A2.1

Question 1 Single best answer Level 1 1 point

Of 509 real high-affinity Spt15 sites, not one is spelled the way the majority is spelled at every position, and the sites differ from one another at many positions. Which term names this variation among real sites?

  1. The consensus
  2. Experimental noise
  3. Motif degeneracy
  4. Information content

Question 2 Single best answer Level 1 1 point

The array’s enrichment score runs from \(-0.5\) to \(+0.5\). What does a value near \(+0.5\) for an 8-letter word mean?

  1. Nearly every probe containing that word ranked among the brightest
  2. The word appears very frequently across the probes on the array
  3. Half of the probes containing the word were among the brightest
  4. The word’s probes reached a specific high raw brightness value

Question 3 Single best answer Level 1 1 point

Every score in the lecture has the form \(\log(\text{observed} / \text{expected})\). Going from one model to the next, what is the part that changes?

  1. The alphabet, and what letters it may be built from
  2. The logarithm, and what base it may take
  3. The observed count, and what it may include
  4. The expected value, and what it may depend on

Question 4 Single best answer Level 2 1 point

The array reads ATATATAT at a median brightness near 14,000 and AAAATTTT near 5,400. A colleague proposes predicting probe brightness from base frequencies alone. Why can no single-letter composition statistic account for the gap between these two words?

  1. The gap comes from the flanking probe context rather than the word, so no statistic of the word could account for it
  2. The alternating word carries the higher A+T content, and composition tracks A+T, so the gap follows from that fraction
  3. Eight letters is too short for a frequency estimate to be stable, so the composition of either word is lost in sampling noise
  4. Both words hold four A and four T, so every single-letter count is identical and composition has nothing left to vary

Question 5 Single best answer Level 2 1 point

The expected count for an adjacent pair multiplies the two single-letter frequencies, \(f_a \cdot f_b\). What has to be true of the sequence for that product to be the right expression?

  1. The letter at one position does not affect the next, so the two probabilities multiply
  2. The adjacent pairs do not overlap one another, so each pair counts as a separate observation
  3. The base composition holds steady along the sequence, so the same two frequencies apply at every position
  4. The two single-letter frequencies are measured from the same sequence, so both describe the material being scanned

Question 6 Single best answer Level 2 1 point

A detector is run on 30 windows, 10 of which are real binders. It returns 16 positive calls, and 8 of those calls are real binders. The false positive rate comes out at \(0.40\) and one minus precision comes out at \(0.50\). Why do the two rates disagree?

  1. The false positive rate divides by the real binders, and one minus precision divides by the positive calls
  2. The false positive rate counts the missed binders while one minus precision counts the wrong flags
  3. The false positive rate divides by the positive calls, and one minus precision divides by the real non-binders
  4. The false positive rate divides by the real non-binders, and one minus precision divides by the positive calls

Question 7 Single best answer Level 2 1 point

Two detectors are scored on the same windows and their ROC curves are drawn on one plot. Curve A runs close to the diagonal along its whole length. Curve B bows toward the upper left and passes through sensitivity \(0.90\) at a false positive rate of \(0.10\). What do the two curves say?

  1. A is set conservatively and B loosely, so the difference between them is a difference of cutoff
  2. A is the safer detector, since a curve close to the diagonal keeps sensitivity and the false positive rate in balance
  3. Nothing can be concluded, because the two curves were drawn without a common cutoff to compare them at
  4. A ranks the windows about as well as chance, and B ranks them usefully at a strict cutoff

Question 8 Single best answer Level 2 1 point

A weight matrix stores \(\log(p_{a,j} / q_a)\) in each cell rather than the plain ratio \(p_{a,j} / q_a\). Why does the scoring formula use the logarithm?

