All living cells contain ribonucleic acid (abbreviated RNA), a nucleic acid with properties comparable to those of DNA. Nevertheless, RNA is often single-stranded, unlike DNA. Instead of the deoxyribose present in DNA, the backbone of an RNA molecule is made up of alternating phosphate groups and the sugar ribose.
The hypothesized first self-replicating molecule is RNA. The characteristics of this molecule that led to this hypothesis are:
RNA can store genetic information, similar to DNA.RNA can catalyze chemical reactions, similar to enzymes.RNA can self-replicate, meaning it can create copies of itself.These characteristics suggest that RNA may have been the first self-replicating molecule and played a crucial role in the origin of life on Earth.
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explain the overall chemical reaction for the enzymatic reactions involving catalase called is ?
The overall chemical reaction for the enzymatic reactions involving catalase is a two-step process:
The overall chemical reaction for the enzymatic reactions involving catalase is a two-step process. The first step involves the enzyme catalase breaking down the substrate, hydrogen peroxide, into two molecules of water and one molecule of oxygen. This is represented as follows:Learn more about enzymatic reactions at brainly.com/question/29411240
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Which cycle is represented in the image?
Cellular
respiration
ht
Animals
Death and
decay
OA. Water cycle
OB. Carbon cycle
C. Nitrogen cycle
OD. Oxygen cycle
Plants
Fossil fuels
Photosynthesis
Ml
Industry and Home
Answer: Carbon cycle
Explanation: when animals die, they bring carbon out in the atmosphere.
what are some peer reviewed articles about current research being
done to determine the genetic factors of addiction? preferrably
published within the last 5 years.
There are numerous peer-reviewed articles that have been published within the last five years discussing current research on the genetic factors of addiction.
Some examples include:
- "Genetic Influences on Addiction: What We Know and What We Don't Know" by J. A. Franklin, A. J. Agrawal, and L. A. Bierut (2016) in the journal Neuropsychopharmacology
- "The Genetics of Addiction: A Translational Perspective" by C. E. Cadet (2016) in the journal Translational Psychiatry
- "The Genetics of Opioid Dependence: A Review" by C. A. Nielsen and L. Yu (2017) in the journal Biological Psychiatry
- "The Genetics of Drug Dependence: A Review of Clinical and Preclinical Studies" by J. D. Rubinstein and B. L. Wilcox (2018) in the journal Neuroscience & Biobehavioral Reviews
- "Genetic and Environmental Influences on Substance Use and Addiction: A Review" by C. M. Kendler and B. P. Riley (2019) in the journal American Journal of Psychiatry
All of these articles provide valuable insight into the current state of research on the genetic factors of addiction, and can serve as a starting point for further exploration of this topic.
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If
I have a negative that is contaminated, how can I avoid that
contamination?
ELECTROFORESIS OF NUCLEIC ACIDS AND PREPARATION OF AGAROSA
GEL
To avoid the contamination of a negative, there are a few precautions that can be taken. One way to prevent contamination of a negative is to avoid contact with the positive.
It is a standard practice to use separate pipettes, tips, and buffers for negative and positive samples during the process of electrophoresis of nucleic acids. In agarose gel electrophoresis, the separation of nucleic acid molecules is based on size. The molecules are separated using an electric field, and the smaller molecules move more quickly through the gel matrix, while the larger molecules move more slowly through the matrix. Agarose gel electrophoresis is widely used to separate DNA fragments, and it is also useful for separating RNA and other nucleic acids. The agarose gel is a matrix made up of a polysaccharide extracted from seaweed.
The matrix is formed by heating the agarose powder in a buffer solution, and then the matrix is poured into a mold and allowed to solidify. Once the gel is solidified, the samples can be loaded into the wells, and the electric current is applied. The samples are loaded into the wells using a pipette, and the buffer solution is poured into the gel chamber. The buffer solution is used to conduct the electric current, and it also helps to maintain the pH of the gel. The gel is then covered with a lid, and the current is applied using electrodes. The smaller nucleic acid molecules move more quickly through the gel matrix and are separated from the larger molecules.
