Answer:
I thik answer is 2 number
Which is the only element in group 1 on the periodic table that forms covalent bonds?
hydrogen
lithium
potassium
sodium
.
Answer:
Hydrogen
Explanation:
Hydrogen is the only nonmetal in group 1, therefore it is the only element in group 1 that forms a covalent bond.
Answer:
A.
Explanation:
What else is produced during the replacement reaction of silver nitrate and potassium sulfate?
2AgNO3 + K2SO4 Ag2SO4 + ________
KNO3
2KNO3
K2
2AgNO3
Answer:
2KNO3
Explanation:
Pls mark it brainliest
hope it helps u
Answer:
The answer is B.) 2KNO3
Explanation:
What is microbiology?
Microbiology is the study of microscopic organisms, such as bacteria, viruses,
Factor affecting crop production
Answer:
pls mark as branilist ans if u like this ans
What do birds produce
Which of the following combinations of phylum and description is correct? (and why?)
a) Nematoda − roundworms, internal skeleton
b) Porifera − gastrovascular cavity, coelom
c) Echinodermata - radial symmetry as a larva, coelom
d) Platyhelminthes − flatworms, gastrovascular cavity, no body cavity
The phylum and description that is correct among the options is Platyhelminthes − flatworms, gastrovascular cavity, no body cavity
Flatworms belongs to the phylum platyhelminthes. They are said to have no true body cavity, but they do have bilateral symmetry. For the lack of a body cavity, it is called acoelomates. They also have an incomplete digestive system. That is, the digestive tract has only one openingsFrom the above we can therefore say that Option d is the correct option that gives the true description of the phylum platyhelminthes
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8. Which of the following statements about plasma membranes is INCORRECT?
A. They are composed of phospholipids.
B. They have both integral and peripheral proteins.
C. They are relatively rigid structures.
D. They function to separate the extracellular and intracellular environment.
CONT
st?
Answer:
C. They are relatively rigid structures.
Explanation:
The plasma membrane is a semipermeable lipid bilayer mainly composed of amphipathic phospholipids, which acts as a barrier to separate the cytoplasm from the extracellular environment. This barrier is considered a fluid combination of proteins, lipids and carbohydrates. The fluid mosaic model states that the plasma membrane is a mosaic of lipids (especially phospholipids and cholesterol), as well as proteins and carbohydrates (glycolipids and glycoproteins) that are constantly moving. This fluidity alters the rotation and diffusion of proteins and other components within the plasma membrane, thereby affecting their functions. Membrane fluidity has been experimentally demonstrated by a variety of techniques. (e.g., X-ray diffraction, labeling techniques, calorimetry, etc).
How are oxygen and carbon dioxide exchanged between the alveoli and the
capillaries?
A. Endocytosis
B. Osmosis
C. Simple diffusion
5
D. Active transport
Answer:
B. Osmosis
Explanation:
Osmosis is the process in which oxygen and carbon dioxide exchanged occur between the alveoli and the capillaries because the oxygen enters the body and the carbondioxide gas leaves the cell through a semi-permeable membrane and we know that Osmosis is a process in which smaller molecules moves from higher concentration to lower concentration through semi-permeable membrane of the cell.
Decide whether the following statement is TRUE or FALSE.
A symporter would function as an antiporter if its orientation in the membrane were reversed (that is, if the portion of the protein normally exposed to the cytosol faced the outside of the cell instead).
Answer:
False
Explanation:
Uniporters, symporters, and antiporters are membrane proteins that transport substances across cell membranes. Uniporters are proteins that carry one specific ion or molecule. Moreover, symporters are carriers that transport at least two substrates/solutes in the same direction, i.e., from one side of the membrane to the other at the same time. On the other hand, antiporters are carrier proteins that catalyze the exchange of two substrates/solutes (ions or molecules) in opposite directions at the same time. In consequence, turning around a symporter would not convert it into an antiporter, only would cause the symporter to bind two solutes on the opposite side of the membrane.
