howdy and welcome back to chapter eight of comprehensive genetics in the previous section we examined how we can predict the progeny and the outcomes of a cross involving two linked genes and one unlinked in this section we're going to examine what how we predict the progeny and the outcomes of a cross when all three genes are linked so we'll start by our having our map in these problems and and generally this is going to be in word form but here i'm showing it in pictorial form where we have echinaceas light and hooked three genes on the same chromosome echinacea echinaceas and light are separated by eight map units and light and hooked are separated by 12 map units we have a fly that is homozygous for echinaceas and hooked is crossed to a fly that is homozygous or light so that's going to be our p1 cross and setting this up correctly is going to give us our f1 female that we need so this is important because it leads to the most important piece which is our heterozygous female so here the first parent is representing the one that's homozygous for echinaceas and hooked but not light cross to the one that is homozygous recessive for light but not echinaceous and hooked and when we put that together when we make that cross to create the f1 female we have to take a whole chromosome from one parent and a whole chromosome from the other parent in the order and the orientation the genes are in to start with we can't start moving alleles from one from top to bottom so that we end up with things all coupled just to make it look pretty it comes from these two parents therefore those alleles are stuck together in the orientation in which they begin so we're going to have the echinaceas with the plush plus and the hooked on one chromosome the plus light and plus on the other chromosome this is very important how we set this up and then we're going to do a test cross to a male that is expressing all three traits so he will express echinacea's light and hooked he's missing a c here from this cross from this f1 cross this is where we're going to get our eight different progeny types that we're going to examine and we want to predict how often each one of those eight different types occurs based on the information that we have here so the first thing we want to do then is to find all those eight gametes that this female produce produces she'll produce eight gametes that will then be the eight phenotypes that we're observing so we start here with our f1 female and the first class that we can find are the parental classes that where we have no crossover and so we should just get the chromosome that's on top as one gamete and the chromosome that's on the bottom as the other gamete so we have our two parental types here when we want to get a crossover we're looking at we have two different regions where crossover can occur and just based on reading from left and right we're going to call the region from echinaceas to light region one and when we do a cross over here we're going to have echinaceas crossover light plus and then the reciprocal of that is plus crossover plus hooked these will represent our single crossovers in region one we'll call the region between light and hooked region two if we have a cross over there we're going to have ec plus crossover plus and then plus lt cross over h k these will be considered single crossovers in region 2. then the last class is our double crossovers where we'll start and actually do this this way we'll start with echinaceas and we'll cross over and pick up light and then hooked and then plus crossover plus crossover plus these will be our double crossover types once we have all eight identified we can start to set up our calculations so the first gamete type for which we can find information is the double crossover type we're going to use whatever we're given as the coefficient of coincidence knowing it is equal to the the observed double crossover divided by the expected we're going to solve for the observed double crossovers so how do we expect to get how do we get the expected double crossovers so this is what we're solving for where does the expected come from so the question is posed here we make it up the sum of the map distances the product of the map distances or the product of the map distance is changed to frequencies answer choice d is the correct answer if you recall when we look at our coefficient of coincidence we have our observed double crossover frequency divided by our expected double crossover frequency if we if we know the the coefficient of coincidence and we have the map we can have our map distance for region 1 over divided by 100 to change it to a frequency times the map distance for region 2 divided by 100 to make it a frequency the observed double crossover frequency is what we're solving for in this case so we'll start there it says calculate the frequency of the double crossovers this is the first thing we're going to do once we know all the gametes our coefficient of coincidence that was given on the first slide is 1 and so therefore the observed double crossover frequency should exactly equal the expected basically you're doing a cross multiplication one times the expected double crossover frequency our observed double crossover frequency is going to include two different things so again going back here when we look at this this is x plus x divided by the total that's how we got it from the chart um so we're gonna we're gonna keep that in mind as we're looking at things so if we're looking at this is our unknown and we assume that both should be equal we can call that a 2x value we're going to multiply the map distance for region 1 divided by 100 times the map distance for region two divided by a hundred so our total double crossover frequency is 0.0096 and that is