Macrophages are immune cells that have the ability to engulf and destroy pathogens through phagocytosis.
The correct answer is D) macrophage. Macrophages are immune cells that have the ability to engulf and destroy pathogens through phagocytosis. They also play a crucial role in presenting antigens to other immune cells, such as T cells, to initiate an immune response.
Macrophages are considered phagocytes because they can engulf and digest foreign substances, including pathogens. They are also antigen-presenting cells (APCs) because they present antigens derived from pathogens to activate other immune cells.
The correct answer is D) plasma cell. Upon re exposure to a pathogen, memory B cells can differentiate into plasma cells. Plasma cells are responsible for the production and secretion of antibodies, which are key players in the adaptive immune response. Memory B cells, formed during the primary immune response, provide a quicker and more robust immune response upon subsequent encounters with the same pathogen.
The correct answer is A) epinephrine. Epinephrine, also known as adrenaline, is the hormone that elicits the "fight or flight" response. It is released by the adrenal glands in response to stress or danger. Epinephrine increases heart rate, dilates airways, and mobilizes glucose for energy, preparing the body to respond to a perceived threat or challenge. It acts as a neurotransmitter and hormone, affecting various physiological processes to enhance physical and mental readiness during stressful situations.
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Human genes responsible for producing complex biological molecules such as hormones, enzymes and cytokines can be inserted into bacterial cells. These cells are easily grown to high cell densities in large volumes and the desired therapeutic materials produced on a large scale. Using a human-derived gene of interest; bacterial DNA as a plasmid vector and Escherichia coli as the host bacterium, Outline and discuss, step by step, how you would make use of the host bacterium machinery as a mechanism to produce the desired therapeutic materials from the gene of interest on a large scale. Include all the necessary enzymes involved and materials. Be guided by the following subheadings. Subheadings: The human DNA; Plasmid vector; Host bacterium; Selection; and Screening
To produce desired therapeutic materials using a human-derived gene of interest in bacterial cells, specifically Escherichia coli, several steps are involved. Let's go through each step in detail:
The Human DNA:
Identify and isolate the human gene of interest responsible for producing the desired therapeutic material. This gene can be obtained from a variety of sources, such as human cells or synthesized artificially.
Plasmid Vector:
Select a suitable plasmid vector, which is a small, circular DNA molecule that can replicate independently within the bacterial cell.
Host Bacterium (Escherichia coli):
Cultivate Escherichia coli cells in a nutrient-rich medium to achieve high cell densities. This can be done by inoculating a small number of E. coli cells into a growth medium and allowing them to multiply under controlled conditions, such as temperature and oxygen availability.
Selection:
Select an appropriate antibiotic for the selective medium that inhibits the growth of E. coli cells lacking the desired plasmid vector.
Screening:
Select colonies from the plates and perform colony PCR or plasmid isolation to confirm the presence of the gene of interest in the transformed E. coli cells.
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Drs. Tsui and Collins discovered the genetic cause of cystic fibrosis (CF) in 1989. However, life expectancy rates of individuals with CF did not start to improve more significantly until about 2003. What might be plausible explanations for why it took about 14 years from the time of the discovery of the genetic cause of CF to seeing greater increases in life expectancy?
There could be several plausible explanations for the time gap between the discovery of the genetic cause of cystic fibrosis (CF) in 1989 by Drs.
Tsui and Collins and the significant improvements in life expectancy that were observed around 2003. Some potential explanations include:
1. Translational research and clinical implementation: Discovering the genetic cause of CF is a crucial first step, but translating this knowledge into effective treatments and therapies takes time. It may have taken several years of research, experimentation, and clinical trials to develop and refine therapies that specifically target the underlying genetic defect in CF.
2. Development of targeted therapies: CF is a complex genetic disorder with multiple genetic mutations. Developing targeted therapies to address the specific genetic variations in different individuals with CF can be challenging. It may have taken time to identify and develop effective treatments that work for a broader range of CF patients.
3. Regulatory approval process: Bringing new therapies to market requires rigorous testing and approval from regulatory authorities. The process of conducting clinical trials, collecting data, analyzing results, and obtaining regulatory approval can be time-consuming. Delays in regulatory processes could have contributed to the gap between the genetic discovery and the availability of improved treatments.
4. Accessibility and adoption of treatments: Even after the development and approval of new therapies, there can be delays in widespread access and adoption of these treatments. Factors such as availability, affordability, healthcare infrastructure, and patient awareness may have influenced the time it took for individuals with CF to benefit from the new therapies.
5. Cumulative effect of advancements: Improvements in life expectancy may not occur immediately after the introduction of a new treatment. It often takes time for advancements in medical care, supportive therapies, and overall management of CF to accumulate and have a significant impact on life expectancy rates.
It's important to note that these are speculative explanations, and the actual reasons for the time gap between the genetic discovery of CF and improvements in life expectancy may involve a combination of factors.
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We want to map the distance between genes A (green color), B (rough leaf), and C (normal fertility). Each gene has a recessive allele (a= yellow, b-glossy and c-variable). Results from the mating are as follow: 1) Green, rough, normal: 85 2) Yellow, rough, normal: 45 3) Green, rough, variable: 4 4) Yellow, rough, variable: 600 5) Green, glossy, normal: 600 6) Yellow, glossy, normal: 5 7) Green, glossy, variable: 50 8) Yellow, glossy, variable: 90 The double crossover progeny can be observed in the phenotype #s 3 (green, rough, variable) with its corresponding genotype ____ and 6 (yellow, glossy, normal) with its Based on the information from the table corresponding genotype _____ and the previous question, the gene in the middle is ____
The results from the mating used to map the distance between genes A (green color), B (rough leaf), and C (normal fertility) are as follow
Green, rough, normal: 85Yellow, rough, normal: 45Green, rough, variable: 4Yellow, rough, variable: 600Green, glossy, normal: 600Yellow, glossy, normal: 5Green, glossy, variable: 50Yellow, glossy, variable: 90Double crossover progeny can be observed in the phenotype
#s 3 (green, rough, variable) with its corresponding genotype GgBbCc and 6 (yellow, glossy, normal) with its corresponding genotype ggBBcc. We can now map the distance between genes A, B, and C:1. Find the parent phenotype that has the most crossovers with the double crossover phenotype:Green, glossy, variable:
50 crossovers.
