The unifying themes of GFP purification by HIC involve protein purification, hydrophobicity, affinity chromatography, and chromatographic separation techniques.
GFP purification by HIC (Hydrophobic Interaction Chromatography) incorporates several unifying themes, such as:
1. Protein purification: The process of GFP purification by HIC is a method of isolating and purifying GFP from a complex mixture of proteins. This theme involves various techniques used to separate proteins based on their unique physical and chemical properties.
2. Hydrophobicity: HIC exploits the hydrophobic nature of GFP by using a stationary phase with hydrophobic ligands. This theme emphasizes the importance of hydrophobic interactions in protein-protein and protein-ligand interactions.
3. Affinity chromatography: HIC is a type of affinity chromatography that selectively binds hydrophobic molecules to a hydrophobic stationary phase. This theme highlights the importance of specific interactions between proteins and their ligands.
4. Chromatographic separation: HIC relies on the differential binding of GFP to the stationary phase, enabling the separation of GFP from other proteins in the mixture. This theme emphasizes the importance of chromatographic separation techniques in protein purification.
Overall, the unifying themes of GFP purification by HIC involve protein purification, hydrophobicity, affinity chromatography, and chromatographic separation techniques.
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The tree-based reduction approach leads to load imbalance. true or false
The given statement "The tree-based reduction approach leads to load imbalance" is false because In parallel computing, a reduction operation involves combining data from multiple processors or nodes into a single result.
The tree-based reduction approach is a common parallel reduction technique where the data is combined in a hierarchical manner, with each level of the tree combining data from the previous level. While it is true that the tree-based reduction approach can lead to load imbalance in some cases, it is not a fundamental property of the technique.
Load imbalance can occur if the data being combined at each level of the tree is not evenly distributed across processors or nodes. However, there are ways to mitigate this issue, such as dynamically reassigning work to balance the load or using load-balancing algorithms. Overall, the tree-based reduction approach is a widely used and effective technique for performing parallel reduction in distributed computing systems.
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Modern researches that study succession find it to be less orderly and more dynamic that previously thought.
The given statement "Modern researchers that study succession find it to be less orderly and more dynamic that previously thought" is true because more recent research has shown that succession is often more complex and dynamic than this traditional view suggests.
Once upon a time, ecological succession was assumed to be a predictable and orderly process in which a community of species evolves over time, with one stable community eventually replacing another.
Recent research, however, has demonstrated that succession is frequently more nuanced and dynamic than this classic understanding predicts.
Fires and floods, for example, can reset the successional clock by eliminating current vegetation and generating new possibilities for colonisation by various species.
Furthermore, interspecies interactions such as competition, predation, and mutualism can have a significant impact on the direction and rate of succession.
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The following question may be like this:
Modern researches that study succession find it to be less orderly and more dynamic that previously thought. True or false.
Refer to the related 10 base pair sequences below to answer the following 3 questions.
Person 1: AGTCTTGACT
Person 2: AGACTGGATT
Person 3: AGTCTGCATA
What is the consensus "wild-type" sequence for this locus?
How many SNPs are present in person 1's shown sequence?
How many SNP sites are present in this three person population?
The consensus "wild-type" sequence for this locus is AGTCTGCACT. The correct answer is person 2.
The consensus "wild-type" sequence is determined by identifying the most common base at each position among the sequences provided. In this case, the consensus sequence is AGTCTGCACT as it appears in 2 out of the 3 sequences.
For person 1's shown sequence, there is 1 single nucleotide polymorphism (SNP) present. The SNP is a difference in the base at position 6, where it is T in person 1's sequence, while it is C in the other two sequences.
In the three-person population, there are a total of 2 SNP sites present. One SNP is at position 6 (T in person 1, C in persons 2 and 3), and the other SNP is at position 9 (G in person 1, A in person 3).
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Compare the colors of the veggies cooked in acid and alkaline cooking water. Which are the predominant pigments in these veggies?
