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Scoring system is a set of values for qualifying the set of one residue being substituted by another in an alignment.
It is also known as substitution matrix.
Scoring matrix of nucleotide is relatively simple.
A positive value or a high score is given for a match & negative value or a low score is given for a mismatch.
Scoring matrices for amino acids are more complicated because scoring has to reflect the physicochemical properties of amino acid residues.
Scoring system is a set of values for qualifying the set of one residue being substituted by another in an alignment.
It is also known as substitution matrix.
Scoring matrix of nucleotide is relatively simple.
A positive value or a high score is given for a match & negative value or a low score is given for a mismatch.
Scoring matrices for amino acids are more complicated because scoring has to reflect the physicochemical properties of amino acid residues.
A diffusion-limited enzyme catalyses a reaction so efficiently that the rate limiting step is that of substrate diffusion into the active site, or product diffusion out. This is also known as kinetic perfection or catalytic perfection. Since the rate of catalysis of such enzymes is set by the diffusion-controlled reaction, it therefore represents an intrinsic, physical constraint on evolution (a maximum peak height in the fitness landscape). Diffusion limited perfect enzymes are very rare. Most enzymes catalyse their reactions to a rate that is 1,000-10,000 times slower than this limit. This is due to both the chemical limitations of difficult reactions, and the evolutionary limitations that such high reaction rates do not confer any extra fitness.
Enzyme kinetics is the study of the chemical reactions that are catalyzed by enzymes. In enzyme kinetics, the reaction rate is measured and the effects of varying the conditions of the reaction are investigated. Studying an enzyme's kinetics can reveal the catalytic mechanism of this enzyme, its role in metabolism, how its activity is controlled, and how a drug or an agonist might inhibit the enzyme. Creative Enzymes is a renowned service provider, supporting many industrial and academic researchers for kinetic studies. https://www.creative-enzymes.com/service/Enzyme-Kinetics_393.html
PAM and BLOSUM are the widely used substitution matrices in the sequence alignment. The mathematical modeling of PAM matrices is explained in these slides.
lead optimization of macrolide drug about
Contents
Introduction
MOA of Antibiotic
Macrolides- General Consideration
Mechanism of Action of Macrolide Antibiotics
Inhibition Of Protein Synthesis
Chemistry of Macrolides
Classification of Macrolide Antibiotics
Structure Activity Relationship
Lead Optimization
Structure of Macrolide Drug
Roxithromycin
Erythromycin
Macrolide Indications
A diffusion-limited enzyme catalyses a reaction so efficiently that the rate limiting step is that of substrate diffusion into the active site, or product diffusion out. This is also known as kinetic perfection or catalytic perfection. Since the rate of catalysis of such enzymes is set by the diffusion-controlled reaction, it therefore represents an intrinsic, physical constraint on evolution (a maximum peak height in the fitness landscape). Diffusion limited perfect enzymes are very rare. Most enzymes catalyse their reactions to a rate that is 1,000-10,000 times slower than this limit. This is due to both the chemical limitations of difficult reactions, and the evolutionary limitations that such high reaction rates do not confer any extra fitness.
Enzyme kinetics is the study of the chemical reactions that are catalyzed by enzymes. In enzyme kinetics, the reaction rate is measured and the effects of varying the conditions of the reaction are investigated. Studying an enzyme's kinetics can reveal the catalytic mechanism of this enzyme, its role in metabolism, how its activity is controlled, and how a drug or an agonist might inhibit the enzyme. Creative Enzymes is a renowned service provider, supporting many industrial and academic researchers for kinetic studies. https://www.creative-enzymes.com/service/Enzyme-Kinetics_393.html
PAM and BLOSUM are the widely used substitution matrices in the sequence alignment. The mathematical modeling of PAM matrices is explained in these slides.