2. Find the percentage of offspring of the parent phenotype that had the double crossover phenotype:
4/50 × 100 = 8%.3. Find the percentage of offspring with a single crossover between the middle gene and the gene nearest the middle gene by subtracting the percentage of offspring with no crossovers from the percentage of offspring with any crossover:
100% - (85 + 45 + 600 + 5) = 100% - 735 = 26.5%.
4. Find the percentage of offspring with a single crossover between the middle gene and the gene furthest from the middle gene by subtracting the percentage of offspring with no crossovers from the percentage of offspring with any crossover:
100% - (85 + 45 + 600 + 5) = 100% - 735 = 26.5%.5. Add the results from steps 2, 3, and 4:
8% + 26.5% + 26.5% = 61%.6. The remaining percentage (100% - 61% = 39%) represents offspring with double crossovers between the gene furthest from the middle gene and the gene nearest the middle gene.
7. The gene in the middle is the gene that has the highest percentage of single crossovers (step 3 and step 4). Therefore, gene B (rough leaf) is in the middle.
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(Hair color in trolls is only produced when the T allele is present. Individuals of the tt genotype have white hair. If color is present, the color is determined by the P locus. PP or Pp results in purple color, while pp results in pink hair color. What is the expected phenotypic ratio from a cross between a white-haired female troll with the genotype Ttpp and a purple-haired male troll with the genotype TtPp?)
The expected phenotypic ratio from the cross between a white-haired female troll with genotype Ttpp and a purple-haired male troll with genotype TtPp is 1:1:1:1, meaning an equal number of offspring with purple hair (regardless of genotype) and offspring with pink hair (regardless of genotype). This results in a balanced distribution of hair color phenotypes.
From the given genotypes, we can determine the possible gametes for each parent:
The white-haired female troll with genotype Ttpp can produce gametes Tp and tp.The purple-haired male troll with genotype TtPp can produce gametes TP, Tp, tP, and tp.Now, let's determine the phenotypic ratio from the cross between these two trolls:
Possible genotypes of the offspring:1/4 of the offspring will have genotype TTPP and exhibit purple hair color.
1/4 of the offspring will have genotype TTpp and exhibit pink hair color.
1/4 of the offspring will have genotype TtPP and exhibit purple hair color.
1/4 of the offspring will have genotype Ttpp and exhibit pink hair color.
Therefore, the expected phenotypic ratio from this cross is 1:1:1:1, meaning an equal number of trolls with purple hair (regardless of genotype) and trolls with pink hair (regardless of genotype).
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If you buy some new bull elk with average antler size of 380 inches, and use them as breeders to try to increase antler size in the herd, your selection differential S between these breeders and the rest of the herd will be (Ignore the fact that we are not dealing with female measurements).
The selection differential (S) between the selected breeders and the rest of the herd, in terms of antler size, is 30 inches.
To calculate the selection differential (S), you need to compare the average antler size of the selected breeders (380 inches) with the average antler size of the rest of the herd. Let's assume the average antler size of the rest of the herd is 350 inches.
The selection differential (S) is calculated by subtracting the average trait value of the unselected group from the average trait value of the selected group. In this case, it would be:
S = Average antler size of selected breeders - Average antler size of the rest of the herd
S = 380 inches - 350 inches
S = 30 inches
This value represents the difference in average antler size between the two groups and can be used to estimate the potential response to selection for increasing antler size in the herd.
The selection differential (S) is the difference in the average trait value between the selected breeders and the rest of the herd. In this case, the average antler size of the rest of the herd is 350 inches, the selection differential would be 30 inches.
A larger selection differential indicates a greater difference in trait value, which can influence the potential for improving antler size through selective breeding.
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What is the correct ecological term for non-synchronous fluctuations in predator and prey populations?
A. A 'time lag'
B. Predator prey dynamics
C. Oscillations
D. All of the above
The correct ecological term for non-synchronous fluctuations in predator and prey populations is time lag.
When the fluctuations in predator and prey populations are not synchronous, there is a time lag between the population cycles of the two species. During this time lag, there is a time delay between the population growth of the two species, leading to fluctuations in the population of one species before the other. In this way, ecological time lag is the time difference between the population cycles of different species within an ecosystem. It's crucial to remember that ecological time lags and synchronous fluctuations are related. Synchronous fluctuations refer to the fact that two populations rise and fall in unison over time, while ecological time lags refer to the time differential between these population cycles.
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1. What is the main difference between gymnoperms and angiosperms? What do they have in common? 2. You remove a cell from a four-cell embryo of a roundworm. Explain what you expect to happen. 3. Describe the life cycle of an insect with complete metamorphosis and provide an example. (3.5 marks)
4. Describe the excretory system of insects. (5 marks)
Here are some facts about plants and animals, including the differences between gymnosperms and angiosperms, the development of roundworms, the life cycle of insects, and the excretory system of insects. Therefore
1. Gymnosperms: uncovered seeds, angiosperms: seeds in fruit.
2. Roundworms: each cell contains complete info, removing a cell = developmental defect.
3. Insect complete metamorphosis: egg-larva-pupa-adult.
4. Insect excretory system: Malpighian tubules, bladder, anus; efficient waste removal.
1. The main difference between gymnosperms and angiosperms is that gymnosperms have uncovered seeds, while angiosperms have seeds that are enclosed in a fruit. Gymnosperms also have pollen cones, while angiosperms have flowers. Both gymnosperms and angiosperms are vascular plants, which means they have xylem and phloem tissues. They also both reproduce by pollination and seed dispersal.
2. If you remove a cell from a four-cell embryo of a roundworm, the embryo will not develop into a complete organism. This is because each cell in the embryo contains all the information necessary to create a complete organism. If you remove a cell, you are essentially removing some of the information that is needed for development. The remaining cells will try to compensate for the missing information, but they will not be able to do so perfectly. This will result in a developmental defect, and the embryo will not develop into a complete organism.
