Auxotroph Yeast and Human DIseases

1. These questions will utilize yeast analysis and provide you with mock experimental data for the phenotypic analysis of yeast growth to infer genotypes. The following symbols indicate growth or no growth on medium lacking particular components (examples in 1A are medium lacking methionine "-MET" and medium lacking tryptophan "-TRP". Growth is indicated by a filled, black circle. No growth on a particular medium is indicated by an unfilled circle. In cases where we look viability of spores, we would use absence of a circle to indicate that a spore did not grow at all (for example, it inherited a lethal mutation). Growth No growth 1A. You have two haploid strains (Haploid of opposite mating types. Their genotypes and phenotypes are shown below. +MET - TRP - MET +TRP Haploid 1 MATA: MET2 trp3 Haploid 2 MATalpha: met2 TRP3 You cross each strain to a WT yeast and observe the following: Haploid 1 x WT= Growth on all three media shown above. Haploid 2 x WT= Growth on all three media shown…

Why is “Agrobacterium-mediated genetic transformation” in plants described as natural genetic engineer of plants?

Before JCVI-syn3.0 was produced, what steps were used to create a synthetic Mycoplasma cell? Put the events below into the correct order. Drag and drop the events into the proper sequence from left to right. ▸ View Available Hint(s) First Step Transformed cells are detected by the presence of lacz, which causes them to cleave Xgal and produce a blue color. Yeast is transformed using Mycoplasma mycoides fragments. The synthetic chromosome is transformed into Mycoplasma capricolum. The synthetic Mycoplasma mycoides chromosome is purified. DNA fragments assemble through homologous recombination in yeast. Reset Help Last Step

Flavr savr Tomato Golden rice, Roundup ready soyabean In order to generate GM foods mentioned above, various gene transfer methods are required to be conducted on these plants. Discuss on both vector-mediated and direct DNA transfer to deliver foreign genes into plants.

Schizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that is unable to synthesize one or more vitamins. The following table represents the key experiments you performed during your genetic screen. Fill in the table with the outcome of each experiment for your mutant strain (using + for growth and - for no growth). Medium Rich media Minimal media Minimal media + all vitamins Minimal media + all amino acids Growth Wild-type + + + + Mutant + + + > > >

Why is Agrobacterium mediated genetic transformation described as Natural Genetic engineering in plants?

Nutritional requirements of white verses red yeast What is the differences in the ability of different strains to grow on different mediums. Propose a hypothesis that helps to explain the presence of red yeast colonies. Why it is more common for cells to be killed by UV light than to become red. Hint: think about how mutagenesis occurs.

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 type

BICD100 HW6 Prokaryotic Genetics Question 1. You are interested in three genes in bacteriophage. The recessive mutant alleles cause plaque phenotypes that were creatively named fuzzy, shaky, and purple. Another lab published the following map of the three genes: fuzzy shaky 2.9 17.4 purple To verify the published map, you cross a purple shaky phage strain with a fuzzy phage strain by co- infecting E. coli at a high multiplicity of infection (every bacterium infected with both types of phage). You plate the resulting lysate and analyze the phenotypes of the plaques caused by the progeny phage. a. State the genotypes of the two parent phage strains: b. List all eight of the possible phenotypes that could result from this cross. If 400 phage plaques were examined from this cross, how many plaques of each of the eight phenotypic classes would you expect?

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…

Streptomycin resistance in Chlamydomonas may result from a mutation in either a chloroplast gene or a nuclear gene. What phenotypic results would occur in a cross between a member of an mt+ strain resistant in both genes and a member of a strain sensitive to the antibiotic? What results would occur in the reciprocal cross?

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?

You have identified five genes in S. cerevisiae that are induced when the yeast are grown in a high-salt (NaCl) medium. To study the potential roles of these genes in acclimation to the growth in high-salt conditions, you wish to examine the phenotypes of loss- and gain-of-function alleles of each. How will you do this?

