In pea plants, the allele for tall plants (7) is dominant over the allele for short
plants (t). The allele for purple flowers (P) is dominant over the allele for
white flowers (p). Two plants that are heterozygous for both traits are
crossed, as shown in the Punnett square.
TP
OA.
O B.
O C.
O D. 16
Tp
tP
tp
TP
TTPP
TTPP
TIPP
TtPp
Tp
TTPP
TIPP
TtPp
Tipp
tP
TtPP
TtPp
ttPP
ttPp
tp
TtPp
Ttpp
ttPp
ttpp
What is the probability of an offspring being short and having white flowers?

Species I has \( 2 n=8 \) chromosomes and species II has \( 2 n=14 \) chromosomes. The expected chromosome numbers in individual organisms with the following chromosome mutations for a. Allotriploidy including species I and II is \( 3n = 2(8) + 14 = 30 \) chromosomes. b. Autotetraploidy in species II is \( 4n = 4(14) = 56 \) chromosomes, and c. Allotetraploidy including species I and II is \( 4n = 2(8) + 2(14) = 44 \) chromosomes.

Allotriploidy is a type of polyploidy that occurs when an organism has three sets of chromosomes from two different species. In this case including species I and II, the expected chromosome number would be \( 3n = 2n_1 + n_2 \), where \( n_1 \) is the haploid number of species I and \( n_2 \) is the haploid number of species II. Therefore, the expected chromosome number in an allotriploid individual including species I and II would be \( 3n = 2(8) + 14 = 30 \) chromosomes.

Autotetraploidy is a type of polyploidy that occurs when an organism has four sets of chromosomes from the same species. In this case in species II, the expected chromosome number would be \( 4n = 4n_2 \), where \( n_2 \) is the haploid number of species II. Therefore, the expected chromosome number in an autotetraploid individual of species II would be \( 4n = 4(14) = 56 \) chromosomes.

Allotetraploidy is a type of polyploidy that occurs when an organism has four sets of chromosomes from two different species. In this case in species I and II, the expected chromosome number would be \( 4n = 2n_1 + 2n_2 \), where \( n_1 \) is the haploid number of species I and \( n_2 \) is the haploid number of species II. Therefore, the expected chromosome number in an allotetraploid individual including species I and II would be \( 4n = 2(8) + 2(14) = 44 \) chromosomes.

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The term you are referring to is "food." Food is a substance that provides the nutrients and energy necessary for the body to function properly.

It is an essential component of human life, as without it, the body would not be able to sustain itself.One of life's fundamental needs is food. Nutrients are compounds that are necessary for the regulation of vital activities as well as the growth, repair, and maintenance of body tissues. The energy our bodies require to function is provided by nutrients. Calories are the units used to quantify the energy in food. A well-balanced diet gives you all the energy you need to stay active all day long. nutrients you require for growth and repair, assisting in keeping you strong and healthy and assisting in the prevention of dietary-related diseases including certain cancers.

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ADDITIVE GENE ACTION refers to the situation where each gene has a small additive effect on the phenotype.

In LARGE WHITE PIG breeds, this means that the more desirable traits are expressed when there are more dominant alleles present. COMPLEMENTARY GENE ACTION, on the other hand, refers to the situation where multiple genes interact to produce a phenotype, with each gene contributing a specific part of the final phenotype.

In LARGE WHITE PIG breeds, this means that certain combinations of genes produce desirable traits, and the presence of both genes is required for those traits to be expressed.

Pleiotropy is the phenomenon where a single gene can have multiple effects on the phenotype. In LARGE WHITE PIG breeds, some genes may exhibit pleiotropy, such as a gene that affects both growth rate and meat quality.

However, the extent to which pleiotropy occurs in these breeds is not well understood and may require further research to identify specific genes and their effects.

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Answer:

it decreases water and air quality

Explanation:

Total 80%  of individuals in this population have the dominant phenotype. (D)

In a given nonevolving population, 20% of the alleles for a given gene are recessive (s). This means that 80% of the alleles are dominant (S).

