A) Identify a signaling molecule from the model present. Explain how receptors play a role in cell differentiation.

B) Identify the dependent variable and two controls the experimenters used when conducting this experiment.

C) Evaluate if the number of Variant 1-Type cells with mating projections was significantly different from those of the Wild Type. Use chi-square analysis.

D) Scientists propose that a mutation has occurred that either changed the mating pheromone or receptor site on the Variant 1-Type yeast cells. Predict where the mutation occurred. Justify your prediction with evidence from the experiment and scientific reasoning, based on your knowledge of cell-signaling pathways.

Answers

Answer 1

Answer:

A) A signaling molecule from the model shown is the pheromone. The pheromone binds to the receptor to create the cellular response in yeast to stop growth and produce shmoo. Shmoo is a nodule that allows the yeast cells to join together. Receptors play a role in cell differentiation, because the signaling molecules bind to the receptor in order to produce a response. When the pheromone binds to the receptor, a series of steps are followed in the transduction pathway in order to create shmoo, a differentiation in the cell. Without the receptor, the signaling pheromone would not be able to trigger the transduction pathway that ultimately results in the differentiation of the cell. Only signaling molecules with a specific shape and size can bind to a specialized receptor and cause a cellular response. Different yeast cell types may have varying receptors, affecting the ability of each pheromone to bind to the receptor to create a mating differentiation and, therefore, the rate of mating.

B) The dependent variable of the experiment is the number of cells that differentiated. The number of cells that differentiated depended on the type of yeast exposed to the pheromones. In this experiment, the experimenters controlled the sample size and the application of the pheromones. Each treated group consisted of 1,000 cells and was given the same concentration of pheromones. The same three pheromones, Wild Type-created, Variant 1-created, and Variant 2-created, were also used for each yeast cell type. These controls allowed the experimenters to observe how yeast cell type affects the rate of mating without the influence of other factors that could have skewed the results.

C) Chi-square analysis can be used to determine if the number of Variant 1-Type cells with mating differentiations significantly differed from those of the Wild Type. The Variant 1-Type yeast cells are being compared to the Wild Type cells, so the Variant-1 Type cells are the observed data and the Wild Type cells represent the expected data. In order to find the chi-square value, the square of the difference between the observed and expected values divided by the expected value must be calculated for each category. For the Wild Type-created pheromone type, there were 450 differentiated cells in the Wild Type cells and 203 in the Variant 1-Type cells. By using these values in the formula, a value of approximately 135.58 results. There were 606 differentiated Wild Type cells and 411 differentiated Variant 1-Type cells in the groups exposed to the Variant 1-created pheromone, showing a value of about 62.75. The value for the Variant 2-created pheromone category can be calculated as 16.82, with 50 differentiated Wild Type yeast cells and 21 differentiated Variant 1-Type cells. Then, these values are added to find the final chi-square value, 215.15, which can be compared to a critical chi-square value to determine the significance of the difference. The critical value with a 95% confidence for three categories is 5.99. The calculated chi-square value is far greater than the critical value, showing a significant variation between the number of cells with mating projections in the Wild Type and Variant 1-Type yeast cells. This also rejects the null hypothesis that there is not an important variation in the values, supporting the alternative hypothesis that a factor is affecting the rates of mating in Variant 1-Type yeast cells.

D) The significant variation between the data values could have resulted from a mutation in the Variant 1-Type cells. This mutation likely changed the receptor site of the cell by affecting its shape. Without the proper shape of specialized receptor sites, the pheromones are inhibited from binding to the receptor. When signals bind to receptors, the signal is received and a sequence of changes occurs throughout the transduction pathway in order to produce a response. Since pheromones cannot bind to the receptor sites to produce a response as easily, the overall cellular response of differentiation cannot be produced as often. The mutation resulted in the inability for signals in the pheromones to be received and communicate the correct response. Therefore, the mutation in the receptor site Variant 1-Type cells explains the significant variation in the values between the Variant 1-Type cells and the Wild Type cells. The data in the experiment shows that different pheromones resulted in differing amounts of cell differentiation in the Variant 1-Type yeast cells. This further suggests that the mutation affected the receptor site, not the pheromones, as the pheromones could still bind in some cells and the differences in the pheromones were not lost. The data indicates that the receptor sites of Variant 1-Type cells were changed by a mutation, creating a significant difference between the number of differentiated cells in the Wild Type and Variant 1-Type yeast cells.

Answer 2

Ligands are called signaling molecules because they bind to receptors and carries information.

A) The signaling molecule in the model is the pheromone. It binds to the receptor to generate a cellular response in the yeast system. It inhibits the growth of yeast cells and secretes shmoo.

Receptors are important in cell differentiation as they bind with receptors and create shmoo via the transduction pathway.

In the absence of a receptor, the transduction pathway will not occur and shmoo will not be produced.

Signaling molecules having specific shapes and sizes can only bind to receptors.

B) In the above experiment, a dependent variable is the number of cells differentiated. The number of cells differentiated depends on its exposure to pheromones.

The application of pheromones and the sample size of cells were in control by the experimenters. The yeast cell types were exposed to Variant 1-created, Variant 2-created and Wild Type-created pheromones equally.

This helped the researchers in determining the rate of mating without any influence of other factors.

C) Chi-square can be used to evaluate the number of variant 1 type cells with that from wild type.

The observed data includes variant type 1 cells and the expected data includes the wild type cells.

[tex]\rm Chi - square = \dfrac{( Observed - Expected \:values)^{2}}{ \:Expected \:values}[/tex]

The required value with a 95% certainty for the three types is 5.99. The calculated chi-square value is greater than that of critical values.

