The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment. Given: In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution. The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time: It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers: Introduction: TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

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Answer 1

Question

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

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Question #3: In the KeyGene paper, the authors state that it would be useful if pollen from an apomict would transmit apomixis-inducing genes to the female in the cross (assuming the pollen is viable). Assuming there was just one gene conferring gametophytic obligate apomixis, and that the two parents are inbreds, what would be the consequences of such a cross if: a) The apomixis was a dominant trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the parental cross. b) The apomixis was a recessive trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the…

Question #5: Assume that two genes are identified that confer gametophytic facultative apomixis in soybean. The genes show independent assortment. Recessive alleles at both loci are required for the facultative apomixis. Facultative apomixis is triggered when the temperature at pollination is above 20 degrees C. At temperatures below 20 degrees C, all reproduction is sexual, independent of genotype. A facultative apomict male, capable of producing viable pollen, was crossed with a sexually reproducing female. Assuming the parents are completely inbred, what are the predicted phenotypic ratios (apomict: non-apomict) for the F1, F2, and DH (F1-derived) generations at each of the following temperatures*: a) 15°C? b) 25°C? *for full credit, show crosses and genotypes where appropriate. Remember to position the female first (left side) in the cross. Type your answer here:

Answer 2

Question

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

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Answer 3

Question

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

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Answer 4

Question

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

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Answer 5

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

Answer 6

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

Answer 7

Chapter 34, Problem 1Q

Summary Introduction

To review:

The role of tonoplast intrinsic protein (TIP) and the importance of the two mentioned conditions by the researchers in the experiment.

Given:

In the first case, researchers injected the oocytes of the frog Xenopus laevis which are large in size with a low osmotic cell permeability with the messenger ribonucleic acid (mRNA) of plant TIP. In the second case, the oocytes were injected with mRNA of a protein which does not help in transport whereas, in the third case, researchers kept the oocytes uninjected. They were observed after incubation of 2 days in isotonic medium and treatment with hypotonic solution.

The following graph was plotted showing the changes in relative volume of the different oocytes of the three cases over time:

It was also observed that the oocytes which were injected by TIP mRNA bursts after sometime while others do not. The following table depicts the osmotic water permeability of the three conditions as calculated by the researchers:

Introduction:

TIP is a transport protein present in the tonoplast of the plant cell. It helps in transport of molecules just like the other aquaporins. Also, hypotonic solution is that solution which has higher water potential than the cell so that due to the difference in water potential, the water moves inside the cell and swells up.

Answer 8

Expert Solution & Answer

Explanation of Solution

From the graph, it can be concluded that with time, there is a significant increase in the volume of the oocytes that were injected with the mRNA of TIP. On the other hand, there is a very little increase in the relative volumes of the oocytes which were injected with mRNA of nontransport protein and the oocytes that were not injected at all. This happened due to the fact that when the oocytes were kept in the hypotonic solution, entry of water inside the cell took place because of the difference in water potential. The movement was rapid and in large quantity in the first case as compared to the second and the third case. The entry in the first case was rapid as TIP which is a transport protein increased the osmotic permeability of the oocytes, allowed the entry of more water and lead to the increase in volume of the oocytes.

Also, for the interpretation of the experiment the other two conditions were important as they proved that in absence of TIP the transportation occurs in a normal way and there is a little change in the volume of the oocytes.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Conclusion

Hence, it can be concluded that the TIP which is a transport protein is responsible for the increment in the relative volume of the oocyte cells as it increases their permeability and hence leads to their bursting.

Answer 12

Ch. 34.1 - Prob. 1R Ch. 34.1 - Prob. 2R Ch. 34.1 - Prob. 3R Ch. 34.2 - Prob. 1R Ch. 34.2 - Prob. 2R Ch. 34.2 - Prob. 3R Ch. 34.2 - Prob. 4R Ch. 34.3 - Prob. 1R Ch. 34.3 - Prob. 2R Ch. 34.4 - Prob. 1R

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