MIC 4124 Study Guide - Final Guide: Mast Cell, Endosome, Adaptive Immune System

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erinraven63

27 Apr 2017

School

University of Ottawa

Department

Microbiology and Immunology

Course

MIC 4124

Professor

L Isa D'ambrosio

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Document Summary

A bacterial strain that is most common among individuals in a population. Can also be a wild type gene or characteristic. To enhance the expression of desirable gene products, reduce the expression of undesirable products through downregulation, and create strains that can grow on lower cost carbon sources. Another reason is to determine the genes responsible for biological processes. This can be done by inducing mutations and screening for a certain phenotype that represents a defect in the desired biological process. Researchers can then determine the gene from that mutant. Explain the process of phenotypic selection and screening. Selection is the process of using media that only permits growth of strains that expression particular gene products. For example, using a medium with antibiotics to select for mutants that expression antibiotic resistance genes. The wild type will not be able to grow on this medium.

Related Questions

18) In your research on gene regulation in E. coli, you produce 2 mutant strains with the same phenotype: they can grow on lactose but not on glucose. To investigate these 2 mutants further, you perform a complementation test using them and discover that the diploid combination strain can grow on glucose or lactose. From this, you can deduce that:

a.)Both mutants have a mutation in the same gene

b.)The 2 mutants have mutations in different genes
c.)Glucose and lactose metabolic pathways in E. coli utilize at least one shared gene in common
d.)At least 2 different genes are required for glucose metabolism in E. coli
e.) b) and d)

19. In your research on the lac operon, you produce a unique regulatory mutatant strain of E. Coli where the lac l gene's repressor protein binding site for the operator is normal (wild-type) but its lactose binding site no longer binds lactose. The expected phenotype of this mutant will be:

a. inability to grow on lactose only medium

b. ability to grow on lactose only medium

c. blue colony color when grown on glycerol plus x-gal medium

d. blue colony oclor when grown on lactose plus x-gal medium

e. b & d

20.

20) In your research on the lac operon, you create a diploid E.coli strain with the following genotype (+) = wild type gene, (-) = mutant gene.

I- /O+ /Z- /Y+

I+ /Oc /Z+ /Y+
Can this diploid mutant grow on medium containing lactose as the only carbon source?

a) Yes

b) No

c) Maybe, it depends on if the Oc mutation is dominant or recessive

salmonporpoise696

In a genetic experiment:

The use and limits of RNA interference as a reverse genetics technique for knocking down gene expression were tested. The wild-type and mutants worms for the strains were selected. The RNAi feeding strains and control bacterial were used for the the RNAi experiment. The eri-1 C. elegan strains (wild-type) were used for RNAi treatment and mutant strains that were used ( unc-22 and dpy-13) C. elegans. Both carbencillin and IPTG wered added to the RNAi feeding plates and E.coli containing the RNAi transgenes were spread on the plates. The antibodic carbacillin only allows growth of bacteria which contains the RNAI construct. IPTG induces the expression of the polymerase which then transcribes the siRNA in the bacteria. 6 NGM plates without bacteria were used and 20ul of 12.5 mg/ml carbencillin concentration were spread on the 6 of the plates. The carbencillin dried and then IPTG was spread (25ug/ml) on each plate. Then 6 plates with the RNAi bacterial strains were used. Theree of plates for strain one and the other three for strain the second strain. When IPTG dried , 150 ul of the appropraite bacteria was added onto each plate. 2NGM plates (without carbenicillin or IPTG) was spread with 150 ul of OP50 bacterial culture. OP50 is the regular E. coli strain for culturing C. elegans and does not contain the RNAi constructs. These plate were used as the control plates6. Then the plates were incubated for two days. Then the adult hermaphrodites from the RNAi feeding plates and the op50 plates were removed. Then from the eri-1 wildtype stock, 2 or 3 L4 hermaphrodites or
young adults were transferred onto each RNAi feeding plate. Then L4 worms were transferred to an intermediate plate first. After they have a minute to move away from each other and the eggs, and transfered to the RNAi plates. On the two control plates seeded with OP50, 2 L4 or young adult hermaphrodites were placed there. Plates were obtain as the reference mutants (unc-22, bli-1 or unc-13). Then the plates were incubated to allow the worm to develop and lay eggs in two days. Then the adult hermaphrodites from your RNAi feeding plates and the op50 plates were removed and then placed back into the incubator.

