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Why do the geneticists use the complementation experiment? Explain complementation with an example that shows the application of this process.
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- What is Complementation Analysis ?Below is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]What is an interrupted mating experiment? What type of experimentalinformation can be obtained from this type of study? Why is it necessary to interrupt conjugation?
- What is the most convenient way of understanding a testcross problem in genetics?How can the concept ofrecombination frequency beused in genetic mapping?Who was Gregor Mendel? Describe the plant that he performed his research with and explain at least two reasons why this plant proved to be an ideal model species for early genetic research.
- T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Why are there no proC− genotypes among the transductants?T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Is there evidence that proA, proB, and proD are located close to proC? Explain your answer.T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Which genotypes represent single transductants and which represent cotransductants?
- What advantages do anonymous DNA markers afford for genetic mapping as opposed to traditional allelic markers associated with visible phenotypes? What are the disadvantages of anonymous DNA markers for mapping?A complex biochemical pathway is shown below, along with the alleles that either promote or inhibit each step of the pathway leading to a phenotype. Gene A has alleles A and a, B has alleles B and b, and so forth. Genes B and C are duplicate dominant epistatic lethal as heterozygotes (i.e. Bb Cc are lethal). Genes D and E are duplicate dominant epistatic (i.e. dd eg = desired phenotype). If I were to cross AA Bb cc Dd Ee with aa BB Cc Dd e, (i) (ii) What proportion of all offspring don't show the phenotype? What proportion of offspring survive? Gene A Gene B B Gene D a Gene C Gene EWhat is an interrupted mating experiment? What type of experimentalinformation can be obtained from this type of study? Whyis it necessary to interrupt mating?