The above question requires a personal answer about your experience with cutting flatworms. I can't tell you what you saw in this cut, but I'll show you what happens when it occurs.
First of all, you must understand that flatworms are worms with a great capacity for cell regeneration even if several pieces have been cut.
In this case, when a flatworm is cut in half, we can see the following events happen:
At the moment of cutting, the two ends of the flatworm body are equally shaped and are straight ends, without deformations.The ends will not be different if the cut is done accurately, but it may differ slightly.After several days, the ends will present some deformations and will lose the straight shape caused by the blade.Deformations are extremely similar and change similarly.When recovering from the cuts, each side of the flatworm body formed a new flatworm and they are no different, they are the same and they are completely recovered.This is because the cellular regeneration of flatworms is very fast and allows any part of the flatworm's body to create exactly the same beings, as they have the same genetic makeup.
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Help meeeeeee lzzzzzz
Answer:
Wrong category
Explanation:
The brain changes which contribute to addiction occur in parts of the brain called the;.
Answer:
basal ganglia, the extended amygdala, and the prefrontal cortex
Explanation:
transcription begins at a promoter. what is a promoter?
Answer:
A site in DNA that recruits the RNA Polymerase
Explanation:
what bones are formed by intramembranous ossification
Answer:
Intramembranous ossification is the process of bone development from fibrous membranes. It is involved in the formation of the flat bones of the skull, the mandible, and the clavicles.
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.
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.
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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What is different about normal cell division and cancer cell division?
In contrast to normal cells, cancer cells don't stop growing and dividing, this uncontrolled cell growth results in the formation of a tumor. Cancer cells have more genetic changes compared to normal cells, however not all changes cause cancer, they may be a result of it
Communities only contain one population of organisms. True or False ?
A community means a group of various different types of species living together in a common location.
A population refers to a group of organisms of the same species living together in a location.
Therefore, the statement communities only contain one population of organisms, is False.
The picture above shows a microscopic view of an animal cell. Which of the following is an observation that can be made about the cell?
A.
The DNA in the cell codes for blue eyes.
B.
The cell is made up of several organelles.
C.
The cell is larger than the cell of a plant.
D.
The cell was taken from the liver of a cat.
Answer:
Awnser
Explanation:
The Awnser is C _________________
Organic phosphate is taken up by producers during photosynthesis and released by cellular respiration.
True or false ?
Answer:
true
Explanation:
pa brainliest po please
#carryonlearning
Which of the following is an example of nonpoint source pollution caused by burning of
fossil fuels such as coal?
O acid rain
eutrophication
O aquifer depletion
O stormwater runoff
Explanation:
Acid rain eutrophication
Acid rain eutrophication is an example of nonpoint source pollution caused by burning of fossil fuels such as coal.
What is meant by eutrophication?Acid rain eutrophication occurs when the environment becomes enriched with nutrients, increasing the amount of plant and algae growth to estuaries and coastal waters.
What is eutrophication and its causes?Eutrophication is characterized by excessive plant and algal growth due to the increased availability of one or more limiting growth factors needed for photosynthesis .
Hence, A is correct option
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The mutation shown in the sequence below can be categorized as which type?
Original DNA sequence:
A T A C G G T A G C A A
T A T G C C A T C G T T
Mutated DNA sequence:
A T C G G T A G C A A
T A G C C A T C G A A(1 point)
insertion mutation
deletion mutation
chromosomal mutation
substitution mutation
Answer:
Deletion
Explanation:
because one base is deleted.
The mutation from A T A C G G T A G C A A to A T C G G T A G C A A is a deletion mutation.
MUTATION:Mutation is referred to as any change in the nucleotide sequence of DNA molecule. Mutation can be of different types namely: deletion mutation, substitution, insertion mutation etc. Deletion mutation is a kind of mutation that involves the removal of one or more nucleotide base from the DNA sequence. According to this question, an original sequence is given as: A T A C G G T A G C A A, mutation occurs and results in the following sequence: A T C G G T A G C A A. Nucleotide base "A" was removed from this sequence, hence, it is an example of deletion mutation.Learn more about deletion mutation at: https://brainly.com/question/6477597?referrer=searchResults
What type of cell or structure stores food within the stem?.
Analyses of the protein ______ can be used to study evolutionary relationships among species because most organisms produce this protein for the electron transport chain in mitochondria.
Evolutionary relationships among species can be studied when similar structures or materials eg the wings of a bat and the arm of a bird or similar proteins between two different species. This proves to show relationship among organisms.
Analyses of the protein Cytochrome C can be used to study this types of evolutionary relationships.
Cytochrome C is an antioxidative enzyme that is a part of the electron transport chain in mitochondria. Mitochondrion is an organelle that majority of living organisms share in common. This protein is therefore found present in almost all organism carrying out cell respiration. Therefore this protein can be used to analyze and study evolutionary relationship among organisms with this protein.Learn more: https://brainly.com/question/22173470
what parts of a cell are most likely involved with inherited traits?