The percentage of individuals with the dominant phenotype can be determined using the Hardy-Weinberg equation: p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele, q is the frequency of the recessive allele, and 2pq is the frequency of the heterozygous genotype.

Since p = 0.8 and q = 0.2, we can plug these values into the equation to find the percentage of individuals with the dominant phenotype:

p^2 + 2pq + q^2 = 1
(0.8)^2 + 2(0.8)(0.2) + (0.2)^2 = 1
0.64 + 0.32 + 0.04 = 1

The term p^2 represents the frequency of the homozygous dominant genotype (SS), and the term 2pq represents the frequency of the heterozygous genotype (Ss). Both of these genotypes result in the dominant phenotype, so we can add them together to find the percentage of individuals with the dominant phenotype:

0.64 + 0.32 = 0.96

Therefore, 96% of the individuals in this population have the dominant phenotype. However, the question asks for the percentage of individuals with the dominant phenotype, not the frequency of the dominant phenotype. To find the percentage, we can multiply the frequency by 100:

0.96 * 100 = 96%

So the correct answer is d) 80%.

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The five factors influencing the nutritional value of plant feed resources are:

We proceed to analyze the factors that directly influence the nutritional value of plant food resources:

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You are a scientist trying to solve global warming. Since plants use CO2 to make carbon compounds, you want to grow plants/crops to sequester carbon from the atmosphere to mitigate global warming. One of the following would be most likely to keep those carbon atoms from re-entering the atmosphere in the long term is D, converting the plant material into a complex carbon compound that cannot be decomposed for respiration or fermentation.

To keep those carbon atoms from re-entering the atmosphere in the long term because it prevents the carbon from being released back into the atmosphere through decomposition or combustion. Options A and B, feeding the plants to college students and animals/livestock, would eventually result in the release of carbon back into the atmosphere through respiration. Option C, using the plants as biofuels for combustion engines, would also result in the release of carbon back into the atmosphere through combustion.

Therefore, option D is the best choice for keeping those carbon atoms from re-entering the atmosphere in the long term.

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Anemia in preterm infants is more prevalent, severe, and protracted than in term infants due to the short life span of their red blood cells (RBCs), and the immaturity of their liver and gastrointestinal tracts.

Preterm newborns have more severe anaemia than term infants due to a variety of causes, including the immaturity of their organs and systems. Preterm newborns have a lower red blood cell (RBC) life span than term infants because they are born with fewer RBCs than term infants and their bone marrow is not completely matured, affecting their ability to create new RBCs.

Anemia in preterm infants can lead to a lower oxygen-carrying capacity of the blood, which can cause various complications such as fatigue, weakness, and difficulty breathing. It is important for preterm infants to receive appropriate medical care and monitoring to prevent or treat anemia and any related complications.

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The species that has the greater geometric mean growth rate is species B.

Given that species A responds to temporal variation as if years were of two types. The annual reproductive rate λA takes two values, with differing probabilities:

Pr[λA = 2/3] = 1/3; Pr[λA = 6] = 2/3.

Species B responds to the same environment as if years were of three types. That is:

Pr[λB = 1] = 1/6; Pr[λB = 4] = 3/6; Pr[λB = 8] = 1/3.

We need to find the species that has the greater geometric mean growth rate. The formula to calculate the geometric mean growth rate of the species is given by;

g = {λ1λ2λ3...λn}1/n

Where g is the geometric mean growth rate of species A and λ1, λ2,... λn are the reproductive rates over n years.

Now, let us calculate the geometric mean growth rate for species A, λA = 2/3 and 6 are the two reproductive rates of species A over two years.

λ1λ2 = 2/3 x 6 = 4 and gA = {4}1/2 = 2

Next, let us calculate the geometric mean growth rate for species B. λB = 1, 4, and 8 are the three reproductive rates of species B over three years.

λ1λ2λ3 = 1 x 4 x 8 = 32 and gB = {32}1/3 = 3.03

Thus, species B has the greater geometric mean growth rate.

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The patterns of butterfly wings form through a combination of genetic instructions and environmental factors. These patterns exist to serve a variety of purposes for the butterflies, including camouflage, warning signals to predators, and sexual attraction. The process of wing pattern formation is complex, involving several different genes and signaling pathways. There are two basic types of butterfly wing patterns: eyespots and stripes/bands.

