library(dplyr)
library(slendr)
init_env(uv = TRUE)
chimp <- population("CHIMP", time = 7e6, N = 5000)
afr <- population("AFR", parent = chimp, time = 6e6, N = 15000)
eur <- population("EUR", parent = afr, time = 60e3, N = 3000)
nea <- population("NEA", parent = afr, time = 600e3, N = 1000, remove = 40e3)
gf <- gene_flow(from = nea, to = eur, proportion = 0.03, start = 55000, end = 50000)
model <- compile_model(
populations = list(chimp, nea, afr, eur),
gene_flow = gf,
generation_time = 30
)
# sampling of two Neanderthals at 70kya and 40kya
nea_samples <- schedule_recording(model, times = c(70000, 40000), list(nea, 1))
# sampling one Chimpanzee sample, 5 African samples, and 10 European samples
present_samples <- schedule_recording(model, times = 0, list(chimp, 1), list(afr, 5), list(eur, 10))
# sampling one European sample between 40kya and 2kya, every 2000 years
times <- seq(40000, 2000, by = -2000)
emh_samples <- schedule_recording(model, times, list(eur, 1))
# binding all sampling schedules together
schedule <- rbind(nea_samples, present_samples, emh_samples)
# simulate a tree sequence with mutations (feel free to increase the amount
# of sequence from 20 Mb to something like 100 Mb if you want)
ts <-
msprime(model, sequence_length = 20e6, recombination_rate = 1e-8, samples = schedule, random_seed = 1269258439) %>%
ts_mutate(mutation_rate = 1e-8, random_seed = 1269258439)Tree-sequence statistics
Introduction
In this session, you are going to build on top of the results from the previous session on programming demographic models. Specifically, you will learn how to compute population genetic statistics on slendr-simulated tree sequences using its interface to the tskit Python module.
TODO — ideas
https://en.wikipedia.org/wiki/Watterson_estimator:
“However, the estimator can be biased by population structure. For example, theta is downwardly biased in an exponentially growing population. It can also be biased by violation of the infinite-sites mutational model; multiple point mutations at a single site will downwardly bias the estimate.”
Exercise 0 — catching up
First, create a new R script popgen-stats.R and paste in the following code. This is one of the possible solutions to Exercise 1 of the session on building demographic models. It’s easier if we all use it to be on the same page from now on, starting from the same model and the same simulated tree sequence:
As a sanity check, let’s use a couple of tidyverse table-munging tricks to make sure the tree sequence does contain a set of samples which matches our intended sampling schedule (particularly the time series of European individuals and the two Neanderthals):
# total number of recorded individuals in the tree sequence
ts_recorded(ts) %>% nrow[1] 38# times of sampling of each recorded Neanderthal
ts_recorded(ts) %>% filter(pop == "NEA") %>% pull(time)[1] 70000 40000# count of individuals in each population
ts_recorded(ts) %>%
mutate(age_group = if_else(time == 0, "present-day", "ancient")) %>%
group_by(pop, age_group) %>%
summarize(n = n()) %>%
select(age_group, pop, n)`summarise()` has regrouped the output.
ℹ Summaries were computed grouped by pop and age_group.
ℹ Output is grouped by pop.
ℹ Use `summarise(.groups = "drop_last")` to silence this message.
ℹ Use `summarise(.by = c(pop, age_group))` for per-operation grouping
(`?dplyr::dplyr_by`) instead.# A tibble: 5 × 3
# Groups: pop [4]
age_group pop n
<chr> <chr> <int>
1 present-day AFR 5
2 present-day CHIMP 1
3 ancient EUR 20
4 present-day EUR 10
5 ancient NEA 2Note: These bits of tidyverse code are extremely helpful when you’re working with large tree sequences with many individuals as sanity checks that your sampling worked as intended. I’m listing them here in case you’ve never worked with the tidyverse family of R packages before (such as the dplyr package where filter(), group_by(), tally(), and pull() come from).
Everything looks good! Having made sure that the ts object contains the individuals we want, let’s move to the first exercise.
