Learn the toughest concepts covered in biology with step-by-step video tutorials and practice problems by world-class tutors

4. Genetic Mapping and Linkage

1

Trihybrid Cross

26m

Play a video:

Was this helpful?

Hi in this video I'm gonna be talking to you about a tri hybrid cross for mapping. So a tri hybrid cross is exactly what it sounds like. It's like a dye hybrid cross. But instead of two instead of two traits you're looking at three traits now you can do a try hybrid cross and just a normal one. You know you're looking at three traits. You're interested in the offspring. So you do a punnett square branch diagram and you ride out all the little eels and figure out what the gina types of the offspring are. But I'm gonna be talking to you and I'm not going to explain that because it's the exact same as that hybrid. You just add another gene. But I'm gonna be talking to you about how to use the results from these crosses to map genetic locations. So for unlinked genes this is going to follow normal Mandali in inheritance and then you're not going to be able to do any kind of linkage and analysis on it. But for linked genes who are all on the same chromosome, the recombination frequencies of the offspring. So the genotype and phenotype sis the offspring that have combined can be used to map the location of the gene lo chi on the chromosome. So instead of just explaining more I just want to go through an example and I think it'll make it clear so I do first want to apologize. So this example is with fruit flies and I'm doing these examples not because I think that they're interesting or necessarily even because I think that they're easy to remember what all the L. E. L. Stand for. I actually think that they're horrible and not very easy to understand and remember but I'm doing this because these are the examples that you're going to find in your book. So the three traits aren't something like height, weight and color instead. It's like very odd things with very odd names. But because you're in genetics you probably need to get used to these names anyway so we'll just go through them. So the first trait is eye color which luckily is kind of easy it's red or vermilion. So red you get red eyed fruit flies. If you have the wild type of deal we know it's wild type because of the plus and you get this vermillion I color in the mutant allele which lacks the plus sign. How you know it's mutant. And vermilion is just like this orange purple color. It's kind of hard to tell unless you study fruit flies but we're not going to be looking at the eye. So I'm just going to give you the genotype. And so therefore it hopefully will be easy. Then the second tray is going to be wing veins. So these are wings on the flies they have veins because their actual organs and they need blood supplies and things. So the wild type of will you have this certain vein called across vein? The mutant. Allele you don't. And then for the third tray it's going to be wing shape? The wild type polio is going to be normal and the mutant is going to be mutated? Which means that it has a cut short wing shape. So the cross that we're gonna do, it starts with these parental als. So you start with wild type for eye color homos I guess wild type and mutant for the vein and the wing shape and then you cross that with the exact opposite. So this is mutant for eye color and wild type for vain and shape. So the types are going to be red which is wild type cross painless which is mutant and cut wings which is mutant. And this cross will be mutant for the eye color and wild type for the other traits. So you can see that these or vice versa and the reason you do that is so that you can easily follow the transmission of alleles and also so that the F. One will be a hetero Zygo. Because these are the gametes produced from the parental, you get one wild type allele into mutants and the other parent you get one mutant into wild types exactly what I wrote right here. Now when you cross these, which is what happens you get F. One and I've already said that it's hetero side I guess right? You can see this here, you have one wild type polio. One mutant the same here and the same here. Now notice my notation for this. Right. Do you remember what all this stands for? Do you remember what the dot stands for? Right. That means it's unknown if it's linked and we're hoping to find out. And in this case you remember what the forward slash means, right? It means it's on different homologous chromosome. So if we were to write this out and a chromosome diagram, what you would see is you have the wild type, the mutant and the mutant across from the mutant wild type and wild type. And this orientation is so so so important. It's like the essential thing for these types of crosses. So make sure that when you see notation like this, you understand how to draw these chromosome diagrams. Because this orientation is essential. If you mess up this orientation, you will not get this problem. Right, guarantee you. So here's our hetero ziggy's F. One and we take that hetero ziggy's F. One here cross it with a tester meaning that it's what Hamas agus dominant. Recessive. It's home as I guess recessive because it's a test. And that is the definition of a test crosses that you're crossing it with the homicide is recessive. And so here are the phenotype, red cross pain, normal wing, which is all wild type Crossing against the three mutations. And so that's the