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Natalka [10]
3 years ago
5

1. How does the principle of independent assortment apply to chromosomes?

Biology
1 answer:
Evgen [1.6K]3 years ago
6 0
<span>1) It is the chromosomes, however, that assort independently, not individual genes.

2) </span> <span>It tells you they're really far apart.
You see, there's this phenomenon called crossing over. Chunks of DNA get randomly swapped between homologous chromosomes. If two genes are close together they're usually swapped together and if they're far apart (say, on opposite ends) they're probably never going to be swapped together because half a chromosome doesn't normally cross over at once.
</span>
3)  <span>Sexual reproduction, because it results in offspring that combine alleles from two different individuals. </span>

<span>(Crossover is fine and dandy and you should mention it, but you'd get a C if you didn't mention that two different individuals are contributing genetic material to the offspring)</span>
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Hope This Helps! :)

Explanation:

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If this doesn't make sense let me restate it:

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Which of these Electromagnetic waves has the most kinetic energy and can be
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I believe

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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

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