We will amplify the LDHA cDNA (LDH-A protein coding sequence) from a cDNA library. Such a library is a large collection of DNA molecules representing the protein coding (mRNA) sequences of (theoretically) all genes significantly expressed in an organism/tissue of choice. How is a cDNA library made

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Answer 1

A cDNA library is made by isolating mRNA molecules from a tissue or organism of interest and converting them into complementary DNA (cDNA) using an enzyme called reverse transcriptase.

The resulting cDNA represents the protein coding sequences of the genes expressed in the tissue or organism. The cDNA molecules are then cloned into a vector, which is a carrier molecule that allows for the replication and expression of the cDNA. The vector is introduced into bacterial or yeast cells, which are grown in culture to generate a large number of copies of the cDNA library. The resulting library contains millions of cDNA molecules, each representing a different mRNA transcript. The cDNA library can then be screened for specific genes of interest, such as the LDHA gene, by using a probe that hybridizes to the cDNA sequence.

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Related Questions

Explain independent assortment and allele segregation. How many gamete possibilities are produced from a parents genotype with 2 traits

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Independent assortment and allele segregation occur during meiosis, leading to genetic variation. For a parent's genotype with 2 traits, 4 gamete possibilities are produced.

Independent assortment and allele segregation are essential processes during meiosis that contribute to genetic variation.

Independent assortment refers to the random distribution of chromosomes to daughter cells during meiosis, ensuring that each gamete receives a unique set of genetic information.

Allele segregation is the separation of alleles for each trait, allowing them to independently combine with other alleles from another parent during fertilization.

For a parent's genotype with 2 traits, there are 4 gamete possibilities (2 possibilities for each trait), generated through the combination of the different alleles from each trait.

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According to the textbook, Cuvier addressed the comparative anatomy of fossils in a theory of life changes called

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According to the textbook, Cuvier addressed the comparative anatomy of fossils in a theory of life changes called Catastrophism.

In this theory, Cuvier proposed that sudden and catastrophic events, such as natural disasters, caused the extinction of species and the appearance of new ones. He used the comparative anatomy of fossils to support his claims, showing significant differences between extinct species and those living today.

Cuvier's approach to comparative anatomy was to carefully analyze the anatomical structures of fossils and compare them to living organisms to determine which species had become extinct due to these catastrophic events. He believed that each catastrophic event had caused the extinction of multiple species, which were then replaced by new species that migrated into the area.

Therefore,  This approach was in contrast to the idea of evolution, which proposes that species gradually change over time due to natural selection and other factors.

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4. A human has 1011 cells and each human cell has about 6 x 109 nucleotide pairs of DNA. What is the length of the double helix that could be formed from this amount of DNA in a human individual

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The amount of DNA in a human individual is truly staggering - with 1011 cells, each containing about 6 x 109 nucleotide pairs, the total amount of DNA in a human individual adds up to over 6 x 1020 nucleotide pairs.

To calculate the length of the double helix that could be formed from this amount of DNA, we need to consider the structure of DNA itself. DNA is a double-stranded molecule made up of nucleotides, which consist of a sugar molecule, a phosphate group, and a nitrogenous base (adenine, guanine, cytosine, or thymine). The two strands of DNA are held together by hydrogen bonds between complementary nitrogenous bases (adenine pairs with thymine, and guanine pairs with cytosine).  The length of a DNA molecule is usually measured in base pairs (bp), which refers to the number of nucleotide pairs in the molecule. It is estimated that each turn of the DNA helix contains about 10 base pairs, and the distance between each base pair is about 0.34 nanometers.

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In the desert, saguaro cacti, owls, horned lizards, and fire ants all share the same space. Which example can be considered a population

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The population in this scenario would be the saguaro cacti, owls, horned lizards, or fire ants.

A population refers to a group of organisms of the same species living in a particular geographic area. In this scenario, there are different species occupying the same space, but each of them has their own population. The saguaro cacti population includes all the individual saguaro cacti in the area, while the owl population includes all the individual owls, and so on. Therefore, any of the mentioned species can be considered as a population.