In this activity you will analyze the results of experiments that investigate nutritional requirement of several mutant strains of yeast. The mutations in these strains cause a nutritional requirement for an amino acid, such that the strains will not grow in media that lack one specific amino acid. Any mutant that has a nutritional requirement is called an auxotroph, and is incapable of growing in a "minimal medium" containing only a carbon source (e.g., glucose), a simple nitrogen source (e.g., ammonium sulfate), and various salts and minerals. Such strains can be supported on a medium supplemented with only the missing nutrient or on a "rich" medium that contains amino acids, vitamins, nitrogenous bases, etc. (often in the form of an extract from yeast). The wild-type individual that can synthesize the metabolic component is a prototroph, and is capable of growth on minimal medium. The mutant strains in this activity are unable to synthesize tryptophan, lysine, or histidine; one…

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?

After preparing cultures and completing a six-hour growth and viability trial, you notice that your lab partner had only used 5 mL of YPD media to prepare your low concentration experimental culture. Your medium and high concentration experimental cultures received 10 mL of YPD as required. What do you think happened to the growth and viability of your yeast during your trials? Why?

A student said, “The approach shown by the Chicago researchers is just conventional gene therapy with biomaterial carrier”. Is he or she correct? Why?      based on  “Targeted polyelectrolyte complex micelles treat vascular complications in vivo”, PNAS 2021, Vol. 118 No. 50 e2114842118;107:110210; https://doi.org/10.1073/pnas.2114842118

21:14 +4G KB/S LTED 41 96 09.00 Vo Microsoft Word - Assignment 02.docx Assignment 02 1. Groups of alleles associated with the lactose operon are as follows (in order of dominance for each allelic series): repressor, IS (superrepressor), I *(inducible), and I "(constitutive); operator, OC (constitutive, cis dominant) and 0+ (inducible, cis- dominant); structural, Z and Y. (a) Which of the following genotypes will produce B-galactosidase and B-galactoside permease if lactose is present: (1) 1*0* Z* Y*, (2) 1 *OCZ+ Y+, (3) OCZ+ Y+, (4) IS O+ Z+ Y+, and (5) 10+ Z+Y+? (b) Which of the above genotypes will produce ß-galactosidase and ß-galactoside permease if lactose is absent? Why? 2. By what mechanism does the presence of tryptophan in the medium in which E. coli cells are growing result in premature termination or attenuation of transcription of the trp operon? 3. Suppose that you used site-specific mutagenesis to modify the trpL sequence such that the two UGG Trp codons at positions…

#22) You have identified five genes in S. cerevisiae that are induced when the yeast are grown in a high-salt (NaCl) medium. To study the potential roles of these genes in acclimation to the growth in high-salt conditions, you wish to examine the phenotypes of loss-and-gain-of-function alleles of each. How will you do this?

Cap, EA1, and Sap are all genes/proteins of interest in this study. For each gene, what gene product is encoded and where is the gene (the literal DNA sequence) located physically in the cell?    I need help fimiding this in the artticle and answer as short as possible  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC106848/

52.  Papaya plants are very susceptible to the papaya ringspot virus (PRSV), which stunts the growth of the fruit and negatively impacts the taste. By 1995, PRSV wiped out the Hawaiian papaya crop completely and made it impossible to produce papaya commercially. In 1998, Cornell University developed a genetically modified papaya called Rainbow that was resistant to PRSV and introduced it to Hawaii. Currently, 80% of the papaya cultivated in Hawaii is genetically modified to be resistant to PRSV.Which of the following rows identifies the effect of PRSV on the original papaya population and the reason why the population recovered? Select one: a. Effect on the Original Population Reason for Recovery Population bottleneck Some papaya had the PRSV-resistance gene. b. Effect on the Original Population Reason for Recovery Increased gene flow The PRSV-resistance gene had to be artificially introduced to the papaya genome c. Effect on the Original Population Reason…

a- https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5866609/pdf/pnas.201800195.pdfb- (Article is no longer available) Here are the articles provided within the question - They are not needed but are available if additional information is warranted - The figures being mentioned are already provided. This is Biochemoistry, please answer each part to the best of your ability. There are a max of 3 parts due to guidelines and please answer each part with clear and efficient work with answers. Thank you

