This lab is really short and easy.start from Question #6NOTE: please use basic English/ Common language no advances and science words as it's necessary. The reason is because my professor would prefer using your own language and not copying out of the internet as he will deduct points. NOTE: answers should be short and simple/ not long just answer for the amount of question is being asked. Thank you In advanceQuestion 6. As the size of the cubes got smaller and smaller, what happened to the surface area : volume ratio?Question 7. Examine Figure 1. For each visible cube in the whole "organism", count the number of sides exposed to the outside. Write these numbers on the cubes in the diagram. Is there a cube with no sides exposed to the environment in Figure 1? Question 8. Locate the cube marked "a" in each of the three figures on page 6. Compare cube "a" in the three figures: how many sides of each "a" cube are actually exposed to the outside surface?Cube "a"Number of Sides ExposedFigure 1. 3 x 3 x 3 Figure 2. 2 x 2 x 2 Figure 3. 1 x 1 x 1 Question 9. Referring the Question 8, if each cube "a" were an equal amount of living tissue, which one would lose heat to a cooler environment fastest? Question 10. Referring to Question 8, if each cube "a" were an equal amount of living tissue, which one would need the fastest heat production to maintain homeostasis of its body temperature? (Remember, Cubus quadrangularis is a mammal.)Question 11. As an animal grows larger, what do you expect to happen to its rate of heat production per gram of body mass? Question 12. Based on the reasoning above, you will formulate a hypothesis for the following experiment. Small mice, medium-sized rats and large rats are available in the laboratory. We will measure their rate of oxygen uptake per gram body mass. What is the dependent variable we will measure?What is the independent variable?What are the groups being compared?Now, write the entire formula using the If…, and…, then… format.Question 13. Which mammal do you expect to have the fastest oxygen uptake per gram body mass? Explain why.Table 1. Metabolic rate measurements for 3 mammalswith different surface area to volume ratios.Group 1MouseYoung RatAdult RatBody weight (g):5 measurements of time required (in seconds) for animal to consume 5000 µL of Oxygen (O2):1.2.3.4.5.Average time (sec)to consume 5000 µL O2 :Step 1: Divide 5000 µL O2 by the average number of secondsrequired by animal to consume it:Step 2: Divide your answer, obtained in Step 1, by the animal’s body weight in grams. This will give you the metabolic rate in µL O2 per 1 second per 1 gram of animal. Group 2MouseYoung RatAdult RatBody weight (g):5 measurements of time required (in seconds) for animal to consume 5000 µL of Oxygen (O2):1.2.3.4.5.Average time (sec)to consume 5000 µL O2 :Step 1: Divide 5000 µL O2 by the average number of secondsrequired by animal to consume it:Step 2: Divide your answer, obtained in Step 1, by the animal’s body weight in grams. This will give you the metabolic rate in µL O2 per 1 second per 1 gram of animal. Group 3MouseYoung RatAdult RatBody weight (g):5 measurements of time required (in seconds) for animal to consume 5000 µL of Oxygen (O2):1.2.3.4.5.Average time (sec)to consume 5000 µL O2 :Step 1: Divide 5000 µL O2 by the average number of secondsrequired by animal to consume it:Step 2: Divide your answer, obtained in Step 1, by the animal’s body weight in grams. This will give you the metabolic rate in µL O2 per 1 second per 1 gram of animal. Group 4MouseYoung RatAdult RatBody weight (g):5 measurements of time required (in seconds) for animal to consume 5000 µL of Oxygen (O2):1.2.3.4.5.Average time (sec)to consume 5000 µL O2 :Step 1: Divide 5000 µL O2 by the average number of secondsrequired by animal to consume it:Step 2: Divide your answer, obtained in Step 1, by the animal’s body weight in grams. This will give you the metabolic rate in µL O2 per 1 second per 1 gram of animal. Table 2. Metabolic rate for all groups. MouseYoung RatAdult Rat Group: Mass (g)Average Metabolic Rate Mass (g)Average Metabolic Rate Mass (g)Average Metabolic Rate1 2 3 4 5XXXXXXXXXXXXXXXXXXXXXXXX6XXXXXXXXXXXXXXXXXXXXXXXX7XXXXXXXXXXXXXXXXXXXXXXXX8XXXXXXXXXXXXXXXXXXXXXXXXAverage Analysis QuestionsFill in the Table 3 with the class average data: Mammal typeAverage Mass (g)Average Metabolic Rate(µL O2 per second pergram of animal)Mouse Young Rat Mature Rate Question 14. Based on the class data, describe the relationship which appears to exist between a mammal's size and its metabolic rate. Question 15. Has your hypothesis been supported by the class data?If not, where did your reasoning go wrong?Question 16. Which would you predict to be higher, the metabolic rate of a sparrow or that of a gull? Explain your answer in terms of thermoregulation. Question 17. To what extent do you think it would be valid to apply this "metabolic rate to body size relationship" to animals other than mammals (clams, insects, fish, frogs, reptiles and birds)? Explain.Question 18. Is there a thermoregulatory advantage for marine mammals such as whales to be so large? Explain your answer.Question 19. Suggest a reason why there are no rat-sized marine mammals.Question 20. If an animal is an ectotherm, which is without internal physiological thermoregulation, how does it manage to survive when the weather is very hot or very cold? Give an example.

