Question

Mathematically speaking, $$||OA|| = ||OB||$$ holds true in Fig-1, so there is a common inference that A and B are same. But the distance itself actually means both A and B have same amount of numeric changes relative to O, without indicating the direction of the change. This is in particular a problem what if $$x_1$$ and $$x_2$$ are two genes driving different biological processes, or two factors suggesting distinct customer preferences.

There is a gap between the numeric difference (or similarity) and identity difference (or similarity).

Goal

Arbitrarily make up cell-type “O”, “A”, “B” such that “A” and “B” generally have same shifts from “O”. And to see how distance-based method decides whether “A” and “B” are the same types.

Create “O”

Suppose there are $$G=450$$ genes in total with three categories of the low (1-10), medium (100-200) and high (800-1000) expressed genes. Each cagegory has 150 genes. The general expression profile is nothing but a vector $$\mathbb E \left(e_1, e_2, e_3, \dots, e_G \right)$$.

num_genes <- 450
expect_apple <- round(c(
# low
runif(n = num_genes/3, min = 1, max=10),
# medium
runif(n = num_genes/3, min = 100, max=500),
# high
runif(n = num_genes/3, min = 800, max=1000)
))
idx_low <- 1:num_genes/3
idx_med <- (num_genes/3 + 1) : (num_genes/3*2)
idx_hig <- (num_genes/3*2 + 1) : num_genes

Observed cells ($$N=300$$) are sampled from the profile. Each gene follows its Poisson distribution where $$\lambda_G = e_G$$. This is consistent to the UMI-filtered single-cell RNA-seq data when the pure RNA spike-ins are the loading input.

num_cells <- 500
# a matrix with cells x genes
mat_apple <- round(sapply(expect_apple,
function(e) rpois(n = num_cells, lambda = e)))

Create “A” and “B”

In order to create new cell type “A” and “B”, each of them has ~40 genes being upregulated. There situations were investigated:

num_up_genes <- round((num_genes/3) / 4)
1. low-expressed genes: $$\lambda$$ ($$[1, 10]$$) adding $$[4, 10]$$
2. medium-expressed genes: $$\lambda$$ ($$[100,500]$$) adding $$[50, 250]$$
3. high-expressed genes: $$\lambda$$ ($$[800,1000]$$) adding $$[400, 500]$$

All the add-ups roughly take half of the expression.

Low

idx_up_1 <- head(idx_low, num_up_genes)
idx_up_2 <- tail(idx_low, num_up_genes)
expect_banana <- expect_apple
expect_banana[idx_up_1] <- expect_banana[idx_up_1] + runif(n=num_up_genes,
min=4, max=10)
mat_banana <- round(sapply(expect_banana,
function(e) rpois(n = num_cells, lambda = e)))

expect_cherry <- expect_apple
expect_cherry[idx_up_2] <- expect_cherry[idx_up_2] + runif(n=num_up_genes,
min=4, max=10)
mat_cherry <- round(sapply(expect_cherry,
function(e) rpois(n = num_cells, lambda = e)))

Bonus: Create “AB”

expect_guava <- expect_apple
idx_down <- idx_up_2
num_down_genes <- num_up_genes
# upregulated genes
expect_guava[idx_up_1] <- expect_guava[idx_up_1] +
runif(n=num_up_genes, min=4* sqrt(2)/2, max=10*sqrt(2)/2)
# downregulated genes
num_down_genes <- num_up_genes
expect_guava[idx_down] <- expect_guava[idx_down] -
runif(n=num_down_genes, min=4*sqrt(2)/2, max=10*sqrt(2)/2)
expect_guava[expect_guava < 0] <- 0
mat_guava <- round(sapply(expect_guava,
function(e) rpois(n = num_cells, lambda = e)))

Visualize the expression profile and the sampled cells

From results of clustering and projection down to 2D PCA space, the different populatons are not considered as different.

Med

Similar procedure is performed again to upregulate med-expressed genes and do sampling cells from expression profile.

Visualize expression matrix of the sample cells

Populations are detected.

High

Not surprisingly, populations are found.

Summary

I did not make a good example to hack distance, but re-learned the following messages:

• Highly expressed genes (with high variance) have more weights when counting the sample distance, which implied the need of variance stablization.

Possible to follow-up:

1. Gene-wise normalize
2. Focus on low-expressed genes