Using an image that has a Creative Commons license

 I'm writing some online training material, and want to include some images that have Creative Commons licenses, so need to figure out how to correctly cite the license information.

 A nice article by Creative Commons expert Cory Doctorow points out that there is a very particular way that you need to cite the license information correctly.

As Cory Doctorow says, all CC licenses (save for the Public Domain dedication) require that users:

• Name the creator (either as identified on the work, or as noted in instructions to downstream users)
• Provide a URL for the work (either as identified on the work, or as noted in instructions to downstream users)
• Name the license
• Provide a URL for the license
• Note whether the work has been modified

The Creative Commons Wiki page gives recommended practices for attribution, including many examples.

Making documentation on readthedocs

I've found the Readthedocs website really useful for making documentation, for example, for the Little Books of R and for the Vibriowatch Tutorial.

 

One very nice thing is that you can keep the versions of your underlying text in github, and when you update the text/figures in github, it triggers readthedocs to update the formatted version on the readthedocs website.

There's a nice tutorial on how you can do this: here.

 

The steps are (as described in the tutorial here):

1. Preparing your repository on github

2. Create an account in readthedocs you don't have one already.

3. Import the new project into readthedocs

4. Checking the first build

5. Configuring the project

6. Triggering builds from pull requests

7. Adding a configuration file

8. Versioning documentation

9. Getting project insights

 

 

Testing whether data follow a uniform distribution

Someone asked me how to test whether a data variable, which has values ranging from 1-1000, follows a uniform distribution.

 Getting some inspiration from Stackexchange, I realised that a Kolmogorov-Smirnov test can be used.

 First we can generate one million random numbers from a uniform distribution that ranges from 1-1000:

> y <- runif(1000000,1,1000)

Let's plot a histogram and check their median:

> hist(y, col="blue")

 

 

 

 

 

 

 

 

 

> median(y) 

[1] 500.1832

It is near 500, as we would expect.

 

Then enter the data that we want to compare to this distribution:

> x <- c(200,100,53,99,77,88,32)

 

Then use a Kolmogorov-Smirnov test:

> ks.test(x, y)

 

    Two-sample Kolmogorov-Smirnov test

data:  x and y
D = 0.80089, p-value = 0.0002518
alternative hypothesis: two-sided

 

The test statistic is 0.80089, and the P-value is 0.002518.

 

The null hypothesis is that the data come from a uniform distribution from 1-1000; the alternative hypothesis is that the data do not.

 

Here the P-value is 0.002518, which indicates strong evidence against the null hypothesis, suggesting that we should reject the null hypothesis in favour of the alternative hypothesis.

 

In other words, we reject the null hypothesis that the 'x' come from a uniform distribution ranging from 1-1000, in favour of the alternative hypothesis (that 'x' does not come from such a distribution).

 

 

 

 

 

 

 

 

 

 

 

Finding SNPs in a core gene alignment using snp-sites

Today I'm using the snp-sites software (by Page et al 2016) to extract SNPs from core gene alignments (output from Panaroo).

It's really easy to run:
% snp-sites -c -o myout aat.aln.fas

where aat.aln.fas is a core gene alignment (in fasta format) from Panaroo for the gene aat, and 'myout' is the name that I want snp-sites to give to the output file.

It outputs all the sites with SNPs, in fasta format.

The option -c tells snp-sites to only look at alignment columns that have just A/C/G/T characters.

Acknowledgements

Thanks to my colleague Lia Bote for advice on snp-sites.

'Practical Statistics for Medical Research' by Douglas G. Altman

I'm doing some Statistics revision, reading the brilliant and classic book 'Practical Statistics for Medical Research' by Douglas G. Altman. It's super clear and well explained!

Just for fun, I'm doing the end-of-chapter exercises using R, and putting my answers here:

 

Chapter 3: Describing Data

Exercise 3.1 (b)

We can enter the data in R using:

> SI_without_adverse <- c(1.0, 1.2, 1.2, 1.7, 1.8, 1.8, 1.9, 2.0, 2.3, 2.8, 2.8, 3.4, 3.4, 3.8, 3.8, 4.2, 4.9, 5.4, 5.9, 6.2, 12.0, 18.8, 47.0, 70.0, 90.0, 90.0, 90.0, 90.0)

