TNBGGA: The No Bullshit Guide to Genetic Analysis

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chapter_09 [2024/09/01 11:47] – [F' is a version of F that carries segments of the $E. coli$ chromosome] mikechapter_09 [2025/03/18 18:04] (current) – [Gene Complementation in Bacteria: F plasmids] mike
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-<typo fs:x-large>Chapter 09. %%Complementation%% in bacteria</typo>+<-chapter_08|Chapter 08^table_of_contents|Table of Contents^chapter_10|Chapter 10-> 
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 +<typo fs:x-large>Chapter 09. %%Complementation in bacteria%%</typo>
  
 In this chapter, we initially touch on some concepts in classical //E. coli// genetics that may not be of practical interest to all students except future microbiologists, such as F plasmids and Hfr mapping. However, it is useful to learn these concepts because later in the chapter we talk about cloning by complementation, which is a critical concept and skill that all serious students of biology (especially those interested in the area of molecular and cellular biology) need to understand.  In this chapter, we initially touch on some concepts in classical //E. coli// genetics that may not be of practical interest to all students except future microbiologists, such as F plasmids and Hfr mapping. However, it is useful to learn these concepts because later in the chapter we talk about cloning by complementation, which is a critical concept and skill that all serious students of biology (especially those interested in the area of molecular and cellular biology) need to understand. 
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 {{ :f_plasmid.png?400 |}} {{ :f_plasmid.png?400 |}}
 <caption> <caption>
-F plasmids. //oriT// (erroneously labeled as OriT in the figure )is the origin of replication for the F plasmid, and the //tra// genes (erroneously labeled as Tra in the figure) are required for forming the pilus needed for transfer of the F plasmid from one cell to another (see Fig. {{ref>Fig2}}). F plasmids are about 10<sup>5</sup> bp long; for comparison, the size of the circular //E. coli// chromosome is 5x10<sup>6</sup> bp, or over 10-fold longer.+F plasmids. //oriT// (erroneously labeled as OriT in the figure) is the origin of transfer (related to how F is transmitted from one host to another) for the F plasmid, and the //tra// genes (erroneously labeled as Tra in the figure) are required for forming the pilus needed for transfer of the F plasmid from one cell to another (see Fig. {{ref>Fig2}}). F plasmids are about 10<sup>5</sup> bp long; for comparison, the size of the circular //E. coli// chromosome is 5x10<sup>6</sup> bp, or over 10-fold longer.
 </caption> </caption>
 </figure> </figure>
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 ===== F' is a version of F that carries segments of the $E. coli$ chromosome ===== ===== F' is a version of F that carries segments of the $E. coli$ chromosome =====
  
-Homologous recombination can sometimes occur at a different position to give excise an F plasmid that carries a part of the //E. coli// chromosome. In the example in Fig. {{ref>Fig6}}, this is the BC segment of the //E. coli// chromosome. This form of F is called an F' (pronounced “F prime”).+Homologous recombination can sometimes occur at a different position to excise an F plasmid that carries a part of the //E. coli// chromosome. In the example in Fig. {{ref>Fig6}}, this is the BC segment of the //E. coli// chromosome. This form of F is called an F' (pronounced “F prime”).
  
 <figure Fig6> <figure Fig6>
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 {{ :r_plasmid.png?400 |}} {{ :r_plasmid.png?400 |}}
 <caption> <caption>
-An R factor that confers multiple drug resistance to bacteria.+An R factor that confers multiple drug resistance to bacteria. Note that $sul^r$, $amp^r$, $kan^r$, and $tet^r$ are erroneously labeled as Sul<sup>r</sup>, Amp<sup>r</sup>, Kan<sup>r</sup>, and Tet<sup>r</sup>, respectively.
 </caption> </caption>
 </figure> </figure>
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 {{ :minimal_r_plasmid.png?400 |}} {{ :minimal_r_plasmid.png?400 |}}
 <caption> <caption>
-Schematic of a modified R plasmid used for cloning. For scale, a typical F plasmid is around 10<sup>5</sup> bp of DNA, but a typical R plasmid used for cloning is just a few thousand bp.+Schematic of a modified R plasmid used for cloning. For scale, a typical F plasmid is around 10<sup>5</sup> bp of DNA, but a typical R plasmid used for cloning is just a few thousand bp. Note that $amp^r$ is erroneously labeled as Amp<sup>r</sup>. 
 </caption> </caption>
 </figure> </figure>
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 </table> </table>
  
-Plaques can be counted to measure how efficient a phage is at infecting bacteria. Phage grown using //E. coli// strain C are much less efficient at infecting //E. coli// strain K (Table {P{ref>Tab2}}). But phage grown using strain C can infect strain C efficiently, just as phage grown using strain K can infect strains C or K efficiently. This phenomenon is known as host restriction, and it behaves like a genetic trait that reverts at a high frequency. +Plaques can be counted to measure how efficient a phage is at infecting bacteria. Phage grown using //E. coli// strain C are much less efficient at infecting //E. coli// strain K (Table {{ref>Tab2}}). But phage grown using strain C can infect strain C efficiently, just as phage grown using strain K can infect strains C or K efficiently. This phenomenon is known as host restriction, and it behaves like a genetic trait that reverts at a high frequency. 
  