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as a forensic analyst/investigator write a policy for drug cases that contain currency. answer the following in your policy
- how will this type of case be worked?
- what are the limitations
- what defines if examiners will process dollar bills or not?
1. Cases will be investigated using scientific analysis to identify drug residue and currency handling.
2. Limitations include the inability to determine the source of currency and the possibility of false positives.
3. Examiners will process dollar bills if there is reasonable suspicion of drug activity or if requested by law enforcement.
As a forensic analyst/investigator, it is important to have a clear policy for handling drug cases that contain currency.
This policy will ensure that all cases are handled in a consistent and thorough manner, and that evidence is collected and preserved in a way that is legally admissible in court.
The policy for handling drug cases that contain currency should include the following guidelines:
- All drug cases that contain currency will be handled by a team of trained forensic investigators who will follow a standardized protocol for evidence collection, analysis, and preservation.
- The team will document all evidence, including the currency, and take photographs of the scene and the evidence.
- The currency will be collected and placed in an evidence bag, labeled, and sealed. The bag will be signed by the investigator who collected it and transferred to the evidence room for storage.
- The limitations of this type of case include the potential for contamination of the currency, the difficulty of linking the currency to the drug crime, and the potential for the currency to be used as evidence in other cases.
- The team will take precautions to prevent contamination, including wearing gloves and using clean evidence bags.
- The team will also carefully document the chain of custody of the currency to ensure that it can be used as evidence in court.
- The decision to process dollar bills will be based on the potential value of the evidence and the likelihood of obtaining useful information from the bills.
- If the currency is believed to be directly related to the drug crime, or if it is believed to contain trace evidence such as fingerprints or DNA, then the examiners will process the bills.
- If the currency is not believed to be directly related to the crime, or if it is unlikely to yield useful evidence, then the examiners may choose not to process the bills.
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Please read this queation and help me to underrestand very well.
Please don't send me other expert's explanations.
in the case of double strand breaks, describe process repair for
this damage?
In the case of double-strand breaks, the process of repair for this damage is called Homologous Recombination (HR).
HR is a mechanism of DNA repair by which two DNA molecules exchange information, producing new recombinant DNA molecules that are not present in either the original or parental molecule. Homologous recombination repairs double-stranded DNA breaks, allowing the broken strand to rejoin the intact, homologous region of another chromosome or the sister chromatid. Homologous recombination involves the use of a template from a homologous chromosome to repair a breakage in the DNA.
It includes the resection of 5′ ends, strand invasion, and heteroduplex DNA formation, DNA synthesis, and resolution. The recombinational repair of DSBs can also bring about genome rearrangements, such as translocations, deletions, and inversions, in addition to restoration of the original structure. So, Homologous recombination plays a key role in maintaining genome stability by repairing double-stranded DNA breaks.
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A student is investigating the process of mitosis with a microscope. The hazard warning label on the chromosome stain they use states: may cause skin irritation, may cause eye irritation and may be harmful if swallowed
Suggest three safety precautions the student should take.
Three safety (3) precautions the student should take:
Safety Precaution 1:
The student should wear gloves to prevent skin irritation from the chromosome stain.
Safety Precaution 2:
The student should wear safety goggles to protect their eyes from irritation caused by the chromosome stain.
Safety Precaution 3:
The student should avoid eating or drinking in the lab to prevent accidentally ingesting the chromosome stain, which could be harmful.
About safety precautions in labLaboratory safety precautions help prevent or avoid accidents and advise what to do in an emergency. As the Boy Scout motto says: Be prepared.
Lists of rules and procedures are common tools for explaining laboratory safety measures. They should be taught to all employees before they start work and reviewed regularly during short refresher days such as safety moments.
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this could occur with hypoxia, intravenous infusion, ventilation, or pneumothorax (lung collapse), capillary rupture could occur; brain anoxia then occurs distal to the rupture.
It is important to understand the potential causes of capillary rupture and the effects that they can have on the body. By understanding these factors, we can better identify and treat potential issues before they lead to serious complications.
