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Quick LabGuided InquiryModeling Genetic Drift Choose 10 candies at random from a bag of colored candies The 10 candies represent a small population of species Mix u...

Question

Quick LabGuided InquiryModeling Genetic Drift Choose 10 candies at random from a bag of colored candies The 10 candies represent a small population of species Mix up the candies and then arrange them in a single row. Record the number of candies of each color. Model the growth of the population For each of the 5 candies on the left side of the rOw, add another candy of the same color. Repeat steps 2 and 3 two more times Now remove 20 candies at random, leaving only 5 candies Then again repeat st

Quick Lab Guided Inquiry Modeling Genetic Drift Choose 10 candies at random from a bag of colored candies The 10 candies represent a small population of species Mix up the candies and then arrange them in a single row. Record the number of candies of each color. Model the growth of the population For each of the 5 candies on the left side of the rOw, add another candy of the same color. Repeat steps 2 and 3 two more times Now remove 20 candies at random, leaving only 5 candies Then again repeat steps 2 and 3 two more imes_ Analyze and Interpret Data Reason Quantitatively How did the distribution of colors change during the procedure? Use Models How does the model represent the evolution and change in allele frequency of the population?



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A scientist is studying the genetics of a population of plants that she suspects is undergoing natural selection. After examining samples of the population's DNA over several years, she finds the following data: Does this provide evidence of natural selection in this population? Why or why not? a. No, because the genotype frequencies, not allele frequencies, have to change for evolution to occur. b. No, because the allele frequencies are changing randomly, suggesting that genetic drift is occurring, not natural selection. c. Yes, because it shows that the previously favorable or neutral allele A is now being selected against in favor of allele B. d. Yes, because it is showing that the frequency of both alleles are changing over time.

In corn kernels. We have the following color possibilities. So we have O for orange P for paint. Let's see for Scarlett and s are for sunrise. And so if we take a look at these crosses, we could determine the hair tints pattern for each of beings A wheels. So the first cross is a son. Read parents with a paint appearance. And in the F one, all of the offspring were son bread. And then the F two. There were 66 son read individuals and 20 pink individuals. So let's take a look. We know that son red is dominant to pink because you in the F one you on Lee had son read individuals So we know that son ran is dominant to pink and our ratio here in the F two of 31 supports that. Okay, so let's look across to cross to Wasn't orange parent with sun red parents All of the individuals in the yet to earn the F one was son read and the individuals in the act to your 998 son read to 314 orange. So let's look at what we have here We know that son Red is dominant to orange because all of the F one is son Brett. So Son Red is dominant to orange ratio in the F two. It's about three toe once. What's that? So again have support of this statement in the F two with a 3 to 1 ratio. Okay, so now let's look across three across three wasn't orange parent with pink parents, and so all of the F one is orange. The offspring were 1300 orange and 429. And so because all of the F one more orange individuals we know that orange is dominant to pink, and again this is supported four times three is 12 to about 1200. This is supported because there is a 3 to 1 ratio in the F two, so we all retained determined that we have a dominant pattern that's going on. So let's make ourselves a list. So we know that this is multiple eagles for one gene and that those a wheels have a dominance hierarchy of sun red. The dominant orange in orange is dominant to pink, so that's the information we could get from the 1st 3 crosses. Now the Fourth Cross is a little bit tricky because it has additional information provided for us. So if you taken Orange Colonel and you breathe out with a starlet parent, all of the offspring are yellow. So let's think about this, because the offspring of the F one are being a typically different than birth parents. That suggests that there to jeans involved in color formation, at least in this example, because the offspring are different in Phoenix type that either parents okay, and so we have a couple of options here. Let's look at the F two in the two, you have three peanut types. 182 were yellow, 80 orange and 58 were starlet. And so, because you have three FINA types, we have a couple of possibilities. Hypothesis one. It's either code dominance or incomplete dominance, and the way we can tell this is completar in complete dominance is you would expect a 1 to 2 to one ratio in the other two. The second possibility, or second hypothesis, is that it's episodic, and then you would expect toe have a modified 9331 ratio in the F to something like 934 or nine for three. And so if we assume yes, the wild, we don't have to assume if we could see that the wild type is present in the offspring and it is we have an orange here and a starlet here. So because they appear in the F to this leads us to believe that this is EPA static and we have complementation that's occurring. And so complementation is where one gene is affecting the expression of the other gene. And because the wild type is president in the F two, I think that the best answer is that this is episodic. And so if we work this out using that logic, we can say we can let the little recessive a sand for Scarlett and that Big A is gonna be the colorless state for Scarlett. And then that's Gene one. If we're looking at complementation, that's one of the a second gene. So if we have a little be the color is expressed. If we have a Capital B, the color is locked. And so if we look across four, your parents are going to be a little be little bit big a big A and that's what the yield orange and starless or big Be big and two little A's and said, This is blocking a color and this is gonna express starlet. And that means in the F one. All of your individuals are headers, I guess. So. They have big be a little bit big, a little a and they're going to be yellow because you're blocking the expression of orange and you can't produce Scarlett concerned missing the homeless, I guess Process of state and therefore in the F to what you have is the following combinations. You have a ratio of nine individuals that are yellow, so they have this combination. You have three individuals that are starlet, so you have big B that's walking the next hour, that big to little A's being expressed for Scarlett. And then you have three more individuals that are little. Be little B with a big A something, and they're gonna be orange was the big yields orange in this state, and then you have one double excessive, which is also going to hear orange because the expression of a is blocked by the expression of B and so this orange here is the episodic one, because this little B is walking the expression of scarlet, which makes the kernel of here orange, because this the up here is dominant. Four. The two little A's are allowed to be expressed, and they make starlet. And so this situation is not a complete dominance, and it's not co dominance. In fact, it's episodic, and so we can tell, because we have a modified 9331 ratio in the two, not a 1 to 2 to one ratio.