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Flor is studying the evolutionary history of house geckos, Pacific gulls, and wolves. All three species share some body structures.
but they also have some differences in their body structures. Below is a table that includes information about the body structures
that Flor is studying

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I'm sorry, but I do not see any table or information provided for me to reference. Please provide the necessary information.

explain why dideoxynucleotide incorporation during extension by dna polymerase terminates elongation

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Dideoxynucleotides (ddNTPs) lack a hydroxyl group (-OH) at the 3' position of the sugar molecule, which is essential for the formation of a phosphodiester bond between adjacent nucleotides in a growing DNA chain.

When a ddNTP is incorporated into the growing DNA chain, the DNA polymerase cannot add another nucleotide to the 3' end of the ddNTP. This results in the termination of DNA synthesis since no further elongation can occur. In contrast, normal deoxynucleotides (dNTPs) have a hydroxyl group (-OH) at the 3' position of the sugar molecule, which can form a phosphodiester bond with the next incoming nucleotide, allowing DNA polymerase to continue elongation.

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In the Avery Experiment, cell extracts treated with __________ were unable to cause a transformation in bacteria. Group of answer choices lipase RNase protease DNase amyla

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DNase treatment of cell extracts failed to transform bacteria in the Avery Experiment.

Using enzymes that break down specific polymers, Avery, MacLeod, and McCarty set out to identify the chemical that was to blame: DNases to digest DNA, RNases to digest RNA, and proteases to digest proteins.

They also found that lipid-digesting enzymes (lipase), protein-digesting enzymes (Peptidase), and RNA-digesting enzymes (RNases) had no effect on transformation.

Oswald Avery's team demonstrated that DNA was the "transforming principle" through a straightforward experiment. DNA was able to change and give characteristics to another strain of bacteria after being isolated from one strain. Hereditary information was in DNA.

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Mendel's focus on ______ genes was pivotal in establishing the science of genetics because it allowed Mendel to formulate basic laws of inheritance.

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Mendel's focus on pea genes was pivotal in establishing the science of genetics because it allowed Mendel to formulate basic laws of inheritance.

Mendel's focus on pea genes was pivotal in establishing the science of genetics because it allowed Mendel to formulate basic laws of inheritance. Gregor Mendel was an Austrian monk who conducted experiments on pea plants in the mid-19th century, and through his work, he discovered the fundamental principles of genetics.

Mendel chose to work with pea plants because they are easy to grow, have a short generation time, and produce large numbers of offspring. He studied seven traits of the pea plant, including seed color, seed shape, and flower color, and observed how these traits were passed down from generation to generation.

Through his experiments, Mendel formulated the laws of segregation and independent assortment, which describe how traits are inherited from parents to offspring. These laws form the basis of modern genetics and have implications for fields such as medicine, agriculture, and evolutionary biology.

By focusing on pea genes, Mendel was able to establish the science of genetics and lay the foundation for future research in the field.

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In 1997, scientists discovered a new species of primate, only 10 cm long, with widely spaced nostrils that open outward and a long prehensile tail. This animal would be a(n) _____. View Available Hint(s)for Part A Old World monkey ape New World monkey tarsier

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A tarsier is a species of primate discovered in 1997 that is only 10 cm long.

This animal has distinctive characteristics that set it apart from other primates, such as its widely spaced nostrils that open outward and its long prehensile tail. Its nostrils help it to locate prey more accurately and its tail allows it to hang from branches, trees, or other surfaces.

The tarsier also has huge eyes, allowing it to see in very low light conditions. Unlike other primates, the tarsier is primarily carnivorous and its diet includes insects, frogs, lizards, and small birds.

It is found in the forests of Southeast Asia, primarily in the Philippines and Indonesia. The tarsier is an endangered species and its main threats are habitat destruction and hunting. Conservation efforts are in place to protect this unique species.

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Researchers have now found that some adult cells can be used as stem cells. However, adult stem cells are not as successful as embryonic stem cells unless they are modified because __________.

