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Instead of moving on to linkage, as originally planned,we're going to use this tutorial to revisit some concepts many ofyou needed more practice on.When genes ...

Question

Instead of moving on to linkage, as originally planned,we're going to use this tutorial to revisit some concepts many ofyou needed more practice on.When genes are on different chromosomes they sort independently,and follow the laws of segregation and independent assortmentregardless of their inheritance pattern or any interactions(epistasis) between them. Basically, genotypes are NOT phenotypes.Inheritance patterns like dominant-recessive, co-dominant, and soon, and any gene-gene interactio

Instead of moving on to linkage, as originally planned, we're going to use this tutorial to revisit some concepts many of you needed more practice on. When genes are on different chromosomes they sort independently, and follow the laws of segregation and independent assortment regardless of their inheritance pattern or any interactions (epistasis) between them. Basically, genotypes are NOT phenotypes. Inheritance patterns like dominant-recessive, co-dominant, and so on, and any gene-gene interactions effect the phenotype. Inheritance Patterns Example: Snap dragons can have seven petals or five petals, and can be red, white or pink. The gene for petal number and the gene for petal colour are on different chromosomes. Five petals is dominant to seven petals, while red and white show incomplete dominance with heterozygotes being pink. Let's designate some alleles: Five petals = F; Seven petals = f Red = R1; White = R2; Pink = R1R2 If we cross heterozygotes for petal number we have - Ff X Ff In the offspring we'd expect 1/4 FF, 1/2 Ff, 1/4 ff, or 1 FF: 2Ff: 1ff (genotypic ratio) F (five petals) is dominant to f (seven petals) so the phenotypic ratio is: 3F:1f (3 dominant trait: 1 recessive trait) If we cross heterozygotes for petal colour we have - R1R2 X R1R2 In the offspring we'd expect 1/4 R1R1, 1/2 R1R2, 1/4 R2R2, or 1 R1R1: 2 R1R2: 1 R2R2 (genotypic ratio) R1 (red) and R2 (white) show incomplete dominance, with heterozygotes (R1R2) being pink. So, the phenotypic ratio is - 1 red: 2 pink: 1 white NOTE: the genotypic ratios in both crosses are the same, the difference in inheritance pattern only changes the phenotype ratios. Two Genes If multiple genes contribute to a trait we can see gene interactions. Let's look at two genes with two alleles each - Gene A (alleles: A, a) and Gene B (alleles: B, b), where A is dominant to a, and B is dominant to b. If we assume the genes sort independently and we perform a dihybrid cross, what phenotypic ratios do we expect? (AaBb X AaBb) Phenotype A = genotype A_ (dominant so second allele can be either A or a) Phenotype a = genotype aa Phenotype B = genotype B_ Phenotype b = genotype bb So for the two genes (sorting independently) we have the following possibilities: Phenotype A;B = genotype A_B_ Phenotype A;b = genotype A_bb Phenotype a;B = genotype aaB_ Phenotype A;b = genotype aabb This is your classic mendelian dihybrid cross, the phenotypic ratios are 9 (A;B): 3 (A;b): 3 (a;B): 1 (a;b) In a complementation test crossing two true breeding mutants produce wild-type offspring, indicating that the two lines represent mutations in different genes. aaBB (mutant line 1) X AAbb (mutant line 2); offspring are AaBb What happens if we cross two F1s; what is the ratio of phenotypes in the offspring? AaBb X AaBb Just like for the incomplete dominance example we started with, the genotypic ratios will remain the same (as no gene interaction). 