Influenza A binds to sialic acid receptors on the surface of host cells, then fuses with the host cell membrane to allow entry of the viral RNA and proteins into the host cell. These proteins and RNA assemble into new virus particles that are released from the host cell by budding.
Part A:
Comparing the culture media type used to grow Influenza A and Streptococcus pneumoniae in the laboratory are:
Influenza A:
Influenza A is grown in two types of media which are egg-based and cell-based. Eggs are preferred for influenza A virus growth because the virus is adapted to grow within eggs, and also embryos are an abundant source of virus-compatible host cells for virus production. Cell-based cultures are also used for growing influenza A because the viruses are capable of infecting many different types of cells.
Streptococcus pneumoniae:
Streptococcus pneumoniae can grow on different types of culture media including blood agar, chocolate agar, and nutrient agar. Blood agar is the most common culture medium used to grow Streptococcus pneumoniae. Blood agar can be made from several types of blood including horse, sheep, rabbit, and human blood. Blood agar provides essential nutrients required for the growth of Streptococcus pneumoniae.
Part B:
Bacteriophage therapy is being investigated for the treatment of an antibiotic-resistant Streptococcus pneumoniae strain.
The bacteriophage enters its host cell in the following ways:
Bacteriophages enter bacterial cells by attaching to specific receptor sites on the surface of the bacterial cell. Once attached to the receptor sites, the bacteriophages inject their genetic material into the bacterium. The genetic material of the bacteriophage takes over the bacterial cell, instructing it to make more copies of the bacteriophage. Once the bacterial cell has made enough copies of the bacteriophage, the cell will rupture, releasing the bacteriophages to infect other bacterial cells.
Influenza A enters its host cell in the following way:
Influenza A binds to sialic acid receptors on the surface of host cells, then fuses with the host cell membrane to allow entry of the viral RNA and proteins into the host cell. Once inside the host cell, the viral RNA is released and used as a template to make new viral proteins and viral RNA. These proteins and RNA assemble into new virus particles that are released from the host cell by budding.
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Amyloid plaques have been identified as one of the causes of
Alzheimer’s disease. Explain briefly how the
amyloid protein causes the disease
In Alzheimer's disease, amyloid plaques form in the brain, and amyloid proteins play a crucial role in their formation. Amyloid proteins are fragments of a larger protein called amyloid precursor protein (APP).
The accumulation of Aβ peptides results in the formation of amyloid plaques, which are clumps of protein that build up between nerve cells. These plaques disrupt normal brain function and are toxic to neurons, causing damage and impairing communication between brain cells.
One theory is that the accumulation of amyloid plaques triggers a cascade of events that leads to the characteristic features of Alzheimer's disease. The presence of amyloid plaques activates immune responses and inflammatory processes in the brain, leading to chronic inflammation and further damage to neurons.
Moreover, the accumulation of Aβ peptides can disrupt the normal functioning of neurons, impairing synaptic transmission and interfering with the brain's ability to form and retrieve memories. This contributes to the cognitive decline and memory loss observed in Alzheimer's disease.
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What advice would you give a friend planning to run a
5-kilometer race about food intake before, during, and after the
event? Your answer must include the goal of the meal as well as
address the size,
When preparing for a 5-kilometer race, it is important to focus on pre-race, during-race, and post-race nutrition. Pre-race, consume a balanced meal that provides energy and includes carbohydrates.
Prior to the race, it is recommended to consume a balanced meal 2-3 hours before the event. This meal should primarily consist of easily digestible carbohydrates such as whole grains, fruits, and vegetables to provide a source of energy for the race. Avoid heavy or high-fat foods that may cause digestive discomfort.
During the race, maintaining hydration is crucial. It is advisable to take small sips of water at regular intervals to prevent dehydration. Depending on the duration and intensity of the race, some individuals may benefit from consuming sports drinks or energy gels to replenish electrolytes and provide additional fuel.
After completing the race, focus on replenishing fluids and nutrients. Hydrate with water or electrolyte-rich beverages and consume a balanced meal within the first hour or two. This meal should include lean proteins like chicken or fish, carbohydrates for replenishing glycogen stores, and a variety of vegetables for essential nutrients.
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Exposure of yeast cells to 2,3,5 triphenyl tetrazolium chloride (TTC) can lead to interaction of the colourless compound with mitochondria where it can be converted to a red form (pigment).
What statement best describes the process in which TTC is converted from its initially colourless form to a red pigment?
A. Initially TTC is colourless however TTC interaction with ATP synthase leads to the ATP-dependent conversion of TTC to TTC-phosphate (where ATP breakdown is coupled to TTC phosphorylation). TTC-P is a red pigment that accumulates in mitochondria.
B. Initially TTC is colourless however TTC interaction with a component of the mitochondrial electron transport system (ETS) leads to transfer of electrons from TTC to the ETS converting TTC to a red pigment.
C. Initially TTC is colourless however TTC interaction with the plasma membrane electron transport system (mETS) in yeast leads to transfer of electrons from the TTC to the mETS converting TTC to a red pigment.
D. The initially the TTC solution used in the method only contains dilute TTC which appears colourless, however TTC becomes concentrated in cells and mitochondria which makes the cells stain red.
E. Initially TTC is colourless however TTC interaction with a component of the mitochondrial electron transport system (ETS) leads to transfer of electrons from the ETS to TTC converting TTC to a red pigment.
The best statement describing the conversion of 2,3,5 triphenyl tetrazolium chloride (TTC) from its initially colorless form to a red pigment in yeast cells is option B.
Initially colorless TTC interacts with a component of the mitochondrial electron transport system (ETS), resulting in the transfer of electrons from TTC to the ETS and the conversion of TTC to a red pigment.
When yeast cells are exposed to TTC, the colorless compound interacts with a component of the mitochondrial electron transport system (ETS). During this interaction, electrons are transferred from TTC to the ETS, leading to the conversion of TTC to a red pigment. This process occurs within the mitochondria of the yeast cells. Option B accurately describes this mechanism of conversion, highlighting the involvement of the ETS in the electron transfer and the resulting formation of the red pigment.