Make sure your response answers all parts of each question. Please make sure to have a core concept introductory paragraph for any of the questions which specifically reference one of our core concepts. 1-Using our core concept of flow-down gradients, explain the movement of glucose in and out of the nephron.2-Using our core concept of homeostasis, explain how the kidneys are involved in controlling fluid osmolarity.3-Compare the structure and functions of the major regions of the nephron.4-Compare the structure and function of cortical and juxtamedullary nephrons.5-Discuss the effect of renin, angiotensin, and aldosterone on blood pressure. (Be thorough, specific, and accurate).6-Explain how the vertical concentration gradient in the medulla and the concentration of vasopressin influences the volume and osmolarity of urine produced.7-Compare the mechanism and effect of loop diuretics, thiazide diuretics, ACE inhibitors, and ARBs (angiotensin II receptor blockers).8-Explain the four stages of external respiration and identify the gradients (driving force) and resistance of each stage. 9-Explain and give the formula for Boyle’s law and how changes in thoracic volume is related to Boyle’s law. 10-Compare tidal volume, expiratory reserve volume, inspiratory reserve volume, residual volume, vital capacity, total lung capacity, inspiratory capacity, and functional residual capacity. List the average values for each volume and capacity.11-Explain how local mechanisms allow for the matching of ventilation and perfusion. 12-Explain how the oxygen dissociation curve is related to the properties and function of hemoglobin.13-Discuss how changes in temperature, PCO2, pH and 2,3 DPG concentration shift the oxygen dissociation curve.14-Explain how oxygen dissociation curve shifts affect oxygen loading in the lungs and unloading at the tissues.

Use one of your three data tables to answer the questions. Inclination =  2°(9 points) 1.  Compare the slope of the distance vs. time graph to the average of all your velocity values. Are they close? Why or why not? What does the slope of a distance (or displacement) vs. time graph mean? Explain the answer using your data and include your Distance vs. Time graph and a chart of distance, time, and average velocity. Answer: The slope of the distance vs. time graph is 0.46 and the average of all of the averages is 0.50. Those two values are close. The slope of a time vs. distance formula indicates the average velocity of the ball. The slope formula (y2 – y1)/(x2 – x1) is pretty much same as calculating change in displacement (y2 – y1) over change of time (x2 – x1) and this gives the average velocity. Those two values are close because the trend line indicates how the data points will continue to be and since it is drawn properly it gives an idea of what the velocity the ball will continue to move in. Distance (m) Average time (s) Velocity (m/s) 0.5 0.89 0.56 1 2.01 0.5o 1.5 3.11 0.48 2 4.19 0.48  Average of the velocities :  0.50 m/s DISTANCE VS TIME GRAPH CHANGE LAYOUT(9 points) 2.  Describe the line on the velocity-time graph. What was the slope of the velocity vs. time graph? What does the slope of a velocity vs. time graph represent? Explain the answer using your data. In doing so, compare and contrast speed and velocity. Include your velocity-time graph in your answer. Answer: The line on the velocity-time graph was a curved one since the velocity was decreasing as time increased. The slope has a negative value of -0.03. This slope represents the average acceleration of the ball. The slope was negative because the ball was decelerating and coming down to a stop as it approached the  to 2m mark. Velocity and speed are different because speed is just a scalar quantity while velocity is a vector quantity that has magnitude and direction. TIME VS VELOCITY GRAPH CHANGE LAYOUT (7 points) 3.  Was the ball moving with uniform motion or did you notice some deceleration or acceleration? From what you have observed, explain what it means for an object to move with uniform motion. How does uniform motion relate to velocity and acceleration? Answer: The ball wasn't moving with a uniform motion. The calculated average velocity as well the the velocity time graph indicated that the ball was decelerating. For an object to move in a uniform motion it has to have either a constant velocity or a constant acceleration. If the object has a constant velocity then it is moving at zero acceleration and this means moving with out slowing down or speeding up. The object can also be said to have uniform motion if it accelerating at a constant rate. This means the velocity of the object changes by a constant number through out the journey. If the object is to continue moving with a uniform motion then the acceleration must not increase or decrease outside that constant rate.