lead optimization of macrolide drug about
Contents
Introduction
MOA of Antibiotic
Macrolides- General Consideration
Mechanism of Action of Macrolide Antibiotics
Inhibition Of Protein Synthesis
Chemistry of Macrolides
Classification of Macrolide Antibiotics
Structure Activity Relationship
Lead Optimization
Structure of Macrolide Drug
Roxithromycin
Erythromycin
Macrolide Indications
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Increasingly Accurate Representation of Biochemistry (v2)Michel Dumontier
Biochemical ontologies aim to capture and represent biochemical entities and the relations that exist between them in an accurate manner. A fundamental starting point is biochemical identity, but our current approach for generating identifiers is haphazard and consequently integrating data is error-prone. I will discuss plausible structure-based strategies for biochemical identity whether it be at molecular level or some part thereof (e.g. residues, collection of residues, atoms, collection of atoms, functional groups) such that identifiers may be generated in an automatic and curator/database independent manner. With structure-based identifiers in hand, we will be in a position to more accurately capture context-specific biochemical knowledge, such as how a set of residues in a binding site are involved in a chemical reaction including the fact that a key nitrogen atom must first be de-protonated. Thus, our current representation of biochemical knowledge may improve such that manual and automatic methods of bio-curation are substantially more accurate.
1 At least 2 questions from this section will be on the .docxmercysuttle
1
At least 2 questions from this section will be on the final exam
SAMPLE QUESTIONS FOR THE FINAL EXAM
Question 1. Ferritin is a protein involved in the storage of iron inside cells. To prevent toxic accumulation of
too much iron inside cells, the intracellular level of ferritin is tightly regulated. To study the regulation of
ferritin synthesis, mammalian cells are grown with or without iron in the culture medium. Note that iron in the
culture medium is rapidly transported inside cells.
a) Upon addition of iron to the culture medium, the intracellular concentration of ferritin mRNA is unchanged
but the concentration of ferritin protein increases. How do you think ferritin expression is regulated? Briefly
explain.
The regulatory sequence given below is found in the ferritin mRNA between the cap structure and the start
codon.
5’-GGGUUUCCGUUCAACAGUGCUUGGACGGAAACCC-3’
Mutations within in this sequence are used to study the regulation of ferritin expression. The following
observation are made:
• ferritin expression is high, independent of the iron concentration, when (i) the entire region is deleted, or
(ii) the region located upstream of the underlined sequence is deleted or (iii) the underlined sequence is
replaced with a random sequence.
• ferritin expression remains iron-dependent when this region is replaced by the following sequence:
5’-GGGCUCAGGUUCAACAGUGCUUGGACCUGAGCCC-3’.
Note that the sequence differences are indicated in bold.
b) Explain why these observations suggest that both sequence and structure of the 5’ end of ferritin mRNA are
important for the regulation of ferritin expression.
c) Ferritin translation becomes iron-independent when the regulatory sequence is moved from the 5’ side
(upstream of the open reading frame) to the 3’ side (downstream of the open reading frame) of ferritin mRNA.
Which step of ferritin translation do you think is affected by the intracellular level of iron?
d) IRP is a protein involved in the regulation of ferritin expression. Anti-IRP antibodies attached to sepharose
beads are added to a cell extract, then the extract is centrifuged to separate the pellet fraction (containing the
sepharose beads ) from the supernatant fraction.
If the cells are cultured in the absence of iron, ferritin mRNA is found together with IRP in the pellet. In
contrast when cells are cultured in the presence of iron ferritin mRNA remains in the supernatant fraction while
IRP alone is found in the pellet. Briefly explain the likely role of IRP in the regulation of ferritin expression.
Question 2. You are studying the development of a newly discovered insect. Like drosophila, it undergoes a
stage in early larval development where the eve gene is expressed in a pattern of 7 stripes. You are particularly
interested in stripes 2 and 5. The following figures show the organization of the cis-acting elements that control
the expression o ...
