Chromosomal Mutations: Aneuploidy

Hi this video, we're gonna be talking about chromosomal mutations focusing on an employee. So remember chromosomal mutations refers to the alteration chromosome structure or the number of chromosome or copies. So what we're talking about now is chromosome copies. There are two types of chromosomal mutation And with reference to copy number and this is a parent employee. And in this video we're not going to talk about. But this is working with the entire set of chromosomes. So if an organism has 40 sets of chromosomes, then there's gonna be abnormal numbers of all 40. But what we're going to focus on is an employee and this changes the part of a single or a few chromosomes. So the organism has 40 chromosome sets. Then in any employee, only one or two of those chromosomes is gonna have abnormal numbers. So an employee. This refers to organisms found in some chromosomes but not all of them. So not the whole set and there are many different types. You can have you can have Trish Chamakh. So I wrote to N plus one. So this is for deployed organisms. Right? So if you have a two in that's gonna say deployed, but then you have one extra chromosome plus one. So, an example of this is down syndrome, right? Where you have an extra chromosome of number 21 or klein felt ear's inhuman where you have that extra X chromosome sex chromosome. Um there's another term. These are just terms you're gonna have to memorize monos o muk. Again, if it's deployed to in is deployed but now it's minus one. So it has one less chromosome than a normal deployed organism. This is Turner syndrome which only has one X instead of two. And if you're not familiar with these these syndromes yet, don't worry about it is just examples if you're familiar with it. The really important thing to know is the terms and these numbers and what those numbers mean, Then you have a null stomach which is two into a deployed organism that has -2. And you can also refer to this as die or die stomach, which is a half Lloyd organism. Notice this is only in and not to in so that suggests hap Lloyd minus one. So it's a Hap Lloyd minus one chromosome. So it's actually lost an entire of one of those chromosomes. So here's an example of what these look like here. We have try stomach. You can see there's two deployed which this organism is normally deployed, but they have this plus one. So they have this extra chromosome here. Here you have Monos O. McM. So again, you're deployed organisms but it's -1. So you have your missing this chromosome here. Noel stomach is two and minus two which might be harder to detect because here you have your two deployed. But because we're dealing with these organisms here, we know they're supposed to be a blue chromosome here but there's not because both are missing minus two dice Tomac. Now we're working with Hap Lloyd. So here are normal Hap lloyds and we have plus one. So here is our plus one. So those are the different types of vocab words you have to memorize. Now here's non disjunction. So non disjunction is the failure of chromosomes to separate properly during division and this is the cause of an employee. It can occur in mitosis, which is most common during the formation of the gametes, but it also concur in mitosis as well. If this is mitosis that's happening very early in development, like within the first few hours or days of development, um, this mitosis is occurring and it just divides the chromosomes wrong. So, um, non disjunction is the cause of these up here and that's chromosomes not separating properly during or during division. Now any employee ids. So the ones we're talking about now, ones with only one chromosome added or missing. These are typically more abnormal than the type we talked about before, which are polyp Lloyd's and these are ones with the entire chromosome set. So all 40 chromosomes for instance, have an extra copy, Whereas any employees would be, there would still be 40, but it would just be 41. So what's why is that why is only having one chromosome extra. So much worse than having 40 extra copies. Well, this actually has to do with a concept called gene balance and this is the ratio of jeans on one chromosomes to jeans on the other chromosome. So if you normally have two chromosomes and you have like three genes on each and you have multiple chromosome sets right here we go. Now it is less balanced to just get one extra copy here. Right? So now I have nine copies of all of these genes, whereas I only have six of these. But if I got an extra chromosome here, then I have nine copies of these. And so the balance of the number of genes on each chromosome is actually really important. So you want to have that balance. So the more so if you only have one extra copy that all of your chromosomes have the same number of alleles except for that one chromosome, which has been added and that gets you extra and that really throws off your gene balance. Whereas if you just have an extra copy of every chromosome, then the number of genes present in that cell is not the same, but it's equal across the chromosomes. And so you want that equal across the chromosome. Even if that means having an extra copy of every chromosome or less or missing a copy of every chromosome because you want that gene balance. Whereas if you have an employee where you only have one set, say chromosome nine, that has an extra copy of chromosome nine now has extra alleles that none of the other chromosomes have and that causes a more severe phenotype and this has to do with gene dosage and that's the relation between number of genes and the amount of gene product. So if um say we go back to uh let me erase these circles here so we can see this better. So we go back to our original example here. We start out with two. We start out with the deployed organisms, these red chromosomes and the black chromosomes and they each have the same number of alleles, Right? Same number of genes. Well, this means say that six gene copies are going to be produced and six of this one say, well, save chromosome number one And chromosome # two. So the red and the black chromosome will each produce six copies. But if you get an extra chromosome, right? So if you get here, then that's going to mean how many are produced. It's going to mean nine are produced. And that means that you're going to have more product than you should compared to every other gene. Right? Because six copies are still going to be produced down here. So you're gonna get more of every gene on this chromosome. And that's going to give you a higher gene dosage and that is really severe. So it's severe dosages, severe when there's an apparent dose of a few two chromosomes versus an apparent dose of all genes. So if every gene had that extra copy of it, then it would equal out, right? Like I said, it typically results in a larger organism, but it's all balanced. It has the same number of copies. I mean it's more but relative to the chromosomes, other chromosomes and other genes, it's the same balance of genes. Whereas if you mess up the dosage of one chromosome that's much more severe than just doing it for all of them. So hopefully hopefully that makes sense. So this is an example of this. So if you have a try stomach organism, so it's deployed. But it has an extra chromosome. Here's your deployed, here's your deployed and here's the extra chromosome. This is more abnormal than having a trip Lloyd organism where all three of the chromosome sets have an extra chromosome on them less abnormal. And this is because this is balanced and you can see it right. All of them have three that's very clearly balanced. And this is not balanced. There's two here, two here and three here, and this will always be more abnormal than this. So with that I think that's it. So with that let's not move on.

Which of the following chromosomal mutations increases the amount of genetic material from only some chromosomes?

• A. Aberrant Euploidy
• B. Aneuploidy
• C. Monoploidy
• D. Tetraploidy

True or False:Aneuploids are more abnormal that polyploids

• A. True
• B. False

A species has 2n = 20. How many chromosomes will be found per mutant cell in an monosomic organism.

• A. 10
• B. 19
• C. 20
• D. 21