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ziro4ka [17]
3 years ago
15

The autorhythmicity of cardiac pacemaker cells is made possible by the reduced permeability of _______. The autorhythmicity of c

ardiac pacemaker cells is made possible by the reduced permeability of _______. magnesium sodium calcium potassium
Biology
1 answer:
vichka [17]3 years ago
7 0

Answer:

<h2>potassium</h2>

Explanation:

The heartbeats of the heart are controlled by the sinoatrial node also known as the pacemaker of the heart. The impulse is generated in the conducting cells of the pacemaker as a result of the movement of sodium, potassium and calcium ions.

The sodium channels allow the movement of sodium into the cell which depolarizes the membrane from -60mv to -40 mv. At this point, the calcium gated channels open which allow the entry of the calcium in the cell which depolarizes the cell up to +5mv.

At this point, the potassium channels open which allows the potassium ions to move out of the cell. This repolarizes the cell and hence the cycle again begins.  Reducing the permeability of the potassium ions help generate the autorhythmicity due to repolarization and thus is the correct answer.

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Which process plays an important role in the biogeochemical cycles of carbon, nitrogen, and phosphorus?
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6 0
3 years ago
In gorillas, the ability to roll the tongue is under the control of 1 gene. The R allele, which confers tongue-rolling ability,
aalyn [17]

Answer:

0.4550

Explanation:

In this example, R allele is dominant, so individuals RR and Rr can roll their tongues. If in a population of 1000 gorillas, there are 575 gorillas who can roll their tongues, they will be RR and Rr.

In this case, there is Hardy Weinberg Equilibrium, so the following will be true:

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

Where:

p^{2}= frequency of RR

2pq= frequency of Rr

q^{2}= frequency of rr

The question is what is the frequency of heterozygotes, or, what is the value of 2pq.

We know that RR+Rr is 575 individuals in a population of 1000, or 0.575.

In other words:

p^{2}+2pq=0.575

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

q^{2}= 1-2pq-p^{2}  \\q^{2}=1-(0.575)\\q^{2}=0.425

Now, there are two alleles in the population, so the following will be true:

p+q=1

It is possible to find q  (the frequency of allele r) and p (the frequency of allele p):

q=\sqrt{q^{2} }\\ q=\sqrt{0.425} \\q=0.652

Therefore:

p=1-q\\p=1-0.652\\p=0.348

Now, the frequency of heterozygotes or 2pq is:

2pq=2*0.652*0.348\\2pq=0.45

7 0
3 years ago
Please help me thank you
trapecia [35]

Answer:

The punnet square would look like this.

R R

R RR RR

B RB RB

6 0
3 years ago
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