Last updated: 2022-04-11
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Knit directory: rare-mutation-detection/
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library(ggplot2)
Q: how many cells (or “cell equivalents”) do we need to sequence to detect variants down to a VAF of X%
Let \(p\) be the probability of selecting a mutant cell from a pool of \(n\) cells.
\(p = v / 2\)
Where:
We assume the probability of selecting a mutant cell is binomially distributed. We want to know the probability of selecting at least one mutant:
\(P(Bin(n, p)) > 0)\) = 0.95
This is equivalent to:
\(P(Bin(n, p)) = 0)\) = 0.05
Where \(n\) is the number of sequenced cells. We note that the number of mutant cells will, on average, will be \(2nv\).
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 / 2),
input_cells=n,
mutant_cells=(n * v * 2))
ggplot(vafs, aes(input_cells, p)) +
geom_point() +
theme_bw() +
geom_hline(yintercept=0.05, alpha=0.4)
Version | Author | Date |
---|---|---|
cde303e | Marek Cmero | 2022-02-17 |
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 600 cells (12 are mutants).
deviation <- abs(0.05 - vafs$p)
print(vafs[which(deviation == min(deviation)),])
vaf p input_cells mutant_cells
6 0.01 0.04941382 600 12
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 / 2),
input_cells=n,
mutant_cells=(n * v * 2))
ggplot(vafs, aes(vaf, p)) +
geom_point() +
theme_bw() +
geom_hline(yintercept=0.05, alpha=0.4)
Version | Author | Date |
---|---|---|
cde303e | Marek Cmero | 2022-02-17 |
We can sequence down to approx 0.6% VAF.
deviation <- abs(0.05 - vafs$p)
print(vafs[which(deviation == min(deviation)),])
vaf p input_cells mutant_cells
30 0.006 0.04956308 1000 12
We 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 / 2),
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:
\((1 – v / 2) ^ n = 0.05\)
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 duplex 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 |
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 28 (i.e., 28x 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 <- 28
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 7 196
69 0.4998370 0.010 69 1932
693 0.4999002 0.001 693 19404
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:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[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):
[1] Rcpp_1.0.7 highr_0.9 pillar_1.6.4 compiler_4.0.5
[5] bslib_0.3.0 later_1.3.0 jquerylib_0.1.4 git2r_0.28.0
[9] tools_4.0.5 digest_0.6.27 jsonlite_1.7.2 evaluate_0.14
[13] lifecycle_1.0.1 tibble_3.1.5 gtable_0.3.0 pkgconfig_2.0.3
[17] rlang_0.4.12 DBI_1.1.1 yaml_2.2.1 xfun_0.22
[21] fastmap_1.1.0 withr_2.4.2 dplyr_1.0.7 stringr_1.4.0
[25] knitr_1.33 generics_0.1.1 fs_1.5.0 vctrs_0.3.8
[29] sass_0.4.0 tidyselect_1.1.1 rprojroot_2.0.2 grid_4.0.5
[33] glue_1.4.2 R6_2.5.1 fansi_0.5.0 rmarkdown_2.11
[37] farver_2.1.0 purrr_0.3.4 magrittr_2.0.1 whisker_0.4
[41] scales_1.1.1 promises_1.2.0.1 ellipsis_0.3.2 htmltools_0.5.2
[45] assertthat_0.2.1 colorspace_2.0-0 httpuv_1.6.3 labeling_0.4.2
[49] utf8_1.2.2 stringi_1.7.5 munsell_0.5.0 crayon_1.4.2