Answer:
It is generally believed that the cells that have the most mitochondria in them are the muscle cells.
Answer:
Muscle Cell
Explanation:
How is real life fingerprinting different from the techniques used on CSI?
CSI used various techniques for fingerprinting like powder dusting, iodine fuming, silver nitrate soaking, etc to determine the fingerprint of criminals.
What do you mean by CSI?CSI stands for Crime Scene Investigation. It is a brach that investigates crime scene evidence.
Real-life fingerprinting does not involve critical techniques for identification. While CSI performs various techniques and is able to find the most similarity between the real criminal involves in any violent activity.
Therefore, CSI used various techniques for fingerprinting like powder dusting, iodine fuming, silver nitrate soaking, etc to determine the fingerprint of criminals.
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Real-life fingerprinting differs from CSI techniques. While CSI portrays fingerprint analysis as rapid and foolproof, reality involves a slower and more nuanced process.
Real-life fingerprinting is a more involved process than the quick and ostensibly error-proof fingerprint analysis seen on CSI. CSI frequently depicts fingerprint matching as a straightforward database search that produces instant identification. However, in the actual world, fingerprint analysis is carried out by skilled experts who painstakingly check ridge patterns, minute spots, and other traits to assure precise matches. To avoid mistakes, this procedure can be time-consuming and requires knowledge.
In addition, the way that huge, networked databases are shown in CSI oversimplifies reality. Although modern technology makes fingerprint comparison easier, the vast network of records that are easily available is more fantasy than reality. Real-world law enforcement databases are subject to stringent guidelines and restrictions.
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What would be the safest way to reduce the impact of the the non-native red fire ant while minimizing the impact on other organisms
Answer:
The red fire ant is indigenous to South America. In recent years, this invasive species has become a pest for both humans and other ants
crop destruction
What would be the safest way to reduce the impact of the non-native red fire ant while minimizing the impact on other organisms?
A pouring gasoline on ant mounds
B
spraying broad-spectrum pesticides
с
release a species of fly that will prey on the fire ants
D
treating all species of ants with ant killer
Answer: Release a species of fly that ill prey on the fire ants
Explanation:
I assume that was the question with the answers and I had the same question thru study island so here ya go
What type of cell or structure stores food within the stem?.
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.
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
The Correct choice is ~
two atoms that share electrons unequallyThe duplicity theory of vision holds that a single receptor system cannot produce both high sensitivity and high resolution. What does this theory explain
Given what we know about the Duplicity theory, we can confirm that this theory attempts to explain the functions of the Rods and Cones present in the human eye.
The Duplicity theory of vision aims to provide insight as to the functions of the Rods and Cones in the human eye and how each of these contributes to our everyday vision. The theory also centers around the difference in activity between these two structures.
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_____ cells are long to carry impulses quickly from one place to another
Answer:
Neuron cells
Explanation:
Neurons use axons to carry nerve impulses
Answer:
i thinks they are called neurons/nerve cells
Explanation:
Which step of cellular respiration uses oxygen.
Answer:
The step of cellular respiration that uses oxygen is oxidative phosphorylation
Explanation:
Oxidative phosphorylation out of three other stages is the only one that uses oxygen directly, but the other two stages can't run without oxidative phosphorylation.
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
when plants are red flowers are crossed with plants with white flowers What proportion of The Offspring will have pink flowers
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:
how 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.
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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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
how do apocrine glands differ from other skin glands
¿Cuales son las funciones vitales de la célula?
Answer:
La célula es el ser vivo más simple y, por lo tanto, realiza sus tres funciones vitales: Función de nutrición. Función de relación. Función de reproducción.
Explanation:
Answer:
Cells can play out the three imperative capacities: nourishment, relationship, and generation.
Explanation:
what are the steps to make beryllium?
Answer:
Beryllium is processed from the ore in four major steps: (1) the beryl ore is converted into a standard grade of beryllium hydroxide, (2) the hydroxide is purified, (3) purified beryllium hydroxide is reacted with ammonium bifluoride for conversion into ammonium beryllium fluoride, and (4) then reduced to beryllium
Answer:
Beryllium is processed from the ore in four major steps: (1) the beryl ore is converted into a standard grade of beryllium hydroxide, (2) the hydroxide is purified, (3) purified beryllium hydroxide is reacted with ammonium bifluoride for conversion into ammonium beryllium fluoride, and (4) then reduced to beryllium ...
Explanation:
hope it helps you
Which of the following can help lessen the effects of inherited peripheral neuropathy?
Answer:
d. All of the above
Explanation: is correct
Glucose and oxygen make what?
What is the genotype of the man?.
the x chromosome is vital for every humans development and the y chromosome is normally male.
Therefore the male genotype is xy
5 A prokaryotic cell is 5.0 um in length. A virus particle is 300 nm in length. How many times larger is the prokaryotic cell compared to the virus particle?
A 2
B 17
C 60
D 167
The prokaryotic cell is 17 times larger as compared to the virus particle.
Both values first need to be converted to the same unit.
5.0 um = 5.0 x [tex]10^-^6[/tex] m
300 nm = 3.0 x [tex]10^-^7[/tex] m
Dividing both values:
5.0 x [tex]10^-^6[/tex] m/3.0 x [tex]10^-^7[/tex] m
= 16.667
Hence, the prokaryotic cell is approximately 17 times bigger than the virus particle.
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1:Can all genetic mutations be observed with a karyotype? Explain:
2:Evaluate the use of karyotyping in genetic testing to determine the health of an unborn baby.
Explanation:
Karyotyping is a test to examine chromosomes in a sample of cells. This test can help identify genetic problems as the cause of a disorder or disease.
[tex] [/tex]
is these correct? ....
Answer:
The first one is correct
Explanation:
Which structure is found in all eukaryotic cells?
flagella
large central vacuole
cilia
Golgi apparatus
Answer:
the Golgi apparatus is the answer
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:
1. Which of the following statements regarding enzymes is TRUE?
A. Are affected by factors such as temperature and pH
B. Are made up of proteins
C. Function as biological catalysts
D. All of the above
Answer:
D. All of the above
Explanation:
Enzymes are proteins, act as biological catalysts, and are sensitive to factors like pH and temperature.
Hope this helps!
6. Challenge: Based on the weathering patterns, guess the rock type shown in each photo.
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 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
PLZZZZZZZ help me and thank you
Answer:
it's in the book hehe 11111111111111
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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