VETERINARY SCIENCE!!!
Why do you think colic is difficult to diagnose correctly?

What consequences would
there be if the problem wasn't diagnosed quickly and correctly?

a) To determine Vmax and KM, we can plot the initial velocity (V0) against substrate concentration ([S]) using the data provided for each trial. From the graph, the x-intercept will give us -1/KM, while the y-intercept will give us Vmax.

Using this method, we can obtain the following values:
- Trial A (no inhibitor): Vmax = 50 µmol/min and KM = 0.5 mM
- Trial B (5 mM inhibitor B): Vmax = 20 µmol/min and KM = 1 mM
- Trial C (5 mM inhibitor C): Vmax = 10 µmol/min and KM = 2 mM
b) To determine the type of inhibition, we can compare the values of Vmax and KM for each trial. We can see that the Vmax decreases as inhibitor concentration increases, while the KM increases. This suggests that both inhibitors are non-competitive. We can confirm this by calculating the inhibition constant (K­I) for each inhibitor using the following equation:
K­I = [I] / (slope x Vmax)
- For inhibitor B: K­I = 1.25 mM
- For inhibitor C: K­I = 0.625 mM
Since both K­I values are different from the substrate concentration, we can conclude that the inhibitors are non-competitive.
c) Based on the kinetic data, we can infer that the inhibitors have similar structures to the substrate, as they bind to the enzyme and affect its activity. This suggests that the inhibitors have a similar shape and size to the substrate, allowing them to fit into the active site of the enzyme. This could be due to the presence of functional groups or chemical bonds that are similar to those found in the substrate. However, without further information about the structure of the enzyme and the inhibitors, it is difficult to determine the exact structural similarities between them. Overall, the data suggests that inhibitors B and C are able to inhibit the enzyme by binding to the active site and interfering with substrate binding and/or catalysis.

To know more about concentration click here:

brainly.com/question/30862330

#SPJ11

Botulism is caused by the ingestion of a proteinaceous exotoxin, which is produced by the bacterium Clostridium botulinum. This bacterium can grow in improperly prepared or preserved foods, such as canned foods or foods that are not cooked thoroughly.

Boiling food for at least 10 minutes can kill the spores of the bacterium, preventing the growth of Clostridium botulinum and reducing the risk of botulism. It is important to note that simply heating or reheating food is not enough to kill the spores; boiling is necessary.

Other measures that can be taken to prevent botulism include avoiding canned or preserved foods that appear to be damaged or bulging, ensuring that home-canned foods are properly prepared and preserved, and preventing fecal contamination of food by washing hands thoroughly and properly storing food.

In terms of administering antibiotics to patients with botulism, this is not recommended as it can actually worsen the symptoms by releasing more toxins into the body. Filtering food may remove some of the toxin, but it does not guarantee that all of the toxins is removed. Therefore, boiling food prior to consumption remains the most effective method of preventing botulism.

In summary, botulism can easily be prevented by boiling food prior to consumption. This is the most effective method of killing the spores of the bacterium Clostridium botulinum, which produces the toxin that causes botulism.

To know more about botulinum visit:

brainly.com/question/15088458

#SPJ11

Explanation:

Between the alternativies the correct steps that must be take in preparation for tropical storms and hurricanes is: before hurricane season determine safe evacuation route. Being all other options not effective for the situation that the question is talking about.

2.The statement "If one microbe has a larger zone of inhibition than another microbe that means the microbe with the larger zone of inhibition should always be considered antibiotic resistant" is false.

3.The statement "If you swab a surface and nothing grows after a 48-hour incubation it is safe to conclude there are no infectious agents on that surface" is false.

4.The statements a,c,d are true.

a. An orange phenol red broth tube indicates the microbe fermented the sugar.

b. The organism did not ferment the sugar The result is negative the result is positive.

d. The environment was slightly basic.

The sample procedure, the type of microbe being tested, and the testing's intended use are all significant elements to take into account when interpreting the results of microbial culture.