Hypoxia, intravenous infusion, ventilation, and pneumothorax (lung collapse) are all potential causes of capillary rupture. When a capillary ruptures, it can lead to brain anoxia, which is a lack of oxygen in the brain. This can occur distal to the rupture, meaning that the area of the brain that is affected is located further away from the site of the rupture.
It is important to note that each of these potential causes of capillary rupture can have different effects on the body. For example, hypoxia can lead to a decrease in oxygen levels in the blood, which can in turn lead to brain anoxia. Intravenous infusion can cause an increase in blood volume, which can put pressure on the capillaries and potentially lead to rupture. Ventilation can also affect capillary rupture by changing the pressure within the lungs, which can affect the blood flow to the brain. Finally, pneumothorax (lung collapse) can lead to a decrease in oxygen levels in the blood, which can also lead to brain anoxia.
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In Maine, a decrease in the gene pool was observed when disease killed off a large portion of the moose population. This is an illustration of
A. Natural selection
B. Genetic bottleneck
C. The Founder Effect
D. Genetic isolation
E. All of the above are correct
The situation stated in the problem illustrates genetic bottleneck. The correct answer is B. Genetic bottleneck.
A genetic bottleneck occurs when a population experiences a drastic reduction in size, leading to a decrease in genetic diversity. This can happen due to natural disasters, disease, or human activities. In the case of the moose population in Maine, the disease that killed off a large portion of the population caused a genetic bottleneck, leading to a decrease in the gene pool.
Natural selection (A) is the process by which certain traits become more or less common in a population due to their impact on an organism's ability to survive and reproduce. The Founder Effect (C) occurs when a small group of individuals establishes a new population, leading to a decrease in genetic diversity. Genetic isolation (D) occurs when a population becomes separated from other populations, preventing gene flow. While all of these processes can impact genetic diversity, the specific scenario described in the question is an example of a genetic bottleneck.
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What is the enzymatic activity of DNA polymerase.
Draw the synthesis of both strands of DNA, show the role of DNA A (ori-binding), DNA B
(Helicase), FIS, histone, HU, IHF, SSB, Gyrase (topoisomerase), Dam (Ori GATC methylase)
The enzymatic activity of DNA polymerase is the process by which a strand of DNA is replicated and/or repaired.
DNA polymerase adds nucleotides, which are the building blocks of DNA, to a single-stranded DNA template in a process known as polymerization. The synthesis of both strands of DNA involves several enzymes, including DNA A (ori-binding), DNA B (helicase), FIS, histone, HU, IHF, SSB, Gyrase (topoisomerase), and Dam (Ori GATC methylase).
DNA A binds to the origin of replication (ori) and opens up the double helix to allow access to the strands of DNA. DNA B, a helicase, then unwinds the DNA and separates the two strands of the double helix. FIS, histone, HU, IHF, and SSB all help maintain the stability of the open single-stranded DNA.
Gyrase (topoisomerase) helps relieve the torsional stress that results from the unwinding of the double helix, and Dam (Ori GATC methylase) helps protect the replication machinery from damage. Finally, DNA polymerase adds nucleotides to the template strands to synthesize the new strands of DNA.
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I have possibly an odd question that I hope can be answered. So I had to centrifuge a leaf of a plant to isolate the chloroplasts. After about 4 rounds of centrifugation, I had more pellet than supernatant (and most of the class got more supernatant which is what I needed more of). Is there reasoning on why I got more pellet than supernatant? Did I do something wrong like maybe needed more chloroplast isolation buffer or maybe I did something else wrong...?
The process of using a centrifuge to isolate chloroplasts from a plant leaf is a delicate one, and there are several factors that could potentially affect the outcome of the experiment.
It is possible that you did something wrong, such as not using enough chloroplast isolation buffer, or not properly preparing the leaf before beginning the centrifugation process. Another potential factor could be the speed at which the centrifuge was operating, as different speeds can affect the amount of pellet and supernatant that is produced.
It is also possible that there were differences in the plant leaves that were used in the experiment, such as differences in size or density, which could have affected the amount of pellet and supernatant produced. Ultimately, it is difficult to pinpoint the exact reason for the difference in results without further investigation, but these are some potential factors that could have played a role.
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What might be an environmental factor in control of gene expression via epigenetic mechanisms?
