mRNA is synthetized during transcription. a) 12 amino acids. b) UAA is the stop codon. c) 2 metionine. d) mutated protein 1: N- Met-Pro -C (STOP) e) mutated protein 2: N- Met-Ala-Met-Thr-Thr-Pro-Val-Ser-Ala-Leu-Ser-Tyr -C (STOP) f) mutated protein 3: N- Met-Ala-Met-Ala-Arg-Gln-Ser-Arg-His-Tyr-Pro-Ile-Lys-Gly-Asn-C
What is transcription?
The first step before protein arrangement is to synthesize messenger RNA, mRNA.
Transcription is the synthesis of the messenger RNA, mRNA. This event occurs in the nucleus and uses fragments of DNA molecules to do it.
When the DNI molecule separates into two strands to form the transcription bubble, we can identify two separate segments: the coding strand and the template strand.
• The coding strand –non-template strand- goes in direction 5' to 3',
• The complementary strand -template strand- grows in direction 3' to 5'.
The template strand is the one that is going to be complemented by the mRNA. RNA polymerase is in charge of reading the original DNI strand for mRNA synthesis.
mRNA molecule grows complementing DNA base sequences. This is,
DNA mRNA
adenine ⇔ uracil
guanine ⇔ cytosine
thymine ⇔ adenine
cytosine ⇔ guanine
Once mRNA synthesis is over, the molecule leaves the nucleus to start the transcription process in the cytoplasm.
5'-ATGGCCTTCACACAGGAAACAGCTATGGCCATGAGCACGC
3'-TACCGGAAGTGTGTCCTTTGTCGATACCGGTACTCGTGCG
CAGTCTCGGCATTATCCTATTAAAGGGAACTGAGGTGA-3' -----+ 80
GTCAGAGCCGTAATAGGATAATTTCCCTTGACTCCACT-5'
template DNA
TCGATACCGGTACTCGTGCGGTCAGAGCCGTAATAGGATAATTTCCCTTGACTCCACT-5'
mRNA
AGCU AUG GCC AUG AGC ACG CCA GUC UCG GCA UUA UCC UAU UAA AGGGAACUGAGGUGA
Protein
N- Met-Ala-Met-Ser-Thr-Pro-Val-Ser-Ala-Leu-Ser-Tyr -C (STOP)
12 amino acidsUAA is the stop codon 2 metioninemutated mRNA 1
AGCU AUG CCA UGA GCA CGC CAG UCU CGG CAU UAU CCU AUU AAA GGG AAC UGAGGUGA
mutated protein 1
N- Met-Pro -C (STOP)
mutated RNA 2
AGCU AUG GCC AUG ACC ACG CCA GUC UCG GCA UUA UCC UAU UAA AGGGAACUGAGGUGA
mutated protein 2
N- Met-Ala-Met-Thr-Thr-Pro-Val-Ser-Ala-Leu-Ser-Tyr -C (STOP)
mutated RNA 3
AGCU AUG GCC AUG GCA CGC CAG UCU CGG CAU UAU CCU AUU AAA GGG AAC UGA GGUGA
mutated protein 3
N- Met-Ala-Met-Ala-Arg-Gln-Ser-Arg-His-Tyr-Pro-Ile-Lys-Gly-Asn-C
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Complete question
Considering the following double stranded DNA sequence of a eukaryotic gene, transcription begins at the A-T base pair at position 21, determine the following:
5'-ATGGCCTTCACACAGGAAACAGCTATGGCCATGAGCACGC
3'-TACCGGAAGTGTGTCCTTTGTCGATACCGGTACTCGTGCG
CAGTCTCGGCATTATCCTATTAAAGGGAACTGAGGTGA-3'
GTCAGAGCCGTAATAGGATAATTTCCCTTGACTCCACT-5'
a. number of amino acid residues in the synthesized peptide?
b. stop codon read in this sequence
c. number of methionine residues present in the synthesized peptide
d. How would the resulting protein change if the G-C base pair at position 27 in the sequence were deleted?
e. How would the resulting protein change if the G-C base pair at position 35 in the sequence were change to a C-G?
f. How would the resulting protein change if the A-T base pair at position 34 were deleted?
What is the cell concentration here? How many μL of cell suspension do you need to seed 10000 cells per well in a 96-well plate?
The required cell concentration to seed 10,000 cells per well in a 96-well plate is 104.16 cells/μL. To prepare the required cell suspension, 96.15 μL of cell suspension is needed per well.
The cell concentration can be defined as the number of cells present in a unit volume of the cell suspension. It is usually expressed in cells/μL or cells/mL. The cell concentration can be calculated by dividing the number of cells by the volume of the cell suspension. In this case, the cell concentration required to seed 10,000 cells per well in a 96-well plate can be calculated as follows:10,000 cells ÷ 96 wells = 104.16 cells/wellTo calculate the volume of cell suspension needed to seed 10,000 cells per well, we can use the following formula: Volume of cell suspension = Number of cells ÷ Cell concentration. Therefore, the volume of cell suspension needed to seed 10,000 cells per well in a 96-well plate can be calculated as follows: Volume of cell suspension = 10,000 cells ÷ 104.16 cells/μL = 96.15 μL/ wellThus, 96.15 μL of cell suspension is needed per well to seed 10,000 cells per well in a 96-well plate.
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Which of the following is true of all organic molecules? They all contain oxygen. They all are composed of water. They all contain carbon. They all have branched skeletons. They are all hydrophobic.
The true statement among the options provided is that all organic molecules contain carbon. Carbon is the defining element of organic molecules and forms the backbone of their structures.
The presence of carbon allows for the formation of a wide variety of organic compounds, as carbon has a unique ability to form covalent bonds with other carbon atoms and with other elements. While oxygen is a common element found in many organic molecules, it is not true that all organic molecules contain oxygen. Organic molecules can include elements such as hydrogen, nitrogen, phosphorus, sulfur, and others, depending on the specific compound.