cross that we set up. So the parental phenotype or gammas, which in this case because we're now in the f one generation is the F one gametes are going to look like this right? It's the exact same thing that I drew up here. It's the wild type mutant mutant alleles across from the mutant wild type wild type alleles. So this is super important. So now we need to go through and figure out. So here are the results. Here's the offspring number and we're gonna go through and figure out what gametes are parental and we're competent. Now I gave it to you right here I'm telling you but I want to explain why I know this to be true. So the first thing is I go through each gammy individually. So we will handle every single one of these individually and figure out whether their parental or recombinant. And then if they were competent which gene or which a leo crossed over the V. N. C. The V. N. C. T. Or the C. B. N. C. T. So the first one we're going to deal with is this one number one and this is mutant wild type wild type. So the first thing you do is you look at the parental, so we know that these two are parental. So is it the same as one of these? Yes or no if yes which it is because this one looks exactly like this one mutant wild type wild type then that is parental. You do the same thing for number two. This is wild type mutant mutant. Is it the same as what we know our parental is to be wild type mutant. Yes. So this is parental. So that's the very first step. Always figure out what your parental chromosomes look like and you can do that before you even get results. Right? If you know it's hetero zig asse, then it's always going to look at it, it's always gonna look like this. Okay, so those two are easy. So the rest of them I know are competent because they're not parental. Right? We already have the two parental ones. There's only two chromosomes that you know from each parent. So um these are the two parental that we're going to have. So the rest are automatically going to be very competent. So you can go ahead and write were competent here just because they're not parental. But the next three these nice columns here we need to fill out. So I'm going to show you how to do this now. This is going to be a little bit of a long explanation. So get ready to be paying attention because it is confusing going back and forth between these alleles. But I think when we get the hang of it it'll kind of be like the new Sudoku um it'll just be um it's a game and you gotta figure it out. So for this one we're going to figure it out. So I'm gonna label this a so this is a gamut number a competent. So we can see here that we have a mutant a mutant and wild type. So how you play this game is the object of the game is to figure out which alil or leal's it could be multiple but which one is different from the parental. So how you do this is you look at this you take a and you compare it to both of the parental. So which one is it most similar to? Well if we take a and we compare it to one we can see that they both share a mutant. Um The real so that's similar. They have different CVS because this one's wild type and this one's mutant and they have the same. Um thirdly so this one is very similar. So what needs to be changed in a in gammy A. To make it the same as gammy one. Well V. And Ct are already the same. So the only thing that you need to change to make it the same is this a little it needs to be wild type. So you know from that there's been a recombination in the C. V. L. L. Right? Because if there was no recombination or no crossing over it would look exactly like the parental. But because it doesn't because the Cv alil is different. Crossing over has occurred in the C. V. L. L. So when we want to figure out whether or not and what genes or what is recombined to make this this way? It was the C. V. L. Liel. So you come over to these three and you figure out okay which one of these lists CV. This one list CV. Which means that there's a possible recombination between B. And C. V. This one does not mention CV at all. So you leave it blank and this one is mentioned CV. So you write are here because there's a combination that occurred there because the CV had to recombine to create that gammy and notice here that these are all parental. So they're just blank. They'll always be blank. Pretzels are blank now let's go through a different allele. Let's do this again. We kind of know the rules now. So let's practice so be so be has wild type wild type mutant. So if we compare this to one or we compare it to two you can see um what needs to be changed. We'll just compare it to the second one. So it has the same via lille. That's great. It has different CV and different see chelios. So what we need to do here is that the way to make this one similar is through oh it has the same ct leo I'm sorry it has different CB So the way to make this similar is you need to change the C. V. Right? Because these two are the same and these two are the same. So we're looking at the Cv again. So where are the CVS located here and here? So these two recombined? So let's do another one. See what do we have? We have mutant mutant mutant. So if let me clear this up a little so you can see it now. So if we look at Gene one and jean to the one with the most mutants is gene to write. It has two mutants. So this is very similar to see. But the only thing that needs to be changed is this one needs to be a mutant. So what we say is