Chromosomes are most likely involved with inherited traits.
What are Chromosomes?A chromosome is defined as the long DNA molecule which contains part or all of an organism's genetic material. Most chromosomes consist of very long thin DNA fibers coated with packaging proteins where the most important of these proteins in eukaryotic cells are histones.
These structures are found in the nucleus of cells which contain long pieces of DNA. DNA is the material that holds genes and is the building block of the human body.
Chromosomes also contain proteins which help the DNA stay in the proper shape.
Thus, Chromosomes are most likely involved with inherited traits.
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Environmental factors typically activate genes in a cell by causing the cell to --
A.produce identical daughter cells through mitosis
B.form haploid gamete cells through meiosis
C.fuse with another cell to increase the size of its genome
D.transcribe specific DNA segments to mRNA for translation
Environmental factors typically activate genes in a cell by causing the cell to transcribe specific DNA segments to mRNA for translation.
This interaction is an example of environmental factors affecting genetic expression. In these situations, the environment causes specific segments of DNA to be expressed, meaning they are transcribed to mRNA and subsequently used to form proteins.
Meanwhile, other parts of the genome remain or are forced to be silent. This can be a result of mutations or simple biochemical reactions between the environment and the genome.
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Answer:
Transcribe DNA to RNA
Saved
When the lac repressor binds to the lac operator site, RNA polymerase is prevented from transcribing which gene(s)?
100!!! Points
Part A
Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.
map of the world's major and minor tectonic plates
Part B
On the map, take a close look at the boundary along southern California and Mexico’s Baja California Peninsula. The arrows on the map show the direction and speed of the plates along this boundary in millimeters per year. Based on the speed and direction of the arrows, identify the type of plate boundary at this location. Which landforms or events are likely to occur there as a result? (Hint: It may be helpful to model the motion of the plates using your hands or pieces of paper.)
a map of North America with the fault lines marked
Answer:
Where's the map?
Explanation:
What is a food chain.
What happens to the peak wavelength in the blackbody spectrum as the temperature of a star increases?
A. The peak wavelength decreases.
B. The peak wavelength increases.
C. The peak wavelength doesn't change.
D. The peak wavelength decreases and then increases
Answer:
D
Explanation: the peak wavelength in the blackbody decreases
How can we make larger drawings ? Can someone tell me the guidelines please help
Answer:
you should have skills of drawing observe the diagram
Answer:
by using a GRID to scale
Explanation:
step by step
Which statement is true?(1 point)
Hair color and IQ are examples of Mendelian traits.
Mendelian traits follows a normal distribution.
Mendelian traits are controlled by multiple genes.
Mendelian traits cannot be shown on a bell curve.
Answer:
A. Hair color and IQ are examples of Mendelian traits.
Explanation:
Answer:
Mendelian traits cannot be shown on a bell curve.
Explanation:
What controls circadian rhythms in humans? Explain how this process works.
help pls pls pls pls
Answer:
2) =b
Explanation:
I don't know others unu
Which of the following choices best describes ‘ovulation?
A. An egg is released from the ovary.
B. An embryo sinks into the uterus lining.
C. The sperm and egg nuclei fuse.
D. The ejection of sperm through the penis.
why is it necessary for the human body to maintain homeostasis?
Answer:
Conditions in the body must be constantly controlled because cells depend on the body's environment to live and function. The maintenance of the conditions by homeostasis is very important because in the wrong body conditions certain processes (osmosis) and proteins (enzymes) will not function properly.
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how do apocrine glands differ from other skin glands
A polar covalent bond is a bond between
a. two polar molecules.
b. two atoms that share electrons unequally.
c. two atoms that share electrons equally.
d. two oppositely charged ions.
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The Correct choice is ~
two atoms that share electrons unequallyhow many rounds of dna replication are there in a meiotic cell cycle?
Answer:
Two rounds
Explanation:
Meiosis is characterized by one round of DNA replication followed by two rounds of cell division, resulting in haploid germ cells. Crossing-over of DNA results in genetic exchange of genes between maternal and paternal DNA.
Which of the following describes a hot spot?
A chain of dormant volcanoes
A place that is prone to earthquakes
A particularly active region of plates
An area where magma is significantly hotter
Answer:
third one
Explanation:
a mammalian stem cell line has a doubling time of 24 hours. if you start with 350,000 of these cells, approximately how many cells would be present after 60 hours of growth?
After 60 hours of growth, there is approximately 2 million cells.
Let y represent the number of cell after x hours.
The bacteria doubles every 24 hours, also there are initially 350000 cells hence:
[tex]y=350000(2)^\frac{x}{24}[/tex]
The number of cells after 60 hours is:
[tex]y=350000(2)^\frac{x}{24}\\\\y=350000(2)^\frac{60}{24}\\\\y = 1979899[/tex]
Therefore after 60 hours of growth, there is approximately 2 million cells.
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