[Music] howdy and welcome to chapter eight of comprehensive genetics in this chapter we're going to be looking at predicting the outcomes of crosses when the map is known so we can use the map set up our cross and then determine the the progeny that will be the outcome of the cross and also how often each of those progeny types occur we're going to start with predicting progeny for two linked genes and here we're still going to have three total genes we're just going to have two of them that are linked and one is unlinked so for example we could have black body and purple eyes which are autosomal recessive mutants and they're 14 map units apart along with held out wings which is on chromosome 3. so if i were to draw this map out i would have black and purple separated by 14 map units and then held out on a different chromosome we're crossing a heterozygous female in a test cross to a male that's expressing all three traits and we want to know what are the progeny expected from this cross and how many of each type of progeny should we expect so because these two these are two different chromosomes we can separate things into our buckets and i usually start with the easy one the one that's the single one by itself and say okay this is a a hybrid individual cross to a homozygous recessive individual so this is basically a monohybrid test cross which gives us a one-to-one ratio and you can see that in the fact that the female gives us either held out or plus and that's paired with the held out from the male we get half a chance that we're expressing held out and half chance that will be wild type for this trait for the other two genes these are linked on the same chromosome as shown in our notation here and they are separated by 14 map units and i'm going to go to another slide just to show what's going on in this female heterozygote here so we have our homologous chromosomes has the b with the plus and one has the plus with the pr i guess it was p u sorry okay so when this goes through a crossover in prophase or phase one specifically we resolve this and we'll have the the b plus on the top on the second line we're going to swap the b and the plus so we're going to get a plus here with a plus here and then we'll have the b now with the purple and then the last line is plus with the purple so when this goes through meiosis we're going to have four completely different gametes that are represented the two in the middle will be our recombinants and the two on the outsides will be our two parentals so if we look back here we have our two parentals and our two recombinants again as the result of a cross over here to where we get the black with the purple and the plus with the plus we are given a map distance of 14 centimorgans which is equivalent to 14 recombination that 14 is all of the recombination so if we have two recombinants the 14 percent has to be divided equally between them so that we get 7 and 7 which totals up to 14. the remaining percent out of 100 has to be divided equally between the two parentals so if we say 1 minus 14 we would have 86 percent for the parentals which again divides equally into 43 for each of them so from this female she can produce four different gametes and i will always write them in this way where the two recombinants are in the middle because if you recall from back here that's how they look when we did the cross here so my two recombinants are here in the middle and that is the 14 percent that gets divided into seven percent for each one as a frequency that is written as 0.07 [Music] for our two parentals where we have the 86 percent that is divided equally to the 43 percent so we have 4.43 for our two parentals when we put that together with what we get from the male the male all of the gametes are going to be the same and they will all be the black allele with the purple allele and we will so we'll say that's 100 if i need something from the female and something from the male that's an and so i'm going to multiply 0.43 times 1 is 0.43 0.07 times 1 is 0.07 and so on and recall we said that what what the gamete the gamete that we get from the female is what we're going to see as expressed this just kind of points that out again we get the b with the plus from the female and the b with the pr from the male so here we have these ones will express black body the next one we get the bpr from the female with the bpr from the male so we will have black body and purple eyes then we get the two wild type alleles from the female we would have a completely wild type here and then the last one is plus pr with the bpr from the male so we will express just the purple eyes so we're going to combine this now with the information that we have for the held out and this is going to be an and situation and we can set up all the different combinations so we have four combinations here and we have two combinations here we can put the one and the one together and that's going to give us black and held out being expressed it's an and so we multiply 0.43 times one-half and that gives us 0.215 if we said out of a thousand progeny we would have 215 and then we could do the same thing with one and two and we would have black only and we would have again point two one five but again these are these are going to be two of our parentals the other two parentals are shown at the bottom here in the middle we have our four recombinants and you can see that data pattern of four large numbers and four small numbers and it's just setting up the different combinations it'll be eight total combinations between these two parts and we get the outcome of each of the different possible genotypes and phenotypes so let's look at a couple of examples here we'll start with white eyes is sex linked and recessive dumpy wings and black body are autosomal recessive mutants on chromosome 2 separated by 20 map units so if i were to draw my map i would have white eyes on the x and then dumpy wings and black body on chromosome 2 separated by 20 map units the first thing i'm going to do is set up my cross and this is going to be the hardest part of these problems is getting things set up to start with it says we're going we're going to cross a male and a female and we have two chromosomes that we're dealing with here we're going to cross a wide-eyed dumpy wing male our female so she's homozygous for white eyes on the x dumpy wings but not black body on chromosome 2. the male is expressing black only so he is not wide-eyed on the x paired with the y not dumpy wings but black body homozygous in the f1 it says we're going to take f1 females and test cross them so what that means is we need f1 females that we're going to test cross to males that are expressing all three traits so they will have white eyes dumpy wings and black body homozygous expressing all three traits the females will get the w on the x from mom and the plus on the x from dad they'll get the whole unit of dp with the plus from mom and the whole chromosome of plus with b from dad this is defined by how the cross starts so don't try to make things coupled if they don't if they aren't coupled to begin with that's a big mistake people often make