TNBGGA: The No Bullshit Guide to Genetic Analysis

User Tools

Site Tools

Trace: • chapter_05
chapter_05

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
chapter_05 [2024/08/19 22:39] – [Questions and exercises] mikechapter_05 [2025/02/19 07:57] (current) – mike
Line 1: Line 1:
-<typo fs:x-large>Chapter 05. Defining genes by position</typo>+<-chapter_04|Chapter 04^table_of_contents|Table of Contents^chapter_06|Chapter 06->
  
-Besides providing experimental evidence for chromosome theory as discussed in Chapter 3, Morgan's research group also demonstrated that genes (usually) have fixed positions on chromosomes. When we think about gene position, the term locus (plural: loci, pronounced "LOW-sigh") is sometimes used as a term to describe a gene in the context of its position rather than its function.+<typo fs:x-large>Chapter 05. Defining %%genes%% by position</typo> 
 + 
 +Besides providing experimental evidence for chromosome theory as discussed in [[chapter_04|Chapter 04]], Morgan's research group also demonstrated that genes (usually) have fixed positions on chromosomes. When we think about gene position, the term locus (plural: loci, pronounced "LOW-sigh") is sometimes used as a term to describe a gene in the context of its position rather than its function.
  
 ===== Recombination between two sex-linked genes ===== ===== Recombination between two sex-linked genes =====
Line 72: Line 74:
 The tiny $Y$ chromosome does not contain either the $cv$ or $w$ genes (in fact, the $Y$ chromosome barely has any genes). Thus, for genes on the $X$ chromosome such as $cv$ and $w$, examining males only makes it effectively a test cross. If $cv$ and $w$ segregated independently from each other (i.e., if they followed Mendel's Second Law) we would expect that the four combinations of progeny classes ($cv^+ \ \ w^+$, $cv^+ \ \ w^-$, $cv^- \ \ w^+$, $cv^- \ \ w^-$) would appear at the same frequency; that is, the four progeny classes would show 1:1:1:1 ratio. However, we find that there are more males that are parental-like ($cv^+ \ \ w^-$, $cv^- \ \ w^+$) than there are non-parental (recombinant)-like ($cv^+ \ \ w^+$, $cv^- \ \ w^-$). Specific numbers are given below in Table {{ref>Tab3}}. Genes on the same chromosome such as $cv$ and $w$ do not assort independently if they are located close to each other; they are biased to assort together most but not all of the time. Such behavior is known as linkage.  The tiny $Y$ chromosome does not contain either the $cv$ or $w$ genes (in fact, the $Y$ chromosome barely has any genes). Thus, for genes on the $X$ chromosome such as $cv$ and $w$, examining males only makes it effectively a test cross. If $cv$ and $w$ segregated independently from each other (i.e., if they followed Mendel's Second Law) we would expect that the four combinations of progeny classes ($cv^+ \ \ w^+$, $cv^+ \ \ w^-$, $cv^- \ \ w^+$, $cv^- \ \ w^-$) would appear at the same frequency; that is, the four progeny classes would show 1:1:1:1 ratio. However, we find that there are more males that are parental-like ($cv^+ \ \ w^-$, $cv^- \ \ w^+$) than there are non-parental (recombinant)-like ($cv^+ \ \ w^+$, $cv^- \ \ w^-$). Specific numbers are given below in Table {{ref>Tab3}}. Genes on the same chromosome such as $cv$ and $w$ do not assort independently if they are located close to each other; they are biased to assort together most but not all of the time. Such behavior is known as linkage. 
  
-We already know that $cv$ and $w$ are both sex-linked; that is, we know they are physically located on the $X$ chromosome. But when we perform the cross shown in Figures {{ref?Fig2}} and {{ref>Fig3}}, we observe non-parental types in the F2 test cross progeny. Based on the phenotypes of the non-parental classes, the alleles appear to have separated and moved from one $X$ chromosome to the other. This implies that there must have been an exchange of chromosomal material (crossing over) in the F1 heterozygous mother when she went through meiosis to form gametes. +We already know that $cv$ and $w$ are both sex-linked; that is, we know they are physically located on the $X$ chromosome. But when we perform the cross shown in Figures {{ref>Fig2}} and {{ref>Fig3}}, we observe non-parental types in the F2 test cross progeny. Based on the phenotypes of the non-parental classes, the alleles appear to have separated and moved from one $X$ chromosome to the other. This implies that there must have been an exchange of chromosomal material (crossing over) in the F1 heterozygous mother when she went through meiosis to form gametes. 
  
