify proliferating cells (green). Tamoxifen indicates exposure to tamoxifen to induce recombination of the TffCre. No Tamoxifen Ki-67 3. What is the consequence of deleting XYZ in gastric organoids, and does this suggest it is an oncogene or a tumour suppressor? Explain your reasoning. DX Equation Editor 4. IWP inhibits the secretion of Wnt ligands from the organoids. Therefore, do Wnt ligands promote or inhibit proliferation of XYZ deficient organoids? Explain your reasoning. X Equation Editor Tamoxifen Q Tamoxifen + IWP A- A I B IU S x, x² Styles Font Size Font Words: 0/250 张 A- A-T BIUS X, x² Styles ▾ SizeHelicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.Chemical Mutagenesis of DNA Bases Show the nucleotide sequence changes that might arise in a dsDNA (coding strand segment GCTA) upon mutagenesis with (a) HNO2, (b) bromouracil, and (C) 2-aminopurine.
- Homologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesn’t the mismatch repair system of E. coli eliminate these mismatches?The Enzymatic Activities of DNA Polymerase I (a) What are the respective roles of the 5 -exonudease and 3 -exonuclease activities of DNA polymerase I? (b) What might be a feature of an E. coli strain that lacked DNA polymerase I 3 -exonuclease activity?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…
- In the following gel showing stained bands of the Alu insertion sequence, what is the genotype of individual 2? 941 bp 641 bp->>> 1 2 3 4 5 6 Homozygous for the 641 bp sequence that does not contain in the Alu insertion Heterozygous, containing one 941 bp sequence and one 641 bp sequence O Homozygous for the 941 bp sequence containing the Alu insertionThe RNA polymerase from bacteriophage T7 diff ers structurally from prokaryotic and eukaryotic RNAPs and is extremely specifi c for its own promoter. Why do these properties make T7 RNAP useful in experiments with recombinant DNA?THE MOLECULAR GENETICS OF SICKLE CELL ANEMIA The following is the base sequence of DNA that codes for first eight amino acids of the ß chain of hemoglobin. The ß chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. mone formed DNA Template Strand: 3'CACGTGGACTGAGGACTCCTC5' 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 4. What amino acids will this mRNA code for? 3. What mRNA will be formed from the template strand of DNA? 5. If the 17th base in the template strand of the DNA is changed from T to A, rewrite the new template strand below. 6. When the template strand of the DNA is changed, this is referred to as a mutation. What kind of mutation is this? 67
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