Cestoda is not a part of the Phylum Cnidaria. Cnidaria is a phylum of aquatic organisms that consists of jellyfish, corals, sea anemones, and hydroids. They are named for their specialized cells, called cnidocytes, which are utilized in predation and defense.
The animals in the phylum Cnidaria are radially symmetric, meaning that their bodies can be divided into equivalent halves by more than one plane through the central axis.There are four classes in this phylum: Hydrozoa, Scyphozoa, Cubozoa, and Anthozoa. Among these classes, Cestoda is not a part of the Phylum Cnidaria. Cestoda is a class of flatworms that includes tapeworms, which are parasitic creatures that reside in the intestines of vertebrates. They have an extremely specialized morphology, which is characterized by a lengthy, flattened shape with a scolex for attachment to the host and a string of proglottids behind it.
Sponges are aquatic animals that belong to the phylum Porifera. They are sessile animals, which means they are permanently fixed in one location. They are the simplest of the three kinds of animals, which also include cnidarians and ctenophores. They have no organs and no true tissues, but they do have specialized cells that work together in order to perform various bodily processes.
They can be found in both freshwater and saltwater habitats, and they vary significantly in size and shape. Many have been utilized for medicinal reasons in traditional medicine.The name "true jellyfish" is used to distinguish the class Scyphozoa from the Hydrozoa, which are commonly referred to as "hydromedusae." Scyphozoans have a gelatinous, umbrella-shaped bell with long tentacles that hang down from it. The bell contracts, propelling the jellyfish through the water, and the tentacles, which are used for feeding and defense, trail behind. True jellyfish feed mainly on plankton, small fish, and sometimes other jellyfish. They are most prevalent in warm waters, but they can be found in a variety of marine environments.
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Question 10 Which alternative correctly orders the steps of the scientific method? O a) making observation - asking question - formulating hypothesis-testing hypothesis in experiment - analyzing results Ob) asking question-making observation - testing hypothesis in experiment-formulating hypothesis - analyzing results c) formulating hypothesis-testing hypothesis in experiment - asking question-making observation - analyzing results d) formulating hypotheses-testing hypothesis in experiment - analyzing results - asking question-making observation Moving to the next question prevents changes to this answer Question 8 of Question 8 0.75 points Save Ar "In 1877, a strange disease attacked the people of the Dutch East Indies. Symptoms of the disease included weakness, loss of appetite and heart failure, which often led to the death of the patient Scientists though the disease might be caused by bacteria. They injected chickens with bacteria isolated from the blood of sick patients. A second group was not injected with bacteria-It was the control group. The two groups were kept separate but under exactly the same conditions. After a few days, both groups had developed the strange disease-Based on the information given here, was the hypothesis supported or rejected? Oa) the data led to supporting the hypothesis bi the data led to relecting the himothori Question 6 What is a variable in a scientific experiment? a) a part of an experiment that does not change Ob) a part of an experiment that changes Question 2 Why is it important to have a control group in an experiment? a) control groups are important to allow for predicting the outcomes of an experiment b) control groups are important to prevent variables from changing during the experiment c) control groups are important to control the outcomes of the experiment d) control groups are important to establish a basis for comparison Why is it important to have a control group in an experiment? a) control groups are important to allow for predicting the outcomes of an experiment Ob) control groups are important to prevent variables from changing during the experiment Oc) control groups are important to control the outcomes of the experiment Od) control groups are important to establish a basis for comparison Dependent variables are: Oa) the part of the experiment that doesn't change Ob) the ones that cause other variables to change c) the ones that respond to other variables in the experiment d) the ones that can stand alone Imagine the following situation: a scientist formulates three different hypotheses for the same question. What should the scientist do next? Oa) test the three hypotheses at the same time in one experiment Ob) test two hypotheses at the same time in one experiment and then perform a second experiment to test the third hypothesis Oc) test each hypothesis separately, one at a time in three different experiments d) nothing, a question that leads to 3 different hypothesis cannot be answered
The correct alternative that orders the steps of the scientific method is: formulating hypotheses-testing hypothesis in experiment-analyzing results-asking question-making observation.The scientific method is a logical, empirical, and systematic method used to determine the accuracy of the observations and theories. Here are the steps involved in the scientific method:Making observations and asking questions Formulating hypotheses Designing experiments to test hypotheses Collecting data Analyze results Communicate results.
The hypothesis is a tentative answer to a question or problem. It is a statement that can be tested. Based on the given information in Question 8, the hypothesis was supported since the chickens in both the control and experimental groups developed the strange disease. Hence, the answer is (a) the data led to supporting the hypothesis.A variable in a scientific experiment is a part of an experiment that changes. It is an element or factor that can change or be changed during the experiment.Control groups are important to establish a basis for comparison. They are used to compare the effects of an independent variable on a dependent variable. Having a control group allows researchers to compare the effects of the independent variable in an experiment on the dependent variable to the other groups in the experiment.
Dependent variables are the ones that respond to other variables in the experiment. They are called dependent variables because they depend on the independent variable to cause a change. The independent variable is the one that causes a change in the dependent variable. For example, in an experiment, the dependent variable could be the amount of sugar consumed by a person each day, while the independent variable is the type of beverage consumed.A scientist should test each hypothesis separately, one at a time in three different experiments, if they have formulated three different hypotheses. Testing all three hypotheses simultaneously may lead to inconclusive or inaccurate results.
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Question 11 2 pts Statetment: It does not matter which DNA polymerase is used when running the PCR. Is the above statement accurate? Defend your answer. Edit View Insert Format Tools Table 12pt Paragraph BIU AV 2²: I 0 words > 2 P
The given statement: "It does not matter which DNA polymerase is used when running the PCR" is not accurate. PCR (Polymerase Chain Reaction) is an important technique used to amplify small fragments of DNA into large amounts that are enough to be analyzed. Thus, it is not accurate to say that it does not matter which DNA polymerase is used when running the PCR.
A polymerase enzyme is used in PCR to amplify the target DNA. There are different types of polymerase enzymes that can be used in PCR. The choice of polymerase enzyme used in PCR is critical as it affects the sensitivity, specificity, accuracy, and yield of the PCR.The Taq polymerase is the first and most widely used polymerase enzyme in PCR. It is derived from the bacterium Thermus aquaticus, which lives in hot springs and geysers, and is ideal for use in PCR as it is stable at high temperatures. The Taq polymerase is used in PCR to amplify DNA fragments from different sources, including human, animal, and plant DNA.
