Novel visual analytics approach for chromosome territory analysis

Mathematical model of the cone of sight

The best and easiest mathematical model to describe a shaft of light is a conical surface Encyclopedia of Math (2020), a quadratic surface that can be described with a mathematical equation, thus making it the most suitable solution for computational purposes. The conical surface is defined as a family of straight lines that cross through a certain point, called the cone vertex, and through any of the points on a flat base, which is on a plane that does not contain the vertex. For the purpose of this paper, the base is circular, i.e., it is a circle made by a cross-section of the sphere by the plane, which passes through a diameter (see Fig. 4). An interior limited by and located inside a conical surface could be considered an appropriate model of the interior of a shaft of light. The sectors of 3D space that are limited by the conical surface will be further referred to as Cones of Sight (CoSes).

Having a set of appropriately arranged spheres we can derive equations for CoS. We wanted to maximize the coverage of space and this led us to the issue of covering space with spheres, which is known as “sphere packing” (WolframAlpha, 2020). This is a known and well-discussed (Conway & Sloane, 1993) problem and will not be covered here in more details. The two types of densest possible space coverage by spheres are hexagonal close packing (HCC) (Weisstein, 2020) and cubic close packing (CCP) (Weisstein, 2021).

To see different sphere arrangements in an interactive way see (WolframAlpha, 2021b) or to observe the difference between HCC and CCP see (WolframAlpha, 2021a). In CCP we have 12 (excluding central) spheres and 12 CoSes. Utilizing the CCP concept in DPC12 we achieved denser space coverage than in DPC, and consequently a more detailed description of the location of surrounding objects. CCP can be easily adjusted to intuitive direction names (Fig. 5).

Packing density, which is defined as a fraction of a volume filled by a given collection of spheres (see Eq. (2), Weisstein (2021)) (2)${\displaystyle \eta =\frac{Vo{l}_{sph}}{Vo{l}_{unit-cell}}}$ is independent of a sphere’s radius and in the case of CCP equals ${\displaystyle {\eta }_{CCP}=\frac{\Pi }{3\sqrt{2}}\simeq 0.74048...}$ which gives us almost 75% of space coverage. The sphere is uniquely defined when we know the coordinates of the center of the sphere and its radius R. To derive the equation of a conical surface, the coordinates of the additional point, which will become the vertex, should also be known.

This point is also assumed to be the geometrical center of the given object (the POV mentioned earlier), in relation to which we want to determine the relative position of other objects. To simplify calculation, it is also assumed that the center of the object is the vertex of the conical surface and is the center of the coordinate system O(0, 0, 0). However, a shift of the center of coordinate system together with all points of the conical surface can be made, if necessary.

The equation of any sphere, with the center in (a, b, c) and the radius R is shown by Eq. (3): (3)${\displaystyle {\left(x-a\right)}^2+{\left(y-b\right)}^2+{\left(z-c\right)}^2=R^2}$

For a given sphere (with a center in a point (a, b, c) and a radius R (denoted as S((a, b, c), R)), it is possible to derive the equation of a desirable conical surface (with a vertex in a given point, here in O(0, 0, 0)) in the following way.

The parametric equation of a line in 3D space that crosses a point P₀(x₀, y₀, z₀) has the following form: (4)${\displaystyle \left\{\begin{array}{c}x=x_{0}t\hfill \\ y=y_{0}t\hfill \\ z=z_{0}t\hfill \end{array}\right.}$ where t is a parameter.

If point P₀ belongs to the line that is also tangent to the sphere, then both equations (of the sphere and the line) have to be fulfilled. So: (5)${\displaystyle {\left(x_0-a\right)}^2+{\left(y_0-b\right)}^2+{\left(z_0-c\right)}^2=R^2}$

Substituting the Eq. (4) for the Eq. (5), after making some transformations (according to Stark, 1974, p. 227), we obtain the conical surface equation (Eq. (6)): (6)${\displaystyle {\left(-2ax-2by-2cz\right)}^2-4\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2-R^2\right)=0}$

Changing sign “=” to “≥” in (Eq. (6)) we have an equation that is fulfilled by points that are on the conical surface or inside it.