An environmental factor that can control gene expression via epigenetic mechanisms is nutrition.
Nutrition can affect epigenetic mechanisms by influencing DNA methylation and histone modifications. Epigenetic modifications, including DNA methylation and histone modifications, are influenced by environmental factors such as diet, pollution, and toxins. Nutrition is one of the main factors that influence epigenetic mechanisms. The dietary intake of nutrients can change the methylation state of DNA and histone proteins, thereby altering gene expression.
It has been shown that maternal nutrition during pregnancy can influence the epigenome of the developing fetus, leading to lifelong changes in gene expression. The amount and type of food we consume can have a profound effect on our health and well-being. In addition to providing the necessary nutrients for our cells to function properly, the food we eat can also influence the epigenetic mechanisms that control gene expression.
By understanding how nutrition affects epigenetic mechanisms, we may be able to develop new strategies for preventing and treating disease.
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Three patterns of genetic diversity were seen, a loss in variability, a gain in variability, or an insignificant change in variability. Describe how selection could lead to each of those patterns in evolution.
Selection is the process of choosing individuals that are better adapted to their environment to produce offspring. This can lead to different patterns of genetic diversity in a population.
1. Loss in variability: This can occur when there is strong selection for a particular trait or allele. If one allele or trait is strongly favored, then individuals with that trait will be more likely to survive and reproduce, leading to a decrease in genetic diversity as the favored allele becomes more common in the population.
2. Gain in variability: This can occur when there is selection for diversity, such as in the case of balancing selection. Balancing selection occurs when there are two or more alleles that are favored in different environments or situations. This can lead to an increase in genetic diversity as different alleles are favored in different situations.
3. Insignificant change in variability: This can occur when there is no strong selection for or against any particular trait or allele. In this case, the genetic diversity of the population will remain relatively constant over time.
In summary, selection can lead to different patterns of genetic diversity depending on the strength and direction of selection. Strong selection for a particular trait can lead to a loss in variability, while selection for diversity can lead to a gain in variability. If there is no strong selection, then the genetic diversity of the population will remain relatively constant.
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PLS HELP XOXO GIVING POINTS
Answer:
A BOY LYING DOWN WITH HIS EAR TO THE GROUND
What is the expected value for each cell in the contingency table for chi-square test to be effective?
In order for the chi-square test to be effective, the expected value for each cell in the contingency table should be at least 5. This is because the chi-square test assumes a normal distribution of data and if the expected value for each cell is too small, the chi-square test will not be able to accurately detect any significant associations or differences.
If the expected value for each cell is too small, the chi-square test will not be able to accurately detect any significant associations or differences. Therefore, it is important that the expected value for each cell in the contingency table is at least 5 for the chi-square test to be effective.
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if the number changes the____ changes?
Explanation:
answer or the sequence has to change
"Can you please help me answer the following questions
please.
9. _______________ is the measurement of antibodies, and other
immunological properties, in the blood serum.
Serology is the measurement of antibodies, and other immunological properties, in the blood serum.
The measurement of antibodies, and other immunological properties, in the blood serum is known as serology. Serology is an important tool in diagnosing and monitoring diseases, as it can help determine the presence of specific antibodies or antigens in a person's blood.
It is also used in the study of immune responses to vaccines and other treatments. Serology is typically performed through laboratory tests, such as enzyme-linked immunosorbent assays (ELISAs) and agglutination tests. These tests measure the number of specific antibodies or antigens in the blood, which can provide valuable information about a person's immune status and potential risk for certain diseases.
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Discuss how we might test whether a university experimental design module has been successful. For information, all students enrolled on a particular degree course have taken this module for the previous six years; previously there was no formal training in experimental design.
To test whether a university experimental design module has been successful, we could compare the performance of students on a particular degree course before and after the implementation of the module.
For example, we could compare the grades of students on experimental design assignments before and after the module was introduced. If there is a significant improvement in grades after the module was introduced, this would suggest that the module has been successful in improving students' understanding and skills in experimental design. Another way to test the success of the module would be to survey the students about their confidence and ability in experimental design before and after taking the module. If there is a significant increase in students' confidence and ability after taking the module, this would also suggest that the module has been successful.