Gene expression can be summarized as
DNA is translated to mRNA that is then transcribed to
make a protein.
O
DNA is transcribed to mRNA that is then translated to
make a protein.
Protein is translated to mRNA that is then transcribed
to make DNA.
mRNA is transcribed to DNA that is then translated to
make a protein.
Answer:
3 one
Explanation:
A student is conducting their science experiment on the effect of caffeine on dogs. He has 3 groups of test subjects. The 1st group of dogs receives plain water. The 2nd group of dogs receives 10 mg of caffeine each, and the 3rd group receives 50 mg of caffeine each. He will measure their activity levels by recording how long each dog runs without stopping, after giving them the pills. What is the dependent variable in this experiment?
Answer:
The amount of activity
Explanation:
In an experiment, the independent variable is the variable that the scientist/investigator purposely changes or manipulates, which isn't changed by other variables in the experiment (in this case, the amount of caffeine). On the other hand, the dependent variable is the variable that is being measured or tested during the experimental procedure (this variable is 'dependent' on the independent variable). Finally, the control group is defined as the group of individuals/subjects who do not receive the experimental treatment (in this case, the dogs that receive water).
Answer: The amount of activity.
Explanation:
The outbreak has rebounded in at least 30 US states in recent weeks, with the three most populous states -- California, Texas and Florida -- seeing a surge in new cases, with the highest daily number of new cases since the outbreak began.
Answer:
whats the question then?
Explanation:
M. magneticum can only survive in low-oxygen environments, which are typically found near the bottom of bodies of water.
a. True
b. False
Which one of the following would be inhibited by a well-designed antiviral drug? Cell wall synthesis Viral binding to human cells Virus assembly outside of the infected cell Translation of host cell RNAs
Viral binding to human cells is inhibited by the antiviral drug.
Well-designed antiviral drug inhibited Viral binding to human cells so that the virus can't get the place of attachment and unable to use the cell's machinery for its growth and multiplication. In this way, the humans can be prevented from having the viral infection. There are some other mechanisms also used by the antiviral drug to inhibit the growth of virus in the human body such as uncoating of virus and synthesis of new viral components.
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Question 5
Not yet answered
Marked out of 1.00
P Flag question
Complete the following sentence: "The interior of living cells is more
than the exterior because more
ions are expelled than ions are taken in by the sodium-potassium pump."
Select one:
O a. electropositive. Nak
O b. electronegative, Na.
O C. electronegative, Na, K
O d. electropositive, Na+, K+
Question 6
Not yet answered
Marked out of 1.00
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The high concentration of protons in the inner mitochondrial space relative to the mitochondrial matrix represent
O
В.
77°F
AGD (0)
10-25 PM
7/28/2021
Answer:
i really really need the brainly points
Explanation:
sry for this answer, i need the answers for myself
A research group discovers that a reagent can be used to form covalent cross-links between various types of atoms in both small molecules and proteins. They add the reagent to a mixture of 14C-lactose and a protein that is thought to have a specific binding site for lactose. What should they do next to draw a conclusion about whether the protein specifically binds lactose
Answer and Explanation:
Ideally, researchers would measure the stoichiometry of the mixture labeled with 14C-lactose and a protein. By measuring this stoichiometry, researchers should be able to find a concentration between lactose and protein of 1:1. This concentration is able to determine that the protein has specifically bound to lactose.
In summary, to determine whether the protein binds lactose, researchers need to calculate whether the concentration of these two elements within the mixture is 1:1.
What is the biological impact of minimum catch sizes on a population of fish?
a. The population comes to be dominated by smaller, slower-growing individuals.
b. Older, less productive adults are removed, improving the population's health.
It applies a selective pressure for larger, faster growing fish.
d. The fish in the population produce more and healthier eggs to compensate.
C.
