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QUESTION #2Fromtthe smallest to the largest; the goals of EXISTANCE of an anlmal are to survive and reproduce: During the course of evolution, the many adaptations ...

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QUESTION #2Fromtthe smallest to the largest; the goals of EXISTANCE of an anlmal are to survive and reproduce: During the course of evolution, the many adaptations that have been established were advantageous and have permitted thelr advancement; Tne establishment of LOCOMOTION has led them to various habilats and t0 difforent means transportatlon; You wlll FIRSTLY discuss among the anlmals studled In thls coursc respcct to the estabilghment ot the locomotlon vla different methods. SECONDLY wit

QUESTION #2 Fromtthe smallest to the largest; the goals of EXISTANCE of an anlmal are to survive and reproduce: During the course of evolution, the many adaptations that have been established were advantageous and have permitted thelr advancement; Tne establishment of LOCOMOTION has led them to various habilats and t0 difforent means transportatlon; You wlll FIRSTLY discuss among the anlmals studled In thls coursc respcct to the estabilghment ot the locomotlon vla different methods. SECONDLY with this Particular locomotion; what type 0f habitat (hls Permitted them t0 oxplore: aquatic, terrestrial;, aerial; THIRDLY, with this movement t0 diffcrent areas,whai change regards to feeding modc had to occur (become prey; predator; parasito, saprophyte etc FOURTHLY, the distinctive acqulred characteristic that was developcd (o faod, FIFTHLY % Inal; ono ol the advantages brought about by (hls now mcans ol dieplacoment The anlmal groups to be dlacussed are the (ollowing: INVERTEBRATE (10 specimens): (1) PROTISTA, (2) PORIFERA, CNiDARIA Or CTENOPHORA: (4 PLATHELMINTHES NEMERTEA or BLASTOCEELOMATES O ANNELIDA OnYCHOPHORA, (51 CHELICERIFORMES, CRUSTACEA, (9) MOLLUSCA (10) ECHINODERMATA (687) INSECTA, VERTEBRATE (5 specimens): (1) FISHES; (2) AMPHIBIA , (3) REPTILIA, (4) AVES; (5) MAMMALIA



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Of all the animals you're likely to see on a summer's day, the most ancient is the dragonfly. In fact, the fossil record for dragonflies extends back over 250 million years, more than twice as long as for birds. Ancient dragonflies could be as large as a hawk, and were surely buzzing around the heads of both $T .$ Rex and Tricentops. Dragonflies belong to the order Odonata ("toothed jaws") and the suborder Anisoptera ("different wings"), a reference to the fact that their hindwings are wider front-to-back than their forewings. (Damselflies, in contrast, have forewings and hindwings that are the same.) Although ancient in their lineage, dragonflies are the fastest flying and most acrobatic of all insects; some of their maneuvers subject them to accelerations as great as $20 \mathrm{g}$. The properties of dragonfly wings, and how they account for such speed and mobility, have been of great interest to biologists. Figure $8-35$ (a) shows an experimental setup designed to measure the force constant of Plexiglas models of wings, which are used in wind tunnel tests. A downward force is applied to the model wing at the tip (1 for hindwing, 2 for forewing) or at two-thirds the distance to the tip $(3$ for hindwing, 4 for forewing). As the force is varied in magnitude, the resulting deflection of the wing is measured. The results are shown in Figure $8-35$ (b). Notice that significant differences are seen between the hindwings and forewings, as one might expect from their different shapes. Treating the model wing as an ideal spring, what is the force constant of the hindwing when a force is applied to its tip? A. $94 \mathrm{N} / \mathrm{m}$ B. $130 \mathrm{N} / \mathrm{m}$ $\mathrm{C} \cdot 290 \mathrm{N} / \mathrm{m}$ D. $330 \mathrm{N} / \mathrm{m}$

