The given question is incomplete due to missing options, however, the options for the question as follows:
A. Nonconservative elements are reactive in seawater and have a long residence time.
B. Nonconservative elements are reactive in seawater and have a short residence time.
C. Nonconservative elements are non-reactive in seawater and have a short residence time.
D. Nonconservative elements are non-reactive in seawater and have a long residence time.
Answer:
The correct answer is option C.
Explanation:
Non conservative components are components that enter the ocean water and have little concentration in the sea and show spatial variations. These components have short home time that is short replacements time and they are non responsive.
Constituents, for example, phosphate, nitrate, and different supplements, and broke up oxygen, carbon dioxide, and so on are non-conservative on the grounds that their concentrations are later altered by chemical reactions in the ocean.
Thus, The correct answer is option C.
What is an advantage of nuclear power plant over coal power plant?
cost and lack of nuclear waste
Explanation:
Use the completed Punnett square in Part B to answer the questions below about the F2 generation.
Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.
Complete question:
Part C: Using the Punnett square to make predictions
Use the completed Punnett square in Part B to answer the questions below about the F2 generation.
Note: You will find the Punnet square in the attached files.
Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.
1) What is the probability that an F2 seed chosen at random will be yellow?
2) What is the probability that an F2 seed chosen at random from along the yellow seeds will breed true when selfed?
3) What is the probability that three F2 seeds chosen at random will include at least one yellow seed?
4) What is the probability that three F2 seeds chosen at random will include one green seed and two yellow seeds?
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
Y is the dominant allele and expresses yellow color y is the recessive allele and expresses green colorSeeds with YY and Yy genotype are yellowSeeds with yy genotype are greenParental) Yy x Yy
Gametes) Y y Y y
Punnet Square)
(1/2) Y (1/2) y
(1/2) Y (1/4) YY (1/4) Yy
(1/2) y (1/4) yY (1/4) yy
F2) 1/4 YY
2/4 = 1/2 Yy
1/4 yy
1) There will be 3/4 of probabilities of getting a yellow F2 seed
1/4 YY + 2/4 Yy = 3/4 Y-
The symbol "-" means that in its position there might be either a Y or y allele.
2) As there are only three yellow possible genotypes for yellow seeds, the fourth genotype for green seeds is not considered. Our pool now is only yellow seeds. So the probability of getting a pure breeding yellow seed among all yellow seeds is 1/3.
1/3 YY
1/3 Yy
1/3 yY
In the pool of yellow seeds, there are three genotypes, and only one of them corresponds to pure breeding yellow seed.
3) To calculate the probability of getting at least one yellow seed among the three seeds randomly chosen, we need to multiply and then perform addition. This is, when we take three seeds from the pool, we have different possibilities of getting seeds with different genotypes. So:
the three chosen seeds could be yellow. In this case, we multiply their probabilities. The probability of getting one yellow seed is 3/4, so, the probabilities of getting three yellow seeds are: 3/4 x 3/4 x 3/4 = 27/64There might also be one green sees and two yellow seed, so we multiply their probabilities in the order in which we might get them. This is:
Two yellow seeds and one green seed: 3/4 x 3/4 x 1/4 = 9/64One yellow seed, one green seed, and one yellow seed: 3/4 x 1/4 x 3/4 = 9/64one green seed, and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64And we could also get two green seeds and only one yellow seed:
Two green and one yellow seed: 1/4 x 1/4 x 3/4 = 3/64one green, one yellow and one green: 1/4 x 3/4x 1/4 = 3/64one yellow seed and two green seeds: 3/4 x 1/4 x 1/4 = 3/64Now, having all the possibilities, we just need to add all the possibilities:
27/64 + 9/64 + 9/64 + 9/64 + 3/64 + 3/64 + 3/64= 63/64.
Note that we did not include the possibility of getting three green seeds (1/4 x 1/4 x 1/4). This is because we were asked to calculate the probabilities of getting AT LEAST one yellow seed.
4) To calculate the probability of getting two yellow seeds and one green seed among the three seeds randomly chosen, we should multiply probabilities and then sum them up.
The orders in which we could get the seeds are:
Two yellows and one green: 3/4 x 3/4 x 1/4 = 9/64One yellow, one green, and one yellow: 3/4 x 1/4 x 3/4 = 9/64one green seed and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64Now, having all the possibilities, we just need to add all the possibilities:
6/64 + 9/64 + 9/64 = 27/64.
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
1) yellow seed could either be GG, Gg or Gg from Punnett square, therefore, there is a 3/4 probability
2) true-breeding = either GG or gg. Therefore, out of the 3 yellow seed options, only 1 is true-breeding, therefore, it's a 1/3 probability
3) "there are 7 possible groupings that have at least 1 yellow seed; the sum of their individual probabilities yields the final answer of 63/64. Note that you could have also calculated this answer by subtracting the probability of the one category that doesn’t fit the criterion (Green, Green, Green) from the total probability (1 – 1/64 = 63/64)"
4) "there are 3 possible groupings that have 1 green and 2 yellow seeds; the sum of their individual probabilities yields the final answer of 27/64"
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150 ml of 54% CaCI2 solution contains how many grams of CaCI2?