The statement that all organic molecules are composed of water is incorrect. Organic molecules can interact with water but are not composed of it. Water, with its polar nature, can form hydrogen bonds with organic molecules, facilitating various biological processes. The statement that all organic molecules have branched skeletons is also incorrect. Organic molecules can have linear, branched, or cyclic structures depending on their composition and arrangement of atoms.
The statement that all organic molecules are hydrophobic is also incorrect. While some organic molecules are hydrophobic, meaning they repel water, others can be hydrophilic, meaning they have an affinity for water and can dissolve in it.
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In the same DNA sequence, present within a gene, a missense mutation occurred that caused deamination of the second C in the top strand; what kind of mutation would be the immediate consequence of this event? (the sequence is broken into triplets only for ease of reading) 5' GGC TAT CTT CGA 3' CCG ATA GCC GCT
Missense mutation due to deamination of the second C in the DNA sequence 5' GGC TAT CTT CGA 3' CCG ATA GCC GCT would be an immediate consequence of the event. A missense mutation is a type of mutation where a change in a single nucleotide of DNA results in a codon that codes for a different amino acid.
A point mutation that causes a codon to code for a different amino acid is called missense mutation.
This is because it alters the amino acid sequence of the protein that the gene codes for. In the given DNA sequence, the deamination of the second C in the top strand would result in a GGC to GAC substitution in the mRNA. This means that the codon that was originally coding for glycine (GGC) would now code for aspartic acid (GAC) in the mRNA sequence. Hence, a missense mutation would be the immediate consequence of this event.
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_________ is a term used to describe abnormal gut function
Irritable bowel syndrome (IBS) is a term used to describe abnormal gut function. It is a common disorder that affects the large intestine and causes symptoms such as abdominal pain, bloating, diarrhea, and constipation.
The exact cause of IBS is unknown, but it is believed to involve a combination of factors including abnormal muscle contractions in the intestine, increased sensitivity to pain, and changes in the gut microbiome. Treatment for IBS usually focuses on managing symptoms through dietary changes, stress reduction, and medication.
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The solubility of peptides in water depends on the relative polarity of their side chain groups, in particular on the number of ionized groups. State which of the three peptides provided in a) and b) below is MORE soluble at the indicated pH and explain your reasoning in both cases. (a) [Lys-Ala] or [Met-Phe] or [Leu-Gln) at pH 7.0 (b) [Ala-Ser-Leu] or [Asn-Ser-His] or [Ile-Phe-Tyr] at pH 6.0
[lys-ala] would have greater solubility due to the presence of the ionized lysine residue.
(a) among the peptides [lys-ala], [met-phe], and [leu-gln] at ph 7.0, [lys-ala] is expected to be more soluble. the solubility of peptides in water is influenced by the relative polarity of their side chain groups and the presence of ionized groups. [lys-ala] contains a lysine (lys) residue, which has a positively charged amino group at physiological ph (ph 7.0). the positive charge makes it more hydrophilic and enhances its solubility in water. in contrast, both [met-phe] and [leu-gln] do not have ionizable groups at physiological ph, so their solubility would depend mainly on the hydrophobicity of their side chain groups. (b) among the peptides [ala-ser-leu], [asn-ser-his], and [ile-phe-tyr] at ph 6.0, [ile-phe-tyr] is expected to be more soluble. at ph 6.0, the solubility of peptides is influenced by the relative polarity of their side chain groups and the presence of ionized groups. [ile-phe-tyr] contains tyrosine (tyr), which has a phenolic hydroxyl group that can be ionized and become negatively charged at lower ph values. this ionization contributes to its solubility in water. on the other hand, [ala-ser-leu] and [asn-ser-his] do not possess ionizable groups at ph 6.0, so their solubility would depend mainly on the hydrophobicity of their side chain groups. hence, [ile-phe-tyr] would have greater solubility due to the presence of the potentially ionizable tyrosine residue.
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Hypothesis: The presence of solute impacts osmosis, causing cells to gain or lose mass. You are given the following materials: 10% sucrose solution, dialysis bags, orange clips, distilled water, beakers, electronic balance, graduated cylinders, weigh boat, timer, a funnel. REMEMBER: SUCROSE IS TOO LARGE TO PASS THROUGH THE PORES OF THE DIALYSIS BAGS. Identify the independent variable (0.5pt): Identify the dependent variable (0.5 pt): State at least 2 confounding variables (1 pts): Identify any controls (1 pt): Now, devise a protocol to test the above hypothesis to demonstrate the gain of mass by a dialysis bag, using the materials listed above. DETAILS MUST BE PROVIDED TO RECEIVE FULL CREDIT. (4 pts) Finally, you construct a graph using data collected from your experiment. What specifically will you put on the X axis? How will label it? (1 pt) What specifically will you put on the Y axis? How will you label it? (1 pt) What type of graph will you construct? (1 pt)
The independent variable in the experiment is the presence of solute in the dialysis bag. The dependent variable is the change in mass of the dialysis bag.
Two potential confounding variables could be the initial mass of the dialysis bag and the temperature of the surrounding environment. The control group would involve using a dialysis bag filled with only distilled water.
To test the hypothesis, the protocol involves filling dialysis bags with different concentrations of sucrose solution, placing them in separate beakers with distilled water, and measuring the change in mass over a specific time period.
The X-axis of the graph will represent the concentration of solute in the dialysis bag, labeled as "Concentration (sucrose %)." The Y-axis will represent the change in mass of the dialysis bag, labeled as "Change in Mass (grams)." A line graph would be suitable for displaying the data.
The independent variable in this experiment is the presence of solute, specifically the concentration of sucrose solution in the dialysis bag. The experiment aims to investigate how the presence of solute impacts osmosis and the resulting change in mass of the dialysis bag.
By varying the concentration of sucrose solution, the effect on osmosis can be observed.
The dependent variable is the change in mass of the dialysis bag. The mass of the dialysis bag before and after the experiment will be measured, and the difference will indicate whether the dialysis bag gained or lost mass.
Two potential confounding variables that should be considered are the initial mass of the dialysis bag and the temperature of the surrounding environment.