9 A_B_: 3 A_bb: 3 aaB_: 1 aabb But the gene interaction has to be taken into account when determining the phenotypic ratio What are the genotypes of mutant offspring? A_bb and aaB_ What are the genotypes of the wild type offspring? A_B_ So, the phenotypic ratio is - 9 wild type: 7 mutant (3 A_bb + 3 aaB_ + 1 aabb) Deviations from Mendelian Ratios (9:3:3:1) If we see deviations from Mendelian ratios we can use them to infer things like patterns of inheritance or interactions between genes. From our snap dragon example, we can see if we have a trait determined by one gene and see a 1:2:1 phenotypic ratio in the offspring of a hybrid cross the alleles likely show incomplete dominance. In crosses with two genes, offspring ratios of 9 (wild type): 7 (mutant) can indicate the genes complement one another. Question : Gryphin eyes can be gold (G), pink (P), or silver (S). Note: letters in brackets represent the phenotype. You cross true breeding lines of the different phenotypes to one-another, then intercross the F1s (F1 X F1) and find the following (letters represent phenotypes) Cross 1 - Parental: G X S; F1 = S; F2 = 102S: 34G Cross 2 - Parental: P X S; F1 = P; F2 = 117P: 39S Cross 3 - Parental: P X G; F1 = P; F2 = 96P: 24S: 8G You are not the only scientist working on gryphin genetics and while you're performing your crosses a paper is published showing that two genes are involved in gryphin eye colour. We will call these genes A, and B. There are two alleles of each - A and a; B and b. Using the cross results above what were the parental genotypes and the genotype of the F1 for each cross (Cross 1, Cross 2 and Cross 3), above. Make sure in your answer it's clear how genotype (genes A and B) and phenotype (pink, gold, silver) are related. Hint: Start with Cross 3. Review how we got a 9:7 phenotypic ratio when two genes complement each other (instead of 9:3:3:1); we calculated 7 by adding multiple categories from the Mendelian 9:3:3:1 ratio together based on how the genes were interacting -> 3 + 3+ 1 = 7. What might a 96:24:8 ratio suggest (95:25:8 = 12:3:1)?



Answers

A three-point testcross was made in corn. The results and a recombination analysis are shown in the display below, which is typical of three-point testcrosses $(p=$ purple leaves, $+=$ green; $v=$ virus-resistant seedlings, $+$ $=$ sensitive; $b=$ brown midriff to seed, $+=$ plain ). Study the display, and answer parts $a$ through $c$ $$\begin{aligned}&P \quad+/+\cdot+/+\cdot+/+\times p / p \cdot v / v \cdot b / b\\&\text { Gametes } \quad+\cdot+\cdot+\quad p \cdot v \cdot b\end{aligned}$$
a. Determine which genes are linked. b. Draw a map that shows distances in map units. c. Calculate interference, if appropriate.
1. Sketch cartoon drawings of the $\mathrm{P}, \mathrm{F}_{1},$ and tester corn plants, and use arrows to show exactly how you would perform this experiment. Show where seeds are obtained. 2. Why do all the $+$ 's look the same, even for different genes? Why does this not cause confusion? 3. How can a phenotype be purple and brown, for example, at the same time? 4. Is it significant that the genes are written in the order $p-v-b$ in the problem? 5. What is a tester and why is it used in this analysis? 6. What does the column marked "Progeny phenotypes" represent? In class $1,$ for example, state exactly what "gre sen pla" means. 7. What does the line marked "Gametes" represent, and how is it different from the column marked "F $_{1}$ gametes"? In what way is comparison of these two types of gametes relevant to recombination? 8. Which meiosis is the main focus of study? Label it on your drawing.
9. Why are the gametes from the tester not shown? 10. Why are there only eight phenotypic classes? Are there any classes missing? 11. What classes (and in what proportions) would be expected if all the genes are on separate chromosomes? 12. To what do the four pairs of class sizes (very big, two intermediates, very small) correspond? 13. What can you tell about gene order simply by inspecting the phenotypic classes and their frequencies? 14. What will be the expected phenotypic class distribution if only two genes are linked? 15. What does the word "point" refer to in a three-point testcross? Does this word usage imply linkage? What would a four-point testcross be like? 16. What is the definition of recombinant, and how is it applied here? 17. What do the "Recombinant for" columns mean? 18. Why are there only three "Recombinant for" columns? 19. What do the R's mean, and how are they determined? 20. What do the column totals signify? How are they used? 21. What is the diagnostic test for linkage? 22. What is a map unit? Is it the same as a centimorgan? 23. In a three-point testcross such as this one, why aren't the $\mathrm{F}_{1}$ and the tester considered to be parental in calculating recombination? (They are parents in one sense.)
