Hello Everyone,
I am trying to do a liftover with mitochondrial sequences using transanno and am running into issues.
I created my own chain file using .fa reference sequences and following this tutorial and it works and gives expected coordinates when I use picard LiftOverVcf.
picard LiftoverVcf -I $VCFIN -O $VCFOUT -C $CHAIN -R $REF --REJECT $REJECTS
Where:
$VCFIN is my vcf called from reads aligned to old build coordinates
$VCFOUT is the lifted over output vcf
$CHAIN is the chain file
$REF is the new build .fa file
$REJECTS is the variants that fail to liftover
My chain file is:
##matrix=axtChain 16 91,-114,-31,-123,-114,100,-125,-31,-31,-125,100,-114,-123,-31,-114,91
##gapPenalties=axtChain O=400 E=30
chain 1556846 chrM 16567 + 0 16567 chrM 16569 + 0 16569 1
309 1 2
212 2 0
15734 2 1
4 0 1
213 2 1
90
Where the first 16567 bp sequence is the target/reference (source) and the second 16569 is the query (destination).
When I try to run transanno:
transanno liftvcf -m -c $CHAIN \
-o $VCFOUT \
-q $REF\
-r $SOURCE \
--vcf $VCFIN \
--fail $REJECTS
Where $SOURCE is the .fa of the source reference genome.
I get this error: "Error: length of chromosome chrM is not equal to length in chain file. Are you using correct reference?"
Does anyone have any suggestions?
I tried changing the names of the target (source) sequence and the query (destination) sequence and reversing reference and query since the naming is somewhat confusing but neither resulted in a successful liftover. I also tried going into the command source code but have not been successful.
Thanks!
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"Are you using correct reference?"
may be your reference is not using the same chrM. see diferent contigs chrM
Thanks. I was wondering if that is the case but was having some issues understanding the transanno code which is written in rust and I don't see any arguments to easily change this because I would like to use the two .fa sequences I input into the liftvcf command for my analysis. I am testing out other liftover programs but was hoping to get transanno liftvcf working.