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Normal Labour/ Stages of Labour/ Mechanism of LabourWasim Ak
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
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Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
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June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
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1. Computation and System Biology Assignment Help
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2. To better understand inborn disorders of metabolism, you isolate a strain of mice that
becomes ill unless fed a diet lacking phenylalanine. You sequence the genome of this
mouse and find several differences from wildtype including a change to a region that
encodes a highly expressed 68 nucleotide RNA which has sequence
5’-UGUACAUGAUGAAGUCAUAGCGAACGGAGAAGGGCCGGCUGAGGAA
ACUGCACGUCACCCUCCUGAAA-3’
in your strain and
5’-UGUACAUGAUGAAAACAGUCUCCCUCUUCUGAAUCUCGCUGAGGAA
ACUGCACGUCACCCUCCUGAAA-3’
in wildtype mice.
Search the sequence in your strain against the mouse genome and transcriptome using
NCBI’s BLASTn: from the BLAST homepage, click on “nucleotide blast” (not “Mouse”)
and use the “Mouse genomic + transcript” (G+T) Database, optimized for “Somewhat
similar sequences”. By expanding the “Algorithm Parameters” box at the bottom, set the
Match/Mismatch scores to
+1/-3.
(A)How many statistically significant hits are there at an E-value of 0.05? In one
sentence, what does an E-value of 0.05 mean? For transcript hits, what are the
maximum reported scores, and are they raw scores or bit scores? (Click on the
hyperlink to view individual hits.) To what parts of your RNA do these hits correspond,
and what is the % match?
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Computation and System Biology Assignment Help
3. (B)Using the E-value and reported score from the result with the highest % identity
match from part (A), calculate the approximate length of the Mouse (G+T) Database.
Using the score S = 50.1 bits and E-value = 2∗ 10!4 along with m = 68nt in the formula
E − value = 𝑚𝑛2! S yields a mouse G+T Database length of 𝑛= 3.55 ∗ 109. Note that the
mouse haploid genome assembly is about 2.7 billion base pairs, so after adding in
transcript sequences, the estimate from the formula is around what we would expect
(various corrections to the simple formula are made for base content, repetitive regions,
and other parameters for the reported BLAST values).
(C)Consider a query sequence Q of length L that matches perfectly to a sequence in the
database, yielding a BLAST E-value E1. How would the E-value change if only the first
half of Q were searched against the database? In particular, would it stay the same, go
up, go down, and how (linearly, exponentially, etc.)?
E1 = 𝑚𝑛2!S and E2 = 𝑚
𝑛2!S/2 ⇒ E12S = 2E22S/2 ⇒ E2=E12(S/2 ! 1).
Thus, the E-value increases essentially exponentially, with an additional decreasing linear
factor of 2 due to halving m. But this latter effect is much smaller than the exponential
increase resulting from the decreased score.
There are two transcript and two genome hits at an E-value of 0.05. The E-value is the
expected number of hits with score at least as high as the hit’s reported score when
searching a query of length 68 nt against the Mouse G+T database. The maximum scores
for the two transcript hits are 54 and 50.1 bits. The hit with score 54 bits corresponds to
positions 38-68 of the query and has 97% identity to its match (matches 30 of 31
positions), while the hit with score 50.1 bits corresponds to positions 14-38 of the query
and has 100% identity.
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4. Intuitively, decreasing the length of query (and therefore match) should make the match
more likely simply by chance and therefore less significant, so we should expect the E-
value to increase. Quantitatively, if the sequence query length were halved (𝑚 → 𝑚/2),
the score S would decrease by a factor of 2 (S → S/2) since there are half as many
positions at which to accumulate positive match scores. Plugging these into equations for
the original query sequence (with score E1) and the half-length query sequence (with
score E2) yields:
(D)Returning to the BLAST results from part (A), to what genes and RNA classes do the
transcript hits with E-values below 0.05 belong? Does your RNA match the sense or
antisense direction of these hits? (Click on the hyperlink of the hit and look at the “Strand”
section, which tells you the DNA strand of the Hit/Query.)
Of the 2 statistically transcript significant hits at an E-value of 0.05, one matches
nucleotides 14- 38 of your RNA complementary to (matching the antisense direction of) an
mRNA that encodes the phenylalanine hydroxylase (PAH) enzyme. Nucleotides 38-68 of
your RNA match the sense direction of Snord100, a C/D Box snoRNA (a type of
noncoding RNA that directs posttranscriptional modifications of other RNAs).
