Rare mutation detection model
Last updated: 2022-05-31
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Knit directory: rare-mutation-detection/
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| File | Version | Author | Date | Message |
|---|---|---|---|---|
| Rmd | e722dc0 | Marek Cmero | 2022-05-31 | Fix typo, minor code update |
| html | b524238 | Marek Cmero | 2022-05-26 | Build site. |
| Rmd | afd79e5 | Marek Cmero | 2022-05-26 | Added revised model, in silico mixtures redone with 1 supporting read, added input cell estimates |
| html | 0b0a0dd | Marek Cmero | 2022-05-12 | Build site. |
| Rmd | 3c84b8f | Marek Cmero | 2022-05-12 | update model for haploid genome |
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| Rmd | f5ca0c0 | Marek Cmero | 2022-04-11 | added mixture experiment calculation |
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| Rmd | 7cfe3e4 | Marek Cmero | 2022-02-17 | Updated/simplified model |
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| Rmd | cf87608 | mcmero | 2021-09-22 | Added equation |
| Rmd | 5cb4f4d | mcmero | 2021-09-22 | Fix probability of selecting mutant fragment |
| html | 5cb4f4d | mcmero | 2021-09-22 | Fix probability of selecting mutant fragment |
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library(ggplot2)Model
Q: how many cells (or “cell equivalents”) do we need to sequence to detect variants down to a VAF of X%
Assumptions
- We consider a haploid genome without SCNAs
- We’re not considering DNA extraction efficiency or ligation efficiency
Probability of sequencing a mutant cell
The probability of selecting a cell with a given variant from a pool of \(n\) cells is the variant allele frequency (VAF) \(v\). This indicates that a VAF of 0.01 means 1 cell in 100 is a mutant (\(nv\)).
We assume the probability of selecting a mutant cell is binomially distributed. We want to know the probability of selecting at least one mutant:
\(P(Bin(n, v)) > 0)\) = 0.95
This is equivalent to:
\(P(Bin(n, v)) = 0)\) = 0.05
Where \(n\) is the number of sequenced cells.
Let’s consider how many input cells (\(n\)) are required to select at least one mutant cell at 1% VAF (95% confidence).
We’ll consider 100 cell increments from \(n = \{100, 200..2000\}\). Using these values, we can plot the probability selecting of sequencing at least one cell.
v = 0.01
n = seq(100, 2000, 100)
vafs <- data.frame(vaf=v,
p=pbinom(0, n, v),
input_cells=n,
mutant_cells=(n * v))
ggplot(vafs, aes(input_cells, p)) +
geom_point() +
theme_bw() +
geom_hline(yintercept=0.05, alpha=0.4)
We can also plot this as mutant cells instead of VAF:
ggplot(vafs, aes(mutant_cells, p)) +
geom_point() +
theme_bw() +
geom_hline(yintercept=0.05, alpha=0.4)
This shows that for a VAF target of 1%, we should sequence at least 300 cells (3 are mutants).
deviation <- abs(0.05 - vafs$p)
print(vafs[which(deviation == min(deviation)),]) vaf p input_cells mutant_cells
3 0.01 0.04904089 300 3Varying the VAF
If we select 1000 input cells, what’s the lowest VAF we can sequence down to, where \(v = {0.0002, 0.0004..0.01}\).
v = seq(0.0002, 0.01, 0.0002)
n = 1000
vafs <- data.frame(vaf=v,
p=pbinom(0, n, v),
input_cells=n,
mutant_cells=(n * v))
ggplot(vafs, aes(vaf, p)) +
geom_point() +
theme_bw() +
geom_hline(yintercept=0.05, alpha=0.4)
We can sequence down to approx 0.3% VAF.
deviation <- abs(0.05 - vafs$p)
print(vafs[which(deviation == min(deviation)),]) vaf p input_cells mutant_cells
15 0.003 0.04956308 1000 3We can look at the relationship between input cells and allele frequency:
cells_vs_vaf = NULL
n = seq(100, 10000, 100)
V = seq(0.001, 0.01, 0.001)
for (v in V) {
toadd <- data.frame(
vaf=as.factor(v),
p=pbinom(0, n, v),
total_cells=n
)
cells_vs_vaf <- rbind(cells_vs_vaf, toadd)
}
ggplot(cells_vs_vaf, aes(total_cells, p, colour=vaf)) +
geom_line() +
theme_bw() +
theme(legend.position = 'bottom') +
geom_hline(yintercept=0.05, alpha=0.4)
We can define an equation based on the binomial probability calculation, of obtaining the number of cells to sequence, to be 95% confident of sequencing the mutation based on the target VAF:
\((1 – v) ^ n = 0.05\)
Coverage
The above thought experiment won’t tell us the probability of sequencing the variant, as sequencing involves a whole host of other factors. Coverage is one of the most important. Given that approximately 30x raw coverage is required to yield 1x of high quality duplex coverage, we can look at the relationship between coverage and VAF (need the variant to be at least 1x to detect).
