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I germinated millet seeds through the process of fermentation.After they germinated, I placed 100 of the seeds in a test tube andplaced a cotton ball saturated in N...

Question

I germinated millet seeds through the process of fermentation.After they germinated, I placed 100 of the seeds in a test tube andplaced a cotton ball saturated in NaOH, and placed the test tubeupside down into a glass beaker full of distilled water. Icompleted the same process with non germinated millet seeds. Iplaced the test tubes (both in the same beaker of distilled water)in front of a 100 watt light bulb for 2 hours and was instructed torecord the oxygen levels at the bottom (opening) of th

I germinated millet seeds through the process of fermentation. After they germinated, I placed 100 of the seeds in a test tube and placed a cotton ball saturated in NaOH, and placed the test tube upside down into a glass beaker full of distilled water. I completed the same process with non germinated millet seeds. I placed the test tubes (both in the same beaker of distilled water) in front of a 100 watt light bulb for 2 hours and was instructed to record the oxygen levels at the bottom (opening) of the test tube. The tube with the germinated seeds had oxygen levels that went about 1.10 cm up the test tube after the two hours were up. How do the processes of fermentation, cellular respiration and photosynthesis all relate to one another? Can you explain what is happening during this entire process? Is cellular respiration and/or photosynthesis occurring during the second portion (light portion) of the experiment?



Answers

In the plant Arabidopsis thaliana, a geneticist is interested in the development of trichomes (small projections). A large screen turns up two mutant plants (A and B) that have no trichomes, and these mutants seem to be potentially useful in studying trichome development. (If they were determined by single-gene mutations, then finding the normal and abnormal functions of these genes would be instructive.) Each plant is crossed with wild type; in both cases, the next generation $\left(\mathrm{F}_{1}\right)$ had normal trichomes. When $\mathrm{F}_{1}$ plants were selfed, the resulting $\mathrm{F}_{2}$ 's were as follows: $\mathrm{F}_{2}$ from mutant $\mathrm{A}: 602$ normal; 198 no trichomes $\mathrm{F}_{2}$ from mutant $\mathrm{B}: 267$ normal; 93 no trichomes a. What do these results show? Include proposed genotypes of all plants in your answer b. Under your explanation to part $a$, is it possible to confidently predict the $\mathrm{F}_{1}$ from crossing the original mutant $A$ with the original mutant $\mathrm{B} ?$

If you have a monitor. Jennet Phoenix Hype. What this means is mono is one, and genic refers to gene, so this is a FINA type that is determined by one gene. So the probability for recess of FINA type in the F two is 1/4. Now, if we have a poly Jinich Tina type, remember, Polly means many and again, genic means gene, So this is a FINA type that's caused by many genes. The F to FINA type will again the 1/4 for the recessive Bina type. But because we don't know how many genes are controlling that particular FINA type, it has to be to the end power. And so we can use this formula to solve part A in our problem. Now, these two sort of rules apply. If you have homes, I guess dominant parents read with each other. It doesn't work if you have other combinations. So the parental caw cross must be all homos, I guess Dominant crossed with all homes, I guess. Process it. And so from the problem in part A. We know there are 256 individuals that were the outcome, and so all we need to do is take this formula and use the information from a part in the question to determine the number of genes that are building a particular FINA type. So here, 1/4 to the end, we're going to convert that to four to the in power equals 256 from the problem. We know that four to the fourth power is 256 so in equals four. The number of genes in this problem that are controlling a given genotype is four. Now for the court beat part. That's good problem. Let's make some notes. We know that the minimum height of the plant is 19 inches, so all recess it plants so they would have too little ace to Little Bee's too little sees. I'm going to underline my sees because I get sloppy and then I won't be able to tell the difference. And neither will you between a big seeking a little sea or if you have a plant that's all recessive. That plant has a height of 19 inches. We also know that a plant that's all dominant so big a big a big, be big B two big seas and two Big D's is 21 inches tall. So then we can subtract to figure out the difference. 21 inches minus nine inches equals 12 inches. We know that there are four genes, right? We know that there are four genes that are controlling this height Tina type, and so we could determine the contribution to height by each gene. And we have to assume that each gene contributes equally to the FINA type. But we have no way of knowing yet, so we have to assume it. So our high differences 12 inches, the number of genes is four, and we divide that, and that tells us that each gene contributes three inches, so three inches her G with violent tackles. See part. So we watch determine what is the probability of hitting in 18 inch plant is. And so if we take the 21 inch plant that tells us that, um, it has to have a peanut time of either all dominance of homeless, I guess dominant or a header Zeiger, but dominant for every trait again, I'm on underlined my see. It's I mean, you could tell him that one, but I do get sloppy, Okay? And then we also know that a nine inch plant is the minimum height for plants, and that's driven by the presence of a recess. It Gina type for all traits. And so we know that each dominant a wheel, it's three inches of heart to the plate. So for 18 inch peanut Time, Way needs a minimum height of nine inches for the recess of trade waas three inches, that's only 12. So that's one dominant Leo, plus three inches that's 15 inches. That's a second dominant over your and then So we need three more inches. Teoh yield an 18 inch plant. That's our goal. And so here this three inches is 1/3 dominant over hell. So what that means is that we have to take this process, it being a time and as one to three dominant A wheels. So here we go. A. There's one. Maybe that's too See, that's three little D little Dean, and that should yield an 18 inch plant. Now it doesn't have to be capital a Capital B Capital C. These can be swapped out so you could have a little a little a pixie make C and Big D that would also yield an 18 inch plant. And so as long as you have three dominant A wheels, you should be able to get an 18 explain, and that's how you would capitalize it. But that's how you would solve this problem is the use of headers like it's and knowing that each dominant olio adds three inches of height to the minimum height plant.