Exercise 1: Computing nucleotide diversity
Our model of ancient human history makes a fairly clear prediction of what should be the nucleotide diversity expected in the simulated populations. Compute the nucleotide diversity in all populations using the slendr function ts_diversity() in your tree sequence ts. Add that command after your simulation code in your new script in popgen-stats.R (do this even in the following exercises, to keep everything in one place). Do you get numbers that (relatively between all populations) match what would expect from the model given the \(N_e\) that you programmed for each of the lineages? Check the figure created by the plot_model(model) command earlier to get a refresher.
NoteClick to see the solution
Population-based nucleotide diversity:
Let’s first get a named list of individuals in each group we want to be working with (slendr tree-sequence statistic functions generally operate with this kind of structure):
sample_sets <- ts_names(ts, split = TRUE)
sample_sets$AFR
[1] "AFR_1" "AFR_2" "AFR_3" "AFR_4" "AFR_5"
$CHIMP
[1] "CHIMP_1"
$EUR
[1] "EUR_1" "EUR_2" "EUR_3" "EUR_4" "EUR_5" "EUR_6" "EUR_7" "EUR_8"
[9] "EUR_9" "EUR_10" "EUR_11" "EUR_12" "EUR_13" "EUR_14" "EUR_15" "EUR_16"
[17] "EUR_17" "EUR_18" "EUR_19" "EUR_20" "EUR_21" "EUR_22" "EUR_23" "EUR_24"
[25] "EUR_25" "EUR_26" "EUR_27" "EUR_28" "EUR_29" "EUR_30"
$NEA
[1] "NEA_1" "NEA_2"We can use such sample_sets object to compute nucleotide diversity (pi) in each population, in a bit of a similar manner to how we would do it with the standard tskit in Python:
pi_pop <- ts_diversity(ts, sample_sets = sample_sets)
arrange(pi_pop, diversity)# A tibble: 4 × 2
set diversity
<chr> <dbl>
1 NEA 0.0000481
2 CHIMP 0.000189
3 EUR 0.000543
4 AFR 0.000593 You can see that this simple computation fits the extreme differences in long-term \(N_e\) encoded by your slendr demografr model, visualized here for reference:
plot_model(model)
Hint: Nearly every slendr statistic function interfacing with tskit accepts a ts tree-sequence object as its first argument, with further arguments being either:
a vector of individual names representing a group of samples to compute a statistic on,
a (named) list of such vectors (each element of that list for a group of samples with that name).
Although you can get all the above information by processing the table produced by the ts_recorded() function, slendr provides a useful helper function ts_names() which only returns the names of individuals as a vector (or a named list of such vectors, one vector per population as shown below).
When you call it directly, you get a plain vector of individual names:
ts_names(ts) [1] "NEA_1" "EUR_1" "NEA_2" "EUR_2" "EUR_3" "EUR_4" "EUR_5"
[8] "EUR_6" "EUR_7" "EUR_8" "EUR_9" "EUR_10" "EUR_11" "EUR_12"
[15] "EUR_13" "EUR_14" "EUR_15" "EUR_16" "EUR_17" "EUR_18" "EUR_19"
[22] "EUR_20" "AFR_1" "AFR_2" "AFR_3" "AFR_4" "AFR_5" "CHIMP_1"
[29] "EUR_21" "EUR_22" "EUR_23" "EUR_24" "EUR_25" "EUR_26" "EUR_27"
[36] "EUR_28" "EUR_29" "EUR_30" This is not super helpful, unless we want to compute some statistic for everyone in the tree sequence, regardless of their population assignment. A bit more useful is to call the function like this, because it will produce a result which can be immediately used as the sample_sets = argument mentioned in the Hint above:
ts_names(ts, split = TRUE)$AFR
[1] "AFR_1" "AFR_2" "AFR_3" "AFR_4" "AFR_5"
$CHIMP
[1] "CHIMP_1"