The op500 was the negative control plate wtih 80 and 300 progeny and unc-22 had around 50. (reference plates)

Mating in the RNAi plates

Dpy (60) x wildtype (15) and unc( 70) x wildtype ( 8) progeny

the phenotype was movement and it seemed similar phenotype within the progeny.

Can you help me understand the results and what is happening genetically in the experiment ? Are the for dyp-15 x unc-22 mating cross and unc-22 x eri-1 wildtype C. elegans. Based on the data:

Briefly discuss the results of the RNAi experiment. How closely did the RNAi-treated worms phenocopy the mutant worms?

Was the gene knockdown 100% effective?

Was the RNAi treatmentequally effective for the two genes? If not, discuss this result and why that could occured.

Please be specific and detailed

Genetics HELP ASAP

Genetics HELP ASAP In a genetic experiment: The use and limits of RNA interference as a reverse genetics technique for knocking down gene expression were tested. The wild-type and mutants worms for the strains were selected. The RNAi feeding strains and control bacterial were used for the the RNAi experiment. The eri-1 C. elegan strains (wild-type) were used for RNAi treatment and mutant strains that were used ( unc-22 and dpy-13) C. elegans. Both carbencillin and IPTG wered added to the RNAi feeding plates and E.coli containing the RNAi transgenes were spread on the plates. The antibodic carbacillin only allows growth of bacteria which contains the RNAI construct. IPTG induces the expression of the polymerase which then transcribes the siRNA in the bacteria. 6 NGM plates without bacteria were used and 20ul of 12.5 mg/ml carbencillin concentration were spread on the 6 of the plates. The carbencillin dried and then IPTG was spread (25ug/ml) on each plate. Then 6 plates with the RNAi bacterial strains were used. Theree of plates for strain one and the other three for strain the second strain. When IPTG dried , 150 ul of the appropraite bacteria was added onto each plate. 2NGM plates (without carbenicillin or IPTG) was spread with 150 ul of OP50 bacterial culture. OP50 is the regular E. coli strain for culturing C. elegans and does not contain the RNAi constructs. These plate were used as the control plates6. Then the plates were incubated for two days. Then the adult hermaphrodites from the RNAi feeding plates and the op50 plates were removed. Then from the eri-1 wildtype stock, 2 or 3 L4 hermaphrodites or young adults were transferred onto each RNAi feeding plate. Then L4 worms were transferred to an intermediate plate first. After they have a minute to move away from each other and the eggs, and transfered to the RNAi plates. On the two control plates seeded with OP50, 2 L4 or young adult hermaphrodites were placed there. Plates were obtain as the reference mutants (unc-22, bli-1 or unc-13). Then the plates were incubated to allow the worm to develop and lay eggs in two days. Then the adult hermaphrodites from your RNAi feeding plates and the op50 plates were removed and then placed back into the incubator. The op500 was the negative control plate wtih 80 and 300 progeny and unc-22 had around 50. (reference plates) Mating in the RNAi plates Dpy (60) x wildtype (15) and unc( 70) x wildtype ( 8) progeny the phenotype was movement and it seemed similar phenotype within the progeny. Can you help me understand the results and what is happening genetically in the experiment ? Are the for dyp-15 x unc-22 mating cross and unc-22 x eri-1 wildtype C. elegans. Based on the data: Briefly discuss the results of the RNAi experiment. How closely did the RNAi-treated worms phenocopy the mutant worms? Was the gene knockdown 100% effective? Was the RNAi treatmentequally effective for the two genes? If not, discuss this result and why that could occured. Please be specific and detailed

MIC 4124 Study Guide - Final Guide: Mast Cell, Endosome, Adaptive Immune System

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