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Stells [14]
3 years ago
7

4. What is viscosity and how is it related to blood flow?

Biology
1 answer:
maria [59]3 years ago
3 0

Increasing evidence supports the importance of hemodynamic forces that are directly related to the work of the heart as the primary triggering event of atherosclerosis and atherothrombosis. The factors that primarily determine the work of the heart include systolic blood pressure (BP), blood viscosity, and the volume of blood the myocardium has to pump.

The relationship between BP and viscosity is such that, given a constant systolic BP, if blood viscosity increases, then the total peripheral resistance (TPR) will necessarily increase, thereby reducing blood flow. Conversely, when viscosity decreases, blood flow and perfusion will increase. Because of the dependence of systemic arterial BP on cardiac output and TPR, if blood viscosity and TPR rise, systolic BP must then increase for cardiac output to be maintained. Consequently, blood viscosity has been established as a major determinant of the work of the heart and tissue perfusion [1]. Since increased viscosity requires a higher BP to ensure the same circulating volume of blood, both the burden on the heart and the forces acting on the vessel wall are directly modulated by changes in blood viscosity.

Three important studies helped establish the relationship between blood pressure and blood viscosity. The earliest study observed 49 normal subjects and 49 patients with untreated essential hypertension, showing a direct correlation between BP and blood viscosity among both normotensive and hypertensive subjects (p<0.001). Systolic blood viscosity was 8 to 10% higher in hypertensive patients compared with normotensive controls, and diastolic blood viscosity was 16 to 28% higher in hypertensive patients. Subgroups each comprised of 25 subjects having matched hematocrits were also compared, and viscosity remained significantly higher in hypertensive subjects (p<0.05). [2]

plz mark me as brainliest :)

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store energy in organic compound

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The smallest among these 4 groups is <em>O</em><em>r</em><em>d</em><em>e</em><em>r</em><em> </em>.

Explanation:

The <em>a</em><em>r</em><em>r</em><em>a</em><em>n</em><em>g</em><em>e</em><em>m</em><em>e</em><em>n</em><em>t</em><em>s</em><em> </em>for classification are <em><u>K</u></em><em><u>i</u></em><em><u>n</u></em><em><u>g</u></em><em><u>d</u></em><em><u>o</u></em><em><u>m</u></em><em> </em><em>,</em><em> </em><em><u>P</u></em><em><u>h</u></em><em><u>y</u></em><em><u>l</u></em><em><u>u</u></em><em><u>m</u></em><em> </em><em>,</em><em> </em><em><u>C</u></em><em><u>l</u></em><em><u>a</u></em><em><u>s</u></em><em><u>s</u></em><em> </em><em>,</em><em> </em><em><u>O</u></em><em><u>r</u></em><em><u>d</u></em><em><u>e</u></em><em><u>r</u></em><em> </em><em>,</em><em> </em><em><u>F</u></em><em><u>a</u></em><em><u>m</u></em><em><u>i</u></em><em><u>l</u></em><em><u>y</u></em><em> </em><em>,</em><em> </em><em><u>G</u></em><em><u>e</u></em><em><u>n</u></em><em><u>u</u></em><em><u>s</u></em><em> </em><em>,</em><em> </em><em><u>S</u></em><em><u>p</u></em><em><u>e</u></em><em><u>c</u></em><em><u>i</u></em><em><u>e</u></em><em><u>s</u></em><em> </em><em>.</em>

Easier way to memorize is :

<em><u>K</u></em><em>i</em><em>n</em><em>g</em><em> </em><em><u>P</u></em><em>h</em><em>i</em><em>l</em><em>l</em><em>i</em><em>p</em><em> </em><em><u>C</u></em><em>a</em><em>m</em><em>e</em><em> </em><em><u>O</u></em><em>v</em><em>e</em><em>r</em><em> </em><em><u>F</u></em><em>o</em><em>r</em><em> </em><em><u>G</u></em><em>o</em><em>o</em><em>d</em><em> </em><em><u>S</u></em><em>p</em><em>a</em><em>g</em><em>h</em><em>e</em><em>t</em><em>t</em><em>i</em><em> </em><em>.</em>

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I’m having trouble understanding this can someone plz explain?
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Answer:

look at the pictures

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1. If we cross the parental generation which is RR x rr, the probability of the F1 generation to get  rough coat is 100% and they will be a carrier of the recessive trait smooth coat.

Since the phenotypic ratio is 100% heterozygous Rr, in crossing the F1 to get the F2, we will use the genotype Rr. To get the F2, use the cross  Rr x Rr.

The phenotypic ration for F2 is 3:1. There is 75% to get rough coat and 25% smooth. The answer is based on the result on the Punnett square. On the other hand, the genotypic ratio is 1:2:1. There is 25% probability to get RR genotype, 50% Rr, and 25% rr.

2. Since the two parental mice got 6 albino offspring and 5 brown mice offspring, it is approximately 50%. it takes a parental mice who is Brown that is carrying an albino trait crossed with an albino to get offspring with almost the same number. Therefore, the genotype of the brown mice is Aa.

8 0
2 years ago
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