BIOL-UA 21 Lecture Notes - Lecture 15: Wild Type, Chromosome, Telomere

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11 Nov 2016

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Lecture 15: molecular genetic analysis i: gene discovery in model organisms: wild type is abbreviated as (+) and mutants are abbreviated sometimes as (-), genetic approach to gene discovery: Generate many mutant individuals using mutagens such as x-rays, uv, or dna damaging chemicals. There are natural mutants, but creation is very slow. This however doesn"t tell you what gene was mutated. This can be done in yeast by functional complementation: screening for recessive mutations: Ems: random mutations of dna of individual sperm. Random mutations; different genes are damaged in different ies. Convenient to expose males; each sperm will have one or more mutations. The sperm then fertilizes eggs of wild type females. In the next generation, males are collected, each with lots of different mutations. A male with an interesting phenotype is then crossed with a wild type homozygous female, and half the progeny would then have a mutation (heterozygotes) and the rest would be wild-type.

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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

1).Which of the following demonstrates the importance of mutation?

A.Mutation is the ultimate source of genetic variation and thus is the foundation of why people are interested in genetics in the first place.

B.Mutation will remove variation from the population.

C.Mutations are always beneficial and thus allow organisms to adapt.

D.Mutation is detrimental to the population and thus is important only because we need to get rid of mechanisms that induce mutations.

E.Mutation is not important.

2).While the SOS repair pathway can repair devastating pyrimidine dimers, it comes at a cost. Which of the following best explains the cost of repairing DNA via the SOS repair pathway?

A.The other strand of DNA is used as a template to repair the damaged strand of DNA.

B.A nucleotide is removed so that DNA polymerase III can move past the pyrimidine dimer.

C.SOS will help DNA replication bypass pyrimidine dimers but will do so by using random nucleotides in place of the lesion.

D.The SOS repair pathway uses analogs in place of nucleotides when repairing damaged DNA.

3.Suppose you have two species of corn. One has sweet kernels and the other has large ears. Both have 20 chromosomes, each of which is homologous in both species. Suppose you want to create a hybrid that has both characteristics of sweet kernels and large ears, so you cross individuals from the two species. The initial hybrid you obtain has 20 chromosomes, but then you establish a viable tetraploid (4n) individual with 40 chromosomes. Which of the following could best explain this observation?

A.		Nondisjunction of a single chromosome caused the doubling of the genome.
		B.The initial F1 hybrid was unable to produce normal haploid gametes and instead produced diploid gametes. As a result, self-fertilization created progeny with 40 chromosomes.
		C.The original F1 hybrid was able to produce normal gametes, thus, the tetraploid was formed by fusing one normal haploid gametes with an abnormal diploid gamete.
		D.Nondisjunction of a random set of chromosomes produced abnormal gametes, and thus the 40 chromosomes represent a random assortment of chromosomes from the two parents.
		E.None of the above can explain the phenomenon.

E.There is no consequence to SOS repair and the damaged DNA is repaired precisely back to the wild-type allele.

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BIOL-UA 21 Lecture Notes - Lecture 15: Wild Type, Chromosome, Telomere

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