Which of the following controls the involuntary actions in the body​

Answers

Answer 1
medulla oblongata

Involuntary actions are under the voluntary control of the medulla oblongata which is generally referred to as the midbrain. The involuntary movements are characterised by relatively slower speed. The best example of this movement is the beating movement of the human heart.

Related Questions

which life cycle stage is found in plants but not animals ​

Answers

Answer:

Multicellular haploid

OAmalOHopeO

Plants have multicellular haploid and multicellular diploid stages in their life cycle.

Gametes develop in the multicellular haploid gametophyte . Fertilization gives rise to a multicellular diploid sporophyte, which produces haploid spores via meiosis.What is multicellular haploid  stage?The haploid multicellular stage produces specialized haploid cells by mitosis that fuse to form a diploid zygote.The zygote undergoes meiosis to produce haploid spores. Each spore gives rise to a multicellular haploid organism by mitosis.

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What is the function of the mitochondria?
A. Stores the cell's DNA
B. Builds proteins
C. Produces energy for the cell by respiration
OD. Stores the cell's glucose
Reset Selection

Answers

C. Mitochondria is like the power plant of the cell and produces its energy.

Answer:

Produces energy for the cell by respiration

Explanation:

The glucose obtained from food is broken down to pyruvic acid in the cytoplasm. This pyruvic acid is broken down into oxygen, water and energy rich ATP molecules in the Mitochondria.

Fertilization, Fruit and Seed Formation x3 3. a. name the two processes that lead to seed formation in flowering

Answers

Answer:

Pollination, the transfer of pollen from flower-to-flower in angiosperms or cone ... In angiosperms, the process of seed development begins with double ...

Missing: x3 ‎| Must include: x3

Ecosystems rely on interdependence between species to keep balance. Which of the following is a threat to a stable
ecosystem?
A. Loss of biodiversity
B. High biodiversity
C. Low biodiversity
D. Increase in biodiversity

Answers

Answer:

loss of biodiversity

Explanation:

Biodiversity- refers to the variety of life on Earth at all its levels, from genes to ecosystems, and can encompass the evolutionary, ecological, and cultural processes that sustain life.

loss in biodiversity affect food chains greatly

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hope it helps

During the performance of the simple staining procedure, you failed to heat fix your E-coli smear preparation. Upon microscopic examination , how would you expect this slide to differ from the correctly prepared slides ?

Answers

OMG I'm a business study Student so I really don't know about this stuff

A small group of mice are released on an island without mice but with abundant food for mice and no predators. After the population size stabilizes for several years, a hurricane drastically reduces it. We can now say that:________.
A) the biotic potential of the population has been reduced.
B) its new population size is a result of density-dependent regulation.
C) its new population size is a result of density-independent regulation.
D) it can now act as a sink metapopulation.

Answers

The correct answer is option C) The new mice population size is a result of density-independent regulation.

The carrying capacity might be affected by different factors, known as limiting factors, which might be a result of the population density (for example, competition) or might be density-independent. This last case refers to dense-independent factors, and among these, we can mention human impact or natural disasters (fires, volcanic eruption, flooding). Natural disaster causes damages in an ecosystem, reducing the available resources such as food or shelter, and consequently decreases the number of individuals. Natural disasters reduce the carrying capacity of the environment

In the exposed example, mice got to stabilize on the island. The population had enough food and no predators. But the occurrense of the huricane reduced drastically the population size. This is an example of a natural dissaster acting as a limiting dense-independent factor affecting the population size.

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A man bought a goldfish in a pet shop. Upon returning home, he put the goldfish in a bowl of recently boiled water that had been cooled quickly. A few minutes later the fish was found dead. Explain what happened to the fish

Answers

Answer:

lack of oxygen in the water

Explanation:

The fish most likely died from lack of oxygen in the water. This is because fishes actually use their gills to extract and breathe in the oxygen from the water while also expelling carbon dioxide from their lungs. Similar to how humans breathe. When the water was boiled it caused the dissolved gases to be expelled, which includes oxygen. Therefore, without the necessary oxygen in the water, the fish ultimately suffocated.

Boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.

What is dissolved oxygen?

Dissolved oxygen is the amount of oxygen present in the water.

The organisms live to consume dissolved oxygen to breathe.

The amount of dissolved oxygen is high in the current water like rivers than in the still water like pond.

If the amount of DO is high in the water, it causes bubble gas disease in the aquatic organisms.

If the amount of DO is low in the water than, fishes and other aquatic organism cant survive due to low oxygen level.

Thus, boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.

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i need help in biology questions please G10?

Answers

Answer:

ok where is it

we can help only if there is something attached

Attach the picture or something so we could respond

The body regulates the amount of hormones are released by using feedback loops. A __ feedback loop increases the response whereas a __ feedback loop decreases the response.

