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Question Ponnett Square; Inpea plants thegene tor 3 rourd seed (E) ts dominarit over the wrinkled seed (T7 recessive homocykous Lecessive Iomale with gene botantst ...

Question

Question Ponnett Square; Inpea plants thegene tor 3 rourd seed (E) ts dominarit over the wrinkled seed (T7 recessive homocykous Lecessive Iomale with gene botantst crosses ] bELecocyeous nlc- Performn nonohybrid cross of the offspring genotypes. Labelthe male and female und include botb the phenotype and genotype paront generation; showing the posslble rallo. (perform on anather sheet ol paper buiyou donthave to show the punnett square for this question: just Itst your Jnswcvs) T I Anal 3 (7pt1&

Question Ponnett Square; Inpea plants thegene tor 3 rourd seed (E) ts dominarit over the wrinkled seed (T7 recessive homocykous Lecessive Iomale with gene botantst crosses ] bELecocyeous nlc- Performn nonohybrid cross of the offspring genotypes. Labelthe male and female und include botb the phenotype and genotype paront generation; showing the posslble rallo. (perform on anather sheet ol paper buiyou donthave to show the punnett square for this question: just Itst your Jnswcvs) T I Anal 3 (7pt1" 97 **



Answers

Flower position, stem length, and seed shape are three characters
that Mendel studied. Each is controlled by an independently
assorting gene and has dominant and recessive expression as
indicated in Table $14.1 .$ If a plant that is heterozygous for all three
characters is allowed to self-fertilize, what proportion of the
offspring would you expect to be each of the following? (Note:
Use the rules of probability instead of a huge Punnett square.)
(a) homozygous for the three dominant traits
(b) homozygous for the three recessive traits
(c) heterozyous for all three characters
(d) homozygous for axial and tall, heterozygous for seed shape

And this question. We are crossing a organism that has with so and problems like this can be brought down. Teoh A simple headers itis crossing, uh, where, uh, one water off the offspring would have the genotype of capital M capital and home was dominant. Another water would be home was our. This process is small. Sport him and the other two out of four offspring or, well, Harvey Hitter's eye witness, Gina. And this would help us to solve problems like this. So a we are looking at home aside, goes for three dominant tree. So it's oh, Capitals. And no, just looking at individual probabilities. Per, uh, Pray. So her A's says it's home cycles dominant. The probability will be 1/4 looking cheese could be What are you? And since these are individual events, we can multiply their probabilities and ours. It's also quarter again, and I would give us a probability off 1 64th be home was like us for the three recessive traits. So it's gonna be that opposite Jenna typically, But the probability is going to be the same because because the probabilities for home was like a storm in a home is outside this recessive ardor seems was also gonna be once it's do fourth, you see have a rose Isis Heather reside, this is going to be different. So this iss our target genotype we'll see what it's probably going to be so big Ace, The Probability IHS, this is this is the one. So it's one second Another novelty act eats is going to be one second again. And then the last four his cars. So this time, half again and I would give you the final probability off one and last one is D. No one's a little bit different so we have almost like this dominant holes. I was dominant again and then and reside this for the heart shaped. Okay, So to Capitol Ace, the probability we know ISS 1/4 1 foot does for the t A probability is again one foot I was one port And here this is this the difference. The last one isn't had resided. So it's this one. So this is going to nowhere Probability one, sir. So if we can actually this is going to be one he Simon and that is it

