Concept explainers
To review:
The hypothesis to explain the formation of two different mutant
Introduction:
Transposable elements in yeast belongs to a family of dispersed and repeated retrotransposons in the genome of Saccharomyces cerevisiae. Saccharomyces cerevisiae is an important model organism to study the LTR (long terminal repeats)-retrotransposons biology. Five retrotransposon families are present in S. cerevisiae, designated as Ty1-Ty5. These Ty elements have a tendency to influence the genome organization. These Ty elements are the causal agents of mutation and genome rearrangment through recombination.
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Chapter 15 Solutions
Biology: The Dynamic Science (MindTap Course List)
- CTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mamma- lian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbal- anced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.arrow_forwardA mutant has no activity for the enzyme isocitrate lyase.Does this result prove that the mutation is in the geneencoding isocitrate lyase?arrow_forwardDescribe how transcription would be affected in the Galactose metabolizing pathway in Yeast in the presence of the following mutations. 1. A mutation that resulted in an inability of Gal80 to enter the nucleus. 2. A mutation that resulted in a lack of ability of Gal3 to bind galactose.arrow_forward
- Let’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.arrow_forwardThe authors state: "Properties of the single mutants K557R, D591V and K617A, a double mutant, D591V/K617A, and a triple mutant, K557R/D591V/K617A, were evaluated to find an allosteric binding site for citrate." The numbers between the letters represent the residue number altered. The letter in front was the wild type and the letter afterwards is the mutant. How would the the triple mutations K557R and D591V and K617A alter the overall protein? O lower the # of charges, increase the pl O increase the # of charges, increase the pl O lower the # of charges, lower the pl O lower the # of charges, increase the pl O no change in the # of charges, increase the pl O no change in the # of charges, lower the plarrow_forwardClary Leonhart used the pET vector system to express her prokaryotic amylase enzyme. She added peptone into her culture broth of BL21 (DE3) Escherichia coli strain. At the end of the experiment, she discovered that her protein was not expressed. She repeated three more times but her protein of interest was still not produced. (i) (ii) Explain the reason why Clary failed to obtain her protein of interest and suggest a solution to troubleshoot this problem. Clary plans to express her protein along with a polyhistidine-tag, or better known by its trademarked name IIis-tag. Explain the importance of His-tag in protein work.arrow_forward
- Termicin is a small antifungal protein in termites that is produced by cells and secreted into termite saliva in response to a pathogen. In vitro translation of the termicin-encoding gene is performed, and the effects of that product are compared to those of termicin extracted from a termite. You see that extracted termicin exhibits more antifungal behavior than in vitro translated termicin. After further analysis, you see that extracted termicin contains 3 disulfide bonds, while in vitro translated termicin contains zero. The addition of microsomes to the in vitro translation reaction results in termicin with all 3 disulfide bonds. What experimental condition is most likely responsible for this difference? A. in vitro translation was not performed at the correct temperature affecting protein folding B. a mutation occurred during in vitro translation, leading to differences in disulfide bond formation O C. the UPR can not be activated in vitro, therefore, this protein can only be…arrow_forwardExplain how the following mutations would affect transcription of the yeast GAL1 gene in the presence of galactose. (a) A deletion within the GAL4 gene that removes the region encoding amino acids 1 to 100. (b) A deletion of the entire GAL3 gene. (c) A mutation within the GAL80 gene that blocks the ability of Gal80 protein to interact with Gal3p. (d) A deletion of one of the four UASG elements upstream from the GAL1 gene. (e) A point mutation in the GAL1 core promoter that alters the sequence of the TATA box.arrow_forwardThe following table lists 4 bacterial strains that are partial diploids for lac operon genes. Given the activity of beta-galactosidase measured for each strain in the absence (-lac) or presence (+lac) of lactose, complete the table by choosing the appropriate symbol (+, -, C, S) to indicate the allele of the gene or site missing from the table (blue numbers). + = wildtype, - = null mutation, c = constitutive, s =super repressor chromosome plasmid B-gal act. strain 10 Z A 1 C B 3 4 + C + 6 D 9 + 1 [Select] 3 [Select] 5 [Select] 7 [ Select] 9 [Select] | 0 2 + + 5 + 7 10 Z + -lac +lac 0.062 0.058 0.003 0.004 0.062 0.117 0.003 0.060 + 8 + 2 [Select] 4 [Select] 6 [Select] 8 [Select] 10 [Select]arrow_forward
- Are genes A, D, and E all under the control of operator O? Explain your reasoningarrow_forwardFive individual His- mutants (hisA, his B, hisC, hisD and his E) were isolated. The genes are all involved in the histidine biosynthetic pathway. A crossfeeding experiment was carried out to determine the order of the enzymes encoded by these genes. The result is shown (A, B, C, D and E represent the mutants hisA, his B, hisC, hisD and his E, respectively. Dark area indicates bacterial growth). From the result shown, what might be the histidine biosynthetic pathway (the order of enzymes)?arrow_forwardThe intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.arrow_forward
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