Hiring. So friends question. The first tip that we have is that the red flowers are recessive, and so we label that as Q skirt and we'll work on finding that purse. So we're given that there are 200 red flowers and 600 blue, so that means there is a total of 800 flowers, the field. So to find out how many Elliott frequency of Red flowers will do, 200 divided by the total, which is 800 and that will give us 0.25. And so we need to find Pia's well. So in that case, we just need Q. So we'll have to find the square root of 0.25 in this case, and that will give us 0.5. So now we have Q or recessive, and we need to find P, which the blue flowers, which are dominant to the red so p we can actually find it by subtracting it one minus. Cute, because people s Q equals one in a party Weinberg equation. And so we do P want to find one minus 10.5 and gun that gives us point by here. So we have our Holmes I guess, and recessive. And to plug it into that equation, the Hardy Weinberg equation. We can just do p squared, and that will give us 0.25 and we have our Q squared, which is 0.25 as well here. And then, let's find R two p cube, which is 0.5. So we have these and we just had to figure out the amount of homes August's hunters like flowers. So to do that, we just multiply these numbers. Um, by the total number of flowers, which is 800 sold to that in blue some world supply these and remember to pick you will give us the header rows. Vegas while the other ones will get us home was, I guess, And the holes, I guess dominant at home was, I guess, recessive. So in that case, we will have 400 Harras i Ghous Flores, which also means flowers that are mixed that are red and blue. And so here we will have, um, 200 read as well as 200 blue Clowers. So Oh, sorry. And this 400 is actually blue header rows. I guess so. Although it might look like this blue is dominant, so it will show up as blue instead of red. So in total, we have 600 blue flowers and 200 red Thanks.

Okay, The results and observed table, which I know I'm not sharing, that you can look at in your textbook, is the number of mutations in each culture and they are so diverse, even though they are grown in the same environmental conditions. Because mutations occur randomly and are not directed during DNA replication processes, the incorrectly paired bases are corrected by DNA repair process. If the mutation escapes these repair processes, the mutation accumulates as the mutations occur randomly, we cannot expect the same number of mutations in each culture.