Eyespots are circular patterns with concentric rings, while stripes and bands can be either vertical or horizontal. The coloration and placement of these patterns can vary widely between different species and individuals within a species. Some of the reasons why these patterns exist and what they do for the butterflies are mentioned below:
Camouflage: Many butterflies use their wing patterns to blend in with their surroundings, making it harder for predators to find them. For example, some species have patterns that resemble the leaves of their host plants or the bark of trees. Warning signals: Other butterflies use bright colors and bold patterns to warn predators that they are toxic or otherwise unpalatable.

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Answer: Education?

Explanation: While I'm not 100% sure, I think that having an average income is going to be most helpful in keeping the infant healthy and alive, while being educated is above all extremely important, I don't think it will necessarily cause death compared to having no food, clean water, or an average income in order to gain those things. Like I said, not 100% sure. Just my take on the question.

A. "Holding playing cards during a preferred card game", represents the use of this approach for this purpose.

A task-oriented approach focuses on improving the client's functional abilities through the use of meaningful and purposeful activities. In this case, holding playing cards during a preferred card game is an activity that is meaningful to the client and also involves the use of the affected upper extremity. This approach helps to improve the client's functional abilities and promotes participation in daily activities.

Option B, moving several weighted items from one shelf to another, is not a task-oriented approach as it does not involve a meaningful or purposeful activity. Similarly, option C, washing dishes while bearing weight through the opposite arm, is not a task-oriented approach as it does not involve the use of the affected upper extremity. Therefore, option A is the correct answer.

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The statement "Assuming steady-state conditions and that water and electrolyte intake remained constant, a 75% loss of nephrons due to chronic kidney disease would cause all of the following except" is incorrect.

A 75% loss of nephrons due to chronic kidney disease would cause a significant decrease in the kidney's ability to filter and regulate electrolytes and water, leading to a large increase in plasma sodium concentration, an increase in plasma creatinine to four times normal, and a significant increase in plasma phosphate concentration.

It would not cause an increase in average volume excreted per remaining nephron to four times normal. In fact, the remaining nephrons would have to work harder to compensate for the lost nephrons, leading to a decrease in the average volume excreted per remaining nephron.

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The probability that their next two children will not have cystic fibrosis is 56.25%.

Cystic fibrosis is an inherited disorder that is caused by a recessive allele. This means that an individual must have two copies of the recessive allele to have the condition. If two parents do not have cystic fibrosis, but have a child with the disease, this means that they are both carriers of the recessive allele.

To determine the probability that their next two children will not have cystic fibrosis, we can use a Punnett square.

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

In this Punnett square, A represents the dominant allele and a represents the recessive allele. The parents are both carriers, so they have one copy of each allele (Aa).

The probability that their next child will not have cystic fibrosis is 75%, since there are three possible genotypes that do not result in the disease (AA, Aa, and Aa). The probability that their next two children will not have cystic fibrosis is 0.75 x 0.75 = 0.5625, or 56.25%.

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In the scenario (a scientist wishes to find out if eating egg yolks increases the risk of heart disease. She takes 50 mice identical in genetic makeup, age, and exercise habits) is true, because the dependent variable is the cholesterol level in mice.

Dependent variables are results that may occur due to the independent variables' influence. In this situation, eating egg yolks or extra mouse-chow is the independent variable. Dependent variables are the variables that the researchers measure to assess the effect of the independent variables. The dependent variable (cholesterol level) in mice is measured every week by the scientist. To determine if eating egg yolks increases the risk of heart disease, she chooses 50 mice with identical genetics, age, and exercise habits. The researcher feeds all mice a diet of mouse chow but adds 30 calories worth of egg yolk to the diet of 25 of the mice. In comparison, she adds 30 calories of extra mouse chow to the diet of another 25 mice. She measures their cholesterol levels every week to assess any changes that may occur as a result of their diet.

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Two common experimental approaches used to confirm the order of substrate addition during DNA polymerase activity are primer extension and gel electrophoresis.