for both of our two different gametes but each one on its own is going to be found at a frequency of half of that 0.0048 because in a perfect world these two gametes should be exactly equal because they are products of the same meiotic event so our total double crossover frequency is 0.0096 divided equally between the two gamete types is 0.0048 once we have the double crossover frequency we can use that to calculate the single crossover frequencies and what we're going to do here is take our map distance formula and rearrange it so that it makes sense so we we're solving for what we're missing we have the map distance we have the total double crossover frequency what we're solving for then are the single crossovers so we're going to rearrange things slightly to get our formula in the orientation that we need it to be in so we we know our map distances the two single crossovers plus the two doubles divide by the total times a hundred the first thing i'm going to do is divide both sides by a hundred to get rid of that then the next thing i'm going to do is set is use the rules that i understand about fractions the numerator has four different things that are added together over a common denominator which means i can separate those things out into four separate fractions and i can have the single crossover over the total plus the single crossover over the total plus the double plus the double the map distance divided by a hundred is the the total frequency for that region a part over a whole is by definition also a frequency so we'll have the frequency of a single crossover the frequency of a single crossover the frequency of a double crossover and the frequency of a double crossover so when we add things that are alike together our total frequency includes the two doubles the two singles plus the two doubles which is basically the same as what we say right here for our map distance the two singles plus the two doubles now what we're solving for is the single crossovers so we can rearrange it further and say two times the frequency of the single crossover should be the total frequency minus two times the frequency of the doubles and this is key we can't just take that map distance and say that's just for the single crossovers it includes both the singles and the doubles so if we want to find the singles we have to subtract out the addition of the doubles okay so this formula right here is really what you need this just shows you how we get to this formula but once we've derived this formula you just need this formula so let's examine that with region one region one in our map is eight map units so if we want to change that to a frequency we're going to divide by a hundred give us 0.08 and we know that equals two times the frequency of the single crossovers we're going to add here our 2x number that we calculated previously notice it's the 0.0096 not the 0.0048 it's 2 times the 0.0048 so now we're going to take our total and subtract out our two doubles and that's going to give us the frequency of the two single crossovers in region one again that's the total for region one if we want one or the other we have to divide both sides by two so we get 0.0352 for each of these again the total is 0.0704 each one gets half of that exactly because in a perfect world these two things should be exactly equal as products of the same meiotic event we can do the same thing for region two using the same formula the only difference here is now we're taking our total map our total frequency for region two is the map distance from region two divided by a hundred we have our two single crossovers in region 2 and again we have our 2x number here so we're going to take our total for region 2 and subtract out those two double crossovers and we get a total of 0.1104 for crossovers that occur in region single crossovers that occur in region 2. again that's for both of them together in a perfect world they should be equal as products of the same meiotic event so we're going to just divide that in half and and really we're just dividing both sides by two so each one has a frequency of 0.0552 so now we have the double crossovers and both single crossovers all that we're missing are the parentals and these are going to be the last type that we can find we know that the two parentals plus the two region ones plus two region twos plus two doubles equals one hundred percent of everything that's possible so if we wanna find the parentals we're gonna take one minus everything else and this is important that we take the two times number for each of these so i always like to have that one written down somewhere as well so we'll subtract out all of those and again remember your order of operations i would add all these together first and then subtract from one we get our parentals at a frequency of 0.8096 so each parental then has a frequency of 0.4048 because again in a perfect world these should be exactly equal once we have the parentals we can fill in our chart we have our eight different gamete types as listed there these are also the phenotypes in the next generation so the f2 phenotypes we have what type of gamete they are shown there and then the frequency with which each one occurs that we have calculated if if we want to find out how many will actually express then we'll do that out of the total number in this case it was a thousand progeny so out of a thousand progeny i'd expect about 405 parentals of each type and about five double crossovers of each type and then we have our singles in our of region one and region two so we can put all that together based on the map um and determine the outcome of a cross of a specific cross that we're given so let's look at a different example to see how how it looks in terms of what you would be asked on a test