However, the Taq polymerase has a major drawback; it lacks 3’-5’ exonuclease proofreading activity, which can lead to errors in the amplified DNA fragments.There are other types of polymerase enzymes, such as Pfu, Phusion, and Platinum, which are more accurate and have proofreading activity. These polymerase enzymes are used in PCR to amplify DNA fragments that are critical for downstream applications such as cloning, sequencing, and mutagenesis. Hence, the choice of polymerase enzyme used in PCR is critical and should be based on the specific application of the amplified DNA fragment. Thus, it is not accurate to say that it does not matter which DNA polymerase is used when running the PCR.
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please help .. thank you
Topic 5: Homeostatic regulation of body systems occurs at three levels - local, neural, and hormonal. Often, similar end results are achieved by actions occurring at each of the three levels. What are
Homeostatic regulation of body systems occurs through local, neural, and hormonal levels. These levels work together to achieve similar end results by maintaining stability at the cellular level, coordinating rapid responses through the nervous system, and releasing hormones to regulate various bodily functions.
Homeostatic regulation of body systems occurs at three levels: local, neural, and hormonal. Each level plays a crucial role in maintaining stability within the body.
At the local level, cells and tissues have intrinsic mechanisms to regulate their immediate environment.
For example, if a tissue becomes acidic, local cells may release chemical signals to increase blood flow, deliver more oxygen, and remove waste products. This ensures a stable environment for cellular function.
The neural level involves the nervous system, which coordinates rapid responses to maintain homeostasis. Sensory receptors detect changes in the body and send signals to the brain or spinal cord.
The nervous system then initiates appropriate responses, such as shivering when body temperature drops or increasing heart rate during physical exertion.
The hormonal level involves the endocrine system, which releases hormones into the bloodstream to regulate various body functions.
Hormones act as chemical messengers, traveling through the blood to target tissues or organs. For instance, the hormone insulin regulates blood sugar levels by promoting glucose uptake by cells.
Although the actions at each level differ, they often achieve similar end results.
For example, if blood glucose levels rise, local cells may take up glucose, neural signals may stimulate the release of insulin, and hormonal actions may enhance glucose uptake by tissues.
This redundancy ensures robust homeostatic control and enables the body to respond effectively to internal and external changes.
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Complete question:
How does homeostatic regulation of body systems occur at three levels (local, neural, and hormonal), and how do these levels collectively achieve similar end results in maintaining stability within the body?
Suppose you want to understand how a model prokaryote regulates its internal pH as the external pH changes. Design an experimental protocol that will allow you to understand the mechanisms involved in such processes. Try to answer, how will you induce the change in pH? what variables will you observe to define the mechanisms by which pH is regulated? what results do you expect to obtain? experimental controls?
To understand how a model prokaryote regulates its internal pH as the external pH changes, the following experimental protocol can be followed.
Inducing pH changeTo induce a change in pH, an acid or a base can be added to the medium in which the prokaryote is grown. By measuring the initial pH of the growth medium, the appropriate amount of acid or base can be added to change the pH to the desired level.
The pH of the medium should be measured periodically over time to ensure that the pH is maintained at the desired level throughout the experiment.Variables to observeTo understand the mechanisms involved in regulating pH, the following variables can be observed:Internal pH of the prokaryote - The internal pH can be measured using a pH-sensitive fluorescent dye.
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References Macrophages, dendritic cells, and B cells Help Save & Ext Subet O All lymphocytes (T and B) O Infected cells only 2. MHC-I molecules normally display "self" proteins, those that are normally produced by a cell. TIME True O False 3. In the case of cancer or viral infection, which MHC class is involved with displaying abnormal proteins to cytotoxic T cells as a signal for destruction? OI Oll 4. MHC-Il molecules are located on what types of cells? O All nucleated cells O Macrophages, dendritic cells, and B cells O Infected cells only All lymphocytes (T and B)
1. Macrophages, dendritic cells, and B cells help save and extend the subset of all lymphocytes (T and B). Macrophages, dendritic cells, and B cells play critical roles in the immune response by presenting antigens to T and B cells.
They capture, process, and present antigens to activate and direct the immune system's response.
2. MHC-I molecules normally display "self" proteins, those that are normally produced by a cell.
This statement is true. Major Histocompatibility Complex class I (MHC-I) molecules are found on the surface of almost all nucleated cells in the body. They present peptides derived from proteins synthesized within the cell. MHC-I molecules help the immune system distinguish between "self" and "non-self" cells, enabling the recognition and elimination of infected or abnormal cells.
3. In the case of cancer or viral infection, MHC class I is involved with displaying abnormal proteins to cytotoxic T cells as a signal for destruction.
In the case of cancer or viral infection, MHC class I is involved in displaying abnormal proteins to cytotoxic T cells as a signal for destruction.
4. MHC-II molecules are located on macrophages, dendritic cells, and B cells. MHC-II molecules are located on macrophages, dendritic cells, and B cells. These cells are considered professional antigen-presenting cells (APCs) and express MHC-II on their surfaces.
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List the names of the genes which are not affected by Doxorubicin and justify your answer. [30%]
Some of the genes that are not affected by Doxorubicin are PTPRO, TFF3, DUSP1, and SLC7A5.
Some of the genes that are not affected by Doxorubicin and justify the answer are:
PTPRO: Protein tyrosine phosphatase receptor type O (PTPRO) is a tumour suppressor gene that is often downregulated in various cancer types. Doxorubicin has been shown to have no effect on PTPRO gene expression in breast cancer cells.
TFF3: Trefoil factor 3 (TFF3) is a gene that is involved in cell proliferation and differentiation. TFF3 is frequently overexpressed in many cancer types, including breast cancer. However, it has been reported that Doxorubicin does not affect TFF3 gene expression in breast cancer cells.
DUSP1: Dual-specificity phosphatase 1 (DUSP1) is a gene that encodes a protein involved in the regulation of cell growth and differentiation. Doxorubicin has been found to have no effect on DUSP1 gene expression in breast cancer cells.
SLC7A5: Solute carrier family 7 member 5 (SLC7A5) is a gene that encodes a protein involved in amino acid transport. This gene has been found to be unaffected by Doxorubicin in breast cancer cells
Doxorubicin is a widely used chemotherapy drug for the treatment of various cancers, including breast cancer. However, the drug has significant side effects and can affect the expression of many genes in cells. The identification of genes that are not affected by Doxorubicin is essential for understanding the drug's mechanism of action and identifying potential targets for combination therapies.