 ==== Tying a sex-linked cross back to meiosis ==== ==== Tying a sex-linked cross back to meiosis ====
Line 94: Line 96:
 $$\text{map distance} = 100 \times \frac{\text{crossover gametes}}{\text{total gametes}}$$ $$\text{map distance} = 100 \times \frac{\text{crossover gametes}}{\text{total gametes}}$$
  
-In essence, map distance between two points is the percent recombination between those two points. We typically use the unit of measures m.u. (which stands for map unit) or cM (centiMorgan, named in honor of Thomas Morgan) for map distances. We can map the distance between $cv$ and $w$ by doing the cross shown in Figs. {{ref>Fig2}} and {{ref>Fig3}} and carefully counting the number of offspring and their different phenotypes (Table {{ref>Tab3}}). We look at males only:+In essence, map distance between two points is the percent recombination between those two points. We typically use the unit of measure m.u. (which stands for map unit) or cM (centiMorgan, named in honor of Thomas Hunt Morgan) for map distances. We can map the distance between $cv$ and $w$ by doing the cross shown in Figs. {{ref>Fig2}} and {{ref>Fig3}} and carefully counting the number of offspring and their different phenotypes (Table {{ref>Tab3}}). We look at males only:
  
 <table Tab3> <table Tab3>
Line 123: Line 125:
   - Maps are internally consistent and concise.    - Maps are internally consistent and concise. 
  
-The first genetic map of any kind was constructed in 1911 by Alfred Sturtevant when he was a sophomore undergraduate student in Thomas Morgan’s lab. It showed the relative positions of several genes on the Drosophila $X$ chromosome. +The first genetic map of any kind was constructed in 1911 by [[wp>alfred_sturtevant|Alfred Sturtevant]] when he was a sophomore undergraduate student in Thomas Morgan’s lab. It showed the relative positions of several genes on the Drosophila $X$ chromosome. 
  
 <figure Fig5> <figure Fig5>
 {{ :first-genetic-map-sturtevant-1913-49.png?400 |}} {{ :first-genetic-map-sturtevant-1913-49.png?400 |}}
 <caption>  <caption> 
-\The first genetic map, created by Alfred Sturtevant in Thomas Morgan's research lab. The gene symbols in the figure are non-standard; they represent genes that we know today as: B, $yellow$; C, $white$; O, an allele of $white$; P, $vermillion$; R, $mini$; M, $rudimentary$. The numbers represent map positions, and map distances between genes can be obtained by subtracting the numbers. Source: Sturtevant, A.H. (1913). J. Exp. Zool., 14:43-59. Licensing: free for scholarly use from the [[https://www.esp.org/|Electronic Scholarly Publishing Project]]. Commercial use is prohibited without permission.+The first genetic map, created by Alfred Sturtevant in Thomas Morgan's research lab. The gene symbols in the figure are non-standard; they represent genes that we know today as: B, $yellow$; C, $white$; O, an allele of $white$; P, $vermillion$; R, $mini$; M, $rudimentary$. The numbers represent map positions, and map distances between genes can be obtained by subtracting the numbers. Source: Sturtevant, A.H. (1913). J. Exp. Zool., 14:43-59. Licensing: free for scholarly use from the [[https://www.esp.org/|Electronic Scholarly Publishing Project]]. Source claims copyright and that commercial use is prohibited without permission; publishing date suggests image is now in the public domain. 
 </caption> </caption>
 </figure>  </figure> 
  
-<figure>+<figure Fig6>
 {{ :drosophila_genetic_map.png?400 |}} {{ :drosophila_genetic_map.png?400 |}}
 <caption> <caption>
Line 145: Line 147:
 The actual relationship between genetic distance in cM and physical distance in base pairs (bp) of DNA depends on the recombination rate, which is different for different organisms. For example, in humans the recombination rate is 1.3 cM/Mbp whereas in yeast it is 360 cM/Mbp (1 Mbp = 10<sup>6</sup> bp). Sometimes recombination rates in the male and female of a species are different. In Drosophila there is no recombination in males so the genetic distance between markers on the same chromosome are always zero when examined by meiosis in the male. In humans the recombination rate (and therefore map distances) in females is twice that of males.  The actual relationship between genetic distance in cM and physical distance in base pairs (bp) of DNA depends on the recombination rate, which is different for different organisms. For example, in humans the recombination rate is 1.3 cM/Mbp whereas in yeast it is 360 cM/Mbp (1 Mbp = 10<sup>6</sup> bp). Sometimes recombination rates in the male and female of a species are different. In Drosophila there is no recombination in males so the genetic distance between markers on the same chromosome are always zero when examined by meiosis in the male. In humans the recombination rate (and therefore map distances) in females is twice that of males. 
  