The initial mass of the dialysis bag may vary between different bags, which could affect the overall change in mass. The temperature can also impact the rate of osmosis, as higher temperatures may increase the rate of molecular movement.
To conduct the experiment, the protocol involves filling multiple dialysis bags with different concentrations of sucrose solution, ranging from 0% (distilled water) to 10%. Each bag will be securely sealed with an orange clip.
The bags will then be placed in separate beakers filled with distilled water. The beakers will be labeled with the corresponding sucrose concentration.
The bags will be left in the beakers for a specific time period, allowing osmosis to occur.
After the designated time, the dialysis bags will be removed from the beakers, gently blotted dry, and weighed using an electronic balance.
The change in mass for each bag will be calculated by subtracting the initial mass from the final mass.
For constructing the graph, the X-axis will represent the concentration of solute in the dialysis bag and will be labeled as "Concentration (sucrose %)." The Y-axis will represent the change in mass of the dialysis bag and will be labeled as "Change in Mass (grams)."
Since the concentration of solute is a continuous variable, a line graph would be suitable for displaying the data and showing any trends or patterns.
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(a) With the aid of a detailed labelled diagram, give an account of the structure of the cell surface membrane, explaining the function of the various components. (b) Explain the role of the major organelles found in an animal cell and explain the importance of their membranes.
(a) The cell surface membrane, also known as the plasma membrane, is a vital component of the cell that separates the intracellular environment from the extracellular space. It is composed of a phospholipid bilayer embedded with various proteins and other components. The phospholipid bilayer consists of two layers of phospholipids, with their hydrophilic heads facing outward and hydrophobic tails facing inward, creating a selective barrier.
The various components of the cell surface membrane include integral proteins, peripheral proteins, cholesterol, and glycoproteins. Integral proteins span the entire phospholipid bilayer, while peripheral proteins are found on the inner or outer surface. These proteins play key roles in transport of molecules, cell signaling, and maintaining the structural integrity of the membrane. Cholesterol molecules are interspersed within the phospholipid bilayer, contributing to membrane fluidity and stability. Glycoproteins, which have carbohydrate chains attached, participate in cell recognition and immune response.
The cell surface membrane functions as a selective barrier, controlling the movement of substances into and out of the cell. It regulates the exchange of ions, nutrients, and waste products, maintaining homeostasis. The proteins embedded in the membrane facilitate cell signaling and communication with the external environment. Additionally, the membrane provides mechanical support, allowing the cell to maintain its shape.
(b) Animal cells contain several major organelles, each with its own specific functions. These organelles are enclosed by membranes that play crucial roles in compartmentalization and maintaining specialized conditions within the organelles.
The nucleus is the most prominent organelle and is surrounded by the nuclear membrane or envelope. It houses the genetic material and controls the cell's activities by regulating gene expression.
Mitochondria are responsible for generating energy in the form of adenosine triphosphate (ATP) through cellular respiration. Their double membrane structure allows for efficient ATP production.
Endoplasmic reticulum (ER) is a network of membranes involved in protein synthesis and lipid metabolism. The rough ER is studded with ribosomes and participates in protein synthesis, while the smooth ER is involved in lipid synthesis and detoxification.
Golgi apparatus consists of a series of flattened membranous sacs. It modifies, sorts, and packages proteins and lipids for transport to specific destinations inside or outside the cell.
Lysosomes contain digestive enzymes that break down cellular waste and foreign substances. Their membrane prevents the enzymes from damaging other cellular components.
The plasma membrane, as mentioned earlier, is also a vital organelle that regulates the exchange of materials between the cell and its environment.
The membranes surrounding these organelles compartmentalize cellular processes, allowing for efficient and specialized functions. They regulate the movement of molecules, facilitate selective transport, and maintain distinct chemical environments necessary for specific cellular processes. Membrane-bound organelles ensure that various metabolic reactions occur in separate compartments, enhancing cellular efficiency and organization.
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Case Study: Part One Saria is at the doctor to get the lab results of the samples she brought in to be tested. From the results, it appears that she is getting the rashes due to Pseudomonas aeruginosa infection that she contracted from the sponge she was sharing with her roommates. Now, we have to run further tests to check for the appropriate antibiotic needed to get rid of the infection. We also need to make sure to protect the normal flora in Saica so only the bad germs die. To do this we will use a gene transfer method to protect her healthy germs from the effects of possible antibiotics we can use. Introduction/Background Material: Basics of Bacterial Resistance: Once it was thought that antibiotics would help us wipe out forever the diseases caused by bacteria. But the bacteria have fought back by developing resistance to many antibiotics, Bacterial resistance to antibiotics can be acquired in four ways: 1. Mutations: Spontaneous changes in the DNA are called mutations. Mutations happen in all living things, and they can result in all kinds of changes in the bacterium. Antibiotic resistance is just one of many changes that can result from a random mutation. 2. Transformation: This happens when one bacterium takes up some DNA from the chromosomes of another bacterium 3. Conjugation: Antibiotic resistance can be coded for in the DNA found in a small circle known as a plasmid in a bacterium. The plasmids can randomly pass between bacteria (usually touching as seen in conjugation) 4. Recombination: Sharing of mutations, some of which control resistance to antibiotics. Some examples are: A. Gene cassettes are a small group of genes that can be added to a bacterium's chromosomes. The bacteria can then accept a variety of gene cassettes that give the bacterium resistance to a variety of antibiotics. The cassettes also can confirm resistance against disinfectants and pollutants. B. Bacteria can also acquire some genetic material through transduction (e.g., transfer through virus) or transformation. This material can then lead to change in phenotype after recombination into the bacterial genome. The acquired genetically based resistance is permanent and inheritable through the reproductive process of bacteria, called binary fission. Some bacteria produce their own antibiotics to protect themselves against other microorganisms. Of course, a bacterium will be resistant to its own antibiotic! If this bacterium then transfers its resistance genes to another bacterium, then that other bacterium would also gain resistance. Scientists think, but haven't proved, that the genes for resistance in Saica's case have been transferred between bacteria of different species through plasmid or cassette transfer. Laboratory analysis of commercial antibiotic preparations has shown that they contain DNA from antibiotic-producing organisms.