Primer extension involves using labeled primer and a DNA template, then running the reaction in the presence of DNA polymerase to form a labeled DNA product.

Gel electrophoresis is a experimental approach used to separate the components of a reaction mixture according to size and charge, enabling one to identify and quantify the product of the reaction.

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An increase in the elk population would lead to an increase in the organisms of upper trophic level organisms such as lion, tiger, etc.

Elk are ecologically important organisms and these can provide an indicator of how well habitats are functioning in the ecosystem. They have a direct role on the vegetation through herbivory and seed dispersal, and serve as prey and carrion for many other wildlife species in the ecosystem.

As the elk population increase, the amount of forage in the ecosystem removed also increases, which affects the plant growth as well as the diversity.

An increase in the elk population would lead to an increase in the predator population, including secondary consumers such as tigers, lions, etc.

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“Mechanisms of Self-Tolerance in T-Cells versus B-Cells. Unique challenges that each cell type faces in achieving self-tolerance.”

The mechanism of self-tolerance for T-cells and B-cells is maintained through processes known as positive and negative selection. T-cells and B-cells also differ in the way they use negative selection to maintain self-tolerance. The unique challenge faced by each cell type is to ensure that they only recognize and respond to foreign antigens.

Positive selection involves the stimulation of the lymphocytes with antigens and self-antigens, whereas negative selection involves the destruction of lymphocytes that recognize self-antigens. T-cells undergo positive selection in the thymus, where they interact with MHC molecules and their receptors. B-cells, on the other hand, undergo positive selection in the bone marrow, where they interact with antigen-presenting cells.

T-cells use negative selection to recognize and destroy self-reactive cells through apoptosis, while B-cells use negative selection to prevent self-reactive B-cells from maturing.

The unique challenge faced by each cell type is to ensure that they only recognize and respond to foreign antigens, while tolerating self-antigens. This is especially important for T-cells, as they are involved in the destruction of cells and tissue. Therefore, it is essential that T-cells are only activated by non-self antigens in order to prevent the destruction of healthy tissue.

In summary, T-cells and B-cells use positive and negative selection to maintain self-tolerance, but there are some key differences between the two cell types. T-cells undergo positive selection in the thymus, while B-cells undergo positive selection in the bone marrow. Furthermore, T-cells use negative selection to destroy self-reactive cells, whereas B-cells use negative selection to prevent self-reactive cells from maturing. The unique challenge faced by each cell type is to ensure that they only recognize and respond to foreign antigens, while tolerating self-antigens.

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Two examples of convergent evolution are The streamlined body shape of dolphins and sharks , The ability to fly in bats and birds .

Convergent evolution is the process by which different organisms independently evolve similar features or adaptations in response to similar environmental pressures.

1) The streamlined body shape of dolphins and sharks: Both dolphins (mammals) and sharks (fish) have evolved a streamlined body shape that allows them to move efficiently through the water.

This is an example of convergent evolution because these two organisms are not closely related, but have independently evolved a similar adaptation in response to the need to move quickly through the water.

2) The ability to fly in bats and birds: Bats (mammals) and birds have both evolved the ability to fly, but they have done so through different means.

Birds have evolved feathers and wings, while bats have evolved elongated fingers and a thin membrane of skin that stretches between them to form wings.

This is an example of convergent evolution because these two organisms are not closely related, but have independently evolved a similar adaptation in response to the need to move through the air.

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Blood flow through the mammalian heart begins with the return of oxygen-poor blood from the systemic circuit through the superior and inferior vena cavae into the right atrium. From there, the blood flows through the tricuspid valve into the right ventricle.

The right ventricle then pumps the blood through the pulmonary valve into the pulmonary artery, which carries the blood to the lungs for oxygenation.

Once the blood is oxygen-rich, it returns to the heart through the pulmonary veins and enters the left atrium. From there, the blood flows through the mitral valve into the left ventricle.

The left ventricle then pumps the oxygen-rich blood through the aortic valve into the aorta, which carries the blood out to the systemic circuit to deliver oxygen to the body's tissues.

