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AY methylation was determincd in control offspring and offspring exposed to ethanol during embryonic development mice (see the data bclow}: Only mice with 100% yell...

Question

AY methylation was determincd in control offspring and offspring exposed to ethanol during embryonic development mice (see the data bclow}: Only mice with 100% yellow and 100% pseudo-agouti coats were tested: Control yellow (CY) offspring were numbered CY1-CY6, control pseudo-agouti (CP) offspring were numbered CP1 CP6, ethanol yellow (EY) offspring are numbered EY1 EY6 and ethanol pseudo-agouti (EP) offspring wcre numbered EP1-EPb_ Methylation of CpG islands near the Aw allele was determined by

AY methylation was determincd in control offspring and offspring exposed to ethanol during embryonic development mice (see the data bclow}: Only mice with 100% yellow and 100% pseudo-agouti coats were tested: Control yellow (CY) offspring were numbered CY1-CY6, control pseudo-agouti (CP) offspring were numbered CP1 CP6, ethanol yellow (EY) offspring are numbered EY1 EY6 and ethanol pseudo-agouti (EP) offspring wcre numbered EP1-EPb_ Methylation of CpG islands near the Aw allele was determined by isolating DNA from the tails of the mice: Each circle represents an individual CpG site; Opcn circles indicate an unmethylated CpG site_ and closed circles represent methylated CpG site. The total percentage of methylated CpGs is shawn above cach individual and group: coytaol talorin @ohtua quuDAaO Based on the data shown; what effect does ethanol consumption have on AW allele expression? Give one answer for the yellow mouse, and one answer for the pseudo-agouti mouse:



Answers

As stated in Solved Problem 2 , recessive mutation in certain mice called waltzers causes them to execute bizarre steps. W. H. Gates crossed waltzers with pure-breeding normal mice and found, among several hundred normal progeny, a single waltzing female mouse. This mouse was mated with a waltzing male, and her offspring were waltzers. When mated with a homozygous normal male, all her progeny were normal. Some of these normal males and females were intercrossed, and, unexpectedly, none of their progeny were waltzers. T. S. Painter examined the chromosomes of some of Gates's waltzing mice that showed a breeding behavior similar to that of the original, unusual waltzing female. He found that these mice had the normal number of 40 chromosomes, In the unusual waltzers, however, one member of a chromosome pair was abnormally short. Interpret these obser vations as completely as possible, both genetically and cytologically.

By observing the data, we find that the four most common classes good wall thing. Rick gree forward. I don't know segregation equally. This'll indicates that Captain W small W on Capitol Be small be jeans on the capital G Small G G is unlinked to either off these genes. Hence the genotype off the to parenting advice capital. W Small blue capital. G capital. G Capital, A capital gain. Cross Rick Small W small blue multiplied by small G small G multiplied by small a small. Now the chromosomes off the parent can be dramatically represented in the following. Now Jean off the blue on we are linked. Any progeny off the test cross that are capital w small B or small capital being are re convenience. The map distance can be calculated in the following way. That is, number off three continents divided by total progeny multiplied by 100 which is four plus five plus five plus six divided by 100 in 200 that is 20 map units

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.

Let us complete the following statement. If a female mouses Hamas, I guess for the A. V. L. L. And mates with a male mouse that is a loss of functional L. A. The food selection to eat by the female during pregnancy would influence her offspring by what Now. Let us go back and finish our statement. If a female mouses homesickness for the A. V. L. L. And mates with a male mouse that is a loss of functional L. A. The food selection to eat by the female during pregnancy would influence her offspring by both dietary agents that promote a lower level of DNA. Methylation would produce offspring with more yellow for and dietary agents that promote a greater level of DNA methylation would produce offspring with darker fur. So both of these are correct for our statement to make it true. So our final answer is both B. And C. Are correct.


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