-Another issue that often causes confusion concerns the map distances of genes that are far apart on the same chromosome. The physical length of a genetic interval is proportional to the frequency of crossovers that occur in that interval during meiosis. But in a cross, we are not actually counting crossovers; rather, we are counting the number of recombinant progeny that are produced. The frequency of recombinants provides a good approximation of distance for short intervals but as the interval length increases, double or even triple crossovers are possible, making the relation¬ship between frequency of recombinants and crossovers not linear. This is discussed further in [[Appendix_A|Appendix A]] on tetrad analysis.+Another issue that often causes confusion concerns the map distances of genes that are far apart on the same chromosome. The physical length of a genetic interval is proportional to the frequency of crossovers that occur in that interval during meiosis. But in a cross, we are not actually counting crossovers; rather, we are counting the number of recombinant progeny that are produced. The frequency of recombinants provides a good approximation of distance for short intervals but as the interval length increases, double or even triple crossovers are possible, making the relationship between frequency of recombinants and crossovers not linear. This is discussed further in [[Appendix_A|Appendix A]] on tetrad analysis.
  
 ===== Unlinked genes ===== ===== Unlinked genes =====
Line 152: Line 154:
 If the measured distance in a cross is statistically indistinguishable from 50 cM then we say that the genes are unlinked. In fact, by definition a map distance of 50 cM is the same thing as saying that the two genes assort independently. But this doesn’t mean that distances greater than 50 cM cannot be obtained. By adding intervals between multiple genes, longer distances that are meaningful can be obtained. For example, if all the intervals between linked genes in the human genome are added together the total length of the genome (in males) is 2,500 cM. Genes that are physically located on the same chromosome can be described as being in the same linkage group, even though they might not be linked. For example, in Figure {{ref>Fig6}}, we can see that $white$, which is at map position 1.5 on chromosome $X$, and $forked$, which is at map position 56.7 also on chromosome $X$, are unlinked even though they are on the same chromosome. But we can calculate their map distance by subtracting their map positions ($56.7 - 1.5 = 55.2$ cM apart).  If the measured distance in a cross is statistically indistinguishable from 50 cM then we say that the genes are unlinked. In fact, by definition a map distance of 50 cM is the same thing as saying that the two genes assort independently. But this doesn’t mean that distances greater than 50 cM cannot be obtained. By adding intervals between multiple genes, longer distances that are meaningful can be obtained. For example, if all the intervals between linked genes in the human genome are added together the total length of the genome (in males) is 2,500 cM. Genes that are physically located on the same chromosome can be described as being in the same linkage group, even though they might not be linked. For example, in Figure {{ref>Fig6}}, we can see that $white$, which is at map position 1.5 on chromosome $X$, and $forked$, which is at map position 56.7 also on chromosome $X$, are unlinked even though they are on the same chromosome. But we can calculate their map distance by subtracting their map positions ($56.7 - 1.5 = 55.2$ cM apart). 
    
- 
- 
 Based on the concepts that Morgan's lab developed, we can now define genes as follows: genes are pieces of hereditary information that have distinct physical locations on chromosomes.  Based on the concepts that Morgan's lab developed, we can now define genes as follows: genes are pieces of hereditary information that have distinct physical locations on chromosomes. 
  
-Exercise 5.4. Let's revisit Fig. 2.3 and yeast. We can imagine that there are four genes in the yeast histidine biosynthetic pathway: his1, his2, his3, and his4. In Chapter 2, we learned that you could determine that these are different genes through complementation testing. In a conceptual sense and without worrying about experimental details, what other way can you use to determine if these are different genes?+
  
 ===== A final note on genetic notation ===== ===== A final note on genetic notation =====
  
