Тесты по инструментам командной строки для анализа геномных данных & Ответы – Курсера
Welcome to our comprehensive guide on Command-Line Tools for Genomic Data Science, a crucial skill set for today’s bioinformatics professionals.
В этом посте рассматриваются фундаментальные концепции и практическое применение инструментов командной строки., providing insights and resources to enhance your genomic data analysis skills. Stay tuned for our upcoming викторины to test your knowledge and reinforce your learning.
Модуль 1 викторина
- Больше
- gzip
- ls
- ср
- rmdir
- CD
- Топ
- gzip
3 квартал. Файл “месяцы” lists each of 12 months on a separate line, and no further lines. What would be result if the following command was run:
- год
- 50
- 12
- месяцы
- Re-direct standard error only
- Re-direct the standard input or standard output of a command
- Act as a character separator between different shell commands, without any effects on the outcome
- Replace the ‘;’ sequencing operator in a complex command
Q5. If typing ‘pwd’ производит “/home/userA/Coursera/L1/”, which of the following commands will list the file content of the current directory?
- listdir .
- more *.txt
- mkdir L1
- ls /home/userA/Coursera/L1
Q6. Suppose you current working directory is “/home/Coursera/L1/”, а также “персик”,”яблоко”, а также “pear” are subdirectories, each containing a single file named “геном”. What would be the current directory, as reported by running the ‘pwd’ команда, after each of the four commands in the sequence below:
”’
- cd apple
- г.м. *
- CD ../..
- mv apple plum
”’
-
- /home/Coursera/L1/apple
- /home/Coursera/L1/apple
- /home/Coursera
- /home/Coursera
-
- L1
- Курсера
- яблоко
- plum
-
- plum
- яблоко
- pear
- strawberry
-
- /home/Coursera/L1
- /home/Coursera/L1/apple
- L1/apple
- /home/Coursera/L1
”’cut -d ‘ ‘ -f1,3 seasons | sort -u | wc -l” а также “cut -f1 seasons | Сортировать | unic -c | wc -l”’ ?
- 4, 6
- 12, 20
- 5, 10
- 12, 12
Q8. Your current working directory is named “Растения”. Its subdirectory “яблоко” contains the files “apple.genome”, “apple.samples” а также “apple.genes”. What would be the result of the command ”’rmdir apple”»?
- All files containing the string “apple” in their names will be removed
- None of these choices
- The command will have no effect, since the directory is not empty
- The “apple” directory and all of its content will be removed
Q9. Suppose that you have two files, A and B, containing experiment data. What would be the sequence of outputs for the commands:
- 3, 2, 2
- 5,2,3
- 3, 1, 3
- 2, 4, 5
Q10. The current working directory contains four subdirectories named “яблоко”, “pear”, “персик” а также “strawberry”, each with the following files: “геном”, “гены”, а также “samples”. Which of the following commands wouls extract the top line from all of the “гены” файлы?
- cat п/genes strawberry/genes | tail -1
- голова -1 п/genes strawberry/genes
- Меньше /г | голова -1
- cat п/genes strawberry/genes | grep –c 1
- APCTSYFPEITHI
- AAAAAAAAA
- AGCTACTACGAGCT
- CCCCCCCCCC
2 квартал. How many lines does it take to specify: я) one fasta sequence? and ii) one fastq sequence? Select the best answer:
- Fasta – 1 линия; fastq – 4 линии
- Fasta – a fasta header followed by any number of sequence lines; fastq – 4 линии
- Fasta – any number of lines, including a fasta header; fastq – 2 линии
- Fasta – 100 линии; fastq – 2 линии
- The SAM format is used to represent alignments.
- The BED format can be used to represent gene features.
- SAMtools flagstats reports the total number of mapped reads.
- The GTF format can be used to represent gene features.
- Soft clipping
- Cut and paste
- Hard clipping
- Padding
- 1
- 2
- 3
- 4
- chr1 516 3312 genA.1 100 + 800 900 0 3 296,115,303 0,485,2494
- chr1 515 3312 genA.1 + 515 3312 0 3 296,115,303 516,1001,3010
- chr1 516 3312 genA + 516 3312 0 2 296,303 0,2494
- chr1 515 3312 genA.1 100 + 515 3312 0 3 296,115,303 0,485,2494
Q7. Determine the number of genes, транскрипты, exons per transcript, gene orientation (strand), and the length of 5′ most exon(s) from the GTF snippet below. Select the correct answer.
- Гены: 1; Стенограммы: 2; Exons: 2,2; Strand: -; Length of 5’ exon(s): 2736, 2194.
- Гены: 1; Стенограммы: 2; Exons: 2,2; Strand: -; Length of 5’ exon(s): 2735, 2193.
- Гены: 1; Стенограммы: 1; Exons: 4; Strand: -; Length of 5’ exon(s): 2736.
- Гены: 1; Стенограммы: 4; Exons: 1,1,1,1; Strand: -; Length of 5’ exon(s): 2736, 1417,2194,795.
- R1 maps uniquely to the genome.
- R2’s mate is unmapped.
- R3 is unmapped.
- The R1 alignment is the primary mapping (hit index 0) for that read.
Q9. For the alignment below, which statements are FALSE? The binary encoding for 97 является 972 знак равно 0000 0110 00012. Select all answers that apply.
- The two mates are identical in sequence.
- The alignment represents a potential PCR or optical duplicate.
- The read and its mate are not properly aligned as a pair.
- Both the read and its mate are mapped.
- This is the first read in the pair.
