BIOL 202 Lecture Notes - Lecture 17: Knockout Mouse, Blastocyst, Chromosome

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25 Jan 2017

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Micropipette with dna solution injected into nucleus of single-cell mouse embryo. Integration of array into chromosome transgenic mous. Insert resistance marker in an exon of gene of interest. Screen for homologous recombination can result in: select for embryonic stem cell clones with gene replacement through homologous recombination by growing in presence of neomycin and. Stem cells are transferred into embryo and implanted (at blastocyst stage) to produce a chimera (made up from cells from two sources) Cross to find mosaic animals (heterozygotes) that express the transgene in their germ-line through pcr, etc. Transgenes can: label cells/tissues, tag proteins, label subcellular structures, cause gain or loss of function phenotypes, act as indicators of cellular events. Entire locus is cloned, as opposed to only the regulatory genes. Gfp is fused to the gene in the proper reading frame (most genes will still remain functional) Can observe endogenous protein levels and localization.

Related Questions

Which of the following statements about pronuclear injection is accurate?

a. DNA is inserted into any specific cell type whose fate has already been determined, such as skin cells.

b. DNA is inserted into a single celled mouse zygote so that the insertion can be passed to most/all cells of the resulting mouse embryo.

c. DNA that is inserted can integrate randomly anywhere in the genome of the mouse.

d. DNA that is inserted integrates at a specific site in the genome via homologous recombination.

e. DNA that is inserted includes an antibiotic resistance marker to be able to select for the embryonic stem cells (growing in dishes) that have incorporated the new DNA.

f. both A and C

g. both B and C

To generate a gene knockout in mice, which of the following is required?

a. a method for screening to verify insertion into the target gene in the mouse chromosome.

b. regions of homology between the researcher-engineered gene and the target gene in the mouse chromosome.

c. mouse Embryonic stem (ES) cells, to increase the chance that many cells in the host mouse including germline cells (which will form sperm and eggs) will carry the change.

d. all of the above

You follow the scheme outlined in to generate a knockout of the mouse gene Myo6, which is involved in motor coordination and movement. You derive ES cells from WT (Agouti-coated) mice, and disrupt one copy of the Myo6 gene in these ES cells. You then you transplant the ES cells into a blastocyst from a black mouse to generate a chimera. As shown in the last step in the figure, you then cross the chimera to a homozygous black mouse.

Some chimeric animals do not produce offspring carrying the gene replacement, while other chimeric animals with the same gene replacement can easily produce offspring carrying the gene replacement. Why?

a. chimeric animals are unable to produce offspring.

b. the gene replacement prevents correct gamete formation.

c. some chimeras have a germline entirely composed of cells from the blastocyst rather than the modified cells whereas others have a germline entirely composed of modified cells rather than cells from the blastocyst.

When you cross the chimera with the black mouse, amongst 20 F1 progeny obtained, you see some Agouti mice and some Black mice. Which of the following statements about the F1 progeny is true?

a. all the F1 that are Agouti will be homozygous for the myo6 knockout allele.

b. all the F1 that are Agouti will be heterozygous for the myo6 knockout allele.

c. 1/2 of the F1 that are Agouti will be heterozygous for myo6 knockout allele.

d. all the F1 that are Black will be homozygous for the myo6 knockout allele.

Q.In the non-9:3:3:1 ratio dihybrid cross below, there are 3 classes of phenotypes: white, pink, red which of the following is NOT true?

9 A_B_ white

3 A_bb pink

3 aaB_ red

1 aabb red

a.) phenotype is dependent on the genotype of gene B

b.) This is an example of recessive episasis

c.) aa is epistatic to B_, and bb

d.) phenotype is dependent on the genotype of gene A

Q,When choosing a selectable marker, you...

a.) choose a enhancer for your gene of interest

b.) choose a rescue gene to save your gene of interest

c.) choose a resistant gene that will allow for colony screening

d.) choose a gene that encodes for DNA replication

Q.The following steps detail an experiment resulting in transgenic mice with a knockout gene for fur color. Recall that agouti (brown) mice are A/A;M/M and black fur coats are a/a;M/M . List the steps in their correct chronological order.

1.Inject transformed agouti embryotic cells into an a/a;M/M embryo to produce A/A;M/m and a/a;M/M altered embryo

2.Self-cross hemizygous mice to produce homozygous transgenic mice (A/-;M/m)

3.Create a a targeting vector containing tk+ for ganciclovir sensitivity, neoR for neomycin resistance, and the gene encoding fur color.

4.Observe mice with chimeric phenotype and cross them with black a/a;M/M mice to produce transgenic hemizygous progeny (A/a;M/m)

5.Isolate all transformed embryonic stem cells that show homologous recombination and inject them into a embryonic stem cells of an agouti mouse.

a.) 3,5,4,1,2

b.) 1,5,3,4,2

c.) 3,5,1,4,2

d.) 5,3,2,4,2

If dog 1 has is black with alleles (A_B_) , and dog 2 is brown with (aaB_), and dog 3 is yellow with (_ _bb) alleles, what kind of epistasis is this and what color would dog 4 be if he was (AAbb).

1. Recessive epistasis, Yellow

2. Recessive epistasis, Black

3. Dominant epistasis, Yellow

4. Dominant epistasis, Black

A gene interaction in which pair of recessive genes at one locus prevents expression of dominant allele at another locus called

a. incomplete dominance

b. epistatis

c. complete dominance

d. polygenic inheritance

BIOL 202 Lecture Notes - Lecture 17: Knockout Mouse, Blastocyst, Chromosome

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