A2.3
Question 1 Single best answer Level 1 1 point
A collaborator sends you a table from an RNA sequencing run and asks what the interior numbers mean. What does one interior number hold?
- The number of reads from that sample that mapped anywhere on the reference
- The number of reads from that sample assigned to that gene
- The number of transcript copies of that gene the sample held
- The number of positions in that gene covered by at least one read
Question 2 Single best answer Level 2 1 point
A synthetic RNA is added to a sample at a known 500 copies. One sequencing run returns 1,840 reads for it and a second run of the same prepared sample returns 2,110. What does the pair of numbers show about a count?
- A count varies with the annotation applied, so the two runs used different releases
- A count tracks the copies present, so the spike degraded between the two runs
- A count is one draw from a sampling process, so it estimates the copy number
- A count varies with how the aligner resolves reads, so the two runs placed them differently
Question 3 Single best answer Level 2 1 point
A skin biopsy and a blood draw are taken from one donor on the same morning. RNA sequencing of the two returns profiles that differ across thousands of genes. Why do the two results differ so widely?
- The two tissues accumulated different mutations, and each profile reflects those changes
- The two samples were counted against different annotations, and each profile lists those genes
- The two tissues carry different sets of genes, and each profile lists the genes present
- The two tissues transcribe different subsets of one shared set of genes
Question 4 Single best answer Level 1 1 point
A counting step fails because the GTF was left out of the command. What information was missing?
- The reference base present at each genome coordinate
- The alignment position reported for each sequenced read
- The coordinate interval belonging to each named gene
- The quality score recorded for each sequenced base
Question 5 Single best answer Level 3 1 point
Gene V has exons at 2,000 to 2,400 and 3,100 to 3,500, with an intron between them. Gene W runs from 3,400 to 4,200, so its start overlaps the second exon of gene V. Seven reads occupy 2,050 to 2,150, 2,600 to 2,700, 3,150 to 3,250, 3,420 to 3,480, 3,600 to 3,700, 3,750 to 3,850, and 4,100 to 4,200. A standard feature counter is run with default settings. What counts does it report?
- Gene V 3, gene W 4
- Gene V 2, gene W 3
- Gene V 2, gene W 2
- Gene V 3, gene W 3
Question 6 Single best answer Level 2 1 point
A lab studying a splicing mutant counts one alignment file twice, under two different conventions, and reports both matrices. For most genes the two numbers are close, and for a few they are far apart. What distinguishes the genes where the two conventions disagree?
- Those genes have close neighbors, which one convention absorbs into them and the other keeps separate
- Those genes have many annotated isoforms, which one convention counts separately and the other merges
- Those genes have many reads inside their introns, which one convention counts and the other drops
- Those genes are unusually long, which one convention corrects for and the other leaves uncorrected
Question 7 Single best answer Level 3 1 point
Gene M spans 4 kilobases and carries one short intron. Gene N spans 90 kilobases, of which introns make up about 80 percent. One alignment file is counted against exons only, then counted again against the whole span of each gene. What happens to the two counts?
- Gene M rises slightly and gene N rises sharply
- Gene M rises slightly and gene N falls sharply
- Gene M and gene N rise by the same proportion
- Gene M and gene N are both unchanged
Question 8 Single best answer Level 2 1 point
A core facility offers to re-sequence a library with much longer paired-end reads. One gene in your study loses most of its unassigned reads after the switch, and a second gene keeps almost all of its own. What distinguishes the two genes?
- The first gene is longer, so more of the new reads fit inside it and fewer crossed its boundary
- The first gene shares a locus with another feature, and the second gene had reads matching a second locus
- The first gene had reads matching a second locus, and the second gene shares a locus with another feature
- The first gene was more deeply covered, and deeper coverage resolves unassigned reads of either kind
Question 9 Single best answer Level 2 1 point
In a plant genome, two gene families are annotated with four members each. One family arose from a whole-genome duplication about 4 million years ago and the other from a duplication roughly 100 million years ago. The recent family reports far more unassignable reads. Why?
- Recent duplicates are transcribed more heavily, so more of their reads arrive to be assigned
- Recent duplicates accumulate more sequencing errors, so reads match several members within tolerance
- Recent duplicates are annotated with overlapping intervals, so a read falls inside two of them
- Recent duplicates have had less time to diverge, so long stretches remain near identical
Question 10 Single best answer Level 2 1 point
A counter run with the common default reports one member of a duplicated family far below its relatives in every sample, and the gene barely moves across a treatment that its relatives respond to strongly. What has the rule done to this gene?
- Its shared reads were removed from every sample, taking down both its level and its response
- Its shared reads were halved between it and its relatives, leaving each with part of the total
- Its shared reads were counted once for the family, leaving individual members with none of them
- Its shared reads were moved to its relatives in every sample, inflating them and lowering it
Question 11 Single best answer Level 4 1 point
A paper reports that one member of a four-gene antiviral family stays flat under infection while the other three rise sharply, and that this member sits far below its relatives in every sample. A reviewer re-runs the same alignment file through a transcript-level quantifier. The member’s counts rise in all samples, and it now rises under infection alongside its relatives. Which account fits both results?
- The first run dropped its shared reads, removing both its level and its response
- The first run used an annotation lacking the gene, so it had no interval to collect reads
- The first run dropped its shared reads, removing its level while leaving its response intact
- The first run was right, and the second credited the gene with reads it did not produce
Question 12 Single best answer Level 2 1 point
A reviewer is sent a manuscript reporting that a transporter gene rose 1.4 fold in drought-stressed roots, supported by raw counts from one control library and one treated library. The reviewer asks for one more number before accepting the claim. What should they ask for, and why?
- The total mapped reads in each library, because a deeper library raises every gene at once
- The transcript length of the gene, because a longer transcript yields more reads at one abundance
- The annotation release that was used, because different releases place different intervals
- The number of replicates in each group, because a single library cannot establish a difference
Question 14 Single best answer Level 2 1 point
A hormone peptide transcript of about 300 bases and a structural transcript of about 4,000 bases are confirmed by a separate assay to sit at the same number of copies in one tissue. In the count matrix the short one reports far fewer reads. Why?
- The short transcript amplifies poorly during library preparation, so fewer of its copies survive
- The long transcript is protected from degradation during extraction, so more of it reaches the library
- The long transcript is cut into more sequenceable fragments, so more of its pieces are drawn
- The short transcript aligns ambiguously more often, so more of its reads are left unassigned
Question 15 Single best answer Level 3 1 point
A student has raw counts for a receptor gene and its ligand from one biopsy, together with that library total, and wants to know which of the two is present at higher abundance. What else do they need, and why?
- Nothing further, because both counts came from one library and share its total
- The ambiguous read counts for each gene, because discarded reads lower the two counts unevenly
- The library totals of the other samples, because a share is meaningful only against the experiment
- The transcript length of each gene, because reads accumulate with length as well as with abundance
Question 16 Single best answer Level 5 1 point
A diagnostics company reports that a long noncoding RNA, LINC-Q, is the most abundant transcript in patient plasma, citing its position at the top of the raw count column in each of 40 patients. The annotated transcripts on the panel run from 400 to 12,000 bases, and the LINC-Q transcript is 9,400. Library sizes across the 40 patients range from 8 to 31 million mapped reads. No normalization is reported. Which judgment does the evidence support?
- Supported, because every transcript in a column shares its library, so depth cannot distort the order
- Supported, because the largest raw count in a column is the largest share of that column
- Unsupported, because ranking transcripts within a column requires dividing out transcript length
- Unsupported, because the library sizes vary and counts from different depths cannot be compared