The basic pattern of breathing is established by the DORSAL RESPIRATORY GROUP located in the MEDULLA OBLONGATA.
Answer:
Control group: volunteer stationary bike riders in a pollution-free chamber
Experimental group: volunteer stationary bike riders in a chamber filled with air pollutants common to Los Angeles
Independent variable: exposure to air pollution
Dependent variable: The heart rates (physical fitness) of the volunteer riders.
Explanation:
In a scientific experiment, data is compared between the control group and experimental group to ascertain the effect of the independent variable on the dependent variable. A control group is the group in an experiment that is not treated with the independent variable i.e. the independent variable is controlled while the experimental group is the group that is treated with an independent variable.
N.B: The independent variable is the variable controlled by the experimenter in order to influence the dependent variable (measurable outcome).
In this experiment, the independent variable is the AIR POLLUTION the volunteer riders are exposed to while the dependent variable is the PHYSICAL FITNESS/ HEART RATES of the volunteer riders, which is dependent on the exposure to air pollution (independent variable).
The control group in this experiment is the volunteer riders placed in a POLLUTION-FREE CHAMBER where the independent variable cannot influence or have any effect on the measurable variable (dependent). The experimental group is the volunteer riders placed in a CHAMBER FILLED WITH AIR POLLUTANTS. The independent variable (air pollution) has been changed in the experimental group.
Answer: option B) Secretin
Explanation:
Secretin is an enzyme formed by the upper intestinal layer and gets transported by blood to the PANCREAS which it stimulates: causing the flow of pancreatic juice, also the flow of bile and intestinal juice but to a lesser extent.
Alongside Gastrin, Secretin is one of the hormones of the Gastrointestinal tract; but their functions differ
Therefore, Secretin is the answer
Answer: 48.93 mL of sprite
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
Explanation:
