It appears that the patient experienced a delayed transfusion reaction after receiving 2 units of red blood cells. The original pretransfusion antibody screening results showed no agglutination, except when IgG-sensitized cells were added. Upon repeat testing, an antibody was detected at the antiglobulin phase.
The most likely explanation for the original results is a false-negative due to low sensitivity of the pretransfusion antibody screening test. This could be attributed to factors such as a low concentration of the clinically significant antibody, weak agglutination reactions, or testing conditions not being optimal for antibody detection. The addition of IgG-sensitized cells possibly enhanced the antibody detection by increasing the sensitivity of the test, leading to the detection of the antibody in the repeat testing.
In summary, the initial antibody screening test may not have been sensitive enough to detect the antibody present in the patient's blood, causing a false-negative result. The repeat testing, with the use of IgG-sensitized cells, increased the test sensitivity and allowed for the detection of the antibody, which was responsible for the delayed transfusion reaction experienced by the patient.
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in labs, b codes for black coat color and is completely dominant to b, which codes for chocolate color. e codes for pigment production and is completely dominant to e, which codes for no pigment production. ee labs are yellow. a chocolate lab that is homozygous for both genes is crossed with a black lab with the genotype bbee. what are the phenotypes and phenotype ratio of the offspring?
Half of the offspring will be black with pigment production and the other half will be black without pigment production. The chocolate lab is homozygous for both b and e genes, meaning it has the genotype bbEE.
When crossed with the black lab with the genotype bbee, the possible gametes produced by the chocolate lab are bE and bE, while the possible gametes produced by the black lab are bE and be.
The offspring genotype ratios are 1:1 for BbEe and Bbee. The BbEe genotype results in black coat color and pigment production, while the Bbee genotype results in black coat color but no pigment production.
The phenotype ratio for the offspring is 1:1 for black labs with pigment production and black labs without pigment production.
Therefore, half of the offspring will be black with pigment production and the other half will be black without pigment production.
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What does it mean to analvze data? O A. To develop a possible answer to a scientific question • B. To support a hypothesis O c. To predict what will happen if a hypothesis is true • D. To examine or interpret observations
To analyze data means to examine or interpret observations.
Therefore option D is correct.
What is data?Data is described as a collection of discrete values that convey information, describing quantity, quality, fact, statistics, other basic units of meaning, or simply sequences of symbols that may be further interpreted.
Data analysis is described as a process of inspecting, cleansing, transforming, and modeling data with the goal of discovering useful information, informing conclusions, and supporting decision-making.
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in what way does glucokinase and hexokinase trap glucose in the cell?
Glucokinase and hexokinase are enzymes involved in the first step of glucose metabolism, which is the conversion of glucose into glucose-6-phosphate. These enzymes are responsible for trapping glucose within the cell by phosphorylating it.
Glucokinase and hexokinase are enzymes involved in the first step of glucose metabolism, which is the conversion of glucose into glucose-6-phosphate. These enzymes are responsible for trapping glucose within the cell by phosphorylating it.
Hexokinase is present in most tissues and has a low Km for glucose (meaning it has a high affinity for glucose). It traps glucose in the cell by phosphorylating it to glucose-6-phosphate, which cannot pass through the cell membrane. This allows the cell to maintain a concentration gradient of glucose, which drives the entry of more glucose into the cell.
Glucokinase, on the other hand, is present mainly in the liver and pancreas and has a higher Km for glucose (meaning it has a lower affinity for glucose). This enzyme helps to regulate glucose uptake in these tissues by phosphorylating glucose when its concentration is high. Glucose-6-phosphate is then used for energy production or stored as glycogen. When blood glucose levels drop, glucokinase activity decreases, allowing glucose to exit the cell and enter the bloodstream to maintain normal blood glucose levels.
Overall, both enzymes play a crucial role in glucose metabolism and help to trap glucose within the cell by phosphorylating it to form glucose-6-phosphate.
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what happens in the lymph nodes as lymph flows through them? debris is filtered out foreign substances are destroyed plasma and mature t cells are added all of these are correct
When the lymph flows through the lymph nodes, the debris gets filtered, plasma as well as matured t cells get added and the foreign substances get destroyed.
The correct option is option d.
Lymph nodes basically perform the function of filtering substances which happen to travel through the lymphatic fluid. These also contain the lymphocytes, which are the white blood cells, which help the body in order to fight infection as well as disease.
When the lymph happens to pass or flow through the lymph nodes, the debris gets filtered, plasma as well as matured t cells get added and the foreign substances get destroyed.
Hence, the correct option is d.
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which answer choice could be inhibited by greatly decreasing extracellular calcium? the fusion of secretory vesicles with the presynaptic plasma membrane the opening of voltage-gated calcium channels on the presynaptic axon terminal the arrival of the action potential at the presynaptic axon terminal the production of neurotransmitter by the presynaptic neuron
Decreasing extracellular calcium can inhibit the opening of voltage-gated calcium channels on the presynaptic axon terminal, which is necessary for the proper functioning of synaptic transmission.
Voltage-gated calcium channels are located on the presynaptic axon terminal and are responsible for the influx of calcium ions into the cell in response to an action potential. This influx of calcium ions triggers the fusion of secretory vesicles with the presynaptic plasma membrane and the subsequent release of neurotransmitters into the synaptic cleft.
However, if extracellular calcium is greatly decreased, there will be a decrease in the concentration gradient of calcium ions between the extracellular and intracellular environments. This will make it more difficult for calcium ions to enter the cell through the voltage-gated calcium channels, and may prevent the necessary influx of calcium ions required for the fusion of secretory vesicles with the plasma membrane and the subsequent release of neurotransmitters.
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what does 3-phosphoglycerate kinase do?
citric acid cycle malfunction can be a cause of human disease. which enzyme malfunction is linked to malignant gliomas?
The citric acid cycle, also referred to as the Krebs cycle or the tricarboxylic acid (TCA) cycle, is a vital metabolic system that generates energy in cells. The failure of citric acid cycle enzymes can result in a range of human illnesses.
The development of malignant gliomas may be connected to a failure in the enzyme isocitrate dehydrogenase (IDH), according to studies. Mutations in the IDH1 and IDH2 genes, in particular, have been found to be common in gliomas along with various types of cancer, and are thought to contribute to tumor creation and growth.
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What would happen to the population estimate if marked animals were either less likely to be captured in the second sample (e.g., fish avoid the boat after being handled) or more likely to be captured in the second sample (e.g., mice return to a trap because they know there is food there)?
The population estimate would most likely be too high if kept animals were less likely to be captured in the second sample. This is due to an overestimation of the total population size as a result of the lower proportion of marked animals in the second sample than anticipated.
For instance, assuming a higher proportion of marked animals in the population than is actually the case, the population estimate would be inflated if 50% of the marked animals were anticipated to be recaptured in the second sample but only 40% were actually recaptured due to avoidance behavior.
On the other hand, the population estimate would probably be too low if marked animals were more likely to be captured in the second sample. This is due to an underestimation of the total population size as a result of the higher proportion of marked animals in the second sample than anticipated. For instance, assuming that there are fewer marked animals in the population than there actually are, the population estimate would be deflated if 50% of the marked animals were expected to be recaptured in the second sample but 60% were actually recaptured due to attraction behavior.
Mark-recapture studies must therefore take into account potential biases and adjust the population estimate accordingly, such as by employing statistical models that take into account the capture probability and the marked animals' behavioral responses.
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: Level I: Reviewing Facts and Terms (Bloom's Taxonomy: Comprehension)
40) As a result of respiratory alkalosis,
A) the respiratory rate increases.
B) the tidal volume increases.
C) the kidneys conserve bicarbonate.
D) the kidneys secrete fewer hydrogen ions.
E) the body retains less carbon dioxide.
As a result of respiratory alkalosis, the body retains less carbon dioxide. The correct option is E.
When the body exhales an excessive amount of carbon dioxide, either as a result of a faster respiratory rate or less carbon dioxide being produced, is known as respiratory alkalosis develops.
which result in the amount of carbon dioxide in the blood decreases, which also lowers the concentrations of carbonic acid and hydrogen ions. A rise in pH follows, which may result in symptoms of discomfort.
To control alkalosis the kidneys can excrete more bicarbonate ions, which helps to increase the hydrogen ion concentration and lower the pH. The correct option is E.
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The bacterial genome was sequenced and a mutation found in a gene adjacent to β-galactoside. How would you best explain these research findings? The:
The sequencing of a bacterial genome permits scientists to denote and focus on the hereditary material of the organic entity. Researchers are able to identify genetic variations, including mutations that may alter the function of genes.
In this specific case, a change has been found in a quality nearby the β-galactoside quality. β-galactoside is a chemical that catalyzes the hydrolysis of lactose into glucose and galactose, and its demeanor is directed by a close-by quality called lacZ. The transformation in the contiguous quality might influence the declaration of lacZ and in this manner the capability of β-galactoside.
To precisely determine the mutation's effect on the functions of the adjacent genes, additional research is required. However, these research results suggest that the mutation may affect the bacterial metabolism of lactose and the bacterium's ability to survive and thrive in its environment.
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The different progestins all have a ___________ affinity for the progesterone receptors
The different progestins all have a varying affinity for the progesterone receptors
The progesterone receptors are more or less responsive to various progestins. Progesterone is a hormone naturally produced by the ovaries. Synthetic progestins are substances that operate similarly to progesterone. Progestins come in a variety of forms, each with a unique chemical makeup and range of progesterone receptor affinities. Levonorgestrel and norgestimate are examples of "strong" progestins since they have a high affinity for the progesterone receptor. Some progestins, such norethindrone and medroxyprogesterone acetate, are regarded as "weak" progestins because of their low affinity for the receptor. Other progestins, like drospirenone and dienogest, have a moderate affinity for the receptor.
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Transcription takes place here (in a eukaryotic cell)
In a typical eukaryotic cell, transcription takes place in the nucleus.
The nucleus is a membrane-bound compartment within the cell that contains the cell's genetic material, including DNA (deoxyribonucleic acid). Transcription is the process by which the DNA sequence is copied into a complementary RNA (ribonucleic acid) molecule.
The process of transcription involves several steps. First, the DNA molecule unwinds and exposes a specific region of the DNA called the gene. Enzymes called RNA polymerases then synthesize a complementary RNA molecule, known as messenger RNA (mRNA), using one of the DNA strands as a template.
The mRNA molecule carries the genetic information from the DNA to the ribosomes, which are cellular structures responsible for protein synthesis.
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--The given question is incorrect, the correct question is
"In a typical eukaryotic cell, where do transcription take place?"--
which two antibody classes are most critical to the maintenance of gut immunity and appropriate tolerance?
The two antibody classes that are most critical to the maintenance of gut immunity and appropriate tolerance are Immunoglobulin A (IgA) and Immunoglobulin G (IgG).
IgA is crucial for gut immunity as it is the main antibody found in mucosal secretions, including those in the gastrointestinal tract. It helps protect the gut lining by neutralizing harmful pathogens and preventing their attachment to the gut lining.
IgG, on the other hand, is important for appropriate tolerance as it can modulate immune responses in the gut by binding to various antigens and inhibiting their potential to cause inflammation. Additionally, IgG can facilitate the clearance of pathogens and immune complexes, further contributing to gut homeostasis.
In summary, both IgA and IgG play essential roles in maintaining gut immunity and appropriate tolerance, with IgA primarily defending the gut lining and IgG regulating immune responses.
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What is chronic atrophic gastritis associated with?
Metaplastic (chronic) atrophic gastritis, also known as gastric atrophy, is the name given to a kind of chronic gastritis that, in addition to inflammation, is characterized by mucosal thinness.
The loss of specialized cells in the gastric glands, and changes in epithelial cell types (ie, metaplasia). When you have atrophic gastritis (AG), your stomach lining becomes chronically inflamed, thins, and the cells that make up your stomach lining begin to resemble those in your intestines.
