Which of the following can lead to decreased venous
return of blood to the heart?
a) an increase in blood volume
b) An increase venous pressure
c) damage to the venous valves
d) increased muscular activity
Answer: c

Answer:

Explanation:

Homeostasis works by creating the opposite effect in order to counteract a change in the body. One of these effects is blood coagulation. This means that blood is thickened and hardened in order to stop blood flow when there is a cut, open wound, or internal bleeding. Hemophilia on the other hand completely prevents blood from coagulating which makes the homeostasis process obsolete. Without blood coagulation, an individual can bleed to death from any small cut which is usually non-life-threatening.

Answer:

20 cm

Explanation:

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Answer:

50 cm

Explanation:

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happy valentines

Because the descending region of the loop of Henle is extremely permeable to water and less permeable to ions, water is easily reabsorbed here while solutes are not.

Water is particularly permeable in the thin descending limb and is reabsorbed due to the existing concentration gradient in the medulla. While being fully impermeable to water, sodium, chloride, and potassium are readily reabsorbed in the thick ascending limb.

Simple squamous epithelium lines the slender limbs (descending and ascending). The thin descending loop is permeable to water (which is easily transported to the interstitium) but not to solutes. The thin ascending limb, on the other hand, is extremely permeable to sodium but impermeable to water.

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Answer:

Yes because we don’t live by an ocean

Explanation:

Answer:

Just say no. If there is a tsunami, the average modern day house would be flooded and sustain significant structural damage. It would not be able to withstand the crushing forces of the water unless it is specially built. Assuming that you don't live in an area where tsunamis are likely, your house was not built that way.

Explanation:

If two genes are on the same chromosome, there will be complete linkage of the genes in the gametes. In crossing over, crossing over will produce both parental and crossover gametes.

When two genes are located on the same chromosome, they are said to be linked. This means that they tend to be inherited together, rather than assorting independently during meiosis.

The degree of linkage between two genes depends on the distance between them on the chromosome.

During meiosis, crossing over can occur between homologous chromosomes, which can result in the exchange of genetic material between the chromosomes.

When crossing over occurs between two linked genes, it can produce parental gametes (which contain the original combinations of alleles) or crossover gametes (which contain recombined combinations of alleles). The frequency of parental and crossover gametes depends on the degree of linkage between the genes.

Complete linkage is a special case of linkage in which two genes are so close together on a chromosome that they always remain together during meiosis and are always inherited together. In this case, crossing over does not occur between the two genes, and only parental gametes are produced.

Overall, the degree of linkage between two genes on the same chromosome affects the frequency of parental and crossover gametes that are produced during meiosis, and crossing over can result in recombined combinations of alleles.

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Answer:

.

Explanation:

.

In this exercise, natural selection is being modeled by simulating the removal of homozygous recessive individuals from a population of foxes with brown and white fur. The white fur is considered deleterious, and thus, all foxes with the white fur phenotype are not allowed to reproduce. By sampling without replacement, the white foxes are eliminated from the population, and the alleles associated with white fur are gradually reduced.

1. The gene pool is initially set up with 100 alleles, consisting of 50 brown beans (representing the B allele) and 50 white beans (representing the b allele). This represents the starting population of foxes with different fur colors.

2. A tally sheet is prepared on a separate sheet of paper, listing the genotypes encountered in the population: BB (homozygous dominant), Bb (heterozygous), and bb (homozygous recessive).

3. Two alleles (beans) are randomly withdrawn from the population. This represents the selection of two foxes for reproduction.

4. The genotype of the selected foxes is recorded on the tally sheet by placing a hash mark next to the appropriate genotype (BB, Bb, or bb). This step helps keep track of the frequencies of different genotypes within the population.

5. Instead of returning the beans to the population, the selected alleles (beans) representing the foxes with the white fur phenotype are placed in a designated area on the workspace. This action simulates the elimination of the white foxes from the population, as they are not allowed to reproduce due to their deleterious phenotype.

By repeating steps 3-5 multiple times and tallying the genotypes encountered, it is possible to observe the changing frequencies of different genotypes in the population over time. As the white foxes are continuously removed from the population, the frequency of the homozygous recessive genotype (bb) associated with white fur decreases, while the frequencies of the homozygous dominant genotype (BB) and the heterozygous genotype (Bb) may remain relatively stable or change depending on the reproductive success of individuals with different genotypes. This modeling exercise helps illustrate how natural selection can shape the genetic composition of a population by favoring individuals with advantageous traits and reducing the occurrence of deleterious traits.

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The tested tripeptide variants most likely have the following characteristics in common:

All three characteristics are required in order for the tripeptide to exist, so the answer is D: I, II, and III.

