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Experimental Design:

Assuming you have two yeast strains HR1 (mating type a) and HR2 (mating type a). Both yeast strains produce red colonies on YED plates, so you know there are ADE1 or ADE2 gene mutations in each of the yeast strain.

How do you conduct a complementation test to determine whether mutations in these two strains of yeast are in the same gene or different gene?

Describe your experimental treatment and the expected outcome for each treatment.

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Nelly Stracke
Nelly StrackeLv2
28 Sep 2019

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Related questions

1. Haploid Saccharomyces cerevisiae yeast cells contain the following alleles at the MAT locus.

a. Allele a

b. Allele alpha

c. Allele a or alpha

d. Alleles a and alpha

2. The following statements are true for the Saccharomyces cerevisiae yeast, except __________.

a. Yeast mating type a produces a-factor.

b. Yeast mating type alpha produces alpha-factor.

c. a-factor or alpha-factor signals the presence of a yeast cell to the neighboring cells.

d. Yeast mating type a-cell responds to a-factor, alpha-cell responds to alpha-factor.

3. A yeast strain can sexually conjugate with another yeast strain of the same mating type.

a. True

b. False

4.Heterothallic Saccharomyces cerevisiae yeast strains commonly used in controlled experiments are haploid, with either mating type a or alpha.

a. True

b. False

5. Complementation test is used to ____________________________.

a. determine the mating type of yeast.

b. determine whether mutations occur at the same gene or different genes.

c. identify mutant organisms.

d. determine functional relationships between genes.

6. Yeast cells with mating type alpha can sexually conjugate with yeast _________.

a. mating type a

b. mating type alpha

c. allele a or alpha

d. alleles a and alpha

7. Complementation test is conducted by ___________ two strains of organisms with different mutations.

a. isolating

b. morphologically identifying

c. growing

d. intercrossing

8. Assuming alleles B or b are found in locus 1; alleles C or c are found in locus 2. Which of the yeast strains below contains recessive mutations at loci 1 and 2?

a. Yeast strain with alleles B and C

b. Yeast strain with alleles B and b

c. Yeast strain with alleles C and c

d. Yeast strain with alleles b and c

9. When mutations occur at different genes, two strains of yeast are considered to be not complement.

a. True

b. False

10. Yeast cells that are ready to mate will produce a mating projecting called _______ to form a pear-shape cell.

a. pilus

b. yeast ring

c. shmoo

d. spore

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

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