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My email address is on the about page of the RUclips channel.

The ribosome translates a single mRNA molecule into a protein at a time. However to do this it uses different ribosomal associated RNA molecules, the ribosome is a complex of several proteins and several rRNAs that work together. In the animation we see the crystal structure of just the ribosome (proteins & rRNA), no mRNA is present. Q1: Only a single mRNA is translated into a protein at a given time Q2: No, see answer Q1 You can learn more at en.wikipedia.org/wiki/Ribosomal_RNA (section: "Subunits and associated ribosomal RNA") for the different types of associated rRNA molecules and how they differ between prokaryotes and eukaryotes.

Thank you! Just to clarify, what is those half helix appeared at 37:14?

Sorry missed the reply, those helix structures are probably ribosomal subunit associated RNAs

​@@DannyArendsNo worries. For me, it takes time to absorb the wiki content. Thanks!

You're welcome, the idea (and software implementation) was the foundation of my PhD thesis

@@DannyArends Yes, it seems like lots of hard work.

You're welcome, got some new things I'm working on.

@@DannyArends Excited about it. Looking forward to it.

Many thanks!

No bother, yeah It seems a newer version was released, you can always get the older versions from the archives, a direct link to the 11.5.0 netinst image: cdimage.debian.org/mirror/cdimage/archive/11.5.0/amd64/iso-cd/debian-11.5.0-amd64-netinst.iso

@@DannyArends thanks a ton

In that case just download trimmomatic v0.39 from here: www.usadellab.org/cms/uploads/supplementary/Trimmomatic/Trimmomatic-0.39.zip and extract it. Make sure to update the script to reflect that you're using 0.39 not 0.40-rc1

@@DannyArends thank you so much and also the virtual box version what you have used in the tutorial and the one in the pdf is different, is it fine?

The version of virtual box should not matter, the important part is to use the same Debian version

This error means that the trimmomatic jar file wasn't found at the path specified. Use the debian file browser to confirm that the file is really there.

For smaller data sets and genomes, 16 Gb will be enough (e.g. Yeast, Bacteria, Bees, some Plants). For Mouse or Human, 16 Gb is probably not going to be enough, and 32 / 64 Gb is going to be the minimum.

Thanks for the info, every Linux flavor is slightly different indeed.

@@DannyArends True, Dr. Arends. Thank you for providing the chance for learning and sharing our experience.

Hi Hector, both versions are essentially similar, the 2022 version has some improvements based on student feedback, but globally topics discussed are the same. You could just switch between lectures, e.g. after 2021 lecture 3, you could jump to 2022 lecture 4 without any issues.

The IMPC generates preferential homozygous knockouts (aa to AA, here big A means the gene is removed), so in a homozygous knockout the gene is removed completely from both genomes. In some cases this isn't possible, since no gene leads to a lethal phenotype (e.g. hemoglobin without it no oxygen can be transported). In those cases heterozygous knockouts are made (aa to Aa), so there will only be 1 working copy of the gene to test its effect on the phenotypes.

The moodle is not available anymore (and was restricted to people with a university account), you can get the assignments, answers, and data from my website: dannyarends.nl/bioinfo/ In case this doesn't work (my website is down sometimes,, and sometimes files get corrupted during download) just send me an email (my email address is on the about page of the channel)...

thank you! 😊

Since every diploid animals has 2 copies of each chromosome, each of these copies have 2 strands of DNA (coding strand and template strand), so in a haploid sperm cell you have 1 copy of a chromosome, which is 2 strand of DNA (which are mirror-identical). However in a diploid individual cells with 2 copies can have SNPs between the two copies of the chromosomes (e.g. on the first chromosome we find A:T, while the second chromosomes has C:G at this position) we always write down SNPs relative to the coding strand so this individual would be genotyped as A/C at this SNP.

So, why do the chromosomes of parent 1 look different?

Because it has 2 different copies of the same chromosome. Unless, I'm misunderstanding your question.

​@@DannyArendsI think I need more biology background information so as to absorb the topic. Could u recommend any youtube channels or books that can help? 😊

​@@sueyue6650 Well a basic introduction to biology / genetics at A-levels could perhaps be advisable if you didn't do biology in high school, for books e.g.: "Introducing Genetics" by Steve Jones "Foundation Biology" by Cambridge Modular Sciences On RUclips, @missangler has very well done introduction videos on Genetics, which might be useful to get into genetics, and @ProfessorDaveExplains has an introduction video on genetics as well.

Hiya, Regarding Q1: independent means separate chromosomes. Q2: we assume 2 copies of each chromosome. Then here, the female only has aa and bb gene versions, as such the female egg cell (contains 1 copy of each chromosome) always has ab in them. The male has aA and bB, so male sperm cells come in 4 flavors (ab, Ab, aB, AB). (2 out of 4 are recombinant: Ab and aB) Since the egg cells are always the same only the male sperm cells are recombinant. Each sperm has a 25% chance to occur when gene a and b are on different chromosomes (i.e independent). If gene a and b would be on the same chromosome then a and b (as well as the A and B) version of sperm would be more common, since the small a,b and big A,B would be physically linked by being on the same chromosome. Hope this helps, try drawing a punnett square with 1 row and 4 columns, and look at the resulting combinations. ab x ab = aabb (looks like mom) ab x Ab = aAbb (recombinant) ab x aB = aabB (recombinant) ab x AB = aAbB (looks like Dad)

why the F1 have to be aabb, instead of AABB, AAbb, or aaBB?

