Answer:
A chromosome consists of DNA and genetic organic molecules.
PLEASE HELP ME IM TIMED!
Answer:
Interphase is your Answer..........
Which one of the following tissues is illustrated by the sketch?
A. xylem C. Collenchyma
B. Parenchyma D. Epidermis
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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what two hydrogen carrying molecules are formed during the krebs cycle
Answer:
Remember that glycolysis produces two pyruvic acid molecules per glucose molecule along with two of the hydrogen-carrying NADH molecules. Remember also that the Krebs cycle produces NADH as well as another hydrogen carrier called FADH2.
hope it helps
are sea snakes venemous?
Sea snakes are usually not aggressive unless provoked or cornered. Although they are highly venomous, only some bites result in significant symptoms of envenomation. The venom is injected by fangs.
Which of these statements correctly describe the result of Mitosis
Answer:
The result of mitosis is two identical daughter cells, genetically identical to the original cell, all having 2N chromosomes. ... A chromatid, then, is a single chromosomal DNA molecule.
Answer:
statements??
Explanation:
What might cause birds in Population A to immigrate to the island?
Answer:
The change in the climate would have made the birds migrate
Explanation:
what is a spygnomanometer ?
Answer:
an instrument for measuring blood pressure
Explanation:
an instrument for measuring blood pressure typically consisting of an inflatable rubber cuff which is applied to the arm and connected to a column of mercury next to a graduated scale, enabling the determination of systolic and diastolic blood pressure by increasing and gradually releasing the pressure in the cuff.
Pocket gophers change the nitrogen cycle in prairies because they:
Primary motor neurons of the ______ motor pathway originating in the primary motor cortex, whereas primary motor neurons of the ______ motor pathway originate in the brainstem.
Answer:
direct; indirect
Explanation:
Why does the moon appear to change shapes in the sky?
A. The gravity on the moon from the Earth changes the shape.
B. The light reflected off the moon from the Earth changes.
C- The amount of sunlight
reflected off the moon changes its shape as seen on Earth.
D- The relative position of the Sun, Moon, and Earth changes the amount of reflected sunlight that we see.
Answer:
D
Explanation:
darwin proposed that homologous structures are evidence of
Answer:
Homologous structures provide evidence for common ancestry
The hypothalamus functions in: A) coordinating movement. B) regulating the autonomic nervous system. C) secreting a hormone called melatonin. D) sending information to the cerebral cortex
Answer:
B is my answer
Explanation:
correct me if im wrong and i hope it helps
Which of the following can help lessen the effects of inherited peripheral neuropathy?
Answer:
d. All of the above
Explanation: is correct
First To Answer Correctly, Gets Brainliest!!!! Connective tissue is present in bones, and blank, muscle tissue makes up muscle, nerve tissue transmits blank and forms nerves, and blank tissue lines organs and covers the body.
Answer:
what
Explanation:
Answer: uh I'm going to say B).
Explanation: Hope this helps? .-.
Which structure is found in all eukaryotic cells?
flagella
large central vacuole
cilia
Golgi apparatus
Answer:
The nucleus
Explanation:
eukaryotic cells have a the nucleus, which means the cells DNA is surrounded by a membrane. Therefore, the nucleus houses the cells DNA
What is added and removed from ATP/ADP
Answer:
ATP can be hydrolyzed to ADP and Pi by the addition of water, releasing energy. ADP can be "recharged" to form ATP by the addition of energy, combining with Pi in a process that releases a molecule of water.
Explanation:
Basically what the other person said
The mitochondria are defective and do not produce sufficient protein to synthesize microtubules in the cilia and flagella.
Answer:
The mitochondria are defective and do not produce sufficient protein to synthesize microtubules in the cilia and flagella. The plasma membrane of the alveoli is not permeable to carbon dioxide during respiration because it is too hydrophobic. The cells do not produce functional motor proteins in flagella and cilia.
Your best friend confides that recently she's been having dental problems. You ask her if it's because she eats a lot of sweets and she denies it, reminding you that she's on the cheerleading squad and has to be very careful about her weight. She says she mostly eats salads and snacks on carrot sticks or other raw vegetables. You have noticed that she has lost a lot of weight over the past six months and that her usually shiny hair is dull. You're concerned because you think your friend may: ________
a. have some type of hormonal problem.
b. be abusing drugs.
c. have developed anorexia nervosa.
d. be binge drinking.
Answer:
c be binge drinking c is the answer
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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the dart thrower was throwing at the bullseye. what is the best description of this attempt?