As mentioned earlier, using spheres with CCP arrangement we have more than the basic six directions shown in Fig. 2. Now we have 12 spheres as shown in Fig. 5A delineating 12 directions with appropriate CoSes. For better readability, we can project the spheres onto a plane as shown in Fig. 5B.

Then, we named directions (and, accordingly CoSes) in the following way (see also Fig. 5B):

• TOP (relative direction: up):

  • top left TL –the corresponding direction of CoS is top, but slightly to the left and slightly back;

  • top front TF –the corresponding direction of CoS is top, in front;

  • top right TR –the corresponding direction of CoS is top, but slightly to the right and slightly back.

• LEFT (relative direction: left):

  • left front LF –the corresponding direction of CoS is left, but slightly to the front;

  • left L –the corresponding direction of CoS is left;

  • left back LB –the corresponding direction of CoS is left, but “over the shoulder” to the back.

• RIGHT (relative direction: right):

  • right front RF –the corresponding direction of CoS is right, but slightly to the front;

  • right R –the corresponding direction of CoS is right;

  • right back RB –the corresponding direction of CoS is right, but “over the shoulder” to the back.

• BOTTOM (relative direction: down):

  • bottom left BL –the corresponding direction of CoS is bottom, but slightly to the left and slightly to the front;

  • bottom behind BB –the corresponding direction of CoS is bottom, behind;

  • bottom right BR –the corresponding direction of CoS is bottom, but slightly to the right and slightly to the front.

To derive the equations of CoSes that correspond to directions it was necessary to determine the centers of spheres in the CCP arrangement. They are presented in Table 1.

Spherical, one-group dataset with uniform distribution

As you can see in Fig. 6, CoSes do not fill the space entirely as there remains a little space in between them. This means that after conification, some points still reside outside the 12 CoSes. Each of these points needs to be assigned to the CoSes. To solve this problem, the topological measure(HD), which in this case means the distance from point x to set A (Kuratowski, 1980), p.140)¹), was used. See Eq. (7): (7)${\displaystyle dist\left(x,A\right)=liminf|x-a|,a\in A}$

Using HD, any points that remained unassigned after the initial conification process were then assigned to the nearest CoS (Eq. (7)). The conification and assigment of the remaining points to the nearest CoS using HD from here on will be referred to as full conification (FC). After FC, each point P_i in the analyzed set is characterized by five parameters P_i(x_i, y_i, z_i, obj_id, CoS_id). Looking at Fig. 6, we can see the difference between the number of points assigned to CoSes after initial conification without applying HD and the number of points assigned to CoSes after FC.

Table 2 presents the relationship between the conification time and the size of the dataset (number of points subjected to conification). It also shows that the space coverage after initial conification equals at least 80%, which means that less than 20% of the points need to be assigned to the CoSes using computationally demanding HD.

Assigning the sets of points to particular CoSes allows us not only to identify the objects located within a given CoS, but also to determine their nearest and farthest points, as well as to compute the amount of points (e.g., as a percentage) of a given object inside a given CoS. This approach provides us with enough information about objects and their relative location to create a “mental map” of the surrounding space. However, what remains problematic is displaying the obtained, often large amounts of data in a user-friendly way that would enable easy and feasible adjacency analysis. We decided that bar charts will be the most convenient and effective form of visually presenting the resulting data. However, in some applications (e.g., computer vision, machine learning algorithms) storing and processing the data as a table or array can be more useful.