24. What is the formula for interference? How are the "expected" frequencies calculated in the coefficient-ofcoincidence formula? 25. Why does part $c$ of the problem say "if appropriate"? 26. How much work is it to obtain such a large progeny size in corn? Which of the three genes would take the most work to score? Approximately how many progeny are represented by one corncob?

So fun. J are generally half Lloyd in nature, with only one copy off. No reasons did odd eat as school schools in the Estes Oh Euros Buddha present as four years. And no, the answer does not match since only one member off each here is presented in the digital. A meeting type in fungi is the same as genders in humans. Do fungi off different meeting dives Can mate do produce? Jeannie mating type is the mind experimentally by seeing the progenitor easels from specific process, The singles E on and a is what really did do don't me name. Oh, this is it nature. They are used to differentiate okay between the different meeting. The symbol in RG one indicates that the organism requires the amino acids Argentine in the media. For its rules, we can test for the genotype by isolating nutritional mutants on later adding Argentine to see if the group is brokering the symbol E. R. G. One plus in the gifts that the organism is quiet type does not require our Jeanne in the media. For groups, the expression violet time refers to the common expression off lease, often organism seen in the natural population. The wood mu Jin's means that the organism has an elite which divorce from the white type. The biological function off the early does not matter, then solving this problem. The expression linear Akhtar analysis refers to the arrangement off. Ask who's force, uh, ask us this one in which reflects the order off to New York tick divisions and also the subsequent New York might Arctic Division. By tracking the position off the ask those schools, we can detect crossing over events which offered at the drug stage.

Okay, So here we're still working on Mendel's experiments and heredity. We're gonna use the probability method to find our gina types. And in this case, we're doing a cross from a head rose ideas and a homeless, I guess. Recessive. So if we actually go ahead and do this cross, we have four possibilities for genes from the head Rose, I guess. But just the one from the Hamas Vegas were having, ah, smaller pool of possibilities here. Okay. And so are total number is going to be out of four. Whereas if we were doing the to a cross between to the, um, homeless, I guess we would, of course, have the 16. And so when we wanna look at the probability method, we were looking for a recessive in a Okay, So how many possibilities of combinations are there for be? Well, there's just too. There's the headers, I guess, or the recessive because there's to these. We would multiply that by the one possibility for a which is what we're looking for. Okay, so it's gonna give us two out of four because we've total numbers. Four here and that would be one half are homeless I guess. Excessive in a So then we wanna look for dominant in both genes. Okay, so in this instance, how many ways can a you know the baby gene be dominant? That only occurs in one way. Multiply that by the number of ways the B gene can be dominant again. They could only be dominant in this headers, I guess. Form. So that means 1/4 are going to be dominant in both, if you want recessive in. Both were basically into the same thing, right? Well, there's one way that they can be excessive in this combination. There's one way that be can be recessive in this combination. So we get 1/4 again. And then just to be a little more complicated, we're looking for recessive in either a or B. Okay, so for this one, because if we start with recessive in a, there would be one way it could be recessive in a and we multiply that by both ways, that be can be, which is gonna be either hundreds, I guess, or recessive we're gonna add that to the two. Is that you know, basically all the ways that it can be mother played by the recessive be Leo. You know, we have to be careful here because we're double counting when, um, both are excessive. To also need to subtract one to make sure we don't double count eso subtracting one time. Basically, we're subtracting the number of times that we get a BB, which in this crosses one. Okay, so that's two plus two minus one is going to be three out of four. Should have recessive features. Um, either a or B. If we go back to our little grid, we can see that that's true. Okay. And something moving on be. In some ways, I was trickier. In some ways, it's easier. We're doing the standard cross, Um, where it's headers. I guess In every Jean Caden, this case we have three genes. There's gonna be a total of 64 that we're working with, and the first thing we want to know is recessive in all traits. So we would we want to find the probability of a B B. C. C. Okay, so there's one way we could get a multiply that by the one way we could get recessive be times the one way we could get recess to see and that would be one out of 64. So next we want recessive in C and dominant and the other two in case you're looking for a anything dominant B and then recess agency. So there's three ways we could have dominance in a Hamas, I guess