Cooking vegetables in acidic or alkaline water can affect their color and flavor due to the changes in the pigments that give them their characteristic hues. When vegetables are cooked in acidic water (pH below 7), the predominant pigments are anthocyanins, which give them red, purple, or blue hues.
Anthocyanins are water-soluble pigments that fade or turn brownish when exposed to acidic pH conditions. Anthocyanin-containing vegetables include red cabbage, red onions, and purple potatoes.
When cooked in acidic water, the colour of these veggies may turn pink or blue-gray, depending on the pH level.
When vegetables are cooked in alkaline water (pH more than 7), the primary pigments are chlorophylls and carotenoids. Chlorophylls are responsible for the green colour of vegetables and are generally stable in alkaline pH levels, whereas carotenoids are responsible for the yellow, orange, or red colours of vegetables and are more stable in acidic pH levels.
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The _____________ separate adjacent vertebrae and they soften forces created by walking
True or False
The large intestine stores and eliminates the waste products of digestion
It is true that the large intestine stores and eliminates the waste products of digestion.
The large intestine is a very important and primary part of the elementary canal. Its primary purpose is to absorb water and electrolytes from the undigested dietary particles, resulting in feces.
The waste material is then stored in the large intestine and then it is removed from the body when the process of defecation is done.. The large intestine also contains a vast population of helpful bacteria that aid in nutritional digestion and absorption, as well as vitamin synthesis.
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Parasympathetic postganglionic fibers of the head travel within theA) facial nerve. B) vestibulocochlear nerve.C) trigeminal nerve. D) accessory nerve.
The correct answer is A) Facial nerve.The parasympathetic nervous system has a craniosacral outflow, meaning that its preganglionic fibers originate from the brainstem or sacral regions of the spinal cord.
In the head, parasympathetic preganglionic fibers travel with various cranial nerves to reach their target organs.The parasympathetic postganglionic fibers of the head that innervate the lacrimal gland, nasal glands, and salivary glands (except the parotid gland) travel within the facial nerve (CN VII). These fibers synapse in the pterygopalatine ganglion or submandibular ganglion, which are located near the target glands.The parasympathetic postganglionic fibers that innervate the parotid gland travel within the glossopharyngeal nerve (CN IX). These fibers synapse in the otic ganglion, which is also located near the target gland.
The vestibulocochlear nerve (CN VIII) is responsible for transmitting auditory and vestibular information from the inner ear to the brainstem. It does not contain any parasympathetic fibers.
The trigeminal nerve (CN V) is a mixed nerve that contains both sensory and motor fibers. It does not contain any parasympathetic fibers.
The accessory nerve (CN XI) is responsible for controlling certain neck muscles, such as the sternocleidomastoid and trapezius muscles. It does not contain any parasympathetic fibers.
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Please select the statements that are true of inducible operons to test your understanding of the differences between inducible and repressible operons.
Inducible operons are typically "off" or repressed by a regulatory protein until an inducer molecule is present is the statements that are true of inducible operons to test you.
The debaser patch attaches to the repressor protein, causing it to alter shape, precluding it from binding to the driver. An inducible operon's genes are generally engaged in catabolic processes, which are utilised to break down nutrients similar as carbohydrates. The lac operon inE. coli is an illustration of an inducible operon that's responsible for lactose breakdown.
Repressible operons vary from inducible operons in that they're typically" on" or active until a corepressor patch attaches to the nonsupervisory protein and stops it from binding to the driver, effectively shutting down gene product. InE. coli, the trp operon is an illustration of a repressible operon that's responsible.
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Explain the relationship between permeability and the flow of water.
Independent assortment of three genesAssuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.- 27:9:9:9:3:3:3:1- 1:1:1:1:1:1:1:1- 1:0:0:0:0:0:0:0- 1:0:1:0:1:0:1:0
Assuming that the three genes undergo independent assortment, the phenotypic ratio of the offspring in the F2 generation is 27:9:9:9:3:3:3:1. Option A is correct.