Here we can see that at VAF = 0.01 (line), we need to sequence to approx 3000x, and this gets exponentially deeper as we go down in target VAF.
v = seq(0.001, 0.1, 0.001)
ccov <- data.frame(d = 30 / v,
v = v)
ggplot(ccov, aes(d, v)) +
geom_line() +
theme_bw() +
geom_hline(yintercept=0.01, alpha=0.4)
| Version | Author | Date |
|---|---|---|
| cde303e | Marek Cmero | 2022-02-17 |
Mixture experiments
How much total coverage will we need, in order to detect a mixture of \(X\%?\) We define this as successfully calling at least 50% of the variants at VAF \(v = \{0.1, 0.01, 0.001\}\) with at least one supporting duplex read. We assume our duplex yield is 30 (i.e., 30x coverage yields 1x duplex coverage).
Our binomial calculation takes the form of:
\(P(Bin(n, v)) = 0)\) = 0.5 where \(n =\) duplex coverage.
df <- NULL
min_reads <- 1
duplex_yield <- 30
duplex_seq_range <- seq(1, 10000, 1)
fraction_we_can_miss <- 0.5
for(vaf in c(0.1, 0.01, 0.001)) {
tmp <- data.frame(
pval=pbinom(min_reads - 1, duplex_seq_range, vaf),
vaf=vaf,
duplex_cov=duplex_seq_range,
total_cov=duplex_seq_range * duplex_yield
)
tmp <- tmp[tmp$pval < fraction_we_can_miss,]
tmp <- tmp[tmp$pval == max(tmp$pval),]
df <- rbind(df, tmp)
}
print(df) pval vaf duplex_cov total_cov
7 0.4782969 0.100 7 210
69 0.4998370 0.010 69 2070
693 0.4999002 0.001 693 20790Revised model
Attempting to combine both coverage and cell input, we can take the following formula:
\((1 - v) ^ {(n * c * d)} = p\)
Where \(v\) = target VAF, \(n\) = number of input cells, \(c\) = coverage per input cell, \(d\) = duplex efficiency and \(p\) = probability of missing a variant.
We set:
- \(c = 10\) (coverage we are aiming for for each genome equivalent)
- \(d = 0.057\) (average efficiency for NanoSeq MB2)
- \(p = 0.05\) (fraction of variants we are okay to miss)
- \(v = \{0.001, 0.01, 0.05, 0.1\}\) (target VAFs)
- \(n = \{100, 200..10000\}\) (range of input cells)
We also need to check whether the number of genome equivalents we take forward for sequencing exceeds the lane capacity (600GB). Our sequenced bases will be:
\(b = g * c * n\) where \(g\) = genome size.
lane_cap <- 6e11
g <- 5e6
pmiss <- 0.05
c <- 10
e <- 0.057
vafs <- c(0.001, 0.01, 0.05, 0.1)
n <- seq(100, 10000, 100)
# calculate duplex coverage
dcov <- n * c * e
# calculate total sequenced bases
bases <- g * c * n
probs <- NULL
min_cells <- NULL
for(vaf in vafs) {
prob <- data.frame(ncells = n,
dcov = dcov,
p = pbinom(0, dcov, vaf),
vaf = vaf,
bases = bases,
within_capacity = bases < lane_cap)
tmp <- prob[prob$p < pmiss,]
tmp <- tmp[tmp$p == max(tmp$p),]
min_cells <- rbind(tmp, min_cells)
probs <- rbind(probs, prob)
}
print(min_cells) ncells dcov p vaf bases within_capacity
1 100 57 0.002465035 0.100 5.00e+09 TRUE
2 200 114 0.002887294 0.050 1.00e+10 TRUE
6 600 342 0.032154114 0.010 3.00e+10 TRUE
53 5300 3021 0.048678808 0.001 2.65e+11 TRUEggplot(probs, aes(ncells, p, colour = factor(vaf))) +
geom_point() +
theme_minimal() +
geom_hline(yintercept = pmiss, alpha=0.4) +
scale_color_brewer(palette = 'Dark2')
| Version | Author | Date |
|---|---|---|
| b524238 | Marek Cmero | 2022-05-26 |
sessionInfo()R version 4.0.5 (2021-03-31)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: CentOS Linux 7 (Core)
Matrix products: default
BLAS: /stornext/System/data/apps/R/R-4.0.5/lib64/R/lib/libRblas.so
LAPACK: /stornext/System/data/apps/R/R-4.0.5/lib64/R/lib/libRlapack.so
locale:
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[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ggplot2_3.3.5 workflowr_1.6.2
loaded via a namespace (and not attached):
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