are kind of both flavors of headers, I guess. Multiply that by the three ways we can get dominance and be just gonna look pretty similar. Then we multiply that by the one way we can ever excessive and see. Okay, so this is gonna give us nine out of 64? Yes, all right. And then we want to do the same thing, But looking at just dominant and a and recessive in both B and C. So we get three ways that they can have dominance times just one for recessive, be times one for recess of C, which is three out of 64. You can see how that's a little bit faster than making this massive 64 square put in square. Then we can move on to part C. So now we're using the Fort line representation eso to start with a right So there's in our if you think about the four ways that, you know, in a crossed with 200 I just ca NBI cross and have the four kinds of offspring. Three of them are gonna have a dominant A, and one will have a recess today. Okay? And then we conduce this same thing for the next one. We could have a three dominant be one recess it be. And it's just that mirror here. We can have three. Dominant be. And when necessity be and three dominant, see one recess, ISI three dominant. See when recessive see same, Same, same over and over again. Three of the dominance c and one recessive. See? Okay, So then the question first is probably that all trades are recessive can again, we know that this one is gonna have a 64 possibilities, but so we say, Oh, are recessive. So we say you go start here and there's gonna be one. Multiply that here. How we get one again and we multiply here and we end up with one out of 64. So that checks The next one is that trades are dominant at each low side. Okay, So we're gonna have to start at the dominant a right, and so get three a. Then we need the dominant. Be there's gonna be times three dominant be and times three dominant. See, that should be 27 for dominance at all. Three low side and then the next one to investigate are dominant at two, but recessive at a third. Okay, so it's gonna be a little bit more complicated. So if we started dominance at a here, we could get three a. Then we can either dio times three b times one little C or we could also do three a times one little B, times three see, Or if we were to have won a the recessive, then we can hit the, um three each for the dominant B and C, sometimes to be times three c. Okay. And if you like this, you were getting three times three times three, right. This is also gonna be 27 and finally trades dominant at one gene but recessive at the other two. This I can use black again. So if we're dominant at one gene and retested at the other two, we could either have three a then times one times one where you have one excessive A. And then, um, times like three B. Then go to recess of C or we could be recessive in a and be but then dominant and see. Okay, so this is just three times three is equal to nine. Right? Then that will actually conclude this problem.

So if we take 42 Hey, if we breathe in a duty mouse with a non beauty mouse, that's one. We'll all be big, a little A and a beauty. And then the two will be free to one ratio, free of beauty, with one individual that has a big A big A in two individuals that are hitters, itis and one individual. It's non of beauty, and that one will have a Gina type of too little ace would be parts. It's similar. So we have a wild type mouse, which is black beauty course with a cinnamon Knauss and then in the F one. They will all be Harris, I guess, because the game it's a big being little B and they all have a wild typos, and then they have to. You'll have three that are wild type with one that 60 Bigbie to you. That hunters, I guess, and one that is cinnamon. And that individual will have the GM type two little beans for C heart. If you take a cinnamon mouse so big a big A to little bee's any time you have to. Little Bee's Cinnamon and too Little A's and two B's. So that's a non beauty mouse. In the F one, you will have a header. Psych it. That is a wild type black beauty and then in the F one star in the yet to When you breathe in, we'll look at what happens in deep part. So in deep part, what we're supposed to do is breed these two individuals together. So deep part, we're going to do a breeding between two headers against from C and what we get. It is this ratio we have. Nine. My set are Black Beauty, which is wild type. We have three mice that are black but Nana Beauty. We have three mice that are cinnamon. So remember cinnamon has these two little bees, and we have one individual that's a double process. It and that individual is chocolate Now. That's really important, because if you have a chocolate individual that shows up, that tells you a lot of information about your F one. So we're going to keep that information sort of handy so that when we move forward, we have it. Okay, so that's deep heart. So let's let it eat. Eat heart, you take. You have parents who are almost like this dominant, almost like a successive. So that individual is cinnamon, right? Because the two little babies read with a mouse that is homeless, I guess. Process it for one gene homicide. Istomin it for the other. And that individual is a black mouse. But he's a non a beauty. And then they asked one. They should all be lacking beauty because you have a