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Trying to understand the concept of ionization . . . an electron
becomes electrically charged with another unstable atom "steal"
it's electron in an ionic bong?
The concept of ionization is when an atom loses or gains an electron, becoming an ion. This can happen when an electron is "stolen" from an atom, as in an ionic bond, creating a positively charged ion and a negatively charged ion.
Ionization is the process of adding or removing electrons from an atom or molecule, which gives it a net positive or negative charge. An ionic bond is a type of chemical bond that is formed when one atom "steals" an electron from another atom, creating two ions with opposite charges that are attracted to each other. The atom that loses an electron becomes a positive ion, and the atom that gains an electron becomes a negative ion. This creates an electrostatic attraction between the two ions, which is the basis of an ionic bond.
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What are three factors that are essential for the optimal
functional lay out of a cell culture room.
The most efficient and effective design of a cell culture chamber depends on a number of elements. Here are the three most crucial considerations:
Cell cultures must be kept clean to avoid contamination. To avoid cross-contamination, the room should have smooth, easy-to-clean surfaces and designated work areas.
Cell cultures need carefully controlled temperature and humidity. This requires a reliable HVAC system, humidifiers, and regular temperature and humidity monitoring.
Cell culture rooms should be organized for efficient workflow and easy access to equipment and supplies. This means there should be designated areas for different types of work, and the layout should be designed with researchers in mind.
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What are the steps of DNA replication? Include all relevant
enzymes within your answer.
The steps of DNA replication are initiation, primer binding, elongation, termination, and proofreading.
1. Initiation: The DNA double helix is unwound by the enzyme helicase, creating a replication fork.
2. Primer binding: The enzyme primase creates a short RNA primer that is complementary to the DNA strand, allowing DNA polymerase to begin adding nucleotides.
3. Elongation: DNA polymerase adds nucleotides to the 3' end of the primer, creating a new strand of DNA. The leading strand is synthesized continuously, while the lagging strand is synthesized in short fragments called Okazaki fragments.
4. Termination: The RNA primers are removed by the enzyme RNase H and replaced with DNA by DNA polymerase. The enzyme ligase then seals the gaps between the Okazaki fragments, creating a continuous DNA strand.
5. Proofreading: DNA polymerase checks for any errors and corrects them to ensure accurate replication.
These steps are carried out by a complex of enzymes and proteins called the replisome, which includes helicase, primase, DNA polymerase, RNase H, and ligase.
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If a plant were growing in an atmosphere with abundant CO2 but with only25%
of the sunlight that it has evolved to use, then how would a sudden doubling of available sunlight impact the Calvin cycle? a. The Calvin cycle would produce less sugar and G3P b. The Calvin cycle would energize fewer electron acceptors c. The Calvin cycle would produce more sugar and G3P d. The Calvin cycle would be unaffected e. The Calvin cycle would energize more electron acceptors
If a plant were growing in an atmosphere with abundant CO₂ but with only 25% of the sunlight that it has evolved to use, then a sudden doubling of available sunlight would impact the Calvin cycle in the following way: The Calvin cycle would produce more sugar and G3P. Therefore, the correct answer is C.
The Calvin cycle is a process that occurs in the chloroplasts of plants and is responsible for converting carbon dioxide into sugar and G3P (glyceraldehyde 3-phosphate). The Calvin cycle requires energy in the form of ATP and NADPH, which are produced by the light-dependent reactions of photosynthesis.
If a plant is receiving less sunlight than it has evolved to use, then the light-dependent reactions will produce less ATP and NADPH, and the Calvin cycle will be limited in its ability to produce sugar and G3P. However, if the available sunlight suddenly doubles, then the light-dependent reactions will produce more ATP and NADPH, and the Calvin cycle will be able to produce more sugar and G3P.
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Consider two populations of bacteria. One lives in a stagnant
the other in your armpit. What are some differences that will
affect the bacteria and how they adapt?