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Adult stem cells have more limited differentiation potential compared to embryonic stem cells. Embryonic stem cells are pluripotent, which means they have the potential to differentiate into any cell type in the body. Adult stem cells, on the other hand, are multipotent, meaning they can only differentiate into a limited number of cell types.

To make adult stem cells more versatile, they can be modified through a process called reprogramming. Reprogramming involves inducing the expression of genes that are normally active in embryonic stem cells, which can then allow the adult stem cells to differentiate into a wider range of cell types.Therefore, the statement that adult stem cells are not as successful as embryonic stem cells unless they are modified is true because of the difference in their differentiation potential.

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Soil will pack down over time which decreases the ability of the soil to hold water. Which animals would best help the soil hold water

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The animals that would best help the soil hold water by preventing soil compaction and increasing water-holding capacity are earthworms and other burrowing creatures.

Earthworms, for example, create tunnels in the soil as they move through it, which helps improve aeration, water infiltration, and overall soil structure. This prevents the soil from packing down over time and increases its ability to hold water.  Additionally, their waste products can help to enrich the soil and make it more fertile.

Similarly, other burrowing animals like moles, voles, and certain insects can also contribute to maintaining good soil structure.

In summary, animals like earthworms and other burrowing creatures help the soil hold water by preventing it from packing down over time and improving its overall structure.

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1. gene in pea plants has a strong influence on plant height. The gene has two alleles: tall (T), which is dominant, and short (t), which is recessive. What are the genotypes and phenotypes of the offspring of a cross between a TT and a tt plant?

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The genotype and phenotype of the offspring of a cross between TT and tt plants would be Tt and tall respectively.

Monohybrid crossing

Allele T, which codes for tallness, is dominant over allele t, which codes for shortness.

Crossing a TT plant with a tt plant:

               TT   x   tt

             Tt  Tt  Tt  Tt

All the offspring have Tt genotype and since T is dominant over t, they will all appear tall.

In other words:

Genotype of offspring = Tt

Phenotype of offspring = tall.

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Looking at the data in this table, this bacterial population is likely in what growth phase between 20 and 24 hours? Time (hours) Live Cell Count (millions) 4 12 8 5 23 12 116 20 72 138 139 24 nces Multiple Choice Exponential phase w a Lag phase O Death phose Stationary phase

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Looking at the data in the table, this bacterial population is likely in the stationary phase between 20 and 24 hours.

This is because the live cell count remains relatively constant between 20 and 24 hours, indicating that the growth rate is equal to the death rate. This is characteristic of the stationary phase, which is the phase where the growth rate slows down and the population reaches a maximum sustainable density. During the stationary phase, the bacterial growth rate slows down and the number of viable cells reaches a plateau due to a balance between cell division and cell death. In other words, the number of new cells being produced is roughly equal to the number of cells dying. Therefore, the bacterial population is likely in the stationary phase between 20 and 24 hours.

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Proliferating centroblasts use the DNA-modifying enzyme activation-induced cytidine deaminase for Select one: a. cell proliferation. b. differentiating into plasma cells. c. apoptosis. d. upregulation of CD40. e. isotype switching.

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Proliferating centroblasts use the DNA-modifying enzyme activation-induced cytidine deaminase for isotype switching.

Activation-induced cytidine deaminase (AID) is an enzyme that initiates somatic hypermutation and class switch recombination of immunoglobulin genes during the process of B-cell activation and maturation. During the germinal center reaction, B-cells differentiate into centroblasts, which undergo rapid proliferation and actively mutate their immunoglobulin genes with the help of AID. This process is essential for generating high-affinity antibodies and producing a diverse repertoire of antibodies that can recognize a wide range of antigens. AID introduces point mutations into the variable regions of immunoglobulin genes, allowing for affinity maturation of the antibody response, and promotes class switch recombination, which enables B-cells to switch from producing IgM to other antibody isotypes, such as IgG, IgA, or IgE.

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In a certain African country 6.00% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population of 1,000 newborn babies, how many offspring would you expect are NOT carrying the allele

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We would expect approximately 570 newborn babies out of a random population of 1,000 to be homozygous dominant (not carrying the allele) for sickle-cell anemia.