To improve the sensitivity and specificity of microbiological identification, it is advised to do several tests and employ complementary techniques.

Learn more about Inhibition

https://brainly.com/question/29744524

#SPJ4

Complete question

2. If one microbe has a larger zone of inhibition than another microbe that means the microbe with the larger zone of inhibition should always be considered antibiotic resistant. True False

3. If you swab a surface and nothing grows after a 48-hour incubation it is safe to conclude there are no infectious agents on that surface. True False

4. Choose all that are true

a. An orange phenol red broth tube indicates the microbe fermented the sugar.

b. The organism did not ferment the sugar The result is negative The result is positive.

c. The organism did not grow in the media.

d. The environment was slightly basic.

Fructose-6-phosphate is one of the molecules that can increase the activity of the glycolytic enzyme phosphofructokinase-1.

phosphofructokinase-1 is an important regulatory enzyme in glycolysis and catalyzes the transfer of a phosphate group from ATP to fructose-6-phosphate to form fructose-1,6-bisphosphate. This reaction is the rate-limiting step of glycolysis, and therefore phosphofructokinase-1 activity is a major determinant of glycolytic flux.

Fructose-6-phosphate binds to the active site of phosphofructokinase-1, resulting in increased enzyme activity. It also increases the affinity for ATP, allowing for more efficient phosphorylation of fructose-6-phosphate.

Fructose-1,6-bisphosphate and fructose-2,5-bisphosphate can also increase the activity of phosphofructokinase-1, however, the binding of fructose-6-phosphate is the most efficient and potent activator of the enzyme. Therefore, fructose-6-phosphate is the molecule that increases the activity of the glycolytic enzyme phosphofructokinase-1.

Know more about Fructose-6-phosphate here

https://brainly.com/question/16094787#

#SPJ11

Homo naledi had a unique shoulder structure and curved finger bones. These features may indicate that they lived in the trees.

Evolution of human beings:

The discovery of homo naledi, a previously unknown species of human being, sheds light on the evolution of our ancestors. While their unique shoulder structure and curved finger bones suggest that they were adapted to climbing, it is not necessarily an indication that they lived in a specific environment such as the water, trees, or desert. It is possible that they lived in groups in a variety of environments, and their physical adaptations allowed them to thrive in those environments.

The study of homo naledi and other early hominids helps us better understand the evolutionary history of our species. The curved fingers would have allowed them to grasp branches more easily, while the unique shoulder structure may have provided more mobility and flexibility for climbing and swinging through trees. This evidence suggests that Homo naledi was well-adapted to arboreal living, which is an important aspect of human evolution.

To know more about the evolution of human beings, visit:

https://brainly.com/question/30590430

#SPJ11

Answer:

two sets of heart valves closing

Answer:

The answer is number 1.

Explanation:

Hope this helps!!

Answer:B

Explanation: just do it

Identical twins would have identical numbers of STRs (Short Tandem Repeats) at all 13 locations since they are genetically identical.

Identical twins arise from a single fertilized egg that splits into two embryos during early development. As a result, they share the same DNA sequence, including the number of Short Tandem Repeats (STRs) at all 13 locations.

STRs are short repeating DNA sequences that vary in length between individuals and are used in DNA profiling to distinguish one individual from another. Because identical twins have identical DNA, they have the same number of repeats at each of these 13 STR locations.

Learn more about the Identical twins at

https://brainly.com/question/20417246

#SPJ4

Hypothesis is an assumption which is made before any research has been done. hypothesis is formed so that it can be tested to see if it is true.

A character which has been lost in a breed and it reappears after a great number of generations, the most probable hypothesis is not that the offsprings are suddenly takes after an ancestor but some hundred generations distant, but that in each successive generation there is some tendency to reproduce the character in question, which at last under unknown favourable conditions, gains an ascendancy

The most common type of virus leading to rhinitis is rhinovirus.

