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Activity 1 You #solate _ piece of tissue from an early embryo and allow it t0 develop Isolatian; You notice that develops into tissue that is very similar t0 tIssue...

Question

Activity 1 You #solate _ piece of tissue from an early embryo and allow it t0 develop Isolatian; You notice that develops into tissue that is very similar t0 tIssue would normally develop into the Intact embryo; this tissue specified? Or determined?When this same piece 0t tissue 5 used t0 make heterotopic gralt the (Issue develops NOT according to its new position, but Into the tissue would have based on its original position. this tissue specified determined?When. different piece of tissue (tha

Activity 1 You #solate _ piece of tissue from an early embryo and allow it t0 develop Isolatian; You notice that develops into tissue that is very similar t0 tIssue would normally develop into the Intact embryo; this tissue specified? Or determined? When this same piece 0t tissue 5 used t0 make heterotopic gralt the (Issue develops NOT according to its new position, but Into the tissue would have based on its original position. this tissue specified determined? When. different piece of tissue (that also develop; Fisolation into similar tissue based uon Its Original location) used Make heteratopic Eralt the tissue develops accordina position; NOT into the Untuc would hov" based On It original posItlon thus tissuc specilted or detetmined?



Answers

Engineering Cloned DNA When joining two or more DNA fragments, a researcher can adjust the sequence at the junction in a variety of subtle ways, as seen in the following exercises. (a) Draw the structure of each end of a linear DNA fragment produced by an EcoRI restriction digest (include those sequences remaining from the EcoRI recognition sequence). (b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates (see Fig. $8-34$ ). (c) Draw the sequence produced at the junction that arises if two ends with the structure derived in (b) are ligated (sce Fig. $25-16$ ). (d) Draw the structure produced if the structure derived in (a) is treated with a nuclease that degrades only single-stranded DNA. (c) Draw the sequence of the junction produced if an end with structure (b) is ligated to an end with structure (d). (f) Draw the structure of the end of a linear DNA fragment that was produced by a PvulI restriction digest (include those sequences remaining from the Pvull recognition sequence). (g) Draw the sequence of the junction produced if an end with structure (b) is ligated to an end with structure (f). (h) Suppose you can synthesize a short duplex DNA fragment with any sequence you desire. With this synthetic fragment and the procedures described in (a) through (g), design a protocol that would remove an EcoRI restriction site from a DNA molecule and incorporate a new BamHI restriction site at approximately the same location. (See Fig. 9-2.) (i) Design four different short synthetic double-stranded DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a Pstl restriction digest. In one of these fragments, design the sequence so that the final junction contains the recognition sequences for both EcoRI and Pstl. In the second and third fragments, design the sequence so that the junction contains only the EcoRI and only the PstI recognition sequence, respectively. Design the sequence of the fourth fragment so that neither the EcoRI nor the PstI sequence appears in the junction.

Those question explains a simple procedure. An experiment. Um, and they were just looking at the fraction of cells and my toes is over time and were given a graph for this. So promotion, eh? It asks what all cells be accepted are expected to contain the radioactive DNA. Um, after they are labeled in the experiment and on lee the cells that are in the S phase of their cell cycle, Um, and the ass phasing that cells are making DNA. Um, during the 30 minute labeling period may contain any radioactive unite number. Question be it asks. Initially, there are no my tonic cells that contain radioactive DNA. We want to know why. This is, um so initially might have excels contained no radioactive Dina, because these cells, um, we're not engaged And, uh, Deena synthesis during living period. It takes about two hours for the 1st 8 will make hot Excel with here. Next, we went to explain the rise and fall, and then the rise again of this curve that we see and just for reference, the curve looks something like this. Besides, what? This question is your friend. So the initial rise of the curve corresponds to cells that were just finishing DNA replication. When the radioactive diamond in was added, a curve rises, more is more labelled. Cells entered my toe, sis, and the peak corresponds to those cells. I just started the s phase in the radioactive gravity was added. So this first arise is, um so starting ass fades. Whether that was they were just finishing it when this, I mean, it was at or they were just starting it when it was added, the labelled cells, then exit for my toes is replaced by unable. Night hot excels which were not in the s phase during the labeling period. So cells then exit my closest and that's the fall and last after 20 hours occur. So it's rising again because it labeled toes. Enter second, my cases and lastly, want to estimate the length of the S phase from this graph. So the initial two hour lag before any label might have excels appear corresponds to the G T phase. Um, which is the time between the end of the S phase and beginning of my toe sis. The first labelled cells seen in my toes is where those just finishing s phase when they radioactive gabardine was added. So in summary, the, um again, the first table cells were those that were just finishing the s phase eso The initial two hour lag corresponds to the G two phase.

All right. So my tongue cycling See DK's? All right, this here, my totting staying plan next with the c. D. K. Are not always active. So we're gonna treat this as are complex and they're not always active because they can have phosphate groups attached to them, which makes them in active. So this is inactive. Inactive. Okay, um Plex. That's because of these phosphates. And so they are actually inactivated by a kindness were by protein and the protein that does this you draw this longer. The protein that does this to these phosphates is called we one in its kindness. And so that's a protein that helps to keep these phosphates on there, and it makes the CTK complex inactive. So in order to help solve this after DNA replication is completed a protein foster taste, which means it can come in and remove phosphates called CDC 25 is activated, so C E C 25 is activated, and it can actually come in here. The bureau here come in and removed. And I'm not gonna write the names over the same over on the left, and it can remove the fast feet so it removed And so now, um, active my ta tick, see? Yep. Yeah. My God! City kids, Because the phosphates have been removed. And so once there is more my tonic See DK's they actually come in. And how? Activate mawr of them. So it's a positive feedback loop. This is a positive Be back blue. All that means is that more active My tonic CD case that are created B'more we won Chinese will be removed.

In this video, we are going to go over the development of some mix. Some might begin developing during the process of no relation, and they are derived from the miz odor. So you can see here on this diagram, the miso Durham is here and some nights form within the method um Here they also form in a wave from the anterior to the posterior. So this means from the head and to the tail end of the organism. So some lines can develop into multiple types of tissue, including skeletal or muscular, and each species of animal will have a different number of semites, so A snake could have 400. A chicken could have 30, so that is how some ice develop.

Things. Question is about triggering a sensitive plant. So they use a capacitor discharge of a capacitor toe trigger the sensitive plant. And there to passing this question in a, um, charge on the capacitor was transferred to the plants. How much how much charge was transferred. So we'll be using to equal to C. V s. So they're given capacitance is 47 micro ferric, and then, uh, who teaches 1.5 and you calculate you get 71 times 10 to the minus €5. And then what? You need to calculate the approximate electric field between the elections, So be using, uh, equals to deal with the Okay, So the electric field sequel to your body 1.5 wars divide by, uh, 2.5. Mm. Okay. And you calculate you get 430 words per meter. Okay, so this is the answer for party, and that's all


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