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inn [45]
4 years ago
12

Which of the following would be considered a limiting resource for a population?

Biology
1 answer:
riadik2000 [5.3K]4 years ago
3 0
All of the above, you need all of those to survive. Especially a community with a big population
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B interest rates are not just the same as we do not like that we have a new account and a dollar on our own account so you don’t want it for the first one and then
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3 years ago
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
What type of molecules commonly end in "-ase"
Leona [35]
Enzymes onto the end of substrates
  Example
preoxidase
telemorase
polymerase
4 0
4 years ago
Which of the following is not a cellular activity associated with microtubules?
sladkih [1.3K]

Answer:

A. maintenance of axons is not a cellular activity associated with microtubules.

Explanation:

Microtubules are hollow, bead-like, tiny tubular structure that helps cells maintain its shapes. Together with microfilaments and intermediate filaments, they form part of the cell's cytoskeleton. Microtubules also contributes to the cell movement or cytokinesis that includes muscle contractions in muscle cells. Microtubules also replicated chromosomes to opposite ends of a cell during cell division. Microtubules also contribute to the parts of the cell that help it move and are structural elements of cilia, centrioles and flagella. A bundle of microtubules makes up an axonemal structure of cilia and flagella.

3 0
4 years ago
To determine the density of a rabbit population, one would need to know the number of rabbits and __________. View Available Hin
serg [7]

Answer:

The growth rate.

Explanation:

To determine the density of a rabbit population, one would need to know the number of rabbits and the growth rate of rabbit's population because both factors allow us to determined the population of rabbit in a specific region. If we know the existing population of rabbits in a particular area as well as the growth rate at which the rabbit population increases so we can find out the density of rabbit population in a specific time period.

7 0
3 years ago
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