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--- chapter_11 [2024/08/25 11:47] – mike
+++ chapter_11 [2025/04/07 20:58] (current) – mike
@@ Line 1: / Line 1: @@
-<typo fs:x-large>Chapter 11. Gene circuits and epistasis</typo>
+<-chapter_10|Chapter 10^table_of_contents|Table of Contents^chapter_12|Chapter 12->
+<typo fs:x-large>Chapter 11. %%Gene circuits and epistasis%%</typo>
 In [[chapter_10|Chapter 10]], we studied regulatory mechanisms in well-known //E. coli// operons to see how mutations in different elements of the system would behave in dominance tests and cis/trans tests. We also presented the information in reverse - we told you the answer first, then discussed how mutant phenotypes were interpreted.
@@ Line 15: / Line 17: @@
 </figure>
-Although loss of function mutations in genes for repressors or activators are generally the most common type of regulatory mutation, Table {{ref>Tab1}} will help you to interpret mutations in sites or more complicated mutations in proteins. With mutants in hand, you can potentially clone them by complementation as discussed in [[chpater_09|Chap. 09]]. You can then sequence your clones as discussed in [[chapter_08|Chap. 08]]. This will allow you identify the amino acid sequence of the protein/enzyme that carries out the function of the gene that is mutated in your mutants. This approach of discovering protein/enzyme function based on random mutants with interesting phenotypes is called forward genetics.
+Although loss of function mutations in genes for repressors or activators are generally the most common type of regulatory mutation, Table {{ref>Tab1}} will help you to interpret mutations in sites or more complicated mutations in proteins. With mutants in hand, you can potentially clone them by complementation as discussed in [[chapter_09|Chap. 09]]. You can then sequence your clones as discussed in [[chapter_08|Chap. 08]]. This will allow you to
+identify the amino acid sequence of the protein/enzyme that carries out the function of the gene that is mutated in your mutants. This approach of discovering protein/enzyme function based on random mutants with interesting phenotypes is called forward genetics.
 <table Tab1>
 <columns 100% *100%*>
 ^  type of mutation  ^  phenotype  ^  dominant/recessive?  ^  cis/trans-acting?  ^
-|  repressor loss-of-function  |  constitutive  |  recessive  |  trans-acting  |
+|  repressor loss of function  |  constitutive  |  recessive  |  trans-acting  |
-|  activator loss-of function  |  uninducible  |  recessive  |  trans-acting  |
+|  activator loss of function  |  uninducible  |  recessive  |  trans-acting  |
-|  operator loss-of-function  |  constitutive  |  dominant*  |  cis-acting  |
+|  operator loss of function  |  constitutive  |  dominant*  |  cis-acting  |
 |  promoter (or initiator) loss-of-function  |  uninducible  |  recessive*  |  cis-acting  |
 |  repressor dominant negative or super activator  |  constitutive  |  dominant  |  trans-acting  |
 |  super repressor or dominant negative activator  |  uninducible  |  dominant  |  trans-acting  |
-</column>
+</columns>
 <caption>
-Analysis of regulatory mutants. *note: operator/promoter mutants need to be analyzed in the context of the coding sequence they regulate, e.g., $lacO$/$lacP$ need to be analyzed in the context of a functioning lacZ. In a sense, $lacO$/$lacP$ and $lacZ$ are different portions of the same gene since they don't complement each other. By comparison, $lacI$ (at least in principle) can be analyzed as a separate entity away from $lacZ$ since $lacI$ and $lacZ$ complement each other and therefore are not in the same gene.
+Analysis of regulatory mutants. *note: operator/promoter mutants need to be analyzed in the context of the coding sequence they regulate, e.g., $lacO$/$lacP$ need to be analyzed in the context of a functioning $lacZ^+$. In a sense, $lacO$/$lacP$ and $lacZ$ are different portions of the same gene since they don't complement each other. By comparison, $lacI$ (at least in principle) can be analyzed as a separate entity away from $lacZ$ since $lacI$ and $lacZ$ complement each other and therefore are not in the same gene.