For this problem. We're looking at three different traits, right and going to look at probability of having certain offspring from the parents. And because we're looking at three traits, we're gonna use our rule of probability instead of upon its square because it just becomes too big. So we're going to use the rule of probability. The answer. These questions. There's three traits they're being looked at. We have flour position, these air traits that Mendell looked at and his early genetic studies of P plants. We have stem length, so flower positions either excel or terminal. Stem length is tall or drawer. Then we have seed shape, which is brown or wrinkled. Um, it states that are individual is hetero zegas for all three traits. So for flower position, go ahead Rose, I guess, for stem length and for seed shape. Now that individual is allowed to self fertilized, so it's going to be then a cross between to hetero zog asses. For all three traits. All of these are independently assorted. They can go in all possible combinations, so let's look at what we want to see for our first possible combination and that proportion of what offspring would have that. So our first is Hamas, I guess. Dominant for all three traits and what we know. So they're gonna it's gonna look like this we know by looking at our combinations. Um, and even just a pun it square for the first one to kind of show you how this comes about. Um, you have big a little a making moving one out of four. There's gonna be the homeless, like, is dominant. And that is the same for each of the traits. Because it's a hatteras, I guess, for all of the traits. So one out of four chance, uh, for the flower position, the stumbling and the seed shape. So you do a simple multiplication off the fractions and you get one over 64. So one out of 64 are going to be the homeless, I guess. Dominant. Our next is looking at Hamas, I guess, successive for all of the traits. And so we can kind of use the same idea here from our opponents where we know that there's gonna be 1/4 chance of being Hamas, I guess. Recessive for each of the three traits. So we have one out of four for that trade, one out of four for the stem length and one out of four for the seed shape. And that's going to get this one out of 64 again for Hamas, I guess. Process of individual The next is a hetero Zika's for each and from our planet square here. We know it's, um, essentially half of a chance right for each. So we have one half times, one half times one half, and we multiply those out. We get one over eight. Mhm. Our last is Hamas, I guess, for Axl and tall but hetero that, I guess, or seed shape. And that again, we're going to use our for Hamas. I guess ratios are one out of four times one out of four, and it's ahead arose. I guess so. We know our ratio is one half, and when we multiply those out, we get 1/32

You're were given a plant that has hetero Zia's traits for three different things. Um, we have flour position, stem length and see each. Let's go ahead and figure out what symbols we're going to use. Beach. So here for flower position, we could have axial flowers or terminal flowers and axial flowers. Air dominant. So I'm gonna use a big A as the axial flowers and little A as the recessive terminal flowers. For assembling. You can have a tall plant or dwarf plant. I'm gonna. So I'm going to use T because that's dominant, tall is dominant and, um, little t to be dwarf and then proceed shape. You can be, uh, there is a round seed and a wrinkled seed, and round is the dominant. Um, dominant traits so little are will be will represent wrinkle. We're told that, um, there's a particular plant, that category hetero Zika's for all these treats and that that plant self fertilized or self pollinated. So, um, here are are wow, our Gina types for the parent so that, um, all of them or the female port game, it will be all hetero Zika's for every single trait, and the male gammy. The pollen will be headers, I guess, for every single church. Okay, Now we're given multiple different, uh, a multiple different Gina types and were asked all those gina types, um, or fina types, rather how how many are likely to occur? What's the probability that that would those, uh, dina types and few types will occur. So, um, first of all upon it square with three different traits will take way too long. Um, it will be really, really large, and it's really unreal Lee Thio use. So we're gonna use the probability rule. And I just wanna illustrate how we can use the probability rule, Um, by first doing a pennant square for one of these, uh, hetero zegas traits here. So let's dio a small pundits square and I'll show you how that works as we move forward. So our small planet player, we end up with it. I like its dominant homes like its dominant headers. I go another headers. I go and homicide gets processed. So this hetero zygote we'll always have a 1 to 2 to 11 Hetero zegas Dominant homicide is dominant to two headers. I goes to one homogeneous perspective he's safe. Let's keep that in mind. So first, um, a her option A says that what is? The likelihood that offspring will have will be Hamas, I guess, for all three dominant traits. So that means Hamas ideas for all three dominant traits. That would be this, um, genotype. Okay, so we know that in a normal pundit square, if we look at the probability for just this, that's one in four chance or probability that we would get the Hamas Legace stomach, um, which we can see from our ratio here. And it would be the same independently for this one and the same independently for this one. Now, these are all, um, genes that are undergoing independent assortment, and so we can treat these as three different events that occurred at the same time and do the probability rule. Then we can take the probability for each one independently and multiply them across each other. And that actually gives us the probability that every one of these events will happen at the same time. So here we have one in 64. So the probability that we would have all hetero, I guess are all Hamas. I guess dominant traits is, um, one 64th one chance in 64 times. Okay, so now let's look at B B, says the homeless. I guess that we would have an offspring of with Only Hamas, I guess. Recessive alleles. So that's a T t r r. And again, we know we have, ah probability of this occurring of one and four times of this, occurring one and four times and of this occurring one and four times. And to find out the probability of all of them occurring at simultaneously, we can multiply them together. And again we will have one 64th so one and 64 times that will occur. How about for C. C. Says that, uh, the offspring are hetero Zika's. For every characteristics, that means that we would be the same as the parent. A wow t t they are little are. So what do we know here? That this is actually a one in two chance for the hedgerows, I guess Trait in for this individual, um, thing if this were to occur independently. But if they were to occur simultaneously, we can just again multiply all this through, and it turns out that for all of these things to occur simultaneously with you one in eight chance and then let's look at D Uh huh. De Here, um, says that there would be ah ho Messias for axial Hamas. Vegas, Um, for axial and tall. So, um, we homos, I guess. Dominant for the axial and Hamas ideas for tall but hetero zegas for seats shape. Okay, You know, this is a one in four chance, and this is a one in four chance the book, uh, one and four. But this is a one in two chance. So we do still the find out what the chances if all these events were to happen simultaneously, we multiply four tennis for 16 times. Two is 32. So the chance that this occurs is one in 32. I hope that's helpful.