 The explanation is that //E. coli// strain K makes a restriction enzyme that cleaves DNA, including λ DNA. The K strain doesn’t destroy its own chromosomal DNA because it also makes a DNA methylase that enzymatically modifies nucleotide bases at the cleavage site and thus prevents the restriction enzyme from working. By rare chance, a small amount of λ phage can grow on strain K because they escaped cleavage long enough to be modified. Subsequently, when λ phage are produced in strain K, the phage DNA is also modified and thus protected from cleavage. λ phage produced in strain C is not modified, and therefore when these phages are used to infect strain C, the infection efficiency is much lower as its DNA is cleaved by the restriction enzyme.  The explanation is that //E. coli// strain K makes a restriction enzyme that cleaves DNA, including λ DNA. The K strain doesn’t destroy its own chromosomal DNA because it also makes a DNA methylase that enzymatically modifies nucleotide bases at the cleavage site and thus prevents the restriction enzyme from working. By rare chance, a small amount of λ phage can grow on strain K because they escaped cleavage long enough to be modified. Subsequently, when λ phage are produced in strain K, the phage DNA is also modified and thus protected from cleavage. λ phage produced in strain C is not modified, and therefore when these phages are used to infect strain C, the infection efficiency is much lower as its DNA is cleaved by the restriction enzyme. 
  
-The //E. coli// genes for restriction enzymes usually come in pairs organized in an operon (more on operons in Chap. {{ref>Fig10}}), with the gene for the restriction enzyme (R) residing next to the gene for the DNA methylase that modifies the same sequence (M) (Fig. {{ref>Fig9}}). +The //E. coli// genes for restriction enzymes usually come in pairs organized in an operon (more on operons in Chap. {{ref>Fig10}}), with the gene for the restriction enzyme ($R$) residing next to the gene for the DNA methylase that modifies the same sequence ($M$) (Fig. {{ref>Fig9}}). 
  
 <figure Fig10> <figure Fig10>
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 </figure> </figure>
  
-Mutants that have a mutated version of the restriction enzyme but a wildtype version of the modifying enzyme ($R^- / / M^+$) are useful for studying phage because they do not show host restriction, but phage grown on these strains are resistant to host restriction. We don't generally consider phage genetics in this course further, even though it's pretty interesting. +Mutants that have a mutated version of the restriction enzyme but a wildtype version of the modifying enzyme ($R^- M^+$) are useful for studying phage because they do not show host restriction, but phage grown on these strains are resistant to host restriction. We don't generally consider phage genetics in this course further, even though it's pretty interesting. 
  
 A large number of restriction enzymes have been isolated from different bacterial species and are commercially available. Most of the enzymes recognize palindromic DNA sequences of 4 or 6 base pairs. A palindrome (when talked about in an everyday sense) is a sentence that reads exactly the same when read forward or backwards (e.g., "A man, a plan, a canal, Panama"). A palindromic sequence is a DNA sequence that is the same on one strand and its reverse complement strand. The 6 bp palindrome below is recognized by a restriction enzyme called //EcoR//I (pronounced "eh-co-ar one"):  A large number of restriction enzymes have been isolated from different bacterial species and are commercially available. Most of the enzymes recognize palindromic DNA sequences of 4 or 6 base pairs. A palindrome (when talked about in an everyday sense) is a sentence that reads exactly the same when read forward or backwards (e.g., "A man, a plan, a canal, Panama"). A palindromic sequence is a DNA sequence that is the same on one strand and its reverse complement strand. The 6 bp palindrome below is recognized by a restriction enzyme called //EcoR//I (pronounced "eh-co-ar one"): 
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 {{ :cloning_by_complementation_e_coli.png?400 |}} {{ :cloning_by_complementation_e_coli.png?400 |}}
 <caption> <caption>
-Cloning by complementation from a DNA library generated from the //E. coli// genome. +Cloning by complementation from a DNA library generated from the //E. coli// genome. Note that the $amp^r$ gene is erroneously labeled as Amp<sup>r</sup>.
 </caption> </caption>
 </figure> </figure>
  