The resistance of bacteria to antibiotics is a major concern for public health. Bacterial resistance to antibiotics can be acquired in four ways; mutations, transformation, conjugation, and recombination.
In this case, Saria contracted Pseudomonas aeruginosa infection through a sponge she shared with her roommates.
To get rid of the infection, the appropriate antibiotic needs to be used while ensuring the healthy germs are protected from the effects of the antibiotic. This bacterium is antibiotic-resistant. Bacterial resistance to antibiotics can be acquired in four ways: Mutations, Transformation, Conjugation, and Recombination. Antibiotic resistance can be caused by random mutations in bacterial DNA. Antibiotic resistance can be coded for in the DNA found in a small circle known as a plasmid in a bacterium. The plasmids can randomly pass between bacteria.
This can be achieved through a gene transfer method.
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Fungicides act to prevent the growth pathogenic fung through the following ways except
a. inhibroion of DNA replication b. inhibition of mitosis c. injury to plasma membrane d. inhibition of peptidoglycan synthesus
Fungicides act to prevent the growth of pathogenic fungi by inhibiting DNA replication, mitosis, and causing injury to the plasma membrane. They do not inhibit peptidoglycan synthesis, as this is a characteristic feature of bacteria rather than fungi.
Fungicides are chemicals specifically designed to target and control fungal pathogens. They work by disrupting essential processes in fungi, inhibiting their growth and reproduction. Several mechanisms of action are employed by fungicides to achieve this goal.
Firstly, fungicides can inhibit DNA replication in fungi. This prevents the synthesis of new genetic material and hampers the ability of the fungi to reproduce and spread.
Secondly, fungicides can interfere with mitosis, the process by which cells divide. By disrupting mitotic processes, fungicides hinder the growth and development of fungal cells.
Additionally, fungicides can cause injury to the plasma membrane of fungal cells. The plasma membrane plays a vital role in maintaining cell integrity and regulating nutrient uptake. Disruption of the plasma membrane leads to cell death and inhibits fungal growth.
However, fungicides do not inhibit peptidoglycan synthesis. Peptidoglycan is a key component of bacterial cell walls, providing structural support and protection. Since fungi have different cell wall structures compared to bacteria, fungicides do not target peptidoglycan synthesis.
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true or false Here is a phylogeny of eukaryotes determined by DNA evidence. All of the supergroups contain some photosynthetic members.
The statement "All of the supergroups contain some photosynthetic members" in reference to a phylogeny of eukaryotes determined by DNA evidence is a true statement.
Supergroups are a collection of phylogenetically related eukaryotes. These lineages, which were once referred to as "Kingdom Protista," are now grouped into the six supergroups that make up the eukaryotic tree of life. In each supergroup, some members engage in photosynthesis.
The six supergroups are as follows:
ExcavataChromalveolataRhizariaArchaeplastidaAmoebozoaOpisthokontaAs a result, it is correct to say that all supergroups contain some photosynthetic members.
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Please explain what each tests indicates when bacteria get tested.
Coagulase is negative which indicates?
Mannitol is negative which indicates?
Oxidase is negative which indicates?
Alkaline Phosphate is positive which indicates?
Sucrose is positive which indicates?
Acetoin production is negative ehich indicates?
Trehalose is negative which indicates?
Coagulase negative indicates the absence of a specific enzyme produced by bacteria.
Mannitol negative indicates the inability of bacteria to ferment mannitol sugar.
Oxidase negative indicates the absence of the oxidase enzyme in bacteria.
Alkaline Phosphate positive indicates the presence of the alkaline phosphate enzyme in bacteria.
Sucrose positive indicates the ability of bacteria to ferment sucrose sugar.
Acetoin production negative indicates the absence of acetoin production by bacteria.
Trehalose negative indicates the inability of bacteria to utilize trehalose sugar.
Coagulase is an enzyme produced by certain bacteria, such as Staphylococcus aureus. A negative coagulase test indicates the absence of this enzyme, which suggests that the bacteria being tested do not belong to the coagulase-positive Staphylococcus species.Mannitol fermentation test is used to determine the ability of bacteria to ferment mannitol sugar. A negative result indicates that the bacteria cannot utilize mannitol as a carbon source for energy production.
Oxidase test is performed to detect the presence of the oxidase enzyme in bacteria. A negative result suggests that the bacteria being tested lack this enzyme, which is commonly found in certain groups of bacteria such as Pseudomonas and Neisseria.Alkaline Phosphate is an enzyme produced by bacteria, and a positive result in the alkaline phosphate test indicates the presence of this enzyme. It is often used in the identification of certain bacteria, including members of the Enterobacteriaceae family.Sucrose fermentation test is used to assess the ability of bacteria to ferment sucrose sugar. A positive result indicates that the bacteria can utilize sucrose as a carbon source, aiding in the identification and differentiation of bacterial species.
Acetoin is a metabolic byproduct produced by some bacteria during the fermentation of glucose. A negative result in the acetoin production test suggests that the bacteria being tested do not produce acetoin.Trehalose is a sugar that some bacteria can utilize as a carbon source. A negative result in the trehalose test indicates that the bacteria being tested cannot utilize trehalose for growth and metabolism.
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23) Cholesterol makes this possible: : : a) testosterone b) ATP c) Glucose d) estrogen e) a and d 24) This accompanied brain enlargement: a) changes in HAR genes b) bipedalism c) quadripedalism d) sagittal crest development e) all of these
Cholesterol makes this possible.Cholesterol makes Testosterone and Estrogen possible. These are sex hormones that regulate various bodily functions.
Both of these hormones are steroid hormones that are synthesized from cholesterol.Cholesterol is a molecule that is vital for the body's normal functioning. It helps to make cell membranes more robust and sturdy. It also aids in the production of hormones, vitamin D, and bile acids.
This accompanied brain enlargement. Changes in HAR genes, bipedalism, and sagittal crest development accompanied brain enlargement. The expansion of the human brain is one of the most significant evolutionary changes that occurred during the course of human evolution. It resulted in a variety of adaptations, including an increase in brain size and complexity.