 </caption>
 </table>
@@ Line 34: / Line 37: @@
 ===== Multiple genes in regulatory pathways =====
-So far, we have been considering simple regulatory systems with either a single repressor (Lac) or a single activator (Mal). Often genes are regulated by a more complicated set of regulatory steps, which together can be thought of as a regulatory pathway. Although there are good methods that can be used to determine the order of steps in a regulatory pathway (as will be discussed shortly), it is usually difficult at first to tell whether a given component identified by mutation is acting directly on the DNA of the regulated gene or whether it is acting at a step upstream in a regulatory pathway. For example, it will often be the case that a recessive trans-acting mutation that causes constitutive expression is not an actual repressor protein, but a protein acting upstream in a regulatory pathway in such a way that the net effect of this proteins is to cause repression of gene function. The best way to represent this situation is to call the gene product a negative regulator and to reserve the term "repressor" for cases in which we know that the protein actually shuts off transcription directly by binding to an operator site. Similarly, the best way to represent a gene defined by a recessive, trans-acting mutation that causes uninducible expression is as a positive regulator until more specific information can be obtained about whether or not the gene product directly activates transcription (Fig. 11.2).
+So far, we have been considering simple regulatory systems with either a single repressor (Lac) or a single activator (Mal). Often genes are regulated by a more complicated set of regulatory steps, which together can be thought of as a regulatory pathway. Although there are good methods that can be used to determine the order of steps in a regulatory pathway (as will be discussed shortly), it is usually difficult at first to tell whether a given component identified by mutation is acting directly on the DNA of the regulated gene or whether it is acting at a step upstream in a regulatory pathway.
+For example, it will often be the case that a recessive trans-acting mutation that causes constitutive expression is not an actual repressor protein, but a protein acting upstream in a regulatory pathway in such a way that the net effect of this proteins is to cause repression of gene function. The best way to represent this situation is to call the gene product a negative regulator and to reserve the term "repressor" for cases in which we know that the protein actually shuts off transcription directly by binding to an operator site. Similarly, the best way to represent a gene defined by a recessive, trans-acting mutation that causes uninducible expression is as a positive regulator until more specific information can be obtained about whether or not the gene product directly activates transcription (Fig. 11.2).
-Figure 11.2. "Activator or repressor" (Fig. 11.1) vs "positive and negative regulator". Compare to Fig. 11.1. The difference is subtle but important. "Activator or repressor" suggests a direct interaction between a protein and a DNA sequence, whereas "positive or negative regulator" makes no assumptions about any intermediate factors that might be involved.
+<figure Fig2>
+{{ :regulator_vs_activator.png?400 |}}
+<caption>
+"Activator or repressor" vs "positive and negative regulator"; compare to Fig. {{ref>Fig1}}. The difference is subtle but important. "Activator or repressor" suggests a direct interaction between a protein and a DNA sequence, whereas "positive or negative regulator" makes no assumptions about any intermediate factors that might be involved.
+</caption>
+</figure>
 An important note about interpreting such diagrams is that the arrow or blocking symbol do not necessarily imply direct physical interaction; they simply mean that the negative regulator or positive activator have a net negative or positive effect, respectively, on gene expression. There may (or may not) be other intermediate factors involved.
@@ Line 44: / Line 53: @@
-Imagine that we are studying the regulation of an enzyme-coding gene and we find a recessive trans-acting mutation in gene A that gives uninducible enzyme expression. The simplest interpretation is that gene A is a positive activator of the enzyme (Fig. 11.3):
+Imagine that we are studying the regulation of an enzyme-coding gene and we find a recessive trans-acting mutation in gene $A$ that gives uninducible enzyme expression. The simplest interpretation is that gene $A$ is a positive activator of the enzyme (Fig. {{ref>Fig3}}):
+<figure Fig3>
+{{ :postive_regulator_a.png?400 |}}
+<caption>
+Model 1 for our "gene $A$" mutant.
+</caption>
+</figure>
-Figure 11.3. Model 1 for our "gene A" mutant.
+Next, say that we find a recessive trans-acting mutation in gene $B$ that gives constitutive enzyme expression. The following model takes into account the behavior of mutations in both $A$ and $B$ (Fig. {{ref>Fig4}}):
-Next, say that we find a recessive trans-acting mutation in gene B that gives constitutive enzyme expression. The following model takes into account the behavior of mutations in both A and B (Fig. 11.4):
+<figure Fig4>
+{{ :ab_model_1.png?400 |}}
+<caption>
+Model 2 for our "gene $A$" mutant.
+</caption>
+</figure>
+The idea is that the gene for the enzyme is negatively regulated by gene $B$, which in turn is negatively regulated by gene $A$. The net outcome is still a positive effect of gene $A$ on enzyme expression. To distinguish between the two models, we will need more mutants to analyze. However, we can also revise Model 1 to fit the new data (Fig. {{ref>Fig5}}).
+<figure Fig5>
+{{ :ab_model_2.png?400 |}}
+<caption>
+Model 1, revised; compare to original model in Fig. {{rev>Fig3}}. This revised model is consistent with our new information on gene $B$. Note that both Figs. {{ref>Fig4}} and {{ref>Fig5}} are consistent with the data.
