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1. What type of DNA repair is most likely to be used to repair a TG mispairing i

1. What type of DNA repair is most likely to be used to repair a TG mispairing in DNA?A. Base excision repairB. Nucleotide excision repairC. Mismatch excision repair2. Choose the pair that does not match.A. Homologous recombination: Holliday junctionB. NHEJ: double stranded break repairC. Dominant mutation: loss of functionD. crossing over: meiosis3. Which of these parts is/are required to be considered a virus?A. capsidB. envelopeC. genomeD. both A and BE. both A and C4. A group of volunteers is screened for mutations leading to a disease that appears at birth and persists through adulthood. 10 volunteers tested positive for the mutation but only 8 have the disease phenotype. Which term best explains this observation?A. codominantB. heterozygousC. variable penetranceD. recessive5. Why are yeast tetrads particularly useful for examining recessive mutations?A. the tetrads have 4 copies of the genome so even if tetrads have 2 defective alleles they can survive.B. the tetrads are haploid so recessive mutations can be observed easilyC. the tetrads divide more rapidly and allow expansion of the colony in a shorter timeD. the tetrads have some chromosomes in 2 copies and some in 1 copy so you can simultaneously examine heterozygous and homozygous loci.6. When would you need to generate a temperature sensitive mutation in yeast?A. When the mutation is lethal.B. When you are studying yeast cellular temperature regulationC. Any time you are doing a tetrad analysisD. When you want to transform the cells using heat shock7. We perform a complementation test on mutations x, y, and z. Individuals with x and y mutations are normal, individuals with x and z mutations have the mutant phenotype, individuals with y and z mutations are normal. What can we conclude from this complementation test?A. x, y, and z are mutations in the same geneB. y and z are mutations in the same geneC. x and y are mutations in the same geneD. x and z are mutations in the same gene8. Two different mutations (A and B) lead to inability to produce tryptophan. Mutation A leads to accumulation of biosynthetic intermediate 1, mutation B leads to accumulation of biosynthetic intermediate 2. When these mutations are combined we see accumulation of biosynthetic intermediate 2. What should we conclude about the epistatic relationship of these mutations?A. mutation A is in a gene whose product acts earlier in the biosynthetic pathwayB. mutation B is in a gene whose product acts earlier in the biosynthetic pathway9. You are trying to map a dominant mutation, T, with a phenotype of no tail and you find linkage to a dominant mutation that causes white coat color in mice, W. You perform the following cross:T w t w——– X ———t W t w You get the following phenotypes. How many centimorgans apart are the T and W loci?24: no tail, black 3: tail, black40: tail, white5: no tail, white A. 11.1 cMB. 12.5 cMC. 62.5 cMD. 88 cM10. When we use homologous recombination to do a gene deletion (knock-out), why do we include the Neo gene?A. To select for cells that have integration of our construct (desired DNA)B. To select against cells that have integrated in a random locationC. To mark the cells and allow screening for cells with integrationsD. To increase the rate of homologous recombination1. What type of DNA repair is most likely to be used to repair a TG mispairing in DNA?A. Base excision repairB. Nucleotide excision repairC. Mismatch excision repair2. Choose the pair that does not match.A. Homologous recombination: Holliday junctionB. NHEJ: double stranded break repairC. Dominant mutation: loss of functionD. crossing over: meiosis3. Which of these parts is/are required to be considered a virus?A. capsidB. envelopeC. genomeD. both A and BE. both A and C4. A group of volunteers is screened for mutations leading to a disease that appears at birth and persists through adulthood. 10 volunteers tested positive for the mutation but only 8 have the disease phenotype. Which term best explains this observation?A. codominantB. heterozygousC. variable penetranceD. recessive5. Why are yeast tetrads particularly useful for examining recessive mutations?A. the tetrads have 4 copies of the genome so even if tetrads have 2 defective alleles they can survive.B. the tetrads are haploid so recessive mutations can be observed easilyC. the tetrads divide more rapidly and allow expansion of the colony in a shorter timeD. the tetrads have some chromosomes in 2 copies and some in 1 copy so you can simultaneously examine heterozygous and homozygous loci.6. When would you need to generate a temperature sensitive mutation in yeast?A. When the mutation is lethal.B. When you are studying yeast cellular temperature regulationC. Any time you are doing a tetrad analysisD. When you want to transform the cells using heat shock7. We perform a complementation test on mutations x, y, and z. Individuals with x and y mutations are normal, individuals with x and z mutations have the mutant phenotype, individuals with y and z mutations are normal. What can we conclude from this complementation test?A. x, y, and z are mutations in the same geneB. y and z are mutations in the same geneC. x and y are mutations in the same geneD. x and z are mutations in the same gene8. Two different mutations (A and B) lead to inability to produce tryptophan. Mutation A leads to accumulation of biosynthetic intermediate 1, mutation B leads to accumulation of biosynthetic intermediate 2. When these mutations are combined we see accumulation of biosynthetic intermediate 2. What should we conclude about the epistatic relationship of these mutations?A. mutation A is in a gene whose product acts earlier in the biosynthetic pathwayB. mutation B is in a gene whose product acts earlier in the biosynthetic pathway9. You are trying to map a dominant mutation, T, with a phenotype of no tail and you find linkage to a dominant mutation that causes white coat color in mice, W. You perform the following cross:T w t w——– X ———t W t w You get the following phenotypes. How many centimorgans apart are the T and W loci?24: no tail, black 3: tail, black40: tail, white5: no tail, white A. 11.1 cMB. 12.5 cMC. 62.5 cMD. 88 cM10. When we use homologous recombination to do a gene deletion (knock-out), why do we include the Neo gene?A. To select for cells that have integration of our construct (desired DNA)B. To select against cells that have integrated in a random locationC. To mark the cells and allow screening for cells with integrationsD. To increase the rate of homologous recombination

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