Rhinovirus is a type of virus that causes the common cold and is highly contagious. It spreads through the air or by touching contaminated surfaces. Rhinitis is the inflammation of the nasal passages, and rhinovirus is one of the primary causes of this condition. The symptoms of rhinitis include a runny nose, sneezing, congestion, and itching. Rhinovirus is typically self-limiting and can be treated with over-the-counter medications to relieve symptoms.

When an infected person coughs or sneezes, respiratory droplets are released, which are how rhinoviruses are transmitted. If someone breathes in these droplets or touches a surface that has been exposed to the virus before touching their eyes, nose, or mouth, the virus can then enter that person's body. Rhinoviruses can also be transmitted through having close physical contact with an infected individual, such as shaking hands or embracing.

To know more about rhinovirus click here:

https://brainly.com/question/28878147

#SPJ11

The zone where most biological productivity of the ocean occurs is called euphotic zone. This zone extends from the ocean's surface to a depth of about 200 meters.

It is characterized by high levels of sunlight, which provides the energy needed for photosynthesis. As a result, there is a high concentration of phytoplankton in this zone, which are the foundation of the ocean's food chain. The euphotic zone is the most important zone for commercial fishing, as it is where most fish and other marine organisms thrive. Overall, the ocean's productivity is essential to the health of our planet, and understanding the different zones and the processes that occur in them is crucial to ensuring their sustainability.

Learn more about phytoplankton here:

brainly.com/question/30933343

#SPJ11

The aim of antiallergy medication is to block the progress of the allergic response between IgE production and the appearance of symptoms. Avoidance of the allergen is crucial for sensitive individuals. Oral anti-inflammatory drugs, such as corticosteroids, can reduce IgE production. IgE can be inactivated by a monoclonal antibody to prevent an allergic response. If IgE persists, Cromolyn can prevent degranulation of mast cells. If degranulation occurs, antihistamines can counteract inflammatory cytokines' effects on target cells.

To review strategies for circumventing allergy attacks, first, avoid the allergen if you're sensitive. Then, oral anti-inflammatory drugs like corticosteroids can reduce the production of IgE from antibody-secreting cells.

In case IgE forms, a monoclonal antibody can inactivate it, preventing an allergic response. If IgE still persists, Cromolyn can be used to prevent degranulation of mast cells, averting the allergic response.

Lastly, if degranulation does happen, antihistamines can be used to counteract the effects of inflammatory cytokines on target cells, reducing allergy symptoms.

To know more about allergy click on below link:

https://brainly.com/question/12893949#

#SPJ11

To find the function for the log model that gives the body weight (g) of a mouse in grams after it is t weeks old, we can use the following formula:
g(t) = a * log10(b * t + 1)

In this function:
g(t) represents the body weight of the mouse in grams at time t (weeks old)
a and b are constants that depend on the specific growth pattern of mice
log10 is the logarithm base 10
To find the exact values for a and b, you would need specific data points of mouse growth. Once you have that data, you can use regression analysis to determine the values of a and b that best fit the data, giving you the final function.

to know more  growth please vist :-

https://brainly.com/question/15133607

#SPJ11

To find the function for the log model that gives the body weight (g) of a mouse in grams after it is t weeks old, we can use the following formula:
g(t) = a * log10(b * t + 1)

In this function:
g(t) represents the body weight of the mouse in grams at time t (weeks old)
a and b are constants that depend on the specific growth pattern of mice
log10 is the logarithm base 10
To find the exact values for a and b, you would need specific data points of mouse growth. Once you have that data, you can use regression analysis to determine the values of a and b that best fit the data, giving you the final function.

to know more  growth please vist :-

https://brainly.com/question/15133607

#SPJ11

Plasmodium cells come from Anopheles mosquitoes that bite humans, then develop into parasites. Plasmodium and Trypanosoma have in common that they location stay in the liver. This has a big affected on red blood cells. Where these two parasites eat red blood cells which causes the body to lack oxygen.