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What are the consequences when prolong periods of drough happen? Select all items that apply. More frequent yet less intense wild fires More frequent and intense wild fires Poor crop & livestock yields Lower aqueduct and groundwater availability Less water restriction mandates
The consequences when prolonged periods of drought occur include:
More frequent yet less intense wildfires
Poor crop and livestock yields
Lower aqueduct and groundwater availability
An unplanned, uncontrolled, and unexpected fire in an area of flammable vegetation is known by several names, including wildfire, forest fire, bushfire, wildland fire, and rural fire. A wildfire may be more precisely referred to as a bushfire (in Australia), brush fire, desert fire, grass fire, hill fire, peat fire, prairie fire, vegetation fire, or veld fire depending on the kind of vegetation that is present. Wildfire is essential to several natural forest ecosystems. Wildfires are separate from controlled or managed burning, which is a good human use of wildland fire, even though controlled burns have the potential to convert into wildfires.
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You are in a lab; you have performed a blue/white screening. You notice that all of your colonies are blue. Were you successful in transforming your bacteria?
(a) Yes, a dysfunctional LacZ will produce all blue colonies.
(b) No, a functional LacZ will produce all blue colonies.
Yes, a dysfunctional LacZ will produce all blue colonies. Option (a) is correct.
In a blue/white screening, it is used to identify and characterize recombinant colonies following transformation. The screening relies on the expression of the E. coli β-galactosidase (LacZ) gene, which encodes an enzyme that cleaves lactose into glucose and galactose. In the absence of lactose, β-galactosidase is repressed by a protein called the lac repressor. This binding of the repressor protein is prevented by the presence of lactose or its analogs such as IPTG. When the lac repressor protein is inhibited, the β-galactosidase gene is expressed, and the colonies will turn blue. If the colonies do not produce functional β-galactosidase, the colonies will be white in appearance because it lacks the ability to cleave the colorless substrate X-gal to produce a blue color. When you see blue colonies, it indicates that the LacZ gene is disrupted or not expressed, so the IPTG induces β-galactosidase expression, and the colonies will turn blue. So, option (a) is correct.
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The two broad classes of cells in the nervous system
include
1. those responsible for information processing, and
2. those providing mechanical and metabolic support.
These two categories of specialized cellss called?
A. microglia and Schwann cells.
B. neurons and glial cells.
C. astrocytes and Schwann cells.
D. schwann cells and glial cells.
The two broad classes of cells in the nervous system are neurons and glial cells. So, the correct answer is B. neurons and glial cells.
Neurons are responsible for information processing in the nervous system. They receive, integrate, and transmit electrical signals, allowing for communication within the nervous system. Neurons are specialized cells that have a cell body, dendrites (which receive signals), and an axon (which transmits signals to other neurons or effector cells).
Glial cells, also known as neuroglia or simply glia, provide mechanical and metabolic support to neurons. They play crucial roles in maintaining the structural integrity of the nervous system, regulating the extracellular environment, and supporting neuronal function. Glial cells include various types, such as astrocytes, microglia, oligodendrocytes, and Schwann cells.
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39. Organic acids are often considered "static" agents because a mechanism of action is to deplete ATP. ATP depletion happens because A. Ribosomes are blocked B. RNA synthesis is inhibited C. Protein synthesis is inhibited D. ATP is used to pump protons out of the cell E. The cell needs ATP to chemically alter the toxin 40. In a low nutrient barrel ageing wine, Brett can get the trace amounts of carbon that it needs from B. diammonium phosphate C. photosynthesis A. wood sugar D. nitrogen fixation E. CO2
Organic acids are often considered "static" agents because a mechanism of action is to deplete ATP. ATP depletion happens because wood sugar.
In a low nutrient barrel aging wine, Brettanomyces (referred to as "Brett") is a type of yeast that can metabolize trace amounts of carbon sources present in the wine. Wood sugars, such as glucose and xylose, are released from the wooden barrels during the aging process. Brettanomyces can utilize these wood sugars as a carbon source for its growth and metabolism.
Diammonium phosphate (option B) is a nitrogen source often used in winemaking but does not provide carbon for Brettanomyces. Photosynthesis (option C) is the process by which plants and some microorganisms convert sunlight into energy but is not relevant to Brettanomyces in a wine barrel. Nitrogen fixation (option D) is a process in which certain bacteria convert atmospheric nitrogen into a usable form, and CO2 (option E) is a byproduct of various cellular processes but is not a direct carbon source for Brettanomyces.
Therefore, the trace amounts of wood sugar present in the low nutrient barrel-aging wine can serve as a carbon source for Brettanomyces growth.
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When considering executive function in the context of the Wisconsin Card Sorting Test, a person who fails to understand the rules have changed after 10 successful trials (lack of flexible thinking) may have damage to:
a.Ventrolateral prefrontal cortex
b.Dorsolateral prefrontal cortex
c. Orbitofrontal cortex
d. Anterior cingulated cortex
Executive functions are the cognitive abilities that help us regulate our thoughts and actions. These functions include reasoning, problem-solving, decision-making, planning, and self-monitoring.
Wisconsin Card Sorting Test (WCST) is an assessment tool that tests executive functions.The Wisconsin Card Sorting Test assesses different aspects of executive function. When a person fails to understand the rules have changed after 10 successful trials (lack of flexible thinking), it indicates a lack of flexibility in thinking. The executive function that controls flexibility in thinking is the dorsolateral prefrontal cortex.
Thus, if an individual fails to understand the rules have changed after 10 successful trials, it suggests that they may have damage to the dorsolateral prefrontal cortex.Option b. dorsolateral prefrontal cortex is the correct answer.
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At her job, Janet accidentally poured a toxic chemical on her foot. As a result, she experienced a mutation in the elastin protein in that area. Thankfully, it was a silent mutation (CGC to CGA). However, a couple of weeks later, Janet notices that although she still has skin, it’s not very tight around her foot- indicating a problem with her elastin in that area. What might be happening and how would scientists test it (describe the process)?
The apparent problem with the tightness of the skin surrounding Janet's foot may not be directly related to the silent mutation in the elastin protein (CGC to CGA) that she encountered.
Silent mutations don't modify the amino acid sequence, therefore in this instance, the switch from arginine (CGC) to arginine (CGA) might not have a big effect on how well the elastin protein functions.Scientists would need to conduct additional research to ascertain the reason why there is an issue with elastin in that region. They might also take into account other things, like potential harm from the poisonous substance or adverse effects from the mutation. They would carry out studies utilising various experimental techniques, such as:1. Histological evaluation: Researchers could take a skin biopsy from Janet's afflicted area.