> length(SI_without_adverse)
[1] 28

> SI_with_adverse <- c(2.0, 2.0, 2.0, 3.0, 3.5, 5.3, 5.7, 6.5, 13.0, 13.0, 13.9, 14.7, 15.4, 15.7, 16.6, 16.6, 16.6, 22.0, 22.3, 33.2, 47.0, 61.0, 65.0, 65.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0, 90.0)

> length(SI_with_adverse)
[1] 37

> hist(SI_without_adverse, col="red")

 

> hist(SI_with_adverse, col="red")

 

 

 

 

 

 

 

 

 

 Exercise 3.1 (d)

> median(SI_without_adverse)
[1] 3.8

> median(SI_with_adverse)
[1] 22.3

Exercise 3.1 (e)

 > SA_with_adverse <- c(360, 2010, 1390, 660, 1135, 510, 410, 910, 360, 1260, 560, 1135, 1410, 1110, 960, 1310, 910, 1235, 2950, 360, 1935, 1660, 435, 310, 310, 410, 690, 910, 1260, 1260, 1310, 1350, 1410, 1460, 1535, 1560, 2050)

> length(SA_with_adverse)
[1] 37
> median(SA_with_adverse)
[1] 1135

Exercise 3.1 (f)

> age_without_adverse <- c(44, 65, 58, 57, 51, 64, 33, 61, 49, 67, 39, 42, 35, 31, 37, 43, 39, 53, 44, 41, 72, 61, 48, 59, 72, 59, 71, 53)

> age_with_adverse <- c(53, 74, 29, 53, 67, 67, 54, 51, 57, 62, 51, 68, 50, 38, 61, 59, 68, 44, 57, 49, 49, 63, 29, 53, 53, 49, 42, 44, 59, 51, 46, 46, 41, 39, 62, 49, 53)

> length(age_without_adverse)
[1] 28
> length(age_with_adverse)
[1] 37

Make a stem-and-leaf plot:

> stem(age_without_adverse)

  The decimal point is 1 digit(s) to the right of the |

  3 | 13
  3 | 5799
  4 | 12344
  4 | 89
  5 | 133
  5 | 7899
  6 | 114
  6 | 57
  7 | 122 

> stem(age_with_adverse)

  The decimal point is 1 digit(s) to the right of the |

  2 | 99
  3 |
  3 | 89
  4 | 1244
  4 | 669999
  5 | 0111333334
  5 | 7799
  6 | 1223
  6 | 7788
  7 | 4

Make stem-and-leaf plots with just one digit to the left of the |:

> stem(age_without_adverse, scale=0.5)

  The decimal point is 1 digit(s) to the right of the |

  3 | 135799
  4 | 1234489
  5 | 1337899
  6 | 11457
  7 | 122

> stem(age_with_adverse, scale=0.5)

  The decimal point is 1 digit(s) to the right of the |

  2 | 99
  3 | 89
  4 | 1244669999
  5 | 01113333347799
  6 | 12237788
  7 | 4 

Exercise 3.2 (b)

> rate_per_100000hr <- c(0.2, 1.5, 1.3, 1.2, 1.8, 1.5, 1.8, 0.7, 1.1, 1.1, 3.2, 3.7, 0.7)

Taking some inspiration from https://www.statmethods.net/graphs/bar.html for plotting the bar plot:

> par(las=2) # make label text perpendicular to axis
> par(mar=c(5,18,4,2)) # increase y-axis margin

> barplot(rate_per_100000hr, names = c("professional_pilots", "lawyers", "farmers", "sales representatives", "physicians", "mechanics and repairmen", "policemen and detectives", "managers and administrators", "engineers", "teachers", "housewives", "academic students", "armed forces members"), col="blue", cex.names=0.8, horiz=TRUE)

 

 

 

 

 

 

> rate_per_1000 <- c(15.9, 11.0, 10.1, 9.0, 8.7, 6.9, 6.6, 6.0, 4.7, 4.2, 3.7, 3.2, 1.6)

> length(rate_per_100000hr)
[1] 13
> length(rate_per_1000)
[1] 13

 

 

 

 

You can see that there is a negative correlation between the two variables.