Some of the genes that are not affected by Doxorubicin and justify the answer are PTPRO, TFF3, DUSP1, and SLC7A5. PTPRO is a tumour suppressor gene that is often downregulated in various cancer types. However, Doxorubicin has been shown to have no effect on PTPRO gene expression in breast cancer cells. TFF3 is a gene that is involved in cell proliferation and differentiation and is frequently overexpressed in many cancer types. However, it has been reported that Doxorubicin does not affect TFF3 gene expression in breast cancer cells. DUSP1 is a gene that encodes a protein involved in the regulation of cell growth and differentiation.
Doxorubicin has been found to have no effect on DUSP1 gene expression in breast cancer cells. SLC7A5 is a gene that encodes a protein involved in amino acid transport and has been found to be unaffected by Doxorubicin in breast cancer cells.
Doxorubicin is a potent chemotherapy drug with significant side effects that can affect the expression of many genes in cells. The identification of genes that are not affected by Doxorubicin is essential for understanding the drug's mechanism of action and identifying potential targets for combination therapies. Some of the genes that are not affected by Doxorubicin are PTPRO, TFF3, DUSP1, and SLC7A5. These genes could serve as potential targets for combination therapies to improve the efficacy of Doxorubicin treatment.
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epidemiology
Short answer questions Question 5 A case series is an example of what kind of study design? O All of the answers listed here are correct. O Analytical Observational O Experimental Descriptive Observat
A case series can be classified as either an analytical observational, experimental study, or descriptive observational study design. Hence option 2, 3, and 4 are correct.
A case series is a type of study design that involves the collection and analysis of data from a group of individuals who share a common characteristic or condition. It is typically used to describe the characteristics, outcomes, and patterns of a specific group of cases, such as patients with a particular disease or those exposed to a certain treatment.
In terms of study design classification, a case series can fall into different categories depending on the nature of the study. It can be considered an analytical observational study design if the data is analyzed to identify associations or relationships between variables.
It can also be an experimental study design if interventions or treatments are applied to the cases. Additionally, a case series can be classified as a descriptive observational study design if it focuses on describing the cases without any interventions. Therefore, all of the answer choices provided are correct options for classifying a case series study design.
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The complete question is:
A case series is an example of what kind of study design?
1. All of the answers listed here are correct.
2. Analytical Observational
3. Experimental study
4. Descriptive Observational
4. None of the answer listed here are correct
Which of the following chromosome abnormalities (assume heterozygous for abnormality) lead to unusual metaphase alignment in mitosis? Why?
I. Paracentric inversions
II. Pericentric inversions
III. Large internal chromosomal deletions
IV. Reciprocal translocation
Among the chromosome abnormalities listed, the main condition that leads to unusual metaphase alignment in mitosis is the reciprocal translocation.
Reciprocal translocation involves the exchange of genetic material between non-homologous chromosomes. During mitosis, when chromosomes align along the metaphase plate, translocated chromosomes can exhibit abnormal alignment due to the altered position of the genes involved in the translocation.
In reciprocal translocation, two non-homologous chromosomes break and exchange segments, leading to a rearrangement of genetic material. As a result, the genes on the translocated chromosomes may not align properly during metaphase. This misalignment can disrupt the normal pairing of homologous chromosomes and interfere with the separation of chromosomes during anaphase, potentially resulting in errors in chromosome distribution and aneuploidy.
It's important to note that paracentric inversions, pericentric inversions, and large internal chromosomal deletions do not directly cause unusual metaphase alignment in mitosis. These abnormalities may lead to other effects such as disrupted gene function or changes in chromosome structure, but their impact on metaphase alignment is less pronounced compared to reciprocal translocations.
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atmosphere had very low oxygen levels, but a to accumulate in the shallow oceans as around 2.4 billion so much that the oxygen was accumulating in the atmosphere peroxides, singlet oxygen, and hydroxyl radicals. Organisms living in thi new oxygen-rich environm Unfortunately, pure oxygen can be converted into reactive oxygen spece (ROS) including superoxide, catalase, to break down ROS. Humans actually have three forms of SOD as las catalase, which is found i the Oxygen Revolution needed to evolve to produce some enzymes, such as superoxide dismutase (500) within the cell as well as damage to DNA and RNA. Bacteria that stayed on and or in shallow oceans during needed mechanisms to convert ROS to a less reactive form in order to prevenciarge-scale oxidation dama peroxisomes. Organisms that didn't already have a mechanism in place to handle the ROS, were either forced a respiration was now possible and highly efficient mitochondria evolved, which allowed early eukaryotes response, the organisms that were able to handle the ROS underwent great diversification. Aer anaerobic refuges or died out in the large extinction event caused by the new oxygen-rich environment. methods organisms become much more complex. Due to the variable environments that existed at different times in Earth's history, highly variable r for ATP regeneration exist - most of which are found in bacteria. Most bacteria and most of the you think of carry out aerobic respiration. As you can see, throughout history, photosynthesis and cellular respiration have been linked. Today, we'll be O, increases as a result of photosynthesis, during respiration the opposite is true: as the plant breaks down exploring that link further by analyzing CO₂ and O; concentrations in spinach leaves. While CO₂ decreases and and photosynthesis by measuring the 0₂ glucose to release stored energy, CO, is released into the surrounding water or atmosphere, i concentrations decrease. Thus, we can estimate rates of respiration or consumption or production of these two gases. Questions (Chapters 9 and 10) to answer the following questions: 1. Oxygen is produced from water in the light reactions in a process called photolysis. What else happens du photolysis? Can the light reactions of photosynthesis continue if water is not available? Explain. 2. Describe the role of oxygen in cellular respiration:
The Oxygen Revolution, which occurred around 2.4 billion years ago, led to the accumulation of oxygen in the Earth's atmosphere. This increase in atmospheric oxygen levels had significant impacts on the evolution of organisms and the development of various mechanisms to handle reactive oxygen species (ROS). Organisms that were able to adapt and produce enzymes like superoxide dismutase and catalase, capable of neutralizing ROS, underwent diversification. However, organisms lacking such mechanisms faced oxidative damage and, in some cases, extinction. The evolution of efficient mitochondria enabled eukaryotes to take advantage of aerobic respiration, leading to their proliferation. The link between photosynthesis and cellular respiration can be observed today through the exchange of CO₂ and O₂ during these processes, allowing us to estimate rates of respiration and photosynthesis.
Around 2.4 billion years ago, the Earth experienced the Oxygen Revolution, during which atmospheric oxygen levels increased significantly. This rise in oxygen resulted from the accumulation of oxygen in the atmosphere due to the activity of early photosynthetic organisms. However, this oxygen posed a challenge for organisms as it could lead to the production of reactive oxygen species (ROS) that could cause cellular damage.