When forming a tripeptide, the sequence of the three amino acids is essential in determining the structure and properties of the peptide. The sequence of the amino acids is determined by the order of the codons in the mRNA strand, which is ultimately determined by the DNA sequence. The primary structure of the tripeptide is then determined by the peptide bond that forms between each of the linked amino acids.

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Answers:

Answer:

NY and PA

Explanation:

Its pretty self explanetory

Carbon makes up almost all of the most common substances in the human body, including proteins, fats, and carbohydrates. Therefore, option (B) is correct.

The human body is made up of a wide variety of substances, including water, proteins, fats, carbohydrates, nucleic acids, and minerals. Water is the most abundant substance in the human body, making up about 60% of its total weight. Proteins are the building blocks of tissues and organs, while fats and carbohydrates provide energy and serve as structural components of cells.

Nucleic acids, such as DNA and RNA, carry genetic information and are essential for cell division and growth. Minerals, such as calcium, phosphorus, and iron, are important for maintaining bone and muscle health, blood circulation, and other physiological functions. Other substances found in the human body include vitamins, enzymes, hormones, and neurotransmitters.

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Answer: The given statement is False.

Explanation: The statement is False.. The words "phenotype" and "genotype" are related to genes, but they mean different things. Genotype is all the genes that an organism has, whether or not you can see them. Phenotype is the traits that you can see, which come from a combination of genes and the environment. So, genotype is about genes, while phenotype is about traits that you can see.

Answer:

its false

Explanation:

need a short answer so you get what your looking for well here i am look at answers. hope it helps

It is important to match the blood types of the donor and recipient.

In this case, the person's blood cells have an agglutinogen A, which is a substance that can cause the blood to clot. They also have anti-A antibodies in their blood, which means that their immune system will recognize and attack cells with agglutinogen A.

The rule that is applied in this case is the ABO blood group system. According to this system, there are four blood types: A, B, AB, and O. People with blood type A have agglutinogen A on their red blood cells and anti-B antibodies in their plasma.

People with blood type B have agglutinogen B on their red blood cells and anti-A antibodies in their plasma. People with blood type AB have both agglutinogen A and agglutinogen B on their red blood cells, but they do not have any anti-A or anti-B antibodies in their plasma. People with blood type O do not have either agglutinogen A or agglutinogen B on their red blood cells, but they have both anti-A and anti-B antibodies in their plasma.

Based on this information, the person you described with cells that have agglutinogen A and anti-A antibodies in their blood has blood type A. They can receive blood from a donor with blood type A or O, but they cannot receive blood from a donor with blood type B or AB. properly in order to avoid a transfusion reaction, which can be serious or even life-threatening.

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Answer:

False

Explanation:

Answer:

ITS (F.) FALSE.

Explanation:

Answer:

Please find the punnet square attached.

Genotype of offsprings: RR(1), RW(2), WW(1)

Phenotype of offsprings: 1 Red: 2Pink: 1White

Explanation:

This question involves a single gene coding for flower color in four oʼclock plants. The alleles for red flower (R) and white flower (W) exhibit incomplete dominance, hence, results in a third phenotype that is pink (RW).

If two pink flowered four oʼclock plants were crossed i.e. RW × RW, the following gametes will be produced by each parent

RW - R and W

Using these gametes in a punnet square (see attached image), the following genotype and phenotype of offsprings will be produced.

Genotype of offsprings: RR(1), RW(2), WW(1)

Phenotype of offsprings: 1 Red: 2Pink: 1White

Answer:

Following the instructions would give the following:

3. B - Minnesota

4. A. Mississippi

   B. Missouri

5. B Iowa

7. A. Louisiana

8. B. Arkansas

9. A. Tennessee

   B. Kentucky

10. A. Wisconsin

     B. Illinois

12.  A. New York

     B. North Carolina

13. A. California

     B. Colorado

Answer:

0.5m/s

The velocity is equal to the distance over time, so we divide 5 by 10

Answer:

CO2 affect the rate of photosynthesis because the increased of the amount of carbon dioxide increase photosynthesis, spurring plant growth.

Answer:

Explanation:

Since there is no extra information about a specific gene and its inheritance patterns, I am going to base my answers of examples I create.

To begin with, in a pair of alleles, dominant alllele is represented by A (Capital letter) and recessive is done by a (small letter). If an organism has AA (homozygous dominant) it will always show the dominant trait. If it has Aa (heterogyzous dominant) it will also show the dominant trait as one of the alleles is dominant, and finally if it as aa then it will show recessive trait

Which allele in mouse is dominant vs recessive is primarily figured out based on inheritance pattern. If majority of offspring represent a certain trait like black fur coat it is likely that the dominant trait for that gene is black fur coat. Allele that is recessive is masked by the dominant allele so if parents are

Aa * Aa ---> you'll get 4 genotypes AA, Aa, Aa,aa (Here the AA, Aa, Aa will all show the dominant trait because the capital letter A represents dominant allele and it masks the small letter a) whereas, aa will show recessive trait because there is no capital letter or dominant allele to mask the recessive trait.