3. The life cycle of an insect with complete metamorphosis has four stages: egg, larva, pupa, and adult. The egg is laid by the adult insect and hatches into a larva. The larva is a feeding stage and grows rapidly. When the larva is mature, it pupates. The pupa is a resting stage during which the insect undergoes metamorphosis. The adult insect emerges from the pupa and begins the cycle again.
An example of an insect with complete metamorphosis is the butterfly. The butterfly lays its eggs on a plant. The eggs hatch into caterpillars. The caterpillars eat leaves and grow rapidly. When the caterpillars are mature, they pupate. The pupae are attached to a plant or other surface. The adult butterflies emerge from the pupae and begin the cycle again.
4. The excretory system of insects is composed of Malpighian tubules, a bladder, and an anus. Malpighian tubules are blind sacs that are located near the junction of the digestive tract and the intestine. The tubules remove waste products from the blood and transport them to the bladder. The bladder stores the waste products until they are excreted through the anus.
The excretory system of insects is very efficient at removing waste products from the body. This is important for insects because they have a very high metabolic rate. A high metabolic rate produces a lot of waste products, so it is important for insects to have a way to remove these waste products quickly.
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1. How did Penicillin rupture the E. coli cells in the video? Or stated another way, what cellular target does the antibiotic attack and what is its mechanism of action? 2. Explain the bacterial cell wall structure and compare/contrast the Gram positive and Gram negative bacterial cell wall.
3. Will Penicillin act equally well on all types of bacteria? If you have answered yes, then explain why? If you have answered no, then which type of cell would be more susceptible to Penicillin? What is it about that one type of cell that allows penicillin to act more effectively??
1-By inhibiting this enzyme, penicillin prevents the proper formation of the cell wall, leading to weakened cell walls and ultimately the rupture of E. coli cells.
2-Gram-positive bacteria have a thick peptidoglycan layer that retains the crystal violet stain, while Gram-negative bacteria have a thinner peptidoglycan layer surrounded by an outer membrane.
3-Penicillin does not act equally well on all types of bacteria.
1. Penicillin primarily targets the bacterial cell wall. It inhibits the formation of peptidoglycan, a crucial component of the cell wall in bacteria. The cell wall provides structural support and protection to the bacterial cell. Penicillin binds to and inhibits the enzyme transpeptidase, also known as penicillin-binding protein (PBP), which is responsible for cross-linking the peptidoglycan strands during cell wall synthesis. By inhibiting this enzyme, penicillin prevents the proper formation of the cell wall, leading to weakened cell walls and ultimately the rupture of E. coli cells.
2. Bacterial cell walls can be broadly categorized into Gram-positive and Gram-negative based on their staining characteristics. Gram-positive bacteria have a thick peptidoglycan layer that retains the crystal violet stain, while Gram-negative bacteria have a thinner peptidoglycan layer surrounded by an outer membrane. In Gram-positive bacteria, the cell wall consists mainly of peptidoglycan, which forms a thick, continuous layer. It provides rigidity and structural support to the cell. In Gram-negative bacteria, the cell wall consists of a thin layer of peptidoglycan sandwiched between two lipid bilayers, forming an outer membrane. The outer membrane acts as an additional protective barrier and contains various proteins, lipopolysaccharides (LPS), and porins that regulate the passage of substances into and out of the cell.
3. Penicillin does not act equally well on all types of bacteria. Gram-positive bacteria are generally more susceptible to penicillin because their cell walls are primarily composed of peptidoglycan, which is the target of penicillin. The thick peptidoglycan layer in Gram-positive bacteria provides more binding sites for penicillin, allowing the antibiotic to have a greater inhibitory effect on cell wall synthesis.
In contrast, Gram-negative bacteria have a thinner peptidoglycan layer, and the presence of the outer membrane acts as an additional barrier for penicillin. The outer membrane limits the access of penicillin to the peptidoglycan layer, making Gram-negative bacteria less susceptible to the antibiotic.
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Which of the statements below best describes the classical pathway of complement?
1) An enzyme expressed by the microbe cleaves a complement protein, which triggers a series of events that lead to C3 cleavage.
2) Antibodies bound to a microbe recruit C1q, which activates a series of events that lead to C3 cleavage.
3) C3 is spontaneously cleaved and remains activated upon interaction with the microbial surface.
4) Lectins bound to a microbe recruit complement proteins, which leads to C3 cleavage.
The classical pathway of complement is best described by option 2, which states that antibodies bound to a microbe recruit C1q, initiating a series of events that lead to C3 cleavage. Option 2 is correct answer.
The classical pathway of complement is one of the three main activation pathways of the complement system. It is primarily initiated by the binding of antibodies, specifically IgM or IgG, to a microbe's surface. In option 2, it states that antibodies bound to a microbe recruit C1q, which is the first component of the classical pathway. C1q, along with other complement proteins (C1r and C1s), form the C1 complex.
Upon binding to the microbe, the C1 complex becomes activated and initiates a cascade of enzymatic reactions, resulting in the cleavage phagocytes of complement protein C3. The cleavage of C3 leads to the formation of C3b, which opsonizes the microbe for phagocytosis and generates the membrane attack complex (MAC) to lyse the microbe.
Options 1, 3, and 4 do not accurately describe the classical pathway of complement. Option 1 describes the alternative pathway, option 3 describes spontaneous cleavage (which is not a characteristic of the classical pathway), and option 4 describes the lectin pathway. Therefore, option 2 provides the most accurate description of the classical pathway of complement.
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How
I can calculated the preload,afterload and contractility ?
To calculate preload, afterload, and contractility, you need to understand the basic principles underlying each concept. Preload refers to the amount of tension or stretch on the myocardial fibers before contraction. It is determined by factors such as ventricular filling pressure and volume. Afterload, on the other hand, refers to the resistance that the heart must overcome to eject blood during systole. It is influenced by factors such as arterial pressure and vascular resistance.
Contractility, also known as inotropy, is the inherent ability of the myocardium to generate force and contract. It is influenced by factors such as sympathetic stimulation and changes in intracellular calcium levels.
