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JoinMap ® List of frequently asked questions

Last modified: 3 November 2023.

1. How do I install my individual license file ?

2. Why do I get the message 'JoinMap is using an evaluation license'?

3. On which versions of MS-Windows runs JoinMap?

4. How do I get results of a JoinMap v4/v4.1 project into a JoinMap v5 project?

5. JoinMap doesn't show my map chart because it didn't fit on the page or it says that it needed 'autofit'; what can I do to fix this?

6. My map has more than 100 loci; which parameters do I adjust for the ML mapping algorithm?

7. My population has more than 4000 loci. What settings do I adjust to let JoinMap v4/v4.1 run more responsive?

8. One of the panels within the program window or the entire program window is not visible; how do I restore this?

9. Why do I get the message 'insufficient linkage in data to complete the map'?

10. What is the best approach to integrating maps from several populations?

11. How is the modified LOD score calculated for the recombination between two loci?

12. How is the chi-square test for heterogeneity calculated?

13. How do I code loci in a CP population that have two alleles (heterozygous) in one parent and one allele (homozygous) in the other?

14. JoinMap does not accept genotypes c or d in my backcross population (BC1). Why?

15. Against what ratio is tested in the 'Locus Genot. Freq.' tabsheet with BCpxFy, IMxFy and RIx populations?


1. How do I install my individual license file ?

JoinMap reads its license file 'JoinMap.lic' in its program file directory (typically: 'C:\Program Files\JoinMap5' for version 5 or 'C:\Program Files (x86)\JoinMap4.1' for version 4.1). This is a protected directory and to place a file in it you need to have 'Administrator' privileges as Windows user. Copy your individual license file to this directory and rename it to 'JoinMap.lic'. With version 4.1 you need first remove the current license file 'JoinMap.lic', as it will be the evaluation license file. The easiest way to install a license is to start JoinMap from within the installation procedure (by placing the appropriate checkmark in the final screen of the installation), and next use the 'Install license' option of the JoinMap Help-menu.

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2. Why do I get the message 'JoinMap is using an evaluation license'?

When you get this message, it means that JoinMap does not have an individual license file installed. This situation only allows evaluation of the software, with limited functionality. If you obtained an individual license file (on your installation CDROM or by e-mail), then you must install this file according to the answer to question 1. The individual license file usually has a file name with the extension '.JM5win' (or '.JM41win').

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3. On which versions of MS-Windows runs JoinMap?

JoinMap 5 runs on 64-bit Windows 10. JoinMap 4 and 4.1 run as 32-bit software on 32-bit and 64-bit Windows 10. All programs are expected to run properly on Windows 11, although this was not tested extensively.

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4. How do I get results of a JoinMap v4/v4.1 project into a JoinMap v5 project?

From a JoinMap v4/v4.1 project, you can export a Data tabsheet as .loc file and load that into a JoinMap v5 project as a population. You can export a 'Map (text)' tabsheet as a .txt file and load that into a v5 project as a map. If you create a multi-group map in v4/v4.1, export its 'Map (text)' tabsheet and import this into the v5 project, then you can use the imported map for creating a grouping and groups from the population in the v5 project. Once you have the groups in v5, then you can copy the fixed orders shown in the session logs of the mapping nodes in the v4/v4.1 project and paste them as fixed order (with the @ character) as well as start order (without the @ character) of the corresponding groups in the v5 project. Next, if you have JoinMap v5 calculate the maps, this will be done quickly and in accordance with the fixed order. Using the final order as start order will make the new map calculation circumvent the gradual map building procedure entirely.

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5. JoinMap doesn't show my map chart because it didn't fit on the page or it says that it needed 'autofit'; what can I do to fix this?

The system prevents writing locus names over each other, so each position will take up a line on the page. Sometimes there are just too many loci to show underneath each other on a page. You may be able to adjust various parameter settings, so that they do allow a fit.

You can increase the available space for drawing by adjusting the Page Setup: choose portrait orientation, take a larger page size and/or use smaller margins.

Next, you can adjust various Map Chart Options: first, reduce all font sizes (locus font, linkage group names font, title font, footer font). If you do not use a title or footer, then their space is available to the actual map (i.e. then their font sizes do not matter).

On the Page Layout tab, reduce the bottom space of the linkage group names (i.e. margin below the group name).

On the Bars tab, JoinMap mentions the currently available height of the chart area, given the page setup, title space (if used), footer space (if used) and group name bottom space, but only after a redraw (!). Set the bar length to use the maximum length and set this to the available height.

On the Loci tab you can increase the number of locus names on one line.

