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Vedmedyk [2.9K]
2 years ago
10

The human circulation system has approximately 1×109 capillary vessels. each vessel has a diameter of about 8 µm . assuming card

iac output is 5 l/min, determine the average velocity of blood flow through each capillary vessel.
Biology
1 answer:
aniked [119]2 years ago
4 0
It seems you mean 1x10^9 because there are a lot of capillaries indeed. The answer I got is v = 1.65788 * 10^9 meters / second

Given: 
vessel diameter = D = 8um
r = D/2 = 4um
v = velocity of blood flow is equal to cardiac output / total capillary area

So, v = Q/A

Where:
Q = cardiac output = 5 liters / minute
A = total capillary area = pi(r^2) multiplied by the number of capillaries

Solve A = pi(r^2) x 10^9
A = pi(4um^2) x 10^9
= pi(16um^2) x 10^9 = 50265000000 um^2

Convert to meters. You will get 50265 m^2 

Since velocity uses a meters/second unit format, you should convert Q to something more applicable. 

Q = 5 Liters / minute
= 5 L/minute x 0.001 cubic meter/ 1L 
= 0.005 m^3/minute 

Combine them all and you will get: 

v = Q/A = (0.005m^3 / min) / 50265 m^2 = 0.005 m^3 / (min)(50265 m^2) 

Then convert minute to seconds, and cancel out m^2 with m^3. 

v = 0.005m / 50265 minutes (1minute / 60 seconds) = 0.005m / [60(50265)] seconds

v = 0.00000000165788 meters / second
= 1.65788 * 10^9 meters / second

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q^2 = (2*17) / 114 = 0.236

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Freq(long plume allele) = p = 1 - q = 0.514

From this North American population we get

0.486 * 114 = 55 long plume alleles

0.514 * 114 = 59 short plume alleles

South American:

252 birds total => 504 alleles

252 - 75 birds = 177 short plume birds.

q^2 = (2*177) / 504 = 0.702

Freq(short plume allele) = q = 0.838

Freq(long plume allele) = p = 1 - q = 0.162

From this South American population we get

0.162 * 504 = 82 long plume alleles

0.838 * 504 = 422 short plume alleles

Blended population has

55 + 82 = 137 long plume alleles

59 + 422 = 481 short plume alleles

137 + 481 = 618 total alleles

p = 137/618 = 0.222

q = 481/618 = 0.778

The new population has the p and q from the blended population. The new population has 1000 individuals. The portion of long plumed will be homozygote long plumage + heterozygote, which is p^2 + 2pq, and you multiply that by the population size 1000 to get the final answer.

population size * (p^2 + 2pq) = 1000 * (0.222^2 + 2*0.222*0.778) = 395 birds (answer)

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Most marine animals live near the surface of the ocean because of ________, which supports photosynthesis by marine algae that f
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The sun they need energy
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In certain bacteria, an oval shape (O) is dominant over round (o) and thick cell walls (T) are dominant over thin (t). Draw on a
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Consider the heterozygous oval, thick cell walled bacteria to have the alleles OoTT and the thin cell walled bacteria to have alleles oott. Results will be 50% oval, thick walled bacteria and 50% round, thick walled bacteria. This will be the F1 progeny.

When the oval, thick walled bacteria from the F1 progeny is cross bred with round, thick walled bacteria then 25 percent of the bacteria will be heterozygous oval, thick walled. 25 percent will be heterozygous oval and heterozygous thick walled. 25 percent will be round and thick walled. 25 percent will be round and thin walled.

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Suppose that a late succession plant that is surrounded by early succession plants grows at a rate of 0.5 cm per day. If the ear
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Answer:

negative, inhibition

Explanation:

The experiment in the illustration shows that early succession plants have a <u>negative </u>effect on the late succession plants, and the observation is most consistent with the <u>inhibition</u> model of succession.

<em>That the late succession plant thrives better in the absence of the early succession plants means that the early succession plant has been impacting the growth of the late succession plant negatively. This is consistent with the inhibition model of succession.</em>

There are 3 different models of succession. These include;

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Answer:

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