H. pylori infection is typically the cause of one kind of atrophic gastritis called environmental metaplastic atrophic gastritis (EMAG). By far, the most typical cause of persistent atrophic gastritis is a stomach infection called H pylori. When H pylori is infected, the chance of developing atrophic gastritis is multiplied by ten.
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A pregnant servicewoman shall remain onboard as long as OB treatment is less than how many hours away
If the workplace is safe for pregnant service women employees, they can work 40 hours per week.
It could be damaging to a pregnant worker's health and the health of the unborn child if she starts working more than 40 hours per week and is under a lot of stress.
The typical amount of hours to work a week in the UK is 40, and employers are not obligated to reduce hours below that
Pregnant employees must be able to perform their assigned hours safely, and if they can't, employers are responsible for taking necessary action.
If the workplace is safe for pregnant employees, they can work 40 hours per week. It could be damaging to a pregnant worker's health and the health of the unborn child if she starts working more than 40 hours per week and is under a lot of stress.
The typical amount of hours to work a week in the UK is 40, and employers are not obligated to reduce hours below that.Pregnant employees must be able to perform their assigned hours safely, and if they can't, employers are responsible for taking necessary action.
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What is a hydrothermal vent?
Why do they have diverse species rich communities?
A hydrothermal vent is a fissure in the ocean floor that releases hot, mineral-rich water from beneath the Earth's crust.
These vents are formed when sea water seeps down into the Earth's crust and is heated by molten rock, known as magma. The hot water then carries minerals up through the cracks in the ocean floor, creating a unique and dynamic environment. This environment is rich in nutrients and minerals that support a diverse and abundant community of organisms.
The hydrothermal vent environment is home to a variety of species, including chemosynthetic bacteria, tube worms, mussels, crabs, and shrimp. These species live in highly specialized niches, taking advantage of the environment’s unique characteristics.
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what is the purpose of the secondary ETC (following PSI) in the light-dependent reactions?
The purpose of the secondary electron transport chain (ETC) following photosystem I (PSI) in the light-dependent reactions of photosynthesis is to generate additional ATP.
After the excited electrons from PSI are passed to the primary electron acceptor, they are transported through a series of electron carriers in the secondary ETC, also known as the cytochrome b6f complex, located in the thylakoid membrane of the chloroplast. This transfer of electrons generates a proton gradient across the thylakoid membrane, which drives the synthesis of ATP through chemiosmosis. The proton gradient is formed by the transfer of protons (H+) across the thylakoid membrane from the stroma (low proton concentration) to the thylakoid lumen (high proton concentration) through the cytochrome b6f complex.
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both prokaryotic and eukaryotic organisms carry out some form of glycolysis. how does that fact support or not support the assertion that glycolysis is one of the oldest metabolic pathways?
The fact that both prokaryotic and eukaryotic organisms carry out some form of glycolysis suggests that this metabolic pathway is ancient and likely originated early in the evolution of life on Earth.
Glycolysis is a fundamental process that occurs in all living cells, and its conserved nature across a wide range of organisms implies that it was present in the last universal common ancestor (LUCA).
This supports the assertion that glycolysis is one of the oldest metabolic pathways. However, it is important to note that the presence of glycolysis in both prokaryotic and eukaryotic organisms does not necessarily provide conclusive evidence that it is the oldest pathway.
Other metabolic pathways may have arisen before or around the same time as glycolysis, but have since diverged and evolved differently in different lineages.
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19. When the carbohydrate portion is attached to a serine residue in a glycoprotein, it is referred to as a(n) _________oligosaccharides
When the carbohydrate portion is attached to a serine residue in a glycoprotein, it is referred to as a O-linked oligosaccharides.
the overall advantage is that _____ photosynthesis can proceed during the day while stomata are closed (reducing _____ loss)
The overall advantage is that CAM (Crassulacean Acid Metabolism) photosynthesis can proceed during the day while stomata are closed (reducing water loss).