-Say we wanted to clone the Lac operon, or genes from the Lac operon (more details on the Lac operon are discussed in [[chapter_10|Chap. 10]]). First, a genomic library would be made from chromosomal DNA from a Lac<sup>+</sup> //E. coli// strain using an R plasmid as our cloning vector (Fig. {{ref>Fig11}}). To make this library, //E. coli// chromosomal DNA (we can also say genomic DNA) is purified and cleaved with a restriction enzyme to fragment the chromosome. An appropriate R plasmid is cleaved with the same restriction enzyme, and the chromosomal fragments are ligated to the R plasmid. This library would then be used to transform a Lac<sup>–</sup> //E. coli// strain. Transformants (i.e., //E. coli// cells carrying a plasmid from the library) would first be selected for by resistance to the antibiotic ampicillin, which is a feature of the R plasmid we are using (Amp<sup>R</sup>). Any cell that is ampicillin resistant must carry a plasmid from the library. (As a side note, R plasmids have a property that make it so that each clone can only carry one type of R plasmid; therefore, each clone will contain a different and unique chromosomal fragment.) The resistant colonies would then be selected for the ability to grow on lactose (Lac<sup>+</sup>). These clones should not only contain R plasmids, but they should also carry an insert that contains a functional Lac operon.+Say we wanted to clone the Lac operon, or genes from the Lac operon (more details on the Lac operon are discussed in [[chapter_10|Chap. 10]]). First, a genomic library would be made from chromosomal DNA from a Lac<sup>+</sup> //E. coli// strain using an R plasmid as our cloning vector (Fig. {{ref>Fig11}}). To make this library, //E. coli// chromosomal DNA (we can also say genomic DNA) is purified and cleaved with a restriction enzyme to fragment the chromosome. An appropriate R plasmid is cleaved with the same restriction enzyme, and the chromosomal fragments are ligated to the R plasmid. This library would then be used to transform a Lac<sup>–</sup> //E. coli// strain. Transformants (i.e., //E. coli// cells carrying a plasmid from the library) would first be selected for by resistance to the antibiotic ampicillin, the phenotype  (Amp<sup>R</sup>) of which is conferred by a gene on the R plasmid we are using ($amp^r$). Any cell that is ampicillin resistant must carry a plasmid from the library. (As a side note, R plasmids have a property that make it so that each clone can only carry one type of R plasmid; therefore, each clone will contain a different and unique chromosomal fragment.) The resistant colonies would then be selected for the ability to grow on lactose (Lac<sup>+</sup>). These clones should not only contain R plasmids, but they should also carry an insert that contains a functional Lac operon.
  
-How many clones would we need to screen? Depending on the type of restriction enzyme used, each plasmid carries about 4 x 10<sup>3</sup> bp of chromosomal DNA (the frequency of recognition sites is {(\frac{1}{4})}^6 for an enzyme like //EcoR//I that recognizes a 6 bp site, or one site every 4096 bp on average). The //E. coli// chromosome is 5 x 10<sup>6</sup> bp; thus, the entire //E. coli// genome will be covered if several thousand clones from the library are screened.+How many clones would we need to screen? Depending on the type of restriction enzyme used, each plasmid carries about 4 x 10<sup>3</sup> bp of chromosomal DNA (the frequency of recognition sites is $(\frac{1}{4})^6$ for an enzyme like //EcoR//I that recognizes a 6 bp site, or one site every 4096 bp on average). The //E. coli// chromosome is 5 x 10<sup>6</sup> bp; thus, the entire //E. coli// genome will be covered if several thousand clones from the library are screened.
  
 All sorts of genes from //E. coli// have been cloned by looking for DNA fragments that can restore function to a mutant. It is also possible to find genes from other bacteria. The following is a dramatic example of a cloning experiment to find an important protein for a pathogenic bacterium. All sorts of genes from //E. coli// have been cloned by looking for DNA fragments that can restore function to a mutant. It is also possible to find genes from other bacteria. The following is a dramatic example of a cloning experiment to find an important protein for a pathogenic bacterium.
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 ===== Questions and exercises ===== ===== Questions and exercises =====
  
-Exercise 1: In Chap. {{ref>5}} we discussed the relationship between recombination frequency (m.u. or cM) and physical distance (measured in bp of DNA) (see also [[chapter_12|Chap. 12]] Fig. 1). What is the relationship between minutes in Hfr mapping and physical distance? In other words, how many bp of DNA is equivalent to 1 minute? See [[chapter_08|Chap. 08]] for some key information you might need. +Exercise 1: In [[chapter_05|Chap. 05]] we discussed the relationship between recombination frequency (m.u. or cM) and physical distance (measured in bp of DNA) (see also [[chapter_12|Chap. 12]] Fig. 1). What is the relationship between minutes in Hfr mapping and physical distance? In other words, how many bp of DNA is equivalent to 1 minute? See [[chapter_08|Chap. 08]] for some key information you might need. 
  
-Conceptual question: why would a strain with a mutated modifying enzyme but a wildtype restriction enzyme (R+ M-) be inviable (incompatible with life)? In other words, why is $R^+ / / M^-$ a lethal mutation?+Conceptual question: why would a strain with a mutated modifying enzyme but a wildtype restriction enzyme ($R^+ M^-$) be inviable (incompatible with life)? In other words, why is $R^+ M^-$ a lethal combination of alleles?
  
  
chapter_09.1725216447.txt.gz · Last modified: 2024/09/01 11:47 by mike