that the V. A little recombined? So we go over here and say where are the village girls? Here's a via lille. So it recombined here here's the villarreal. So it recombined here. There's no V. A little here so we leave it blank. Let's do one more. We'll go through G. We have wild type wild type wild type. This is most similar to one right? Because there's two wild types. The only difference is this V. Here so you do the same thing. Where are the V's? They're here and they're here but not here. And then let's let's keep going. I'll use a boulder one this time. Um Here's E. So we have wild type or we have mutant wild type mutant. So which one is this most similar to? Well this one's a little bit more difficult because there's a couple of different things that are happening here. But if we compare we can say that this one very similar is the same to here. This one is the same to this one. And so the thing that needs to be changed is the C. T. So then you come here and you say where's C. T. Not here here's C. T. So are here C. T. R. So finally hoping you're getting the how this works. So I'm gonna give you I suggest if you haven't already just pause now and figure out if you can figure out which leal is recombining here and once you figure it out replay and then see if you got it right. So I'll give you just one second. Okay so hopefully you pause you guess and now you're back. Um So with this one you have a wild type mutant. Wild type. So the wild type here is the same as this. The mutants parental is the same as this one. And this one needs to change the C. T. So you go and you say well there's no C. T. Here here's A. C. T. And here's A. C. T. So now we have all the recombining types and these are where the recombination occurred between these genes and you have to figure this out. You're gonna have to figure this like you're gonna have to go through this all on an exam at least once. So I suggest you practice this game figuring out which one is different. So you can figure out which one recombined. And the reason is is because you're going to need these numbers. And what these numbers are are the total number of offspring that recombined with these two genes. So because I know there are these four here, all recombined, then I add them together. 45 plus 40 plus 89 plus 94. And that gets me to 68. Do the same here. So I do 89 plus 94 plus three plus five. I get 1 91 and then the same thing here, 45 43 plus five and I get 93. And the reason that I need these numbers is because these numbers are what I use to map, I use these for mapping. So practice this game a lot. Go online. Look at table. See if you can figure out which one think. You're often see this pattern where you have these four, these four these two and these two this pattern is very common. Um So you'll probably see this pattern on a test but it's not guaranteed. Sometimes professors like to mix you up and switch these around so that they'll trick you and make sure that you know what you're doing. So make sure you can double check this and don't just assume it's this pattern. So now that we have these numbers, we have 2 68 1, 91 93. What do we do with them? Well we calculate recombination frequencies now do you remember the recombination frequency formula it's our f. Equals um number of recombined offspring over total offspring. So because there's three genes we can't just add them all together. We actually need to calculate them individually. So for the V. And C. V. Gene we do 2 68/14 48. And that's gonna get us 18.5%. Remember I got that here for the V. And C. T. You do the same thing. This is 1 91 from here. You get 13.2 and for the C. V. And C. T. Use 93 93 40 48 6.4. And you say okay here are my recombination frequencies. So if that's all the question asked for then you can stop but remember you can use recombination frequencies to figure things out. And we so what can we determine from these recombination frequencies? The first thing that we know from these frequencies is that all three genes are linked? Because the the recombination frequencies are less than 50. Remember if it's less than 50% gold linkage Equal to 50 is no linkage And nothing greater than 50 doesn't exist. And then so we know if they're linked. So if the question asked you know are these genes linked or not you would say yes. And the reason is because they're less than 50%. Um If it asked for the recombination frequencies for C B N C T, then you would give them this number, right? But if it asks you to go one step further and calculate or write the map of the gene. And what you would do is you would take these re combinations and put them onto a chromosome map. So here is your chromosome, Your chromosomes. So the first thing you do is you take the genes with the highest recombination frequency, which in our case is v. CV at 18.5. And you put them farthest away from each other. The very first thing you do, that's the largest number that are farthest away. Then you take the third gene. So we have V. And C. V. So what what are we missing? We're missing the C. T. Leo. So where does this go? Well, it goes in between them. But is it closer to V or is it closer to C C. V. And the way that you answer this is you look at the lowest frequency. So first we took the highest. Now we're taking the lowest, the lowest is 6.4. And that is C. V. And C T. So C. T. Is going to be closer to C. V. Than it is to V. So our maps here are going to be V two, C T is 13.2 and C T