The follicle left after an egg is released during ovulation is called a(n)

In a population of 150 toads, poisonous (P) is dominant over non-poisonous (p). According to the Hardy-Weinberg equation, the ex

aalyn [17]

Answer: Χ²=8.18

Explanation: In this population, the frequency of homozygote poisonous toads is:

$p^{2} =\frac{100}{150}$

$p^{2} = 0.667$

$p = \sqrt{0.667}$

p=0.82

For the homozygote non-poisonous toads:

$q^{2} = \frac{5}{150}$

$q^{2} =0.034$

$q = \sqrt{0.034}$

q = 0.18

The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: $p^{2} + 2pq + q^{2} = 1$ , in which, heterozygote frequency is given by 2pq

2pq = 2*0.82*0.18 = 0.3

Now, to compute the chi-square test, follow the instructions:

1) Find the observed values: in this case, they are the found frequency:

0.82 0.3 0.18

2) Find the expected values: As the question mentioned, the expected proportion is:

16 8 1

3) Subtract the observed value from the expected value:

0.82 - 16 = - 0.184 0.3 - 8 = -7.7 0.18 - 1 = - 0.818

4) Square each value from above:

(-0.184)² = 0.034 (-7.7)² = 59.3 (-0.818)² = 0.77

5) Divide each value by expected value:

$\frac{0.034}{16}$ = 0.0021 $\frac{59.3}{8}$ = 7.41 $\frac{0.77}{1}$ = 0.77

6) Add all the values and we will have the chi-square test:

Χ² = 0.0021 + 7.41 + 0.77 = 8.18

The chi-square test is Χ² = 8.18

8 0

3 years ago

Simon has been researching the B vitamins and was reading information on how a niacin deficiency is common with those who are po

earnstyle [38]

Answer:

A

Explanation:

Chicken, Mushrooms, and Potatoes all have Niacin

7 0

3 years ago

What can be made from glycerol

I am Lyosha [343]

Soap can be made from glycerol :D Hope this helps!

7 0

3 years ago

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Circulatory systems compensate forCirculatory systems compensate fortemperature differences between the lungs and the active tis

erma4kov [3.2K]

Answer:

B) the slow rate at which diffusion occurs over large distances.

Explanation:

The complexity of the respiratory system is direct proportional with the size of the organism. As an organism size increases, diffusion begins to take place over a larger distance and the ratio of surface area to volume is seen to decrease. In unicellular organisms, diffusion across the cell membrane is adequate for distributing oxygen to the cell unlike in multicellular organisms.

Diffusion is known to be a slow, gradual and passive transport activity. It is important that for diffusion to be a pratical way of supplying oxygen to the cell, the amount of oxygen intake must be the same as the amount of diffusion across the membrane. This implies that, if the cell happens to be very large or thick, diffusion would not serve as the best means to distribute oxygen swiftly and in the right quantitiy to the inner region of the cell.

We can say that reliance on diffusion as a tool of supplying or distributing oxygen and extracting carbon dioxide is actually attainable only for small size organisms or the ones that possess a highly-flattened bodies, e.g. flatworms (Platyhelminthes).

Thus, this accounts for why Larger organisms had to develop specialized respiratory tissues, such as gills, lungs, and respiratory passages in conjunction with a complex circulatory systems, to disburse oxygen all through their entire body and to compensate for the slow rate at which diffusion occurs over large distances.

6 0

3 years ago

Describe how to find the hardness of an unknown mineral.

dsp73

The scratch test.

The scratch test is used by scientists and people who study rocks and minerals to determine the strength of the rock. They have the scratchers numbered one through ten and scratch the rock or mineral until the scratcher makes a mark in the mineral then they know the strength of the rock.

5 0

4 years ago

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