APPENDIX B
`
`
`
`
`
`APPENDIX B
`
`
`
`eyatrs
`sAFATNSNS
`
`Consider two DNAtargets T1 and T2 corresponding to two dyes FAM andVIC, respectively
`Let T1 and T2 be always on separate DNAfragments. Let the number of DNA fragments
`with T1 and T2 targets be M1 and M2 respectively.
`
`S
`
`X
`x
`x
`X
`8
`s
`SON
`NY
`x
`s
`RS
`X
`§
`§
`Q
`wt
`s
`§
`Y zp"
`s
`x
`seh
`Xx
`é
`‘
`ye”
`\
`:
`SS
`X
`Ss Yaar
`s
`3
`
`oe
`3
`
`=
`§
`=
`=
`=
`s
`s
`§
`§
`8
`X
`‘
`
`S
`
`x
`s
`s
`s
`S
`8
`§
`8
`\
`SPAY
`X
`SN QF
`iy
`SHH
`y
`X
`8
`.
`
` SSN
`
`S
`§ \
`=
`x
`§
`\
`;
`=
`yy
`s
`S
`x
`=
`
`5
`=
`<
`SS
`
`~~
`
`SOOHpmpses»58h
`&
`$
`=
`§
`
`Maat
`
`SEAN,
`Ro
`aN
`SS
`5
`
`‘
`
`Let the counts of FAM and VIC positive partitions be N1 and N2, respectively. Note that N1
`and N2 will be smaller than M1 and M2,respectively, as there can be multiple DNA
`fragments in a partition. Let the total number of partitions be N. We will refer to digital
`PCR partitions simply as partitions. In this case, we can expectto see the following table of
`countsof partitions.
`
`S
`
`S
`=
`Sy
`
`cS
`ss
`
`=
`=$
`=
`§
`\
`y
`Wag
`Sy
`
`| VIC Negative
`FAM Positive
`N1.(N-N2)/N
`FAM Negative
`
`
`
`N1 .N2/N
`
`N1
`
`N -N1).(N-N2)/N_|(N-N1).N2/N oo
`[No
`
`If we denote the probability of seeing a partition to be FAM positive as p1=N1/N, and of
`seeing a partition to be VIC positive as p2 = N2/N, then the corresponding probability table
`is:
`
`|CIC Negaattive Probabili
`
`
`FAM Positive
`
`
`FAM Negative
`L-p1).(1-p2
`iap! p2
`
`Probabili
`
`1-p2
`
`In this case, we can say that 100% fragmentation exists.
`
`Appendix B
`
`
`
`We can compute the number of T1 and T2 molecules, M1 and M2, respectively as follows
`M1 =-N log(1 - p1)
`M2 = -N log(1 - p2)
`
`(Given N digital partitions in which P are positive, the numberof molecules is
`- N log (1 -P/N).)
`
`
`
`Nowconsider the other extreme. Both the targets T1 and T2 are on the same DNA fragment
`always. Theyare linked, perhaps because their loci are quite close to each other on the
`same part of a chromosome, andrestriction enzymedid not cut these into separate parts
`Therefore N1 = N2.
`
`RX
`X
`X
`\
`XK
`YK
`
`rpg
`&
`
`Tes
`
`TANT
`X
`\
`x
`x
`SS
`8
`§
`se
`S
`§
` \
`s
`s
`x
`RS
`Re
`X Ns
`s
`Yass
`s
`
`«
`oe
`>
`
`x
`N
`=
`s
`8
`s
`§
`§
`x
`\
`Y
`Ss
`Ss
`
`
`
`RS
`§
`<
`§
`§
`§
`
`8
`§
`x
`\
`
`g
`8
`8
`§
`§
`s
`‘
`s
`S
`S
`N
`yg A
`Wg
`y
`SSS
`
`§
`s
`8
`§
`S
`S
`m&
`SN
`;
`& SonSS
`X
`\
`x
`s
`
`SN
`.
`< \
`sy
`Kg
`,
`
`OAH
`RSS
`8
`§
`é
`=
`
`Sw
`
`oe
`
`SABA
`
`$
`<=gs
`=
`=
`g
`Nay
`Dye
`
`Ne
`
`
`
`|FAMNegative|N-N1_JONNT
`
`
`
`FAM Positive
`
`FAM Negative
`
`Probabili
`
`a Negative
`a p1
`
`1-p1
`
`Probabili
`
`In this case, we can say that 0% fragmentation exists
`
`Wecan compute the number of T1 and T2 molecules as follows, where p1=N1/N
`M1 =-N log(1 - p1)
`M2 = -N log(1 - p1)
`
`Appendix B
`
`
`
`
`
`In intermediate situation, when on the targets are together on the some fragments,but also
`happen to be on separate fragments, then wehave partial fragmentation.