two. Cv is 6.4 math units. And we got these directly from here. So V. And C T. V N C T. R. 13.2 and uh C. T. And C V right here is 6.4. So this is great. So but you'll notice is that in all of our experiments to this point we've been writing the alleles V, C V and C. T. But we've actually now figured out that the order is different. So we now know that this is the order and this is important because this is the order your answer needs to be written in not the order that it was the question proposed in because the question wasn't even sure if it was linked. Right? You because it was using the dots to begin with. So we didn't even know if these genes were linked. But now we know that they're linked. We know their math distance and we know their order. And so when you write your answer, your answer has to be written in the correct order. So that's the first thing we figured out. But for the astute ones of you, you may be asking, well why does 13.2 plus 6.4 equal 19.6 but not 18.5. Right? We use our recombination frequency. We said the distance between B and C. V. Was 18.5 but it can't be if the distance between V. And C. T. Or 13.2 and C. T. And C V or 6.4 because that's 19.6. So why is it 19.6 and not 18.5. Well the reason is because there are such things as double crossovers and double crossovers means that there was a crossover here and across over here which created the gammy. And we didn't account for that, right? So when we calculated up here, we just said that there was one recombination each for each one. But there are two things here, these two which are usually the double crossovers. And I'll show you how I know that but generally a good trip to remember is it's going to be the lowest one. So but I'll show you directly how I know that. But that's also a good trip to have that we didn't account for, we didn't account for that double. So because we know now the order and we know that V and CVR farthest apart. Then when we go to do our gametes and here the parental again. And we calculated these before with the game. But when we do this one we can say, okay well there's here's A V A C T and C V. But what happened is that this one crossed over and this one crossed over. And so we calculate that and if you do it again we use a different color then what you get is that this one crossed over and this one crossed over and you get recombination here and you don't only calculate these twice when they're double crossovers, you have to calculate them twice. So we missed this entire allele because we didn't know the order before. But now we know the order. And so we can't figure out that this is what happened. It was a double crossover and not just a single way. Whereas before with everything else, we were just looking for the one allele that was different. But because there's two potential crossover areas, we now need to think of the fact that it actually may not just have one allele that crossed over. But another way that it could have happened is the two adjacent alleles crossed over. So that's what we take in here. So if you add these up now. 45 40 89 94 3 plus three and five plus five. Because you do it twice for double crossovers, you get to 84 to 84 plus 14, 48 is 19.6. And so that gets us our map here, 13.2, and 19.6. So the things too, it's a long video. I know and I'm really sorry for it, but there's nothing really I can do about it. Things you have to know here is you have to be able to take this and turn it into a chromosome map and you have to play this game correctly. You have to figure out which alil is different and figure out and then write the correct are in the correct location. With a well that's different when you do that, you add them up and you figure out your recombination frequencies using the correct formula. And when you do that, you then can figure out are they linked? Yes or no, less than 50% they are. And then the map using the two longest here and then the shortest value to figure out which one is closer to which If there are double crossovers, which they're going to be essentially um most likely then you need to be able to calculate this number and understand why this number 19.6 is different from the one that you calculated above and how you do that is you add the double crossovers twice. So that's kind of the overview. I realize it's confusing and you're probably gonna have to watch this very long video again. But it's super important because I guarantee you I'm 100% positive about this. You will see questions on this or a very similar experiment on a test. So please please please make sure that you understand everything that I talked about. Ask me questions if you need to because this is super important. So with that let's move on

2

Problem

The following table shows data from a cross (ABC x abc) examining three genes (a, b, and c). Calculate the recombination frequency for A and B

A

20%

B

32%

C

37%

D

9.8%

3

Problem

The following table shows data from a cross examining three genes (a, b, and c). Calculate the recombination frequency for A and C

A

32%

B

37%

C

20%

D

9.8%

4

Problem

The following table shows data from a cross examining three genes (a, b, and c). Calculate the recombination frequency for B and C

A

20%

B

32%

C

37%

D

9.8%

5

Problem

The following table shows data from a cross examining three genes (a, b, and c). Determine the order of genes

A

A B C

B

B C A

C

C A B

© 1996–2023 Pearson All rights reserved.