The principle of independent assortment states that during gamete formation, the alleles of different genes segregate independently of each other. In this scenario, there are three genes that undergo independent assortment, and each gene has two alleles (A and a, B and b, C and c).
When two individuals heterozygous for all three genes (AaBbCc x AaBbCc) are crossed, the resulting F1 generation will all be heterozygous for each gene (AaBbCc). During the formation of gametes in the F1 generation, the alleles of each gene will segregate independently of each other, resulting in eight possible gamete combinations (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc) with equal probability.
When these gametes combine randomly in the F2 generation, the resulting phenotypic ratio will be 27:9:9:9:3:3:3:1, which corresponds to the number of individuals with each possible combination of genotypes (AAA, AAB, AaB, AaBb, AaBc, Aabc, aaBc, aabc). Option A is correct.
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What type of species, if removed from the community, could lead to the collapse of the entire community?A. Flagship speciesB. Indicator speciesC. Endemic speciesD. Keystone species
Keystone species of species, if removed from the community, could lead to the collapse of the entire community
Keystone species are species that have a disproportionately large impact on the community relative to their abundance. If a keystone species is removed from the community, the community structure and function can change dramatically, potentially leading to the collapse of the entire community.
Keystone species can have a variety of roles in the community, such as predators, pollinators, or ecosystem engineers, and their removal can cause cascading effects throughout the food web. Examples of keystone species include sea otters, which help maintain kelp forest ecosystems, and beavers, which can alter river systems and create habitats for other species.
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Which buffers in GFP purification by HIC have the highest salt content and how can you tell
The buffers with the highest salt content in GFP purification by HIC are typically the elution buffers.
In GFP purification by HIC (hydrophobic interaction chromatography), the buffers with the highest salt content are usually the elution buffers. This is because higher salt concentrations are needed to disrupt the hydrophobic interactions between the GFP and the stationary phase of the HIC column. Typically, the salt concentration in the elution buffer is higher than that in the equilibration buffer and wash buffers. One way to determine the salt content is by looking at the ionic strength of the buffer, which is a measure of the concentration of all ions in solution. A higher ionic strength indicates a higher salt content. Another way to determine the salt content is by measuring the conductivity of the buffer. Conductivity is a measure of the ability of a solution to conduct electricity, which is influenced by the presence of ions. A higher conductivity indicates a higher salt content.
In summary, the buffers with the highest salt content in GFP purification by HIC are typically the elution buffers, which can be identified by their higher ionic strength and conductivity compared to the other buffers used in the process.
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(c) A type AB individual carries both A and B oligosaccharides, and therefore will not recognize either A or B as foreign in donated blood.
That is correct. Due to carrying both A and B toligosaccharides, type AB individuals can receive blood from individuals with blood types A, B, AB, and O without any adverse reactions as they do not recognize any of the ABO antigens as foreign.
However, they can only donate blood to other individuals with blood type AB as their blood contains both A and B oligosaccharides.When receiving donated blood, this person's immune system will not recognize either A or B oligosaccharides as foreign, allowing them to safely accept blood donations from A, B, AB, or O blood types.
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What happen when the thyroxine hormone is not secreted in sufficient quantities
Whenever the thyroxine hormone is not secreted in a sufficient amount, it causes hypothyroidism.
Thyroid gland is present in our neck and is a very important gland as it is involved in a number of important bodily functions which include regulating body temperature, metabolism etc.
Hypothyroidism is basically a very commonly condition in which the thyroid gland basically doesn't create as well as release enough amount of thyroid hormone in the bloodstream. This causes the metabolism to slow down. Hypothyroidism can make the patient feel tired, gain weight as well as make them be unable to tolerate the colder temperatures.
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After they are formed, lymphocytes are housed in _____ lymphatic structures.
After they are formed, lymphocytes are housed in lymphoid tissues and lymphatic structures.
Lymphocytes are housed in lymphoid tissues and lymphatic structures after they develop. Lymph nodes, the spleen, tonsils, adenoids, and Peyer's patches in gut-associated lymphoid tissue (GALT) are examples of these structures.
