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3 years ago

REALLY NEED HELP

1 answer:

4 0

T cell activation is initiated by an activated B cell.

The B cell presents antigens on its MHC receptors. The T cell recognizes the antigen bound to the MHC. There are other cosignals that fully activate the at cells but the antigen on the MHC receptor on the B cell is where it all starts.

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PLEASE HELP ME ASAP!!!!!!!!!!!! GIVING 15 POINTS

m_a_m_a [10]

Contemporary landscapes result from many causes, including variability in abiotic conditions, such as climate, topography, and soils; biotic interactions, such as competition, mutualism, herbivory, and predation, that can generate spatial pattern even when environmental conditions are homogenous; natural disturbances. I hope this helps :)

6 0

3 years ago

assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar

svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

$E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}} \Big )$

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

$E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}$

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

$100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\ 53.57 = \dfrac{4}{[K^+]_{in}}$

$[K^+]_{in} = \dfrac{4}{53.57}$

$[K^+]_{in} = 0.0476$

For [Cl⁻]:

$100 \times 10^{-3} = -0.0257 \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)$

$-3.981 = \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)$

$0.01867 = \dfrac{120}{[Cl^-]_{in}}$

$[Cl^-]_{in} = \dfrac{120}{0.01867}$

$[Cl^-]_{in} =6427.4$

For [Na⁺]:

$100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}} \Big)$

$53.57= \Big ( \dfrac{145}{[Na^+]_{in}} \Big)$

$[Na^+]_{in}= 2.70$

6 0

3 years ago

Explain why the gametes have only one allele for a trait

zzz [600]

A gamete contains one set of chromosomes

6 0

3 years ago

Suppose there are 500 units of energy available at the producer level of the energy pyramid. Approximately how many units of ene

Salsk061 [2.6K]

With every other lever 10% of the energy with be transferred. In this case:

Primary Producers (500)
1) First- level consumer. (50)
2) Second-level consumer. (5)
3) Third- level consumer. (0.5)
Remember the formula is:
(Primary Producer) __units X 10%= __
In this case: 500 units X 10%= 50

5 0

3 years ago

What is the name of the process by which decomposers convert organic nitrogen to ammonia? A) Ammonification. B) Denitrification.

Soloha48 [4]

Answer:

Ammonification

Explanation:

6 0

3 years ago