Hacktoberfest 2026: los issues que los mantenedores marcaron para octubre, abiertos y aptos para principiantes. Explorar issues de Hacktoberfest

modkit dmr results- question

Abierto
#591 4 comentarios 0 reacciones 0 asignados Ver en GitHub

Nadie ha tomado este issue todavía.

Evaluación

Dificultad
4/5
Tiempo estimado
3-5 días
Aptitud para principiantes
25/100
Tipo de issue
Documentación
Claridad
Necesita aclaración
Estado de actividad
Estancado
Stack tecnológico
rust
Área
cli, data

Línea de trabajo

Start with the modkit dmr pair command and the six referenced pileup BED inputs, comparing the reported percentages with the DMR result for ENSMUST00000174287. The response is complete when it explains how the per-sample values contribute to the differential result and whether one sample can materially influence it.

Escrito por el modelo de indexación a partir del texto del issue.

Descripción

troubleshooting

Dear authors,
Thankyou for making ModKit, it is a really nice tool. I have some questions on the interpretation of the results I got from the DMR pair between two conditions after m6A profiling and I would be grateful to you if you could explain to me what it means.

What I did:
  • Dorado basecalling for m6a and pseudouridine, alignment to transcriptome, followed by modkit pileup to get bed methylfiles
  • modkit dmr pair between two conditions with the following command to get results:
    dmr_result=differential_tx_CT20vsPerKO20.bed modkit dmr pair \ -a ${A1} \ -a ${A2} \ -a ${A3} \ -b ${B1} \ -b ${B2} \ -b ${B3} \ --header \ --min-valid-coverage 20 \ -o "$out_dir"/dmr/${dmr_result} \ --base A \ --ref ${ref} \ --base T \ --threads ${threads} \ --log-filepath dmr_2.log
  • Filter based on effect size and balanced p-value.
issue:

One of the positions is different significantly on the transcript, ENSMUST00000174287, showing its percentage going from 80% in A to 38% in B, meaning a decreased modification in condition B.

Image

However, in the input pileup bed files (condition A: KM4,5,6, condition B: KM7,8,9), the percentage values is A are, 80, 20, 20 and in B are 66, 40, 38. Based on just the % values, it seems that there only one sample in A has higher % than in B
Image

Question

I am confused about how the differential marks are calculated and if one sample can cause so much effect. If that is the case, could you please tell me how I can interpret this?
Any input from you would be really helpful, as I might have also misunderstood the results.

Thankyou in advance.
Khushi.

Lenguaje dominante
Rust
Estrellas
278
Forks
34
Métricas de merge de PR
Sin PR fusionados en 30 d

Preparar el entorno

Este proyecto no incluye contenedor de desarrollo, Dockerfile ni guía de contribución, así que la configuración corre por tu cuenta: empieza por su README y consulta nuestra guía para la primera contribución para los pasos generales.

Primeros pasos

  1. Lee el issue completo y luego la guía de contribución del proyecto.
  2. Comenta en el issue que vas a ocuparte — evita que dos personas hagan lo mismo.
  3. Haz un fork del repositorio y trabaja en una rama.
  4. Abre un pull request que haga referencia al número del issue.

Más de nanoporetech/modkit

Todos los issues de nanoporetech/modkit

Issues similares

Más issues de Rust

Recibe los nuevos issues en tu correo

Un resumen breve de issues de GitHub para principiantes.