These structures are important in the immune system because they provide a location for lymphocytes to encounter and respond to antigens such as infections or foreign substances.
The lymphatic system is a network of veins, tissues, and organs that helps the body maintain fluid balance and fight infections. Lymphocytes are an essential component of the lymphatic system, and their presence in lymphatic structures aids in the immunological response to invading pathogens.
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What is the least number of tissues an organ can consists of? A) 2B) 4C) 3D) 5E) 1
The least number of tissues an organ can consist of is one. Therefore the correct option is option B.
The smallest number of tissues that an organ can have is one. This is due to the fact that an organ is defined as a collection of tissues that execute a specific function.
The lens of the eye, which is made up of just lens fibres, and the adrenal medulla, which is made up of only chromaffin cells, are two examples of organs that are made up of a single tissue.
Most organs, on the other hand, are made up of several tissues that work together to fulfil their duties. As a result, the correct answer is E) 1.
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If a tiger and a lion were to mate, what may be a result?
1. Which of these limiting factors is density independent?
a. Predation
b. Natural disasters
c. Competition
d. Disease
2. The natural greenhouse effect keeps the Earth’s temperatures warm enough for life.
a. True
b. False
3. The more biodiversity in an ecosystem….
a. the lower the stability of that ecosystem.
b. the more disturbances that ecosystem experiences.
c. the greater the stability of that ecosystem.
4. Which one of the following human activities is the greatest threat to biodiversity?
a. Habitat destruction
b. Climate change
c. Introduced species
d. Pollution
e. Overharvesting
5. Which of the following pairs of events was most responsible for the rapid increase in human population?
a. Black plague and development of fire
b. The industrial revolution and energy from coal
c. Steam power and energy from burning wood
d. Agricultural green revolution and antibiotic availability
6. What might happen if man continues to burn fossil fuels and put more carbon dioxide into the atmosphere?
a. Overall global temperatures will be warmer.
b. There might be more frequent and more severe droughts.
c. There might be more frequent and more severe storms.
d. All of the above are possible if man continues to add carbon dioxide to the atmosphere.
7. Ecosystem ________________ is anything that disrupts the homeostasis of an ecosystem.
a. disturbance
b. resistance
c. resilience
d. respiration
8. ___________________ diversity describes the variety and abundance of different types of species that inhabit an area.
a. Genetic
b. Species
c. Ecosystem
d. Terrestrial
Natural disasters are density-independent factors because they affect every individual in a population independent of their density. The natural greenhouse effect is essential for life on Earth, as it helps to maintain a temperature range that is suitable for living organisms.
Thus, greater the diversity of species in an ecosystem, greater the stability of that ecosystem because a diverse ecosystem is better to recover from disturbances and protect the life of more living organisms.
Habitat destruction, such as deforestation is the greatest threat to biodiversity that lead to the loss of habitats and the extinction of species. The agricultural green revolution and the availability of antibiotics increases the food production and a increases the life expectancy of living organisms.
The increased concentration of carbon dioxide can lead to global warming, which can cause more frequent and severe droughts and storms, thereby, threatening the life of various living organisms. Ecosystem disturbance is anything that disrupts the homeostasis of an ecosystem. Species diversity describes the variety and abundance of different types of species that inhabit an area.
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The phase of hemostasis that involved clotting of blood is called
Answer:
The answer is Secondary hemostasis (coagulation cascade)
Question 51 Marks: 1 Halon-1211 is used primarily in ______.Choose one answer. a. portable fire extinguishers b. fixed fire suppression systems c. printed circuit board cleaning d. vapor degreasing
Halon-1211 is used primarily in portable fire extinguishers. The correct option is a.
The Halon 1211 is used primarily in the portable fire extinguishers and this is the streaming agent. The Halon 1211 is majorly used in the streaming agent in the portable, and the hand be held the fire extinguishers.