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Which of the following statements regarding manatee oropharyngeal anatomy is true?
Group of answer choices
1) The downward sloping angle of the jaw does not impact the ability to visualize the oropharynx
2) Nasogastric or nasotracheal intubation offers the most likely success give anatomical positioning.
3) The angle from the mouth opening to the level of the esophageal opening and tracheal opening is a straight line
4) Manatees have an abbreviated soft palate.
The true statement regarding manatee oropharyngeal anatomy is: Manatees have an abbreviated soft palate. Therefore, answer is option 4, that is Manatees have an abbreviated soft palate.
Manatee Oropharyngeal Anatomy is basically the study of the mouth, pharynx, and esophagus of a manatee. Manatees are large, fully aquatic, herbivorous marine mammals. They have no hind limbs, and their forelimbs are paddle-like flippers. They use their tail as the primary means of propulsion in water.Manatees have a very unique anatomy, and understanding their oropharyngeal anatomy is essential for providing appropriate medical care. One of the important aspects of manatee oropharyngeal anatomy is its soft palate. Manatees have an abbreviated soft palate, which makes their airway more prone to collapse during anesthesia. Additionally, manatees have a very small oropharynx, which makes intubation more challenging than in other species.
The other given options are incorrect because the downward sloping angle of the jaw does impact the ability to visualize the oropharynx, nasogastric or nasotracheal intubation does not offer the most likely success given anatomical positioning, and the angle from the mouth opening to the level of the esophageal opening and tracheal opening is not a straight line.
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This blood smear is abnormal. It shows that: there are not enough platelets there are too few red blood cells there are too many platelets erythrocytes are sickle-celled there are too many basophils
The blood smear is abnormal because there are too few red blood cells and too many platelets. An abnormal blood smear can indicate various health conditions or abnormalities in the blood.
A blood smear is a laboratory test that involves examining a sample of blood under a microscope to evaluate the different blood cells and their characteristics.
In the given scenario, the blood smear shows two abnormalities: too few red blood cells (erythrocytes) and too many platelets.
Too few red blood cells may indicate a condition called anemia, where the body has a decreased number of red blood cells or a decreased amount of hemoglobin, the oxygen-carrying protein within red blood cells. Anemia can result from various causes, including iron deficiency, vitamin deficiencies, blood loss, or certain diseases.
On the other hand, an increased number of platelets, known as thrombocytosis, may be indicative of several conditions, such as infection, inflammation, bone marrow disorders, or certain cancers. Platelets are involved in blood clotting and their excess can lead to an increased risk of abnormal clot formation.
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The possible existence of E.H.1 (extinct hominin 1) was a concept introduced in class lectuer. This species is significant because
E.H.1, if confirmed, would be significant as it could provide crucial insights into human evolutionary history, potentially representing a previously unknown branch of the hominin family tree.
Its existence would challenge our current understanding of human evolution and shed light on the diversity of early human ancestors. Studying E.H.1's anatomy, behavior, and genetic makeup could offer valuable information about the complex process of hominin speciation and the factors that influenced our evolutionary trajectory.
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In a species the haploid number of chromosomes is 4. Independent assortment has the possibility of producing ______________ different gametes.
O 24 O 4*2 O 2*4
O 8
In a species, the haploid number of chromosomes is 4. Independent assortment has the possibility of producing 8 different gametes. Independent assortment is the separation of homologous chromosomes into daughter cells that occurs during meiosis.
This separation occurs randomly, which results in genetically diverse gametes being produced. Independent assortment is a principle of genetics that states that the alleles of different genes segregate independently of each other during meiosis. In other words, each allele has an equal chance of ending up in a gamete, regardless of what other alleles are present.
Independent assortment occurs due to the arrangement of homologous chromosomes in the middle of the cell during meiosis.
The chromosomes are randomly sorted and separated, resulting in genetic variation in the gametes produced by the organism.
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compare and contrast plants, fungi, invertebrate animals, and
chordate animals.
Plants use photosynthesis, fungi decompose organic matter, invertebrates lack a backbone, and chordates have a vertebral column, showing their differences in nutrition, structure, and complexity.
Plants, fungi, invertebrate animals, and chordate animals represent different branches of the tree of life, each with its own unique characteristics and evolutionary adaptations.
Plants:
Plants are multicellular organisms that belong to the kingdom Plantae. They possess chlorophyll and can perform photosynthesis, converting sunlight, carbon dioxide, and water into glucose and oxygen. This process allows them to produce their own food and is essential for their survival.
Plants have a rigid cell wall made of cellulose, which provides structural support. They also have specialized tissues for conducting water, nutrients, and sugars throughout the plant. Examples of plants include trees, flowers, grasses, and ferns.
Fungi:
Fungi belong to the kingdom Fungi and are eukaryotic organisms that include mushrooms, molds, and yeasts. Unlike plants, fungi cannot perform photosynthesis and obtain nutrients through decomposition or absorption. They secrete enzymes to break down organic matter and then absorb the released nutrients.
Fungi have cell walls made of chitin and reproduce through spores. They play important roles in ecosystems as decomposers, symbiotic partners with plants, or as pathogens causing diseases in animals and plants.
Invertebrate Animals:
Invertebrate animals constitute a diverse group of organisms that lack a backbone or vertebral column. They make up the majority of animal species on Earth. Invertebrates include insects, worms, mollusks, arachnids, and many others.
They exhibit a wide range of anatomical and physiological adaptations to their environments. Invertebrates can have exoskeletons (such as insects), hydrostatic skeletons (like worms), or shells (such as mollusks). They display a variety of feeding strategies and have adapted to numerous habitats.
Chordate Animals:
Chordate animals belong to the phylum Chordata and are characterized by the presence of a notochord at some stage in their development. Chordates also possess a dorsal hollow nerve cord, pharyngeal slits.
This group includes vertebrates such as fish, amphibians, reptiles, birds, and mammals. Chordates exhibit a wide range of adaptations for locomotion, and feeding. Vertebrates have an internal skeleton made of bone or cartilage, which provides structural support and protects their internal organs.
In summary, while plants and fungi differ in their modes of nutrition, with plants performing photosynthesis and fungi obtaining nutrients through decomposition or absorption, both are multicellular eukaryotes.
Invertebrate animals, on the other hand, lack a backbone and encompass a wide variety of species with different adaptations. Chordate animals, including vertebrates, possess a notochord and exhibit more advanced characteristics, such as a dorsal nerve cord and well-developed organ systems.