If 6.00% of the newborn babies in a certain African country have sickle-cell anemia, which is a recessive trait, then the frequency of the recessive allele causing sickle-cell anemia (q) can be calculated using the following formula:

q = square root of (0.06)

q = 0.245

Since sickle-cell anemia is a recessive trait, the frequency of the dominant allele (p) can be calculated by subtracting the frequency of the recessive allele (q) from 1:

p = 1 - q

p = 0.755

Using the Hardy-Weinberg equation, we can calculate the expected proportion of carriers in the population:

2pq + [tex]p^{2}[/tex]+ [tex]q^{2}[/tex]= 1

where:

[tex]p^{2}[/tex] is the frequency of homozygous dominant individuals (not carrying the allele)

[tex]q^{2}[/tex] is the frequency of homozygous recessive individuals (with sickle-cell anemia)

2pq is the frequency of heterozygous carriers

We know that q = 0.245, so:

2pq = 2(0.755)(0.245) = 0.369

[tex]p^{2}[/tex]= (0.755)² = 0.570

[tex]q^{2}[/tex] = (0.245)² = 0.060

Therefore, the expected proportion of individuals who are not carrying the allele (homozygous dominant) is [tex]p^{2}[/tex] = 0.570.

To find the number of individuals out of 1,000 newborn babies that are expected to be homozygous dominant (not carrying the allele), we can multiply the proportion by the total population:

0.570 x 1,000 = 570

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"Consider an E. coli gene whose promoter contains -10 and -35 sequences and is transcribed by the housekeeping RNA polymerase, i.e. the core polymerase associated with the sigma 70 subunit. Given the DNA sequence of this promoter, how would you predict whether it is a strong or a weak promoter? Explain your reasoning."

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When considering whether a promoter is strong or weak, we need to look at the consensus sequences of the -10 and -35 regions. A strong promoter will have sequences that closely match the consensus sequences, while a weak promoter will have deviations from the consensus.



The consensus sequence for the -10 region in E. coli is "TATAAT", and the consensus sequence for the -35 region is "TTGACA". If the promoter sequence of the gene in question closely matches these consensus sequences, then it is likely a strong promoter. If there are deviations from the consensus, it may be a weaker promoter.


Therefore, to predict whether the promoter is strong or weak, we need to compare the promoter sequence with the consensus sequences for the -10 and -35 regions. If the promoter sequence closely matches the consensus, it is likely a strong promoter. If there are deviations from the consensus, it may be a weaker promoter.

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Coal is derived from living organisms, and some of their amino acids contained sulfur which persists in the coal. A possible outcome of this is:

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A possible outcome of sulfur persisting in coal due to amino acids in living organisms is that when coal is burned, sulfur dioxide is released into the atmosphere.

This can lead to acid rain, which can harm plants, animals, and infrastructure. Additionally, sulfur dioxide is a contributor to air pollution and can have negative effects on human health. The presence of sulfur in coal is due to the fact that the original organisms that formed the coal had sulfur-containing amino acids in their bodies. As these organisms died and were buried, heat and pressure over millions of years transformed them into the coal we use today. However, the sulfur from their amino acids remained in the coal.

When coal is burned, the sulfur in the coal reacts with oxygen to form sulfur dioxide gas. This gas can then react with water vapor in the atmosphere to form sulfuric acid, which falls to the earth as acid rain. Acid rain can lower the pH of soil, making it difficult for plants to grow, and can also harm aquatic ecosystems.

The release of sulfur dioxide from burning coal is a major contributor to air pollution, particularly in areas where coal-fired power plants are common. Sulfur dioxide can react with other pollutants in the air to form small particles, which can be harmful to human health when inhaled. These particles can exacerbate respiratory problems such as asthma and can even contribute to heart disease.

To address these negative outcomes, regulations have been put in place in many countries to limit the amount of sulfur dioxide that can be emitted from coal-fired power plants. Technologies such as scrubbers can also be used to remove sulfur dioxide from the emissions before they are released into the atmosphere.