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Which of the following is not considered a deadenylation-independent degradation pathway? a. Histone mRNA pathway b. All of these are considered deadenylation-independent degradation pathways c. miRNA pathway d. Endonucleolytic pathway e. Deadenylation-independent decapping
The correct answer is: b. All of these are considered deadenylation-independent degradation pathways.
All of the options listed are considered deadenylation-independent degradation pathways. Let's briefly explain each pathway:
a. Histone mRNA pathway: Histone mRNAs undergo a specific degradation pathway that does not involve deadenylation. These mRNAs lack a poly(A) tail and are degraded through a specialized mechanism.
c. miRNA pathway: MicroRNAs (miRNAs) are small non-coding RNAs that can target specific mRNA molecules for degradation. The degradation of targeted mRNAs by miRNAs occurs independently of deadenylation.
d. Endonucleolytic pathway: In the endonucleolytic pathway, mRNA degradation occurs through the cleavage of the mRNA molecule at internal sites by endonucleases. This pathway bypasses the deadenylation step.
e. Deadenylation-independent decapping: In some cases, mRNA degradation can occur through the removal of the protective cap structure at the 5' end of the mRNA, leading to its rapid degradation. This decapping-mediated degradation can occur independently of deadenylation.
Therefore, all of the listed pathways (a, c, d, e) are considered deadenylation-independent degradation pathways, making option b incorrect.
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Thin-layer chromatography (TLC) is a commonly used technique to separate macromolecules in biochemistry labs worldwide. There are many other types of chromatography techniques. Perform some research and describe one other type of chromatography technique and one application where this technique might be used, e.g. example in medicine, research or biotechnology.
Thin-layer chromatography (TLC) is a separation technique that is used to identify the presence of different components in a sample. This process involves the separation of a mixture into its individual components and the analysis of these components in terms of their chemical and physical properties.
There are various types of chromatography techniques used in the field of biochemistry labs other than TLC.
Two-dimensional chromatography is one of the popular methods used for separating complex mixtures.
2D chromatography is a type of separation method that employs two chromatographic procedures to separate a sample.
One of the commonly used 2D chromatography techniques is gas chromatography (GCxGC).
GCxGC separates samples based on the boiling points of their individual components.
In addition, GCxGC separates different components based on their individual retention times.
2D chromatography is used in many applications in biotechnology, research, and medicine.
In the biotechnology field, 2D chromatography is used to separate complex proteins, peptides, and other macromolecules that cannot be separated using traditional chromatography techniques.
In the medical field, 2D chromatography is used to separate complex biological samples, such as blood samples, for analysis and diagnosis.
In the research field, 2D chromatography is used to separate complex organic samples for chemical analysis and identification.
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Biol 128
Week 10 Worksheet
1. Based on discussions in Chapter 14a/15, would someone with poorly
managed diabetes be more prone to acidosis, or to alkalosis? Why?
2. What is the role of the respiratory system in controlling acid-base balance?
(Specifically, how does the respiratory system make the blood pH more basic,
and more acidic?)
Poorly managed diabetes can lead to a condition called ketoacidosis. When the body can't use glucose for energy, it starts to break down fat instead. This process produces ketones, which can cause the blood to become more acidic.
Therefore, someone with poorly managed diabetes would be more prone to acidosis than alkalosis.2. The respiratory system helps regulate the pH of the blood by controlling the levels of carbon dioxide (CO2) in the body. Carbon dioxide is a waste product of cellular respiration and is produced in the body continuously. When CO2 levels in the blood increase, the respiratory system responds by increasing the rate and depth of breathing. This causes more CO2 to be exhaled from the body, which helps to lower the concentration of CO2 in the blood. This, in turn, makes the blood more alkaline. Conversely, when CO2 levels in the blood decrease, the respiratory system responds by decreasing the rate and depth of breathing. This causes less CO2 to be exhaled from the body, which helps to increase the concentration of CO2 in the blood. This, in turn, makes the blood more acidic. In summary, the respiratory system plays an important role in regulating the pH of the blood by controlling the levels of carbon dioxide in the body.
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Please help
Requirement
1. TRUE: Write the statement only; do not explain why the
statement is
true.
2. FALSE:
A. Original statement: Write the statement as written
(below).
B. Corrected statement: Wr
TRUE: The Earth is round.
FALSE:
Original statement: The Earth is flat.
Corrected statement: The Earth is not flat.
What does a true statement look like?The Earth is a sphere, which means that it is round like a ball. It is not flat, as some people believe. The Earth's round shape is supported by scientific evidence, such as the fact that ships disappear over the horizon as they sail away, and that the Earth casts a round shadow on the moon during a lunar eclipse.
The belief that the Earth is flat is a relatively recent phenomenon. It is often associated with conspiracy theories and pseudoscientific beliefs. There is no scientific evidence to support the claim that the Earth is flat.
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Complete question:
Please help
Requirement 1:
* TRUE: Write the statement only; do not explain why the statement is true.
* FALSE:
* Original statement: Write the statement as written below.
* Corrected statement: Write the corrected statement.
A molecular clock uses changes in the DNA sequences of a common gene to measure the time since related organisms shared a common ancestor. To construct a molecular clock, we need to determine (calculate) how the DNA sequence for common genomic regions have changed (diverged) over a known period of time for the organisms (individuals) being studied. Known divergence times are often inferred from the fossil record, or can be estimated based on known mutation rates. The percent sequence divergence is a straightforward calculation. First, determine how many differences there are between two DNA sequences from the same gene in different individuals or species. Figure 1 below shows some hypothetical DNA sequences from three primate species over four gene regions. Use the information in Figure 1 to complete Table 1. (1 pt)
Molecular clocks are a set of measures used to determine the divergence time between organisms based on the differences in the nucleotide sequences of the same gene.
The method of estimating the time since two species diverged is based on the premise that mutations accumulate in nucleotide sequences over time at a relatively constant rate. The number of nucleotide differences in a given gene between two species can be used to estimate how long ago the two species diverged from a common ancestor. These differences are often referred to as sequence divergence.