  1. Raw probability ratios range over many orders of magnitude, and the logarithm compresses them into a range where differences between candidate windows are easier to read by eye
  2. The logarithm makes the distribution of scores approximately normal, which is needed so that a valid threshold can be computed from the mean and standard deviation of the score distribution
  3. The logarithm turns a product of per-position ratios into a sum, so the window score is a stable addition rather than a fragile multiplication of many small numbers
  4. The ratio already removes the background by dividing out each base frequency, and taking its log simply converts the result into bits of information so the total can be compared directly against the genome-search threshold

Question 9 Single best answer Level 2 1 point

One strand of a genome of twelve million bases is scanned for every exact match to a 6-base word, and then for every exact match to a 10-base word. The 10-base scan returns far fewer chance matches. What sets the size of that reduction?

  1. Each added base multiplies the number of possible words by four, so four added bases make a specific word four to the fourth times rarer
  2. A longer word samples more bases, so it must match the genome’s base composition more tightly, and that constraint is what makes it rarer
  3. The four added bases each impose a one-in-four requirement, so the reduction is the four bases times the factor of four that each one contributes
  4. A 10-base window leaves four fewer starting positions than a 6-base window, so the scan reports fewer matches because it tests fewer windows

Question 10 Single best answer Level 3 1 point

A 20-base sequence has eight A, five T, four C, and three G. A substitution converts all three G to T, leaving the length unchanged. Using \(f_a = n_a / L\), what is the A+T content of the converted sequence?

  1. \(0.50\)
  2. \(0.65\)
  3. \(0.80\)
  4. \(1.00\)

Question 11 Single best answer Level 3 1 point

A 36-base probe is scanned with a window that moves one base at a time. The width is set first to 8 bases and then to 12. How many windows does the scan produce at each width?

  1. 36 windows at width 8, and 36 at width 12
  2. 28 windows at width 8, and 24 at width 12
  3. 29 windows at width 8, and 25 at width 12
  4. 4 windows at width 8, and 3 at width 12

Question 12 Single best answer Level 3 1 point

A column of a training stack has \(n = 96\) rows and holds zero C. With Laplace pseudo-count \(\alpha = 1\) and \(p_{a,j} = (n_{a,j} + \alpha)/(n + 4\alpha)\), what probability does C get in that column?

  1. \(1/100 = 0.010\)
  2. \(4/100 = 0.040\)
  3. \(1/96 = 0.010\)
  4. \(1/4 = 0.250\)

Question 13 Single best answer Level 3 1 point

An exact 10-base consensus is scanned along one strand of a genome of \(1.2 \times 10^7\) bases. With \(4^{10} \approx 1.0 \times 10^6\) and a uniform base composition, roughly how many chance matches do you expect?

  1. \(\approx 1\)
  2. \(\approx 120\)
  3. \(\approx 1 \times 10^{13}\)
  4. \(\approx 12\)

Question 14 Single best answer Level 4 1 point

In identical flanks, the matrix scores AAAAAAAA above TATATATA, while the assay puts TATATATA among its strongest words and AAAAAAAA well below it. Each core column of the matrix favors A or T about equally. What produces this inversion?

  1. The two flanking columns at each end are included in the total, and their near-empty distributions add noise that swamps the core comparison
  2. The background uses uniform base frequencies instead of the genome’s AT-rich composition, so every A or T is overweighted
  3. The matrix scores each column on its own and adds, so it cannot represent that alternation between neighbors is what the protein rewards
  4. The core columns were estimated from too few sites, so the matrix misjudges how strongly each position prefers A over T

Question 15 Single best answer Level 5 1 point

A weight matrix trained on real Spt15 sites separates sites from non-sites almost perfectly on held-out probes. Scanned across the yeast genome, the same matrix returns about fifteen flags for every real site it recovers. A colleague proposes each fix below. Which best explains the failure and names what has to be added?

  1. The columns rest on 254 training rows, so estimating them from many more real sites would close the gap
  2. The uniform background flatters an AT-rich motif, so scoring against the genome’s own composition would close the gap
  3. The site carries too little information, so several weak signals have to be combined under constraints on their arrangement
  4. The cutoff is too strict for a genome this large, so a looser one would recover the sites it rejects
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