so next we're going to go to the f2 which means we're going to separate our two chromosomes and look at them each independently where we have our two genes that are linked separate from our single gene and we'll look at the single gene first we can get w for mom with w from dad w from mom with the y chromosome from dad plus from mom with w from dad and plus from mom with the y chromosome from dad next we'll look at the two genes that are linked from the male all we can give is dumpy and black together on the same chromosome from the female we can give four different gametes however we can have dumpy with plus and plus with b those are our two parentals that i'll put on the outside if we have a crossover here we can have the dumpy with the black or the plus with the plus and these are our two recombinants we're given that there's 20 percent recombination therefore each of those two recombinant types has half of that or 10 percent and we'll say 0.1 as a frequency for the two parentals those would occur eighty percent of the time to add up to a hundred and divided equally between the two we get forty percent or point four as a frequency when you add all four of these numbers up it should equal one when we put things together we always get the dp with the b from dad and so it's going to be whatever comes from mom that's going to be the determining factor here so if we get dp from mom with the plus we're going to express dumpy if we get both recessive alleles we're going to express both if we get both wild types we'll be wild type if we get plus b we're going to express b 0.4 times 1 is 0.4 0.1 times 1 is 0.1 and so on and the question asks at what frequency would you expect to observe wild-type progeny from this cross so we it does not specify male or female so from here we can get females that are wild type or males that are wild type so that would be one half and then we need wild type from our chart here times 0.1 which is 0.05 so in this case we crossed an f1 female and test crossed her to a male expressing all three traits but this is not the only thing that we could do we could also take the f1 female and cross it to the f1 male so let's see what happens when that occurs because that's going to be a little bit that's going to make things a little bit more complicated instead of having a chart with just a single row column here we're going to actually double that and make two columns so let's look at what that looks like we have blackbody and curved wings are separated by 12 map units on chromosome 2 and then we have white eyes that's sex linked so this is our map and we're going to set up a cross between a female looking at two genes two chromosomes three genes and the same thing for the male we have a female that is homozygous for white so white eyes not black body not curved wings cross to a male that is homozygous for black and curved so not wide eyes on the x paired with the y and he's recessive for both black body and curved wings and here it says the f1 progeny or inner bread so that's important to make sure you read and see whether you're doing inbreeding or you're doing the test cross in this case we're going to take f1 females and cross them to f1 males so in both cases we're going to get the w for mom on the x the females will get the plus on the x from dad the males get the y chromosome in both cases we're going to get the plus plus chromosome from mom and the bcu chromosome from dad and again we're going to separate things out looking at the x chromosome separate from chromosome 2. look at the x chromosome first we get w for mom with w from dad w from mom with y from dad plus from mom with w from dad and plus from mom with the y from dad and again all of these have an equal chance of occurring on the other side this is where it gets a little more tricky we have the female and we have the male and in this case the male is capable of giving two different gamete types again the the drosophila notation represents the homologous pair at metaphase so those are will segregate and split and half the time we're going to get a plus plus from the male half the time we'll get the bcu from the male from the female we still have the potential of getting four different gametes we'll get either the plus plus or the bcu as our parentals when we have a crossover we could have plus with cu or b with plus so again these are our recombinants we have 12 map units so 12 recombination that 12 percent is divided equally between the two recombinant types so each one gets six percent or 0.06 as a frequency the other 88 goes to the two parentals divided equally each one gets 44 or 0.44 as a frequency so now when we put things together we're in this column we'll always get plus plus from dad and in the other column we're always going to get the bcu from dad from mom we'll get plus plus in the first row plus cu and then b plus and finally bcu since these are recessive traits you still have to have two copies to express so we'd have homozygous for curved or homozygous for black and then homozygous for both everything else is going to be considered wild type again we're multiplying because we're taking a gamete from the female and the male so 0.44 times a half is 0.22 0.06 times a half is 0.03 and you will notice if you set things up like this every time these two boxes will always equal 0.25 if they don't then something's wrong and then we have 0.03 and 0.22 another 0.25 and then the same thing here 0.25 between these two and .25 between these two so that you have when you add them all together you have 100 of everything so now it's just a matter of picking out what we need from our buckets the first one says at what frequency would you expect to find a phenotypically white curved but not black progeny of either sex so white of either sex is going to be either one of these again or means we're going to add so we would have one half and we need curved but not black from the other bucket so that's going to be our 0.03 so that will be 0.015 the next one says at what frequency do you expect wild-type progeny from this cross again it doesn't specify male or female so we can say our ore situation here and we want one half and now we need everything that's wild type from the other bucket and that includes five different boxes in this case and if you remember the trick that i told you that this is 0.22.03 that equals 0.25 then that whole column equals 0.5 plus the 0.22 on the other side and we get 0.72 so again i'm just adding up all the numbers in those five boxes that i've circled there in blue so one half times point seven two is point three six and that's for the progeny that are wild type from this cross this completes looking at just the the two linked genes and again the things to look out for or whether they're doing the test cross or whether we're inbreeding the f1 progeny separating things out and then using what you understand about map distance directly equated to that percent recombination we're going to do the same type of thing in the next segment that involves three linked genes again working backwards from the map and we'll see how all of that comes together in the next section [Music] you