Genetic recombination and conditional knockouts
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Genetic recombination and conditional knockouts
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[ID] => 553814
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[post_date] => 2024-12-23 07:43:57
[post_date_gmt] => 2024-12-23 12:43:57
[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
A "conditional knockout" defines an animal model in which a gene of interest is either: (1) inactivated only in specific cell types such that other cell types exhibit unmodified, functional gene expression (tissue-specific knockout model); or (2) temporarily suppressed at a given time-point in development (inducible knockout model).
Researchers can use enzymes such as Cre recombinase to engineer tissue-specific knockouts. Cre recombinase cleaves DNA at palindromic consensus sequences called loxP sites. If researchers synthesize a gene flanked by loxP sites (called a “floxed” locus), this locus is susceptible to elimination in the presence of Cre recombinase, as shown in Figure 1. A key step in generating conditional knockouts is selecting a promoter that will drive the expression of the inserted Cre gene coding for CRE recombinase but only in certain cell types.
Figure 1 Cre-Lox Conditional Knockout Technology
Researchers used this technology to examine the effects of myeloid-specific Caveolin 1 (CAV1) knockouts. CAV1 is a ubiquitously expressed gene that codes for a protein involved in endocytosis and phagocytosis. Researchers engineered mice with floxed CAV1 alleles and used a LysM promoter to express Cre recombinase. Myeloid cells (including monocytes, macrophages, and neutrophils) express LysM, the gene coding for lysozyme M.
Figure 2 LysM promoter specifies Cre expression (CDS = coding sequence)
[post_title] => Genetic recombination and conditional knockouts
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[question] => Rank mice of the following genotypes in order of decreasing vulnerability to bacterial infection.
I. Heterozygous for LysMCre; CAV1flox/flox
II. Homozygous for LysMCre; CAV1+/–
III. Wild-type LysM; CAV1+/+
IV. Heterozygous for LysMCre; CAV1 flox/–
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
Decreasing vulnerability to bacterial infection means that the “weakest” innate immune system should come first.
According to the passage, CAV1 is important for endocytosis and phagocytosis. Therefore, vulnerability to bacterial infection is proportional to a loss of CAV1 expression, and cells with the least CAV1 expression should come first; cells with the most CAV1 expression should come last.
The presence of a floxed CAV1 gene means that CAV1 expression will be turned off in the presence of Cre recombinase, which would only be expressed in cells expressing LysMCre and therefore containing this gene.
Cells with the wildtype genotype III. Wild-type LysM; CAV1+/+ have the maximum CAV1 expression and should be ranked last (have the least vulnerable immune system).
Here is what would happen with the others:
I. Heterozygous for LysMCre; CAV1flox/flox: CAV1 would be fully expressed but completely floxed and therefore significantly reduced by LysMCre expression (due to the heterozygous LysMCre genotype).
II. Homozygous for LysMCre; CAV1+/–: CAV1 would be reduced by 50% regardless of LysMCre expression.
III. Heterozygous for LysMCre; CAV1 flox/–: CAV1 would be reduced by 50% due to one gene knockout (the “–”), and the expressed CAV1 (floxed) would be further reduced by LysMCre expression (due to the heterozygous LysMCre genotype).
It is probably easiest to see that option IV would have lower CAV1 expression than the other two options, and the order must be IV > I > II > III.
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[answers] => Array
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[each_answer] => A. I > II > IV = III
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[1] => Array
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[each_answer] => B. IV > I > II > III
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[2] => Array
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[each_answer] => C. III > I > IV > II
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[3] => Array
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[each_answer] => D. II > IV > III > I
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[1] => Array
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[quiz_unique_key] => 3243476205
[question] => Why would a vascular endothelial cell express the same level of CAV1 regardless of whether it came from a LysMCre; CAV1flox/flox mouse or from a wild-type mouse?
[value] => Array
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[answer] => 4
[description] => Reason for Correct Answer:
The passage states that Cre recombinase can generate conditional knockouts if animal models contain the genes of interest, like CAV1, that are surrounded by loxP sites (“floxed”), where Cre recombinase can cleave.
Paragraph 2 states that part of the technology is “selecting a promoter that will drive the expression of the inserted Cre gene coding for CRE recombinase but only in certain cell types.” This will ensure that floxed genes are only cut out in these cell types.
Paragraph 3 explains that “myeloid cells… express LysM, the gene coding for lysozyme M,” and using the LysM promoter allows models to express Cre recombinase in these myeloid cells only. (Also see Figure 2.)
This means that, in LysMCre + animals, myeloid cells selectively receive the proper transcriptional activation signals to express LysM and synthesize Cre recombinase, which reduces CAV1 expression, whereas endothelial cells do not activate the LysM promoter and therefore do not express Cre recombinase and do not exhibit decreases in CAV1 expression**.
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[answers] => Array
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[each_answer] => A. Cre recombinase would inhibit the translation of CAV1 mRNA only in cells with an active LysM promoter, which does not include endothelial cells.
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(
[each_answer] => B. Endothelial cells do not have floxed alleles in this system, so the presence of Cre recombinase does not matter.
)
[2] => Array
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[each_answer] => C. The coding sequence for Cre is only in myeloid cells, so none of this enzyme is produced in endothelial cells.