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If there are 3 sufficient causes for diabetes and obesity is in all 3 , then obesity is a cause. 2 points QUESTION 25 10,000 people are screened for hypertension. 2,000 tested positive for hypertension using a new screening test and 1,750 of these people really did have hypertension. 500 of the people who tested negative using the new screening test really did have hypertension. 1) How many false positives are there? 2) Calculate the prevalence of hypertension in the sample 3) Calculate and write a plain English interpretation of the (3a) sensitivity, (3b) specificity, (3c) predictive value positive, and (3d) predictive value negative for the screening test (1 point each).
1) The number of false positives can be calculated by subtracting the true positive from the total number of positive tests. In this case, false positives = Total positive tests - True positives = 2000 - 1750 = 250.
2) The prevalence of hypertension in the total number of people in the sample. In this case, prevalence = (True positive + False negatives) / Total sample = (1750 + 500) / 10000 = 0.225 or 22.5%.
3a) Sensitivity: Sensitivity measures the ability of the screening test to correctly identify individuals with the condition. It is calculated as True positives / (True positives + False negatives). In this case, sensitivity = 1750 / (1750 + 500) = 0.777 or 77.7%. This means the screening test correctly identifies 77.7% of people with hypertension.
3b) Specificity: Specificity measures the ability of the screening test to correctly identify individuals without the condition. It is calculated as True negatives / (True negatives + False positives).
3c) Predictive value positive: Predictive value positive measures the probability , It is calculated as True positives / (True positives + False positives). In this case, predictive value positive = 1750 / (1750 + 250) = 0.875 or 87.5%. This means that if a person tests positive for hypertension, there is an 87.5% chance that they actually have the condition.
3d) Predictive value negative: Predictive value negative measures the probability that a person does not have the condition. It is calculated as True negatives / (True negatives + False negatives). Since the number of true negatives is not given.
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If the attack rate for a given organism (disease) is 25% and the case fatality rate is 3%, this suggests that
Group of answer choices
a. this organism has high infectivity and low virulence
b. this organism has low infectivity and high virulence
If the attack rate for a given organism (disease) is 25% and the case fatality rate is 3%, this suggests that the organism has low infectivity and high virulence.Therefore, the correct option is (b) this organism has low infectivity and high virulence.
In epidemiology, the term attack rate refers to the proportion of people who are affected by a disease in a given time period within a particular population. The case fatality rate, on the other hand, refers to the proportion of people who die due to a disease after contracting it. So, in this case, the attack rate is 25%, which means that out of the total population, 25% of people are affected by the disease in a given time period.The case fatality rate is 3%, which means that out of the total number of infected people, 3% of people die because of the disease. Since the case fatality rate is low, this suggests that the disease is not very deadly. However, since the attack rate is high, this suggests that the disease spreads quickly in the population. Therefore, the organism has low infectivity and high virulence.
So, the attack rate for a given organism is the proportion of people who are affected by a disease in a given time period within a particular population. The case fatality rate refers to the proportion of people who die due to a disease after contracting it. In this case, the attack rate is high (25%), indicating that the disease spreads quickly in the population. The case fatality rate is low (3%), indicating that the disease is not very deadly. Thus, the organism has low infectivity and high virulence. It is essential to know the infectivity and virulence of a disease to control its spread. Epidemiologists use various measures to study the patterns of diseases and their spread to prevent or manage outbreaks.
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Mitocondrial disease is one of the important disease that may be possibly cause multisystem disorders with vast clinical signs and symptoms. Based on your understanding on the function of mitochondria, discuss how the disease affect aerobic respiration and outline the major signs and symptoms.
Mitochondrial diseases are a group of genetic disorders that primarily affect the function of mitochondria, the powerhouses of our cells responsible for producing energy through aerobic respiration. These diseases can lead to multisystem disorders with diverse clinical signs and symptoms.
Mitochondrial diseases interfere with the normal functioning of aerobic respiration, which is the process by which cells convert glucose and oxygen into usable energy in the form of adenosine triphosphate (ATP). Mitochondria play a crucial role in this process, as they house the electron transport chain and produce ATP through oxidative phosphorylation.
In mitochondrial diseases, there are defects in the mitochondrial DNA or nuclear DNA that encode proteins essential for mitochondrial function. These defects impair the efficiency of oxidative phosphorylation, resulting in reduced ATP production. As a consequence, cells and tissues throughout the body that rely heavily on energy, such as the brain, muscles, heart, and kidneys, can be significantly affected.
The signs and symptoms of mitochondrial diseases can vary widely, as different organs and tissues may be affected to varying degrees. Some common symptoms include muscle weakness, exercise intolerance, fatigue, developmental delays, neurological problems (such as seizures, migraines, or movement disorders), poor growth, gastrointestinal issues, respiratory problems, and cardiac abnormalities. Other manifestations may include hearing or vision loss, hormonal imbalances, and increased susceptibility to infections.
Due to the broad range of signs and symptoms, mitochondrial diseases can be challenging to diagnose accurately. Genetic testing, muscle biopsies, and specialized metabolic evaluations are often used to aid in diagnosis. While there is currently no cure for mitochondrial diseases, management strategies focus on alleviating symptoms, improving quality of life, and supporting affected organ systems.
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a blast produces a peak overpressure of 47,000 n/m2 . a. what fraction of structures will be damaged by exposure to this overpressure? b. what fraction of people exposed will die as a result of lung hemorrhage?
The level of damage caused by a blast depends on several factors, including the distance from the blast, the duration of the overpressure, and the strength of the structures or materials involved.
However, it is possible to provide some general information about blast injuries based on the peak overpressure of 47,000 N/m2.
At this level of overpressure, individuals who are within close proximity to the blast (i.e., within the "lethal radius") are likely to experience significant injuries, including trauma to the lungs, ears, and other internal organs. The severity of these injuries can vary depending on the individual's distance from the blast and other factors.
In terms of fatalities, the risk of death from a blast injury is also influenced by several factors, including the intensity and duration of the overpressure, the location of the individual relative to the blast, and the individual's health status and other demographic factors. Without more detailed information about the specific circumstances of the blast and the population at risk, it is not possible to estimate the fraction of people who would die as a result of lung hemorrhage.
Overall, blast injuries are complex and multifactorial, and their severity and impact depend on many different variables. It is important to take appropriate precautions to prevent exposure to blasts and to seek medical attention immediately if blast-related injuries occur.