Plasmodium cells are one of the parasites that cause malaria. This disease is widely developed in tropical areas. Meanwhile, trypanosoma cells also include parasites originating from the tsetse fly and can cause Chagas disease. This disease is common in Africa. Both of these parasites belong to a group of protozoa that attack red blood cells. They live in liver cells and then start eating red blood cells until the body can't circulate oxygen throughout the body. Thus resulting in death for the sufferer.

Learn more about Plasmodium cells: https://brainly.com/question/8984426

#SPJ11

If the concentration of the calf liver DNA preparation is 50 ng/ml, then we can calculate the amount of DNA in a certain volume by using the formula:

amount of DNA (ng) = concentration (ng/ml) x volume (ml)

To find the volume needed to obtain between 50 and 100 ng of DNA, we can rearrange the formula:

volume (ml) = amount of DNA (ng) / concentration (ng/ml)

For 50 ng of DNA, the volume required would be:

volume = 50 ng / 50 ng/ml = 1 ml

Since we need to use μl instead of ml, we can convert 1 ml to μl by multiplying it by 1000:

volume = 1 ml x 1000 = 1000 μl

Therefore, we need to use 1000 μl (or 1 ml) of calf liver DNA preparation to obtain 50 ng of DNA.

For 100 ng of DNA, the volume required would be:

volume = 100 ng / 50 ng/ml = 2 ml

Converting 2 ml to μl:

volume = 2 ml x 1000 = 2000 μl

Therefore, we need to use 2000 μl (or 2 ml) of calf liver DNA preparation to obtain 100 ng of DNA.

Therefore, if you need a mass of DNA between 50 and 100 ng in order to have successful electrophoresis, you should use between 1000 and 2000 μl (or 1 and 2 ml) of your calf liver DNA preparation.

To know more about DNA, click here:-

https://brainly.com/question/264225

#SPJ11

Contraction of the superficial muscles in the gluteal region results in extension of the thigh.

The gluteal region is the area of the body that includes the buttocks, and it contains several muscles that are involved in movement of the hip joint. The three major muscles in this region are the gluteus maximus, gluteus medius, and gluteus minimus.

When the superficial muscles in the gluteal region contract, they produce extension of the thigh, meaning that the thigh moves backward relative to the hip joint. This movement is important for activities like walking, running, and climbing stairs. The gluteal muscles also play a role in other movements, such as abduction (moving the thigh away from the midline of the body) and rotation (turning the thigh inward or outward), but extension is their primary action.

To know more about the hip joint

brainly.com/question/13687028

#SPJ11

The beardless trait is dominant, and the bearded trait is recessive. Let's represent the beardless trait as "B" and the bearded trait as "b."

The P1 generation can be represented as:

Bearded female (bb) x Beardless male (BB)

Since the female goat is bearded, we know she must be homozygous recessive (bb). And since all the offspring of the F1 generation are bearded males and beardless females, we can deduce that the male goat must be homozygous dominant (BB). Therefore, the F1 generation can be represented as:

Bearded male (Bb) x Beardless female (BB)

When we cross the F1 individuals, we get the following Punnett square:

   | B  | b

----|----|----

B   | BB | Bb

----|----|----

b   | Bb | bb

The genotypic ratio of the F2 offspring is 1 BB : 2 Bb : 1 bb. The phenotypic ratio of the F2 offspring is 3 bearded: 3 beardless: 1 bearded female.

To test the hypothesis that beardless is dominant and bearded is recessive, we can perform a test cross between an F1 individual and a homozygous recessive individual. Let's use an F1 male goat and a bearded female goat as an example. The cross can be represented as:

Bearded male (Bb) x Bearded female (bb)

The Punnett square for this cross would be:

   | B  | b

----|----|----

b   | Bb | bb

The expected genotypic ratio of the offspring would be 1 Bb : 1 bb, and the expected phenotypic ratio would be 1 bearded: 1 beardless.