To cope with the presence of ROS, organisms needed to evolve mechanisms to handle and neutralize these reactive molecules. One crucial enzyme involved in this process is superoxide dismutase (SOD), which converts superoxide radicals into less harmful hydrogen peroxide. Humans possess three forms of SOD. Another enzyme, catalase, helps break down hydrogen peroxide into water and oxygen.
The ability to handle ROS became essential for survival in an oxygen-rich environment. Organisms that already had mechanisms in place to neutralize ROS were able to adapt and diversify. On the other hand, organisms lacking these mechanisms were susceptible to oxidative damage and faced challenges in their survival and reproduction.
Aerobic respiration, which is highly efficient in energy production, evolved in response to the increased availability of oxygen. Efficient mitochondria played a vital role in aerobic respiration, enabling early eukaryotes to thrive in oxygen-rich environments and undergo further diversification.
Today, the link between photosynthesis and cellular respiration can be observed by analyzing the exchange of CO₂ and O₂. During photosynthesis, plants take in CO₂ and release O₂, while during respiration, the opposite occurs as glucose is broken down to release energy, resulting in the release of CO₂ and the consumption of O₂. By measuring the concentrations of these gases, we can estimate the rates of respiration and photosynthesis in organisms.
Overall, the Oxygen Revolution and the subsequent evolution of mechanisms to handle ROS played a significant role in shaping the diversity and complexity of life on Earth.
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Photolysis is the process by which water molecules are split into hydrogen ions, electrons, and molecular oxygen during the light reactions of photosynthesis. Oxygen is essential in cellular respiration as it serves as the final electron acceptor in the electron transport chain.
Explanation:Oxygen is produced from water in the light reactions of photosynthesis through a process called photolysis. During photolysis, water molecules are split into hydrogen ions, electrons, and molecular oxygen. The light reactions of photosynthesis cannot continue without water, as water provides the source of electrons needed to replace those lost during the conversion of light energy to chemical energy.
Oxygen plays a crucial role in cellular respiration. During cellular respiration, glucose is broken down to release energy that is used to produce ATP. Oxygen acts as the final electron acceptor in the electron transport chain, accepting electrons from complex IV and combining with hydrogen ions to form water. Without oxygen, the electron transport chain cannot function, and ATP production is severely impaired.
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Q) An older 50 ml of (MW) access How Cell biology protocal requires a o·gº Nacl solution 58.44 g/mole). You only have 650 ml of 3 M Nad. to much of the Stock do you use?
1.67 mL of the stock solution to make the required NaCl solution
Given:
Molecular weight of NaCl = 58.44 g/mole
Volume of NaCl solution required = 50 mL = 0.05 L
Concentration of NaCl solution required = 0.1 M
Volume of 3 M NaCl solution available = 650 mL = 0.65 L
We can use the formula,C1V1 = C2V2, where C1 and V1 are the concentration and volume of the stock solution and C2 and V2 are the concentration and volume of the diluted solution.
Let's calculate the volume of the stock solution required to make the diluted solution.
C1V1 = C2V2V1 = (C2V2)/C1V1
= (0.1 M × 0.05 L)/(3 M)V1
= 0.00167 L
= 1.67 mL
Therefore, we need 1.67 mL of the stock solution to make the required NaCl solution.
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In the last step of secretion, proteins or ions made by a cell
are delivered to the cell membrane in a vesicle so that exocytosis
can deliver the contents to the extracellular space. True/false
Its True, In the last step of secretion, proteins or ions made by a cell are delivered to the cell membrane in a vesicle so that exocytosis can deliver the contents to the extracellular space.
Exocytosis is a type of active transport in which a cell transports molecules (such as proteins) out of the cell by secreting them through an energy-dependent process. It is a process in which a cell releases materials from its intracellular space to the extracellular space. The materials being secreted are typically large molecules such as proteins, lipids, and carbohydrates, and they are packaged into vesicles for transport to the cell surface.
The process of exocytosis is tightly regulated by a variety of intracellular signals that control the release of vesicles from the cell membrane. When a vesicle reaches the cell membrane, it fuses with the membrane and the contents of the vesicle are released into the extracellular space. The proteins or ions are then delivered to the cell membrane in a vesicle so that exocytosis can deliver the contents to the extracellular space.
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35 A section of the coding strand of the DNA sequence of a gene that is expressed in a healthy human liver cell is 5'-ATGCGCCGTAT-3'. A microRNA (miRNA) regulates this gene by signaling an enzyme to c
The mRNA molecule transcribed from this gene. The complementary sequence of the coding strand provided is 3'-TACGCGGCATA-5'.
Based on this information, the microRNA (miRNA) would bind to the mRNA molecule through base pairing interactions. miRNAs are small non-coding RNA molecules that play a crucial role in post-transcriptional gene regulation. They typically bind to the 3' untranslated region (UTR) of target mRNA molecules, leading to gene silencing or degradation of the mRNA. In this case, the miRNA would recognize and bind to the complementary sequence on the mRNA molecule. The binding occurs through base pairing interactions between the miRNA and the mRNA, where complementary nucleotides pair up. This binding can interfere with the translation of the mRNA into protein or lead to the degradation of the mRNA molecule. The specific binding of the miRNA to the mRNA sequence would signal the enzyme responsible for mRNA degradation or repression, ultimately regulating the expression of the gene in the liver cell. This regulation can control the amount of protein produced from the gene, influencing various cellular processes and functions in the liver cell.
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What is stress and stress tolerance in plants?
ii. What is the difference between abiotic and biotic stress?
What is the difference between acclimation and adaptation?
iv. What are the main abiotic stresses worldwide?
V. What are the main abiotic stresses in Bahrain?
i. Stress in plants refers to any adverse external factor or condition that disrupts the normal physiological processes and growth of plants. It can include various factors such as extreme temperatures, drought, salinity, nutrient deficiency or toxicity, heavy metals, pollutants, radiation, and physical damage.
ii. The difference between abiotic and biotic stress lies in the nature of the stressors affecting plants:
Abiotic stress refers to the adverse effects caused by non-living factors in the environment. Examples include temperature extremes (heat or cold stress), water scarcity (drought stress), excessive or insufficient light (light stress), high salinity (salt stress), and toxic substances (chemical stress).
iii. Acclimation and adaptation are two concepts related to how plants respond to environmental challenges:
Acclimation refers to the reversible physiological and biochemical adjustments that plants make in response to changes in their immediate environment. It involves short-term responses that allow plants to cope with specific environmental conditions.
iv. The main abiotic stresses worldwide include:
- Drought: Lack of water availability or water scarcity.