Steps for creating DPC12

The process of creating DPC12 can be described in six simple steps:

1. selecting one fixed object (referred to as central object because it becomes the center of the coordinate system);

2. shifting the center of the coordinate system to the (geometrical) center of central object. This point, now having the coordinates O(0, 0, 0), is the vertex of all cones;

3. making a full conification of the 3D space and assigning all objects to appropriate CoSes;

4. computing what percentage of a given object belongs to a given CoS;

5. computing the distances between the central object and the nearest and farthest points of other objects in each CoS;

6. constructing a visual representation of object locations and distances relative to the central object in the form of a Distance-Profile Chart (DPC12).

DPC12 design

The design of the DPC12 chart was intended to provide the information necessary to create a mental map of all objects surrounding the central object using data computed during steps 4 and 5. During AA, a DPC12 chart can be created for every object in the analyzed scene (space) or only for the objects selected by the user. For example, if there are three objects in the space, up to three DPC12s can be created, one for each object. Using rice nucleus as an example, we could generate twenty-four DPC12 charts, one for each of the twenty-four chromosomes of rice; or we could choose any other number of chromosomes for which we would want to create DPC12s.

The level of detail of the chart depends on the number of spheres from which the CoSes are derived. However, one should take into account that increasing the number of CoSes also increases the complexity of the chart and the difficulty in its reading and interpretation.

The chart consists of a given number of subcharts that correspond to the number of CoSes, so in the case of DPC12 the number of subcharts equals 12. The subcharts are placed on a grid in order to present information in a clear and concise manner, and their names parallel the names of the CoSes (see Fig. 5). The layout of the subcharts of DPC12 is shown in Fig. 7B. The subcharts are patterned after bar plots from which users can read information about the minimum (min) and maximum (max) distances from the central object (Fig. 7A). The color of the bars corresponds to an intensity scale that represents the percentage(fraction) of an object inside a given CoS. The scale used in this article is discrete, made from the sequential palette RColorBrewer Zeileis, Hornik & Murrell (2009) and is shown in Fig. 7C. In the next section we will present a simple example of DPC12 and explain how to interpret it.

In summary, we can divide the DPC12 interpretation process into three steps that involve:

• identifying the objects in each CoS and their position (direction) relative to central object;

• determining the distances of these objects from central object and the range of distances they span;

• reading the fraction or percentage of object(s) inside a given CoS (using the color intensity scale).

Instructional reading of DPC12 –artificial 3D Mouse Set

In this section we will demonstrate how to interpret DPC12 using a relatively simple set of objects in 3D space. It consists of three spherical objects positioned in a way that they resemble the head of a mouse with asymmetrically positioned ears (see Fig. 8). We will conveniently refer to this set as a “3D Mouse Set”, abbreviated M3DS.

In Fig. 8 we see three objects: the head, abbreviated H, and two ears, left (LE) and right (RE). Due to their fixed positions, they can be rotated as a whole relative to the external coordinate system because such rotation has no impact on AA. In this example the ears are situated above the head. Starting from the left, LE is the first object, H is the second object and RE is the third object.

Since we have three objects, three DPC12s need to be created: one for LE, the second for H, and the third for RE. We need to perform three full conification; each one will feature a difference central object. Thus, the three CoSes will be: LE, H, and RE.

The effects of the three full conifications are shown in Fig. 9. The different colors correspond to different conses oriented in various directions according to Fig. 5. When LE is used as a co, the LE itself occupies several CoSes. In contrast, the entire RE is located in just one CoS, together with a part of H. The rest of H is located in several CoSes (Fig. 9A). By analogy, the same applies to RE, when RE is central object (Fig. 9C). When H is central object, it also occupies a few CoSes, while LE is almost entirely in one CoS and all of RE is in another CoS (Fig. 9B).

When the set is fully conified and the distances in all CoSes for all objects are calculated, we can then construct the DPC12 (chart). The three central objects and their respective DPC12s are shown in Fig. 10. More specifically, DPC12 for LE is shown in Fig. 10A, for H is shown in Fig. 10B and for RE is shown in Fig. 10C. Although the data concerning central object could be removed from the chart, we have decided to display them since it allows us to gain some insight into the features of central object. In all DPC12subcharts within each of the Figs. 10A–10C, the leftmost bar on the horizontal axis of the subchart corresponds to LE, the second (middle) one to H, and the rightmost to RE.