This Halon 1211 is also used in the large wheeled extinguishers and it is called as the "Flight Line" units to be combat the airplane fires at the runway. The Halon 1211 fire extinguisher is the liquefied gas, and it pressurized with the nitrogen, that will discharges as the vapor causing the no cold. The option a is correct.
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What do Lycophytes look like?
A. Flowers
B. Pine Trees
C. Botry
Lycophytes look like Botry, the correct option is (C).
Lycophytes, also known as clubmosses, are a group of primitive vascular plants that have been around for over 400 million years. They are characterized by their small, herbaceous stature and spore-bearing structures that are arranged in club-shaped clusters.
Lycophytes have a unique reproductive structure called a strobilus or botryoid (botry). The strobilus is a cone-like structure that contains sporangia, which are structures that produce and release spores. The sporangia are arranged in a spiral pattern on the strobilus, giving it a distinctive appearance. They also do not resemble pine trees, which are a type of gymnosperm. Instead, they have small, scale-like leaves that are arranged spirally around the stem, giving them a moss-like appearance, the correct option is (C).
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what event initiates an adaptive immune response? group of answer choices the production of mhc class i or ii molecules the phagocytosis of a pathogen by a macrophage the interaction of a naive t cell with an antigen-presenting cell the expression of cytokines cd4 or cd8 the interaction of a b cell with a th cell
The interaction of a naive T cell with an antigen-presenting cell initiates an adaptive immune response. Therefore the correct option is option B.
When a pathogen enters the body, phagocytic cells such as macrophages engulf it, breaking it down into little pieces and presenting those fragments (antigens) on their surface utilising MHC molecules.
After then, the antigen-presenting cell connects with a T cell, which recognises the antigen as foreign and initiates an immunological response.
The contact between the T cell and the antigen-presenting cell is usually what triggers the adaptive immune response.
T cells grow and differentiate into effector cells that can directly fight the pathogen or drive other immune cells, such as B cells, to make antibodies once activated. Therefore the correct option is option B.
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gibberellin stimulates the production of α-amylase in germinating seeds, an enzyme that converts ______ into ______.
Germinating seeds. α-amylase is an enzyme that converts starch into glucose, which is a simple sugar that can be used by the growing plant for energy.
Gibberellin is a plant hormone that plays an important role in the growth and development of plants.
When a seed germinates, it starts to grow and develop into a seedling. In order to do this, it needs energy, which it obtains by breaking down its stored reserves of starch. α-amylase is the enzyme that is responsible for breaking down the starch into glucose, which can then be used by the growing plant.
Gibberellin stimulates the production of α-amylase by activating the genes that code for this enzyme. This allows the plant to rapidly break down its stored reserves of starch and obtain the energy it needs to grow and develop.
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PLS HELP ME ASAP I WILL MARK THE BRAINLIEST!!
Sexual reproduction and meiosis go hand-in-hand. Meiosis is the process responsible for gamete (sex cell) production and ensures genetic variation. But how does it do this?
Drag the appropriate label/explanation to the correct location on the meiosis diagram.
Diploid parent cell: This is the cell that meiosis begins with, containing two copies of each chromosome.
What is chromosome?A chromosome is a structure found in the nucleus of cells that contains genetic material. It is made of two strands of DNA, wound around proteins called histones, that are tightly packed together. Chromosomes are essential components of the cell and are responsible for the transmission of hereditary information.
Independent assortment of chromosomes: During the first stage of meiosis (prophase I), the chromosomes line up and exchange genetic information, leading to the random assortment of genetic material and increasing the variation of the daughter cells.
Crossing over: During the second stage of meiosis (prophase I), the chromosomes cross over, or exchange genetic material, further increasing the variation of the daughter cells.
Haploid daughter cells: The two haploid daughter cells, containing one copy of each chromosome, are the result of meiosis. These cells are used for sexual reproduction and contain a unique combination of genetic information.
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The extent to which the body can respond or adapt to the demands and stress of physical elfort denotes physical fitness. physical activity.exercise. antaerobic capacity.