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if, in a seperate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount

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In an experiment where one-third as much enzyme and twice as much substrate had been combined, it would take a longer amount of time for the same amount of product to be formed compared to the original experiment.

This is because enzymes are biological catalysts that speed up chemical reactions by lowering the activation energy required for the reaction to occur. In this case, having less enzyme would result in a slower reaction rate. The rate of an enzymatic reaction depends on several factors, including the concentration of the enzyme and substrate, the temperature, and the pH. In this particular case, the decreased enzyme concentration and increased substrate concentration would result in a lower reaction rate.

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_______ are lipoproteins made of dietary fat surrounded by a shell of cholesterol, phospholipids, and protein; their role is to transport fat that has beenabsorbed from the GI tract. Group of answer choices

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Chylomicrons are lipoproteins made of dietary fat surrounded by a shell of cholesterol, phospholipids, and protein; their role is to transport fat that has been absorbed from the GI tract.

Chylomicrons are the lipoproteins responsible for transporting dietary fat.  Chylomicrons are formed in the small intestine after absorption of dietary fat and are transported via the lymphatic system to the bloodstream where they can deliver the fat to cells throughout the body. Chylomicrons are responsible for transporting fats that have been absorbed from the gastrointestinal (GI) tract to various tissues in the body, such as adipose tissue, muscle, and liver, where they are either stored or used for energy production.

To sum up, chylomicrons are the lipoproteins that play a crucial role in transporting dietary fats absorbed from the GI tract throughout the body.

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The active form of vitamin D is produced when metabolized in two different organs; the enzyme 25-hydroxylase adds a hydroxyl group in the

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The active form of vitamin D, also known as calcitriol, is produced when metabolized in two different organs: the liver and the kidneys. In the liver, the enzyme 25-hydroxylase adds a hydroxyl group to vitamin D, forming 25-hydroxyvitamin D (25(OH)D), which is then transported to the kidneys. In the kidneys, the enzyme 1-alpha-hydroxylase further modifies 25(OH)D by adding another hydroxyl group, producing the active form of vitamin D, calcitriol. Calcitriol plays an important role in regulating calcium and phosphorus levels in the body, as well as promoting bone health and immune system function.

First, the enzyme 25-hydroxylase adds a hydroxyl group in the liver, converting vitamin D to 25-hydroxyvitamin D. Then, in the kidneys, another enzyme called 1-alpha-hydroxylase further hydroxylates the molecule to form the active calcitriol. This active form plays a crucial role in calcium regulation and bone health.

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Trace the circulation of aqueous humor from the site of production to the site of where it is reabsorbed. 1. posterior chamber 2. anterior chamber 3. ciliary body 4. scleral venous sinu

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The aqueous humor is a clear fluid that is produced by the ciliary body in the eye. Here is the circulation of aqueous humor from the site of production to the site of reabsorption:

Aqueous humor is produced by the ciliary body, which is a structure located behind the iris.From the ciliary body, the aqueous humor flows into the posterior chamber of the eye, which is located between the iris and the lens.The aqueous humor then passes through the pupil into the anterior chamber of the eye, which is located between the cornea and the iris.Once in the anterior chamber, the aqueous humor circulates around the front of the eye, providing nutrients and oxygen to the cornea and lens.The aqueous humor is reabsorbed into the bloodstream via the scleral venous sinus (also known as the canal of Schlemm), which is located at the junction between the sclera and the cornea.From the scleral venous sinus, the aqueous humor drains into the bloodstream, and any excess fluid is carried away by the lymphatic system.

In summary, the aqueous humor is produced in the ciliary body, circulates through the posterior and anterior chambers of the eye, and is reabsorbed into the bloodstream via the scleral venous sinus.

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A person could not see a single thread from 100 feet away, but if you wound thousands of threads together in a rope, you could see it. Explain how this statement relates to our DNA extraction.