CAM photosynthesis is a special type of photosynthesis that is adapted to arid or water-limited environments. In CAM plants, such as many succulents, cacti, and some orchids, the stomata, which are tiny openings on the surface of leaves that allow for gas exchange, remain closed during the day to reduce water loss through transpiration, which is the process of water vapor escaping from plant tissues into the atmosphere.
By keeping the stomata closed during the day, CAM plants are able to conserve water in their tissues and reduce water loss, which is a significant advantage in arid or water-scarce environments.
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in a peptide bond formation condensation reaction, what group is the nucleophile and what group is the electrophile?
In a peptide bond formation condensation reaction, the amino group (NH₂) is the nucleophile and carboxyl group (COO⁻) is the electrophile.
A peptide bond is a type of covalent bond which is formed between two amino acids. These bonds are formed as a result of condensation reaction of carboxylic group of amino acid and the amino group of another amino acid, followed by the elimination of a water molecule. Condensation reaction is a reaction involving dehydration synthesis.
On hydrolysis of a peptide bond , a free amino group and the carboxyl functional groups are produced.
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Describe what the results of experiments in wet and dry meadow regarding the effect of the availability of nutrients on NPP indicate.
The results of experiments in wet and dry meadows indicate that the availability of nutrients has a significant effect on net primary productivity (NPP).
In the wet meadow, the availability of nutrients resulted in higher NPP than in the dry meadow, which had lower nutrient availability.
This is likely due to the fact that higher nutrient availability in the wet meadow allowed for more efficient photosynthesis and respiration, leading to higher NPP.
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the first evidence for nucleosome formation came from digesting chromosomal dna with a non-specific nuclease. gel electrophoresis of the dna after the reaction revealed:
The first evidence for nucleosome formation came from digesting chromosomal DNA with a non-specific nuclease. Gel electrophoresis of the DNA after the reaction revealed a ladder-like pattern of DNA fragments, indicating the presence of repeating units, which were later identified as nucleosomes.
Gel electrophoresis is a laboratory technique for separating DNA, RNA, or protein mixtures based on molecular size. An electrical field pushes the molecules to be separated through a gel with microscopic holes in gel electrophoresis. The molecules move through the pores in the gel at a rate that is proportional to their length. This indicates that a little DNA molecule will move further across the gel than a larger DNA molecule will. The gel electrophoresis of the DNA after digestion with a non-specific nuclease revealed a ladder-like pattern, indicating that the DNA was fragmented into multiple sizes. This pattern was consistent with the formation of nucleosomes, which are comprised of DNA wrapped around histone proteins, and suggested that the chromosomal DNA had been organized into repeating units of nucleosomes.
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great (type II) alveolar cells
Type II alveolar cells, also known as great alveolar cells, are specialized cells found in the alveoli of the lungs.
These cells are responsible for producing and secreting surfactant, a substance that helps to reduce surface tension in the alveoli and prevent them from collapsing during exhalation. Surfactant is composed of lipids and proteins and is crucial for maintaining normal lung function.
In addition to producing surfactant, type II alveolar cells also play a role in the immune response of the lungs. They are capable of producing and secreting cytokines, which are signaling molecules that help to recruit immune
cells to the site of infection or injury.
Type II alveolar cells are critical for maintaining proper lung function and are susceptible to damage by various environmental insults, such as cigarette smoke or pollution. Damage to these cells can lead to a decrease in surfactant production and an increase in lung stiffness, which can result in breathing difficulties and other respiratory problems.
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Explain
great (type II) alveolar cells
Chlorophyll molecules are embedded within the ___________ ____________, which is the most extensive membrane system
The most extensive membrane system, the thylakoid membrane, contains chlorophyll molecules embedded within it.
The thylakoid membrane houses the green pigment chlorophyll, and the stroma is the space in between the thylakoid and chloroplast membranes.
A chlorophyll molecule is embedded in the chloroplast's thylakoid membrane. Special proteins bind the chlorophyll molecules to the thylakoid membrane in clusters of several hundred molecules known as antenna complexes.