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Function and Evolution of Membrane-Enclosed Organelles The endomembrane system consists of the Endoplasmic Reticulum (ER), the Golgi apparatus, Lysosomes, Peroxisomes and Endosomes. The ER membrane is continuous with the nuclear envelope and the ER lumen directly communicates with the space between the outer and inner nuclear envelope membranes. . Additionally, for each of the components of the endo membrane system listed above the luminal facing lipid monolayer (See Ch 11, pages 367-368; Fig. 11-17] is different in composition from the cytosolic facing layer and the contents of the organelle (the lumen) is treated by the cell as something extracellular." a) How are these observations explained by the endomembrane origin story (the theory of how endomembrane compartments evolved through cl toplasmic membrane invaginations) depicted in Figure 15-3, page 491, b) The theory specifically refers to the formation of the nuclear envelope but it is thought that the Golgi complex arose in a similar fashion What might that have looked like? Draw a sketch (or series of sketches) depicting a possible scenario.
Therefore, all of these organelles are composed of phospholipid bilayers, and the lumen of these organelles is treated by the cell as something extracellular due to differences in composition from the cytosolic facing layer. b) Thus, the evolution of the Golgi complex through the endomembrane origin story is likely to have involved multiple rounds of plasma membrane invagination, leading to the formation of the ER, followed by ER invagination and formation of the Golgi complex
a) Explanation of observations by endomembrane origin story:
The endomembrane system consists of the Endoplasmic Reticulum (ER), the Golgi apparatus, Lysosomes, Peroxisomes and Endosomes.
The ER membrane is continuous with the nuclear envelope and the ER lumen directly communicates with the space between the outer and inner nuclear envelope membranes.
Additionally, for each of the components of the endo membrane system listed above the luminal facing lipid monolayer (See Ch 11, pages 367-368; Fig. 11-17] is different in composition from the cytosolic facing layer and the contents of the organelle (the lumen) is treated by the cell as something extracellular.
These observations can be explained by the endomembrane origin story (the theory of how endomembrane compartments evolved through cl toplasmic membrane invaginations) depicted in Figure 15-3, page 491.
The endomembrane system is thought to have originated from the plasma membrane. It happened by invagination of the plasma membrane, which separated the cytosol and extracellular environment.
The invagination formed vesicles that pinch off and then fused to form the ER, the Golgi complex, and lysosomes, in addition to other organelles like peroxisomes and endosomes.
b) Sketch depicting a possible scenario of the Golgi complex evolution through the endomembrane origin story:
The Golgi complex arose in a similar fashion to the formation of the nuclear envelope through the endomembrane origin story. This is shown in Figure 15-3, page 491.
As per this theory, it is thought that the Golgi complex evolved through cl toplasmic membrane invaginations, which subsequently developed into the complex membranous system.
The Golgi complex likely started as a series of flattened sacs derived from the plasma membrane by invagination.
As depicted in the figure, the first step involved the invagination of the plasma membrane, which then led to the formation of vesicles that fuse together to form the ER.
Then the invagination of the ER gave rise to the Golgi complex.
The vesicles formed by this process fuse together to form the Golgi cisternae, which mature into the Golgi complex.
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Epigenetic mechanisms control the expression of genes (Chapter 27828). a. Explain how DNA mothylation is inherited mitotically (specify the onzyme involved)? b. What is dosage compensation? c. Describe the process of X inactivation. Ensure you specify the name and type of epigonotic molecules involved in this and what they do
The above question is asked in three sections from the chapter Epigenetic mechanisms control the expression of genes - 272828.
a. DNA methylation is inherited mitotically through maintenance methylation, which is carried out by the enzyme DNA methyltransferase 1 (DNMT1). During DNA replication, DNMT1 recognizes hemimethylated DNA, which has one methylated and one unmethylated strand, and adds methyl groups to the newly synthesized unmethylated strand. This process ensures that the methylation pattern is faithfully replicated and inherited by daughter cells during cell division.
b. Dosage compensation is a mechanism that equalizes gene expression between males and females, particularly for genes located on sex chromosomes. In mammals, females have two X chromosomes while males have one X and one Y chromosome. To balance gene dosage, one of the X chromosomes in females undergoes X inactivation. This process is mediated by non-coding RNA molecules such as Xist, which coats the inactive X chromosome and leads to its transcriptional silencing. By equalizing gene expression between the sexes, dosage compensation ensures proper development and functioning of cells and organisms.
c. X inactivation is the process of inactivating one of the two X chromosomes in female mammals. It is initiated by the long non-coding RNA Xist, which is transcribed from the X chromosome to be inactivated. Xist spreads along the chromosome and recruits chromatin modifiers that lead to gene silencing and structural changes, forming a condensed structure called a Barr body. Another non-coding RNA called Tsix regulates Xist expression and prevents X inactivation on the active X chromosome.
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The greenhouse effect is bad. Without the greenhouse affect life
on Earth would be better off because it would mean no climate
change
true
or
false
The greenhouse effect is not bad but is a necessary phenomenon that allows life to exist on Earth is False. Therefore, correct option is False.
Without the greenhouse effect, Earth would be much colder, making it difficult for life to survive. The greenhouse effect happens when certain gases in the atmosphere trap heat from the sun and radiated heat from the Earth’s surface, keeping the planet warm.The issue of climate change is caused by an enhanced greenhouse effect. Human activities have led to an increase in the amount of greenhouse gases in the atmosphere, which traps more heat and causes the planet to warm up. This leads to changes in the Earth’s climate, such as rising temperatures, melting ice caps, and changes in precipitation patterns.
These changes can have negative impacts on ecosystems and human societies. So, in conclusion, the greenhouse effect is not bad, but an enhanced greenhouse effect caused by human activities is leading to climate change, which can have negative impacts.
Hence correct option is False.
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What kind of unethical issues might rise due to human
participation in COVID-19 treatment approaches? Explain at least 3
of them in details.
Unethical issues may arise due to human participation in COVID-19 treatment approaches, including inequitable access, exploitation of vulnerable populations, and informed consent violations.
1. Inequitable Access: One ethical concern is the unequal distribution of COVID-19 treatments, where limited resources are disproportionately available to certain groups based on socioeconomic status or geographical location. This can perpetuate health disparities and deprive disadvantaged communities of life-saving interventions.