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

Okay, so I actually learned something new today. Apparently, there's a formula that kind of tells you when a fledgling bird is going to be able to fly on its own. And it's the ratio of two functions of time, one of which kind of gives an indication for how long their wings are and the other one is their body mass. Okay, so whenever these ratios kind of approach one, whenever FFT approaches one thing, the fledging fledgling is able to fly on its own. You didn't know all this question is asking us to do is to interpret the physical meaning behind those and described the units associated with them. Okay, so the 1st 2 shouldn't be too difficult, right? Because M prime of tear is the time derivative. It's the time derivative of em of tea. It's a rate of change. It's how fast this function changes with respect to time. And since we're given that the average body mass is measured in grams, this is gonna be essentially grams per per unit time and that the time is in weeks, by the way, since grams per weeks and just like I said, ah, the interpretation of physical meaning is how fast the body mass is changing with respect to time. It's the rate of change of body mats. Okay, for w prime of tea, that's again the time derivative of W T. Which is the length of the wings. And since they said that wing length was gonna be measured in millimeters, this rate of change is millimeters per week. And again, it's just how fast the length of the ones you're changing F prime of tea has to have special analysis because, um, f prime of TIA is the time derivative of f of tear. But f of tia is the ratio. Oh, the length of the wings and the average body mass. These air two functions that change. So we actually do have to use the quotient rule, which is lo de I minus high. Do you low swearing the bottom and we're gonna determine the units of this function As far as a physical meeting goes out, I said half of Tia as FFT approaches one, then the fledgling is gonna be more able to fly. This is a rate of change of that. So if this is a really, really positive number, then it's going to rapidly go towards, and it's gonna take a lot less time for to be able to fly. Maybe it's some sort of growth spurt. So all that remains to do is to find the units of this I'm of Tia. Ah was measured in grams w prime of tea we said was millimeters per week. Okay, W of tea waas um millimeters and M prime of tea is grams per week, divided by AM of T squared M of T is measured in grants. This is just grams squared. Okay, so we have a grams times of millimeters minus of grams, times of millimeters, all divided by a week. Okay, so this is gonna be some sort of of grams times millimeters. We don't know. Of course you know how exactly these variables are changing, so they're obviously not gonna be the same necessarily. So when you subtract two things that are like each other, you're gonna get something else that's like each but that's like those two things. So this is gonna be some grams millimeters divided by weeks, all divided by Graham Sward, which is gonna be grams times, millimeters times, weeks times one over. Graham squared. One of these grams cancels with one of those, and then we're left with millimeters Her grands week. Here we go.

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.

Now this problem were asked to investigate the difference between means for different bird types on whether the mean length of the coup eggs is different for different species. So what you're gonna want to do is plug all of this data into your statistical program or your calculator to get some basic statistics here. So we want our sample sizes. So I'm gonna say N s is for Sparrow. Well, Dio Robin will be in our on dwa AG tail will be in w. So I have here 15 for my sample set 17 and 16 were also going to need the mean value for those eggs. So the mean for the sparrow was about 23.2 The mean for the robin was about 22.53 on the mean for the wag tail came to about C 22.8 four. We're then going to need the standard deviation. Cistern deviation came to 0.979 Standard for the robin came to 0.686 and the standard for the wag tail deviation came Teoh 1.6 five. All right, so from here, if you have statistical software, you can go ahead and run analysis and get your confidence interval. If you need to know how to do it by hand, I will do it for one example. So if we wanted to compare the Sparrow to the robin with a 95% confidence interval, we're going to use the formula the mean of the first sample minus the mean of a second sample to get that difference, plus or minus R T statistic based on degrees of freedom times the standard era. So a T statistic based on degrees of freedom. We're going to look up in a tea table on this degrees of freedom. You need to use your computer to find you're gonna plug in this basic information that we've already found and we get for the first example that degrees of freedom for Spiro versus Robin comes to 20 four. If you look up in a tea table Alfa of 0.5 for two tailed T test, This is coming from our 95% confidence, right? So one minus 10.95 gives us at 0.5 Alfa with degrees of freedom of 24 R t statistic is 2.63 nine. So we've gone ahead and found this first part of our formula. We also need to find the standard error. You could do this by hand by taking the square root of the standard deviation of your first sample squared, divided by and for your first sample, plus the standard deviation for your second sample square divided by the end for that second sample for us. If I was to plug in these values, it would look like 0.979 squared, divided by 15 plus 0.686 squared, divided by 17. This value altogether gives you 0.302 six. Now we're able to actually construct are 95% confidence interval. So we get our first mean 23.0 to minus our second me and 22.53 plus or minus that t statistic 2.639 times 0.30 to 6. And if we were to solve all this out, get plus or minus, we wind up with a confidence interval of negative 0.565 20 Oh, I'm sorry. That is the in Kratch Interval. I'm getting ahead of myself here. You would wind up with negative 0.114 to to 1.962 So the really important part of this confidence interval, whether you found it by hand or you ran a test in your program, is that we're ranging from a negative to a positive. So what that means is that if we were to place this on a number line, our range of possibilities includes zero meaning with 95% confidence. There might not be any difference between these two egg sizes. The same is true if you compare the sparrow doing the exact same process the sparrow to the wag tail. You mind up whether you do this in your program where if you do it by hand like we just did with an interval of negative 0.565 to 0.936 one. All right, so that again we have zero included in that confidence interval. So what that really means is that here we go. I ran out of room. There is that there might not be any difference at all. So is 95% confidence intervals. We say that these do not have any statistically significant difference between them


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