$EUR
[1] "EUR_1" "EUR_2" "EUR_3" "EUR_4" "EUR_5" "EUR_6" "EUR_7" "EUR_8"
[9] "EUR_9" "EUR_10" "EUR_11" "EUR_12" "EUR_13" "EUR_14" "EUR_15" "EUR_16"
[17] "EUR_17" "EUR_18" "EUR_19" "EUR_20" "EUR_21" "EUR_22" "EUR_23" "EUR_24"
[25] "EUR_25" "EUR_26" "EUR_27" "EUR_28" "EUR_29" "EUR_30"
$NEA
[1] "NEA_1" "NEA_2"As you can see, this command returned a normal R list, with each element containing a vector of individual names in a population. This means that we can use standard R named-list indexing to get subsets of individuals like this:
# this saves the named list of individual samples into a variable `names`
names <- ts_names(ts, split = TRUE)
# this is how we can then access the names of all recorded Neanderthals
names["NEA"]$NEA
[1] "NEA_1" "NEA_2"# this is how we can access the lists of names of two populations
names[c("EUR", "NEA")]$EUR
[1] "EUR_1" "EUR_2" "EUR_3" "EUR_4" "EUR_5" "EUR_6" "EUR_7" "EUR_8"
[9] "EUR_9" "EUR_10" "EUR_11" "EUR_12" "EUR_13" "EUR_14" "EUR_15" "EUR_16"
[17] "EUR_17" "EUR_18" "EUR_19" "EUR_20" "EUR_21" "EUR_22" "EUR_23" "EUR_24"
[25] "EUR_25" "EUR_26" "EUR_27" "EUR_28" "EUR_29" "EUR_30"
$NEA
[1] "NEA_1" "NEA_2"Many of the following exercises will use these kinds of tricks to instruct various slendr / tskit functions to compute statistics on subsets of all individuals sub-sampled in this way, so you can refer back to this Hint later whenever you need help to compute some statistic.
After you’ve computed nucleotide diversity per-population, compute it for each individual separately (so, not for whole populations across all individuals) using the same function ts_diversity() (which will in this setting give you the heterozygosity for each individual). If you are familiar with plotting in R, visualize the individual-based heterozygosities across all populations.
Hint: Following the previous hint on specifying individual names, you can do this by giving the ts_diversity() function a vector of names as sample_sets = instead of an R list of vectors like you did in the previous exercise. One way to make plotting of the result easy is to first get a table of all individuals using pi_df <- ts_recorded(ts) and then add heterozygosities in each of those individuals as a new column.
NoteClick to see the solution
Per-individual heterozygosity:
We can do this by passing the vector of individual names directory as the sample_sets = argument, rather than in a list of groups as we did above.
For convenience, we first get a table of all individuals (which of course contains also their names) and in the next step, we’ll just add their heterozygosities as a new column:
# get the table of all sample individuals
pi_df <- ts_recorded(ts)
pi_df# A tibble: 38 × 3
name time pop
<chr> <dbl> <chr>
1 NEA_1 70000 NEA
2 EUR_1 40000 EUR
3 NEA_2 40000 NEA
4 EUR_2 38000 EUR
5 EUR_3 36000 EUR
6 EUR_4 34000 EUR
7 EUR_5 32000 EUR
8 EUR_6 30000 EUR
9 EUR_7 28000 EUR
10 EUR_8 26000 EUR
# ℹ 28 more rows# this is how we get names of all these individuals
pi_df$name [1] "NEA_1" "EUR_1" "NEA_2" "EUR_2" "EUR_3" "EUR_4" "EUR_5"
[8] "EUR_6" "EUR_7" "EUR_8" "EUR_9" "EUR_10" "EUR_11" "EUR_12"
[15] "EUR_13" "EUR_14" "EUR_15" "EUR_16" "EUR_17" "EUR_18" "EUR_19"
[22] "EUR_20" "AFR_1" "AFR_2" "AFR_3" "AFR_4" "AFR_5" "CHIMP_1"
[29] "EUR_21" "EUR_22" "EUR_23" "EUR_24" "EUR_25" "EUR_26" "EUR_27"
[36] "EUR_28" "EUR_29" "EUR_30" # and we can create a new `diversity` column of this table like this