If your changes are insufficient to resolve the problem, then you can make JoinMap split up the map according to the setting 'Split map interval (cM)' on the bottom of the Bars tab.

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6. My map has more than 100 loci; which parameters do I adjust for the ML mapping algorithm?

The spatial sampling thresholds seem to work more efficient with the next set of values: 0.100, 0.0300, 0.010, 0.003, 0.001. The map order optimization needs longer chains (e.g. chain length > 10,000) and a more stringent stopping criterion (e.g. Stop after # chains > 20,000). Experiment with the values and observe any differences.

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7. My population has more than 4000 loci. What settings do I adjust to let JoinMap v4/v4.1 run more responsive?

JoinMap v4 and v4.1 were not developed to deal with such large numbers of loci. Basically, the program has problems with displaying the very large tables obtained with so many records. First, you should load the data using a loc-file, as the JoinMap Dataset tabsheet will be slow and sometimes act strange. Next, uncheck the checkbox 'Load Data tabsheets' on the right hand side of the toolbar (if you occasionaly wish to inspect the genotypes, then check the box temporarily). Of the Calculation Options you can uncheck the checkboxes 'Show weak linkages' and 'Show strong linkages' and reduce the 'Number of maximum linkages' to 1 or 0 if you are not interested in them.

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8. One of the panels within the program window or the entire program window is not visible; how do I restore this?

JoinMap stores various view settings of its main window (position and size on the screen, partitioning of panels) in the file 'Documents\JoinMap5\JoinMap.Settings.sqlite' or 'in the 'Documents\JoinMap4.1\JoinMap.ProgSetup'. You should close the program, remove this file and start the program again; it will reset all view settings of the program. For versions 4 and 4.1 this will also remove your default Calculation, Environment and Map Chart options.

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9. Why do I get the message 'insufficient linkage in data to complete the map'?

The LOD grouping procedure uses linkage to any single locus already in a group to determine whether that locus belongs to that group. The mapping procedure first has LOD and REC thresholds determining what data are used, and next any marker fitted onto the map during the building process must have at least two distinct links (i.e. links to loci that themselves have r>0, so not r=0, between them) in order to establish a direction in the localization. So if there are not two distinct links due to the stringent thresholds then the program reports that there is insufficient linkage in the data, even though the locus is in the LOD grouping node.

You can set the LOD and recombination frequency thresholds for mapping to a lower stringency, so that all information necessary to obtain sufficient linkage will come available.

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10. What is the best approach to integrating maps from several populations?

Generally, the best approach is:
a) calculate maps for each population separately;
b) use fixed orders based on these separate maps to calculate the integrated map; when there are no conflicts between fixed orders, you are ready;
c) when there are conflicts between fixed orders, there are several things you can do to decide which order is the most acceptable:
1) remove the conflicting fixed orders when calculating the integrated map, and see which order is the dominant one;
2) impose the conflicting fixed orders on the separate populations, and see whether these result in acceptable solutions, i.e. reasonable chi-squares and no negative distances;
3) it may be impossible to resolve the conflicts; this can be caused by errors in the data, but also by a natural cause, i.e. there is a chromosome inversion. Of course, it is impossible to join maps in the latter case.

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11. How is the modified LOD score calculated for the recombination between two loci?

For each pair of loci a contingency table is produced of the genotypes. The dimensions of the table depend on the population type and the segregation types of the loci (unknown genotypes are ignored). From this table the G statistic (-2 times the logarithm of the likelihood ratio for the Poisson distribution, see e.g. Fienberg, 1979, The analysis of cross-classified categorical data, MIT Press) is calculated to test for independence; the expected number (E) in each cell is calculated from the row-total (R), the column-total (C) and the grand-total (T):

E = R*C/T .

The G statistic then is a summation (SUM) over all cells (O is the observed number, ln() is the natural logarithm):

G = 2 * SUM [ O*ln(O/E) ] .

The G statistic (Gd) has an approximate chi-square distribution with the number of rows in the table minus 1 multiplied by the number of columns minus 1 as the degrees of freedom (d). When the loci have different numbers of genotypes in their segregation, this would present a problem, because this number affects the degrees of freedom in the G test and in the testing of linkage one would need to take account of the degrees of freedom. In order to remove this problem and to ensure the comparability of data coming from different population or segregation types the G statistic with d degrees of freedom, Gd, is transformed approximately to a G statistic, G1, that would have been obtained if there was just a single degree of freedom (as if in a backcross). This approximate transformation is an empirically determined formula (exp() is the exponential function):

e = exp( -Gd/(2*(d-1)) ) ,
G1 = ((4-e)*e - 3)*(d-1) + Gd .