Answer:
mass of CaCl₂ = 67.4 g of CaCl₂
Explanation:
A 45% CaCl₂ Solution contains 45 g of solute per 100 mL of solution.
mass concentration = mass of solute (g)/ volume of solution (L)
100 mL of solution = 100 * 1L /1000 = 0.1 L
Mass concentration of solution = 45 g/0.1 L = 450 g/L
Molar concentration of solution = mass concentration / molar mass
molar mass of CaCl₂ = 111 g/mol
molar concentration = 450 g/L / 111 g/mol = 4. 05 mol/L
Number of moles of CaCl₂ present in 150 ml of 4.05 mol/L solution = molar concentration * volume (L)
number of moles of CaCl₂ = 4.05 * 150 * 1 L/1000 = 0.6075 moles
mass of CaCl₂ present in 0.6075 moles = number of moles * molar mass
mass of CaCl₂ = 0.6075 * 111 = 67.4 g of CaCl₂
If the scientist had collected 20 dung samples, would you expect more bands, fewer bands, or the same number of bands on the gel?
Answer:
The expectations of the bands will be more on the gels.
Explanation:
One sample provides several sizes and thickness bands in the gel electrophoresis according to the size of fragments that run in it. An individual would expect more bands due to the increase in the number of samples which is 20 samples now.
An increase in the number of samples produces more bands. So one can easily expect more bands from the 20 samples. Scientists would get more bands on the gels.
Thus, the expectations of the bands will be more on the gels.
If the scientist collects 20 dung samples ; we should expect more band on the gel
Given that each dung sample has varying sizes and varying thickness of bands in the gel electrophoresis which depends on the size of fragments contained in it. Therefore when a number of dung samples ( 20 ) are collected it is only logical that the number of bands on the gel will increase ( i.e. you should expect more band on the gel ).
Hence we can conclude that If the scientist collects 20 dung samples we should expect more band on the gel .
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Although your question lacks some data a general answer within the scope of your question is provided
Which center bears the primary responsibility for generating the respiratory rhythm, but is influenced by several other centers
Answer:
This is the Ventral Regulatory Group-VRG
Explanation:
In the medulla,there are four groups of neurons which forms the exhalation parts of the respiratory control.These 4 groups are called the VRG.They occupy 5mm anterior in length, and laterally positioned to the dorsal respiratory group.They are therefore said to occupy the ventrolateral part of the medulla.
The dorsal respiratory group(DRG) consists of the inspiration neurons,contrary to the VRG. They do not extend to the DRG,but the DRG, extends to them
They(VRG) function in propagating action potentials needed for the mechanisms of respiration.(exhalations), and Inspirations
difference between plant and animal cyanide resistance respiration.
Answer:
The cyanide resistant respiration differs in plants as oxidation of reduced coenzyme continues even in the presence of cyanides.
Explanation:
The mitochondrial is the electrons through which the animals and plants perform the aerobic exercise and is blocked by the release of the cyanides that stops the activity and is known as cyanide sensitive respiration Plant mitochondria is different from the animal as they both have different oxidase system pathways. Cyanide resistant is responsible for the climacteric in fruits and for the generate heat in thermogenic tissues.A compound synthesized by bacteria or fungi that destroys or inhibits the growth of other microbes is a/an:__________.
a. synthetic drug
b. antibiotic
c. antimicrobial drug
d. competitive inhibitor
Answer:
antimicrobial drugs
Explanation:
An antimicrobial is a substance that kills or inhibits the growth of microorganisms such as bacteria, fungi, or protozoans.
Answer:
[tex]\boxed{Option C}[/tex]
Explanation:
A compound synthesized by bacteria or fungi that destroys or inhibits the growth of other microbes is an anti-microbial drug. Anti means "against" so anti microbes mean against microbes. Antimicrobial drugs are synthesized by bacteria and fungi and are used against microbes.
which describes a meta phase plate
Answer:Metaphase plate refers to sister chromatids lining up in the center of the cell. This event occurs during the metaphase of the cell division and it is the indication of this phase. Metaphase plate is a moment when all of the chromosomes organize, line up literally in a row and after that begin to.
Which statement describes one aspect of an earthquake's magnitude?
It is measured by the Mercalli scale.
It is based on the size of seismic waves.
O It is not usually affected by the amount of fault movement.
It is used to determine damage caused by earthquakes.