Calculating these parameters requires various techniques and measurements. For preload, common methods include estimating the central venous pressure or using echocardiography to assess left ventricular end-diastolic volume. Afterload can be calculated by measuring arterial pressure or using invasive techniques such as cardiac catheterization.
Contractility is often evaluated through indices such as the ejection fraction, fractional shortening, or the dP/dt max (the rate of pressure increase during systole). These parameters provide insights into the strength of myocardial contraction.
It's important to note that the actual calculations and specific methods used may vary depending on the clinical setting, available resources, and the context in which these parameters are being assessed. Consulting medical textbooks, guidelines, or seeking expert advice would be beneficial in accurately determining preload, afterload, and contractility in a given scenario.
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4. Discuss the reactions and events of glycolysis indicating substrates, products, and enzymes - in order! I did the first for you. Substrate Enzyme Product i. glucose hexokinase/glucokinase glucose-6-phosphate ii. iii. iv. V. vi. vii. viii. ix. X.
Glycolysis is a multistep process involving the breakdown of glucose into pyruvate for the generation of energy.
The steps involved in glycolysis are as follows:
1. Glucose → (enzyme hexokinase) → glucose-6-phosphate
2. Glucose-6-phosphate → (enzyme phosphoglucose isomerase) → Fructose-6-phosphate
3. Fructose-6-phosphate → (enzyme phosphofructokinase-1) → Fructose-1,6-bisphosphate
4. Fructose-1,6-bisphosphate → (enzyme aldolase) → Dihydroxyacetone phosphate (DHAP) and Glyceraldehyde-3-phosphate (G3P)
5. DHAP → (enzyme triose phosphate isomerase) → Glyceraldehyde-3-phosphate (G3P)
6. Glyceraldehyde-3-phosphate → (enzyme glyceraldehyde-3-phosphate dehydrogenase) → 1,3-bisphosphoglycerate
7. 1,3-bisphosphoglycerate → (enzyme phosphoglycerate kinase) → 3-phosphoglycerate
8. 3-phosphoglycerate → (enzyme phosphoglycerate mutase) → 2-phosphoglycerate
9. 2-phosphoglycerate → (enzyme enolase) → Phosphoenolpyruvate (PEP)
10. Phosphoenolpyruvate (PEP) → (enzyme pyruvate kinase) → Pyruvate
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Concerning homing of effector T cells to the gut, which of the following is not true?
O Interaction with gut epithelium is enhanced by integrin AEB7 binding to cadherin once in the lamina propria
O Antigen-activated T cells in the GALT effector T cells, enter the blood, and then populate mucosal tissues.
O T cells are guided by chemokine CCR9
O Homing is mediated by an interaction between the integrin A4B7 on the T cell and MACAM1 on the endothelial cell
Option (B), Antigen-activated T cells in the GALT effector T cells, enter the blood, and then populate mucosal tissues is not true.
Effector T cells are a subtype of T cells that are primarily responsible for the actual immune response to an antigen. Effector T cells can be present in numerous tissues and are often referred to as tissue-specific. These effector T cells are tissue-specific because they are produced and activated in response to antigens in specific tissues.
Homing of effector T cells to the gut is an essential part of the immune response. It is mediated by an interaction between the integrin A4B7 on the T cell and MACAM1 on the endothelial cell. The chemokine CCR9 guides T cells to the small intestine. It was discovered that binding to gut epithelium is improved by integrin AEB7 binding to cadherin once in the lamina propria. Hence, we conclude that antigen-activated T cells in the GALT effector T cells, enter the blood, and then populate mucosal tissues is not true.
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Assume that the frequency of gene E in populations Breed-A and Breed-B are 0.9 and 0.1, respectively. Given Genotype Phenotypic Value (kg)
EE 60 Ee 48 ee 10 compute the value of dominance (d), difference of gene frequencies (y) between the parental populations, heterosis in F1, and heterosis in F2.
Frequency of gene E in populations Breed-A = 0.9Frequency of gene E in populations Breed-B = 0.1Genotype Phenotypic Value (kg)EE60Ee48ee 10Compute:
Dominance difference of gene frequencies between the parental populations, heterosis in F1, and heterosis in F2. Dominance (d)It is given that EE is dominant, therefore the difference in gene frequencies is given is the gene frequency in and is the gene frequency.
Heterosis in F1Heterosis in F1 is given by: Heterosis mean of parental populations)/mean of parental populations Heterosis mean of parental populations)/mean of parental populations Heterosis Therefore, the values of dominance (d), difference of gene frequencies between the parental populations.
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About 70% of the salt in our diet typically comes from _______ a. meals prepared at home b. peanut butter, ketchup, mustard, and other condiments c. prepared or processed food from the grocery store or restaurants d. potato chips and similar salty/crunchy snacks
About 70% of the salt in our diet typically comes from prepared or processed food from the grocery store or restaurants. The correct option is c).
Processed and prepared foods from grocery stores or restaurants contribute to about 70% of the salt in our diet. These foods often contain high amounts of added salt for flavoring and preservation purposes.
Common examples include canned soups, frozen meals, deli meats, bread, and savory snacks. Additionally, condiments like ketchup, mustard, and salad dressings can also add significant salt content to our diet.
It is important to be mindful of our salt intake as excessive consumption can increase the risk of high blood pressure and other related health issues. Therefore, the correct option is c).