(E)After performing an RNA-protein affinity purification (pull-down) from mouse cell
lysates followed by mass spectrometry, you determine that your RNA interacts with the
product of the ADAR1 gene. What does this enzyme do, and what type of RNA does this
enzyme act on? Looking back at the function and strand of the gene hit to the second part
of your RNA, state a hypothesis as to how your RNA might function to cause your
mouse’s metabolic disorder. (Hint: on the BLAST hit entry corresponding to the mRNA,
click on the “Graphics” link to see the hit in red and how your query at the bottom overlaps
with it. If ADAR1 acts at the UAU codon, what is the resulting change during translation?)
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5. The ADAR1 enzyme catalyzes A-to-I editing, post-transcriptionally deaminating adenosine
in double-stranded RNA duplexes, yielding inosine. Since I is interpreted as G during
translation, A-to-I changes in protein-coding sequences may lead to codon changes and
altered functional properties of the proteins. In addition, A-to-I editing can play important
roles in regulating gene expression, such as by altering alternative splicing, miRNA
sequences, or miRNA target sites in the mRNA.
The PAH gene product is a critical enzyme in phenylalanine metabolism and catalyzes the
rate- limiting step in its complete catabolism. Nucleotides 14-38 of your RNA overlap a
region of the
PAH ORF antisense to the mRNA, including Tyrosine 414 encoded by the codon UAU.
Deamination of this adenosine by ADAR would result in the ribosome interpreting a UGU
codon, which encodes for the much smaller Cysteine. Thus, your mutant snoRNA provides
an RNA duplex for ADAR1 to cause a missense mutation, which could resulting in reduced
activity of the PAH enzyme and contribute to your mouse’s metabolic disorder. Indeed,
genetic Y414C mutations have been observed in human Phenylketonuria patients, and the
mutation has been shown to induce global PAH conformational changes (Gersting et al.
Am. Journ. Human Genetics 83 2008
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2443833/pdf/main.pdf). Note that the RNA
found in the wildtype mouse is very similar to the normal Snord100 snoRNA, which directs
2'O-ribose methylation of rRNA and does not affect PAH.
The example in this problem was inspired by SNORD115 (HBII-52), a human brain-specific
C/D box snoRNA that exhibits sequence complementarity to an alternatively spliced
transcript of the serotonin receptor. For more details of how SNORD115 regulates serotonin
processing through A-to-I editing and alternative splice products, see Kishore and Stamm
Science 2006 (http://www.sciencemag.org/content/311/5758/230.full.pdf).
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6. Problem 2. Gapped sequence alignment
In this problem, you will use the algorithms discussed in class to find the optimal
alignment for a pair of short peptides.
(A)In order to perform this alignment, you must first choose a scoring matrix. For
example, you could use a constant match and mismatch penalty of 1 and -1, respectively,
so that Sij = 1 if i = j and Sij = −1 otherwise. Is this a good idea? Why or why not? In one
sentence, briefly describe how you might obtain a better scoring matrix for protein
comparison.
No - not all amino acid substitutions are equally (dis)favored. Some changes will more
heavily impact protein structure and function than others, and will therefore evolve less
frequently, and so they should be scored differently. For example, changing from one
medium-sized hydrophobic residue to another (e.g., Val to Ile or Leu) within a signal
peptide or transmembrane helix is often tolerated, but changing a hydrophobic to a
charged residue could disrupt function in these contexts, and changing a buried medium-
sized hydrophobic residue like Val to a much larger residue (e.g., Trp) could disrupt
packing. Instead, commonly used scoring matrices are created by comparing related
protein sequences and seeing how often evolution has allowed particular substitutions
occur - these matrices better capture proteins’ functional constraints than this simple +1/-
1 scoring scheme.
(B) You decide to explore more commonly used protein alignment scoring matrices
instead. Compare the score for aligning two tryptophans (W) to the score for aligning two
alanines (A) in the PAM250 scoring matrix. Both of these alignments are “matches”, so
why are these scores so different?