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[each_answer] => D. Only myeloid cells receive the proper transcriptional activation signals to synthesize Cre recombinase, so endothelial expression of CAV1 is unaffected.
)
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[2] => Array
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[quiz_unique_key] => 2187790141
[question] => Cre recombinase cuts DNA at loxP sites. What other enzymatic process is required after this cleavage in order for recombination to occur?
[value] => Array
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[answer] => 4
[description] => Reason for Correct Answer:
DNA polymerization is the synthesis of DNA from a DNA template.
DNA helicase action is the unwinding of the double helix in order to make a replication bubble.
In DNA replication, DNA ligase seals Okazaki fragments to make a new DNA strand.
The same thing (ligation) would need to happen here, joining the ends of DNA cleaved by Cre.
Note DNA topoisomerases unravel twists in DNA during DNA transcription and replication, to help relieve torsional stress. Topoisomerases I and II act through cutting the DNA backbone on one or two strands, passing either the other strand of the same helix or another double strand through the break, and finally resealing the DNA backbone.
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[answers] => Array
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[0] => Array
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[each_answer] => A. DNA polymerase action
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[1] => Array
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[each_answer] => B. DNA topoisomerase action
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[2] => Array
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[each_answer] => C. DNA helicase action
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[3] => Array
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[each_answer] => D. DNA ligation
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[3] => Array
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[quiz_unique_key] => 1579828684
[question] => What might be the effect of a floxed premature stop codon in the coding sequence of a wild-type CAV1 allele in a LysMCre mouse?
[value] => Array
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[answer] => 3
[description] => Reason for Correct Answer:
Premature stop codons cause protein synthesis to be terminated prematurely. This results in the production of a truncated, and usually nonfunctional, protein.
In LysMCre mice, Cre recombinase will be present in myeloid cells and will cleave DNA at loxP sites.
If the loxP sites surround the premature stop codon, Cre recombinase would remove the premature stop codon.
The floxed premature stop codon would therefore prevent proper CAV1 translation in all cells except those that contain Cre recombinase. Since Cre is specifically expressed in myeloid cells, Cre would remove the floxed premature stop codon and CAV1 would be expressed properly in myeloid cells only.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Without the floxed premature stop codon, RNA polymerase would synthesize mRNA with extraneous, nonsense nucleotides at the 3’ end.
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[each_answer] => B. The floxed premature stop codon would prevent RNA polymerase from synthesizing mRNA from the DNA template strand.
)
[2] => Array
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[each_answer] => C. Cre would remove the floxed premature stop codon and CAV1 would be expressed in myeloid cells only.
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[each_answer] => D. Since this would be in an intron of the CAV1 gene, there would be no effect on the protein product.
)
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[4] => Array
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[quiz_unique_key] => 1498560436
[question] => Which sequence most likely allows Cre recombinase to bind and cleave DNA at a loxP site?
[value] => Array
(
[answer] => 1
[description] => Reason for Correct Answer:
The passage states that Cre recombinase cleaves at palindromic sequences.
Palindromic sequences are sequences of nucleotides, amino acids, or other units that read the same backward as forward.
In the context of DNA, RNA, or protein sequences, palindromic sequences refer to sequences of nucleotides or amino acids that are the same when read from the 5′ to 3′ direction on both strands.
For these choices, see which complementary strand would read the same in the 5’ to 3’ direction (start on the right side of the strand given).
Only the strand in Choice A is palindromic. You can probably eliminate the other answer choices just by looking at the beginning of each strand as highlighted here.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 5′-ATAACTTCGTATAGCATACATTATACGAAGTTAT-3′
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[1] => Array
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[each_answer] => B. 5′-CGTGATCTAATGACATGCAGTGCATGATACCTA-3′
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[2] => Array
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[each_answer] => C. 5′-TACGTAAGTGCTAATTGTCCAATGCGGGCCACC-3′
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[3] => Array
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[each_answer] => D. 5′-GCTTAGAGTCTGTGTGCTCTGTCTGAGATTCG-3′
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[5] => Array
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[quiz_unique_key] => 3044259542
[question] => Which of the following are likely reasons why researchers generate conditional knockouts?
I. Deleting a gene from an organism completely can be lethal if that gene influences vital, ubiquitous processes.
II. Researchers are testing the influence of a given gene on the expression of other genes in different cell types.
III. Researchers are interested in assessing the role of a gene in a given developmental process.
IV. Researchers want to assess the function of a gene in a specific cell type or tissue.
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
In a conditional knockout model, a specific gene is inactivated in a certain cell type or tissue.
By observing the phenotype of the knockout model, the function of this gene can be determined (IV is correct).
By observing the development of the knockout animal, the role of the gene in development can be determined (III is correct).
If the gene of interest is critical in certain tissues, its global inactivation could be lethal; therefore, conditional knockout allows the gene to be inactivated only in tissues where it is not essential (I is correct).
If the expression of genes in several tissues is compared between wild-type and knockout animals, the influence of the gene of interest on other genes in different cell types could be determined (II is correct).
)
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[each_answer] => A. I, II, and IV only
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[each_answer] => B. I, II, III, and IV
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[each_answer] => C. II and IV only
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[each_answer] => D. I and III only
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[553814|2] => D
[553814|3] => D
[553814|4] => C
[553814|5] => A
[553814|6] => B
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Figure 1
Figure 2