For peace with yellow pods and one p we green. And then he says that two of the peace randomly selected so and is equal to to to the peace around them. And so the replacements and then the first question, um, the sample size is the sample size is 25 and then it says after 25 possible samples find the proportion of peace, we yellow pause in each of them and then construct the table. So we're supposed to find the proportions of peace we yellow pods in each of them. So we're gonna let, um, we're gonna let let let white represents yellow and G represent green. So now that you've represented that, and then we know the number of samples or 25 then the end is five because we know that we have four piece and one for peace and one for peace with yellow at one p weeks queen. So we know that the end is five. So we're gonna on We know that for every for every selected to samples, because two are picking at random as I kind of random for every two samples is either one yellow, one green to greens one green, one yellow or one green, one yellow, one green. So the possible without possible outcomes are Sweeney. And then, um, the outcome is going to be why I'm sorry about that. The outcome is gonna be Why? Why? And I am y y Y g um, g y or you have G. And then, um, we know that the pop, we're gonna create a table and then we're gonna have the proportions off piece yellow boards, the perp questions provision of why sun embedded. And then we're gonna have the probability, the probability. And since they're 25 possible outcomes off peeking y or picking yellow and green pot, the total probabilities quote 25 and each of them we want about 25. So we don't have to worry about this table, but for why, Why? The probability of 50 proportions of why that you can get to yellow is gonna be Since it's two yellows on, we want to yellows. So it's gonna be to write to and you can get two yellows, 16 possible times. Remember, the question stays for peace. We ghetto called So in every possible five on pigs there has to be four yellows and one green. So that's gonna be tight 16, then the Y g The y G is gonna be one. What? To times four. The g y is gonna be one about to, because you can pick you for what's a good time. What about sometimes full and then g zero, we don't need G. So you're gonna get all 11 time and then the all probability of picking each every one of them is one of 25. Oh, sorry is gonna be 16 about 25 because it's 16 times and then this is gonna be forward 25 and then there's gonna be 4175 Yeah, this is gonna be one of about 25 and then the proportions off the correct. The question also wants us to have the proportions off piece we yellow parts. So we're gonna create that table. So the professions, the proportions off the yellow on the left side and the right side. I'm sorry about that. And the other side, the probably etc. And then the proportions. Or you can either have zero wise one y or two wise and the probability of having zero y. He's one of a 20 wife like we can see here. They probably have everyone wise is four plus four. That's 8/25. You probably have having to. Wise is 16. It was 25. I'm sorry about that. 16 of 25. And that's the answer for the first one. Now, the second one, the second question wants us to find the me off the sampling distribution of yellow piece. So the mean and then that has a formula off. So mission exp times eggs. And that's gonna be equals two. We're gonna find this hotel expedient. Are we know that they're probably cheese or zero wants to. So we're gonna have cereal times 1/25 like you can see it. 1/25. They will close the black hair loss one of us to times 8/25. But you can see you then. We're gonna have clothes. One, but two. I'm sorry to over to. And that's one one times 16 or 25 and we use your calculator. We're gonna have on zero source for over 25. Yes, 16 over 25. Loss. Sweeney Tom. So when you're about to fry, and that's going to be able to zero points. Eight. And that's the me off the sampling distribution. Now. The next question, the next question asked, is the mean of the sampling distribution from B, that zero point it is equal to the population proportion off piece with yellow ports. For us to know the answer, we already know the first one. So we're gonna have to figure out the proportions off piece yellow we yellow the population proportions. So revelation for pushing off yellow. And this is gonna be caused. X over n. And we know that ass is 4/5. Because remember, we have in the first question indeed say that you should have four piece yellow at one p with green and the total outcomes five. So definitely things proportion the population proportions of yellow before over five on, when you complete that is gonna give you 0.8. So, yes, the question is, the answer is yes, because as you can see, the mean off the sample distribution is equal to the population proportions off on the what the piece of yellow part and then the other part of the question say does. Does the mean off the sampling distribution off proposition always equal the population proportion? The answer is going to be yes. Also, the means of the sampling distribution off it's always equal to is always equal to the population proportion, and that's the answer.


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