What is the average life span of juvenile upland game birds

Answers

Answer 1

Answer:

90 percent

Explanation:

Based on average estimated juvenile and adult survival rate for each species


Related Questions

Which of the following example(s) does NOT cause bias in a research study.
A) Measuring body temperature with an oral thermometer in half of subjects and an ear thermometer in others
B) Counting the number of firefighters at your house to predict if it will be destroyed
C) Using a set of standardized questions to measure food intake in children
D) Studying workers to see the effects of a new health care law on unemployed

Answers

(A) The internal body temperature can be taken from any area of the body without a difference in the type of temperature, resulting in NO bias. Of course this is completely reliant on the type of illness, but out of all options this is correct.

Illnesses affect the throat/mouth and ear.
checking the thermometer underneath a tongue will result in the same temperature as the ear, in an ill individual.

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.

Answers

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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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.

Answers

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 green

Parental) 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 = 27/64

There 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 = 9/64

And 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 = 3/64

Now, 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 = 9/64

Now, 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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What best describes the behavior of nonconservative elements in seawater?

Answers

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.

150 ml of 54% CaCI2 solution contains how many grams of CaCI2?

Answers

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?

Answers

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

A chemical reaction is an interaction that always causes:
O A. matter to be changed into energy.
B. macromolecules to be formed from glucose.
O C. molecules to break apart into atoms.
O D. atoms to be rearranged into new substances.

Answers

Answer:

D. atoms to be rearranged into new substances.

Explanation:

Chemical reactions involve interaction between chemicals such that all reactants are changed into new materials. The properties of the new materials are different from those of the reactants.

Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules). The number of atoms before and after the chemical change is the same but the number of molecules will change.

Specialized cells such as white blood cells interact with foreign particles in the body. A particular cell belonging to this category recognizes bacteria and forms a covering around it, pinching off the vacuole. The cell then performs . The foreign particles recognized by the particular type of cells differ from the ones recognized by another type of white blood cell. This situation indicates that the receptors are .

Answers

Answer:

specific

Explanation:

The immune system contains different types of cells that act in both innate and acquired (adaptive) immune responses to destroy pathogenic microorganisms that invade our body. The immune cells include B and T cells, dendritic cells, macrophages, neutrophils, eosinophils, monocytes, etc. These cells contain specific receptors that play different roles in the immune response. For example, macrophages and dendritic cells contain receptors required for antigen presentation, while white blood cells (B and T cells) have receptors required for antigen recognition.

Answer: I just took the test

Explanation:

Here's the answer

Uncellular eukaryotic organisms belong to?

Answers

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 :)

Which activity involves a chemical change?
A. Ice melting into liquid water
B. A metal rod being bent into a ring
C. Colorless egg becoming a white solid
D. Salt dissolving in water
SUBMIT
Need help now pls!!

Answers

option C is correct........

.. Please follow me and mark as brainliest

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

Answers

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

Answers

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.

difference between plant and animal cyanide resistance respiration.

Answers

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.

which describes a meta phase plate

Answers

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 effect is most likely to occur as a result of clear-
cutting?
extinction of species
death of organisms
speciation
global warming

Answers

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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What was the population density of the original sample? What would have happened if you had inoculated an agar plate with 1 mL of the original sample?

Answers

Answer:

The sample was at a population density of 8.75x10^4 CFU/mL originally.

Explanation:

Had it been inoculated in the way as described in the question statement, the bacterial growth would have covered all of the agar plate and we would not have been able to distinguish individual colonies of bacteria. 1 mL is too much of a sample volume to use in cases like these.

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