This shows the difference in the mating of variant and wild type varieties. It also repudiates the null hypothesis.

D) The mutation likely occurred on the receptor site of the variant type that changed the shape of the site binding.

The change in the receptor site will inhibit the pheromones from binding. This will affect the cell differentiation and transduction pathway.

Therefore, mutation on the receptor site indicated the variant and the wild type have different cell differentiation and mating rates. The mutation is responsible for the varied data and not pheromones.

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Answer:

(C) A gated channel that allows glucose, a very large molecule, to cross the cell membrane, and (D) An ion channel that allows ions to diffuse across the cell membrane, are both examples of facilitated diffusion.

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The cell cycle consists of a series of steps during which the chromosomes and other cell material double to make two copies. The cell then divides into two daughter cells, each receiving one copy of the doubled material.

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Answer:

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

The parasympathetic division handles things that stimulate digestion.

Sympathetic nerve fibers are not associated with situations involving digestion.

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Answer: that carbon cycles through the atmosphere whereas phosphorus does not

Explanation:

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Answer:

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Answers

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Somatostatin from hypothalamus inhibits the glands growth hormone

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Answer:

The Mucus that is around the inner lining

Explanation:

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Answer:

excess volatiles

Answer:

Excess volatiles is found in the oceans and atmosphere.

Not sure if its correct but hope it helps!!

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Answer:

Theory of Measurements:

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Answers

Answer:

sorry big answer Answer with Explanation:

Coriolis Force - refers to the fictitious force that acts perpendicularly to the direction of a rotating motion.

Air parcel - refers to a body of air that is "imaginary."

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However, when the air parcel starts to be in motion, its direction will be changed with the help of the Coriolis force. Thus, it moves to the right side of the Northern hemisphere.

Once the speed of the wind increases, the change in direction of the air parcel increases. This happens until the pressure gradient force and the Coriolis Force are equal in magnitude. When this happens, the wind will start blowing parallel to the points of equal pressure. The wind will now then be referred to as in "geostrophic balance."

When friction happens, the geostrophic balance breaks. The flow of the wind will be slowed down. This means that the Coriolis force will also be lessened. The air parcel will then move towards the lower region.

Explanation:

Why does DNA need to replicate before cells divide?

Answers

Answer:

Below:

Explanation:

Replication is an essential process because, whenever a cell divides, the two new daughter cells must contain the same genetic information, or DNA, as the parent cell. ... Once the DNA in a cell is replicated, the cell can divide into two cells, each of which has an identical copy of the original DNA.

whenever a cell divides, the two new daughter cells must contain the same genetic information, or DNA, as the parent cell.

Four friends were walking on a nature trail in their local park. They began to question whether or not the trees need to breathe, or need food like humans do. Here are their thoughts
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Hugo:I think that plants need to eat and need to breathe.
Danielle:I think that plants need to breathe but don’t need to eat.
Liam:I don’t think plants need to eat or breathe.
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Answers

Answer: I agree with Hugo that plants do need to breathe and eat because Plants do require oxygen to respire, the process in return gives out carbon dioxide and eat thorugh the process of phoyosynthesis

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1. All cells would become blood cells.


2. Meristem cells would form.


3. All cells would be exactly the same.


4. Cell differentiation occurs.

Answers

Answer: 2. Meristem cells would form.

Explanation:

Permanent cells are cells that are incapable of regeneration. These cells are considered to be terminally differentiated and non-proliferative in postnatal life. This includes neurons, heart cells, skeletal muscle cells and red blood cells.

in turkeys the allele for brown feather is dominant to the allele for a yellow feather. if two heterozygous turkeys mate and produce offspring, what is the probability of an individual offspring having yellow feather?


50%
0
25%
100%

Answers

Answer:

50%

Explanation:

trust me because the heterozygous is so weird but I know this.

In turkeys the allele for brown feather is dominant to the allele for a yellow feather. if two heterozygous turkeys mate and produce offspring, than the probability of an individual offspring having yellow feather is 50%. Thus, option A is correct.

What will be produced in turkeys a dominant gene?

In turkeys a dominant gene, R, produces the familiar bronze color; its recessive allele, r, results in red. Another dominant gene, H, results in normal feathers; its recessive allele, h, produces feathers without webbing, so that they resemble tufts of hairs.

Two bronze turkeys with normal feathers were mated, and their offspring consisted of eight bronze with normal feathers, three bronze with hairy feathers, two red with normal feathers, and one red with hairy feathers.

Since, this offspring received a recessive red, r, allele from each parent and a recessive hairy, h, allele from each parent, both parents must heterozygous for both characters: RrHh.

Therefore, In turkeys the allele for brown feather is dominant to the allele for a yellow feather. if two heterozygous turkeys mate and produce offspring, than the probability of an individual offspring having yellow feather is 50%. Thus, option A is correct.

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Answer:

Red blood cell has lot of haemoglobin to carry more oxygen

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EZ

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Answers

Answer:

Photosynthesis and cellular respiration are connected through an important relationship. ... Notice that the equation for cellular respiration is the direct opposite of photosynthesis: Cellular Respiration: C6H12O6 + 6O2 → 6CO2 + 6H2O. Photosynthesis: 6CO2 + 6H2O → C6H12O6+ 6O

Explanation:

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Answers


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Answers

Answer:
Genetic variation is caused by:


mutation
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random fertilization
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Answers

Answer:

Pyruvate is a charged molecule, so in eukaryotic cells it must enter the mitochondrion via active transport, with the help of a transport protein. ... Pyruvate is broken down into three CO2 molecules, including the molecule of CO2 released during the conversion of pyruvate to acetyl CoA

Hope it helps :D

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Answers

Answer:

Explanation:

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