Please select the best answer from the choices provided
Ο Α
OB
Ос
OD
Answer:
answer is A.) The population comes to be dominated by smaller, slower- growing individuals
As you read in this chapter, fungi have long formed symbiotic associations with plants and with algae. How may these two types of associations lead to emergent properties in biological communities
Answer:
Fungi show symbiotic association with algae and plants. With plants, they thrive as endophytes in a form of the symbiotic association. This symbiotic association results in the emergence of novel characteristics in the world of biology.
The lichens function as a tool in finding the quality of air, as they grow in the environment containing good air quality. The tolerance towards heat is another characteristic. Some of the endophytes are witnessed in the plants, which grow in very hot conditions.
At such conditions, no fungi or plant can thrive, however, in the symbiotic association, they possess the tendency to thrive. If one tries to separate them, it results in the death of both.
Explanation:
A postsynaptic neuron has three presynaptic inputs - from neurons X, Y and Z. When X, Y, and Z are stimulated simultaneously, the postsynaptic neuron reaches threshold and undergoes an action potential. What is this process called?
Answer:
Summation
Explanation:
Presynaptic neurons release chemical signals called neurotransmitters into the synaptic cleft in order to trigger graded potentials in the postsynaptic neuron. Some neurotransmitters produce excitatory post-synaptic potentials, whereas other neurotransmitters generate inhibitory post-synaptic potentials. Summation refers to the process that determines whether an action potential is generated by the combined effects of excitatory and/or inhibitory signals. The summation can be temporal or spatial regarding the number of cells communicating with the neuron. Temporal summation occurs when a presynaptic terminal is stimulated repeatedly in rapid succession. On the other hand, spatial summation occurs when excitatory potentials fire from different presynaptic neurons.
In this excerpt, a reader can conclude that Lizzie is playful based primarily on her
Answer:
based on Lizzie's words
Indicate whether each statement is true or false: 1. Compliance is the tendency for blood vessel volume to increase as blood pressure decreases. True 2. Blood vessels with a large compliance exhibit a small increase in volume when the pressure increases a small amount. True 3. Venous compliance is approximately 24X greater than arterial compliance, so as venous pressure increases the volume of veins greatly increases. True
Answer:
1. False
2. False
3. True
Explanation:
1. Compliance is the capacity of a container to increase in size to allow it hold more content. Blood vessel, arteries and veins expand (increase in volume) to be able to accommodate a surge in blood flow, which is as a result of an increase in pressure of the blood from the heart pumping of the blood
Therefore, compliance in the tendency for blood vessel volume to increase as the blood pressure increases not decrease
The statement is false
2. A large compliance is indicative of being highly sensitive to changes in pressure
Compliance, C = ΔV/ΔP
From the above equation, a blood vessel with a large compliance, exhibit a large increase in volume when the increase in pressure is small
Therefore, the statement 'Blood vessels with a large compliance exhibit a small increase in volume when pressure increases a small amount; is false
3. The compliance of the vein ranges from 10 to 20 times (30 times in some literature) greater than arteries. A factor which can be affected by the vascular smooth muscle contraction or relaxation
Therefore, the statement, 'venous compliance is approximately 24 times larger than arterial compliance, so as venous pressure increases the volume of veins greatly increases' is true
A 17-year-old student has experienced reversible, periodic attacks of chest tightness with coughing, wheezing, and hyperpnea. She states that expiration is more difficult than inspiration. She is most comfortable sitting forward with arms leaning on some support. X-rays revealed mild overinflation of the chest. Results from laboratory and pulmonary function tests are as follows:
• Frequency 20 breaths/min
• Vital capacity (VC) 2.9 L
• FEV1.0 1.4 L
• FEV1.0/FVC 56%
• Functional residual capacity (FRC) 3.89 L
• Total lung capacity (TLC) 6.82 L
• PaO2 70 mm Hg
• PaCO2 26 mm Hg
• Pulse 108 beats/min
• BP 120/76 mm Hg
Intermittent use of a bronchial smooth muscle dilator (1:1000 epinephrine by nebulizer) for several days caused marked improvement, resulting in the following laboratory and pulmonary function tests:
• VC 4.15 L
• FEV1.0 3.1 L
• FEV1.0/FVC >75%
• FRC 3.7 L
• TLC 5.96L
• PaO2 89 mm Hg
• PaCO2 38 mm Hg
• Pulse 129 beats/min BP 122/78 mm Hg
1. What is the disorder of this 17-year-old student?
2. Is this primarily a restrictive or an obstructive disorder? Why?
3. Write the formula for determining residual volume (RV).
4. Determine the residual volume (RV) before and after the use of the bronchodilator.
a. RV before using the bronchodilator:b. RV after using the bronchodilator:
5. Why is expiration more difficult than inspiration in this person?
6. What does the change in pulmonary function after the bronchodilator therapy indicate?
7. Why does the bronchodilator exaggerate the tachycardia?
8. What causes the hypoxemia and the hypocalcemia in this person?
9. A beta2-adrenergic agent was prescribed for further use because it has less cardio stimulatory (beta1) effect. Based on your knowledge of beta1 and beta2 receptors, why is this a good suggestion?
10. An anticholinergic agent was also suggested as a possible nebulizer agent. How might this helps the breathing problem?
Answer:
Frecuencia 20 respiraciones / min
Explanation:
Water droplets are pulled towards earth by ______________.
Starch and protein digestion in a single stomach?
Answer:
Explanation: Protein digestion occurs in the stomach and the duodenum through the action of three main enzymes: pepsin, secreted by the stomach, and trypsin and chymotrypsin, secreted by the pancreas. During carbohydrate digestion the bonds between glucose molecules are broken by salivary and pancreatic amylase.
In a hydrogen ion pump, the energy is used to join small molecules together
to
make larger ones. Which factor most likely has the greatest effect on the
number of molecules mitochondria can produce?
Answer: The number of H+ ions moving down the channel
Explanation:
What do you understand by each of the following terms
a.DNA nucleotide
b.RNA nucleotide
Which of these is a covalent bond in which the electrons are not evenly shared?
Answer:
polar covalent bond
Explanation:
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.
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), q² (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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Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L
Answer:
The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.
Explanation:
Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.
Nernst equation:
E = 58 millivolts/z. [Log₁₀ (C-out/C- in)
Where,
• E = Equilibrium potential
• 58 millivolts/z = Constant
• z = Ion charge + positive or negative symbol
• C-out = Ion concentration out of the cell
• C-In = Ion concentration inside the cell
By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.
Now let us see the provided values,
• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+
• C-out = Ion concentration out of the cell ⇒ 5 mEq/L
• C-In = Ion concentration inside the cell ⇒ 150 mEq/L
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)
E = 58 millivolts (Log₁₀ 30)
E = 58 millivolts (1.477)
E = 85.67 millivolts
85.7 mV is the absolute value of equilibrium potential.
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)
E = 58 millivolts (Log₁₀ 42.85)
E = 58 millivolts (1.63)
E = 94.65 millivolts
94.7 mV is the absolute value of equilibrium potential.
If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.
The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.
Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.
Nernst equation:
[tex]E = 58 millivolts/z. [Log₁0 (C-out/ -in)[/tex]
Where,
E = Equilibrium ability.58 millivolts/z =Constant.z=lon charge + advantageous or terrible symbol. C - out = Ion awareness out of the cell. C-ln= ion awareness in the cell.For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,
[tex]z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)[/tex][tex]E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.[/tex]
What happens if the extracellular attention of potassium is modified from 5.0 to 3.5?If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.
Thus it is clear from this that the potassium equilibrium potential is affected.
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How do endophytes affect grazing animals?
A. Endophytes pull nutrients from the soil, making plants more nutritious for grazing animals.
B. Endophytes that produce toxins can be poisonous to grazing animals.
C. Endophytes fix nitrogen from the air, making plants more nutritious for grazing animals.
D. Endophytes are poisonous to plants, depriving grazing animals of forage.
Answer:
B. Endophytes that produce toxins can be poisonous to grazing animals.