pi_df$diversity <- ts_diversity(ts, sample_sets = pi_df$name)$diversity
pi_df# A tibble: 38 × 4
name time pop diversity
<chr> <dbl> <chr> <dbl>
1 NEA_1 70000 NEA 0.0000446
2 EUR_1 40000 EUR 0.000594
3 NEA_2 40000 NEA 0.0000416
4 EUR_2 38000 EUR 0.000488
5 EUR_3 36000 EUR 0.000631
6 EUR_4 34000 EUR 0.000544
7 EUR_5 32000 EUR 0.000541
8 EUR_6 30000 EUR 0.000555
9 EUR_7 28000 EUR 0.000594
10 EUR_8 26000 EUR 0.000555
# ℹ 28 more rowsLet’s plot the results using the ggplot2 package (because a picture is worth a thousand numbers!)
library(ggplot2)
ggplot(pi_df, aes(pop, diversity, color = pop, group = pop)) +
geom_boxplot(outlier.shape = NA) +
geom_jitter() +
theme_bw()
Exercise 2: Computing pairwise divergence
Use the function ts_divergence() to compute genetic divergence between all pairs of populations. Similarly to the previous exercise, do you get results compatible with our demographic model in terms of the expectations given by the split times between the populations in the model? Is there any indication of Neanderthal introgression in our results (a bit of a spoiler for later)? What is missing to be able to confidently say something about this in this case?
Hint: Again, you can use the same concept of sample_sets = we discussed in the previous exercise. In this case, the function computes pairwise divergence between each element of the list given as sample_sets = (i.e., for each vector of individual names in that list).
Exercise 3: Outgroup \(f_3\) statistic
Use the function ts_f3() to compute the outgroup \(f_3\) statistic between African-European, African-Neanderthal, and European-Neanderthal pairs versus a Chimpanzee outgroup. For instance, the first statistic can be computed as ts_f3(ts, B = "AFR_1", C = "EUR_1", A = "CHIMP_1").
Note: The \(f_3\) statistic is traditionally expressed as \(f_3(A, B; C)\), where C represents the outgroup. Unfortunately, in tskit the outgroup is named A, with B and C being the pair of samples from which we trace the length of branches towards the outgroup, so the statistic is interpreted as \(f_3(B, C; A).\)
Note: As a reminder, the higher the \(f_3\) value, the more evolutionary time has been shared between samples B and C.
How do the outgroup \(f_3\) results compare to what you would expectat given the phylogenetic relationships between populations in your model?
Do you see any impact of introgression on the \(f_3\) value when a Neanderthal is included in the computation (again, more on that later)?
Exercise 4: mode = "branch" vs mode = "site"
Repeat the exercise 3 by setting mode = "site" in the ts_f3() function calls (this is actually the default behavior of all slendr tree-sequence based tskit functions). This will switch the tskit computation to using mutation counts along each branch of the tree sequence, rather than using branch length themselves. Why might the branch-based computation be a bit better if what we’re after is investigating the expected values of statistics under some model?
Exercise 5: Outgroup \(f_3\) statistic as a linear combination of \(f_2\) statistics
You might have learned that any complex \(f\)-statistic can be expressed as a linear combination of multiple \(f_2\) statistics (which represent simple branch length separating two lineages). Verify that this is the case by looking up equation (20b) in this amazing paper and computing an \(f_3\) statistic for any arbitrary trio of individuals of your choosing using such linear combination of \(f_2\) statistics.
Hint: Here’s that equation in a schematic form
f3(A, B; C) = f2(A, C) + f2(B, C) - f2(A, B) / 2Exercise 6: Detecting Neanderthal admixture in Europeans
Let’s now pretend its about 2008, we sequenced the first Neanderthal genome, and we are working on a project that will change human evolution research forever. We also have the genomes of a couple of people from Africa and Europe, which we want to use to answer the most burning question of all evolutionary anthropology: “Do some people living today carry Neanderthal ancestry?”