Answer:
A. Accurate but not precise
Step-by-step explanation:
The illustration given shows that the was far from each other which means it is not precise because precision is that the given data must be close to each other. However, it is accurate because the darts are closer to the bullseye which can still be considered as accurate. Therefore, I believe this is accurate but not precise.
The best description of this attempt is that the dart thrower is aiming to hit the center of the target.
What is Dart thrower?A person who throws darts at a dartboard is known as a dart thrower. In the game of darts small missiles commonly referred to as darts are thrown at a circular target called a dartboard. Apart from being played in international competitions, the game is commonly played in bars or pubs. Each of the 20 numbered parts on a typical dartboard corresponds to a different score.
The goal of the game is to collect points by throwing darts at the board and hitting special spots to do so. The bullseye, located in the middle of the dartboard, awards the most points when hit. The popular sport and pastime requires precision, concentration and expertise.
Therefore, the best description of this attempt is that the dart thrower is aiming to hit the center of the target.
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10. Which of the following statements about the leading strand is true?
A. It is synthesized discontinuously and uses 1 RNA primer.
B. It is synthesized discontinuously and uses multiple RNA primers.
C. It is synthesized continuously and uses multiple RNA primers.
D. It is synthesized continuously and uses 1 RNA primer.
The statement 'the leading strand is synthesized continuously and uses 1 RNA primer' is TRUE. It is replicated in the 3' – 5' direction.
The leading strand refers to the DNA template strand that is synthesized in the 3' – 5' direction during DNA replication.
The 3' – 5' direction that occurs during DNA replication is the direction of the replication fork.
RNA primers are short RNA sequences used for the initiation of the synthesis of DNA (DNA replication) both in the leading DNA strand and the lagging DNA strand.
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autonomic reflexes enable the ans to control what kinds of visceral functions? check all that apply.
Given what we know about the Autonomic Nervous System we can confirm that autonomic reflexes enable the ANS to control functions such as heart rate, urination, and breathing rates.
These are among the many visceral functions that the Autonomic Nervous System controls. As per the pattern that is being established, we can conclude that the Autonomic Nervous System functions at a subconscious level, controlling the visceral functions that do not require conscious thought.
Despite this, some organs can also be controlled consciously, such as the lungs and bladder. Other visceral functions such as heart rate, digestion, and even ocular responses, are not able to be controlled consciously.
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What features distinguish annelids from roundworms?
Answer:
Annelids have a well developed mesoderm while round worms lack a mesoderm.
Annelids are segmented worms whereas round worms are not segmented.
Annelids have a true coelom while round worms have a pseudo coelom.
What is photorespiration.
If you help me I'll mark you as a brain list
Answer:
picture is of a heart
superior venacava bring deoxy blood to left ventricle which goes to lungs becomes oxygenated carried by atrium to heart which pumps blood to body
The purpose of transcription is
a. to make a strand of messenger RNA.
b. to make two copies of the DNA molecule.
c. to line up amino acids to form protein molecules.
d. to allow the cell to grow.
Answer:
A
Explanation:
I'm pretty sure I took this quiz in K12.
A vaccine against HIV proteins made by a genetically-engineered vaccinia virus that has infected a eukaryotic cell line is a(n)
Which processes contribute to the formation of chemical sedimentary rocks?
Organisms decay in a solution.
Minerals dissolve and crystalize.
Rock fragments squeeze together.
The remains of plants and animals
Explanation:
Rock fragments squeeze together
The basic pattern of blood flow from the lungs to the rest of the body is:
pulmonary vein; left ventricle; left atrium; vena cava.
O pulmonary artery; left atrium; left ventricle; aorta.
O vena cava; aorta; right ventricle; right atrium.
O pulmonary vein; left atrium; left ventricle; aorta.
Answer: pulmonary vein; left atrium; left ventricle; aorta
Explanation: I just did this question in the final exam of A&P and it was correct, hope this helps:)
The basic pattern of blood flow from the lungs to the rest of the body is: pulmonary vein; left atrium; left ventricle; aorta. The option (B) is correct.
What are pulmonary vessels?Pulmonary veins are blood vessels that carry oxygen-rich blood from your lungs to your heart. Your pulmonary veins are part of your body's pulmonary circuit. This is a system of blood vessels that moves blood between your heart and your lungs. This circuit also includes your pulmonary arteries.
For the exchange of gases to occur effectively, the lungs require a dual vascular supply known as pulmonary and systemic circulations. The pulmonary circulation is responsible for bringing deoxygenated blood from the body to the lungs for gaseous exchange and then returning oxygenated blood to the heart.
There are 4 total pulmonary veins—with 2 pulmonary veins coming from each lung, left and right—that empty into the left atrium of the heart.
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