Robert Bost and Richard Cribbs studied a strain of E. coli (araB14) that possessed a nonsense mutation in the structural gene that encodes Lribulokinase, an enzyme that allows the bacteria to metabolize the sugar arabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From the araB14 strain, they isolated some bacteria that possessed mutations that caused them to revert back to the wild type. Genetic analysis of these revertants showed that they possessed two different suppressor mutations. One suppressor mutation (R1) was linked to the original mutation in L-ribulokinase and probably occurred at the same locus. By itself, this mutation allowed the production of L-ribulokinase, but the enzyme produced was not as effective in metabolizing arabinose as the enzyme encoded by the wild-type allele. The second suppressor mutation (SuB) was not linked to the original mutation. In conjunction with the R1 mutation, SuB allowed the production of L-ribulokinase, but SuB by itself was not able…

Robert Bost and Richard Cribbs studied a strain of E. coli (araB14) that possessed a nonsense mutation in the structural gene that encodes Lribulokinase, an enzyme that allows the bacteria to metabolize the sugar arabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From the araB14 strain, they isolated some bacteria that possessed mutations that caused them to revert back to the wild type. Genetic analysis of these revertants showed that they possessed two different suppressor mutations. One suppressor mutation (R1) was linked to the original mutation in L-ribulokinase and probably occurred at the same locus. By itself, this mutation allowed the production of L-ribulokinase, but the enzyme produced was not as effective in metabolizing arabinose as the enzyme encoded by the wild-type allele. The second suppressor mutation (SuB) was not linked to the original mutation. In conjunction with the R1 mutation, SuB allowed the production of L-ribulokinase, but SuB by itself was not able…

Cystic fibrosis (CF) is a genetic disorder affecting a number of organs, including the lung airways, pancreas, and sweat glands. Mutations in both copies of the CFTR gene causes cystic fibrosis. Imagine that you have sweat gland samples from several Cystic Fibrosis patients (A-C) with unknown mutations in CFTR. You also have normal (+) sweat gland sample to use as a positive control. А В С А В С Choose which mutation would explain the RNA and protein results in A, B, & C: 1. Promoter/Regulatory mutation 2. Silent mutation 3. Missense mutation 4. Deletion mutation 5. Splice site mutation 6. Nonsense mutation RNA gel Protein gel

In the early 1990s, Carolyn Napoli and her colleagues were working on petunias, attempting to genetically engineer a variety with dark purple petals by introducing numerous copies of a gene that encodes purple pigment in the flower petals (C. Napoli, C. Lemieux, and R. Jorgensen. 1990. Plant Cell 2:279–289). Their thinking was that extra copies of the gene would cause more purple pigment to be produced and would result in a petunia with an even darker hue of purple. However, much to their surprise, many of the plants carrying extra copies of the purple gene were completely white or had only patches of color. Molecular analysis revealed that the amount of mRNA produced by the purple gene was reduced 50-fold in the engineered plants compared with wild-type plants. Somehow, the introduction of extra copies of the purple gene silenced both the introduced copies and the plant’s own purple genes. Provide a possible explanation for how the introduction of numerous copies of the purple gene…

Experiment: Page 6 de can exist stably in either a haploid or a diploid state. A haploid S. cerevisiae mes. When certain haploids come into contact, they fuse their cell walls and membranes, followed by the fusion of their nuclear membranes. The single nucleus now has 32 chromosomes, 16 from each parent strain, and is thus a diploid. Haploid yeast strains divide mitotically to give rise to haploid progeny, and diploid strains divide mitotically to give rise to diploid progeny. Certain haploids can fuse to form diploids. Haploid S. cerevisiae exists in two "mating types," called a and a Mating occurs only between a and a cells; no mating occurs between cells of identical mating type. We have a collection of eight a haploid mutant strains and eight & haploid mutant strains of yeast unable to synthesize tryptophan (trp). These will be combined (mated) in all possible combinations to yield diploid strains. If the diploids can grow on minimal medium, complementation occurred between the…

What is “translocation”? Why it is essential for plants.

Human Biology- MILESTONE 6 X P genetics-and-biotechnology-glo O https://app.sophia.org/spcc/human-biology-2-mile -> Sophia Pathways M Gmail Purdue Global - Sig... Welcome - Login.gov UNIT 6 – MILESTONE 6 Which of the following is a gene abnormality in which two nonhomologous parts rearrange and fuse together? O Translocation O Deletion None of these, O Duplication SAVE & CONTINUE Report an issue with this question Type here to search