If we observe the expected ratios in the offspring, it would support our hypothesis that the beardless trait is dominant and the bearded trait is recessive.

To know more about homozygous

brainly.com/question/30622664

#SPJ11

Eukaryotic DNA replication is a complex and tightly regulated process that involves multiple origins of replication, bidirectional replication forks, and careful coordination with the cell cycle checkpoints and other cellular processes to ensure timely and synchronous replication of the large eukaryotic genome.

Eukaryotic DNA replication initiates from multiple origins of replication along the DNA molecule. These origins of replication are specific DNA sequences that are recognized by a complex of proteins called the pre-replication complex (pre-RC), which assembles at these sites prior to the start of DNA replication. The pre-RC includes proteins such as the origin recognition complex (ORC), Cdc6, and Cdt1, among others.

Once the pre-RC is assembled, it recruits additional proteins, including DNA helicases, which unwind the DNA double helix to create a replication fork. The replication fork is the point where DNA replication actually occurs, with the two parental DNA strands being separated and new daughter strands being synthesized by DNA polymerases.

Eukaryotic DNA replication is a bidirectional process, with replication forks moving in opposite directions from the origins of replication along the DNA molecule. This allows for efficient and timely replication of the large eukaryotic genome. The replication forks move along the DNA strands, unwinding the DNA ahead of them and synthesizing new DNA strands behind them.

Importantly, eukaryotic DNA replication is tightly regulated to ensure that it occurs in a coordinated and synchronous manner. The initiation of DNA replication is controlled by a series of cell cycle checkpoints, which ensure that DNA replication occurs only once per cell cycle and is completed before the cell enters mitosis.

These checkpoints monitor the progress of DNA replication and prevent the firing of new origins of replication until ongoing replication is completed.

In addition, eukaryotic DNA replication is facilitated by multiple proteins that help to stabilize and restart stalled replication forks, repair DNA damage, and coordinate the replication process with other cellular processes, such as transcription and chromatin remodeling.

Learn more about DNA here

https://brainly.com/question/31454823

#SPJ11

A) Ensures a renewable resource stays renewable.

Low flow shower heads, toilets, and faucets reduce the amount of water being used, which can help to conserve water resources. This is particularly important in regions where water is scarce or where there is high demand for water due to population growth or other factors.

By reducing water consumption, low flow fixtures help to ensure that water resources remain renewable and sustainable for future generations. This is especially important as global population growth and climate change put increasing pressure on water resources.

Learn more about water consumption here : brainly.com/question/31051878

#SPJ1

The pathway of blood from the left ventricle of the heart to a cell in the left flexor carpi radialis muscle and back to the right atrium of the heart begins with the left ventricle.

From there, the blood travels through the aortic arch and eventually reaches the capillary bed in the left flexor carpi radialis muscle. Once in the capillary bed, the blood exchanges nutrients and gases with the surrounding tissues and then makes its way back to the heart through the venous system.

The blood returns to the heart through the venous system and enters the right atrium. From the right atrium, the blood is then pumped into the right ventricle, which completes the circuit of blood flow through the heart and body.

Learn more about pathway here:

https://brainly.com/question/20308520

#SPJ11

If two genes are on the same chromosome, there will be complete linkage of the genes in the gametes. In crossing over, crossing over will produce both parental and crossover gametes.

When two genes are located on the same chromosome, they are said to be linked. This means that they tend to be inherited together, rather than assorting independently during meiosis.

The degree of linkage between two genes depends on the distance between them on the chromosome.

During meiosis, crossing over can occur between homologous chromosomes, which can result in the exchange of genetic material between the chromosomes.

When crossing over occurs between two linked genes, it can produce parental gametes (which contain the original combinations of alleles) or crossover gametes (which contain recombined combinations of alleles). The frequency of parental and crossover gametes depends on the degree of linkage between the genes.

