WT A B с D E F G H WT A + + B + C DE + + + + F + G H + + + The above complementation table analyzes seven mutants in the phenylalanine (amino acid) synthesis pathway. A "+" indicates the resulting diploid was able to grow on media lacking phenylalanine, while a "-" indicates a failure to grow on media lacking penylalanine. Determine the complementation groups of the above mutants A through H.
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PLEASE ANSWER BOTH PARTS OF THE QUESTION
a.) The Question on the image b.) From this experiment, in how many genes have we identified mutations?
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- The pathway for arginine biosynthesis in Neurospora crassa involves several enzymes that produce a series of intermediates as shown. O O O O ornithine citrulline ARG-E arginosuccinate arginine N-acetylornithine arginine You did a cross between ARG-E ARG-H* and ARG-E* ARG-H¯¯ Neurospora strains and identified an Arg- strain from an NPD tetrad. (Assume that Neurospora forms tetrads in the same way yeast do.) Which compound would rescue growth of this Arg- spore? N-aceltylornithine ARG-F ornithine citrulline ARG-G ARG-H → argininosuccinateThe intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +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 + + + > > >
- The 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.Assume that a series of compounds has been discovered in Neurospora. Compounds A–F appear to be intermediates in a biochemical pathway. Conversion of one intermediate to the next is controlled by enzymes that are encoded by genes. Several mutations in these genes have been identified and Neurospora strains 1–4 each contain a single mutation. Strains 1–4 are grown on minimal media supplemented with one of the compounds A–F. The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth). Which biochemical pathway fits the data presented? Media Supplement Strain A B C D E F 1 o o o + + + 2 o o o o + + 3 o o o o + o 4 o o + + + + A) A → B → C → D → E → F B) A → B → C → F → D → E C) F → B → C → D → A → E D) A → B → C → D → F → E E) A → B → F → E → C → DThree 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
- For each of the following genotypes, explain how mutation (identified by a (-) will affect the organism grown in lactose medium. Indicate whether a) B-galactosidase will be synthesized or not, b) synthesis of B-galactosidase is inducible (1) or constitutive (C) and c) growth of the organism will occur or not. i. I+P+0-Z+Y+A+/l+P+O+Z-Y-A- ii. I-P+0+Z+Y+A+/I+P-O+Z+Y+A+In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?You generate mutants in the metabolic pathway for starlase. You conduct some complementation tests (after testing for dominance of course) and come up with the following results: 1 2 3 4 5 6 1 I 2 + 3 + 4 + + LO 5 a. How many complementation groups are there? [Select] 6 + + + + +
- 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.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…