The term percent sequence divergence is used to describe the proportion of nucleotides that differ between two sequences from a given gene in different individuals or species. To use molecular clocks, it is essential to determine how the DNA sequence for common genomic regions has changed (diverged) over a known period of time for the organisms being studied. Known divergence times can be inferred from the fossil record, or they can be estimated based on known mutation rates.
For example, scientists know that some DNA regions are more conserved than others and evolve more slowly, while other regions of the genome are more prone to change (or evolve more rapidly).
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What structure is necessary for the reversible binding of O2
molecules to hemoglobin and myoglobin? At what particular part of
that structure does the protein-O2 bond form?
The structure that is required for the reversible binding of O2 molecules to hemoglobin and myoglobin is known as heme. Heme is a complex organic molecule consisting of a porphyrin ring that binds iron in its center, which is the binding site for O2.
The iron atom is held in a fixed position by four nitrogen atoms that form a planar structure. The fifth position is occupied by a histidine residue, which is supplied by the protein. The sixth position is where O2 binds in the presence of heme. The binding of O2 to heme is an electrostatic interaction between the positively charged iron atom and the negatively charged O2 molecule.
This interaction causes the O2 molecule to be slightly bent, which enables it to fit more tightly into the binding site. The strength of this bond is affected by various factors such as pH, temperature, and pressure, which can cause the bond to weaken or break. The protein-O2 bond forms at the sixth position of the heme structure.
The sixth position is where the O2 molecule binds to the iron atom, forming a complex that is stabilized by the surrounding amino acids. The histidine residue in the protein provides one of the nitrogen atoms that hold the iron in place. The other three nitrogen atoms are provided by the porphyrin ring.
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Calculate the maximum tension of bone with a cross section area was 2.4 cm2 could withstand just prior to fracture long of bone is 35cm. the tensile breaking is 120N/mm^2.
The maximum tension a bone with a cross-sectional area of 2.4 cm² could withstand just before fracturing is 28,800 Newtons
To calculate the maximum tension a bone can withstand, we need to use the formula:
Maximum Tension = Tensile Breaking Strength × Cross-Sectional Area
Given that the tensile breaking strength is 120 N/mm² and the cross-sectional area is 2.4 cm², we need to convert the units to be consistent. Since 1 cm² equals 100 mm², the cross-sectional area can be converted to mm² by multiplying it by 100:
Cross-Sectional Area (mm²) = 2.4 cm² × 100 mm²/cm² = 240 mm²
Now we can calculate the maximum tension using the formula:
Maximum Tension = 120 N/mm² × 240 mm² = 28,800 N
Therefore, the maximum tension the bone can withstand just prior to fracture is 28,800 Newtons. This calculation assumes that the bone is under pure tension and that other factors such as fatigue, bending, or impact are not considered.
It is important to note that bones have complex structures, and various factors can influence their strength and resistance to fracture in real-life scenarios.
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limitation of filtration membrane model in compared of the function
in the human body
4. Materials: filter paper funnel. beaker/ cup Solvent X Solute A . Solute B . Wooden sticks/stirrer
Filtration membrane model is one of the biological models that is used to explain the ultrafiltration process in the kidney.
It is used to understand how the kidney removes waste products from the body and maintains the balance of water, salts, and other substances in the body. However, this model has some limitations when compared to the function of the kidney in the human body.Limitations of Filtration Membrane Model when compared to the function in the human body are as follows:The filtration membrane model only explains the physical process of ultrafiltration and does not take into account the complex biochemical processes that occur in the kidney. The model does not include the role of different cells and enzymes in the kidney that are involved in the filtration process. These cells and enzymes help to regulate the filtration rate and maintain the balance of fluids and electrolytes in the body.The filtration membrane model only considers the filtration of small molecules and ions, but in reality, the kidney filters a wide range of molecules including proteins and other macromolecules. The model does not explain how these larger molecules are removed from the body.The filtration membrane model does not take into account the dynamic nature of the kidney and how it adjusts to different physiological conditions.
The kidney has the ability to adjust the filtration rate in response to changes in blood pressure, hormone levels, and other factors. The model does not explain how this is achieved.The filtration membrane model is a simplified representation of the complex process of ultrafiltration that occurs in the kidney. It has some limitations when compared to the actual function of the kidney in the human body. However, it provides a useful framework for understanding the basic principles of ultrafiltration and how the kidney maintains the balance of fluids and electrolytes in the body.
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Hello, I have a question in terms of Hemolysis that I need an in-depth answer to if possible. I conducted an experiment using different Urea compounds with Urea causing almost instantaneous Hemolysis while a compound such as Ethyl Urea takes 30 seconds longer in relation to it. I need to describe why Urea causes hemolysis, the polarity of the four compounds tested (Urea / Methyl Urea / Dimethyl Urea / Ethyl Urea), and how tonicity works in the case of urea, and just what are all the factors that affect transport across the cell membrane.
Hemolysis is the rupturing of red blood cells (RBCs) with the release of hemoglobin, which is a protein present in RBCs. The compound Urea is known to cause Hemolysis by disrupting the osmotic balance of the RBCs. It causes water to move out of the cell leading to cell shrinkage.
The presence of a hypertonic solution causes the cell to lose water via osmosis resulting in cell shrinkage. The greater the osmotic pressure, the more the cell shrinkage. This is the reason why urea causes hemolysis.Tonicity is the ability of a solution to cause a cell to gain or lose water molecules.The transport across the cell membrane depends on several factors. The factors that affect the transport across the cell membrane include the following:
Concentration gradient: The concentration gradient is the difference in solute concentration across the membrane. The molecules move from the region of higher concentration to lower concentration.
Osmotic pressure: The pressure generated due to water flow is osmotic pressure. The higher the concentration of solute, the greater the osmotic pressure.
Membrane permeability: It is the extent to which a membrane allows the molecules to pass through it. Membrane permeability varies for different molecules.