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In your study group you are describing the feeding and nutrition profiles of the unicellular eukaryotes. Which of the following are accurate statements? Check All That Apply There are two types of heterotrophs in the unicellular eukaryotes, phagotrophs and osmotrophs. Phagotrophs are heterotrophs that ingest visible particles of food. Osmotrophs are heterotrophs that ingest food in a soluble form Both phagotrophs and osmotrophs are generally parasitic unicellular eukaryotes Contractile vacuoles are prominent features of unicellular eukaryotes living in both freshwater and marine environments. True or False True False In your study group you are considering the unicellular eukaryotes and discussing specific aspects of their biology. Which of the following statements are accurate regarding the role of contractile vacuoles? Check All That Apply Contractile vacuoles are primarily present on freshwater unicellular eukaryotes because they live in a hypoosmotic environment. Contractile vacuoles are primarily present on marine unicellular eukaryotes because they live in a hyperosmotic environment Contractile vacuoles are primarily used to remove excess water from the cytoplasm Contractile vacuoles are only found in multicellular eukaryotes, not in the unicellular eukaryotes Plasmodium reproduction involves a complex series of steps. Which of the following statements are accurate representations of this complex process? Check All That Apply Plasmodium reproduction requires both sexual and asexual phases of the life-cycle. Sexual reproductive phases of the Plasmodium lifecycle occur in both the mosquito and the human. Asexual reproductive phases of the Plasmodium lifecycle occur in the mosquito only Sporozoites form in the body of the mosquito and infect humans by reproducing asexually, first in liver cells and then in red blood cells
The accurate statements regarding the feeding and nutrition profiles of unicellular eukaryotes are:
- There are two types of heterotrophs in unicellular eukaryotes, phagotrophs and osmotrophs.
- Phagotrophs are heterotrophs that ingest visible particles of food.
- Osmotrophs are heterotrophs that ingest food in a soluble form.
- Contractile vacuoles are prominent features of unicellular eukaryotes living in both freshwater and marine environments.
Unicellular eukaryotes exhibit various feeding and nutritional strategies. Among these, there are two types of heterotrophs: phagotrophs and osmotrophs. Phagotrophs are organisms that actively ingest visible particles of food, while osmotrophs absorb nutrients in a soluble form. These strategies allow unicellular eukaryotes to obtain the necessary nutrients for their survival and growth.
Contractile vacuoles are specialized organelles found in many unicellular eukaryotes. They play a vital role in maintaining osmotic balance by regulating water content within the cell. Contractile vacuoles are particularly prominent in unicellular eukaryotes living in both freshwater and marine environments, where osmotic conditions may fluctuate. They function by actively pumping excess water out of the cell, preventing it from swelling or bursting.
It's important to note that the given statements accurately describe the feeding and nutrition profiles of unicellular eukaryotes, including the distinction between phagotrophs and osmotrophs and the role of contractile vacuoles in maintaining osmotic balance.
the diverse feeding strategies and adaptations of unicellular eukaryotes to different environments to gain a deeper understanding of their biology and ecological roles.
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What is the relationship between ΔG and ΔG‡?
What properties distinguish enzymes from other catalysts?
The relationship between ΔG (free energy change) and ΔG‡ (activation energy) is that ΔG‡ represents the energy barrier that must be overcome for a reaction to proceed, while ΔG represents the overall change in free energy during the reaction.
Enzymes possess specific properties that distinguish them from other catalysts, including their ability to be highly specific, their efficiency in catalyzing reactions, and their regulation through factors like temperature and pH.
The relationship between ΔG and ΔG‡ can be understood in the context of chemical reactions. ΔG represents the difference in free energy between the reactants and products of a reaction. It indicates whether a reaction is thermodynamically favorable (ΔG < 0) or unfavorable (ΔG > 0). On the other hand, ΔG‡, also known as the activation energy, represents the energy barrier that must be overcome for the reaction to occur. It is the energy required to reach the transition state, where the bonds are breaking and forming. ΔG‡ is not directly related to the overall change in free energy (ΔG) but influences the rate at which the reaction proceeds.
Enzymes are specialized catalysts that facilitate biochemical reactions in living organisms. They possess several properties that distinguish them from other catalysts. Firstly, enzymes exhibit high specificity, meaning they can selectively bind to particular substrates and catalyze specific reactions. This specificity is crucial for the regulation of metabolic pathways and cellular processes. Secondly, enzymes are highly efficient, enabling them to catalyze reactions at a faster rate than non-enzymatic catalysts. Their efficiency is due to their ability to lower the activation energy required for the reaction to occur, thus increasing the reaction rate. Lastly, enzymes can be regulated by factors such as temperature and pH, allowing for precise control of biochemical reactions within cells. This regulation ensures that enzymes are active under optimal conditions and can be turned off or modulated as needed.
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Which of the following statement about genetic drift is true? a. Genetic drift can cause a population to adapt to its environment. b. Genetic drift cannot fix alleles in a population without the action of natural selection. c. Genetic drift is unbiased: the frequency of an allele in a population is equally likely to go up or down. d. When populations are large, genetic drift is not invoved in causing them to differentiate. e. Genetic drift causes non-random loss of alleles from a population.
Genetic drift is a mechanism of evolution that affects the genetic structure of populations. It refers to the random fluctuations in allele frequencies that occur due to chance events rather than natural selection. Genetic drift is more pronounced in small populations, where chance events can have a significant impact on the genetic composition of the population.
In response to your question, option (e) is true about genetic drift. Genetic drift causes non-random loss of alleles from a population. This is because genetic drift refers to random fluctuations in allele frequencies, which can lead to the loss of alleles from the population. This can occur due to various chance events, such as mutations, migrations, or the death of individuals carrying particular alleles.
Genetic drift can also result in the fixation of alleles in a population, whereby one allele becomes the only allele present in the population. This can occur in small populations where chance events can have a significant impact on the genetic composition of the population. In summary, genetic drift is an important mechanism of evolution that can cause random fluctuations in allele frequencies, leading to the loss or fixation of alleles in a population.
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The following question is about the citric acid cycle. Select all the enzymes that catalyze oxidation reactions. O citrate synthase O aconitase O isocitrate dehydrogenase O a-ketoglutarate dehydrogenase complex O succinyl-CoA synthetase O succinate dehydrogenase O fumarase O malate dehydrogenase
The citric acid cycle (CAC) is a complex metabolic pathway that occurs in the mitochondria of eukaryotic cells and the cytosol of prokaryotic cells.
The pathway is used to break down acetyl-CoA, generated from the oxidation of glucose and other molecules, and generate energy in the form of ATP. The enzymes that catalyze oxidation reactions in the citric acid cycle include isocitrate dehydrogenase, a-ketoglutarate dehydrogenase complex, succinate dehydrogenase, and malate dehydrogenase. Isocitrate dehydrogenase catalyzes the oxidation of isocitrate to a-ketoglutarate, producing NADH in the process.