Some of the main differences in the media that will impact the batteries and their adaptation are:
The amount of oxygen in each mediumHumidity and temperature levelsThe two populations of bacteria will have different adaptations based on the environments they live in. See:
Bacteria living in a stagnant environment, such as a pond or a pool of water, will have adaptations to help them survive in low oxygen conditions. This may include the ability to use alternative forms of respiration or the ability to form biofilms to protect themselves from harsh conditions.Bacteria living in your armpit will have adaptations to help them survive in a warm, moist environment. This may include the ability to tolerate higher temperatures, the ability to use sweat and oils on the skin as a source of nutrients, and the ability to resist the body's immune system.Overall, the differences in these two environments will lead to different adaptations and survival strategies for the bacteria populations.
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i've had a stubborn cough for weeks that just won't go away. The doctor takes a throat sample and results come back all blue when viewed in the microscope during an acid-fast stain. is this good or bad news? explain uour answer
This is bad news. The acid-fast stain is a test done to check for the presence of bacteria that are resistant to antibiotics. These bacteria, called acid-fast bacilli, which can cause illnesses such as tuberculosis, leprosy, and other infections.
When the sample is viewed under a microscope, the bacteria will appear blue. This indicates that the bacteria are present and that the infection is resistant to antibiotics.
Treatment for these infections can be difficult and time consuming, often requiring long courses of antibiotics and other medications. It is important to seek medical advice as soon as possible for treatment and to prevent the spread of the infection.
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explain obesity problem and how
to solve it.
tell us more about the
problem
what will solve the problem
what supports the solution
Obesity is a medical condition in which excess body fat accumulates to the point of negative health impacts. It is typically caused by an unhealthy diet and lack of physical activity. To solve the problem of obesity, a combination of healthy diet, regular physical activity, and lifestyle changes is needed.
A healthy diet includes eating whole grains, lean proteins, fruits, vegetables, and healthy fats. Regular physical activity can include any type of physical activity such as walking, running, cycling, or swimming. Making lifestyle changes such as reducing stress levels and getting adequate sleep can also help.
Support for these solutions come from studies showing that these measures can help reduce and even reverse obesity. For example, the Centers for Disease Control and Prevention (CDC) recommends an individualized approach to losing weight that includes a combination of healthy eating, physical activity, and lifestyle changes.
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Neuron X has a glutamatergic synapse one length constant (the distance it takes for a signal to decay to 37% - or 1/e in size) from the AIS. The resting potential of the neuron is -70 mV, and threshold is -50 mV (to simplify this problem, just assume threshold is fixed). When activated, the synapse produces an EPSP (Excitatory PostSynaptic Potential) with an initial amplitude of 35 mV that decays with a time constant of 10 ms (meaning in 10 ms the signal will be 37% or 1/e of its original size).
A. If the presynaptic neuron innervating this synapse starts to fire at 100 Hz (once every 10 ms), how many stimulations will it take to make Neuron X fire an action potential? Assume there is no time delay between EPSP initiation at the synapse and the signal reaching the AIS. First figure out how big would the depolarization at the synapse have to be to make Neuron X fire. Then try to figure out how many EPSPs coming 10 ms apart would give you more than that level of depolarization.
B. If you decreased the presynaptic firing rate to 10 Hz, do you think the synapse could make Neuron X fire?
C. Let’s go back to 100 Hz stimulation of the synapse. If you doubled the size of Neuron X, but kept the input resistance the same, would it take more or fewer synapse activations to make the Neuron X fire an action potential? Answer qualitatively and explain your reasoning.
It would take 8 stimulations of the synapse at 100 Hz for Neuron X to fire an action potential. To figure this out, first you need to determine the threshold depolarization that Neuron X needs to reach to fire an action potential. Since the resting potential is -70 mV and the threshold is -50 mV, the neuron needs a depolarization of 20 mV to reach threshold. The EPSP initial amplitude is 35 mV, so it would take 8 stimulations of the synapse at 100 Hz, with no time delay, for the depolarization to reach 20 mV and fire an action potential.
If you doubled the size of Neuron X while keeping the input resistance the same, it would take fewer synapse activations to make Neuron X fire an action potential. This is because the threshold depolarization to reach firing remains the same, but the larger neuron would be more sensitive to the EPSPs coming from the synapse. Therefore, fewer stimulations of the synapse would be required to reach a threshold.
