Answer:
<em>when</em><em> </em><em>an </em><em>organism</em><em> is</em><em> </em><em>covered</em><em> by</em><em> </em><em>sediment</em><em>,</em><em>it </em><em>begins</em><em> to</em><em> </em><em>decay,</em><em> when</em><em> </em><em>decaying</em><em>,</em><em>the </em><em>organism</em><em> </em><em>releases</em><em> </em><em>a </em><em>gas</em><em>.</em><em>T</em><em>h</em><em>i</em><em>s</em><em> </em><em>process</em><em> </em><em>in </em><em>turns </em><em>leaves</em><em> </em><em>a </em><em>thin </em><em>film </em><em>of </em><em>carbon</em><em> </em><em>that</em><em> </em><em>preserves </em><em>part </em><em>of</em><em> the</em><em> </em><em>organisms.</em><em>T</em><em>h</em><em>e</em><em> </em><em>hallow </em><em>that</em><em> </em><em>forms</em><em> </em><em>on </em><em>the</em><em> </em><em>sediment</em><em> </em><em>is </em><em>the</em><em> </em><em>carbon</em><em> </em><em>film </em><em>that</em><em> </em><em>contain</em><em> </em><em>fossils</em><em> </em><em>of </em><em>the</em><em> </em><em>organism</em><em>. </em>
Answer:
The correct answer is C) 8 exons.
Explanation:
If we assume that each domain matches with an exon, and we have a 6 domain protein (domain= the functional regions of the protein that include the active site), there must be at least 6 exons in the gene. Also, remember that exons include the 5'- and 3'- untranslated regions or UTR. So, assuming that each unstraslated region is equivalent to an exon, there must be at least 8 exons in the gene (6 exon for each domain and one 2 exons for UTR).
Anyway, be careful, because this is a simple exercise where we assume a lot of things. In the real world, the first exon tends to include the 5'-UTR and it is already proved that one exon is not always equivalent to one protein domain.
Answer:
24555 years
Explanation:
Data provided in the question:
Half life of carbon 14 = 5700 years
carbon present in the dead man = 5% of the original amount of carbon 14
Now,
let the initial Amount of carbon be 'A'
Therefore,
after half life
Amount of carbon = 
Thus,
= 
here,
k is the decay constant
t is the time = 5700 years
therefore,
1 = 
or
0.5 = 
taking natural log both the sides
-0.69314 = 5700k
or
k = - 0.000122
therefore,
for dead man Amount of carbon = 5% of A = 0.05A
thus,
0.05A = 
or
0.05 = 
taking natural log both the sides
we get
-2.995 = - 0.000122 × t
or
t = 24555.18 ≈ 24555 years
<span>One way classification is used at school is the separation of different subjects and subject materials. Kind of like, one day you may go to a math class, while another day you may go to a french class.
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C; Earthquakes and volcanoes
Faith xoxo