Complete linkage is a special case of linkage in which two genes are so close together on a chromosome that they always remain together during meiosis and are always inherited together. In this case, crossing over does not occur between the two genes, and only parental gametes are produced.

Overall, the degree of linkage between two genes on the same chromosome affects the frequency of parental and crossover gametes that are produced during meiosis, and crossing over can result in recombined combinations of alleles.

Learn more about crossing over here:

https://brainly.com/question/22173626

#SPJ11

Answer:

the trachea widens and lengthens

Answer:

The answer is the trachea widens and lengthens.

Explanation:

Hope this helps!!

The best explanation for why applying constant pressure to an edematous part of the body reduces swelling is The applied pressure adds to the pressure generated by the protein concentration gradient across capillary walls.(B)

When an area of the body is swollen, fluid accumulates in the interstitial spaces due to an imbalance in the forces that regulate fluid movement across capillary walls. The main forces at play are the hydrostatic pressure within the capillaries and the oncotic pressure created by the protein concentration gradient.

By applying constant external pressure to the swollen area, you are effectively increasing the hydrostatic pressure outside the capillaries. This added pressure helps to counterbalance the forces causing fluid to leave the capillaries, promoting fluid movement back into the capillaries and reducing the edema.

Essentially, the applied pressure aids in reestablishing the balance between the hydrostatic and oncotic pressures, which helps to alleviate the swelling.(B)

To know more about hydrostatic pressure click on below link:

https://brainly.com/question/28206120#

#SPJ11

Answer:LOOK

Explanation:LOOK

Based on the given information, the most likely mechanism of action for Wambe is that it is an M-d (muscarinic-d) receptor antagonist.

Muscarinic receptors are a type of acetylcholine receptor found in smooth muscle, including the smooth muscle in the digestive system. Antagonists of muscarinic receptors block the action of acetylcholine, which is a neurotransmitter that can stimulate smooth muscle contraction.

By inhibiting the M-d receptors, Wambe would decrease the contraction rate of smooth muscle in the digestive system, thereby stopping stomach muscle cramps.

The other options, including being an M-h (muscarinic-h) receptor agonist, M-d receptor agonist, or an alpha-1 receptor antagonist, are less likely based on the given information. M-h receptor agonists would increase smooth muscle contraction, which is contradictory to the desired effect of Wambe.

M-d receptor agonists would also increase smooth muscle contraction, which is not consistent with Wambe's reported function of decreasing the contraction rate.

Alpha-1 receptor antagonists, on the other hand, target a different type of receptor (alpha-1 adrenergic receptors) and are typically used for different purposes, such as treating conditions like hypertension and benign prostatic hyperplasia, and are not typically associated with smooth muscle contraction in the digestive system.

Learn more on receptors here

https://brainly.com/question/11985070

#SPJ11

The amount of caffeine in Francisco's body 2.42 hours after consuming the energy drink is 75.5 mg.

Assuming that the rate of caffeine metabolism in Francisco's body is constant, we can use the formula:

C = Co × [tex]e^{(-kt)}[/tex]

where C is the amount of caffeine in Francisco's body at time t, Co is the initial amount of caffeine, k is the rate constant for caffeine metabolism, and t is the time elapsed since consuming the energy drink.

We can solve for k by using the information given:

158 = Co × [tex]e^{(-k*1.42)}[/tex]

Co = 158 / [tex]e^{(-k*1.42)}[/tex]

We can then use the value of Co to find the amount of caffeine in Francisco's body at 2.42 hours:

C = Co × [tex]e^{(-kt)}[/tex] = (158 / [tex]e^{(-k*1.42)}[/tex]) * [tex]e^{(-k*2.42)}[/tex]

C = 158 × [tex]e^{(-k*t)}[/tex]

≈ 75.5 mg

Therefore, there would be approximately 75.5 mg of caffeine in Francisco's body 2.42 hours after consuming the energy drink.