- Heat stress: High temperatures that exceed the optimal range for plant growth.
- Cold stress: Low temperatures that can cause chilling injury or frost damage.
- Salinity stress: High concentration of salts in the soil or irrigation water.
- Flooding: Excessive waterlogged conditions that limit oxygen availability to plant roots.
v. The main abiotic stresses in Bahrain may vary based on the specific environmental conditions of the region. However, some potential abiotic stresses in Bahrain could include:
- High temperatures and heat stress due to the country's arid climate.
- Water scarcity and drought stress, as Bahrain faces limited freshwater resources.
- High salinity levels in the soil and irrigation water due to the surrounding saltwater environment.
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Explain the roles of key regulatory agencies within the United
States in the safe release of bioengineered organisms in the
environment and in regulating food and food additives produced
using biotech
The key regulatory agencies in the United States for the safe release of bioengineered organisms and regulation of biotech food and additives are the EPA, USDA, and FDA.
The key regulatory agencies within the United States that play important roles in the safe release of bioengineered organisms in the environment and in regulating food and food additives produced using biotech include the U.S. Environmental Protection Agency (EPA), the U.S. Department of Agriculture (USDA), and the Food and Drug Administration (FDA).
The U.S. Environmental Protection Agency (EPA) is responsible for regulating bioengineered organisms that are intended to be released into the environment. The EPA evaluates the potential risks associated with these organisms and assesses their potential impact on ecosystems and human health. They ensure that appropriate measures are in place to minimize any potential adverse effects and to protect the environment.
The U.S. Department of Agriculture (USDA) plays a role in regulating bioengineered crops and organisms. The USDA's Animal and Plant Health Inspection Service (APHIS) is responsible for assessing the potential risks and impacts of genetically modified crops and organisms on agriculture and the environment. They oversee the permitting process for field trials and commercialization of genetically modified crops.
The Food and Drug Administration (FDA) is responsible for regulating food and food additives produced using biotechnology. The FDA ensures that these products are safe for consumption and accurately labeled. They evaluate the safety and nutritional profile of genetically modified crops, as well as the safety of food additives derived from biotech processes.
These regulatory agencies work together to establish and enforce regulations and guidelines to ensure the safe release of bioengineered organisms and the regulation of biotech-derived food and food additives in the United States. Their collective efforts aim to protect the environment, safeguard public health, and provide consumers with accurate information about the products they consume.
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Autosomal Recessive Trait. For this example, we’ll use albinism as our trait. Albinism results from the homozygous occurrence of the autosomal recessive allele a (genotype aa), which prevents the body from making enough (or any) melanin. For this example, use A for the normal pigmentation allele, and a for the albinism allele.
a) Consider two phenotypically non-albino parents, who have some children with albinism. What would be the possible genotypes of both the parents and the offspring? (Use a Punnett square to show your work.)
b) What genotypes would we expect from a family consisting of a non-albino man and a woman with albinism who have two children with albinism and two non-albino children? Provide genotypes for all six family members. You may find it useful to draw a Punnett square.
c) What genotypes would we expect for a family consisting of two parents with albinism who have only children with albinism? Again, provide the genotypes for both parents and children.
a. The Punnett square shows that there are four possible genotypes for the offspring: AA, Aa, Aa, and aa.
b. The genotypes for the family members are as follows:
Non-albino man: Aa
Woman with albinism: aa
Child 1 (albino): aa
Child 2 (albino): aa
Child 3 (non-albino): Aa
Child 4 (non-albino): Aa
c. The expected genotype of all their children will be aa.
What are the possible genotypes?a) If two phenotypically non-albino parents have children with albinism, it means that both parents must be carriers of the albinism allele (Aa) because albinism is an autosomal recessive trait.
Let's use the genotypes A and a to represent the normal pigmentation allele and the albinism allele, respectively.
Possible genotypes of the parents:
Parent 1: Aa
Parent 2: Aa
A a
A AA Aa
a Aa aa
The genotypes AA and Aa represent individuals with normal pigmentation, while the genotype aa represents individuals with albinism.
b) If a non-albino man (genotype Aa) and a woman with albinism (genotype aa) have two children with albinism and two non-albino children, let's create a Punnett square to determine the genotypes:
A a
a Aa aa
a Aa aa
The Punnett square shows the following genotypes for the family members:
Non-albino man: Aa
Woman with albinism: aa
Child 1 (albino): aa
Child 2 (albino): aa
Child 3 (non-albino): Aa
Child 4 (non-albino): Aa
c) If both parents have albinism (genotype aa) and they have only children with albinism, the Punnett square would look like this:
a a
a aa aa
a aa aa
In this case, both parents have the genotype aa, and all their children will also have the genotype aa, resulting in albinism in all offspring.
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describe the relationship in chemical and physical the sturcture of L-Dopa and the decarboxylase? how do they interact with eachother?
L-Dopa, a chemical compound, interacts with the enzyme decarboxylase, which removes a carboxyl group from L-Dopa, converting it into dopamine. This interaction is significant for increasing dopamine levels in the brain and is essential in the treatment of Parkinson's disease.
L-Dopa, also known as Levodopa, is a chemical compound that serves as a precursor for the neurotransmitter dopamine. It is used as a medication for treating Parkinson's disease. L-Dopa has a specific chemical structure that allows it to cross the blood-brain barrier, where it is converted into dopamine by the enzyme decarboxylase.
Decarboxylase is an enzyme that catalyzes the removal of a carboxyl group from a molecule. In the case of L-Dopa, decarboxylase removes the carboxyl group, converting it into dopamine. This interaction between L-Dopa and decarboxylase is crucial for increasing dopamine levels in the brain, as dopamine deficiency is a characteristic feature of Parkinson's disease.
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Two Factor Cross Practice Problem You are a tomato breeder with an extensive collection of red tomato lines. You recently received seeds for a true-breeding line with delicious yellow tomatoes, but it is susceptible to tobamovirus. You want to produce a true-breeding tobamovirus-resistant yellow tomato line for your collection. You have a true-breeding red tomato line that is resistant to tobamovirus. You know that resistance is due to a dominant allele of the Tm-2 gene (or T-locus). You also know red coloration is due to a dominant allele at the R-locus, and yellow coloration is the recessive R-locus trait. 1. What are the genotypes of the true-breeding a) susceptible yellow tomato line and b) resistant red tomato line? These are your parental lines. 2. If you cross the two parental lines, what will the F, genotypes and phenotypes be? Is this the final tomato line you want? Why or why not? 3. If you cross F, with Fy, what will the phenotypic ratio be in the Fz population? What proportion of the F, will have the phenotype you desire? Of those that have the phenotype you desire, how many possible genotypes can they have? 4. Now working only with the Fplants that have your desired phenotype, what kind of plant will you cross them with to determine their genotype? We will call these test crosses. What will the results be in the testcross progeny for your desired F,? What will the results be in the testcross progeny for F, with the non-desirable genotype?
