Answer:
the pathway will be under-expressed.
- the alpha subunit helps to bind with either GDP or GTP. when the α subunit is bound with GDP, it will be bound to β and γ subunits and thus forms an inactive state for G-protein.
- when the alpha subunit binds with the GTP, it becomes activated and dissociates β and γ subunits.
if G-protein Coupled Receptor is unable from dissociating β and γ subunits, then the pathway will go under expression.
The chemical qualities of the alpha subunit allow it to bind easily to one of two guanine subunits, GDP or GTP. The protein thus has two functional formations. When GDP is bound to the alpha subunit, the alpha subunit remains bound to the beta-gamma subunit to form an inactive trimeric protein.
G-proteins, cAMP, and Ion Channel Opening. The alpha subunit activates adenylate cyclase, in purple, and loses GTP. Adenylate cyclase converts ATP to cyclic AMP, which then activates Protein Kinase, shown in blue. Protein Kinase phosphorylates an ion channel, letting sodium ions rush into the cell.
As a result of the ligand binding to its site on the G-protein-linked receptor, A) the G-protein changes conformation and GTP replaces the GDP on the alpha subunit. ... Inactivation of the alpha subunit occurs when its own phosphorylase activity removes a phosphate from the GTP.
I would go with B & D because they change over time and they can go extinct like those big octopus in the oceans, they are rare to see in my opinion because they can blend in with other things that keeps them away from danger.
Answer:
You will measure the rate of your pules and how fast its going.
You need to keep the thing you are doing for example( if your running you have to make the people run the same length as every one else.
I hope this helps :)
Explanation:
Answer:
d.0.48
Explanation:
When a population is in Hardy Weinberg equilibrium the <u>genotypic </u>frequencies are:
freq (AA) = p²
freq (Aa) = 2pq
freq (aa) = q²
<em>p</em> is the frequency of the dominant <em>A</em> allele and <em>q</em> is the frequency of the recessive <em>a</em> allele.
In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.
Therefore:
q²=16/100
q² = 0.16
q=√0.16
q = 0.4
And p+q=1, so:
p = 1 - q
p = 1-0.4
p = 0.6
The frequency of heterozygotes is:
freq (Aa) = 2pq = 2 × 0.4 × 0.6
freq (Aa) = 0.48