Hi there. First of all, to answer the question, did you create life? So you had this flask with your yeast extract and amino acids and inorganic salts. All of these things are in your flask. Check this one person. Your flask is open to the air. So no, you most likely did not generate your own life form because things like spores or bacteria or any type of microorganisms are found in the air. So, more than likely, some of these found their way into your mixture and then started to reproduce, which were the cells that you saw there. So to redo this, um, so that air can get in oxygen in particular and yet exclude spores and microorganisms from your substance. One thing you could dio is what pasture did He had a flask with a swan type neck on it sort of something along this order. Okay? And he heated up. He boiled his mixture probably something stronger than a candle. But candles are easiest to draw. So he did this up and then extinguish the candle chicken and let it cool back down. Well, the water vapor condensed and accumulated in the neck, effectively blocking out things from the outside. So things from the outside can only get this far. But then they'll trap get trapped in the water. Meanwhile, you still have oxygen in here, but it is sterile. Since you heated this up, boiled it. That is sterile. So if you ran your experiment this way, there is oxygen present and you can see if anything will grow. Probably another thing you can do adjusting mawr to modern day is you could potentially you some sort of filter with very tiny pore openings. So put some sort of filter over this so air can still get in being very small molecules. Oxygen molecules could still get in. And yet it would exclude spores and cells because they would be too large to pass through that filter. So that would be a something else that you could do using something a little more modern day, Um, thes filters that will capture spores and bacterial cells. All right, so those are some options of things that you could do this set up with oxygen available and see if you get any growth in your solutions. Thanks so much for watching

Already. So we're in one of late. We have another hypothetical experiment here and here. We have a student who, uh, take some tobacco sees and puts them in a petri dish. Right. A petri dish, a in a petri dish. Be okay. And there's 20 seats in each, even amount. And then the student wraps. Uh, picture this a in some sort of cover that would exclude light. Right. Then he places them. Ah, three equal distance from a light source. Okay, It's another equal distance from a light source. And he says the cycle to there's 14 hours of light and then followed by 10 hours of darkness. Okay? And for one week, this experiment goes on for one week. Lengthy end takes some some measurements. Right? So we removed the cover from a and say what's going on under the hood here. Curious and heinz Ah, A few interesting things, right? So the first thing to note is that they had less germinating seeds, right? So there is just 12 germinating scenes and this be all 20 Germany. All right? And then, uh, for dish be 15 of them where green and we had five that were yellow, whereas in dish, eh? All 12 were actually this yellow leaved ceiling. But more interesting is the length of the stem was quite different. So and, b, they were shorter. The average length of this time was three millimeters. But the average length of this stem for dish A was actually a little bit taller. They were closer to eight millimeters. Okay? And so then the question of this problem is why? Why would that be longer? Okay. And so just broadly speaking, we could say, Well, there was an elongation of this time, right, because it's it's longer and a that and be in its longer due to a lack of light because that was the conditions of a. But why that might be is fury in sort of plants where they are under shades that there's a plants were covering them right? The response might be to to try to grow and find light. They're going to try to seek it out. They're not getting it now, but maybe if they move a little bit in one direction or another, get taller, they'll find it. And so being in the dark stimulated this elongation response right? So than the correct answer is going to be choice B

So here we have a few different parts to this podcast. In the first part, we're taking a look at sucrose Grady in central allegation and they're separated by molecules out by size, where we have a larger biomolecules further down the radiant compared to lights biomolecules. So we do have three distinct pinks here. So the Eukaryotic 80 s rapper zone has two subunits. That's the 40 in the sixties sub unit. We have a small amount of art and ace will that will particularly separate out the ribosomes into the sub units. So therefore we have the first peak. This is the smallest sub unit. This is 40 s. We have the second Pete, that is a 60 s. And finally, we have the third peak on this corresponds to the S. Robertson. So now taking a look at the second part here. So the circus Grady in shows more peaks near the bottom of the radiant meaning that the we have biomolecules that are larger than the single Reiber zone mixture. So therefore the larger peaks must be polio zones. This with each of the small peaks corresponding to the police homes containing a certain number of red prisms. So the only difference between this mixture and the previous one is up is the R and A's treatment. So therefore, the Dilute Arnas treatment associates, holly ISMs and the lack of our Chinese leaves Polly ISMs intact, moving on to the last part here. So if we can compare this graph to the previous one, we have many uh, fewer peaks for larger biomolecules, meaning there are many fewer Polly ISMs. So conversely, the peaks for the single vibe resumes are much larger, correlating with the evidence that there are fewer Polly ISMs. So the lack of policies, um, suggest that the protein synthesis is decreased, he asked. Protein synthesis is decreased in hypoxic cells, but this experiment alone does not give any information for why this occurs exactly.


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