There are different approaches to estimate Runs-Of-Homozygosity (ROHs) from genome sequencing reads mapped to a reference genome (BAM format) or from variant and invariant called sites (GVCF format). In this tutorial for the Workshop on Population and Speciation Genomics 2025 we will use the output of a software called ROHan.

First, log into your AWS instance and move to the folder activity ~/workshop_materials/30_genetic_load/ROH

Take your time to answer the questions we left for you along the tutorial. We will then discuss them during the final wrap-up.

ROHan is a Bayesian framework to estimate local rates of heterozygosity, infer runs of homozygosity (ROHs) and compute global rates of heterozygosity. ROHan can work on modern and ancient samples with signs of ancient DNA damage Renaud G et al (2019).

Alternative approaches: BCFtools/RoH using also allele frequency to improve ROH inference; Darwindow.

(as it takes too long for this lab session)

rohan --rohmu 1e-5 -t 12 -o $IND --size 100000 --tstv 1.55 --auto ../autosomes.list $FASTA_FILE $BAMfile/$IND.sorted.dedup.bam

$IND: the IDs of the sample you are running

$FASTA_FILE: the reference genome sequence

$BAMfile: the folder containing the BAM files to be analyzed

Typing ROHan in the terminal will output the full list of options.

Options to think about:
--rohmu : heterozygosity threshold to call a ROH
What do you think is a good value? How to decide?

--size : size of the non-overlapping genmic windows where local heterozygosity will be estimated
Any idea about what could be a relevant ROH length to analyze? Below you will find more on this

--tstv : transition/transversion ratio. This can be estimated on the sample of individuals from the population/species of interest
How would you do it? Which software? Which input file?

--auto : list of the chromosomes/scaffolds to use
Why do we have to exclude sexual chrs and too short chrs/scaffolds?

The coalescent times of ROH of a certain length is estimated as g = 100/(2rL), where g is the expected time (in generations) back to the parental common ancestor where the IBD haplotypes forming a ROH coalesce, r is the recombination rate in cM/Mb, and L is the length of the ROH in megabases (lots of references for this, eg. Kardos et al 2018; Khan et al. 2021).

Examples using a (recombination rate of ~2.8 cM/Mb as in the lizard)
ROH(100kb) -> g = 100/(2 x 5 x 0.1) ~ 180 generations
ROH(200kb) -> g ~ 90 generations ago
ROH(500kb) -> g ~ 35 generations ago
#ROH(1Mb) -> g ~ 18 generations ago

Full output of ROhan from 15 lizard samples (10 Podarcis raffonei from the small islets of La Canna (LC) and Strombolicchio (ST) and 5 Podarcis wagleriana from the main island Sicily (DB)) are provided (one folder per individual) inside the activity folder 30_genetic_load/ROH/lizard.
More details about ROHan output files can be found in ROHan github page.

Check the pdf files of one of the individual from each population.

What are they showing?

Can you find the summary text file of the ROHan analysis for each sample?

What is the minimum ROH length we are recoding in this ROHan run?

How can we get longer ROHs?

We will use HMM midpoint estimates (files with extension .mid.hmmp) to merge adjacent ROHs and summarize their length distribution.

Get the genomic windows with probability of being a ROH > 0.9 for the midpoint estimates of the HMM (.mid.hmmp).
Every time you see $IND in the command line, it stands for one of the individuals in the lizard sample.

Move inside one of the individual folders and run the following commands replacing $IND with the folder sample name.

zcat $IND.mid.hmmp.gz | grep -v "NA" | awk '$5 < 0.1' > $IND.roh100kb

Using the command above we selected ROHs using the probability cutoff of 0.1 of not being a ROH, meaning 0.9 of being a ROH. This saves us a bit of sanity check of the ROHan output.

Do you agree with this 9:1 threshold?

What is the meaning of a Bayes Factor 9:1?

(After running the whole pipeline using this cutoff, you can come back, change the cutoff and run the pipeline again to check the effect of a different cutoff on your estimates)

We ran ROHan using a rather small window size (100kb) but we can merge adjacent small regions into larger ones starting from the output in the hmmp file. There is also another ROHAN output file (extension .hmmrohl) summarizing the results. Have a look a that too if you are interested.

To be sure we are not introducing any bias by merging short ROHs into larger ones, we can go back to ROHan, run it with larger window sizes, and compare the results with the current run after merging.

Merge adjacent regions with bedtools merge.

First, we need to sort the regions by chr and position.

sort -k1,1 -k2,2n $IND.roh100kb > $IND.roh100kb.sorted

Then, we can merge the ROHs

bedtools merge -i $IND.roh100kb.sorted > $IND.roh100kb.sorted.merged

And check the distribution of ROHs by length

awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' $IND.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c

Execute the command above for a couple of individuals from the three different populations of lizards.

How long are the longest ROH in the individual from LC?

And in individuals from the Sicilian wall lizard (DB population)?

What is the cumulative length of ROHs larger than 5Mb in the LC01 individual?

Try the following to answer the question above:

awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' LC01/LC01.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c | awk '$2 > 5000000  {sum+=$2} END {print sum}'

Can you retrieve the patterns we saw earlier today during the lecture?