header. Rows ICUs opens a library a little be there should have a headers, I guess. Which exhibits the wild type. Remember? Which is black beauty? No. And then in the F two, you should have a 9331 ratio Where nine. Exhibit the wild type color. So a something be something wild type and that distribution of Gina types looks like this. There should be one big a big a big, big me. There should be two headers, I guess a home is like a Solomon in beans to almost like a song to a header. Rows like this be and one header is up Food. I'm sorry. Four headers, I guess, for both traits. Now, if you're uncertain, just simply do it. Di hybrid cross between two headers like these individuals from the F one. Three of these individuals will be black Nana Beauty, and they will look like 12 little A's dominant, almost like this thing. And two homes, I guess. Process it A's headers, I guess. Fees. Three individuals will be cinnamon so big a something to little bee's and they will be like this. Want homicide installment A homeless like this process of B and two headers. I just a homeless, I guess processing creamy. And then the last individual will be chocolate. And that Gina type will be, too. It's a double recesses, so it's two homes, I guessed deans. Now we continue and we look It s it says You know what happens if you read a wild type mouse with the cinnamon Knauss and what happens is you get 1/4 homos, I guess. Dominant for a headers August for B 1/4 headers, I guess for both and both of those are wild types that there's 50% of your mice will be wild type color and then 50% will be cinnamon. And so they will have a big a big A into little bees and then headers. I guess for a but still to little bee's, because that's the only way you get cinnamon and then the other reading for yes, we take a wild type mouse, and we read it with a black Nana beauty mouse. And so here we have a little a 50 a little bit house with two little A's and two B's. And so remember, this is wild type. And this is Black Nana Beauty and in the offspring What it is. 1/4 big. A little ANC, too Big B's and that's wild type power. 1/4 big a little a big be a little bit. And so that's 50% wild type our and then 1/4. Too little a stick. Be big B and 142 Little Ace had ours. I kiss fee. And so these air 50% of your offspring and they're gonna be flat. Nana beauty. Okay. And then let's work on G part. Jeanne is what happens when you take a wild type males and you read it with a chocolate mass. And so what happens is you get 14 that are big, a little a big, beady little being, and that's gonna be wild type 14 will be big. A little a to little bee's that cinnamon 14 will be to little a speak, be a little baby. And because this tube a laser there, that's Black Nana Beauty and then 14 will be chocolate. What? And then H Part has four parts to it. So H. Ross one. The information we need here is that to be albino, they must have too little C's. Now I underline my recess of seeds because you can't tell the difference between my upper and lower case sees. So an underlying CD is excessive. So across one, the parents are albino somewhat underlined this recessive traits. We don't know what they are for A or B, so we'll leave them blank when they're crossed. With a parent as to its home is, I guess, dominant for all three James in the F one. We know that they were there. No albino, but they carry the albino gene. They have at least one big A and one big green in the F to you get 87 my slender wild type in power. 32 better cinnamon. He is 39. Better Alvina and so there's a couple things we could make a point of, and I'm going to just write them over here. Or maybe I'll just give it up in a minute. Difficulty where? Next to it. We know that cinnamon has to be too little beings. Oh, so in the F one, that means that the F one parents must have at least one piece that I have to be headers. I just will be. And so because the parent here is be Bigbie, this parent must be with the situation where they can give a little B so they haven't least one little being because that's where that one pain from. And so they're probably homeless, I guess, successive or be, because that's the only way you could get 100% headers like this for being in the F one. And so now we know that the F one parent is big. See little. See a question mark the big me a little bit. 1/4 of your progeny are expected to be albino, and 3/4 of them must be either a beauty or non 1/4 would be cinnamon a beauty or chocolate If Nana Beauty, if you notice we don't have any chocolate, so the F one parent must not carry the alil for non abuse so that F one parent must be with two big A's. Otherwise, you would have some Nana beauty offspring or chocolate offspring. So the original parent money off, I know Must Be Home is like it's recess it for seen homeless, I guess. Dominant for a homeless. I just process it for be read with this second parent who's homeless, I guess. Element for all three traits. Producing an offspring that looks like this in the F one. Okay, that's cross one. Let's look at a cost to across to This is still a church across two is an albino parent, and we don't know what A or B is. Cross the parent who is homeless Eyepiece. Dominant for all three traits. The F one we know it's fixed seat. Little seen a something be something. The F two. It's 