A-ketoglutarate dehydrogenase complex catalyzes the conversion of a-ketoglutarate to succinyl-CoA, producing NADH in the process. Succinate dehydrogenase catalyzes the oxidation of succinate to fumarate, producing FADH2 in the process. Malate dehydrogenase catalyzes the oxidation of malate to oxaloacetate, producing NADH in the process. The enzymes that catalyze non-oxidation reactions in the citric acid cycle include citrate synthase, aconitase, succinyl-CoA synthetase, and fumarase.
Succinyl-CoA synthetase catalyzes the formation of succinyl-CoA from succinate and CoA, producing ATP in the process. Fumarase catalyzes the conversion of fumarate to malate.
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Which of the following statements about viruses is FALSE? Viruses have a nucleus but no cytoplasm. а Viruses can reproduce only when they are inside a living host cell. Viruses cannot make proteins on their own. Some viruses use RNA rather than DNA as their genetic material.
The option that is untrue of the ones offered is "Viruses have a nucleus but no cytoplasm."
Acellular infectious organisms with a fairly straightforward structure are viruses. They are made up of genetic material, either DNA or RNA, that is encased in a protein shell called a capsid. A virus's outer envelope may potentially be derived from the membrane of the host cell.However, biological organelles like a nucleus or cytoplasm are absent in viruses. They lack the equipment needed to synthesise proteins or carry out autonomous metabolic processes. In place of doing these things themselves, viruses rely on host cells.
The remaining assertions made are accurate:
- Only when a virus is inside a living host cell can it proliferate. They use the host cell's biological machinery to stealthily copy their genetic material.
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Plants store glucose as starch because ... a. Starch is easier to store because it's insoluble in water b. Starch is more calories per gram than glucose c. Starch is a simpler molecule and therefore easier to store d. All of the above
Plants store glucose as starch because starch is easier to store because it is insoluble in water. Plants are autotrophic organisms that use photosynthesis to create glucose to store energy. Glucose is the primary source of energy in all living cells.
Plants store glucose as starch because starch is easier to store because it is insoluble in water. Plants are autotrophic organisms that use photosynthesis to create glucose to store energy. Glucose is the primary source of energy in all living cells. However, the glucose produced through photosynthesis is not immediately used. It is stored within the plant cells for later use. Storing glucose as starch is the most common way of preserving it. Starch is a polysaccharide, or a complex carbohydrate that can be found in plants, that is stored as a food reserve in plants, and can also be extracted and used commercially as a thickening agent in cooking.
Plants store glucose as starch for a variety of reasons, including its insolubility in water, which makes it easier to store. Starch is also a more compact form of energy storage since it can store more calories per gram than glucose. Furthermore, it is less reactive than glucose and has a lower osmotic pressure, which can prevent damage to the plant cells. Therefore, plants store glucose as starch because it is easier to store and more convenient for later use.
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Many females prefer to mate with territorial males and NOT with males that hold no territories. Why?
Females prefer mating with territorial males due to resource access, genetic superiority, parental care, and a competitive advantage, ensuring higher survival and reproductive success for themselves and their offspring.
The preference of females for mating with territorial males can be attributed to several factors, many of which are rooted in evolutionary biology and reproductive strategies. Here are some reasons why females may show a preference for territorial males:
Resource availability: Territorial males often have access to more resources within their territories, such as food, nesting sites, or shelter. By choosing a territorial male, females can gain access to these resources, which can enhance their own survival and the survival of their offspring.Good genes hypothesis: Territorial males may demonstrate higher genetic quality, indicating their ability to survive and succeed in acquiring and defending a territory. Females can benefit from mating with such males as it increases the likelihood of their offspring inheriting advantageous traits, including better disease resistance, physical prowess, or cognitive abilities.Parental care: Territorial males are more likely to invest in parental care, as they have a stake in protecting and providing for their offspring within their territories. By selecting a territorial male, females increase the chances of receiving support and assistance in raising their young, leading to higher survival rates for their offspring.Competitive advantage: Mating with a territorial male can also confer a competitive advantage to the female. Territorial males often engage in aggressive behaviors to defend their territories from other males, reducing the chances of infidelity and ensuring the offspring's paternity.It's important to note that while these preferences may be observed in many species, including some primates and birds, mating preferences can vary across different animal groups, and not all females exhibit the same preferences. Additionally, social and ecological factors can influence the extent to which these preferences are expressed in a given population or species.
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Make a simple dichotomous key for taxonomic identification
all 13 7:58 Instructions: How to make a simple dichotomous key for taxonomic identification Dichotomous keys are based on the use of pairs of contrasting statements. That is, the pairs of statements a
To make a simple dichotomous key for taxonomic identification, follow the instructions given below: Step 1: Choose an organismSelect the organism that you want to identify.
For example, let's choose an insect.Step 2: List characteristicsList a few characteristics of the organism you selected. For instance, an insect has six legs, two wings, and compound eyes.Step 3: Group the characteristicsGroup the characteristics into two categories based on their similarities. For example, legs and wings can be grouped under one category, while compound eyes can be grouped under another. Step 4: Create a contrast statement Create a statement that contrasts the two categories.
For example, the contrast statement for the categories created in step 3 can be "Does the organism have legs and wings or compound eyes?"Step 5: Create more categories and statementsAdd more categories and contrast statements until there are no more characteristics left to differentiate the organism. For instance, more categories like "has antennae or not" and "more than 100 legs or less than 100 legs" can be added to differentiate insects further.Step 6: Label the categoriesLabel each category, starting with category 1 and ending with the last category added.
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Discussion Unit 22 A Describe the flow of air from the nose to the alveoli, name all structures in the pathway and one abnormal condition associated with it.
An abnormal condition associated with this pathway is asthma. Asthma is a chronic respiratory disorder characterized by inflammation and narrowing of the airways. This can lead to difficulty in breathing, wheezing, coughing, and chest tightness.