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DNA extraction from a large number of cells and amplifying it through polymerase chain reaction (PCR), we can generate a large amount of DNA that can be seen

DNA extraction is a process used to isolate DNA from cells, tissues, or organisms for further study or manipulation. This technique is a fundamental step in many areas of biological research, including genetics, genomics, and forensic science. The extraction process typically involves breaking open the cell membrane and nuclear envelope to release the DNA, and then separating the DNA from other cellular components, such as proteins, lipids, and RNA.

The specific extraction protocol used depends on the source of the DNA, as well as the downstream applications for which the DNA will be used. Common methods for DNA extraction include using chemical reagents to lyse cells and separate the DNA, or using mechanical disruption methods, such as sonication or grinding, to break open cells. Once the DNA is extracted, it can be purified and quantified and then used for downstream applications, such as PCR, sequencing, and genetic engineering.

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What can we observe in order to Drag labels of Group 1 to indicate the genotypes of the parents and offspring. Drag labels of Group 2 to indicate the genetic makeup of the gametes (sperm and egg). Labels can be used once, more than once, or not at all.Mendel's Law of Segregation

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In order to understand the genotypes of parents and offspring as well as the genetic makeup of the gametes, we can apply Mendel's Law of Segregation.

Mendel's Law of Segregation states that during the formation of gametes (sperm and egg), each parent's two alleles (versions of a gene) for a particular trait separate, and each gamete receives only one allele. This means that offspring inherit one allele from each parent, resulting in their own unique genotype.

For example, let's consider a trait with two possible alleles, A and a. If the genotypes of the two parents are Aa and Aa, we can predict the possible genotypes of their offspring using a Punnett square.

Step 1: Identify the alleles in the parents' genotypes (Aa and Aa).

Step 2: List the possible gametes that can be formed by each parent: Parent 1 - (A, a) and Parent 2 - (A, a).

Step 3: Create a Punnett square, and fill in the boxes by combining the gametes from both parents:

|---|---A-|---a-|
| A | AA | Aa |
| a | Aa | aa |

Step 4: Analyze the Punnett square to determine the offspring's genotypes. In this case, we can observe:
- 1 offspring with genotype AA
- 2 offspring with genotype Aa
- 1 offspring with genotype aa

To summarize, we can use Mendel's Law of Segregation to predict the genotypes of offspring by observing the genotypes of the parents and the possible genetic makeup of the gametes (sperm and egg). In this example, the genotypes of the parents were Aa and Aa, and the possible genetic makeup of their gametes was A or a. The offspring's genotypes were determined as AA, Aa, and aa.

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_______________ salts help in the digestion of fats, cholesterol and fat-soluble vitaminsbyformingmixed micelles (clumps of _____________ fat)which allowthem to be accessible to water and digestive enzymes.

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By aggregating long-chain fats into mixed micelles, which are then accessible to water and digestive enzymes, bile salts aid in the digestion of fats, cholesterol, and fat-soluble vitamins. Bile once more facilitates this procedure.

In order for fats to be absorbed, they must be close enough to the microvilli of intestinal cells to be surrounded by micelles, which are structures made of bile salts that form a cluster around the byproducts of fat digestion.In the gastrointestinal tract (GIT), bile salts (BS) are bio-surfactants that are essential for nutritional digestion and absorption.

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The large diversity of shapes of biological molecules is possible because of the extensive presence of _____ in the molecules. See Concept 4.2 (Page)

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Due to the abundance of carbon in molecules, it is possible for them to have a wide range of shapes.

Carbon is responsible for the variety of biological molecules that have allowed for a wide range of living things to exist. Carbon atoms bonded to each other and to atoms of other elements make up proteins, DNA, carbohydrates, and other molecules that separate living things from inorganic materials.

Isomers. The three-layered arrangement of particles and synthetic bonds inside natural atoms is fundamental to figuring out their science. Isomers are molecules that have the same chemical formula but have different atom placement (structure) or chemical bonds.

Four highly reactive oxygen atoms make up the phosphate group. The transfer of the phosphate group from one molecule to another provides energy for a chemical reaction. Three phosphate groups make up ATP, the cell's primary energy carrier.

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Bacteria, pollen, and transplanted tissue are examples of Group of answer choices foreign antibodies. chemotactic chemicals. foreign antigens. self antigens. pyrogens.