60 to wild type. It's 18 albino, so we know the wild type is big. See something and albino is too little sees. This is a 3 to 1 ratio, so one genus, hetero Xigris and so the other has to be homeless, I guess. Dawn or processes And so well, smart legs. So the albino parent must be We'll see you, little c a little, a little, A little bit. A little being to make sure that these genes are headers, I guess because that's the only way you get a 3 to 1 ratio. So the albino parent must be homeless, like a successive for all of its rights and the other parent must be what we have it. Okay, across three, this is still a church. We haven't albino parents. We don't know what A and B are. Okay, And then that parent is red with another parents and the F one. Yes, A something and being something in the F to you have 96 wild type. You have 30 black and 41 albino. So we know the wild type has to have the capital cities, and albino has to have the two little A's. We also know that black is too little A's, and that's being wild type. You have to have one big A in one big B. And to be black, you have to have to recess. If the wheels for a and one dominant being, and we don't know for A and B for the albino. So for a black Nana beauty Bina, type two here in the act to the F one must have been headers, I guess. Who were the A gene. So here therefore it's genotype could be with this week. See little C make a little AP question Mark and the albino parent must be too little A's in order to get the big A little a combination have in love that one. So among the non albino offspring, which is wild type in the black, this is a 3 to 1 ratio of color. His 30 tons three is 90. And so that means that one of the two genes in the F one is header rows. Argus. So the F one must be big. See, little seen big A little a big be they be That means that the albino parent must be This header is a home is I guess process it to see homos I just processing for a home. So I guess dominant will be And homicide is dominant for all three teens in the other parent. All right, almost there. Let me erase this so that I have a different person out of whiteboard Space will do cross four. So across four, um, has a chocolate, so that shouldn't be too bad. Just one second, actually. No, this up since we'll be racing anyway. But I'm out of whiteboards face. That's how big and monstrous this question waas almost there. Oh, dear. Okay, Cross for last one. We made it across four. The parent is an albino and we don't know about A and B and the other parent is homeless, I guess. Solid it for all of the other trend of jeans. So we know in the F one. Have a letters I get. See something A and something be in the F two. We have 287 wild type. We're 86 black, we have 92 cinnamon, 29 chocolate and 164 albino. So let's write what we know. We know the wild type has to have a big C a big A in the big D. We know that Black has to have a big see homeless. I guess recessive a and home is I guess something or headers like this for being cinnamon has to be big C make a and too little bees. Always chocolate is big. See something and then a double home is like a successive. And then albino is too little. Sees something something we don't know. So they get chocolate offspring. The F one parent must be header rows I vis for all three traits. So we know this has to be head resigns. Otherwise, chocolate is not possible with that devil excessive. And so that means that excuse me, the albino parent must be homeless, I guess processed ID for all three genes because the other pair is homeless, I guess Dominant for all engines.

So for this problem, we are to use Thebe given pc FD three vector to express single guide RNA in Drosophila to create a knock out of this n I PP one gene given in the problem statement. So the first part of this problem part a asked us to find the two Pam sites within the secrets we know from the blurb before this problem that the canonical Pam site is five crime and G g and being any nuclear tied. So all we need to do is scan through the sequence to find any NGS. The two that are present in this sequence are near the end eso at the end of the Exxon, we have tea. Oops, T g and then in the first intron, we have another t g. This first part of the problem also asks us which site we would use to produce, Annul, illegal and why we would prefer that site. So this first Pam site would be the ideal one for us to designer cas nine system around because it would cause a bubble within this Exxon for it to cut. And hopefully we would get a break, causing a framed shift somewhere within this coding sequence and leave us with a no Leo. The next part of this problem part be asked us to determine the percentage of imprecisely repaired genes that we could say with confidence would be no Leal's. So the first possibility is that it prepares exactly where it broke so right at zero and it repairs without any addition. However, this problem statement said that we're only to consider the percentage of the imprecisely repaired genes, so that's eliminated as a possibility. Um, the next option is that it can add up to six nucleotides. Um, however, if it adds three or if it had six thes could possibly still maintain function because this would cause a frame shift of a full three nucleotides, which would be a single amino acid. Now