Exercise 3.3

> IgM <- c(rep(0.1, 3), rep(0.2, 7), rep(0.3, 19), rep(0.4, 27), rep(0.5, 32), rep(0.6, 35), rep(0.7, 38), rep(0.8, 38), rep(0.9, 22), rep(1.0, 16), rep(1.1, 16), rep(1.2, 6), rep(1.3, 7), rep(1.4, 9), rep(1.5, 6), rep(1.6, 2), rep(1.7, 3), rep(1.8, 3), rep(2.0, 3), rep(2.1, 2), 2.2, 2.5, 2.7, 4.5)

> length(IgM)
[1] 298

> quantile(IgM, probs=c(0.025, 0.25, 0.50, 0.75, 0.975))
 2.5%   25%   50%   75% 97.5%
  0.2     0.5     0.7      1.0    2.0
 

Chapter 4: Theoretical Distributions

Exercise 4.1

> pnorm(2, lower.tail=FALSE)
[1] 0.02275013

Exercise 4.2

We can use a binomial distribution to calculate this:

> dbinom(x=0, size=100, prob=0.08) + dbinom(x=1, size=100, prob=0.08) + dbinom(x=2, size=100, prob=0.08)
[1] 0.0112728

Or we can use:
> pbinom(q=2, size=100, prob=0.08, lower.tail=TRUE)
[1] 0.0112728

Exercise 4.3

The probability of a boy is 0.52 so the probability of a girl is 0.48. 

> 0.48 * 0.52 * 0.48 * 0.52 * 0.48 * 0.52

[1] 0.01555012

> 0.52 * 0.52 * 0.52 * 0.48 * 0.48 * 0.48

[1] 0.01555012

> 0.48 * 0.52 * 0.52 * 0.52 * 0.52 * 0.52

[1] 0.01824979

Exercise 4.4(a)

We can use a binomial distribution:

> dbinom(x=6, size=10, prob=0.15) +  dbinom(x=7, size=10, prob=0.15) + dbinom(x=8, size=10, prob=0.15) + dbinom(x=9, size=10, prob=0.15) + dbinom(x=10, size=10, prob=0.15)

[1] 0.001383235

Or we can use:

 > pbinom(q=5, size=10, prob=0.15, lower.tail=FALSE)

[1] 0.001383235

Exercise 4.4(b)

The probability of 6 or more miscarriages out of 10 pregnancies is 0.001383235 from the previous question.

We can calculate the expected number of clusters using:

> 20000*0.001383235

[1] 27.6647 

Exercise 4.5(a)

The probability of a child having the infection is 0.10, if it is present in the school.

The probability of a child not having the infection is 0.90, if it is present in the school.

If test m children, and the infection is present in the school, the probability of m positive tests is (0.10)^m and the probability of m negative tests is (0.90)^m.

We want the probability of >0.95 of detecting the infection if it is there, ie. we want (0.9)^m < 0.05.

log(0.9^m) = log(0.05)

m * log(0.9) = log(0.05)

So m = (log(0.05)) / (log(0.9))

> (log(0.05)) / (log(0.9))

[1] 28.43316

So we need sample size m = 29.

Exercise 4.6

> pnorm(q = 172.0, mean=175.8, sd=5.84, lower.tail=TRUE)

[1] 0.2576249

> pnorm(q = 172.0, mean=179.1, sd=5.84, lower.tail=TRUE)

[1] 0.1120394

Exercise 4.8(a)

> 0.75 * 0.75

[1] 0.5625

Exercise 4.8(b) 

0.75

Exercise 4.8(c)

The probability of both parents being heterozygous, and their child having cystic fibrosis is:

> (1/22)*(1/22)*(0.25)

[1] 0.0005165289

If there are 3500 live births a year, we expect to see this number of children with cystic fibrosis:

> 0.0005165289*3500

[1] 1.807851

This is about 2.

 

Chapter 7: Preparing To Analyse Data

 

 

 

 

 

 

 

 

We're hiring - Training and Events Coordinator

 We are currently recruiting in Nick Thomson's group at the Wellcome Sanger Institute for a 'Training and Events Coordinator' to join our team to provide administrative support for developing cholera genomics training, including overseas training courses and an online symposium on cholera genomics.              

The application deadline is 28th January 2024 and you can see the job advert here.

We are ideally looking for someone with excellent administrative skills and attention to detail, who is a great communicator and has experience organising events. 

This can be a part-time position, minimum 2.5 days/week.              

 Please feel welcome to email me at alc@sanger.ac.uk if you'd like more details.

 I'll be very grateful if you can share with anyone you think may be interested!             