Perhaps you’ve heard earlier about an \(f_4\) statistic (or its equivalent, the \(D\) statistic) which can be used as a test of “treeness”. As a recap, for some “quartet” of individuals or population samples (whose names can be giving as characters such as “EUR_1” to the ts_f4() function, as you can see in the help page of ?ts_f4), they can be used as a hypothesis test of whether the history of those samples is compatible with there not having been an introgression.
Compute the \(f_4\) test of Neanderthal introgression in EUR individuals using the slendr function ts_f4() (make sure to set mode = "branch" argument to minimize stochastic noise).
Note: By default, each slendr / tskit statistic function operates on mutations, and this will switch them to use branch length (as you might know, \(f\)-statistics are mathematically defined using branch lengths in trees and mode = "branch" does exactly that).
Hint: If you haven’t learned this earlier, to solve this exercise, you want to compute and compare the values of these two statistics:
\(f_4\)(<some African>, <another African>; <Neanderthal>, <Chimp>)
\(f_4\)(<some African>, <a test European>; <Neanderthal>, <Chimp>),
here <individual> can be the name of any individual recorded in your tree sequence, such as names you saw as name column in the table returned by ts_recorded(ts) (i.e. "NEA_1" could be used as a “representative” <Neanderthal> in those equations, similarly for "CHIMP_1" as the fourth sample in the \(f_4\) quarted representing the outgroup).
The first \(f_4\) statistic above is not expected to give values “too different” from 0 (even in case of Neanderthal introgression into Europeans) because we don’t expect two African individuals to differ “significantly” in terms of how much alleles they share with a Neanderthal (because their ancestors never met Neanderthals!). However, the other statistic should be “significantly negative” if there Neanderthal introgression into Europeans happened some time throughout their history.
Is the second of those two statistics “much more negative” than the first, as expected assuming introgression from Neanderthals into Europeans?
Why am I putting “significantly” and “much more negative” in quotes in the previous sentence? What are we missing here for this being a true hypothesis test as you might be accustomed to from computing \(f\)-statistics using a tool such as ADMIXTOOLS? (We hinted at this issue a couple of times earlier.)
Exercise 7: Detecting Neanderthal admixture in Europeans v2.0
The fact that we don’t get something equivalent to a p-value in these kinds of simulations is generally not a problem, because we’re often interested in establishing a trend of a statistic under various conditions, and understand under which conditions and in which way its expected value behaves in a certain way. If statistical noise is a problem, we work around this by computing a statistic on multiple simulation replicates or even increasing the sample sizes.
Note: To see this in practice, you can check out a paper in which I used this approach quite successfully on a related problem.
On top of that, p-value of something like an \(f\)-statistic (whether it’s significantly different from zero) is also strongly affected by quality of the data, sequencing errors, coverage, etc. (which can certainly be examined using simulations!). However, these are aspects of modeling which are quite orthogonal to the problem of investigating the expectations and trends of statistics given some underlying evolutionary model, which is what we’re after in these exercises.
That said, even in perfect simulated data, what exactly does “significantly different from zero compared to some baseline expectation” mean can be blurred by noise with simple single-individual comparisons that we did above. Let’s increase the sample size a bit to see if a statistical pattern expected in \(f_4\) statistic from our Neanderthal introgression model becomes more apparent.
Compute the first \(f_4\) statistic (the baseline expectation between a pair of Africans) and the second \(f_4\) statistic (comparing an African and a European), but this time on all recorded Africans and all recorded Europeans, respectively. Plot the distributions of those two sets of statistics. This should remove most of the evolutionary stochastic noise and make a statistical trend stand out much more clearly.
