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HauiaQuesnon_ arrt etoratan studed ( fodio geno? being otthe ganos recessio E Ior Ie phenatypa ( nomarnal Oomntnt = 07do>s Exvreni tnat ( Eacnt inbt nas {he calo...

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HauiaQuesnon_ arrt etoratan studed ( fodio geno? being otthe ganos recessio E Ior Ie phenatypa ( nomarnal Oomntnt = 07do>s Exvreni tnat ( Eacnt inbt nas {he calod tnu {oj"cro5s ameni non-Ceatodrs narent being jud"J Cird lechnique usetul = [03tcr039` Icuton Inat & fallaw"g nol u Which 5 0i4ii? wnon atldyng the Anhenanga pominanii phenotype indlvldual ! Itn to3tcros? Cin Detenura hetorozygous alle 63, Wne homazyqous anty ncessivo Dyrenut Can contreuta Hecaule Ing loatcioss

Hauia Quesnon_ arrt etoratan studed ( fodio geno? being otthe ganos recessio E Ior Ie phenatypa ( nomarnal Oomntnt = 07do>s Exvreni tnat ( Eacnt inbt nas {he calod tnu {oj"cro5s ameni non-Ceatodrs narent being jud"J Cird lechnique usetul = [03tcr039` Icuton Inat & fallaw"g nol u Which 5 0i4ii? wnon atldyng the Anhenanga pominanii phenotype indlvldual ! Itn to3tcros? Cin Detenura hetorozygous alle 63, Wne homazyqous anty ncessivo Dyrenut Can contreuta Hecaule Ing loatcioss oxpressad in tna otfspring Aie4 contributed by the other parent vril tencica determinud Irom the: gen= pinduced by tha tastcrogs parent &re Ie3 orosS progeny: among the testcross FA relatiiy frequenc 9s 0f the d Iferent phonotypes MogHly cin be used (0 map Ilnked Genas {owowing inlormatian (Q answer Queston Ujot0 In Drosapnwa (rurt flesl; Ank e 8328 (Pl Gurled wngs (C)a haln body and en abony bady dotenined bY genes located on tht same chromosome. Gona Comblnation EJH Cik Aecambination Frequency 706 9,.296 220/ HiP Cip 5%6 The dlslance butween genes € and E18 (ep units; Record vour answer 94 & valye rourded to ene decimal place;



Answers

As discussed in this chapter, crossing over within a pericentric inversion produces chromosomes that have extra copies of some genes and no copies of other genes. The fertilization of gametes containing chromosomes with duplications or deletions often result in children with syndromes characterized by developmental delay, intellectual disability, abnormal development of organ systems, and early death. Maarit Jaarola and colleagues examined individual sperm cells of a male who was heterozygous for a pericentric inversion on chromosome 8 and determined that crossing over took place within the pericentric inversion in $26 \%$ of the meiotic divisions (M. Jaarola, R. H. Martin, and T. Ashley. $1998 .$ American Journal of Human Genetics $63: 218-224$ ). Assume that you are a genetic counselor and that a couple seeks counseling from you. Both the man and the woman are phenotypically normal, but the woman is heterozygous for a pericentric inversion on chromosome $8 .$ The man is karyotypically normal. What is the probability that this couple will produce a child with a debilitating syndrome as the result of crossing over within the pericentric inversion?

It is observed that the myself knock out for either of the genes need religion to no needle into have no finna topic. Editors. This is because the other type of negligence can replace the first one. Ah, the result here shows the for no types off mice with genetic defects and components off basil lemina, NATO than one. Have gene or garden minus minus. You know type none isn't to. 10. Gino Cortez minus minus fennel type. None lemon on comma one needed in binding side dilation, plus minus none on a lemon in gamma. One. NATO gin binding site minus minus. Direct birth Lemon gamma one is essential in need. Agent to bind. If Nadine could not bind, then there would be, uh, there would not be formation off basil lemina in the mice and the mice will die. That's the my sector Homo jazz. I goes mutants for lemon on gamma. One have severe fanatic. From this observation, it is found that the mice, which are Hamas egas for knockouts off both NATO agents will have more severe phenotype than obtained in the case off lemon in gamma one mutants here, um, the plus minus stands for Head Rodallega's and minus minus stands for home

