Answer:
a. inhibits cAMP phosphodiesterase, the enzyme that converts cAMP to AMP.
Explanation:
The phosphodiesterases are a group of enzymes that are capable of breaking phosphodiester bonds such as those between nucleotides in nucleic acids. The Cyclic adenosine monophosphate (cAMP) is a second messenger involved in intracellular signaling pathways. This molecule (cAMP) is hydrolyzed by the cAMP-dependent phosphodiesterase, which catalyzes the enzymatic breakdown of phosphodiester bonds (i.e., hydrolyzing cAMP to 5-AMP). The cAMP levels are modulated by the balance between its generation and the degradation via cyclic nucleotide phosphodiesterase.
I’ll like to answer but Where the question ?
It is A. Lung tissue i believe.
The magnitude of the work done by the electric field of the membrane is <u>W = 1.28 × 10⁻²⁰ Joules</u>.
We start with the necessity to take into account a value for the voltage present there in order to solve this problem by first considering that the membranes have two layers, one internal and one external, each responsible for producing a potential difference between the two levels.
As a result, in order to find a solution, it is necessary to take into account the potential difference between the two surfaces. In this instance, we'll assume a particular value for the load, but the recipient is free to substitute a different value if they prefer.
The product of the potential difference and the charge is used to define the work that an electric field performs. The charge of the potassium ion will be equal to that of its electron, so,
q = 1.6 × 10⁻¹⁹ Coulombs
Then the Work would be:
W = Vq
Here,
v = Potential difference
q = Charge
The 80mV potential difference we will have is quantified as follows:
W = (80mV (1V/1000mV))( 1.6 × 10⁻¹⁹ C)
W = 1.28 × 10⁻²⁰ Joules is the amount of work that the membrane's electric field has produced.
Find more on work done at : brainly.com/question/25573309
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