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Agata [3.3K]
4 years ago
14

Why is a key bed used?

Biology
1 answer:
Anna11 [10]4 years ago
6 0
In geology, a key bed (syn marker bed) is a relatively thin layer of sedimentary rock that is readily recognized on the basis of either its distinct physical characteristics or fossil content and can be mapped over a very large geographic area.[1] As a result, a key bed is useful for correlating sequences of sedimentary rocks over a large area. Typically, key beds were created as the result of either instantaneous events or (geologically speaking) very short episodes of the widespread deposition of a specific types of sediment. As the result, key beds often can be used for both mapping and correlating sedimentary rocks and dating them. Volcanic ash beds ( and bentonite beds) and impact spherule beds, and specific megaturbidites are types of key beds created by instantaneous events. The widespread accumulation of distinctive sediments over a geologically short period of time have created key beds in the form of peat beds, coal beds, shell beds, marine bands, black  in cyclothems, and oil shales. A well-known example of a key bed is the global layer of iridium-rich impact ejecta that marks the Cretaceous–Paleogene boundary (K–T boundary). Please let me know if it works.
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How many teeth are present in four months old child mouth​
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If a human somatic cell is in metaphase, it has __________ chromatids
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8 0
4 years ago
In trout, Gene A affects body color and yellow color (allele a) is recessive to red. In a particular river system you find that
hjlf

Answer:

The frequency of the dominant A allele is 0,4 or 40%.

Explanation:

In this population, we have:

- "A" red body color Dominant

- "a" yellow doy color Recessive

So, individuals AA and Aa will be red and trouts aa will be yellow.  In this case, 64% of the fish are red, so they are AA or Aa.

If the population is in Hardy-Weinberg equilibrium the following will be true:

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

Where:

p^{2}=frequency of AA

2pq= frequency of Aa

q^{2}=frequency of aa

It is known that 64% of population is red, in other words:

p^{2}+2pq=0,64

Or AA+Aa=0,64

So, it is possible to find the value of q^{2} (aa):

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

(0,64)+q^{2}=1

q^{2}=1-0,64q^{2}=0,36

Now, there are two alleles in this population, therefore their frequencies will be:

p+q=1

So, from q^{2} it is possible to find q and p:

q=\sqrt{q^{2} }

q=\sqrt{0,36}

q=0,6

And:

p+q=1\\p=1-q\\p=1-0,6\\p=0,4

7 0
3 years ago
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