TNBGGA: The No Bullshit Guide to Genetic Analysis

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chapter_11 [2024/08/25 12:05] – [Questions and exercises] mikechapter_11 [2025/04/07 20:58] (current) – mike
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-<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.  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. 
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 </figure> </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> <table Tab1>
 <columns 100% *100%*> <columns 100% *100%*>
 ^  type of mutation  ^  phenotype  ^  dominant/recessive?  ^  cis/trans-acting?  ^ ^  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  | |  promoter (or initiator) loss-of-function  |  uninducible  |  recessive*  |  cis-acting  |
 |  repressor dominant negative or super activator  |  constitutive  |  dominant  |  trans-acting  | |  repressor dominant negative or super activator  |  constitutive  |  dominant  |  trans-acting  |
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 </columns> </columns>
 <caption> <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> </caption>
 </table> </table>
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 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. {{ref>Fig4}}):
  
-<fig Fig4>+<figure Fig4>
 {{ :ab_model_1.png?400 |}}  {{ :ab_model_1.png?400 |}} 
 <caption> <caption>
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 </caption> </caption>
 </figure>  </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}}). 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}}).
  
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 </figure> </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.+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> 
 + 
  
 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]].  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. +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. 
  
-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.  
  
 ===== Stable regulatory circuits ===== ===== Stable regulatory circuits =====
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 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. 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.
  
-1. 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. 
-2. 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 ===== ===== Questions and exercises =====
  
-Discussion Box: As a general principle, making double mutants in obligate diploids is much easier than making double mutants in bacteria. Why?+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.  
  
chapter_11.1724612751.txt.gz · Last modified: 2024/08/25 12:05 by mike