Because in genetics one is used to LOD scores, which are likelihood ratio statistics using the 10-base logarithm instead of the natural logarithm multiplied by -2, the modified LOD score (mLOD) is simply derived from G1:

mLOD = G1 / (2*ln(10)) .

It can be shown that for the case of two loci each segregating in two genotypes (e.g. population types BC1 or DH1) when there is no segregation distortion this mLOD equals the 'normal' LOD score. The modified LOD score is not sensitive to segregation distortion, in contrast to the 'normal' LOD score.

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12. How is the chi-square test for heterogeneity calculated?

From the pwd-file the locus pairs occurring more than once are collected. The pwd-file contains for each pair the recombination frequency and the modified LOD score. In this modified LOD score essentially two important corrections are contained: (1) the segregation data are insensitive to segregation distortion, and (2) the segregation data are transformed as if obtained from a backcross (BC1) under no segregation distortion; this ensures the comparability of data coming from different population or segregation types. For each pair (with two identical loci) the hypothetical number of recombinant (Nr) and non-recombinant (Nn) plants in a backcross (BC1) are calculated based on the recombination frequency (r) and the LOD score (LOD) (r and LOD are read from the pwd-file for each pair) (Nt is the total number of plants, log() is the 10-base logarithm, the numbers Nr, Nn and Nt are treated as 'real's and not as 'integer's):

since:

LOD = Nt * log(2)  +  r * Nt * log(r)  +  (1-r) * Nt * log(1-r) ,

we get:

Nt = LOD / (log(2)  +  r * log(r)  +  (1-r) * log(r)) ,

and hence:

Nr = r * Nt

and

Nn = (1-r) * Nt.

This way a contingency table is created with for each pair a row with two columns, i.e. the number of recombinants and the number of non-recombinants. From this table the G statistic is calculated to test for independence; the expected number (E) in each cell is calculated from the row-total (R), the column-total (C) and the grand-total (T):

E = R * C / T .

The G statistic then is a summation (SUM) over all cells (O is the observed number, ln() is the natural logarithm):

G = 2 * SUM [ O * ln( O / E ) ] .

The G statistic has an approximate chi-square distribution with the number of pairs minus 1 as degrees of freedom. For each pair the contribution to the G test is given in the output, so that it is sometimes possible to locate the most deviant pair.

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13. How do I code loci in a CP population that have two alleles (heterozygous) in one parent and one allele (homozygous) in the other?

Loci heterozygous in one parent and homozygous in the other must be coded as loci with a segregation type <lmxll> or <nnxnp>, depending on which of the parents is heterozygous. The allele in the homozygous parents does not need to be identical to either one of the alleles in the heterozygous parent, genetically the situation <abxaa> is identical to <abxcc> as is <aaxab> to <ccxab>.

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14. JoinMap does not accept genotypes c or d in my backcross population (BC1). Why?

Loci coded as c and d represent dominant genotype observations. Code c means the genotype underlying the observation may be h or b, while d means a or h. This can only occur, if one of the two alleles involved in the cross is a null allele.

In a backcross it is assumed that P1 and P2, the parents of the F1, are fully homozygous. If either P1 or P2 is homozygous for two null alleles, then the F1 will be heterozygous with a non-null allele (+) and a null allele (−).

As you can see in the following scheme, there is no visible segregation in the backcross to P1, BC1a: (++) cannot be distinguished from (+−). Whereas there is visible segregation in the backcross to P2, BC1b: (+−) can be distinguished from (−−). Of course, the situation is opposite if P1 holds the null alleles.

            P1 a (++)  x  P2 b (−−)
                       |
                       v
         P1 a (++)  x  F1 h (+−)  x  P2 b (−−)
                    |             |
                    v             v
                   BC1a          BC1b
           (++) or (+−)          (+−) or (−−)
Thus, if a marker is visibly segregating (BC1b in this scheme), then it is always the two genotypes (+−) or (−−), i.e. either h or b, respectively, in this scheme. If P1 holds the null alleles, then only the BC1a can have visible segregation, (−−) or (+−), i.e. a or h, respectively.

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15. Against what ratio is tested in the 'Locus Genot. Freq.' tabsheet with BCpxFy, IMxFy and RIx populations?

For RIx and IMxFy population types, the expected frequencies of genotypes a and b are always equal. In the classification (a,b) any observation of genotype h is ignored in the test. In the other classifications the expected frequencies of h, c, or d are based on the expected frequency of h, which depends on the generation numbers x and y. For population type BCpxFy the expected frequencies are always based on the backcross parent and the generations x and y. The classification type can be changed for loci marked in the tabsheet using the Population menu option 'Set X2-Test Classification for Marked Loci'.

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