Answers

Please mark brainliest

Answer

The answer for first fill in the blank is “ positive”
The answer for second fill in the blank is negative

Positive feedback loop increases the response whereas a negative feedback loop decreases the response.

What is positive feedback?

Positive feedback is the amplification of a body's response to a stimulus. For example, in childbirth, when the head of the fetus pushes up against the cervix (1) it stimulates a nerve impulse from the cervix to the brain (2).

A feedback mechanism resulting in the inhibition or the slowing down of a process.

Examples of processes that utilise negative feedback loops include homeostatic systems, such as thermoregulation (if body temperature changes, mechanisms are induced to restore normal levels), blood sugar regulation (insulin lowers blood glucose when levels are high ,glucagon raises blood glucose when levels are low).

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The nitrogen cycle is the using and reusing of nitrogen in an ecosystem. True or false?

Answers

Answer:

True

Explanation:

Nitrogen is a fundamental component of both inorganic and organic compounds, where it is the main constituent of biomolecules such as nucleic acids (DNA, RNA) and proteins. The nitrogen cycle refers to the biogeochemical processes by which nitrogen circulates between the components of an ecosystem, i.e., between organisms (like plants and decomposers), and non-living things (i.e., soil, water, air). This cycle consists of several processes which include, among others, nitrogen fixation (i.e., the process by which nitrogen in the atmosphere is converted into ammonia), nitrification (i.e., the oxidation of ammonia is oxidized into nitrite and subsequent transformation of nitrites into nitrates), denitrification (where nitrate is reduced), anaerobic ammonia oxidation and putrefaction.

Carnivore that feeds on primary consumers
Question 3 Multiple Choice Worth 3 points
(01.01 LC)
Which of the following is an example of a decomposer?

Answers

we need the options in order to answer

Digestion is primarily controlled by the _____.

Answers

Stomach and large and small intestine
gastron,secretin and cholecystokinin

What else is produced during the replacement reaction of silver nitrate and potassium sulfate?

2AgNO3 + K2SO4 Ag2SO4 + ________

KNO3
2KNO3
K2
2AgNO3

Answers

Answer:

2KNO3

Explanation:

Pls mark it brainliest

hope it helps u

Answer:

The answer is B.) 2KNO3

Explanation:

Human being get energy from

Answers

Are you asking how humans get energy?
Any food they consume or sleep. Hope this helps

QUESTION 11 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the allele's frequencies experimentally derived, calculate the frequency of the H4/H5 genotype that would be expected if the class were a population in Hardy-Weinberg equilibrium. 1. 0.28 2. 0.51 3. 0.19 4. 0.72 5. 0.14 6. 0.24 7. 0.41 QUESTION 12 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the genotype frequencies derived assuming that the class were a population in Hardy-Weinberg equilibrium, calculate the number of H4/H4 individuals that would be expected in the class (rounded numbers). 1. 19 2. 57 3. 72 4. 147 5. 171 6. 120 7. 96 QUESTION 13 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic 1. 2.69 2. 0.05 3. 28.67 4. 14.59 5. 0.50 6. 22.31 7. 3.84 QUESTION 14 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium, which is the (correct) null hypothesis tested by Chi-square? 1. The whole class represents a population that is in Hardy-Weinberg equilibrium 2. The whole class represents a population that may not be in Hardy-Weinberg equilibrium 3. The whole class represents a population that is not in Hardy-Weinberg equilibrium 4. The whole class represents a population that may be in Hardy-Weinberg equilibrium QUESTION 15 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom, I conclude that P>0.05. Hence, I fail to reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis.

Answers

According to Hardy-Weinberg, when a population is in equilibrium, it will have the same allelic frequencies generation after generation, meaning that they are stable, they are not evolving.

When any evolutive force is acting on the population, this equilibrium breacks, and allelic and genotypic frequencies change through generations, differing from the expected ones.  

A) Option 7 is the correct answer ⇒ 0.41

B) Option 6 is the correct answer ⇒ 120

C) Option 7 is the correct answer ⇒ 3.84

D) Option 1 is the correct answer ⇒ The class represents a population that is in H-W equilibrium

E) Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for  the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.

-------------------------------------------

Allelic frequencies in a locus are represented as p and q, referring to the

allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (H0m0zyg0us  dominant), 2pq (H3ter0zygous), (H0m0zyg0us recessive). Populations in H-W equilibrium will get the same

allelic frequencies generation after generation.

The sum of the allelic frequencies equals 1, this is p + q = 1.

In the same way, the sum of genotypic frequencies equals 1, this is

p² + 2pq + q² = 1

Being

 p the dominant allelic frequency,

 q the recessive allelic frequency,

 p² the h0m0zyg0us dominant genotypic frequency

 q² the h0m0zyg0us recessive genotypic frequency

 2pq the h3ter0zyg0us genotypic frequency

 

Situation: Through PCR, we have determined the PER3 genotypes for a class of students as follows:

 H4/H4 = 125 individuals;

 H4/H5 = 85 individuals;

 H5/H5=24 individuals.