2. Exploitation of Vulnerable Populations: The pandemic creates opportunities for exploitation, particularly regarding clinical trials and experimental treatments. Vulnerable populations, such as marginalized communities or individuals in desperate situations, may be coerced or manipulated into participating in risky interventions without adequate protection or benefit.
3. Informed Consent Violations: Informed consent is essential in medical interventions, but in the urgency of the pandemic, there is a risk of compromised consent processes. Patients may not receive sufficient information about the potential risks and benefits of treatments, or they may be pressured into consenting without fully understanding the implications. This compromises their autonomy and right to make informed decisions.
Addressing these ethical issues is crucial to ensure that COVID-19 treatment approaches are conducted with fairness, respect for human rights, and adherence to ethical principles.
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1) Mention the reactions catalyzed by Ribulose 1, 5 bis phosphate carboxylase /oxygenase (Rubisco) mentioning the chemical structures of the end products of both reactions. Regulation of the Calvin Cycle: Iodoacetate reacts irreversibly with the free -SH groups of cysteine (Cys) residues in proteins. Predict which Calvin cycle enzymes would be inhibited by iodoacetate & name them. Discuss with diagram the regulation of any one of the above mentioned enzymes.
2) Discuss with diagram the roles of the triose phosphate-phosphate translocator of chloroplast.
1) Ribulose 1,5-bisphosphate carboxylase/oxygenase (Rubisco) is a key enzyme involved in the Calvin cycle, which is the primary pathway for carbon fixation in photosynthesis.
2) The triose phosphate-phosphate translocator (TPT) is an essential protein located in the chloroplast inner membrane. It plays a crucial role in the transport of triose phosphates, specifically dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (G3P), between the chloroplast stroma and the cytosol of plant cells.
1) Rubisco catalyzes two distinct reactions:
Carboxylation: Rubisco catalyzes the addition of carbon dioxide (CO2) to ribulose 1,5-bisphosphate (RuBP), resulting in the formation of two molecules of 3-phosphoglycerate (3-PGA). This reaction is crucial for carbon fixation and the subsequent synthesis of carbohydrates.Oxygenation: Rubisco can also react with oxygen (O2) instead of CO2, leading to the production of one molecule of 3-phosphoglycerate and one molecule of 2-phosphoglycolate. This reaction is known as photorespiration and can result in the loss of fixed carbon.The end products of the carboxylation reaction are two molecules of 3-phosphoglycerate, which are subsequently converted to other compounds through the Calvin cycle.
Inhibition of enzymes in the Calvin cycle by iodoacetate:
Iodoacetate irreversibly inhibits enzymes that contain free sulfhydryl (SH) groups, such as cysteine residues. In the Calvin cycle, several enzymes can be inhibited by iodoacetate, including glyceraldehyde 3-phosphate dehydrogenase (GAPDH) and phosphoribulokinase (PRK). Both of these enzymes contain cysteine residues in their active sites.
Regulation of GAPDH by iodoacetate:
GAPDH is an important enzyme in the Calvin cycle that catalyzes the conversion of glyceraldehyde 3-phosphate (G3P) to 1,3-bisphosphoglycerate (1,3-BPG). The active site of GAPDH contains a cysteine residue that plays a critical role in its catalytic function.
When iodoacetate reacts with the free SH group of the cysteine residue in GAPDH, it forms a covalent bond, inhibiting the enzyme's activity. This leads to the disruption of the Calvin cycle and reduces the production of carbohydrates.
2) The roles of the TPT are as follows:
Export of triose phosphates from the chloroplast: The TPT facilitates the export of DHAP and G3P, which are the products of the Calvin cycle, from the chloroplast stroma to the cytosol. These triose phosphates serve as substrates for various metabolic pathways outside the chloroplast, including the synthesis of sugars, lipids, and amino acids.Import of inorganic phosphate (Pi) into the chloroplast: The TPT also facilitates the import of inorganic phosphate into the chloroplast in exchange for triose phosphates. This is important for maintaining the supply of phosphate required for ATP synthesis and other cellular processes within the chloroplast.Diagrammatically, the TPT is represented as a protein embedded in the chloroplast inner membrane. It spans the membrane and contains binding sites for triose phosphates and inorganic phosphate.
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please explain how an increase of CO2 in the blood will effect the
URINE pH (explain the mechanism, will it increase or decrease urine
pH?)
When an increase in carbon dioxide (CO2) occurs in the blood, it leads to an increase in the acidity of the blood. This is because when CO2 levels in the blood rise, carbonic acid levels also rise, causing a decrease in blood pH. In response, the body will try to eliminate the excess CO2 by increasing respiration.
When an increase in carbon dioxide (CO2) occurs in the blood, it leads to an increase in the acidity of the blood. This is because when CO2 levels in the blood rise, carbonic acid levels also rise, causing a decrease in blood pH. In response, the body will try to eliminate the excess CO2 by increasing respiration. The increase in respiration leads to the elimination of carbon dioxide from the body via exhalation. This process restores blood pH to its normal level.
The kidneys also play a role in regulating blood pH. They excrete excess hydrogen ions (H+) into the urine to maintain the balance of acids and bases in the blood. When blood pH falls, the kidneys excrete more H+ into the urine, causing urine to become more acidic.
Conversely, when blood pH rises, the kidneys excrete fewer H+ into the urine, causing urine to become more alkaline. Therefore, an increase in CO2 levels in the blood will lead to an increase in acidity, causing the kidneys to excrete more H+ into the urine, which will decrease the urine pH.
This is because the excess CO2 in the blood leads to an increase in the production of carbonic acid, which dissociates into bicarbonate ions (HCO3-) and H+ ions. The H+ ions are then excreted by the kidneys into the urine, causing a decrease in urine pH. Hence, an increase in CO2 levels in the blood will cause a decrease in urine pH. This mechanism helps to maintain the body's acid-base balance.
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1. Organism is a regular, non-sporing Gram-positive rod 2. Cell morphology - short rods, often short chains and filaments 3. Diameter of rods (um) - 0.4-0.5 Genus: 4. B-hemolysis negative 5. Acid production from mannitol - positive 6. Acid production from soluble starch - positive 7. Reduction of nitrate - positive Genus/species:
The organism is a non-sporing, Gram-positive rod, with short rod morphology, often forming short chains and filaments. It belongs to the genus Lactobacillus, specifically Lactobacillus plantarum, exhibiting negative B-hemolysis, positive acid production from mannitol and soluble starch, and positive reduction of nitrate.