In summary, the major vessels involved in blood flow through the mammalian heart are the vena cavae, pulmonary artery, pulmonary veins, and aorta. The heart chambers involved are the right atrium, right ventricle, left atrium, and left ventricle.

The valves involved are the tricuspid valve, pulmonary valve, mitral valve, and aortic valve. The locations in which the blood is oxygen-poor are the vena cavae, right atrium, right ventricle, and pulmonary artery. The locations in which the blood is oxygen-rich are the pulmonary veins, left atrium, left ventricle, and aorta.

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1. The false statements are Epsilonproteobacteria; H. pylori; and chickens often harbor C. jejuni in their gastrointestinal tract and feces.

2. The correct statements are traditionally, the classification of prokaryotes; Gram-positive bacteria; Gram-negative bacteria; and Gram-negative bacteria possess a thick peptidoglycan cell wall.

Thus, the correct answers are

1. A, B, and D.

2. A, C, D and E.

Prokaryotes can be classified into two domains based on phylogenetic and genetic evidence: Bacteria and Archaea. Prokaryotes, unlike eukaryotes, do not have a nucleus or other membrane-bound organelles, and their DNA is circular rather than linear. Gram-positive and Gram-negative bacteria are the two major bacterial types based on their cell wall structure, which is determined by the Gram stain.

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The correct answer is 25% homozygous dominant; 25% homozygous recessive; 50% heterozygous.

We can use a Punnett square to determine the genotypes and phenotypes of the offspring from this cross. Here is the Punnett square for the cross between two heterozygous red flowers:

|  | R | r |
|---|---|---|
| R | RR | Rr |
| r | Rr | rr |

From this Punnett square, we can see that there are four possible genotypes for the offspring: RR, Rr, Rr, and rr. This means that the genotypes of the offspring are 25% homozygous dominant (RR), 25% homozygous recessive (rr), and 50% heterozygous (Rr).

The phenotypes of the offspring are determined by their genotypes. The homozygous dominant (RR) and heterozygous (Rr) offspring will have red flowers, while the homozygous recessive (rr) offspring will have white flowers. This means that the phenotypes of the offspring are 75% red flowers and 25% white flowers.

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Transcription-coupled nucleotide excision repair is a DNA repair process that occurs when damage is detected in the strand of DNA that is currently being transcribed.

Transcription-coupled nucleotide excision repair (TC-NER) is a process that repairs damaged DNA in actively transcribed genes. It is a sub-pathway of nucleotide excision repair (NER) that specifically targets lesions that block transcription.
The process begins when RNA polymerase II (RNAPII) stalls at a DNA lesion during transcription. This recruits the Cockayne Syndrome B (CSB) protein, which then recruits other proteins, including the Cockayne Syndrome A (CSA) protein and the Xeroderma Pigmentosum (XP) proteins. These proteins work together to remove the damaged DNA and fill in the gap with new, undamaged DNA.
If this process stopped functioning properly, the immediate consequences to a cell would be an accumulation of DNA damage in actively transcribed genes. This could lead to mutations and potential cell death. Additionally, the cell's ability to produce the proteins encoded by the damaged genes would be impaired, potentially leading to further cellular dysfunction.

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In the microbiology lab, the function of an incubator chamber is to maintain optimal temperature.

The incubator is a chamber used to regulate the temperature, humidity, and other environmental conditions in order to promote optimal growth of the bacteria being studied. It can also be used to maintain a desired temperature range in order to prevent bacterial overgrowth.

Аn incubаtion chаmber is а device used to grow аnd mаintаin plаnts аnd microbiologicаl cell. The incubаtor chаmber is designed to mаintаins optimаl temperаture, humidity, light, pressure, vаcuum аnd other conditions such аs the cаrbon dioxide ([tex]CO_{2}[/tex]) аnd oxygen content of the аtmosphere inside the chаmber.

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"RecBCD is the prokaryotic equivalent to MRX/N complex in eukaryotes. The both bind to the DSB. RecA is the prokaryotic equivalent of RAD51 in eukaryote. They both promote strand invasion."