Answer:
b. It is based on the size of seismic waves
Explanation:
just did the test
The statement that describes one aspect of an earthquake's magnitude is: "It is based on the size of seismic waves."
What is an earthquake's magnitude?An earthquake's magnitude is a measure of the amount of energy released by the earthquake, and it is determined by measuring the amplitude (height) of the seismic waves generated by the earthquake. The size of the seismic waves is directly related to the amount of energy released by the earthquake, and this is used to calculate the earthquake's magnitude.
The most commonly used scale for measuring earthquake magnitude is the Richter scale, which is based on the amplitude of the largest seismic wave recorded on a seismogram. Other scales, such as the moment magnitude scale, are based on the seismic moment, which takes into account the size of the fault rupture and the amount of slip on the fault.
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Which effect is most likely to occur as a result of clear-
cutting?
extinction of species
death of organisms
speciation
global warming
Answer:
Death of organism
Explanation:
Human activities include the clear-cutting of areas of forest to make space for farming.
Which effect is most likely to occur as a result of clear-cutting?
Is this your question?
Death of organisms effect is most likely to occur as a result of clear-cutting.
What do you mean by clearcutting?Clearcutting is an extreme logging method in which resilient natural forests are harvested and replaced with man-made tree plantations that do not replicate the ecosystem services of a healthy forest.
Clearcutting is one of the most publicly maligned and misunderstood forest regeneration treatments. It is the only way to grow trees that require full sunlight to reproduce, such as tulip poplar, black cherry, and aspen, and the fastest way to provide/create shrubland habitat.
Clear-cutting often contributes to reductions in root strength and soil water-holding capacity, due to soil compaction and reduced transpiration. Moreover, the removal of the forest cover exposes the soil surface to heavy precipitation and large variations in temperature.
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Uncellular eukaryotic organisms belong to?
All eukaryotic unicellular organisms belong to Protista.
Answer:
Unicellular eukaryotic organisms belong to kingdom protista.
Explanation:
Unicellular eukaryotic organisms belong to kingdom protista. They depend on other organisms for their food since they're not able to produce their own.
Hope this helps :)
An athlete has an enhanced genotype for carbohydrate utilization, however they feel sluggish after they eat carbohydrates, which of the following should the trainer consider?
a. The type of carbohydrates being consumed
b. Decrease the carbohydrate intake
c. Add more caffeine to the daily diet
d. They are likely over trained and tired
Answer:
Carbohydrates’ bad reputation ignores the benefits they can offer
sooo i would say it is B. i think....
The first glycolysis rate-limiting enzyme is under various allosteric regulations. This protein is nearly inactive because of inhibition by ( ) under physiological conditions. The activity is restored by ( ), its most potent allosteric activator.
a. AMP :::: citrate
b. AMP :::: Fru-2,6-P2
c. ATP :::: citrate
d. ATP :::: Fru-2,6-P2
e. All of these
Answer:
d. ATP; Fructose-2,6-bisphosphate
Explanation:
Phosphofructokinase-1 is the enzyme that catalyzes the formation of fructose-1,6-bisphosphate from fructose-6-phosphate and ATP. The phosphofructokinase step is the first rate-limiting step of glycolysis.
Phosphofructokinase-1 activity is allosterically regulated. Its activity is increased whenever the cell's ATP supply is depleted or when its breakdown products, ADP and AMP accuulates in he cell. However, it is inhibited when the cell is amply supplied with ATP.
The activity of phosphofructokinase-1 is restored by fructose-2,6-bisphosphate, its most potent activator.
From the given options:
a. AMP :::: citrate is wrong because AMP increases the activity of phosphofructokinase-1 and citrate is not its activator but an inhibitor
b. AMP :::: Fru-2,6-P2 is wrong because AMP increases its activity same as fructose-2,6-bisphosphate
c. ATP :::: citrate is wrong because both citrate and ATP are inhibitors
d. ATP :::: Fru-2,6-P2 is correct as explained above
e. All of these is wrong because of the other wrong options above.
As Hite and Seitz (2016) discuss in Chapter 5: Climate Change, reputable scientific studies continue to show that the Earth is warming at an unsustainable rate because of a rise in the release of CO2 and methane gases. What is the relationship between population increases and climate change
Answer:
See the answer below
Explanation:
The relationship between climate change and increased population of humans is a direct one. As the population of humans increases, their activities leading to the release of greenhouse gases such as CO2 and methane also increases. The greenhouse gases are primarily responsible for the warming of the earth.
As more and more greenhouse gases are released into the atmosphere, more radiation from the sun is trapped in the atmosphere and the earth receives more warmth. This extra heat received leads to increase in the global temperature, a phenomenon known as global warming. As the average global temperature increases, climate becomes more erratic and unpredictable.
The greenhouse gases-related activities of humans that are bound to increase with the increase in population include deforestation, fossil fuel burning, industrialization, etc.