AA for chromosome 5′ of rice

The chromosome 5′, which is visible in Fig. 1 in orange, is our central object for current analysis. We would like to assess the distance of other chromosomes in relation to 5′. Obviously, the maximum possible distance between two chromosomes cannot exceed the diameter of the nucleus. In the given example of the modeled rice nucleus its diameter equals 7.3 µm.

Looking at Fig. 1 it is easy to notice that the:

• nearest CT, which is pink and represents chromosome 9, is situated on the right side of the chromosome 5′;

• farthest visible chromosome, which is cyan and represents chromosome 10, is situated across from chromosome 5′ on the opposite side of nucleus;

• chromosome 12′, which is blue, is seemingly located farther away from 5′ than chromosome 9 and situated closer to the center of the nucleus.

In order to determine whether we are able to obtain the same information from DPC12 for the chromosome 5′, let us take a look at Fig. 11. We first notice that a few CoSes are almost empty as they contain no or only a small fraction of CTs. These CoSes are:

• TL, TF, RF, R, LF. Based on this, we can deduce that the 5′ chromosome is situated rather close to the nuclear envelope as there are areas around it where no other chromosomes can be found;

• TR, where we can see chromosome 5 and chromosome 5′. The bar’s color intensity indicates that only small portions of their CTs are located in that CoS.

The majority of chromosomes are situated in four CoSes: RB, L, LB, and BB. To determine which chromosomes are the nearest and farthest in relation to 5′, we should compare the minimum values (min) on the y axis of the bars in all subcharts of DPC12 created for 5′. We should also pay attention to the color of the bar.

Identifying the chromosome nearest to rice chromosome 5′

In order to identify the chromosome nearest to 5′, we need to find the chromosome bars with the lowest min value on the y axis. We can see on the BL subchart that min on y axis for chromosome 4′ is about 0.7 µm. However, on the RB subchart the min for chromosome 9 is even lower and equals about 0.5 µm. The bar spans about 2 µm above its min, reaching its maximum value (max) of 2.5 µm. Moreover, the bar for chromosome 9 is the darkest colored, which means that, based on the intensity scale in Fig. 7C, almost all of chromosome 9 is located in the RB. We could then infer that chromosome 9 is the nearest to 5′. This calculation agrees with the model of chromosome territories presented in Fig. 1, where chromosomes 5′ and 9 seem to be tangent. Reading the DPC12 provides not only the same information about the nearest chromosome in an objective way, but also numerical, comparable values of distance. Moreover, using DPC12 does not entail the need to run special software that requires powerful hardware for manipulating 3D objects. Furthermore, we do not have to make manual rotations of the nucleus in order to view the chromosomes of interest, but instead we can read the distances between all of them in one chart.

Finding the nearest and furthest CT s based on DPC12 for rice chromosome 10

The previous section presented a DPC12 interpretation when the CT arrangement in a 3D model was already known. Now let us see if we can determine the distance between CTs and their spatial distribution based exclusively on DPC12 analysis (when the CT arrangement is not visualized in 3D). For that purpose we will analyze another DPC12 (presented in Fig. 13) in which chromosome 10 shown in Fig. 1 in cyan is the central object. Like in the previous section, we will determine which CTs are the nearest to and farthest from chromosome 10.

As we established in the previous section, our initial statement that chromosome 10 is the farthest away from 5′ was wrong, therefore, we can also assume that 5′ is not the farthest from 10. Thus, we must start our analysis from scratch. Relying solely on the values presented in DPC12, like before, we must search for bars that have the lowest and the highest min value on the y-axis in all subcharts. Also like before, we have to take into account the percentage of a given CTsituated in a particular CoS. In the case of DPC12 for chromosome 10, we can see that almost all other chromosomes are situated in the top three CoSes (TL, TF, TR). Hence, we can assume that this chromosome is situated near the nuclear periphery.