⇒ Total number of individuals= 125 + 85 + 24 = 234

⇒ Genotypic frequencies, F(xx):

 F(H4/H4) = 125/234 =0.534

 F(H4/H5) = 85/234 = 0.363

 F(H5/H5) = 24/234 = 0.102

⇒ Allelic frequencies, f(x):

 f(H4) = p = F(H4/H4) + 1/2 F(H4/H5) = 0.534 + 0.363/2 = 0.534 + 0.182 = 0.716

 f(H5) = q = F(H5/H5) + 1/2 F(H4/H5) = 0.102 + 0.363/2 = 0.102 + 0.182 = 0.284

Questions:

A)  According to the theoreticall frame, we know that 2pq is the h3ter0zygous genotypic frequency. So,  

F(H4/H5) = 2pq = 2 x 0.716 x 0.284 = 0.408 ≅ 0.41 ⇒ Option 7 is the correct answer.

-------------------------------------------------------------------------------------------------------------

B)  According to the theoreticall frame, we know that p² is the h0m0zyg0us genotypic frequency. So,

p = 0.716

p² = 0.5126 ≅ 0.513 ⇒ This is the genotypic frequency.

To calculate the number of individuals carrying this genotype, we need to multiply it by the total number of

individuals.

H4/H4 individuals = p² x total number of individuals = 0.513 x 234 = 120

Option 6 is the correct answer.

-----------------------------------------------------------------------------------------------------------

C)  Up to here we know that 2pq = 0.41 and p² = 0.513

Now we need to calculate q ²

q = 0.284, then q² = 0.284² = 0.08

These are the expected frequencies if the population was in H-W equilibrium.

The expected number of individuals with each genotype are:

 H4/H4 = 0.513 x 234 = 120 individuals

 H4/H5 = 0.41 x 234 = 96 individuals

 H5/H5= 0.08 x 234 = 18 individuals

The observed number of individuals with each genotype are:

 H4/H4 = 125 individuals

 H4/H5 = 85 individuals

 H5/H5=24 individuals

X² = ∑ (Observed - Expected)²/Expected)

X² = ((125-120)²/120) + ((85 - 96)²/96) + ((24-18)²/18)

X² = 0.21 + 1.26 + 2 =

X² = 3.47

The clossest option is option 7 = 3.84. The difference might be related to decimals and rounding.

-------------------------------------------------------------------------------------------------------------

D)  The correct answer is  1 ⇒ The whole class represents a population that is in Hardy-Weinberg equilibrium

The null hypothesis always predict that populations are in H-W equilibrium.

-----------------------------------------------------------------------------------------------------------

E)  

 X² = 3.47

 Freedom degrees = n - 1 = 3 - 1 = 2

 Table p value: 7.82

 Significance level, 5% = 0.05

 Table value/Critical value = 5.991

5.991 > 0.347

Meaning that the difference between the observed individuals and the expected individuals is statistically  significant. Not probably to have differe by random chances. There is enough evidence to reject the null

hypothesis.

Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for  the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.

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H. pylori cannot grow in other microenvironments of the human body because the conditions are unsuitable for its growth, but other species require different conditions.

a. True
b. False

Answers

Answer:

it should be an true statement

Our body needs both vitamin and mineral in a small quantity ,still they are important why?​

Answers

Answer:

Vitamins and minerals are considered essential nutrients—because acting in concert, they perform hundreds of roles in the body. They help shore up bones, heal wounds, and bolster your immune system. They also convert food into energy, and repair cellular damage.

How does pollution affect biodiversity

Answers

Answer:

All forms of pollution pose a serious threat to biodiversity, but in particular nutrient loading, primarily of nitrogen and phosphorus, which is a major and increasing cause of biodiversity loss and ecosystem dysfunction. ... In addition, nitrogen compounds can lead to eutrophication of ecosystems.

functions of insulin

Answers

Insulin is a hormone that controls blood glucose (sugar) levels by signaling the liver and muscle and fat cells to take in glucose from the blood. Insulin therefore helps cells to take in glucose to be used for energy. If the body has sufficient energy, insulin signals the liver to take up glucose and store it as glycogen.

Answer:

Insulin helps control blood glucose levels by signaling the liver and muscle and fat cells to take in glucose from the blood. Insulin therefore helps cells to take in glucose to be used for energy. If the body has sufficient energy, insulin signals the liver to take up glucose and store it as glycogen.