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7. W-W pairings have a large positive score, while A-A pairings have a small positive
score. This means that tryptophan residues are generally highly conserved, and
changes from tryptophan to another amino acid are rare (and therefore generally
evolutionarily unfavorable). Conversely, alanine is not as strongly conserved and
changes relatively frequently. From a biochemical perspective, this makes sense since
alanine is very small and won’t generally have a big impact on protein structure (and is
similar to many other nonpolar amino acids), while tryptophan is very big and changing
it to almost anything else could dramatically alter protein structure.
(C)Perform a global alignment of the two peptides ATWES and TCAET, using the
Needleman-Wunsch algorithm to fill out the alignment matrix below. Use the
BLOSUM62 scoring matrix and a linear gap penalty of 2.
After filling out the matrix, circle the traceback path and write the final alignment. If
there are multiple traceback paths, write out all top-scoring alignments.
Using the BLOSUM62 matrix in the textbook or commonly found online:
Gap A T W E S
Gap 0 -2 -4 -6 -8 -10
T -2 0 3 1 -1 -3
C -4 -2 1 1 -1 -2
A -6 0 -1 -1 0 0
E -8 -2 -1 -3 4 2
T -10 -4 3 1 2 5
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8. The traceback is highlighted in gray above. The final alignment is:
A T W - E S
- T C A E T
Note: There was a slightly different version of the BLOSUM62 matrix on the lecture slides
(the scoring matrix was created from a different set of aligned sequences). This does not
change the traceback or final alignment, only a few scores as shown below. Full credit was
given for either answer. Using the BLOSUM62 matrix in the lecture slides:
Gap A T W E S
Gap 0 -2 -4 -6 -8 -10
T -2 0 3 1 -1 -3
C -4 -2 1 3 1 0
A -6 1 -1 1 3 1
E -8 -1 1 -1 6 4
T -10 -3 4 2 4 8
(D)Different scoring matrices and gap penalties can give very different alignment results.
Below is the alignment of the peptides from part (C) using the PAM250 scoring matrix
(same gap penalty). The traceback path is shaded.
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9. What is the resulting alignment?
A - T W E S
T C A - E T
Compare the optimal alignments obtained using the BLOSUM62 and PAM250 scoring
matrices. Why are they different?
The main reason the alignments are different is because of how strongly the C-W
mismatch is penalized under the PAM250 matrix (score = -8), compared to in the
BLOSUM62 matrix (score
= -2). This means that under BLOSUM62 the C-W mismatch is tolerated without
producing a gap, whereas under PAM250 a gap is preferred over the strong -8 penalty.
Additionally, under PAM250, A-T pairings are more favorable (score = +1 vs. 0 for
BLOSUM62).
Gap A T W E S
Gap 0 -2 -4 -6 -8 -10
T -2 1 1 -1 -3 -5
C -4 -1 -1 -3 -5 -3
A -6 -2 0 -2 -3 -4
E -8 -4 -2 -4 2 0
T -10 -6 -1 -3 0 3
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10. Problem 3. Sequence similarity search statistics (7 points)
You are conducting local nucleotide sequence alignments with your favorite local
alignment tool (e.g. BLAST) with match and mismatch scores of +1 and -1 respectively.
You align a 100bp query sequence to a 1Mbp genome and find that a 20-nt subsequence
from your query is a perfect match.
For each of the following cases, calculate the significance of a 20-nt perfect match
(assume K = 1 in each case):
Note: The Gumbel distribution is continuous, so the P-value for a score x, P(S ≥ x), is
equal to the formula P(S > x) on the lecture slides for continuous x since a single point
P(S = x) has no probability mass. However, we are applying this continuous distribution
to a scoring system that only takes on discrete values, so the P(S = x) values in our
scoring system have nonzero mass (a reasonable value for P(S = x) would be CDF(x+1)
– CDF(x), where CDF is the cumulative
distribution function given on the lecture slides). Thus, our intention was that the P-value is
P(S
≥ 20) = P(S > 19), so 19 would be plugged into the Gumbel CDF formula; however,
since the lecture slides and the textbook have different wording regarding P(S ≥ x) vs.