Homozygous dominant, heterozygous, and homozygous recessive genotypic frequencies are 36%, 48%, and 16%, respectively:

Here: homozygous dominant genotype, AA

Aa for the genotype of heterozygotes

Homozygous recessive genotype for aa, 16 out of the 100 people in the population have the recessive trait, hence there are 16 in the population. Thus, the genotype frequency for aa is 16/100, or 0.16.

The frequency of the A allele plus the frequency of the an allele must equal one because there are only two potential alleles at this locus (A and a). This fact, along with the frequency of the aa genotype, can be used to determine how common the A allele and the aa genotype are:

frequency of aa genotype = q² = 0.16,

frequency of A allele = p = 1 - q

frequency of Aa genotype = 2pq,

q = 0.16 = 0.4

p = 1 - q = 0.6

frequency of Aa genotype = 2pq = 2(0.6)(0.4) = 0.48

frequency of AA genotype = p² = (0.6)² = 0.36

Therefore, the genotypic frequencies are:

AA: 0.36 or 36%

Aa: 0.48 or 48%

aa: 0.16 or 16%

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Answer:

recessive

Explanation:

Some scientists have reported that handedness is due to a single gene with right handedness dominant and left handedness recessive.

I hope that this helped you :D

Derivative classifiers are individuals who are responsible for re-classifying or downgrading previously classified information. All of the following responsibilities are expected of derivative classifiers EXCEPT having original classification authority. Here option 3 is the correct answer.

Access to classification guidance: Derivative classifiers must have access to the appropriate classification guidance in order to properly re-classify information. Original classification authority: Original classification authority refers to the ability to classify information at the highest level of classification.

Understanding derivative classification policies and procedures: Derivative classifiers must be familiar with the policies and procedures for re-classifying information, including the requirements for protecting classified information and marking it with the proper security classifications.

Possession of subject matter expertise and classified management and marking techniques: Derivative classifiers must possess the necessary subject matter expertise and be familiar with the techniques for managing and marking classified information.

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If phytoplankton populations grow more quickly in cooler water, then the herbivorous fishes will have abundant food resulting in an increase in their population.

Phytoplankton are the aquatic organisms that are autotrophic in nature. They can be found in both freshwater as well as marine water. These are one of the most important species of aquatic food chains as they form the first trophic level of most of the good chains.

Herbivorous are those animals that feed on plants and plant parts for their food and energy requirements. Since they cannot synthesize their own food, herbivorous animals belong to the category of heterotrophs.

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Answer:100% black with rough for both phenotype and genotype;4/4,1

Explanation:

so if a black coat and rough coat are dominant and a guinea pig that is a purebred(homozygous)black coat with rough coat and a homozygous white coat with smooth coat breed the dominant allele will show up for both so as a result since the black coat and rough coat are dominant the result will be 100% black with rough coat.

Answer:

Sandstone does not react with Hydrochloric acid. The acid soaks into the rock because the rock is porous, filled with tiny holes. It is made up of tiny grains of sand cemented together by intense pressure.

Explanation:

the amount of carbon dioxide released into the atmosphere as a result of the activities of a particular individual, organization, or community.

Answer:

1: carbon footprint is the amount of carbon dioxide released into the atmosphere as a result of the activities of a particular individual, organization, or community.

2: A carbon footprint is a measure of the impact your activities have on the amount of carbon dioxide (CO2) produced through the burning of fossil fuels and is expressed as a weight of CO2 emissions produced in tonnes.

The answer is (b) 10. To limit inbreeding to a maximum of 14.678481% after 5 generations, we need to calculate the expected amount of inbreeding in each generation and then adjust the number of males to add to the population accordingly.

The expected amount of inbreeding in each generation can be calculated using the formula:

Expected inbreeding = (1 - (1/2)^(generation/2)) * (1 - (1/2)^(generation/2)^2) * ... * (1 - (1/2)^(generation/2)^(14))

Substituting the value of generation=5 gives:

Expected inbreeding = (1 - (1/2)^5) * (1 - (1/2)^5)^2 * ... * (1 - (1/2)^5)^(14)

= 0.81924005762932

The maximum amount of inbreeding allowed is 14.678481%. Substituting this value gives:

Maximum inbreeding = 14.678481 / 0.81924005762932

= 17.5937535396166

Therefore, to limit inbreeding to a maximum of 14.678481% after 5 generations, we need to add at least 17.5937535396166 males to the population.

Solving for the number of males needed if we have 20 females, we get:

17.5937535396166 = (20 * 1 - 14.678481) / 20

= 3.45344278981875

Rounding to the nearest whole number, we get:

The number of males needed is 3.

Therefore, the answer is (b) 10.

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