Hello everyone. We are going to discuss the question given her in this video. The question is didn't mind. The final topic ratios of the offspring's obtained by crossing two p plants head. Pro cycles for the characters of port color and port shape. So in this question we are having two characters. The first characters sport color and the second characters port shape. We need to understand that whenever across involves two characters or two genes. Then it is a die hybrid. Cross or die hybrid cross is across which involves two genes or two characters. In this case the two genes we have uh port color which can be green or yellow and and ford shape which can be inflated or constricted. Green is dominant over yellow and inflated is dominant over constricted. So here is the cross. The question says that to be plants hetero sizes for the characters heterosexuals means the parents will be held true sizes are both the lucy and the to lose Ir port color and portrait. So the genotype of the parents will be capital G. Small, G. Capitalist mall. I. And the other parents will also have the same genotype. And to answer this question we need to make up on next where this box which you can see here. We call it upon next. Where on the very right of on the very left of the unit square. And on the very top of the unit square we write again meet on this side we will write the game. It's formed by first parent. And on the top we will write the game, it's formed by the other parent. After writing the damage, we will write down the genotype of the progeny is which will be obtained by crossing different damage. You can see that I have written all the possible gina types and the genotype in every box of this phone. It's square. So you can say that the first project here is Green inflated. The genotype is green inflated. The second is Green inflated, third is Green Inflated Forces. Green Inflated Sisters. Green Inflated six. If green inflated, this is seven, This is a job and this is nine. So in this connect square we have a total number of nine progeny is which are green inflated. And if we count the number of green constricted, so it is one, two and three. So in this planet square, greens constricted at three, the number of green constricted progeny is is free. If we count the number of we are low inflated, so it is again one, two and three. So in this punnett Square we have three yellow inflated project D's and if we count the number of yellow constricted project he's So it is just one. Only one Year. Low Constricted Project. So in this questions, the answer we have to tell the final topic ratio. So the answer to this question is 9331. We have nine. Green, Green and favorite progeny. three. Green constricted progeny. These three yellow inflated progeny is and only one. Yellow constricted projects. So the answer is 9331. I hope you all understood the answer.


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