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Fill out 2 the empty columns based off of the rest of the table, please show work on how it was done.
Hint: activity assayed = convert the absorbance value into activity units measured in the assay (mmoles PNP produced per min)
total activity = convert the activity assayed (column 6) to total enzyme activity in the entire 2 ml fraction collected.
Fraction #
A410nm /min
e (M-1)
Assay volume (ml)
Activity assayed
mmol PNP /min
Vol added to assay (ml)
Total Fraction Volume (ml)
Total Activity
7
y = 0.1514x + 0.1018
18300
3.1
0.1
2
8
y = 0.0335x + 0.0291
18300
3.1
0.1
2
9
y = 0.0069x - 0.0005
18300
3.1
0.1
2
Given the table Fraction #A410nm /min (−1)Assay volume (ml)Activity assayed /Vol added to assay (ml)Total Fraction Volume (ml)Total Activity7
100.1514+0.1018183003.10.120.128.11188.116 800.0335+0.0291183003.10.060.128.1164.180 900.0069−0.0005183003.10.020.128.1164.174
The first empty column is the Vol added to assay (ml), and the second empty column is the Total Activity. =Fraction volume x 0.1
For example, in fraction 7: =28.1 x 0.1=2.81.We will repeat this calculation for each fraction to find the Vol added to assay (ml). Total Activity=Activity Assayed (mmol PNP/min) x Total Fraction Volume (ml)
For example, in fraction 7:Total Activity=0.12 x 28.1=3.372
We will repeat this calculation for each fraction to find the Total Activity. The final table after filling the two empty columns looks like Fraction #A410nm /min (−1)Assay volume (ml)Activity assayed /Vol added to assay (ml)Total Fraction Volume (ml)Total Activity7100.1514+0.1018183003.10.122.813.3728.116 800.0335+0.0291183003.10.061.622.9454.180 900.0069−0.0005183003.10.020.582.9876.174
Therefore, Vol added to assay (ml) and Total Activity are filled based on the formulae described above.
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if 9% of the population exhibit signs and symptoms of the disease what percent of the population is a carrier
Answer:
Without additional information about the disease and its mode of inheritance, it is not possible to determine the percentage of the population that is a carrier. The carrier frequency is dependent on the specific genetics of the disease.
Explanation:
1. Increased expression of RNA methyltransferase BCDIN3D will result in
A. Increased expression of miR-21
B. Increased expression of miR-145
C. Decreased expression of miR-145
D. Decreased expression of miR-21
Increased expression of RNA methyltransferase BCDIN3D will result in C. Decreased expression of miR-145.
This is because BCDIN3D is a negative regulator of miR-145 expression. Therefore, when there is increased expression of BCDIN3D, there will be a decrease in the expression of miR-145.
This is because BCDIN3D is a negative regulator of miR-145 expression. When there is increased expression of BCDIN3D, the expression of miR-145 is decreased, which can be seen in studies that have examined this phenomenon.
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What are the physiological adaptations of a shark?
Some of these adaptations includ Streamlined body , Strong jaws and teeth , Electroreceptors , Lateral line system , Efficient respiratory system , Buoyancy control.
Sharks have several physiological adaptations that allow them to survive and thrive in their aquatic environment.
1. Streamlined body: Sharks have a streamlined body that allows them to swim efficiently and quickly through the water.
2. Strong jaws and teeth: Sharks have powerful jaws and sharp teeth that allow them to catch and eat their prey.
3. Electroreceptors: Sharks have electroreceptors on their snouts that allow them to detect the electrical fields of their prey.
4. Lateral line system: Sharks have a lateral line system that allows them to detect changes in water pressure and movement in the water.
5. Efficient respiratory system: Sharks have an efficient respiratory system that allows them to extract oxygen from the water and breathe efficiently.
6. Buoyancy control: Sharks have a large liver that produces an oily substance called squalene, which helps them maintain buoyancy in the water.
These physiological adaptations allow sharks to be efficient hunters and survive in their aquatic environment.
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