Hint: Whenever you need to compute something for many things in sequence, looping is very useful. One way to do compute, say, an \(f_4\) statistic over many individuals is by using this kind of pattern using R’s looping function lapply():
# Loop over vector of individual names (variable x) and apply a given ts_f4()
# expression on each individual (note the ts_f4(..., X = x, ...) in the code)
X <- c("ind_1", "ind_2", ...)
list_f4 <- lapply(
X,
function(x) ts_f4(ts, W = "AFR_1", X = x, Y = "NEA_1", Z = "CHIMP_1", mode = "branch")
)
# The above gives us a list of data frames, so we need to bind them all into a
# single table for easier interpretation and visualization
df_f4 <- do.call(rbind, list_f4)
NoteClick to see the solution
This gives us list of vectors of the names of all individuals in each population…
sample_sets <- ts_names(ts, split = TRUE)
# ... which we can then access like this
sample_sets$AFR # all Africans[1] "AFR_1" "AFR_2" "AFR_3" "AFR_4" "AFR_5"sample_sets$EUR # all Europeans [1] "EUR_1" "EUR_2" "EUR_3" "EUR_4" "EUR_5" "EUR_6" "EUR_7" "EUR_8"
[9] "EUR_9" "EUR_10" "EUR_11" "EUR_12" "EUR_13" "EUR_14" "EUR_15" "EUR_16"
[17] "EUR_17" "EUR_18" "EUR_19" "EUR_20" "EUR_21" "EUR_22" "EUR_23" "EUR_24"
[25] "EUR_25" "EUR_26" "EUR_27" "EUR_28" "EUR_29" "EUR_30"Let’s compute the \(f_4\) statistic for all Africans…
f4_afr_list <- lapply(
sample_sets$AFR,
function(x) ts_f4(ts, W = "AFR_1", X = x, Y = "NEA_1", Z = "CHIMP_1", mode = "branch")
)
# ... and Europeans
f4_eur_list <- lapply(
sample_sets$EUR,
function(x) ts_f4(ts, W = "AFR_1", X = x, Y = "NEA_1", Z = "CHIMP_1", mode = "branch")
)Bind each list of data frames into a single data frame:
f4_afr <- do.call(rbind, f4_afr_list)
f4_eur <- do.call(rbind, f4_eur_list)
# add population columns to each of the two results for easier plotting
f4_afr$pop <- "AFR"
f4_eur$pop <- "EUR"
# bind both tables together
f4_results <- rbind(f4_afr, f4_eur)Now we can visualize the results:
f4_results %>%
ggplot(aes(pop, f4, color = pop)) +
geom_boxplot() +
geom_jitter() +
geom_hline(yintercept = 0, linetype = 2) +
ggtitle("f4(AFR, EUR; NEA, CHIMP)") +
theme_bw()
We can see that the \(f_4\) statistic test of Neanderthal introgression in Europeans indeed does give a much more negative distribution of values compared to the baseline expectation which compares two Africans to each other.
This is why simulations are so powerful — we establish models which generate data under known population history, and then we can investigate what would be the values of our statistics of interest under those models (and perhaps guess which models are then closest to the hidden truth behind our empirical data).
Exercise 8: Trajectory of Neanderthal ancestry in Europe over time
There used to be a lot of controversy about the question of whether or not did Neanderthal ancestry proportion in Europeans decline or not over the past 40 thousand years (see Figure 1 in this paper and Figure 2 in this paper).
Your simulated tree sequence contains a time-series of European individuals over time. Use the slendr function ts_f4ratio() to compute (and then plot, if you’re familiar with data visualization in R) the proportion of Neanderthal ancestry in Europe over time, commonly designated alpha in the literature. Use \(f_4\)-ratio statistic of the following form:
ts_f4ratio(ts, X = <vector individual names>, A = "NEA_1", B = "NEA_2", C = "AFR_1", O = "CHIMP_1")If a Neanderthal introgression happened at 3% tens of thousands of years ago, what would you expect the proportion to be in the present day? Lower? About the same?