+</caption>
+</figure>
+The best way to distinguish between the two possible models is to test the phenotype of a double mutant. In revised Model 1 (Fig. {{ref>Fig5}}), the $A^–; B^–$ double mutant is predicted to be uninducible and in Model 2 (Fig. {{ref>Fig4}}) it is predicted to be constitutive. This experiment represents a powerful form of genetic analysis known as an epistasis test. In the example above, if the double mutant were constitutive, we would say that the mutation $B^–$ is epistatic to $A^–$. Such a test allows us to determine the order in which different functions in a regulatory pathway act. If the double mutant in the example were constitutive, we would deduce that gene $B$ functions after gene $A$ in the regulatory pathway.
+<WRAP center round important 60%>
+If mutants $A^-$ and $B^-$ have different mutant phenotypes, and an $A^-; B^$ double mutant exhibits the mutant phenotype of $A^-$, then we say that $A$ is epistatic to $B$. We interpret this to say that $A$ functions after $B$.
+</WRAP>
-Figure 11.4. Model 2 for our "gene A" mutant.
-The idea is that the gene for the enzyme is negatively regulated by gene B, which in turn is negatively regulated by gene A. The net outcome is still a positive effect of gene A on enzyme expression. To distinguish between the two models, we will need more mutants to analyze. However, we can also revise Model 1 to fit the new data (Fig. 11.5).
-Figure 11.5. Model 1, revised; compare to original model in Fig. 11.3. This revised model is consistent with our new information on gene B. Note that both Figs. 11.4 and 11.5 are consistent with the data.
-The best way to distinguish between the two possible models is to test the phenotype of a double mutant. In revised Model 1 (Fig. 11.5), the A– B– double mutant is predicted to be uninducible and in Model 2 (Fig. 11.4) it is predicted to be constitutive. This experiment represents a powerful form of genetic analysis known as an epistasis test. In the example above, if the double mutant were constitutive, we would say that the mutation B– is epistatic to A–. Such a test allows us to determine the order in which different functions in a regulatory pathway act. If the double mutant in the example were constitutive, we would deduce that gene B functions after gene A in the regulatory pathway. To perform an epistasis test, it is necessary that the different mutations under examination produce opposite (or at least different) phenotypic consequences. When the double mutant is constructed, its phenotype will be that of the function that acts later in the pathway. Constructing double mutants in E. coli uses techniques that are somewhat esoteric and probably not of general interest to most students unless they plan to become molecular microbiologists. For the purposes of this book, we can just assume that "we made the E. coli double mutant" without worrying about the technical details. If you are interested in those details, a good reference to look up would be a textbook on molecular microbiology. We will, however, be examining how to make double mutants in yeast using tetrad analysis in Chap. 14.
-Discussion Box: As a general principle, making double mutants in obligate diploids is much easier than making double mutants in bacteria. Why?
+To perform an epistasis test, it is necessary that the different mutations under examination produce opposite (or at least different) phenotypic consequences. It doesn't matter if mutations are dominant or recessive; either will work in the epistasis test. When the double mutant is constructed, its phenotype will be that of the function that acts later in the pathway. Constructing double mutants in //E. coli// uses techniques that are somewhat esoteric and probably not of general interest to most students unless they plan to become molecular microbiologists. For the purposes of this book, we can just assume that "we made the //E. coli// double mutant" without worrying about the technical details. If you are interested in those details, a good reference to look up would be a textbook on molecular microbiology. We will, however, be examining how to make double mutants in yeast using tetrad analysis in [[chapter_14|Chap. 14]].
-Epistasis tests are of very general utility. If the requirement that two mutations have opposite phenotypes is met, almost any type of hierarchical relationship between elements in a regulatory pathway can be worked out. For example, the lacOc mutation is in a regulatory site, not a coding sequence, but it is still possible to perform an epistasis between lacOc and lacIs since these mutations satisfy the basic requirement for an epistasis test. One mutation is uninducible while the other is constitutive for Lac gene expression. When the actual double mutant, lacOc lacIs, is evaluated it is constitutive. This makes sense given what we know about the Lac operon, since a defective operator site that prevents repressor binding should allow constitutive expression regardless of the form of the repressor protein. Formally, this result shows that a mutation in lacO is epistatic to a mutation in lacI. Even if we did not know the details of Lac operon regulation beforehand, this epistasis test would allow us to deduce that the operator functions at a later step than the repressor.