`
`\
`x
`\
`\
`XY
`y
`
`SATA,
`\
`:
`\
`g
`SK
`<
`§
`s
`8
`s
`nN
`try
`s
`s
`X
`Si
`8
`§
`Ry
`ok
`§ Yeu
`s
`
`“
`
`ow
`SS
`me
`Ther
`Ist
`
`s
`
`s
`=
`§
`§
`=
`=
`§
`
`x
`\
`
`8
`s
` §
`s
`§
`s
`s
`
`XS
`
`‘
`
`MII
`
`
`
`Ss
`
`AA,
`S
`s
`<= \
`§
`.
`iN
`§
`<
`Xy
`§
`KY
`§
`
`oo
`er
`reaasao
`SE
`SRHaserr’c
`
`‘
`s
`S
`§
`8
`8
`8
`TR ose
`& DeggghP DE '\Y
`x
`S
`
`~~
`oe
`“Dy
`we
`
`Sy
`
`s
`
`yy
`Sy,
`Ny
`Ss
`
`*
`
`
`
`S
`
`§§
`=
`§
`.
`\
`Ny
`Nye
`
`Suppose we have M3 molecules of linked T1 and T2 fragments, M1 molecules of separate
`T1 fragments, and M2 molecules of separate T2 fragments.
`
`Wemakea table of counts ofpartitions.
`
`|CCC VIC Negative
`FAM Positive
`FAM Negative
`
`
`
`Problem Statement
`
`How can wefind number of molecules M1, M2 and M3, and therebygetting extent of
`fragmentation? For example if M1 = M2 = M3, then we would saythat there is 50%
`fragmentation, as 50% of linked molecules got fragmented into separate fragments and
`50% remainedintact.
`
`
`
`Suppose wehave a solution to the above problem. Then we have anotherinteresting
`application of this solution. Using this algorithm and by using FAM probe fora target T1,
`VIC probe for target T2, and both FAM andVIC probes placed close to each other for third
`target T3, we can achieve multiplexing of 3 targets by using 2 colors.
`
`Basically, we geta third color for “free” using the abovealgorithm.
`
`Nowconsider the case if there are 3 dyes. Thus, wewill have 2x2x2 table of 8 observed
`counts.
`
`Appendix B
`
`
`
`There are 7 different kinds of targets: T1, T2, T3, T12, T23, T13, T123. Here, for example,
`T12 means target in which weplace Dye 1 and Dye 2 probes nextto each other on the
`amplicon region, and T123, meanstarget in which putall the three dyes together on the
`same amplicon,andlikewise for others.
`
`If we have 4 dyes, then we have 24 = 16 counts, and we can now multiplex quantitation of
`2* -1 = 15 target genes. In general, with n colors, we have 2" observed counts and we can
`have 2"-1 targets.
`
`b
`£3
`at
`S
`atSeetad of
`SYSS ANG
`
`Suppose weare given the following counts:
`
`Total|N= N2
`
`| VIC Negative
`FAM Positive NorNIEND
`
`FAM Negative
`
`Consider the following three cases:
`1. Weturn off VIC, as if VIC can not be seen atall.
`2. We turn off FAM, as if FAM can not be seen atall.
`3. We consider both FAM and VIC asif they are really 1 color.
`
`Wewill have the three observations:
`1. Turning off VIC: We will be able to see T1 or T12, together andindistinguishably, as
`if there were one target. It gives the total numberof molecules of T1 and T12, asif
`there were one target species.
`2. Turning offFAM: We will be able to count T2 or T12, indistinguishably,as if there
`were one target. It gives the total numberof molecules of T2 and T12, as if there
`were one target species.
`3. Considering both FAM and VICindistinguishably: We will be able to count T1, T2 or
`T12, indistinguishably.It gives the total numberof molecules of T1, T2 and T12, as if
`there were one target species.
`
`This allows us to step 3 equations in 3 unknowns. We can showthese three cases in the
`form ofa table.