The flow of air from the nose to the alveoli involves several structures in the respiratory pathway. It begins with the inhalation of air through the nostrils or nasal passages. The air then passes through the following structures:
Nasal cavity: The nasal cavity is the hollow space behind the nose. It is lined with mucous membranes and contains structures called turbinates that help filter, warm, and moisten the air.
Pharynx: The pharynx, also known as the throat, is a muscular tube located behind the nasal cavity. It serves as a common passage for both air and food.
Larynx: The larynx, or voice box, is located below the pharynx. It contains the vocal cords and plays a role in speech production.
Trachea: The trachea, commonly known as the windpipe, is a tube that connects the larynx to the bronchi. It is lined with ciliated cells and cartilaginous rings, which help maintain its shape and prevent collapse.
Bronchi: The trachea branches into two bronchi, one leading to each lung. The bronchi further divide into smaller bronchioles, which eventually lead to the alveoli.
Alveoli: The alveoli are small air sacs located at the ends of the bronchioles. They are the primary sites of gas exchange in the lungs, where oxygen is taken up by the bloodstream, and carbon dioxide is released.
An abnormal condition associated with this pathway is asthma. Asthma is a chronic respiratory disorder characterized by inflammation and narrowing of the airways. This can lead to difficulty in breathing, wheezing, coughing, and chest tightness. In individuals with asthma, the airway inflammation and increased sensitivity to certain triggers result in the constriction of the bronchial tubes, making it harder for air to flow freely. Proper management and treatment of asthma are important to maintain normal airflow and prevent respiratory distress.
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Solar energy is a renewable energy source because: a. No storage of the energy is required. O b. It can be used directly as a fuel. O c. A new supply is constantly available. O d. It is no affected by air pollution
Solar energy is a renewable energy source because a new supply is constantly available. Solar power is harvested from the sun, which generates a vast amount of energy each day. Due to the immense amount of energy that the sun produces, it is considered a renewable energy source.
Since the sun's energy is infinite and will never be depleted, it is referred to as a renewable resource.The sun’s light is harvested through photovoltaic cells that convert the sun's light into electricity. There is no need to store solar energy since the sun is constantly producing it. Additionally, solar energy is a clean and environmentally friendly option since it does not produce any emissions or pollutants that can contribute to air pollution. Furthermore, it is simple to use solar power directly without the need for fuel or storage, making it an excellent option for powering homes, businesses, and even large-scale power plants.
Solar energy is rapidly gaining in popularity as a renewable energy source, with solar panels being installed on residential and commercial buildings worldwide. Solar energy is cost-effective, reduces reliance on nonrenewable fossil fuels, and has the potential to revolutionize the way we produce and consume electricity.
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13. Which of the following represents the correct order of stages during Drosophila development?
Select one:
a.
zygote, syncytial blastoderm, cellular blastoderm, gastrula, larva, pupa, adult
b.
syncytial blastoderm, cellular blastoderm, zygote, gastrula, larva, pupa, adult
c.
zygote, larva, gastrula, pupa, syncytial blastoderm, cellular blastoderm, adult
d.
cellular blastoderm, syncytial blastoderm, zygote, gastrula, pupa, larva, adult
and.
zygote, cellular blastoderm, syncytial blastoderm, gastrula, larva, pupa, adult
14.The following protein represents an inductive signal for the creation of lens tissue:
Select one:
a.
FGF8
b.
BMP4
c.
crystalline
d.
all of the above
and.
a and b are correct
15.The following molecule acts as a paracrine factor:
Select one:
a.
wnt
b.
hedgehog
c.
Delta
d.
all of the above
and.
a and b are correct
14. The protein that represents an inductive signal for the creation of lens tissue is: c. crystalline.
15. The molecule that acts as a paracrine factor is: d. all of the above (a. wnt and b. hedgehog).
these are correct answers.
Crystalline is a protein involved in the development and function of the lens in the eye. It plays a crucial role in the formation of lens tissue during development.
Both Wnt and Hedgehog molecules are examples of paracrine factors. Paracrine signaling refers to the release of signaling molecules by one cell to act on nearby cells, affecting their behavior or gene expression. Both Wnt and Hedgehog molecules function as paracrine signals in various developmental processes and tissue homeostasis.
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For each group of life history classifications below list as many of the characteristics that would been seen for organisms in each group. You may simply write the number of the life history classification and then list as many letters of characteristics that are associated with the life history classification (i.e. 9) a, b, g, i)
Characteristics:
a) Long Life
b) Short Life
c) Rapid Development
d) Slow Development
e) Many potential offspring produced (lifetime)
f) Few potential offspring produced (Lifetime)
g) Large adult size
h) Small adult size
i) Stable habitat (low stress and low disturbance)
Life History classifications:
1) R-strategists
2) K - strategists
3) Ruderal
4) Stress-Tolerant
5) Competitive
6) Opportunistic
7) Equilibrium
8) Periodic
1.R-strategists: b, c, e, h 2.K-strategists: a, d, f, g 3.Ruderal: b, c, e, h 4.Stress-Tolerant: a, d, f, h 5.Competitive: a, d, f, g 6.Opportunistic: b, c, e, g 7.Equilibrium: a, d, f, g 8.Periodic: a, d, f, h
1.R-strategists are characterized by short life span, rapid development, high reproductive output, and small adult size. They produce many potential offspring during their lifetime, as they invest little energy in individual offspring and rely on quantity over quality to ensure survival.
2.K-strategists have long life spans, slow development, low reproductive output, and large adult size. They produce few potential offspring during their lifetime but invest more energy in each individual offspring, prioritizing quality over quantity.
3.Ruderal organisms have a short life span, rapid development, high reproductive output, and small adult size. They produce many potential offspring during their lifetime and are adapted to disturbed and unpredictable environments.
4.Stress-Tolerant organisms have long life spans, slow development, low reproductive output, and small adult size. They produce few potential offspring during their lifetime and are adapted to stressful and stable habitats.
5.Competitive organisms have long life spans, slow development, low reproductive output, and large adult size. They produce few potential offspring during their lifetime and are adapted to competitive and resource-rich environments.
6.Opportunistic organisms have a short life span, rapid development, high reproductive output, and large adult size. They produce many potential offspring during their lifetime and exploit favorable conditions as they arise.