The genotypes of the true-breeding a) susceptible yellow tomato line are rr and tt and that of b) resistant red tomato line is RR and Tt.
On crossing two parental lines, the F1 genotypes will be Rr and Tt and phenotypes will be red and resistant to tobamo virus. No, this is not the final tomato line that is required. 3. If we cross F1 with Fy, the phenotypic ratio will be 9:3:3:1 in the F2 population. 1/16 or 6.25% of the F2 population will have the desired phenotype. Out of those who have the desired phenotype, 2 possible genotypes can be there.
The test cross plant for determining the genotype of F1 with the desired phenotype will be rr and tt genotype with yellow coloration and susceptible to tobamo virus. The results in the test cross progeny for the desired F1 would be all red and resistant to tobamo virus. The results in the test cross progeny for F1 with a non-desirable genotype would be 1:1:1:1.
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You engineered a new gene which includes GFP fused to a cytosolio protein. You then added a non-specific promoter and incorporate this new gene into the genome of a mouse. When you examine cells from these mice in the fluorescent microscope: O a. You will see the fluorescence throughout the cytoplasm of all the cells of the mouse. Ob. You will see the fluorescence throughout the cytoplasm of all cardiac cells in the mouse. Oc. You will see the fluorescence from the protein in the membrane of all cardiac cells in the mouse. Od. You will see the fluorescence in the membranes of all the cells of the mouse. Oe. None of the above will be seen.
You engineered a new gene which includes GFP fused to a cytosolic protein. You then added a non-specific promoter and incorporate this new gene into the genome of a mouse.
Option A is correct
When you examine cells from these mice in the fluorescent microscope: O a. You will see the fluorescence throughout the cytoplasm of all the cells of the mouse. Ob. You will see the fluorescence throughout the cytoplasm of all cardiac cells in the mouse. Oc. You will see the fluorescence from the protein in the membrane of all cardiac cells in the mouse. Od. You will see the fluorescence in the membranes of all the cells of the mouse. Oe. None of the above will be seen.When a new gene is engineered that includes GFP (green fluorescent protein) fused to a cytosolic protein and a non-specific promoter is added, and then the new gene is incorporated into the genome of a mouse, the fluorescence in the cells from these mice in the fluorescent microscope will be visible. The question is, where will the fluorescence be seen?Option A: You will see the fluorescence throughout the cytoplasm of all the cells of the mouse.This answer choice is incorrect.
The fluorescence will not be visible throughout the cytoplasm of all the cells of the mouse. Option B: You will see the fluorescence throughout the cytoplasm of all cardiac cells in the mouse. This answer choice is incorrect. The fluorescence will be seen in some parts of the mouse cells. Thus, the correct answer is none of the answer choices presented. Instead, the correct answer is that the fluorescence will be visible in the cytoplasm and not in any specific region.
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Eventually, you are able to grow the chemolithoautotroph as well. Given what you know about the organism’s metabolism and the environment it came from, what should you change about the standard culturing conditions to promote the growth of this organism?
A) Lower the pH
B) Add more anaerobic electron acceptors
C) Expose the cells to sunlight
D) Add glucose
E) Grow the cells anaerobically
The metabolic pathway of chemolithoautotrophs is unique in the fact that these bacteria are able to survive without light, organic compounds, or oxygen as they gain their energy through the oxidation of inorganic compounds like nitrate, ammonia, and sulfur.
In order to promote the growth of chemolithoautotrophs, a few modifications can be made to the standard culturing conditions. The options are provided below:
1) Lower the pH: This condition won't be helpful in promoting the growth of the chemolithoautotrophs as most of the chemolithoautotrophs are found to grow at a neutral or an alkaline pH.
2) Add more anaerobic electron acceptors: This condition could be useful in promoting the growth of chemolithoautotrophs as most of these organisms require electron acceptors like CO2, NO2-, SO4-2, Fe2+, etc for their metabolism.
3) Expose the cells to sunlight: As chemolithoautotrophs are known to survive without light, this condition is not applicable.
4) Add glucose: This condition is not applicable as chemolithoautotrophs do not rely on organic compounds for their metabolism.
5) Grow the cells anaerobically: This condition could be useful in promoting the growth of chemolithoautotrophs as most of these organisms are found to grow in anaerobic conditions.
Therefore, growing the cells anaerobically could help in promoting the growth of the chemolithoautotroph.
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Alzheimer's disease can be sporadic and familial . what is the
difference ?
There are two basic types of Alzheimer's disease: sporadic and familial. The underlying causes and inheritance patterns are different.
The majority of cases of Alzheimer's disease are sporadic, which is the most prevalent type. There is no obvious family history or genetic predisposition associated with it. Although the precise origin of sporadic Alzheimer's is unknown, it is thought that a mix of genetic, environmental, and lifestyle factors may play a role.On the other hand, familial Alzheimer's disease is relatively uncommon and has a distinct hereditary component. Certain genes, including the amyloid precursor protein (APP), presenilin 1 (PSEN1), and presenilin 2 (PSEN2) genes, are mutated to cause it. As a result of the autosomal dominant pattern of inheritance for these mutations, an individual is
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I have a mantle that protects my internal organs, and a calcareous shel for protection. I accomplish locomotion using my foot, and scrape algae off of rocks using my radula. To what animal phylum do I belong? a. Arthropoda b. Platyhelminthes c. Porifera d. Cnidaria e. Mollusca f. Echinodermata
The animal phylum that includes animals with a mantle that protects their internal organs, a calcareous shell for protection, foot locomotion, and a radula for scraping algae off rocks is Mollusca. Therefore option (E) is the correct answer.
Mollusca is a phylum of invertebrate animals that includes snails, slugs, mussels, octopuses, and squids. This phylum is the second-largest animal phylum, with over 100,000 known species. They have a diverse range of forms, including snails, octopuses, squids, and mussels. Molluscs are present in a variety of environments, including saltwater, freshwater, and terrestrial environments.
They have a radula, a rasping tongue-like structure that aids in the consumption of food. The foot of a mollusk is used for movement, while the mantle is used to protect the internal organs and produce a shell. In conclusion, the animal phylum that includes animals with a mantle that protects their internal organs, a calcareous shell for protection, foot locomotion, and a radula for scraping algae off rocks is Mollusca. Option (E) is the correct answer.