  
-In Chap. 2 we used the symbol "dot" (ꞏ) to separate gene symbols between which we did not know the linkage status. For instance, writing "shi ꞏ vg" meant that we don't know if shibire and vestigial are linked or unlinked. In this chapter, we saw that linked genes should be written without any separating symbol. For instance, "b vg" indicates that black and vestigial are linked. We can write genes that are on the same chromosome this way as well to indicate that they belong to the same linkage group (that is, all the genes in a linkage group are directly or indirectly linked to each other), even if they are over 50 cM apart. For instance, aristaless (al) is at map position 0.0 on chromosome II whereas vg is at position 67.0 on chromosome II; thus, al and vg are unlinked (that is, they are over 50 m.u. apart) but you could write the genotype as al vg. Finally, we use a "semicolon" to separate gene symbols to indicate that two genes are not on the same linkage group. For instance, "shi; w" would indicate that shibire and white are unlinked and also on different chromosomes.+In [[chapter_02|Chapter 02]] we used the "dot" symbol ($\cdot$) to separate gene symbols between which we did not know the linkage status. For instance, writing "$shi \cdot vg$" meant that we don't know if $shibire$ and $vestigial$ are linked or unlinked. In this chapter, we saw that linked genes should be written without any separating symbol. For instance, "$b \ \ vg$" indicates that $black$ and $vestigial$ are linked. We can write genes that are on the same chromosome this way as well to indicate that they belong to the same linkage group (that is, all the genes in a linkage group are directly or indirectly linked to each other), even if they are over 50 cM apart. For instance, $aristaless$ ($al$) is at map position 0.0 on chromosome II whereas $vg$ is at position 67.0 on chromosome II; thus, $al$ and $vg$ are unlinked (that is, they are over 50 m.u. apart) but you could write the genotype of a double mutant as "$al \ \ vg$". Finally, we use a semicolon (;) to separate gene symbols to indicate that two genes are not on the same linkage group. For instance, "$shi\text{; } w$" would indicate that $shibire$ and $white$ are unlinked and also on different chromosomes.
  
 ===== Questions and exercises ===== ===== Questions and exercises =====
  
  
-Exercise 5.1. Why do we only look at males in the above experiment? What kinds of female F2 progeny would you expect to get? Can you predict their phenotypes? Can you predict their genotypes? Would that information be helpful in measuring recombination frequency between cv and w? Why or why not?+Exercise 1. Why do we only look at males in the experiment shown in Figs. {{ref>Fig2}}-{{ref>Fig3}} and Table {{ref>Tab3}}? What kinds of female F2 progeny would you expect to get? Can you predict the female phenotypes and genotypes? Would that information be helpful in measuring recombination frequency between $cv$ and $w$? Why or why not? 
 + 
 +Exercise 2: Design and write out a genetic cross between $black$ ($b$) and $vestigial$ ($vg$) to map their distance. You will probably want to use a test cross strategy (you can assume that you have a true breeding $b \ \ vg$ double mutant in the lab). Draw the tetrad and the crossing over similar to Fig. 5.2. Based on the information in Fig. 5.4., what kind of F2 test cross progeny will you get, and what will their frequencies be?
  
-Exercise 5.2: Design and write out a genetic cross between black (b) and vestigial (vg) to map their distance. You will probably want to use a test cross strategy (you can assume that you have a true breeding b vg double mutant in the lab). Draw the tetrad and the crossing over similar to Fig. 5.2. Based on the information in Fig. 5.4., what kind of F2 test cross progeny will you get, and what will their frequencies be?+Exercise 3: Design and write out a genetic cross between white and forked to measure their linkage, using information from Fig. 5.4. Draw the tetrad and the crossing over similar to Fig. {{ref>Fig4}}. Calculate the ratios of the test cross progeny based on the idea that map distance = recombination frequency. It's relatively easy to come up with a number, but it's a little harder to determine if an answer makes sense. Does your answer make sense? 
  
-Exercise 5.3. Design and write out a genetic cross between white and forked to measure their linkage, using information from Fig. 5.4. Draw the tetrad and the crossing over similar to Fig. 5.2. Calculate the ratios of the test cross progeny based on the idea that map distance = recombination frequency. It's relatively easy to come up with a number, but it's a little harder to determine if an answer makes sense. Does your answer make sense? +Conceptual question: "You can have map distances greater than 50 cM, but you can't have recombination frequencies greater than 50%". Does this sentence make sense? Explain in your own words. 
  
-Discussion box: "You can have map distances greater than 50 cM, but you can't have recombination frequencies greater than 50%". Does this sentence make sense? +Conceptual question:  Let's revisit [[chapter_02#fig3|Chapter 02 Figure 3]] and yeast. We can imagine that there are four genes in the yeast histidine biosynthetic pathway: $his1$, $his2$, $his3$, and $his4$. In [[chapter_02|Chapter 02]], we learned that you could determine that these are different genes by using the complementation test. Without worrying about experimental details, what other method can you now use to determine if these are different genes after studying this chapter?
  
  
chapter_05.1724132376.txt.gz · Last modified: 2024/08/19 22:39 by mike