`
`Invisible|Distinct Indistinct|Target Positive Count|Molecules
`
`
`Dyes
`Dyes
`Dyes
`Detected
`
`
`
`aaN1/N
`aaN2/N
`
`4
`
`Appendix B
`
`
`
`Visible Invisible|Distinct Indistinct|Target Positive Count|Molecules
`
`
`
`Dyes
`Dyes
`Dyes
`Dyes
`
`Detected FAM,VIC
`
`{FAM, {T1,T2,T12}|NO1+N10+N11|-log(1 -
`
`VIC}
`(NO1+N10
`
`The 3 linear equations in 3 unknowns M1, M2 and M12 are:
`
`101
`tit
`
`011
`
`1
`12
`
`2 =
`
`=
`
`|
`
`— log 1-—
`
`|
`*
`— log 1-—
`|
`9
`01+ 10+ |
`|- log(1 - —————}
`
`Wesolve the above equationsto get values of M1, M2 and M12.
`
`Wecan then computethe extentof fragmentation in % as follows:
`M = (M1 + M2)/2
`F = M/(M + M12) * 100
`
`8
`A
`Aig
`
`Now wewrite downthesteps of the algorithm clearly.
`
`
`
`ieee TRV ¥INPU
`
`[VIC Negative
`FAMPositive
`
`[N12
`
`N2
`
`
` FAMNegative N21 N22 ¥Aa___
`
`
`N-N2
`
`Step 1. Compute the three entities:
`Mioriz = -N . log(1 - N1/N)
`Moor12 = - N. log(1 - N2/N)
`Mior2or12 =- N. log(1 -(N01+N10+N11)/N)
`
`5
`
`Appendix B
`
`
`
`Step 2. Solve the following linear equations:
`
`101
`
`tid
`
`1
`
`12
`
`[
`
`1
`
`— log 1-—
`2
`
`|
`
`01+ 10+ 11
`— log(1 —- —————_)
`
`Step 3. Compute extent of fragmentation.
`M = (M1 + M2)/2
`F = M/(M + M12) * 100
`
`Step 4. Compute confidence intervals based on the concentration and expected number of
`positive counts. To computeconfidence interval of F, we notethat it is ratio of two random
`variables. Then we apply techniques to estimate confidence interval of ratio of two random
`variables for F.
`
`Now wepresent analternative solution to the problem of partial fragmentation phrased in
`termsof optimization of an objective criterion. First let us make the following table.
`Depending upon what type of molecules wehavein a partition we will have corresponding
`FAM andVIC colors of the partition.
`
`Table mapping molecules into partition color
`
`oFNeg Neg
`jooegPos
`aCPosNeg
`
`
`
`Suppose we make a “guess” of M1, M2 an M12. Wecan compute the probabilities ofa
`partition having T1, T2 or T12 target using inverse equations:
`p12 =1- exp(-M12/N)
`p1=1-exp(-M1/N)
`p2 =1-exp(-M2/N)
`
`6
`
`Appendix B
`
`
`
`From these probabilities we can compute the predicted countsas follows.
`
`|i VIC Negative
`
`FAM Positive|p1(1-p2)(1-p12)N (1 - (sum of other 3 cells
`
`in this table))N FAM Negative|(1-p1)(1-p2)(1-p12)N
`
`(1-p1)p2(1-p12)N
`
`The probabilities above can be filled using the table shown above which maps presence of
`molecules into partition colors.
`
`If our “guess'is really correct, then the predicted counts will “match” closely with our
`expected counts. Thus an optimization algorithm underthe above objective criterion that
`needs to be minimized could also be used to solve the problem.
`
`M1, M2, M12 = best guess= least deviation of predicted counts and actual counts
`
`Wecan then computethe extent of fragmentation in % as follows:
`M = (M1 + M2)/2
`F = M/(M + M12) * 100
`
`rs
`y 2
`LY F
`So ap
`3 ay ye
`SAARATAANA
`WAG § LArSess
`
`To solve the problem for greater number of dyes, we need more equationsas there are
`more unknowns. For 3 dyes and 7 targets, we need 7 equationsto find the concentration of
`these 7 targets.
`
`Wehavethe following table with 2° = 8 rowsof counts of partitions depending upon which
`dyes are positive.
`
`Denote the three dyes by D1, D2 and D3.
`
`Presence ofcolor ina pso CountofPartitions
`Rego
`
`
`
`Neg
`
`Neg
`
`NO10
`
`N100
`NOO1
`NO11
`
`N101
`
`7
`
`Appendix B
`
`
`
`Consider the following table.
