TNBGGA: The No Bullshit Guide to Genetic Analysis

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<-chapter_04|Chapter 04^table_of_contents|Table of Contents^chapter_06|Chapter 06->

<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 =====


One of the goals of genetic analysis is to figure out where genes are physically located on chromosomes relative to one another. This is called mapping. In modern times, mapping is done by DNA sequencing, but classically it was done through genetic crosses. Even though we rarely use this technique anymore in its original form, it is instructive to learn about it because it illustrates important concepts about recombination, which is still important even in modern mapping approaches. 

==== More sex-linked mutations ====


After the discovery of $white$, Morgan's lab found mutant alleles of many other genes that are also sex-linked. Consider another sex-linked mutation on the $X$ chromosome of Drosophila: $crossveinless$ ($cv$):

<figure Fig1>

{{ :drosophile_normale_et_cross-veinless.jpg?400 |}}
<caption> 
Wild type Drosophila wing (left) and a wing from a $crossveinless$ ($cv$) mutant. Source: [[https://commons.wikimedia.org/wiki/File:Drosophile_normale_et_cross-veinless.jpg|Wikimedia]]. Licensing: [[https://creativecommons.org/licenses/by-sa/4.0/deed.en|CC BY-SA 4.0]]. 
</caption>
</figure>
<table Tab1>
<columns 100% *100%*>
^  Male genotype  ^  Phenotype  ^
|  $\frac{+}{\rightharpoondown}$  |  wildtype (normal wings and eyes)  |  
|  $\frac{cv}{\rightharpoondown}$  |  crossveinless wings  |
|  $\frac{w}{\rightharpoondown}$  |  white eyes  |
</columns>
<caption>
Genotypes and phenotypes of $crossveinless$ and $white$ mutants. See [[chapter_04|Chapter 04]] for Drosophila genotype writing conventions.  
</caption>
</table>

Let's do a cross between a $crossveinless$ male and $white$ female:

<figure Fig2>
$$P: \frac{cv \ \ +}{\rightharpoondown}\text{ ♂}  \times \frac{+\ \ \ w}{+\ \ \ w} \text{ ♀}$$
<caption>
A cross between two sex-linked mutants: $crossveinless$ ($cv$) and $white$ ($w$). What do you think will happen?
</caption>
</figure>   

<WRAP center round tip 60%>
Note how we write the genotype of the flies in Figure {{ref>Fig2}} using fractional notation - by putting $cv$ and $w$ over the same "fraction bar", it indicates that they are on the same chromosome. Can you easily convey this information using Punnett squares?
</WRAP>


All of the F1 daughters from Figure {{ref>Fig2}} will have two different $X$ chromosomes, which differ at two loci: $\frac{cv\ \ +}{+\ \ \ w}$. (Note that this genotype is not the same thing as $\frac{cv\ \ w}{+\ \ \ +}$!) We want to follow these $X$ chromosomes into the next generation, so we cross these F1 females to wildtype males, and we look at F2 male flies:

<figure Fig3>
$$F1: \frac{cv\ \ +}{+\ \ \ w}\text{ ♀} \times \frac{+}{\rightharpoondown}\text{ ♂}$$
<caption>
Test cross of the F1 females shown in Figure {{ref>Fig2}}.
</caption>
</figure>

The possible F2 outcomes from this cross are shown in Table {{ref>Tab2}}. 

<table Tab2>
<columns 100% *100%*>
^  progeny class  ^  phenotypes  ^  inferred genotypes of males  ^
|  parental  |  crossveinless wings, red eyes  |  $\frac{cv\ \ +}{\rightharpoondown}$  |
|  parental  |  normal wings, white eyes  |  $\frac{+\ \ \ \ w}{\rightharpoondown}$  |
|  recombinant  |  crossveinless wings, white eyes  |  $\frac{cv\ \ \ w}{\rightharpoondown}$  |
|  recombinant  |  normal wings, red eyes  |  $\frac{+\ \ \ +}{\rightharpoondown}$  |
</columns>
<caption>
Possible F2 outcomes in a test cross between $w$ and $cv$. Note that "non-parental" is synonymous for recombinant. We use these two terms interchangeably. 
</caption>
</table>

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. 

==== Tying a sex-linked cross back to meiosis ====


To see what's really going on we need to look at the chromatids arranged as a tetrad in prophase I of meiosis in the F1 heterozygous mother.