Explanation:

Environmental scientists cite several challenges to protecting endangered species. Which of the following is NOT a difficulty faced by those working to protect an endangered species? Scientists who are focused on a single species may not recognize another species in peril. Extinction is natural in nature and its rate of occurrence has remained steady. Protection efforts may be slowed because of difficulty enforcing new regulations. Fundraising can be more difficult if the species is not perceived as valuable.

Answers

Answer:

Extinction is natural in nature and its rate of occurrence has remained steady.

Explanation:

As of right now we are going through what is known as the sixt extinction, and human acticity is one of the Main causes. (which has never happened before our time) Species are going extinct at a rapid rate, and considering how all life is connected it makes it easy to understand why the extinction rates can increase due to a biotic factor (niche) being "removed". (:

State whether the following statements are true or False. Hormones in plants travel by the vascular bundle.​

Answers

Answer:

True

Explanation:

In plants, hormones travel large throughout the body via the vascular tissue (xylem and phloem) and cell-to-cell via plasmodesmata. In contrast, many animal hormones are produced only in specific glands. Plants do not have specialized hormone-producing glands.

Which of the following is an example of an enzymatic cycle?

Answers

Answer:

Catabolism

Explanation:

The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.

A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.

Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.

The most basic organization level of life is a ____________. A. membrane B. tissue C. cell D. organ

Answers

[tex]Hello[/tex] [tex]There![/tex]

The answer is...

C. Cell.

Hopefully, this helps you!!

[tex]AnimeVines[/tex]

DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.Which part of the nucleotides' structure is responsible for the incredible variation that exists amongst all types of organisms

Answers

Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).

The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits; here are the arrangements:

-Adenine is paired with Thymine (think of A for apple and T for tree)

-Cytosine is paired with Guanine (think of C for car and G for garage)

search "DNA base pairs" and go to images for better understanding

The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).

Who is responsible for passing down of genetic traits from one generation to the next?

DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.

The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits and here are the arrangements are mentioned below:

Adenine is paired with Thymine (think of A for apple and T for tree)Cytosine is paired with Guanine (think of C for car and G for garage)

Therefore, The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).

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Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors

Answers

Parasites is correct

You are studying an enzyme that is inactivated by phosphorylation and create a mutant in which the threonine that is normally phosphorylated is replaced with glutamate. Predict the impact of this change on the activity of this enzyme. Group of answer choices

Answers

Answer:

always active  

Explanation:

Phosphorylation is a posttranslational modification that consists of the addition of phosphate groups to specific amino acids on the protein. Phosphorylation acts as a molecular switch for proteins that are phosphorylated (i.e., in some situations phosphorylation acts to activate protein function, whereas in other situations phosphorylation can inactivate protein function). Phosphorylation modifies the three-dimensional structure of the protein, thereby affecting, for example, the accessibility of the active site of a phosphorylated enzyme to its substrate. Phosphorylation can occur only at the side chains of three amino acids: Serine, Threonine and Tyrosine. In this case, the enzyme is inactivated by phosphorylation on the Threonine residue, so it is expected that the mutant enzyme cannot be phosphorylated, remaining in an active state.

What about the structure of DNA allows it to copy itself?
A. It is held together mostly by hydrogen bonds, which are easy to open for replication
B. It has enzymes built into the helix that make a copy when needed.
C. It is double stranded, so each side serves as a template for a new strand.
D. DNA is not copied, it is only passed down through each generation.

Answers

Answer:

C. It is double stranded, so each side serves as a template for a new strand

Explanation:

The structure of DNA is double stranded, so each side serves as a template for a new strand. The correct option is C.

What is DNA replication?

Replication is the procedure by which the DNA of the genetic code is copied in cells.

Before dividing, a cell must duplicate (or replicate) its a whole genome so that each eventually results daughter cell has its own entire genome.

The DNA replicates itself multiple times during the process of replication. It is a biological polymerization that goes through the steps of commencement, elongation, and termination.

The reaction is catalyzed by enzymes. The main enzyme in the process of replication is DNA Polymerase.

DNA replication occurs in eukaryotic cells' interphase nuclei. At the S-stage (synthesizing) of the cell cycle, DNA replication occurs prior to mitosis.

Thus, the correct option is C.

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Describe the disadvantages of an open circulatory system relative to a closed circulatory system.

Answers

Explanation:

The disadvantage of a closed circulatory system is that with the blood always contained in vessels, it's under higher pressure. This means it can flow faster, which is necessary for supporting the metabolic rate of more complex animals such as vertebrates. Open circulatory systems are low-pressure systems that cannot meet the needs of a fast metabolism.

how much water was retained by soil C​

Answers

Answer:

we dont know sorry but i dont know

3. When recording medication administered to a client on the medical administration record, always add
next to each medication you deliver.
A. the client's weight
B. the client's initials
C. your initials
D. the client's resting heart rate

Answers

Answer:

option B

Explanation:

Answer:

your initials

are the ones that should be in the MAR. since your the one who gave the medication.

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