Learn more about caffeine: https://brainly.com/question/9255669

#SPJ11

The amount of caffeine in Francisco's body 2.42 hours after consuming the energy drink is 75.5 mg.

Assuming that the rate of caffeine metabolism in Francisco's body is constant, we can use the formula:

C = Co × [tex]e^{(-kt)}[/tex]

where C is the amount of caffeine in Francisco's body at time t, Co is the initial amount of caffeine, k is the rate constant for caffeine metabolism, and t is the time elapsed since consuming the energy drink.

We can solve for k by using the information given:

158 = Co × [tex]e^{(-k*1.42)}[/tex]

Co = 158 / [tex]e^{(-k*1.42)}[/tex]

We can then use the value of Co to find the amount of caffeine in Francisco's body at 2.42 hours:

C = Co × [tex]e^{(-kt)}[/tex] = (158 / [tex]e^{(-k*1.42)}[/tex]) * [tex]e^{(-k*2.42)}[/tex]

C = 158 × [tex]e^{(-k*t)}[/tex]

≈ 75.5 mg

Therefore, there would be approximately 75.5 mg of caffeine in Francisco's body 2.42 hours after consuming the energy drink.

Learn more about caffeine: https://brainly.com/question/9255669

#SPJ11

The cyclic light reaction is a process that occurs during the light-dependent stage of photosynthesis. It involves the flow of electrons through a series of electron carriers in the thylakoid membrane of chloroplasts.

Here's a step-by-step description of the process:

1. Light energy is absorbed by photosystem I (PSI) in the thylakoid membrane.

2. This energy excites an electron in the reaction center of PSI, causing it to jump to a higher energy level and leave the chlorophyll molecule.

3. The electron is captured by the first electron carrier in the electron transport chain (ETC), called ferredoxin (Fd).

4. The electron is passed along a series of carriers in the ETC, including cytochrome b6f and plastocyanin (PC), before it returns to the reaction center of PSI.

5. As the electron moves through the ETC, it releases energy that is used to pump protons (H+) from the stroma into the thylakoid lumen, creating a proton gradient.

6. The proton gradient is used to power ATP synthase, which produces ATP from ADP and inorganic phosphate.

7. The electron is returned to PSI, where it can be excited again by another photon of light and continue to cycle through the ETC.

Overall, the cyclic light reaction generates ATP through the action of ATP synthase, but does not produce any NADPH or oxygen. It is called "cyclic" because the electron is returned to PSI and can cycle through the ETC multiple times, rather than being passed on to photosystem II as in the non-cyclic electron flow.

To learn more about photosynthesis click here

brainly.com/question/29775046

#SPJ11

Both the exons and introns of a eukaryotic gene are transcribed. However, during RNA processing, the introns are removed from the pre-mRNA transcript before it becomes mature mRNA.

mRNA stands for messenger RiboNucleic Acid and is the single-stranded molecule that carries the instructions to make proteins. It has a fundamental and essential role that makes our bodies function and is found in all living cells.

This is done through a process called splicing, which results in the mature mRNA only containing the exons. Therefore, the final protein product is only composed of the exons of the gene. The promoter and terminator sequences are also transcribed as they are essential for regulating the transcription process.

To know more about exons and introns please click:-

https://brainly.com/question/31494556

#SPJ11

Both the exons and introns of a eukaryotic gene are transcribed. However, during RNA processing, the introns are removed from the pre-mRNA transcript before it becomes mature mRNA.

mRNA stands for messenger RiboNucleic Acid and is the single-stranded molecule that carries the instructions to make proteins. It has a fundamental and essential role that makes our bodies function and is found in all living cells.

This is done through a process called splicing, which results in the mature mRNA only containing the exons. Therefore, the final protein product is only composed of the exons of the gene. The promoter and terminator sequences are also transcribed as they are essential for regulating the transcription process.

To know more about exons and introns please click:-

https://brainly.com/question/31494556

#SPJ11