We're looking at the development of Basil Amina, and specifically, we're looking at two proteins nitrogen and lemon in grammar and the gamma want. So we've got a table in the textbook, but, um, summarizes this, but basically. So if you have a knockout in 1921 uh, you're okay if you have a knockout in addition to your okay, if it's in, um, Lemon in gamma one, um, you die. You get a severe severe phenotype, but only if it's home asparagus. If its head for Vegas on this one, you're okay. So visible Hamas. I guess I should probably likes to go. Why can we explain these results? And what would we expect if you had a knockout in both nature and one and two? Right. So the best explanation for why this is happening is that there's some redundancy here. So, redundancy. Yeah, the German one or two combines lemon camera woman. So in this case, if you look at your knockout for 12 is stepping in and carrying out the tasks for it and vice versa. Whereas if you have the knockout in Lebanon, neither of them combined. So you're getting a severe phenotype. And from this, we can guess what we would see if we had a knock out for both of one and two. And in that case, we would expect a similar genotype to the knockout for laminate. So knock out in both decision. One answer should, uh, have severe insight because just like in the lemon in knockouts, uh, this knockout would have no binding between in intelligence and the lamb in, whereas both of you still get binding because it was because of redundancy.

So we first in 1/2 to draw a cell in meta Phase one. And that cell has to have the chromosomes. Ah, with their genes in a particular way. And so let's see if I could do this. So question says there's going to be no crossing over, so I'm not gonna have them very close to touching, but not quite touching. So that's my first pair of homologous chromosomes. And here is my second pair, homologous chromosomes. I'm just trying the chromosomes. And first, and then they will leave on my wheels. Right. And the wheel could be put sort of any place on the chromosomes. Um, but I'm going to have, uh my big P Leo. That's a big P and my big peel here. And then my big Emma Leo here and there's gonna be another big Emily ill. They're kind of hard to see that one. Um, And in this chromosome, we're gonna have our little P Leo and our little Emma Leo. And of course, they'll be little peas and a little Emily Oldham that committed as well. And then this one, the crime some on the left will have the big are Leal's and the Kremlin. So on the right, we'll have the little our Leos. So this is to a This is to be This is one A and this is one. Be so that's the answer toe question. A keep in mind, things be, could be mere images and stuff like that. That's sort of just upto. However you chose to draw it for be then were asked to produce Gandhi's the GAM meats that can come out of this assuming there's no crossing over and that makes a big difference. So we're gonna hand up with four game meats and I will start with two here. Um, and remember, this is the end of my Oh, sis, But we're going to have to. Chromosomes are gonna be employed, um, and and they don't have to be in my ASUs our meta face so they could just be sort of hither and young, right? And this will be a big P. That's when we'll have a big M and this one will have a big are. So this game he has all the dominant Jalil's and then this one well again have the big P and the Big M well but it will have the lower case. R a Leo. All right, so that's two of our for a game meats. Look to the other two. So this these air big p began big r and big p begin. Little are than we need to switch to new little P and little M So so little p and little M. And this 1st 1 will have the big Our Leo on the second chromosome. And then finally, we will again have little p little m. And finally, little are so this one is as the recess of all three recess. It feels so, therefore, gaming's that can be made from this set up. Assuming that there's no crossing over, keep in mind if there's crossing over that, it would get more complicated.

From the first cross, we can deduce the distance between Jeanne e and be so it can be deduced by the formula. 100% off number off re confidence divided by total progeny. Therefore, it will be 100% off 300 divided by 1000 which is 30 map unit. Now. The distance between B and D can also be deduced by using the same formula, and it will be 100% off 100 divided by 1000 which will be then mark unit. Now from the second cross, the distance between Jeanne E. On C can be deduced using the same formula and it will be 100 course in it off 200 divided by 1000 which is 20 map units on the distance between C and E will be 100% off 400. That is total number off re confidence divided by total progeny and that will be for deep map units. So the map You can be sure in a way that this is e that this bc that this be a on Let this be big. So the distance between E and C will be 40 map unit. The distance between A and C will be 20 month unit on the distance between E and B will be 30 map unit on the distance between B and E will be then map units. Now the indifference can be calculated using the formula indifference equal to one minus observed double cross over, divided by expected double cross over so far across one, I will be equal to one minus 30 divided by 30 which is zero. And for cross to I will be equal to one minus 80 divided by 18 which is equal to zero. So therefore none off the grosses show any individual it's.


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