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Which of the following chordate characteristics is incorrectly matched? a) dorsal hollow nerve cord-spinal nerve cord. b) pharyngeal slits-mouth. c) notochord-spine. d) Cendostyle-thyroid.
The incorrectly matched chordate characteristic is:
d) Cendostyle-thyroid.
What are chordates?Chordates are a diverse group of animals that belong to the phylum Chordata. Chordates have a notochord at a stage of their lives.
Considering the above:
The correct term that should be matched with the thyroid is "endostyle."
The endostyle is a glandular groove found in the pharynx of some chordates, such as invertebrate chordates and early embryonic stages of vertebrates. It produces mucus and plays a role in filter feeding and thyroid hormone production in vertebrates.
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Describe the process of fertilization.
a. Indicate the two cells involved.
b Indicate the resulting cell that is produced at
fertilization.
c. Indicate the location in which this process takes place.
Fertilization is the process by which a sperm cell and an egg cell combine to form a new individual. It is a crucial step in sexual reproduction.
a. The two cells involved in fertilization are the sperm cell and the egg cell (also known as the ovum). The sperm cell is produced in the male reproductive system, specifically in the testes, while the egg cell is produced in the female reproductive system, specifically in the ovaries.
b. The resulting cell produced at fertilization is called the zygote. The zygote is formed when the sperm cell fuses with the egg cell during fertilization. This fusion combines the genetic material from both parents, resulting in a single cell with a complete set of chromosomes.
c. Fertilization typically takes place in the fallopian tubes of the female reproductive system. After ovulation, the released egg cell travels through the fallopian tube. If a sperm cell successfully reaches and penetrates the egg cell in the fallopian tube, fertilization occurs. The fertilized egg, or zygote, then continues its journey towards the uterus, where it implants itself in the uterine lining and develops further during pregnancy.
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Name some of the organs in the digestive system. Can you name the order of the organs? What are the functions of the organs? 2. A Please name the organ affected by the following diseases/disorders hepatitis, cheilitis, gingivitis, gastritis, colitis. 3. Many terms end in-uria'to describe urinary conditions. Give five examples of terms ending in-uria and explain their meaning 4 Identify three urinary system disorders and identify which structure in the system is dysfunctional? Briefly explain each disorder
The digestive system is made up of many organs that help break down food and extract nutrients. Here are the organs and their functions in order: Mouth: The mouth is where digestion begins.
Teeth break food down into smaller pieces, while enzymes in saliva begin to break down carbohydrates. Esophagus: The esophagus is a muscular tube that carries food from the mouth to the stomach. Stomach: The stomach churns food, mixing it with enzymes and acid that help break it down further.
Small intestine: The small intestine is where most of the nutrients from food are absorbed into the bloodstream. Liver and pancreas: The liver produces bile, which helps digest fats.
The pancreas produces enzymes that help break down proteins, carbohydrates, and fats. Large intestine: The large intestine absorbs water and electrolytes from the remaining food, turning it into solid waste that can be eliminated through the rectum and anus.
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Match the four common fungal diseases and their causative agents. Histoplasma capsulatum [Choose ] Tinea species [Choose] Candida [ Choose] Aspergillus [Choose ]
Match the four common fungal diseases and their causative agents. Histoplasma capsulatum - Histoplasmosis, Tinea species - Dermatophytosis (ringworm), Candida - Candidiasis, Aspergillus - Aspergillosis.
Diseases are abnormal conditions or disorders that affect the normal functioning of the body, leading to physical or mental impairments. There are numerous types of diseases, including infectious diseases caused by pathogens like bacteria, viruses, or parasites (e.g., influenza, malaria); chronic diseases characterized by long-term persistence or recurring symptoms (e.g., diabetes, hypertension); genetic disorders caused by inherited genetic mutations (e.g., cystic fibrosis, sickle cell anemia); autoimmune diseases where the immune system attacks the body's own tissues (e.g., rheumatoid arthritis, lupus); and many others affecting various organs and systems in the body. Accurate diagnosis, treatment, and preventive measures are vital in managing diseases and promoting overall health.
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Which of the following is mismatched? a) albumin transport cholesterol. b) globulin- make antibodies. c) albumin - regulate osmotic balance. d) globulin - lipid transport. e) fibrinogen -blood clotting.
The mismatched option is globulin - make antibodies. So, option B is appropriate.
The correct association between globulin and its function is globulin - lipid transport. Globulins are a group of proteins found in the blood plasma and they have various functions, including lipid transport. Examples of globulins involved in lipid transport are low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs) that transport cholesterol and other lipids in the bloodstream.
On the other hand, antibodies, which are proteins involved in the immune response, are produced by a specific type of globulin called immunoglobulins. They are not directly responsible for making antibodies.
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6. What is the most likely cause of exfoliation in granite rock? a) The lowering of pressure exerted on the rock as it gets nearer to the earth's surface b) The uniform carbonation of the outermost layer of the rock c) Little elves with chisels d) Salt accumulation at the rick's surface 7. An earthquake can cause a) ground rupturing, liquefaction, and landslides b) landslides c) liquefaction d) ground rupturning 8. The minimum number of seismograph stations. required to determine the epicenter of an earthquake is a) 3 b) 2 c) 1 d) 4 9. Mass wasting is most likely to occur a) after heavy rains b) on steep slopes and after heavy rains c) on steep slopes d) on flat land 10. In the system of stream orders, the streams found at the highest elevations of a watershed that have no tributaries are a) 3rd order streams b) 1st order streams c) 2nd order streams d) 10th order streams 11. An increase in stream gradient causes a) a decrease in erosional capacity b) an increase in stream velocity c) deposition to occur d) calm pools to form
6. The most likely cause of exfoliation in granite rock is the lowering of pressure exerted on the rock as it gets nearer to the earth's surface.
7. An earthquake can cause ground rupturing, liquefaction, and landslides.8. The minimum number of seismograph stations required to determine the epicenter of an earthquake is 3.9. Mass wasting is most likely to occur on steep slopes and after heavy rains.10. In the system of stream orders, the streams found at the highest elevations of a watershed that have no tributaries are 1st order streams.11. An increase in stream gradient causes an increase in stream velocity.The explanation of the above answers are as follows:6. The most likely cause of exfoliation in granite rock is the lowering of pressure exerted on the rock as it gets nearer to the earth's surface.7. An earthquake can cause ground rupturing, liquefaction, and landslides. Earthquakes occur due to the sudden release of energy stored in rocks, leading to the shaking of the ground surface. This shaking can lead to ground rupturing, liquefaction, and landslides.8. The minimum number of seismograph stations required to determine the epicenter of an earthquake is 3.