NoteClick to see the solution
# Extract table with names and times of sampled Europeans (ancient and present day)
eur_inds <- ts_recorded(ts) %>% filter(pop == "EUR")
eur_inds# A tibble: 30 × 3
name time pop
<chr> <dbl> <chr>
1 EUR_1 40000 EUR
2 EUR_2 38000 EUR
3 EUR_3 36000 EUR
4 EUR_4 34000 EUR
5 EUR_5 32000 EUR
6 EUR_6 30000 EUR
7 EUR_7 28000 EUR
8 EUR_8 26000 EUR
9 EUR_9 24000 EUR
10 EUR_10 22000 EUR
# ℹ 20 more rows# Compute f4-ration statistic (this will take ~30s) -- note that we can provide
# a vector of names for the X sample set to the `ts_f4ratio()` function
nea_ancestry <- ts_f4ratio(ts, X = eur_inds$name, A = "NEA_1", B = "NEA_2", C = "AFR_1", O = "CHIMP_1")
# Add the age of each sample to the table of proportions
nea_ancestry$time <- eur_inds$time
nea_ancestry# A tibble: 30 × 7
X A B C O alpha time
<chr> <chr> <chr> <chr> <chr> <dbl> <dbl>
1 EUR_1 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0741 40000
2 EUR_2 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0202 38000
3 EUR_3 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0731 36000
4 EUR_4 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0354 34000
5 EUR_5 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0132 32000
6 EUR_6 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0401 30000
7 EUR_7 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0295 28000
8 EUR_8 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0476 26000
9 EUR_9 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0261 24000
10 EUR_10 NEA_1 NEA_2 AFR_1 CHIMP_1 0.0350 22000
# ℹ 20 more rowsnea_ancestry %>%
ggplot(aes(time, alpha)) +
geom_point() +
geom_smooth(method = "lm", linetype = 2, color = "red", linewidth = 0.5) +
xlim(40000, 0) +
coord_cartesian(ylim = c(0, 0.1)) +
labs(x = "time [years ago]", y = "Neanderthal ancestry proportion") +
theme_bw() +
ggtitle("Neanderthal ancestry proportion in Europeans over time")`geom_smooth()` using formula = 'y ~ x'
# For good measure, let's test the significance of the decline using a linear model
summary(lm(alpha ~ time, data = nea_ancestry))
Call:
lm(formula = alpha ~ time, data = nea_ancestry)
Residuals:
Min 1Q Median 3Q Max
-0.027358 -0.011309 -0.002650 0.007912 0.034803
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.569e-02 4.153e-03 11.001 1.13e-11 ***
time -1.588e-07 2.123e-07 -0.748 0.461
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.01589 on 28 degrees of freedom
Multiple R-squared: 0.0196, Adjusted R-squared: -0.01541
F-statistic: 0.5598 on 1 and 28 DF, p-value: 0.4606In conclusion, assuming neutral drift, we don’t expect the Neanderthal trajectory to change, even after tens of thousands of years.
Exercise 9 (hard bonus challenge): The ADMIXTURE controversy
In the discussion about ADMIXTURE, you have heard about very different models giving exactly the same ADMIXTURE result (see Figure 2 in this paper).
Write a simulation script for each of the three model topologies in the figure linked above. Then use the function ts_vcf() to save a VCF file with genotypes from each of those models, convert those VCF files into the PLINK file format, and run ADMIXTURE to confirm the results of that paper. Of course, the PLINK conversion and ADMIXTURE analysis is something you’re going to have to do on the command-line, using the skills you’ve learned earlier in this course.
Exercise 10 (hard bonus challenge): How many unique f4 quartets are there?
If you’ve learned about \(f\)-statistics before, you’ve probably seen various symmetries in \(f_4\) (but also other \(f\)-statistics) depending on the arrangement of the “quartet”. As a trivial example, an \(f_3(A; B, C)\) and \(f_3(A; C, B)\) will give you exactly the same value, and the same thing applies even to more complex \(f\)-statistics like \(f_4\).
Use simulations to compute how many unique \(f_4\) values involving a single quartet are there.
Hint: Draw some trees to figure out why could that be true. Also, when computing ts_f4(), set mode = "branch" to avoid the effect of statistical noise due to mutations.