Identifying the chromosome nearest to rice chromosome 10

In order to find the chromosomes closest to chromosome 10, we first need to identify the subcharts with the lowest situated bars. In Fig. 13 these are TF, RF, LF, TR, and LB. To facilitate comparison between the bar min values, we added blue baselines that serve as a ruler-guide (see Figs. 14 and 15).

We will begin by examining the CoSes: TF, RF, and LF (zoomed-in in Fig. 14). In TFthe lowest min value belongs to the bar corresponding to chromosome 4; it equals approximately 0.7 µm. Based on the intensity scale in Fig. 7C, we can determine that about 70% of chromosome 4 is situated in this CoS. In both subcharts RF and LF the lowest min value of the bar also belongs to chromosome 4 and equals approximately 0.9 µm. The weak color intensity of the chromosome 4 bars in both CoSes means that only a small part of chromosome 4 is near chromosome 10, c. 10% in the case of RF and 20% for LF.

Next, we will examine subcharts TR and LB (Fig. 15). In TR the lowest min value of 1.1 µm can be read for chromosome 3′ and 80% of this chromosome is located within the CoS. In LB chromosome 3′ has the lowest min value of 1.3 µm and c. 20% of this chromosome is located in this CoS. The results of the analysis presented above are summarized in Table 3. Comparing the parameters allows us to state that the closest neighbor of chromosome 10 is chromosome 4, while 3′ is slightly farther away, but still very close. In the next section we will show readers how to find the farthest chromosome from chromosome 10.

Table 3:

Summary of min bar values in DPC12 and percentage of CT in CoS.

	4	3′
TF	0.7 µm (70%)	NA
RF	0.9 µm (10%)	NA
LF	0.9 µm (20%)	NA
TR	NA	1.1 µm (80%)
LB	NA	1.3 µm (20%)

DOI: 10.7717/peerj.12661/table-3

Identifying the chromosome farthest from rice chromosome 10

Now that we have identified the chromosomes nearest to chromosome 10, we can proceed by analogy to find the ones that are most distant from 10. To do that, we must first select the DPC12subcharts containing the bars with the highest min value. In this case these are TLand TFsubcharts, zoomed-in and presented in Fig. 16.

We can read in Fig. 16 that in both subcharts the farthest chromosome from chromosome 10 is chromosome 9. The bars indicate that in TLthe chromosome 9 bar spans the distance from 4.9 µm to 6.1 µm, whereas in TFthe same chromosome spans the distance from 4.9 µm to c. 5.8 µm. Up to 40% of chromosome 9 CT is located in TL, and the remaining part of its territory is in TF.

After analyzing DPC12 charts with chromosome 10 as the central object we can definitively state that:

• chromosome 10 is situated near the edge of the nucleus, which is indicated by the almost empty top CoSes (TL, TF, TR)

• no other chromosomes separate chromosome 10 from the nucleus edge, which is indicated by the completely empty bottom CoSes (BL, BB, BR)

• the nearest CTs to chromosome 10 are the CTs of chromosomes 3′ and 4

• the farthest CT from chromosome 10 is the CT of chromosome 9.

We can now compare the data derived from DPC12 analysis to the information provided by the ChroTeMo model of rice nucleus (Fig. 17). As we can see, the conclusions drawn from reading DPC12 are in full agreement with the 3D nucleus model concerning the spatial positioning of chromosomes 10, 3′, 4, and 9. Furthermore, this full agreement proves that DPC12 is sufficient not only for identifying the nearest and farthest CTs in regard to a given chromosome, but also for performing detailed adjacency analyses of CTs without the necessity of visualizing them in 3D.