Based on the provided characteristics, the genus/species of the organism described is likely to be Lactobacillus plantarum.
Lactobacillus is a Gram-positive rod-shaped bacterium commonly found in various environments, including the human gastrointestinal tract, dairy products, and fermented foods.
The organism's short rod morphology, often forming short chains and filaments, aligns with the typical appearance of Lactobacillus species.
The diameter of the rods, ranging from 0.4 to 0.5 micrometers, is consistent with the size of Lactobacillus bacteria.
The identification of the organism as B-hemolysis negative indicates that it does not cause complete lysis of red blood cells on blood agar plates. This is a characteristic feature of Lactobacillus species.
The positive acid production from mannitol and soluble starch is indicative of the organism's ability to ferment these sugars, producing acid as a metabolic byproduct.
Lactobacillus species, including L. plantarum, are known for their fermentative abilities.
The positive reduction of nitrate indicates that the organism possesses the enzyme nitrate reductase, which reduces nitrate to nitrite or other nitrogenous compounds.
This characteristic is commonly found in Lactobacillus species, including L. plantarum.
Therefore, considering all the provided characteristics, the most probable genus/species of the organism described is Lactobacillus plantarum.
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Vince Sandra two children with down syndrome. Vince brother has down syndrome and vince Sister has two kids with down Syndrome. vince and another Sandra plan to have baby, what is probability child wont have down syndrome? a.1/4 b.1/3 c.2/3 d.0
e.1
Since Vince and Sandra already have two kids with Down's syndrome and as Vince's siblings have the disease, there is a high probability of having the disease. The probability that another child will have Down's syndrome will be 1/4. Thus, the correct option is A.
Down's syndrome is an autosomal recessive disease. Lets assume the allele for the disease to be denoted by d. The dominant, unaffected allele would be denoted by D.
Since Vince and Sandra have children affected with Down's syndrome, it means both are carriers of the disease. Thus, the genotype of both Vince and Sandra would be Dd.
A cross between these two would result in genotypes DD, Dd and dd in the ratio of 1:2:1. Thus, the genotype of the affected offspring will be dd.
Therefore, probability of affected offspring = [tex]\frac{Number of affected offspring}{Total number of offspring}[/tex] = [tex]\frac{1}{4}[/tex]
Thus, the correct option is A.
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A double-stranded DNA molecule with the sequence shown here produces, in vivo, a polypeptide that is five amino acids long. -TACATGATCATTTCATGAAATTTCTAGCATGTA- -ATGTACTAGTAAAGTACTTTAAAGATCGTACAT- a. Which strand of DNA is transcribed and in which direction? b. Label the 5' and the 3' ends of each strand. c. If an inversion occurs between the second and the third triplets from the left and right ends, respectively, and the same strand of DNA is transcribed, how long will the resultant polypeptide be? d. Assume that the original molecule is intact and that the bottom strand is transcribed from left to right. Give the base sequence and label the 5' and 3' ends of the anticodon that inserts the fourth amino acid into the nascent polypeptide. What is the amino acid?
a. In this case, the template strand will be 5'-TACATGATCATTTCATGAAATTTCTAGCATGTA-3' and will be transcribed from 3' to 5'.
Transcription is the process by which DNA information is copied into RNA. During transcription, the double-stranded DNA helix is opened, and RNA polymerase moves along one of the strands to generate an RNA transcript. The strand that is copied by RNA polymerase is referred to as the template or antisense strand. The sequence is complementary to the RNA transcript.
b. The 5' end of the template strand is T and the 3' end is A. The 5' end of the sense strand is A, and the 3' end is T.
The direction of DNA synthesis is typically referred to as the 5' to 3' direction. The two DNA strands have a 5' end and a 3' end. The strand running from 5' to 3' on the template will be referred to as the antisense strand, and the complementary strand running from 3' to 5' will be referred to as the sense strand.
c. If an inversion occurs between the second and the third triplets from the left and right ends, respectively, the new sequence will be:
- TACAGTACTTTAAAGATCGTACATGATCATTTCATGAAATTTCTAGCATGTA -
The new sequence has the codons TAC, AGT, ACT, TTT, AAA, GAT, CGT, ACA, TGA, and TTC. As a result, the new polypeptide would be a 10-amino-acid-long chain.
d. The template strand in this case is 5'-TACATGATCATTTCATGAAATTTCTAGCATGTA-3', which would generate a complementary RNA transcript 3'-AUGUACUAGUAAAGUACUUUAAAGAUCGUACAU-5'.
Assuming that the fourth amino acid is coded by the codon AGA, the anticodon sequence will be UCU. The 5' end of the anticodon is the first nucleotide of the anticodon, and the 3' end is the last nucleotide. Therefore, the anticodon sequence will be 5'-UCU-3', which corresponds to the 3' end of the RNA transcript. The amino acid encoded by the AGA codon is arginine. Thus, the fourth amino acid in the nascent polypeptide will be arginine.
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Match each definition with the most correct term. Killer of insects [ Choose ] Killer of pests [Choose ]
Mite and tick killers [ Choose ] Weed Killers [ Choose ] Fungus killers [ Choose] Nematode killers [ Choose ]
Answer Bank:
- Insecticide
- Fungicide - Nematicide - Pesticide - Herbicides - Acaricides
It is a type of pesticide that is specifically designed to eliminate or control insect pests in agriculture, forestry, and households. Examples of insecticides include organophosphates, pyrethroids, and carbamates.A pesticide is any substance that is used to kill, repel, or control pests.
A pesticide can be a herbicide, fungicide, or insecticide depending on the target organism. Some common examples of pesticides include glyphosate, malathion, and permethrin.Acaricides are chemicals used to control or eliminate ticks and mites. They are commonly used in agriculture to prevent damage caused by these pests to crops. Some examples of acaricides include carbaryl and chlorpyrifos.
Herbicides are chemicals used to kill or control weeds. They are commonly used in agriculture to prevent weeds from competing with crops for water and nutrients. Some examples of herbicides include atrazine and glyphosate.Fungicides are chemicals used to kill or control fungi. They are commonly used in agriculture to prevent fungal diseases from damaging crops. Some examples of fungicides include copper sulfate and chlorothalonil.