Finding core genes shared by a bacterial species using Panaroo

This week I learnt how to use the Panaroo software for finding core genes (genes present across all isolates of a species) shared across a bacterial species.

There is nice documentation for Panaroo available here.

Panaroo has been described in a paper by Tonkin-Hill et al (2020).

What does Panaroo do?

Panaroo is a graph-based pangenome clustering tool. It tries to identify the 'core' genes shared across isolates of a species (or shared across a set of related species), while taking into account errors in gene predictions (e.g. caused by missing genes, or fragmentation of the genes due to assembly fragmentation).

Running Panaroo

I found Panaroo  easy to run, I used the command:

% panaroo -i prokka_results/*.gff -o panaroo_results --clean-mode strict --remove-invalid-genes

where prokka_results was a folder containing gff file outputs from Prokka for a set of assemblies for my species of interest, and panaroo_results was the name I wanted to give to the output directory.

The  '--clean-mode strict' option is recommended in the Panaroo documentation here. It means that Panaroo needs quite strong evidence (presence in at least 5% of genomes) to keep likely contaminant genes.

The Panaroo documentation here says that the '--remove-invalid-genes' option is also a good idea, as it ignores invalid gene predictions in the input gff files (e.g. with premature stop codons, or invalid lengths).

 I was running Panaroo for about 4500 input assemblies (ie. 4500 gff files), for the bacterium Vibrio cholerae, and found that it needed quite a lot of time to run (about 12 hours), and lots of memory (RAM; about 20,000 Mbyte).

Making a core gene alignment using Panaroo

 If you want Panaroo to produce a 'core gene alignment' (alignment of all the core genes), you can use a command like this:

 % panaroo -i prokka_results/*.gff -o panaroo_results --clean-mode strict --remove-invalid-genes -a core --aligner clustal --core_threshold 0.95

which will  align all genes present in at least 95% of isolates using clustal.

I found that Panaroo is quite slow to run if it has to make a core gene alignment. For 2573 input assemblies (i.e. 2573 input gff files), for the pandemic lineage (7PET lineage) of the bacterium Vibrio cholerae, it found 3239 core genes, and took 3 days to run, requesting 150000 Mbyte of memory (RAM) and running it in the 'week' queue on the Sanger farm, with 30 CPUs. Here is the command I was running, using a core_threshold of 1.00, so asking for core genes present in all genomes:

% panaroo -i prokka_results/*/*.gff -o panaroo_results_with_core_aln --clean-mode strict --remove-invalid-genes -a core --aligner clustal --core_threshold 1.00 -t 30

and here is how I submitted it to the Sanger farm:

% bsub -o /lustre/scratch125/pam/teams/team216/alc/000_Cholera_SNPCalling/myscript3.o -e /lustre/scratch125/pam/teams/team216/alc/000_Cholera_SNPCalling/myscript3.e -q week -n30 -R "select[mem>150000] rusage[mem=150000]" -M150000 /lustre/scratch125/pam/teams/team216/alc/000_Cholera_SNPCalling/myscript3

This found me 1239 core genes using a core_threshold of 1.00.

Panaroo outputs

These are the outputs that Panaroo made for me in my output folder. 

The descriptions of the output files are found on the Panaroo documentation here

gene_presence_absence.csv => describes which gene is in which assembly 

combined_DNA_CDS.fasta => DNA sequences of the genes in gene_presence_absence.csv               

combined_protein_CDS.fasta  => protein sequences of the genes in gene_presence_absence.csv        

gene_presence_absence.Rtab => a binary, tab-separated version of  gene_presence_absence.csv

final_graph.gml => the final pangenome graph made by Panaroo, which can be viewed in Cytoscape

struct_presence_absence.Rtab => describes genome rearrangements in each assembly

pan_genome_reference.fa => a linear reference of all the genes found in the data set (collapsing paralogs) 

gene_data.csv => mainly used internally by Panaroo     

summary_statistics.txt => says how many core genes were found

    

If you ask Panaroo to make a core gene alignment file (see above, and the       

Panaroo documentation here), it will also make a 'core gene alignment' file core_gene_alignment.aln, that has an alignment of the genes present in at least 95% (by default) of the input assemblies (input gff files).

Acknowledgements

 

Thank you to my colleague Lia Bote, who helped me get started with Panaroo, and to my colleagues Mat Beale and Stephanie McGimpsey for advice on running Panaroo on the Sanger compute farm.