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Bacteria, pollen, and transplanted tissue are examples of foreign antigens. Foreign antigens are substances that are recognized as non-self by the immune system, triggering an immune response. When foreign antigens, such as bacteria or pollen, enter the body, the immune system produces specific proteins called antibodies to neutralize or eliminate them.

Pyrogens, on the other hand, are substances that can induce a fever. They can be produced by microorganisms like bacteria or by the host's immune system in response to an infection or inflammation. Pyrogens are not classified as foreign antigens, but they can be related to the immune response against them.

In contrast, self-antigens are molecules found on the surface of the body's own cells. They are recognized as self by the immune system and do not trigger an immune response. Foreign antibodies are antibodies produced in response to foreign antigens, whereas chemotactic chemicals are substances that attract immune cells to a specific site, like the site of an infection.

In summary, bacteria, pollen, and transplanted tissue are examples of foreign antigens that can trigger an immune response, while pyrogens are fever-inducing substances related to the immune response but not classified as antigens themselves.

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Serological methods Question 57 options: A) are useful in identifying unknown bacteria. B) rely on the specificity of an antibody-antigen interaction. C) may be simple and rapid. D) use cellular proteins and carbohydrates as markers.

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Serological methods rely on the specificity of an antibody-antigen interaction and are useful in identifying unknown bacteria.

These methods may be simple and rapid, but they do not use cellular proteins and carbohydrates as markers. Instead, they use specific antibodies that bind to the antigens on the surface of the bacteria, allowing for their identification.

Serological methods are useful in identifying unknown bacteria (A), rely on the specificity of an antibody-antigen interaction (B), may be simple and rapid (C), and use cellular proteins and carbohydrates as markers (D).

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A negative feedback system is used until the blood ____________ and osmolarity return to normal levels.

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A negative feedback system is used until the blood volume and osmolarity return to normal levels.

A negative feedback system is a biological mechanism that helps to maintain homeostasis by counteracting any deviation from the normal set point.

In the context of blood regulation, a negative feedback system is utilized to regulate blood volume and osmolarity.

When there is an increase in blood volume and osmolarity due to dehydration or excess salt intake, the body responds by activating the hypothalamic-pituitary-adrenal axis (HPA axis) to release antidiuretic hormone (ADH) from the posterior pituitary gland.

ADH acts on the kidney tubules to reabsorb water, thus decreasing urine output and increasing blood volume.

Once the blood volume and osmolarity return to normal levels, the negative feedback system is turned off, and ADH secretion decreases, leading to increased urine output and a return to normal blood volume and osmolarity.

Therefore, the answer to the question is that a negative feedback system is used until the blood volume and osmolarity return to normal levels.

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Suppose a mutation in separase allows it to recognize but not cleave cohesin. What is the most likely outcome for cells expressing this mutant form of separase

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If separase is unable to cleave cohesin, the most likely outcome is that the sister chromatids will remain attached to each other even after the onset of anaphase. This will prevent the proper separation of chromosomes and ultimately result in aneuploidy, a condition where cells have an abnormal number of chromosomes.

Aneuploidy can lead to various genetic disorders, such as Down syndrome, and can also cause cancer. The failure of sister chromatid separation during cell division is a critical step in the development of cancer, as it can lead to the accumulation of genetic abnormalities that promote tumor formation. Therefore, a mutation in separase that disrupts its ability to cleave cohesin can have serious consequences for the normal functioning of cells.

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A student examines a section of skin under the microscope. She observes a region composed of dense irregular connective tissue. Which portion of the skin is she observing

Answers

The deep layer is the reticular layer, which forms a thick layer of dense connective tissue that forms the bulk of the dermis.

What is a tissue?

Tissue is a collection of cells with similar structures that work together as a unit. The intercellular matrix is a nonliving substance that fills the spaces between the cells. This may be plentiful in some tissues while being scarce in others.

Tissue is classified into four types:

connective tissue, epithelial tissue, muscular tissue, and nerve tissue.

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