Methyl Urea is polar due to the presence of the carbonyl functional group. Dimethyl Urea is polar due to the presence of the carbonyl functional group.Ethyl Urea is polar due to the presence of the carbonyl functional group.
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Do you think we are at greater risk for vitamin deficiencies in
the US or vitamin toxicities?
The United States has seen an increase in vitamin deficiencies and vitamin toxicities, and both can have serious consequences. The country has experienced both increased and decreased intakes of specific vitamins over the years, leading to various health concerns.
Vitamin D deficiency is widespread in the US, and vitamin B12 deficiency is becoming increasingly prevalent. Overconsumption of certain vitamins such as Vitamin A and Vitamin D can lead to toxicities. There is no simple answer to whether the US is at greater risk for vitamin deficiencies or toxicities because it depends on which nutrient you are looking at and in which population.
Those with specific medical conditions, such as digestive disorders, may have difficulty absorbing certain vitamins, resulting in deficiencies. In contrast, people who take high doses of certain vitamins, particularly fat-soluble vitamins like Vitamin A and Vitamin D, are more likely to experience toxicities.High-quality diets, in which a variety of foods from all food groups are consumed, are recommended to promote nutrient adequacy and reduce the risk of nutrient deficiencies and toxicities. Furthermore, individuals should consult their healthcare providers before taking dietary supplements to determine if they are required and what amounts are appropriate.
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I am a member of the phytoplankton community that is covered with calulose plates called a theca dominate the phytoplankton in late summer in mid-lattudes, and am almost always dominant in the tropics I am also bioluminescent To which group do I belong? a. diatoms b. coccolithophores c. cyanobacteria d. dinoflagellates
I belong to the Dinoflagellates group.
Dinoflagellates are a group of single-celled organisms that belong to the Protista kingdom. Dinoflagellates have two flagella that help them move in the water column. These organisms are the largest group of marine phytoplankton. Dinoflagellates are important members of the food chain in the ocean. They are also known for producing bioluminescence, which means they emit light. A member of the phytoplankton community that is covered with calcite plates called a theca is a coccolithophore. They are a group of single-celled algae that have calcified external coverings. Coccolithophores are also dominant in the tropics and have bioluminescence. But, they are not the dominant phytoplankton in late summer in mid-latitudes. Diatoms are another type of phytoplankton. They are single-celled organisms that have cell walls made of silica. However, diatoms are not bioluminescent and do not have theca. Cyanobacteria are also known as blue-green algae. They are a group of photosynthetic bacteria that are typically found in freshwater. They do not have a theca and are not bioluminescent. Therefore, the correct option is (d) dinoflagellates.
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Rapidly-dividing stem cells exhibit a major difference in their
mode of cell division compared with slower-dividing stem cells.
Explain what this difference is and a possible reason for it.
Rapidly-dividing stem cells exhibit a significant difference in their mode of cell division compared to slower-dividing stem cells. The major difference is that rapidly dividing stem cells undergo asymmetric division while slower dividing stem cells undergo symmetric division.
Asymmetric division occurs when a stem cell divides into two daughter cells with different fates, one daughter cell being a stem cell and the other differentiating into a specialized cell. On the other hand, symmetric division occurs when a stem cell divides into two daughter cells that are identical to each other and are also stem cells.
A possible reason for this difference is the rate of DNA replication. Rapidly dividing stem cells need to produce a large number of cells in a short period of time. During DNA replication, there is a possibility of mutations and errors, which are a cause of cell death or abnormalities.
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Match the following terms with their description: Column A 1. Oats rich in soluble fiber Bran rich in insoluble fiber Sugar replacer Wheat flour White jasmine rice Satiety Artificial sweeteners Fiber
Fiber is a vital nutrient for the human body, which helps maintain normal digestion and is also essential for reducing the risk of chronic diseases such as heart disease, stroke, cancer, and diabetes.
Soluble fiber is known to bind with water and slows down digestion, which in turn makes us feel full longer. Insoluble fiber is not easily digestible and helps prevent constipation by adding bulk to the stool. Here is how the terms are matched with their description.1. Oats rich in soluble fiber - Soluble fiber2. Bran rich in insoluble fiber - Insoluble fiber3. Sugar replacer - Artificial sweeteners4. Wheat flour - Fiber5. White jasmine rice - Satiety6. Artificial sweeteners - Sugar replacer7.
Fiber - Wheat flour8. Satiety - White jasmine riceA healthy diet is the key to good health. Whole foods, fruits, vegetables, nuts, and legumes are all good sources of dietary fiber. To increase fiber intake, you should aim to eat at least 25 grams of fiber per day.
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You are studying a virus that causes a severe form of gastroenteritis. The spike protein on the outside of a virus appears to bind to a specific receptor on the outside of cells of the gastrointestinal tract that it infects. When the spike protein binds to this receptor, it causes the receptor to change shape and distort the shape of the plasma membrane, effectively pulling the virus into the cell. A. Does this reflect an 'induced fit' or 'conformation stabilisation and selection' model of receptor activation? Explain briefly. B. An active site histidine is found to be crucial for the conformational change in this receptor and is located close to a glutamate residue that also appears to be essential for activity. The pka of the side chain of this His residue is ~9 rather than ~6. Explain why the pka of the histidine is unusually high. (Question total: 4 marks)
A. The above mentioned phenomena reflect an 'induced fit' model of receptor activation. The induced fit model of the enzyme-substrate complex describes the association between the substrate and enzyme during the reaction. The enzyme is thought to change its shape to better fit the substrate as it binds to it.
Substrates are thought to be in rapid motion, colliding with each other, and the enzyme's active site. The term "induced fit" refers to the process by which the enzyme adjusts its shape to better fit the substrate, creating a strong, complementary fit. The concept of induced fit is used to explain the selective nature of enzyme-substrate binding. B.
The pka of histidine is high because the imidazole ring on the histidine side chain is stabilized by the positive charge from the side chain nitrogen. Histidine is considered an amphoteric amino acid because of its unusual pKa value. The side chain pKa of histidine is between 6 and 7, which makes it ideal for use in the active sites of enzymes.