7.Equilibrium organisms have long life spans, slow development, low reproductive output, and large adult size. They produce few potential offspring during their lifetime and are adapted to stable and predictable habitats.
8.Periodic organisms have long life spans, slow development, low reproductive output, and small adult size. They produce few potential offspring during their lifetime and are adapted to cyclic or periodic environments.
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Breast cancer involves several proteomic modifications. A surgeon has operated on a breast cancer patient and provided you with a sample from the breast tissue of the patient containing a piece of the tumor to analyze its proteome. Design the experiment. Which method are you going to use and why? which approach and why? Don't forget to mention the controls you will use, and the different steps in your workflow, and where will you deposit your results.
The experiment involves using mass spectrometry-based proteomics to analyze the proteome of a breast cancer tumor sample.
The chosen method, mass spectrometry-based proteomics, allows for comprehensive analysis of proteins in the tumor sample. Label-free quantitative proteomics approach will be employed to compare protein abundances between the tumor sample and controls. The workflow includes sample preparation, protein digestion, mass spectrometry analysis, data analysis, and potential validation of selected proteins. Controls such as a positive breast cancer control and a negative healthy tissue control will be used for comparison. The results will be deposited in public proteomics databases for accessibility and further research.
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1. When you stand on a foam pad with eyes closed in a
BESS test, the primary sensory input for balance is ______ .
a. olfaction
b. vestibular
c. somatosensation
d. vision
2. Olfaction affects the accu
The BESS test:When standing on a foam pad with closed eyes in the BESS (Balance Error Scoring System) test, the primary sensory input for balance is somatosensation. This is defined as the body’s internal and external sensory systems that help control balance and movement.
The somatosensory system comprises cutaneous and proprioceptive receptors located in the skin, muscles, joints, and bones of the body.
Olfaction affects the accuracy of taste: Olfaction (sense of smell) affects the accuracy of taste. Olfaction and gustation (sense of taste) are interconnected senses that work together to produce the perception of flavor. While the tongue is responsible for detecting taste, the nose is responsible for identifying smells. These two senses work together to produce a complete picture of flavor.
When the olfactory system is damaged, the sense of taste may be compromised, making it difficult to distinguish between different flavors. For example, without olfaction, foods may taste bland, and it may be challenging to differentiate between salty, sweet, bitter, or sour tastes.Hence, we can conclude that somatosensation is the primary sensory input for balance in the BESS test, and olfaction affects the accuracy of taste.
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1. List sugar, galactose, and glucose in order of
efficiency of fementation. (Describe reasons as well)
2. How temperature can affect ethanol fermentation?
1. List sugar, galactose, and glucose in order of efficiency of fermentation along with their explanation:Galactose: Galactose is a monosaccharide, similar to glucose, that can be converted to glucose-1-phosphate before being used in glycolysis,
Galactose is converted into glucose-6-phosphate in the liver. The sugar, which is an epimer of glucose, is not a key sugar used in fermentation. The efficiency of fermentation of galactose is less than that of glucose.Glucose: Glucose is the primary fuel for glycolysis, and it has the highest efficiency of fermentation among sugars. Glucose, unlike other sugars, does not need to be converted into a different type of sugar before being used in glycolysis. Glucose is broken down into pyruvate, which is a critical product of glycolysis, during glycolysis. Glucose fermentation is highly efficient.
Sugar: Sugar is a disaccharide consisting of fructose and glucose molecules, which is hydrolyzed into glucose and fructose before being used in fermentation. As a result, fermentation efficiency is less than glucose.2. How temperature can affect ethanol fermentation?Ethanol fermentation, like other enzymatic reactions, is influenced by temperature. Fermentation's optimal temperature range is between 20°C and 35°C. Lower temperatures reduce enzyme activity, and hence fermentation rate, while higher temperatures can cause enzyme denaturation or destruction, which will prevent ethanol fermentation from occurring. Therefore, the temperature can affect the ethanol fermentation.
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Conservationists are translocating 250 fi sh (taken at random) from a lake to a smaller pond that has recently been restored after a chemical spill, to repopulate the pond. After two days of the translocating of the fish, conservationists determined that 40 fish have died due to being homozygous for a recessive allele at a detoxify cation gene that has made the fish susceptible to residual pollutants still left in the pond.
QUESTION:
(A) Considering the gene pool, what percentage of the zygotes produced by these translocated fish will be expected not to survive(assume there is random mating amongst surviving fish, they mate monogamously and all mate-pairs produce the same number of fertilised eggs)? (Two decimals)
(B) After 5 generations (assuming the residual level of toxins remains the same), what will the allele frequency of the non-susceptible allele be in the pond population? (Two decimals)
The frequency of the dominant allele is represented as p and the frequency of the recessive allele is represented as q. According to the Hardy-Weinberg equilibrium, the frequency of homozygous dominant individuals is p², the frequency of homozygous recessive individuals is q², and the frequency of heterozygous individuals is 2pq.
Therefore, the frequency of homozygous recessive individuals (q²) is equal to 0.25, as we know that 40 fish have died out of 250 translocated fish and the deaths have occurred due to homozygous for a recessive allele at a detoxify cation gene, so 0.16 is the frequency of the recessive allele (q).The frequency of the dominant allele (p) can be calculated by subtracting the recessive allele (q) from one(1), i.e., 1 - 0.16 = 0.84.The frequency of zygotes that will not survive can be found by multiplying the homozygous recessive frequency by itself(q²). Therefore, the percentage of zygotes that will not survive is 0.0256 * 100 = 2.56%.Hence, the answer is 2.56%.B) After 5 generations (assuming the residual level of toxins remains the same), (Two decimals)The frequency of the recessive allele (q) is 0.16, which was previously calculated.
The Hardy-Weinberg equilibrium can be used to calculate the frequency of the non-susceptible allele.The sum of the frequency of the two alleles is equal to one(1).The frequency of the non-susceptible allele = p= 1-q= 1- 0.16= 0.84.In each generation, the frequency of the non-susceptible allele remains the same, i.e., 0.84.
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