P(S > x), we will accept P-values with either 19 or 20 used in the Gumbel formula.
(A)Query sequence and genome both have approximately balanced base composition
A=C=G=T=25%).
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11. 4 4
solve 1
e! + 3
e! ! = 1, which has solutions λ = 0 or ln(3) (by substituting in y = e!). Since
λ
must be positive, we use λ=ln(3). The score for the perfect 20nt match is x=20, so
using the distribution of the scores P(S > x) = 1 - exp[-KMNe!!!], we obtain the P-
value:
P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!19𝑙𝑛(3)] = 0.0824.
(0.0283 for x = 20)
(B)Query sequence and genome are both highly A-T rich (A=T=40%, C=G=10%).
A/A and T/T matches occur with probability 16/100 while C/C and G/G matches occur
with probability 1/100. There are also two mismatches each with probability 16/100 (A/T
and T/A) and two with probability 1/100 (C/G and G/C). The remaining 8 pairs are all
mismatches with probability 4/100. Overall, the total probability of a match is 34/100
and probability of a mismatch is 66/100. We need to solve (0.34) e! + (0.66)e!! = 1,
which has nonzero solution λ = 0.6633. The corresponding P-value is:
P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!19(0.6633)] ≈1.
(also ≈1 for x = 20)
Since every pair of nucleotides occurs with equal probability, the probability of a match
(A/A, T/T C/C or G/G) is ¼, and the probability of a mismatch is therefore ¾. So to find
λ, we need to
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12. (C)Query is moderately A+T-rich (A = T = 30%, C = G = 20%) but genome is
moderately C+G-rich (A = T = 20%, C = G = 30%).
In this case, all matches are equiprobable with probability (0.3)(0.2) = 0.06. Therefore the
probability of a match is 4(0.06) = 0.24, and the probability of a mismatch is 1-0.24 =
0.76. Solving (0.24)e! + (0.76)e!! = 1, we obtain nonzero solution λ = 1.153, and the P-
value is:
P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!19(1.153)] = 0.0301.
(0.0096 for x = 20)
(D) Briefly explain why the ordering of the P-values from (A) - (C) makes sense.
Since in (B) we are searching a highly A-T rich query against a highly A-T rich genome,
we expect to see more similarity between the query and the genome by chance than in
(A).
Therefore, the match becomes much less significant than in (A). When the query is A-T
rich and the genome is G-C rich as in (C), however, a match becomes less likely than if
both query and genome had equiprobable base compositions as in (A), and so the P-
value in (C) is smaller than
in (A).
(E)Design a new scoring system for application to searching a 20 nt query of unbiased
composition against a highly A+T-rich genome (as in (B) above) that will increase the
sensitivity for detection of matches to that genome by drawing lines from each box on the
left to its new score in the right box (+1, 0, or -1 for different types of
matches/mismatches). What would the P- value of a perfect match to this query (with 5
A’s, 5 C’s, 5 G’s, 5 T’s) be using your new scoring system?
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13. Since C/C and G/G matches are unlikely by chance due to their low genome content,
observing these matches provides the most evidence of a true alignment; they should
therefore be given a score of +1. In contrast, because A/A and T/T matches will occur
fairly often simply by chance due to their high genome content, these matches provide
less evidence of a true alignment and should be given a score of 0. Mismatches generally
provide evidence against a true alignment, so they should be given a score of -1.
With a query of unbiased content (A=C=G=T=25%) against the biased genome
(A=T=40%, C=G=10%), there is 0.05 total probability of C/C or G/G match (score = +1),
0.2 probability of A/A or T/T match (score = 0), and 0.75 probability of a mismatch (score
= -1). The equation
For a perfectly matched 20 nt query of unbiased content, there will be 10 matches of
score +1 (C/C and G/G) and 10 matches of score 0 (A/A and T/T), for an overall score
of +10. The P- value is therefore:
P(S ≥ 10) = P(S > 9) = 1 - exp[-(20)(1000000)e!9(2.708$)] = 5.1988∗ 10!4
(3.4665∗ 10!5 for x = 10)
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