Physical fitness refers to the ability of the body to respond and adapt to the demands and stress of physical effort. This includes the ability to perform physical activity and exercise with ease and efficiency, as well as to recover quickly from physical exertion.
Physical fitness encompasses a range of factors, including aerobic and anaerobic capacity, muscular strength and endurance, flexibility, and body composition.
Aerobic capacity refers to the body's ability to use oxygen to produce energy during exercise, while anaerobic capacity refers to the body's ability to produce energy without oxygen. Both are important components of physical fitness, and can be improved through regular exercise and training.
Overall, physical fitness is essential for maintaining good health and reducing the risk of chronic diseases. Regular physical activity and exercise can help to improve cardiovascular health, increase strength and endurance, and enhance overall quality of life.
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Question 14
The majority of the world's water is used in:
a. Households
b. Industry
c. Agriculture
d. Sewage disposal
The majority of the world's water is used in: Agriculture
Option C is correct
The majority of the world's water is used for agricultural purposes, such as irrigation and livestock production. In some regions, agriculture can account for up to 80% of water use.
Industrial and household water use, while significant, typically account for a smaller portion of overall water use. Sewage disposal, while an important aspect of water management, does not involve the consumption of large amounts of water.
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The Minnesota Twin Study conducted its research only on identical twins, not fraternal twins. T F
Answer:
False
Explanation:
MISTRA researched identical and fraternal twins to determine how genetics and environment affect psychological development. Researchers compared identical and fraternal twins to determine how genetic and environmental variables affect physical and mental health.
By looking at both sets of twins, Mistra tried to figure out how much genetics and how much the world affected the way people's minds developed. Researchers were able to learn more about how nature and the environment affect a person's overall health and happiness by using this method.
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Which system do you find in a human, but not in a flatworm?
The circulatory system is found in humans but not in flatworms.
The circulatory system is in charge of transporting blood and nutrients throughout the body. This system in humans is made up of the heart, blood vessels (arteries, veins, and capillaries), and blood.
However, because flatworms lack a circulatory system, nutrients and gases are exchanged directly between cells and the environment via diffusion over the body surface. Humans have a circulatory system, but flatworms do not.
Flatworms have a rudimentary excretory system, neurological system, and digestive system, but they lack many of the specialised organs and systems seen in more sophisticated organisms such as humans.
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How does a test for protein work?(Biuret's Assay)
The absorbance of the solution is measured using a spectrophotometer at a wavelength of 540 nm. The intensity of the color is proportional to the amount of protein present in the sample.
The Biuret assay is a commonly used test to detect the presence of proteins in a sample. The assay is based on the ability of copper ions to form a complex with the peptide bonds in proteins, resulting in a color change that can be detected using a spectrophotometer.
Here's how the Biuret assay works: First, a small amount of the sample containing the protein of interest is added to a test tube or cuvette. Next, a solution of copper sulfate (CuSO4) is added to the sample. The copper ions in the solution react with the peptide bonds in the protein, forming a complex with a characteristic blue-violet color. Then, a solution of sodium hydroxide (NaOH) is added to the mixture. The NaOH solution increases the pH of the solution, which helps to stabilize the Cu-protein complex and intensify the color.
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A B-DNA molecule has 1 million nucleotide pairs. How many complete turns are there in this molecule?
The number of complete turns in a B-DNA molecule with 1 million nucleotide pairs is 10.
B-DNA is a right-handed double helix structure, meaning that it twists in a clockwise direction as it extends along its axis. In B-DNA, there are approximately 10 base pairs per turn of the helix.
Since the molecule in question has 1 million nucleotide pairs, we can calculate the number of turns as follows:
Number of turns = (number of base pairs) / (number of base pairs per turn)
Number of turns = 1,000,000 / 10
Number of turns = 100,000
Therefore, the B-DNA molecule with 1 million nucleotide pairs has approximately 100,000 complete turns.
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