-Exercise 11.1: Revisit Figure 2.3. Is it possible to analyze yeast his mutants using epistasis? Why or why not? What question would you be answering with epistasis? What additional information might you need to know first? You may want to look up some classic genetic experiments by Beadle and Tatum using the bread mold Neurospora to help with answering this question.
+Epistasis tests are of very general utility. If the requirement that two mutations have opposite phenotypes is met, almost any type of hierarchical relationship between elements in a regulatory pathway can be worked out. For example, the $lacO^c$ mutation is in a regulatory site, not a coding sequence, but it is still possible to perform an epistasis between $lacO^c$ and $lacI^s$ since these mutations satisfy the basic requirement for an epistasis test. One mutation is uninducible while the other is constitutive for Lac gene expression. When the actual double mutant, $lacO^c; lacI^s$, is evaluated it is constitutive. This makes sense given what we know about the Lac operon, since a defective operator site that prevents repressor binding should allow constitutive expression regardless of the form of the repressor protein. Formally, this result shows that a mutation in $lacO$ is epistatic to a mutation in $lacI$. Even if we did not know the details of Lac operon regulation beforehand, this epistasis test would allow us to deduce that the operator functions at a later step than the repressor.
 ===== Stable regulatory circuits =====
@@ Line 70: / Line 100: @@
 The best understood case of a stable switch is the lysis vs. lysogeny decision made by bacteriophage λ. When λ infects cells there are two different developmental fates of the phage.
-.	In the lytic program the phage replicates DNA, make heads, tails, packages DNA, and lyses host cells.
+  - In the lytic program the phage replicates DNA, make heads, tails, packages DNA, and lyses host cells.
-.	In the lysogenic program the phage integrates into the host DNA and shuts down phage genes. The resulting quiescent phage integrated into the genome is known as a lysogen.
+  - In the lysogenic program the phage integrates into the host DNA and shuts down phage genes. The resulting quiescent phage integrated into the genome is known as a lysogen.
-The decision between these two options must be made in a committed way so the proper functions act in concert. The switch in the case of phage λ hinges on the activity of two repressor genes cI and cro. The cI and cro genes have mutually antagonistic regulatory interactions (Fig. 11.6):
+The decision between these two options must be made in a committed way so the proper functions act in concert. The switch in the case of phage λ hinges on the activity of two repressor genes $cI$ and $cro$. The $cI$ and $cro$ genes have mutually antagonistic regulatory interactions (Fig. {{ref>Fig6}}):
-Figure 11.6. Bistable gene circuit in bacteriophage .
-After an initial unstable period immediately after infection, either cro expression or cI expression will dominate.
-•	Mode 1: High cro expression blocks cI expression. In this state, all of the genes for lytic growth are made and the phage enters the lytic program.
+<figure Fig6>
-•	Mode 2: High cI expression blocks cro expression. In this state, none of the genes except for cI are expressed. This produces a stable lysogen.
+{{ :lambda.png?400 |}}
+<caption>
+Bistable gene circuit in bacteriophage λ.
+</caption>
+</figure>
-In gene regulation, as in good circuit design, stability is achieved by feedback. The result is a bi-stable switch that is similar to a “flip-flop”, one of the basic elements of digital electronic circuits. Other genes participate in the initial period to bias the decision to one mode or the other. These genes act so that the lytic mode is favored when E. coli is growing well and there are few phage per infected cell, whereas the lysogenic mode is favored when cells are growing poorly and there are many phage per infected cell.
+After an initial unstable period immediately after infection, either $cro$ expression or $cI$ expression will dominate.
+  * Mode 1: High $cro$ expression blocks $cI$ expression. In this state, all of the genes for lytic growth are made and the phage enters the lytic program.
+  * Mode 2: High $cI$ expression blocks $cro$ expression. In this state, none of the genes except for $cI$ are expressed. This produces a stable lysogen.
+In gene regulation, as in good circuit design, stability is achieved by feedback. The result is a bi-stable switch that is similar to a “flip-flop”, one of the basic elements of digital electronic circuits. Other genes participate in the initial period to bias the decision to one mode or the other. These genes act so that the lytic mode is favored when //E. coli// is growing well and there are few phage per infected cell, whereas the lysogenic mode is favored when cells are growing poorly and there are many phage per infected cell.
 ===== Questions and exercises =====
+Exercise 1 (challenge question): Revisit [[chapter_02|Chapter 02]] Figure 3. Is it possible to analyze yeast $his$ mutants using epistasis? Why or why not? What question would you be answering with epistasis? What additional information might you need to know first? You may want to look up some classic genetic experiments by [[wp>One_gene–one_enzyme_hypothesis|Beadle and Tatum]] using the bread mold Neurospora to help with answering this question.