Here is a loop to iterate the commands above across individuals.
It has to be executed wihtin the 30_genetic_load/ROH/lizard main folder.

Example for La Canna population:

for IND in LC*
do
zcat $IND/$IND.mid.hmmp.gz | grep -v "NA" | awk '$5 < 0.1' | sort -k1,1 -k2,2n > $IND/$IND.roh100kb.sorted
bedtools merge -i $IND/$IND.roh100kb.sorted > $IND/$IND.roh100kb.sorted.merged
rm $IND/$IND.roh100kb.sorted
done

The inbreeding coefficient F(ROH) is calculated as the average fraction of the autosomal genome in ROH across individuals of a population. 95% confidence intervals can be estimated by jackknife resampling among individuals within each population.

To estimate F(ROH), we first need to get the total length of the genome that has been checked for ROHs.

(ROHan is expected to be robust to low coverage data (that is the risk of calling a ROH because we do not have data to call a SNP) classifying regions as "Unclassified" if the probability of being ROH/nonROH is not enough to take a position.)

We can use the following command on one of the individuals and assume that the genome length is the same for all individuals.

zcat LC01/LC01.mid.hmmp.gz | sort -k1,1 -k2,2n > LC01/LC01.mid.hmmp.sorted
bedtools merge -i LC01/LC01.mid.hmmp.sorted | awk '{sum+=$3} END {print sum}'

The result should be 1429961685 bp

Calculate F(ROH) using all ROHs equal/longer than 100kb.

awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' LC01/LC01.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c | awk '$2 > 99999 {sum+=$2} END {print sum/1429961685}'

What is the F(ROH) for this individual?

Make a loop to estimate the F(ROH) per population: It has to be executed wihtin the 30_genetic_load/ROH/lizard main folder.

for IND in LC*
do
echo $IND
awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' $IND/$IND.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c | awk '$2 > 99999 {sum+=$2} END {print sum/1429961685}'
done

We can use these results to calculate the mean F(ROH) per population.

What is the mean F(ROH) for the population from LC, ST and DB?

F(ROH) estimates can also be limited to the fraction of ROHs which are longer than a certain length to estimate more recent inbreeding events

Can you check the proportion of the genome in ROHs per population for 99999 < ROHs < 500000, 499999 < ROHs < 5 Mb, and ROHs > 5 Mb?

For example, to get the proportion of the genome in ROHs longer the 100kb and shorter than 500kb in the population of lizard from La Canna:

for IND in LC*
do
awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' $IND/$IND.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c | awk '$2 > 99999 && $2 < 500000 {sum+=$2} END {print sum/ 1429961685}'
done

In the folder 30_genetic_load/ROH/fish you can find 20 samples which have been already analyzed with ROHan.
This is an invasive species of which we collected data from one population in the source range and from three populations in the invasive range.

We know the invasion happened just a few generations in the past and we expect there could have been quite an intense bottleneck at introduction.

Which one is the population from the native range and which are the invasive populations?

If the estimated coalescent times for ROH are unbiased, then the average FROH based on ROH with estimated maximum coalescent times less than t generations back in time is an estimator of the inbreeding accumulated in the population from the time of sampling back to t generations ago. Ne is estimated from the following expression of mean expected individual inbreeding as a function of Ne over t generations:

F(ROH),t = 1-(1-1/2Ne)^t

This approach assumes that the probability of inferring an ROH segment due to several smaller homozygous segments is very low.

What was the population size when the most recent inbreeding event occurred for our populations?

Do not forget the confidence intervals! See below.

It would be better to add 95% confidence intervals for these estimates by jackknife resampling (we have five individuals per population only, so the following is meant to be just an example).

What are the 95% confidence intervals of total F(ROH) > 5Mb in ST?

Make one list per population (use the prefix of each population)

for IND in ST*
do
echo $IND >> ST_list
done

Resample each population by leaving-one-out approach (100 times here, but 10000 when it matters)

echo 'ST' >> ST_FROH_size5Mb_95perc

for i in {1..100}
do
cat ST_list | shuf | tail -n +2 > ST_list_shuffled

while read IND
do
printf $IND >> F_ROH_size5Mb_shuffled
printf '\t'>> F_ROH_size5Mb_shuffled
awk -F'\t' 'BEGIN {OFS=FS} {print $0, $3-$2}' $IND/$IND.roh100kb.sorted.merged | cut -f 4 | sort -n | uniq -c | awk '$2> 5000000 {sum+=$2} END {print sum/1429961685}' >> F_ROH_size5Mb_shuffled
done < ST_list_shuffled

awk '{sum+=$2} END {print sum/NR}' F_ROH_size5Mb_shuffled >> ST_FROH_size5Mb_95perc
rm F_ROH_size5Mb_shuffled
done

Get the mean of the distribution

grep -v "ST" ST_FROH_size5Mb_95perc | awk '{sum += $0} END{print sum/NR}'

Get the 95% confidence intervals as the 0.025 and 0.975 percentiles of the distribution.

grep -v "ST" ST_FROH_size5Mb_95perc | sort -n | awk '{all[NR]= $0} END{print all[int(NR*0.025)]}'

grep -v "ST" ST_FROH_size5Mb_95perc | sort -n | awk '{all[NR]= $0} END{print all[int(NR*0.975)]}'