<figure Fig4> 
{{ :crossing_over_cv_and_x.jpg?400 |}}
<caption>
Crossing over between cv and w during F1 female gamete formation in Cross 5.1. Blue chromatids represent paternal contribution from the P generation; similarly, red chromatids represent maternal contribution. After crossing over, chromosomal material is exchanged, as shown by the color coding. At the end of meiosis, four gametes containing one each of the four chromatids in the tetrad are formed, labeled (1) though (4). (1) and (4) are parental types, and (2) and (3) are recombinant (non-parent) types. (1) and (2) are sisters, and (3) and (4) are sisters. "Completing meiosis" is simplifying the fact that two cell divisions are occurring. Note that the map positions of cv and w are not drawn to scale. Credit: M. Chao.
</caption>
</figure>

During prophase I, structures called chiasmata form between non-sister chromatids. The term chiasmata (kai-as-MAH-tah; singular form is chiasma) comes from the Greek letter χ, which is pronounced "kai" and is shaped like the letter X. In Figure {{ref>Fig4}}, a chiasma forms between chromatids (2) and (3), but chiasmata can form between any two non-sister chromatids (for instance, you could also form a chiasmata between chromatids (1) and (3)). A chiasma is simply a structure where physical breaks have occurred on the non-sister chromatids and they exchange material with each other. More commonly, we use the term "crossover" both as a noun and a synonym for chiasmata, and as a verb to describe the process. 

Crossovers between non-sister chromatids occur at random places during meiosis. In Figure {{ref>Fig4}}, a crossover is shown to occur between $cv$ and $w$, but it can happen anywhere. For instance, it can happen between the centromere and $cv$, or it can happen between $w$ and the end of the chromosome. The likelihood that a crossover will happen at any specific location on the chromosome is roughly the same across the entire chromosome. In most organisms, there is an average of one crossover per chromosome arm per meiosis, although there can be more than one crossover per chromosome arm per meiosis (these are called double crossovers, triple crossovers, etc.).

Because the location of the crossover is random, the frequency of crossover occurring between two points on a chromosome depends on the distance between those two points. Crossovers between two points that are close together will occur rarely, whereas crossovers between points that are far apart will occur more frequently. Geneticists can thus use these random crossovers as a tool to measure map distance.

$$\text{map distance} = 100 \times \frac{\text{crossover gametes}}{\text{total gametes}}$$

In essence, map distance between two points is the percent recombination between those two points. We typically use the unit of 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>
<columns 100% *100%*>
^  inferred genotype based on phenotype  ^  number of male progeny  ^
|  $cv^- \ \ w^+$  |  430  |
|  $cv^+ \ \ w^-$  |  450  |
|  $cv^- \ \ w^-$  |  52  |
|  $cv^+ \ \ w^+$  |  68  |
^  total  ^  1000  |
</columns>
<caption>
Table 5.3. Example data of F2 male progeny frequency from $cv \times w$ (Figs. {{ref>Fig2}} and {{ref>Fig3}}). Note that these numbers are pooled results from several crosses. Each Drosophila female can produce at most a few hundred progeny.
</caption>
</table>

In this cross, the recombinant classes are $cv^- \ \ w^-$ and $cv^+ \ \ w^+$. The number of crossover gametes = $52 + 68 = 120$. Therefore, map distance = $100\times\frac{120}{1000} = 12$ m.u., or 12 cM.

When recombinant classes appear as 50% of the F2 offspring of a test cross, we say the two genes are unlinked (this is discussed further below). When recombinant classes appear frequently but are less than 50% (for example, 41%), we say that two genes are weakly linked. Conversely, if recombinant classes appear very rarely then we say that two genes are tightly linked (for example, 5%). It is important to note that once a map distance between two genetic markers has been established, this distance can be used to predict the expected numbers of each type of progeny. For example, if you know that two mutations are 12 cM apart then you should expect that about 6% of the gametes from a cross will be of each of the two recombinant classes. 
 
===== Map distances can be used to generate a genetic map =====


Things get interesting when we make several pairwise crosses between genes on the same chromosome. We can use this data to construct a genetic map. Genetic maps have the following properties: 

  - Physical distance is proportional to frequency of crossovers (this approximation actually only holds for short distances of <20 cM);
  - Distances are approximately additive: mapped points fall on a line; 
  - Maps are internally consistent and concise. 

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>
{{ :first-genetic-map-sturtevant-1913-49.png?400 |}}
<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]]. Source claims copyright and that commercial use is prohibited without permission; publishing date suggests image is now in the public domain. 
</caption>
</figure> 

<figure Fig6>
{{ :drosophila_genetic_map.png?400 |}}
<caption>
A partial but more modern genetic map of Drosophila showing all four chromosomes. Source: [[https://www.labxchange.org/library/items/lb:LabXchange:11fa2f5a:lx_image:1|LabXchange]]. Licensing: [[https://creativecommons.org/licenses/by/4.0/|CC BY 4.0]].
</caption>
</figure>

It is important to remember that genetic distances are measured using a property of meiosis (genetic recombination) that varies from one organism to another. The relationship between genetic distance and actual physical distance can be summarized in this way: 

$$\text{Genetic distance = physical distance } \times \text{recombination rate}$$

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 relationship between frequency of recombinants and crossovers not linear. This is discussed further in [[Appendix_A|Appendix A]] on tetrad analysis.

===== Unlinked genes =====


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. 



===== A final note on genetic notation =====


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 =====


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 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? 

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. 

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?