Due the the genotype of the parents used in a 2-point cross (AaBb x aabb), the offspring genotypes cannot be AABB (since the mother does not have the big A and big B alleles. only the 4 possibilities listed above can occur, the frequency at which they occur teaches us about the genetic linkage between locus a and b.

The assignments should be available from my website. It might be down, please drop me an email (on the about page) and I'll send them to you by email.

Ahh, I see how this can be confusing. I made a highlight of the image which can be found here: 1drv.ms/i/s!AtYWSYRMmSHZh7x1ExuYBku3hA_pvQ?e=U053HN to be able to talk about the example more easily The child flowers are inside the orange square, there are 4 of them. The "2 flowers" in the yellow circles aren't flowers, they are "parental alleles" so inside each box we see 2 alleles which belong to 1 parent flower, e.g. the father flower is on top (represented by it's 2 alleles), the mother flower is on the left (also represented by the two alleles). Since we have 2 copies of each gene. In the additive case, both parents will look orange (since they have both a white and a red allele) In the dominance case, both parents will look red (since they have a while and a red allele, but red is dominant) In the next slide at 46:00 this is explained, but indeed the graphic is confusing, since I use the same image for flower and allele For a more detailed explanation see: en.wikipedia.org/wiki/Punnett_square

Thank you for the reply. I never expect such detailed explanation. It help a lot for a 0 beginner. Now i understand more about the Punnett square. The 1st row and 1st column are parent genotype (circled in yellow). The offspring genotypes are in orange circle.

Indeed, I'm happy it's clear now. If I present the slide again in the future I'll update it to avoid confusion.

Thanks for leaving a comment, and I'll put your request on the list of topic ideas for the future

Hey Samuel, this sometimes happens due to connection issues. Normally another browser can help. Alternatively, drop me an email (my address is on the about page), then I can send you a copy.

Very good to hear it was helpful, thanks for leaving a comment

R will be for you then, combining stats, programming, and data visualization

Thanks for leaving a comment, R is an interesting language to learn

Indeed 👍

Hope it's not too loud 📢

Yes you can use the heatmap function on the normalized expression data after filtering down to the probes which are differentially expressed

when you want to select from a list, which at each entry in the list has vectors of elements, you can use: lapply(list, "[", 1) # for all vectors in the list return the first element of the vector GPL201 is just the variable holding the GPL201 annotation file, you could change it to another affy annotation file, since all of them use the same way of storing GO data.Probably would have been better to call the variable GPL (without the 201), or AffyAnnot to be more generic.

Hmm, the error indicates that the file is not a valid.cel file. My suggestion would be to delete the files downloaded and make sure that there are no other files in the working directory except the cel files. The ReadAffy function generally is smart enough to figure out the files are gzipped when the extension is .CEL.gz but perhaps one of the files got corrupted during downloading. If you extract them you're going to need 7zip or another tool that deals with gz giles under windows, but plain cel files can be opened in an editor since they're plain text.

Glad it helped!

For a human genome it would be advisable to run it in Linux, or even better on a university high performance compute cluster. The data size is going to be orders of magnitude more than the example S. Cerevisiae RNA sequencing dataset.

You're welcome, thanks for taking the time to comment really appreciated 👍

Good to hear that you enjoyed it!

Thanks, and thanks for taking the time to leave a comment

The idea behind CTL Analysis is that it can help you identify epistatic phenotypes (they will share a CTL). The second/follow-up step is to combine it with QTL data to figure out the causality between the two phenotypes.

@@DannyArends thank you prof

Glad it was helpful! thanks for leaving a comment

That is a good idea, to have someone to keep you on track.

You are welcome

You're welcome, thanks for taking the time to leave a comment.

Thanks for the compliment, making more is on the schedule. But with the new job time is lacking.

My pleasure!

I'm not quite sure what you mean by testing package in R ? You can use the command line R CMD check command to test packages for a number of issues regarding data, code, and documentation issues. If you plan on submitting your package to CRAN, add the --as-cran flag to make sure that additional requirements for CRAN are tested.

In python we have UnitTest or pytest libraries to test our code. Do we have something similar in R? For example I develop R package and now I want to test it, and instead of manually write everything I can utilize some other packages to perform testing.

Ahh, no R provides two ways of testing code, the examples section in the R documentation are run every time you compile your package as kind of a unit test. Integration tests can be written in the test/ folder in your package and are executed on R CMD check. So R has a test runner integrated but you'll have to write the tests yourself (or have chatGPT write them for you)

Thank you! I really appreciate your responses! :)

You're welcome

Nope, it's just the threshold at which the number of hits is manageable. Better would be to do a proper permutation to get 1% and 5% thresholds by randomly reassigning labels.

You're welcome, thanks for the comment and enjoy learning bioinformatics/R

Thank you for the kind words, and for leaving a comment.

My pleasure, thanks for taking the time to leave a comment.

You're welcome, I hope you'll enjoy the lectures and learning about bioinformatics.