we're going to start with our map and here the map is a little bit hidden but we can find it we have ebony and hooked are separated by eight map units and then we have scarlet is ten map units from hooked and this is the order so we have eight map units and ten map units this is the map that we're starting with we're gonna set up our cross between a female and a male and we only have one chromosome here because all three genes are linked it says we have an ebony bodied hooked bristled male our male is homozygous for ebony and hooked but not scarlet is crossed to a scarlet eyed female not ebony not hooked expressing scarlet and again this is key to set this up correctly because this tells us what the f1 female looks like we can't we can't assume the f1 female looks like this because that is not what this particular problem sets up okay we have to go we have to take one whole chromosome from mom which is the plus plus scarlet and one whole chromosome from dad which is the ehk plus this is the most common thing that in the setup is mist so make sure that you're not assuming what that female looks like because it does give you different outcomes if you set it up incorrectly and then we're going to do a test cross to a male that's expressing all three traits so once we have that female we can set up our eight different gamete types we can get plus plus scarlet or ebony hooked and plus as our two parentals if we have a crossover between ebony and hooked starting at the top we can have plus and then we cross over h k plus and then the reciprocal of that starting at the bottom is e cross over plus st these two would be our single crossovers in region one we can have the crossover between hooked and scarlet starting at the top we have plus plus crossover plus starting at the bottom the reciprocal is e h k s t these are single crossovers in region 2. the last class is the double crossover class again starting at the top we have plus cross over h k cross over scarlet and then the reciprocal e cross over plus cross over plus these are our double crossovers the first thing that we can find is the double crossovers we are given a coefficient of coincidence of 0.7 and we know that is equivalent to our observed double crossover frequency divided by our expected our observed is what we're trying to solve for the expected comes from the map we'll take region 1 as 0.08 times region 2 as 0.10 when we do the cross multiplication our observed double crossover frequency which we're calling 2x because we're adding two things together there will be 0.7 times 0.08 times 0.1 when we plug this into the calculator we get 0.0056 just double check my math that i did in my head yes 0.0056 so that's the 2x number if i want one or the other i'm going to divide by 2 and i get 0.0028 for each of these next i can use this to find my single crossovers in region one we had our formula that said two times the frequency of a single crossover in region one is equal to the total frequency for region one plus or minus the two doubles our total for region 1 is 8 divided by 100 0.08 and then we're going to subtract out the 0.0056 from the doubles again we're taking the 2x number and just plugging that in there our two times the frequency of a single crossover in region one is then 0.08 minus 0.0056 or 0.0744 again that's both of them together i want one or the other i divide by two and i get point o three seven two for each one follow the same logic for single crossovers in region two our two times the frequency of a single crossover in region 2 is equal to that total so 0.10 minus the same two doubles 0.0056 so 0.1 minus 0.0056 is 0.0944 and again this is the two times number for region two so if i want one or the other type i'm going to divide by 2 and for these it's 0.0472 then the last class is the parentals and we said our two parentals is going to be one minus our two region ones plus our two region twos plus our two doubles so when we plug these in we're going to want to plug in the 2 times number for region 1 was 0.0744 for region 2 we had 0.0944 and for our double two doubles we had 0.0056 so i'll add those together first 0.0744 plus 0.0944 plus 0.0056 and we'll subtract that from one our two parentals equals 0.8256 again that's both of them together if i want one or the other i divide by two and i have point one four point four one two eight now i know this is a lot of work to answer these questions generally you'll have two or three questions that you can answer pretty quickly once you once you have all this set up and you'll find ways that make it faster it's a little slower when i'm saying it out than when you're actually doing the math but it's the same process over and over again once you get the hang of it so the first question is asking about progeny that are completely wild type that's going to be our plus plus plus across the board so that's going to be 0.0472 out of 1000 progeny that's 47.2 the next one asks about ebony scarlet but not hooked that's going to be where we have the e plus and the scarlet which is one of our single crossovers in region one so .0372 times 1 000 is point is 37.2 so again it's a lot of work but you can answer the questions pretty quickly because you basically have a chart that you're pulling directly from to answer those questions again the biggest thing to look out for is how we set up the original cross and how that leads to the f1 female once you have that situated everything else falls in place and it's just plugging in the numbers and that concludes chapter eight again here we're just we're just looking at predicting crosses using everything we learned from chapter seven in reverse taking our map and predicting the outcome of crosses and we're going to look at different types of mapping in the next chapter [Music] you