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?

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:

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:

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:

And we could also get two green seeds and only one yellow seed:

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:

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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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

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.

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 :)

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.

Answer:

The cyanide resistant respiration differs in plants as oxi­dation of reduced coenzyme continues even in the presence of cyanides.

Explanation:

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.

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."

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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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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