Answer:
Cardiac output = 20 L/min
Explanation:
The cardiac output of a person refers to the volume of blood the heart pumps in a minute. The normal range for cardiac output is about 4 to 8 L/min. However, this can vary depending on the metabolic needs of the individual at any instant of time, for example, it is higher when exercising than when resting.
Cardiac output is calculated from the product of the stroke volume in liters/min and the heart rate in beats/min
Cardiac output = stroke volume * heart rate
For the individual who is running; stroke volume = 100 mL/beat or 0.1 L/beat
heart rate = 200 beats/min
Cardiac output =0.1 L/beat * 200 beats/min
Cardiac output = 20 L/min
The sun rotates faster at its equator.
Answer:
Radiolabeled carbon atom in CO2
Explanation:
Photosynthesis is the process by which green plants fix the atmospheric CO2 into glucose. The process includes carbon fixation during which RuBisCo enzyme catalyzes the reaction of CO2 and a five-carbon compound called RuBP to form 3-phosphoglycerate (3-PGA). The 3-PGA enters the reduction phase of the Calvin cycle wherein it is reduced into glyceraldehyde 3-phosphate. Two molecules of glyceraldehyde 3-phosphate make one molecule of glucose.
To test the hypothesis that glyceraldehyde 3-phosphate from photosynthesis is used by plants to synthesize lipids, radiolabeled CO2 must be used. The radiolabeled carbon atom in the CO2 would be fixed in the form of glyceraldehyde 3-phosphate. If the plant uses glyceraldehyde 3-phosphate as a precursor for lipid synthesis, the synthesized lipid molecules would carry the radiolabeled carbon atom.
Answer:
b. the use of DNA as the information storage molecule
Explanation:
Prokaryotic cells are the ones that lack the membrane-bound organelles and well-defined nucleus. Eukaryotic cells have the nucleus and other membrane-bound organelles. Most of the prokaryotic cells are smaller in size as compared to eukaryotic cells. Despite these differences, both prokaryotic and eukaryotic cells have DNA as their genetic material. DNA serves to store genetic information in both types of cells.
Answer:
% of wolves have normal fur.
Explanation:
Given , the allele for white fur is recessive and the allele for normal fur is dominant
Let "N" represents the allele for normal fur and "n" represents the fur for white fur.
As per Hardy Weinberg's principle, the frequency for dominant allele is represented by "p"
Given ,
Then frequency for dominant genotype will be "
"
So, Frequency for wolves with normal fur is
Percentage of the wolves with normal fur is
%