The biosphere's most extensive membrane system is represented by photosynthetic membranes, also known as thylakoids. In the cytosol of cyanobacteria and the stroma of chloroplasts, they form membrane cisternae that are flattened.
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Root Tissues
Identify the root tissues. Record your answer under "Slide 5" on your lab report.
The epidermis, cortex, endodermis, pericycle, xylem, and phloem are the root tissues on Slide 5. The specific tissues may differ depending on the slide's sample.
What is the root tissue?From the outermost to the innermost part of the root, the epidermis, cortex, and vascular cylinder are the primary tissues. The epidermis typically only has one cell layer of thickness and is made up of cells with thin walls.
What are the root tissues and their capabilities?System expansion and root tissue. The root's vascular system is within the pericycle. The phloem, which transports photosynthesis products from the leaves to the roots, and the xylem, which transports water and minerals from the roots to the rest of the plant, are the two types of tissues found here.
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what catalyzes the splitting of Fructose 1,6-bisphosphate > dihydroxyacetone phosphate and glyceraldehyde 3-phosphate?
The splitting of Fructose 1,6-bisphosphate into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate is catalyzed by the enzyme aldolase.
This enzyme breaks the carbon-carbon bond between the two carbon atoms in the molecule, resulting in the formation of two three-carbon molecules. This reaction is an important step in the glycolysis pathway, which is a metabolic pathway that breaks down glucose into pyruvate and ATP.
The resulting dihydroxyacetone phosphate and glyceraldehyde 3-phosphate can then be further metabolized to produce ATP, which is the primary energy source for cells. Overall, aldolase plays a crucial role in the regulation of cellular energy metabolism.
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The Conservation Reserve Program pays farmers to ________.
A) grow native crops such as corn and beans
B) grow soybeans and other kinds of harvestable groundcover
C) stop cultivating highly erodible land
D) grow cattle feed instead of commercial crops
E) stop growing tobacco
The Conservation Reserve Program pays farmers to stop cultivating highly erodible land.
The correct option is option C.
The Conservation Reserve Program is basically a land conservation administered by the FSA which is the Farm Service Agency. In exchange for a rental payment which is yearly, the farmers who are basically enrolled in this program agree to not include the environmentally sensitive land for the purpose of agricultural production as well as the plant species that will improve the health and quality of the environment.
Contracts for land which are basically enrolled in the CRP range from 10 to 15 years in length. The goal of this program is basically to be able re-establish the valuable land cover in order to help prevent soil erosion, improve water quality, and reduce loss of wildlife habitat.
Hence, the correct option is option C.
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What is the standard dose range for ethinyl estradiol?
A synthetic oestrogen called ethinyl estradiol is frequently used in hormone replacement therapy and hormonal contraception. Depending on the indication and the particular product being used, ethinyl estradiol has a defined dose range.
Ethinyl estradiol is commonly dosed for oral contraceptives at 20–35 micrograms per day, frequently in conjunction with a progestin. Ethinyl estradiol dosages in hormone replacement treatment might vary from 0.5-2 milligrammes per day, again depending on the particular product and indication. The right dosage of ethinyl estradiol for a specific person should be decided by a healthcare professional and may vary on elements including age, weight, and medical history.
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Termination (3rd step of Prokaryotic Translation)
The third step of prokaryotic interpretation is the end, which is the cycle by which the ribosome perceives the stop codon on the mRNA and deliveries the recently blended polypeptide chain.
There are two proteins involved in termination: the first release factor (RF1) and the second release factor
When the ribosome reaches the stop codon (UAA, UAG, or UGA) during termination, it does not recognize any tRNA that has an anticodon that is compatible with the codon. Instead, RF1 or RF2 binds to the A site of the ribosome after recognizing the stop codon. The newly synthesized polypeptide chain from the tRNA in the P site is released as a result of this binding.
The ribosome splits into its two subunits, the small subunit, and the large subunit when the polypeptide chain is released. With the help of chaperone proteins, the newly synthesized protein is folded into its final three-dimensional structure and can function as a protein in the cell.
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