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VARVARA [1.3K]
2 years ago
10

During ____, atoms arrange in a pattern over an over.

Biology
1 answer:
erma4kov [3.2K]2 years ago
5 0

Answer:

crystallization

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Answer: Animals undergoing examination should be restrained with just enough force to calm them and keep them still so that the veterinarian can provide proper veterinary care. Each animal's restraint needs will be different and dependent on their behaviors and anatomies.

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3 years ago
Give SOME EXAMPLES OF NATURAL CYCLE CHECKPOINTS .
Shalnov [3]

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For example, delays in mitosis are often ascribed to 'activation' of the mitotic checkpoint, a descriptor that fails to recognize that the checkpoint by definition is active as the cell starts mitosis. Conversely, the completion of mitosis in the presence of misaligned chromosomes is often automatically interpreted to indicate a defective checkpoint, even though in the absence of critical testing alternative interpretations are equally likely. In this article, we define the critical characteristics of checkpoints and illustrate how confusion generated by the inconsistent use of terminology may impede progress by fostering claims that mean very different things to different researchers. We will illustrate our points with examples from the checkpoint that controls progression through mitosis

Explanation:

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3 years ago
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The answer please will give brainliest
Talja [164]

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true

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3 years ago
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small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

p^{2} + q^{2} + 2pq = 1\\

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

7 0
4 years ago
Question 15 (5 points)
-BARSIC- [3]

Answer:

Both Anti A and Anti B bodies are present in the blood stream of person with blood group O

I hope this will help you:)

7 0
3 years ago
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