`
`Detected
`Dyes
`Dyes
`Dyes
`Dyes
`
`
`D1,D2,D3
`T12, T13, T23,
`
`- {D1,D2,D3}|{T1,T2,T3, C1 =N -NO000
`
`
`
`D1
`
`D2, D3
`
`D1, D3
`
`D1, D2
`
`{T1,T12,T13,
`T123}
`
`{T2,T12,T23,
`T123}
`
`{T3,T13,T23,
`T123}
`
`D3
`
`D1,D2
`
`{T1,T13}
`
`D2
`
`D1,D3
`
`{T1,T12}
`
`D1
`
`D2,D3
`
`{T2, T12}
`
`C2 =
`N100+N101+N110+
`N111
`C3 =
`N0O10+N011+N110+
`N111
`C4 =
`N0O01+N011+N101+
`N111
`Compute
`concentration
`Mior13 by solving 2
`dye problem*
`Compute
`concentration
`Mior12 by solving 2
`dye problem*
`Compute
`concentration
`Moaor12 by solving 2
`dye problem*
`
`*Note that for row 5, we could have also detected {T2,T23} or {T12, T123}, and for row 6,
`{T3, T23}, and for row 7, {T3, T13} and {T23, T123}. This choice does not matter andall
`lead to same results.
`
`We make 3 recursivecalls, in rows 5, 6 and 7.
`
`For example, in row 5, we are really solving 2 dyes, D1 and D2, case with 3 targets: {T1,
`T13} as one target, {T2, T23} as second target, and {T12, T123} as third linked target.
`
`Similarly, in rows 6 and 7, we makerecursivecalls to solve simpler problem of 2 dyes and 3
`targets.
`
`8
`
`Appendix B
`
`
`
`The system of equation lookslike as follows:
`
`1
`log 1-—
`
`11111147
`
`/1001101||
`}0101011]|
`0010111]|
`1000100]|
`1001000||
`
`0101000)L
`
`1
`
`|
`2 |
`3
`|
`12 =|
`13
`23
`
`123
`
`2
`
`— log 1-—
`|
`I 3 |
`yo og 1 |
`4 |
`|
`j- log 1-—|
`
`|
`
`|
`
`SA 8Aneantin
`
`
`
`Now wegivethe steps to generalize the algorithm to arbitrary number of dyes. This isa
`recursive algorithm.Since we have a solution for the case when R = 2, we cansolvefor any
`number of R through recursion.
`
`For R dyes, we are given a table of counts of partitions which has 28 rows,for all possible
`combinationsof presence or absenceof colorsin a partition.
`
`
`
`oy
`STEP 1. Set up a system of 28-1 linear equations. To obtain these equations consider
`different ways of making somecolors invisible. Also consider the case when all colors are
`indistinguishable, whichgivesthefirst row. It can be shownthat we can obtain 2 - 1
`equations in 28 - 1 target concentrations (unknowns). When we makecertain colors
`invisible, then we reduce the problem to a case with fewer colors, which can be solved
`recursively, all the way downto case when there are 2 dyes and 3 targets.
`
`STEP2. Solve the equations.
`
`STEP 3. Compute confidence intervals based on the concentration and expected number of
`positive counts.
`
`ns
`vyOUTPUT:
`At
`Se
`pond
`
`Concentrations of 28- 1 targets along with confidenceintervals.
`
`9
`
`Appendix B
`
`
Accessing this document will incur an additional charge of $.
After purchase, you can access this document again without charge.
Accept $ Charge
This document could not be displayed.
We could not find this document within its docket. Please go back to the docket page and check the link. If that does not work, go back to the docket and refresh it to pull the newest information.
Your account does not support viewing this document.
You need a Paid Account to view this document. Click here to change your account type.
Your account does not support viewing this document.
Set your membership
status to view this document.
With a Docket Alarm membership, you'll
get a whole lot more, including:
- Up-to-date information for this case.
- Email alerts whenever there is an update.
- Full text search for other cases.
- Get email alerts whenever a new case matches your search.
One Moment Please
The filing “” is large (MB) and is being downloaded.
Please refresh this page in a few minutes to see if the filing has been downloaded. The filing will also be emailed to you when the download completes.
Your document is on its way!
If you do not receive the document in five minutes, contact support at support@docketalarm.com.
Sealed Document
We are unable to display this document, it may be under a court ordered seal.
If you have proper credentials to access the file, you may proceed directly to the court's system using your government issued username and password.
Access Government Site
We are unable to display this document.
Connectivity issues with tsdrapi.uspto.gov. Try again now (HTTP Error 429: ).
