Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.


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

Chi Square and Linkage


Chi Square and Linkage

Play a video:
Was this helpful?
Hi in this video, I'm gonna be talking about using the chi square test for linkage. So um if you haven't seen the other chi square analysis video, I suggest you watch that first because that video is going to explain how to walk through the formula and everything of like why you do certain things where this video is actually just going to assume that you have been through that video. You know what I'm talking about? And just use the formula for a linkage problem. So remember that a chi square test can be used to identify the likelihood of something that you're tested. So in this case we're looking at gene linkage. So it's going to be the likelihood that two genes are linked. And so because the chi square test evaluates whether you're experimental values are different from the predicted values. And it uses this formula here where this is your chi square and this is some then you have observed and expected. Now again, if you don't understand this formula, go back and watch the other high square video first, because you really have to understand this before we can go on. So with this example, this says that I want to know if genes A and jeans B are linked. So I have this. You know, I have I know about two genes and I want to see if they're linked or not. And so I do an experiment and I cross to heterocyclic organisms and this is super important. You have to do hetero ziggy's organisms for these crosses. Um but I'm giving that to you in the problem. So it doesn't really matter. Just make sure if you see a problem like this, you make sure it's hetero sickos. So they get 50 offspring, There's 31 parental and 19 for competence. So there's 31 that looked like the parents. That's supposed to be a highlight and 19 that look like a mixture of the parents. So is it likely that A and B are linked? Okay, so the first thing you do when you're answering a chi square question is you figure out what are the expected numbers? So I did this cross. So if A and B are linked, what would my expected numbers be? Um and if they're not linked, which is the question we're actually going to ask. The two genes are not linked. What are the expected number is going to be? So to not linked, genes have a recombination frequency of what do you remember? So if the genes are not linked, their recombination frequency is going to be around 50 or lower right, it's going to be around 50%. So that means that if I have 50 offspring and the genes are not linked, then I would expect to get 50% parental and 50% recombination of the mixture because the recombination frequency is 50%. And so 50% of 50 is going to be 25. So there would be 25 parental and 25 for confidence. Now these are my expected number. So if everything's working perfectly, these two genes are not linked, I would get 25 parental and 25 or confidence. Now the second thing is to actually calculate the chi square and that's going to look at the difference between what I experimentally got which was 31 and 19 compared to what I is predicted to get via the recombination frequency, which was 25 and 25. So here's a chart. I typically like to use these charts. Here's the parental so those look like the parents and the competence of the mixtures. This is what I observed was 31 19. This is what I expected 25 25. And so if I put this into the chi square formula, which is this right? If you go back up, if you go back up this formula right here Then I get 1.44,. And that gives me a 2.88 calculation. Now you feel free to go back and do this math if you'd like. But these should be the this should be your chi square value. So what do you do with that kite? What is 2.88 mean? Well, really nothing. But it's just a value that you have to find on a chart. So we start looking at the chart to determine what the p value is, which is always the third step. So how do you use this chart? The first thing you always do is figure out what your degrees of freedom are this case are degrees of freedom are one. We know this because we have two traits we have the parental and the recombinant that we're looking at, right? So that is the number of variables minus one equals degree of freedom. Which in this case is 12 minus one equals one easiest math you're going to do in genetics. So we have our degrees of freedom. So we now know we're gonna use this row, right degrees of freedom here, there's one. So that's the row we're gonna use. Then we have to find our chi square value on this row. And so it's between um I don't know why this box got moved over here, but actually it's between me actually use a brighter color. So here This is our this is our real box here. It's between 2.77 and 3.84. How do I know that? Because 2.71 is less than 2.88 and 2.88 is less than 3.84. And that's our chi squared value. So here's our chart And we know it's between here. So we find our p value. Remember that was the purpose of step three? And so our p value is between .10 and .05. So if we turn that times it by 100 or multiplied by 100 and that gets you the percentage of your p value. So 10% to 5%. So now we have 10% to 5%. What does that mean when we use those numbers to figure out whether we accept or reject the null hypothesis. So take a second. What do you think the null hypothesis was in this question? You have an idea. So the null hypothesis always says that the values are not different so that the expected and the observed values are not different. So if we reflect this back to the question that we were asking, we were asking about linkage. So if the expected and the observed are not different, that means that genes would not be linked. So how do we determine that? We take that 10 5 to 10% that we just found with the p value. And we say that if the p values are greater than five, we accept the null hypothesis, which they were right. Our lowest was 5% and our highest was 10%. So these are greater than 5 to 10% or 5%. So we accept the null hypothesis which stated that these values are not different and therefore they're not linked. So we can say with 95% confidence if you don't know how I got that number, remember this has to do with the 5% value that we chose here for the P value. But I can say with 95% confidence that the genes are not linked. Now. The important thing to remember here is I haven't proved or confirmed linkage at all. Right. I haven't actually looked at the genes and sequenced it and I don't know 100%. I only can say with 95% confidence that the genes are not linked. So it states a likelihood. Not an exact measure, right? I would have to do more tests to confirm that they're not linked. But I'm pretty confident 95%. It's pretty good. I'm pretty confident that the genes are not linked. So that is the chi square test for linkage. Let's not move on.

Black(B) rabbit coat colors are dominant to white(b) coat colors. Long hair (H) is dominant to short hair (h). A breeder crosses a rabbit homozygous for white, short hair with a homozygous black rabbit with long hair. The F1 is backcrossed to the rabbit with white, short hair and the following progeny are produced. Use the chi-square test to answer the following questions.

What are the expected offspring numbers if the two genes are not linked, and therefore assort independently? 


Calculate the chi-square value for the above problem. 


In this example, how many degrees of freedom should be used?


Using the appropriate chi-square value and degrees of freedom, do the coat color and hair length genes assort independently?