Nematicides are chemicals used to kill or control nematodes. They are commonly used in agriculture to prevent damage caused by nematodes to crops. Some examples of nematicides include fenamiphos and carbofuran.
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Studying an interesting new unicellular organism (Cellbiology rulez (C.rulez)), you Identify a new polymer which you name Cables and discover the protein that makes up its subunits, which you name Bits. You reflect on your knowledge of actin and microtubules to try to better understand how Cables might be put together Question 6 1 pts Which of the following statement about microtubules and actin is TRUE. Choose the ONE BEST answer. O Microtubules and actin are each made up of monomer subunits that connect together in a head to tail fashion to make protofilaments which come together to form a polymer Nucleotide hydrolysis in actin and microtubule polymers is an essential component of their dynamic behaviors O Microtubules and actin polymers rely on strong bonds between subunits O Microtubules and actin preferentially add subunits to their minus ends A microtubule or actin polymer exposes the same part of its subunit on the plus and minus end You next do a turbidity assay to determine the steady state or critical concentration of Cables, which you determine to be 8UM. In another experiment, you determine that the critical concentration of the D form of Cables is 1uM. Question 7 1 pts Based on what you know for microtubules and actin, which of the following statement is TRUE. Choose the ONE BEST answer. At a subunit concentration below 1 UM. both ends of the Cable will be shrinking At a subunit concentration above 1 UM both ends of the Cable will be shrinking At a subunit concentration below 1 UM. both ends of the Cable will be growing O At a subunit concentration above 1 uM, both ends of the Cable will be growing Question 8 1 pts If Cables behave like microtubules, which of the following do you expect to occur in the presence of non-hydrolyzable GTP? Choose the ONE BEST answer. O Cables would exhibit dynamic instability Cables would increase in polymer mass O Cables would treadmill None of the above
The correct answer is "Nucleotide hydrolysis in actin and microtubule polymers is an essential component of their dynamic behaviors."
Microtubules and actin filaments are both composed of monomer subunits that connect together to form polymers. However, the arrangement and behavior of these polymers differ. Microtubules are composed of α-tubulin and β-tubulin heterodimers that assemble in a head-to-tail fashion to form protofilaments. Multiple protofilaments come together to form the microtubule polymer. Microtubules exhibit dynamic behavior and undergo constant assembly and disassembly, a process known as dynamic instability. Nucleotide hydrolysis of GTP (guanosine triphosphate) bound to β-tubulin is a crucial component of microtubule dynamics. Actin filaments, on the other hand, are composed of monomers called globular actin (G-actin) that polymerize to form filamentous actin (F-actin) in a head-to-tail manner. Actin filaments also exhibit dynamic behavior, and their assembly and disassembly are regulated by ATP (adenosine triphosphate) hydrolysis. Therefore, the correct statement is that "Nucleotide hydrolysis in actin and microtubule polymers is an essential component of their dynamic behaviors." Question 7: The correct answer is "At a subunit concentration above 1 uM, both ends of the Cable will be growing." Based on the behavior of microtubules and actin filaments, the critical concentration of a polymer corresponds to the concentration at which the polymer ends are in a dynamic equilibrium between growth and shrinkage. If the subunit concentration of Cables is below 1 uM (critical concentration), it means that the concentration is too low for the polymer to efficiently assemble, and both ends of the Cable will be shrinking.
Conversely, at a subunit concentration above 1 uM (above the critical concentration), it means that the concentration is sufficient for polymer assembly, and both ends of the Cable will be growing.
Therefore, the correct statement is that "At a subunit concentration above 1 uM, both ends of the Cable will be growing." Question 8: The correct answer is "None of the above." If Cables behave like microtubules, the presence of non-hydrolyzable GTP (guanosine triphosphate) would not cause Cables to exhibit dynamic instability, increase in polymer mass, or undergo treadmill-like movement. These behaviors are specific to microtubules and not necessarily shared by other polymers. The effects of non-hydrolyzable GTP on Cables would depend on the specific mechanisms and properties of Cables, which are currently not described in the given information. Therefore, based on the information provided, none of the given options can be determined as an accurate expectation if Cables behave like microtubules in the presence of non-hydrolyzable GTP.
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genetics (proteins, genes and mutations) Genetics (Proteins, genes & mutations) a) Which of the following types of bond is responsible for primary protein structure? (1 mark) A. Hydrogen bonds B. Ionic bonds C. Covalent bonds D. Glycosidic bonds b) The coding segment of an mRNA molecule contains 873 bases. How many amino acids will be found in the polypeptide chain it codes for? (1 mark) c) Why do silent mutations have no effect on protein structure? (2 marks) d) Name.two types of mutation that can result in a frame shift. (2 marks) e) What is protein denaturation and what could be a potential reason for it to occur in living cells? (2 marks)
a) C. Covalent BONDsb) The coding segment of an mRNA molecule contains 873 bases. Since each codon consists of three bases and codes for one amino acid, the number of amino acids can be determined by dividing the number of bases by three.
In this case, 873 bases divided by three gives 291 amino acids.
c) Silent mutations have no effect on protein structure because they do not change the amino acid sequence encoded by the gene. These mutations occur when a nucleotide substitution in the DNA sequence does not result in a change in the corresponding amino acid due to the degeneracy of the genetic code. Different codons can code for the same amino acid, so even if a nucleotide is changed, the same amino acid is incorporated into the protein during translation.d) Two types of mutations that can result in a frame shift are:
1) Insertion: When one or more nucleotides are inserted into the DNA or mRNA sequence, shifting the reading frame. 2) Deletion: When one or more nucleotides are deleted from the DNA or mRNA sequence, also causing a shift in the reading frame.
e) Protein denaturation refers to the disruption of the protein's native structure, resulting in the loss of its functional shape and activity. It can be caused by various factors, such as high temperature, changes in pH, exposure to chemicals or detergents, or extreme conditions. Denaturation involves the unfolding or disruption of the protein's secondary, tertiary, and quaternary structures, while the primary structure (amino acid sequence) remains intact.
In living cells, potential reasons for protein denaturation include exposure to stressors such as high temperature, extreme pH levels, or the presence of denaturing agents. Environmental changes or cellular stress can lead to the unfolding and misfolding of proteins, disrupting their normal structure and function. Denatured proteins may lose their enzymatic activity, ligand binding ability, or structural integrity, impacting cellular processes and potentially leading to cellular dysfunction or disease.
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