- The sequence of the read’s mate is reverse complemented in its alignment.
- 5, 5, 5
- 3, 4, 2
- 9 , 2, 2
- 3, 2, 2
- Differences in the genomes of individuals are strong contributors to their phenotypic variations.
- Different versions of a gene resulted from genomic mutations are called alleles.
- SNV refers to a Single Nucleotide Variant.
- SNP refers to a Single Non-defined Polymorhism
- The VCF format shows the changes in amino acid resulting from the nucleotide mutation, in column 3.
- The VCF INFO lines describe characteristics of the variant, included in column 8.
- The BCF format is a binary compressed version of VCF.
- VCF stands for Variant Call Format.
3 квартал. What program can be used to generate a list of candidate sites of variation in an exome data set:
- samtools
- мкдир
- bcftools
- bedtools
4 квартал. In a comprahansive effort to study genome variation in a patient cohort, you sequence and call variants in the exome. whole genome shotgun and RNA-seq data from each patient. Which of the following is FALSE when comparing these three types of resources:
- Exome sequencing comprehensively captures variants in the 3’ and 5’ UTRs of genes.
- Exome sequencing can capture variants in a pre-defined set of coding exons and their immediate surrounding area.
- Exome sequencing cannot determine variants in novel polymorphic alternative splicing events.
- Exome sequencing captures fewer variants than whole genome sequencing.
- –местный
- -D
- –ignore-quals
- –чувствительный
-
Only site 2 shows potential variation;
the alternate letter for site 2 is ‘.’;
сайт 1 имеет 8 supporting reads, and site 2 имеет 16
-
Only site 2 shows potential variation;
the alternate letter for site 2 is G;
сайт 1 имеет 8 supporting reads, and site 2 имеет 16
-
Only site 2 shows potential variation;
the alternate letter for site 2 is A;
сайт 1 имеет 8 supporting reads, and site 2 имеет 16
-
Only site 2 shows potential variation;
the alternate letter for site 2 is A;
the alternate allele for site 2 is supported by 9 reads
- Average mapping quality for variant 3 является 40
- The sample contains only the alternate allele for variant 1
- The sample contains only the alternate allele for variant 3
- The sample contains both alleles for variant 2
-
Run bowtie2 with a set of single-end reads, reporting the best alignment only;
then determine the number of matches on each genomic sequence
-
Run bowtie2 with a set of single-end reads, reporting up to 5 alignments per read; then determine the number of matches on each genomic sequence
-
Run bowtie2 with a set of paired-end reads, allowing for local matches;
then report the numbers of alignments containing insertions and deletions, соответственно;
-
Run bowtie2 with a set of paired-end reads, allowing up to 10 matches per read;
then report the number of matches on each genomic sequence
- Produce a 7-column intermediate mpileup file that is piped to ‘cut’
- Report an empty column
- Report in the intermediate mpileup output the qualities of all read bases aligned at that position
- Require a sorted BAM file
- Write the output to file out.vcf.gz
- Report all candidate sites
- Take input from the file in.vcf.gz
- Take input from a VCF compressed file
- Alternative splicing is a common phenomenon in both animals and plants.
- The coding region with a protein-coding gene is used as the template for forming a protein.
- A codon is a nucleotide triplet that is translated into one amino acid.
- A human gene can express at most 12 splice variants.
- Genes that have only one exon are not alternatively spliced
- Some eukaryoyic genes are single exon
- The length of the coding region in a transcript must be a multiple of 3
- The length of intron cannot be a multiple 3
3 квартал. What programs could you use to align RNA-seq reads to: я) a reference genome, and ii) a transcript database?
- cufflinks, bowtie
- tophat, расколоть
- tophat, bowtie
- tophat, cufflinks
- As measures of gene expression, RPKM is determined at the level of reads and FPKM is determined at the level of fragments.
- FPKM stands for fragments-per-kilobase of cDNA sequence-per million reads.
- The sums of FPKM values of all genes in a sample is 1,000,000.
- The sums of FPKMs of all transcripts of a gene is equal to the gene’s expression level.
Q5. What programs could be used to i) assemble transcripts from RNA-seq reads, and ii) identify potentially novel transcripts and genes
- cufflinks, cuffcompare
- tophat, cuffcompare
- tophat, bowtie
- tophat, samtools
- The two transcripts for gene MG051951 overlap on the genome.
- It contains only one gene, MG051951.
- Gene MG051951 has two transcripts, MT162897 and MT070533.
- Transcript MT162897 has a single exon.
- Report spliced reads with at most 6 mismatches in the anchor site
- Create the output in the /home/me/SRR100000 directory
- Run multi-threaded, с 10 threads
- Report only reads with 10 or fewer alignments on the genome
- Label cufflinks transcripts with the prefix ‘Test1’
- Use the default reference transcript annotation to guide assembly
- Run cufflinks to assemble transcripts
- Create a soft link to the BAM read alignment file in the Test1 directory
- 94.0% of the mate 2 reads were mapped
- Of the mapped mate 1 reads, 11.7% had multiple matches on the genome
- The library was strand-specific
- Of the mapped mate 2 reads, 5.0% had multiple matches on the genome
Q10. Which of the following is NOT TRUE about the output below, obtained from a cuffdiff differential expression analysis:
- Locus XLOC_000004 corresponds to gene AT1G01073
- There are too many alignments for testing for differential expression at locus XLOC_000004
- Locus XLOC_000042 corresponds to gene AT1G01580
- There are not enough alignments for testing for differential expression at locus XLOC_000004
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