RecBCD is a protein complex found in prokaryotes that plays a role in DNA double-strand break repair. It is structurally similar to the MRX/N complex found in eukaryotes, which also plays a role in the early stages of DNA double-strand break repair. Both complexes recognize and bind to DNA double-strand breaks, initiating the process of end resection.

RecA is a protein found in prokaryotes that plays a role in homologous recombination. It is structurally similar to RAD51, a protein found in eukaryotes that also plays a role in homologous recombination. Both proteins promote strand invasion, which is a key step in homologous recombination. They bind to single-stranded DNA and facilitate the search for homologous DNA sequences, leading to accurate repair of DNA breaks and maintenance of genome stability.

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Each immunoglobulin class is distinguished by amino acid sequences in the c. constant region of the light chain.

Immunoglobulins, also known as antibodies, are proteins that play a crucial role in the immune system. They are produced by B cells and help to recognize and neutralize foreign substances such as bacteria, viruses, and other pathogens. Immunoglobulins are divided into five different classes, each with distinct structural and functional properties. These classes are IgA, IgD, IgE, IgG, and IgM.

The distinguishing feature of each class is the amino acid sequences in the constant region of the light chain. The constant region is the part of the immunoglobulin molecule that does not vary between different antibodies within a given class. The variable regions, on the other hand, are the parts of the molecule that are unique to each individual antibody and are responsible for recognizing specific antigens. In summary, each immunoglobulin class is distinguished by amino acid sequences in the constant region of the light chain.

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Based on cladogram, an illustration of a hypothetical relationship between groups of organisms, including their shared ancestors:

11.) A bony skeleton is derived in ray-finned fish.

12.) The trait that is common to primates and rodents is opposable thumbs.

13.)  Amphibians differ from ray-finned fish in that they are tetrapod with four limbs.

14.) Sharks differ from ray-finned fish in that their skeletons are made of cartilage rather than bone.

15.) The most basic characteristic that all of these organisms share is the presence of a notochord during embryonic development.

11. The bony skeleton derived because of the distinct characteristics of the ray-finned fish, such as their ray-like fins, which are supported by bony spines, and a highly evolved swim bladder that helps them to maintain buoyancy.

12.) Primates and rodents share the trait of opposable thumbs, which are essential for grasping and manipulating objects.

13.) Amphibians differ from ray-finned fish in that they have four limbs instead of fins, they undergo a metamorphosis from larvae to adults, and they breathe through lungs, skin, or gills.

14.) Sharks differ from ray-finned fish in that they have a cartilaginous skeleton rather than a bony one, have a modified gill structure, and lack a swim bladder.

15.)  The most basic characteristic that all of these organisms share is the presence of a notochord during embryonic development. The notochord is a flexible rod-like structure that is present in the embryos of all chordates, including vertebrates like ray-finned fish, sharks, amphibians, rodents, and primates.

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The blood is composed of two portions - a cellular portion made up of the various groups of blood cells and a liquid portion in which the cells are suspended.

The cellular portion is made up of red blood cells, white blood cells, and platelets. The liquid portion, also known as plasma, is made up of water, electrolytes, proteins, and other substances. Together, these two portions make up the blood and allow it to carry out its vital functions within the body.

Blood is a bodily fluid in the circulatory system of humans and other vertebrates that distributes vital chemicals such as nutrition and oxygen to the cells, and moves metabolic waste products away from those same cells.

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Answer:

Carbon in the atmosphere is carbon dioxide (CO2).

Explanation:

Hope this helps.

Answer:

Carbon in the atmosphere exists as carbon (iv) oxide CO2

Scientific Method and Laboratory P 5, the conversion of 4.5% to a fraction is: 4.5/100

the conversion of 30% to decimal is: 0.30,

the conversion of ½ to percent is: 50%

a. To convert 4.5% to a fraction, divide the percentage by 100 to get the decimal equivalent, which is 0.045. Then, rewrite the decimal as a fraction by moving the decimal point two places to the left and adding zeroes to the end to make it a whole number. The fraction is 4.5/100.

b. To convert 30% to a decimal, divide the percentage by 100 to get the decimal equivalent, which is 0.30.

c. To convert ½ to a percentage, multiply the fraction by 100 to get the percentage equivalent, which is 50%.

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