the same goes for the removal of three or the removal of six um, now that could still maintain some function. Um, it's unlikely, but it could be possible. So now we know that with the addition of 1 to 4 or five and thesis obstruction of 1 to 4 or five nucleotides, we would have a frame shift that would most certainly disrupt function So that means we had 12 possibilities. Eso the addition of six or the subtraction of six nucleotides and then eight, uh, known Knowles. So now it is just simple math. 8/12 is equal to 66.6 percent of the imprecisely repair genes would be null alleles seventies. Next, we are to diagram the cut pc FD three vector, um, and where to ignore the blue segment that would be removed. Now this would be cut using the BBS won recognition site. So we have are five prime and and three prime ends. Eso this would be the left site or the orange side eso we've got t t Uh huh, A c and then we've got our matches and then we've got an overhang. So see a g c. And so that is where it would cut on the left side. On the right side, we would have our overhang G t t t. And then we have our pairs. So t a g a chief. And here is our overhang and that would be the factor. So the next part of this problem Part D assess designed to 24 nucleotide pieces of DNA that could a Neil together and fit inside the cut plasma. Um, that would be useful for expressing single guide Arne. So to design a single guide RNA, we have to go back to Thebes, blurb before this problem and look at how cast nine bubbles. Um, the genomic target and now single guide RNA fits. So we've got our five prime and three prime here, and the figure looks something like this with our genomic target site, our Pam. And then we've got our compliment. And then we've got our single guide RNA that fits here. And so this is our genomic target. So if our single guide RNA is a compliment to the complement of the genomic target than our 24 um, blip piece of DNA essentially just needs to be the genomic target. Um, what? We have to include four overhanging pieces. So if you look here, I have already typed it out. But these first and last four are compliments to the cut portion of the plasma, which is C A, g C and G T t t eso. We know that that would Aneel and then this 24 20 base pair section of DNA. On the top side is simply the last, uh, 20 nuclear tides of the coding Exon. And this is because we want the single guide RNA to bubble the genomic tart, um, genomic target just upstream of the Pam site. And so we have the single guide RNA paired to that region. So next part E asks us to show exactly where cast nine would cut in the N I PP one gene. Now, if we've done our design correctly, then according to the figure before this problem, our cast nine should cut three base pairs or nucleotides upstream of the Pam. And this is simply shown in the figure before the problem where we have our n g g. And then we've got our bubble and our genomic target site and they show a base pair another base pair. Excuse me? Nuclear tied and then we see after three cast nine cuts. So if we look back to our ah and a P p one Jean um, and we find our and nucleotides towards the end of the Exxon that account for our Pam, we've got a t a G. And then we've got the intro on with another G eso if we go three nucleotides back from the T, we've got another T A g and and A and it would cut between the A and the C before it, um, specifically between the history and and the Syrian coat on. So part f of this problem asked us to outline how we would go about making this nor mutation with a competent plasma. Um, one common way is to inject a newly fertilized EG, and you would inject it with the plasma expressing the single guide RNA as well as another plasma expressing cast nine. And once they're injected, um, the placements are expressed. Andi, the cast nine protein and the single guide RNA are expressed. And then if the design process is gone, um, as desired, the genomic target is altered into a no. Um, and then you have a fertilized egg with the no alil that will grow into a fully functioning organism. Now, apart G is very similar. Eso They ask how you would modify the technique to create a knockin, um, to change the, uh um, the three men right after the initiative, initiating met to an al Ani um now to do this essentially, you would just have to create a complementary piece of DNA. So, um, some sort of piece of DNA that could be expressed on a plasma that iss complimentary to the gene eso the NYPD one gene. Except that in this piece of DNA, you have a few base pairs of overhang on each direction. So essentially you would have something like a t g representing the Met. Um, and then the A in the three ning would be changed to a G Wow. And then you could have c t completing the sequence and a pair to make this a complementary piece of DNA, and then you would need to get this expressed within the cell, likely via plasma. But now, if you have this chunk of DNA expressed, hopefully you would get a, um hm ology directed repair instead of no, no homologous and joining. And if you've done the design correctly, the HDR would lead to uh huh 80 g g c t. In the beginning of the Exxon which would change the 3. 19 to ah, excuse me to Anel Anin, and that is the end of the problem.


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