Because it is a weak acid, it can donate protons to or accept protons from a variety of different functional groups. When it is a proton donor, it is said to be in the "acidic" form, while when it is a proton acceptor, it is in the "basic" form.
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Match the descriptions in the question items to the appropriate selection in the answer items from A-C Answer items may be selected more than once. Decrease Sympathetic activity Decrease somatic motor nerve activity Decrease parasympathetic activity Increase somatic motor nive activity Increase Parasymathetic activity A Detrusor contracts, internal urethral sphincter opens B.External urethral sphincter opens C. Inhibit micturation 2
The correct matching would be:
A - C
B - A
C - B
What is the matchup?Decrease Sympathetic exercise: When concerned activity decreases, it restricts micturition, that means it forestalls pouch contraction. This admits the pouch to hire urine and avoids requirement to urinate.
B) Increase bodily engine nerve activity: When bodily engine nerve activity increases, it leads to the gap of the extrinsic urethral sphincter. This sphincter is under willing control and its gap admits urine expected freed from the bladder all the while willing the act of excreting.
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Different types of cancer have different combinations of characteristics. There are some characteristics that characterize cancer cells in general and make them different from normal cancer cells.
Explain what properties this is.
Different types of cancer have different combinations of characteristics.
However, there are some properties that characterize cancer cells in general and make them different from normal cells.
Cancer cells usually divide uncontrollably.
Here is a detailed explanation of the properties of cancer cells:
Properties of cancer cells
Cancer cells usually divide uncontrollably, and they are different from normal cells in several ways.
Here are the main properties of cancer cells:
Uncontrolled growth:
Cancer cells don't respond to the signals that regulate cell growth.
This means that they divide uncontrollably and form tumors.
Avoidance of apoptosis:
Apoptosis is the programmed cell death that occurs in normal cells.
Cancer cells have a mechanism that allows them to avoid apoptosis and survive.
Angiogenesis:
Cancer cells need a blood supply to grow and divide.
They secrete signals that promote the growth of new blood vessels around the tumor site.
Metastasis:
Cancer cells can spread to other parts of the body through the bloodstream or lymphatic system.
This is known as metastasis.
Genetic instability:
Cancer cells have unstable genomes.
They accumulate genetic mutations that can lead to changes in the properties of the cell.
Cancer cells have properties that make them different from normal cells, and these properties contribute to the development and progression of cancer.
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Strenous exercise should cause an increase in systemic capillary blood flow due to the sympathetic nervous system. True False QUESTION 7 In myocardial contractile cells, the action potential will occu
The given statement is false.
Strenuous exercise causes an increase in systemic capillary blood flow primarily due to vasodilation of arterioles, not the sympathetic nervous system. The sympathetic nervous system plays a role in regulating heart rate and cardiac output during exercise, but its effect on capillary blood flow is limited. Vasodilation of arterioles is mediated by factors such as metabolic demands, local factors (e.g., nitric oxide release), and hormonal responses (e.g., epinephrine), which increase blood flow to active tissues during exercise.
Solution of Question 7:
In myocardial contractile cells, the action potential occurs as a result of a series of electrical changes. The action potential begins with the depolarization phase, initiated by the influx of sodium ions through fast voltage-gated sodium channels. This rapid depolarization leads to the opening of calcium channels, resulting in a plateau phase, where calcium influx balances potassium efflux, thus prolonging the action potential and allowing for sustained contraction. Finally, repolarization occurs as potassium channels open, leading to potassium efflux and restoring the resting membrane potential. This sequential pattern of electrical changes allows for coordinated contraction and relaxation of the myocardium, enabling the heart to pump blood effectively.
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Conclusion about importance gut microbiome in health
and disease
I need adequate answer
The gut microbiome plays an essential role in maintaining overall health and well-being. It is involved in various physiological functions, including digestion, immune system regulation, and metabolism. Therefore, it is vital to maintain a healthy gut microbiome through healthy eating habits, regular exercise, and stress management,
The gut microbiome is an essential component of the human body that is involved in maintaining and regulating a healthy immune system and other physiological functions.
Microbiome imbalances have been linked to numerous diseases and disorders, such as obesity, inflammatory bowel disease, and cancer, among others.
Therefore, maintaining a healthy gut microbiome is essential for overall health and well-being. Here is a more detailed explanation of the importance of the gut microbiome in health and disease:Importance of the gut microbiome in health1. Digestion: The gut microbiome is responsible for breaking down complex carbohydrates and dietary fibers that the human body cannot digest on its own.
This process allows the body to extract essential nutrients, such as vitamins and minerals, from food.2. Immune system: The gut microbiome plays a vital role in regulating and supporting the immune system. It produces immune cells, such as T cells and B cells, and helps maintain a healthy balance of beneficial bacteria that protect against harmful pathogens.3. Metabolism: Studies have shown that the gut microbiome plays a crucial role in regulating metabolism and maintaining healthy blood sugar levels.
It also contributes to the regulation of hormones such as insulin and ghrelin, which regulate appetite and metabolism.
Importance of the gut microbiome in disease1. Obesity: Studies have shown that imbalances in the gut microbiome can lead to weight gain and obesity.
An unhealthy gut microbiome may cause metabolic changes that alter the body's ability to store and burn fat.2. Inflammatory bowel disease: The gut microbiome is believed to play a critical role in the development and progression of inflammatory bowel disease (IBD).
A healthy gut microbiome can prevent inflammation in the gut, whereas an unhealthy gut microbiome may trigger chronic inflammation and the development of IBD.3. Cancer: The gut microbiome can play a crucial role in the development and progression of certain cancers, such as colon cancer.
An unhealthy gut microbiome can lead to inflammation and DNA damage, increasing the risk of developing cancer.
In conclusion, the gut microbiome plays an essential role in maintaining overall health and well-being. It is involved in various physiological functions, including digestion, immune system regulation, and metabolism.
Therefore, it is vital to maintain a healthy gut microbiome through healthy eating habits, regular exercise, and stress management, among other lifestyle factors.
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