The epicenter of an earthquake can be located by using the data collected from at least three seismograph stations.9. Mass wasting is most likely to occur on steep slopes and after heavy rains. Mass wasting refers to the downhill movement of rock, soil, or sediment under the influence of gravity. It is more likely to occur on steep slopes and after heavy rains when the soil is saturated and less stable.10. In the system of stream orders, the streams found at the highest elevations of a watershed that have no tributaries are 1st order streams. The Strahler Stream Order system is used to classify streams based on their position in the drainage network. The smallest streams in the network are classified as 1st order streams.11. An increase in stream gradient causes an increase in stream velocity. Stream gradient refers to the slope or steepness of a stream channel. An increase in stream gradient leads to an increase in stream velocity, as the water flows downhill faster.
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A cross between two true breeding lines produces F1 offspring that are heterozygous. When the F1 progeny are selfed a 1:2:1 ratio is observed. What allelic interaction is manifested with this result? Select the correct response(s): Overdominance Co Dominance None of the choices Complete Dominance Incomplete Dominance All of the choices
The observed 1:2:1 ratio in the F2 generation suggests an allelic interaction known as incomplete dominance.
Incomplete dominance occurs when the heterozygous condition (F1 generation) exhibits an intermediate phenotype between the two homozygous parental lines. In this case, neither allele is completely dominant over the other, resulting in a blend or mixture of the traits in the F1 offspring.
During selfing of the F1 generation, the possible genotypes and phenotypes of the F2 offspring are as follows: 1/4 will be homozygous for one allele and display the phenotype of one parent, 1/4 will be homozygous for the other allele and display the phenotype of the other parent, and 1/2 will be heterozygous and exhibit an intermediate phenotype between the two parents.
This pattern of inheritance, where the heterozygotes show an intermediate phenotype, is characteristic of incomplete dominance. It is important to note that incomplete dominance is different from complete dominance, where one allele completely masks the expression of the other, and also differs from co-dominance, where both alleles are fully expressed in the heterozygous condition.
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What type of cells possess unlimited proliferation potential, have the capacity to self- renew, and can give rise to all cells within an organism? Question 2. Which laboratory method can be used to quantify levels of mRNAs expressed in samples of two different types of stem cells? Question 3. A cell that can differentiate into any cell within the same lineage is known as: Question 4. How did the researchers Kazutoshi Takahasi and Shinya Yamanaka accomplish cellular reprogramming of mouse fibroblasts in their 2006 publication in Cell?
The cells that possess unlimited proliferation potential, have the capacity to self-renew, and can give rise to all cells within an organism are known as stem cells.
1. The laboratory method that can be used to quantify levels of mRNAs expressed in samples of two different types of stem cells is known as Reverse transcription polymerase chain reaction (RT-PCR).
2. The cell that can differentiate into any cell within the same lineage is known as a multipotent stem cell. Multipotent stem cells have the capacity to differentiate into various cell types within the same lineage or tissue, but not all cell types.
3. The researchers Kazutoshi Takahashi and Shinya Yamanaka accomplished cellular reprogramming of mouse fibroblasts in their 2006 publication in Cell by inducing the expression of four transcription factors: Oct4, Sox2, Klf4, and c-Myc.
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**The answers are D and C please explain why with work.
two genes show redundant gene action, where the presence of at least one wild type allele at one of the two genes will lead to normal heart-shaped fruits, while a homozygous recessive genotype at both genes leads to cylindrical fruits.
If an inbred line with heart-shaped fruits (A/A;B/B) is crossed to an inbred cylindrical fruit individual (a/a;b/b), and the F1 generation is selfed, what fraction of the F2 progeny will be heart-shaped? Assume independent assortment.
A)1/16
B)1/4
C)3/4
D)15/16
How would the answer to the previous question change if you discovered that the two genes were completely linked?
A)7/16
B)1/4
C)3/4
D)The answer would not change.
On the off chance that the genes are not linked, 3 out of 4 F2 progeny will be heart-shaped. In case linked, as it were 1 out of 2 will be heart-shaped.
What fraction of the F2 progeny will be heart-shaped if the two genes were completely linked?To illuminate the issue, let's begin with analyzing the cross between the innate line with heart-shaped natural products (A/A; B/B) and the innate round and hollow natural product person (a/a;b/b).
Since the qualities appear repetitive quality activity, the nearness of at slightest one wild-type allele at either quality will result in typical heart-shaped natural products. In this way, the genotype A/A will contribute to heart-shaped natural products notwithstanding the genotype at the B quality, and the genotype B/B will contribute to heart-shaped natural products notwithstanding the genotype at the A quality.
When these two people are crossed, the F1 generation will have the genotype A/a; B/b. Presently, in the event that the F1 era is selfed, it experiences free collection, meaning that the alleles from each quality are isolated arbitrarily amid gamete arrangement.
To decide the division of heart-shaped natural products within the F2 offspring, we ought to consider the conceivable genotypes coming about from the F1 cross. These are:
A/A;B/b
A/a;B/b
A/A;b/b
A/a;b/b
Three, Out of these four genotypes (A/A; B/b, A/a; B/b, A/A;b/b) have at slightest one wild-type allele at either quality and will yield heart-shaped natural products. As it were one genotype (A/a;b/b) features a homozygous latent genotype at both qualities and will deliver round and hollow natural products.
In this manner, the division of the F2 offspring that will be heart-shaped is 3 out of 4, which can be spoken to as 3/4.
In the event that it was found that the two qualities were totally connected, meaning they are found near together on the same chromosome and don't experience free combination, the reply to the previous address would alter.
Total linkage implies that the two qualities are continuously acquired together as a unit, and their alleles don't group autonomously amid gamete arrangement. In this case, the genotypes A/A and B/B would continuously be acquired together, as well as a/a and b/b.
In case the two qualities were totally connected, the conceivable genotypes within the F2 offspring would be:
A/A;B/B
A/a;b/b
Out of these two genotypes, as it were one (A/A; B/B) will result in heart-shaped natural products, whereas the other (A/a;b/b) yields round and hollow natural products.
Therefore, within the case of total linkage, the division of the F2 offspring that would be heart-shaped is 1 out of 2, which can be spoken to as 1/2 or 50%. The proper reply would be A) 1/2 or B) 50%.
In outline, in the event that the qualities are not totally linked, the division of heart-shaped natural products within the F2 offspring is 3/4 (reply choice C). On the off chance that the qualities are totally connected, the division would be 1/2 or 50% (reply choices A or B).
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