A shuttle vector is a vector constructed so that it can propagate in two different host speci One of the most common types of shuttle vectors is the yeast shuttle vector. (i) Name the THREE (3) most common yeast shuttle vectors. (ii) Based on the answer above, differentiate these THREE (3) vectors.
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- The figure below shows the life cycle of the fungus Neurospora. The adult stage of the Neurospora is a multicellular haploid. b) Neurospora has an arginine amino acid synthesis pathway shown below. Suppose I take the strain above that only grows with arginine supplements and cross it to a different mutant Neurospora strain that grows with arginine and citrulline supplements but not with ornithine supplements. Assuming gens A, B, and C are unlinked and there is only one mutation per stain: What percentage of the progeny will grow on ornithine? What percentage on citrulline? What percentage on arginine?Three haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typeWith respect to F+ and F- bacterial matings, answer the following questions: (a) How was it established that physical contact between cells was necessary? (b) How was it established that chromosome transfer was unidirectional? (c) What is the genetic basis for a bacterium’s being F+?
- In a process of production of a recombinant protein by E. coli cells, it was observed accumulation of acetate in the culture medium. In this situation, it can be said that: (a) certainly the process in question was being conducted in anaerobiosis (B).Acetate accumulation is advantageous for the process as the acetate formation reaction generates 1 molecule of ATP (c)Knowing that decreasing the temperature of the process causes a reduction in the rate of glycolysis, this could be a strategy to reduce the accumulation of acetate (d).the acetate formed can be re-assimilated by the cell if the glyoxylate pathway is activated at some point in the cultureWe have two specific strains of E. coli that have shown horizontal gene transfer (HGT) when mixed. To experimentally determine the method of HGT that is happening, the following conditions are set up in different tubes of culture media: A) Donor and recipient strain mixed together (control - no treatment). B) Donor and recipient strains mixed together, DNase added (can digest DNA in solution, not within cells).C) Special tube containing a membrane filter (with pores that allow DNA and viruses to pass through, but not bacterial cells) that separates two compartments. Donor strain is added on one side, the recipient strain on the other (they are separated by the filter).D) Donor and recipient strains mixed together, with chemical that inactivates viruses (chemical affects bacteriophages in solution so they are unable to attach to cells). The results: Tubes A, B, and D: HGT was observed. Tube C: HGT was NOT observed. Based on this, which type of HGT was occurring? Conjugation,…Nine rII− mutants of bacteriophage T4 were used inpairwise infections of E. coli K(λ) hosts. Six of themutations in these phages are point mutations; theother three are deletions. The ability of the doubly infected cells to produce progeny phages in large numbers is scored in the following chart.1 2 3 4 5 6 7 8 91 − − + + − − − + +2 − + + − − − + +3 − − + − + − −4 − + − + − −5 − − − + +6 − − − −7 − + +8 − −9 −The same nine mutants were then used in pairwise infections of E. coli B hosts. The production of progenyphages that can subsequently lyse E. coli K(λ) hosts isnow scored. In the table, 0 means the progeny do notproduce any plaques on E. coli K(λ) cells; − meansthat only a very few progeny phages produce plaques;and + means that many progeny produce plaques(more than 10 times as many as in the − cases).1 2 3 4 5 6 7 8 91 − + + + + − − + +2 − + + + + − + +3 0 − + 0 + + −4 − + − + + +5 − + − + +6 0 0 − +7 0 + +8 − +9 −a. Which of the mutants are the three deletions? Whatcriteria did…
- Which statement is true regarding a bacteriophage in the lysogenic cycle? 111 A) O The bacteriophage can exit the lysogenic state depending on the condition of host cell B) O While in this cycle, it would cause no harm to the host cell. O OAs the host cell divides, the bacteriophage genome will be present in all of the host cell's progeny. |D) OIt would become a prophage. E) O All are true statements.Two mutations that affect plaque morphology in phages (a and b ) have been isolated. Phages carrying both mutations (a b) are mixed with wild-type phages (a* b*) and added to a culture of bacterial cells. Once the phages have infected and lysed the bacteria, samples of the phage lysate are collected and cultured on plated bacteria. The following numbers of plaques are observed: Plaque phenotype Number a* b* 2043 a* b- 320 a b* 357 2134 What is the frequency of recombination between the a and b genes?Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…
- Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3? After crossing the F1 generation of the cross between mutant strains 1…Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3?Two mutations that affect plaque morphology in phages (a− and b −) have been isolated. Phages carrying both mutations (a− b−) are mixed with wild-type phages (a+ b+) and added to a culture of bacterial cells. Once the phages have infected and lysed the bacteria, samples of the phage lysate are collected and cultured on plated bacteria. The following numbers of plaques are observed: Plaque phenotype Number a+ b+ 2043 a+ b− 320 a− b+ 357 a− b− 2134 What is the frequency of recombination between the a and b genes?