Can one use EDTA genome.mod.pass.list and genome.mod.EDTA_TEanno.out output in LAI about ltr_retriever

[oushujun]

Hi Chris,

Yes you can! Even further, you may use the EDTA TEanno.gff3 file to calculate LAI if this is all you have. See this wiki:
https://github.com/oushujun/EDTA/wiki/Calculate-LAI-from-EDTA-GFF3-files

Best,
Shujun

[chaimol]

I am follow the https ://github.com/oushujun/EDTA/wiki/Calculate-LAI-from-EDTA-GFF3-files to caculate the LAI.
But the LAI value is big more than 100.
I run the command like this:

##从EDTA的输出结果计算LAI
genome="genome.fa" #这个是用于EDTA计算的基因组文件的名称
EDTA_path="/share/home/chai/soft/EDTA" #EDTA的安装路径
LAI="/share/home/chai/soft/LTR_retriever/LTR_retriever/LAI" #LAI的绝对路径 
threads=16

for genome in ${genome}; do perl ${EDTA_path}/util/gff2bed.pl $genome.mod.EDTA.TEanno.gff3 structural > $genome.mod.EDTA.TEanno.struc.bed ; done
for genome in ${genome}; do grep LTR $genome.mod.EDTA.TEanno.struc.bed|grep struc|awk '{print $1":"$2".."$3"\t"$7}' > $genome.mod.EDTA.TEanno.LTR.pass.list ; done
for genome in ${genome}; do perl -nle 'my ($chr, $s, $e, $anno, $dir, $supfam)=(split)[0,1,2,3,8,12]; print "10000 0.001 0.001 0.001 $chr $s $e NA $dir $anno $supfam"' $genome.mod.EDTA.TEanno.struc.bed > $genome.out.EDTA.TEanno.out ; done
for genome in ${genome}; do ${LAI} -genome $genome -intact $genome.mod.EDTA.TEanno.LTR.pass.list -all $genome.out.EDTA.TEanno.out -q -t ${threads} ; done

In the end ,The output info as follow:

######################################
### LTR Assembly Index (LAI) beta3.2 ###
######################################

Developer: Shujun Ou

Please cite:

Ou S., Chen J. and Jiang N. (2018). Assessing genome assembly quality using the LTR Assembly Index (LAI). Nucleic Acids Res. gky730: https://doi.org/10.1093/nar/gky730

Parameters: -genome genome.fa -intact genome.fa.mod.EDTA.TEanno.LTR.pass.list -all genome.fa.out.EDTA.TEanno.out -q -t 16


Tue Sep  5 09:02:15 CST 2023    Dependency checking: Passed!
Tue Sep  5 09:02:15 CST 2023    Calculation of LAI will be based on the whole genome.
                                Please use the -mono parameter if your genome is a recent ployploid, otherwise high identity between LTR homeologues will overcorrect raw LAI scores and result in low LAI.
Tue Sep  5 09:02:15 CST 2023    Estimate the identity of LTR sequences in the genome: quick mode
Warning: LOC list genome.fa.out.EDTA.TEanno.out.LAI.LTRlist is empty.
genome.fa.out.EDTA.TEanno.out.LAI.LTR.fa is empty, please check the genome.fa file and the genome.fa.out.EDTA.TEanno.out.LAI.LTRlist file
Tue Sep  5 09:02:16 CST 2023    The identity of LTR sequences: %
Argument "" isn't numeric in numeric lt (<) at /share/home/chai/soft/LTR_retriever/LAI line 143.

                                【Warning】 The identity drops below the safe limit. Instead, identity of 92% will be used for LAI adjustment.

Tue Sep  5 09:02:16 CST 2023    Calculate LAI:

                                                Done!

Tue Sep  5 09:02:20 CST 2023    Result file: genome.fa.out.EDTA.TEanno.out.LAI

                                You may use either raw_LAI or LAI for intraspecific comparison
                                but please use ONLY LAI for interspecific comparison

The output file genome.fa.out.EDTA.TEanno.out.LAI.LTR.fa and genome.fa.out.EDTA.TEanno.out.LAI.LTRlist were empty.
and the LAI file LAI value is more than 100.
head genome.fa.out.EDTA.TEanno.out.LAI

whole_genome    1       118716139       0.0740  0.0740  100.00  105.63
A01     1       3000000 0.0350  0.0350  100.00  105.63
A01     300001  3300000 0.0368  0.0368  100.00  105.63
A01     600001  3600000 0.0452  0.0452  100.00  105.63
A01     900001  3900000 0.0454  0.0454  100.00  105.63

[oushujun]

The total LTR estimation is incorrect. It appears that it only contains structural LTRs but it should also contain homology LTRs. Maybe you need to update your EDTA. I calculate on my end and it works.

Shujun