BILD 2 Fall 2008 Dr. Towb Homework #1 10/13/08 1) Of the carbohydrates, fats, and proteins in a hamburger, what has been partially digested prior to entry into the small intestine? Carbohydrates are partially digested in the mouth by salivary amylase, and proteins are partially digested in the stomach by pepsin. 2) Why are there different proteases during the different stages of digestion? Won’t just one do the trick? There are different proteases because proteins are comprised of a combination of various amino acids. Proteases as enzymes each recognize a specific substrate which is generally a specific region (specific amino acid residues) on the protein substrate. In order to fully digest the protein, multiple different proteases that can cleave a protein at different amino acid residues are required. Also, different proteases are required to digest proteins in the different environments of the digestive organs (proteases that function optimally in acidic environments in the stomach and proteases that function optimally in the more neutral environment of the small intestine) 3) What does the acidic environment of the stomach do to ingested proteins? How does this aid protein digestion? The acidic environment aids in protein digestion by denaturing proteins (unfolding them so that they can be easily cleaved by proteases). 4) What causes the pH change in the “digestive broth” that is released from the stomach to the small intestine? Why is there a pH change at this point in human digestion? (What purpose does it serve?) Bicarbonate (HCO3-) in an alkaline solution secreted by the pancreas neutralizes the acid chyme released from the stomach to the small intestine. The neutralization raises the pH from 2 to around 7, and allows various digestive enzymes secreted by the pancreas and lining of the small intestine to function (they work at a pH near 7). 5) Cells must be bathed continuously in an aqueous medium to take in oxygen and nutrients and get rid of waste products via diffusion. Diffusion is efficient only over short distances. In fact, diffusion is efficient only for a distance of about three cell diameters maximum (approx. 200-300µm). Note the following times required for oxygen to diffuse specific distances: Diffusion Distances (µm) Time required for diffusion 1 0.5 msec 10 50 msec 100 5 sec 1000 (1mm) 8.3 min 10,000 (1cm) 14 hr a. From the table calculate the mathematical relationship between the increase in distance and the corresponding increase in time required to diffuse that distance.For each tenfold increase in diffusion distance there is a corresponding hundred-fold increase in the time required for diffusion to occur. Therefore, the time required for diffusion increases as the square of the distance diffused. b. How much time would be required for oxygen to diffuse 5µm? 200µm? 52= 25 time required to diffuse 1 µm is 0.5 msec, so for 5 µm it is 25(0.5 msec) = 12.5 msec similarly, it takes 20,000 msec or 20 sec to diffuse 200 µm 6) What happens to the surface area-to-volume (SA/V) ratio of a three dimensional object (such as a cell) as its linear dimension increases? For example, how does the SA/V ratio a sphere or cube change as the linear dimensions increase? (Hint: choose some sizes like 1mm in diameter, 1cm, and 10cm and calculate the ratios.) Formulas for a sphere: surface area = 4r2; volume = 4/3r3 1mm diameter: r = 0.05cm SA = 4(0.05cm)2 V = 4/3(0.05cm)3 = 0.01cm2 = 1.67 x 10-4cm3 SA/V = 60cm-1 1cm diameter: r = 0.5cm SA = 4(0.5cm)2 V = 4/3(0.5cm)3 = 1cm2 = 0.167cm3 SA/V = 6cm-1 10cm diameter: r = 5cm SA = 4(5cm)2 V = 4/3(5cm)3 = 100cm2 = 166.67cm3 SA/V = 0.6cm-1 The surface area-to-volume ratio decreases as by 10 as the linear dimensions increase by 10. 7) Since heat is generated by all cells in an organism, but is lost to the outside world only through the surface in contact with the outside, which of these two imaginary organisms would maintain their internal temperature with the least expenditure of energy – one that is a 1cm in diameter ball of cells or one that is 100cm in diameter? Why? The organism that is a 100cm in diameter ball of cells would maintain its internal temperature with the least expenditure of energy per unit mass because in these spherical organisms, the larger diameter corresponds to a smaller surface area-to-volume ratio and thus less surface area from which to lose heat, making it easier to maintain its internal temperature. 8) Since oxygen and nutrients have to be absorbed from the outside world, would either of the above organisms be able to supply these materials effectively to all of their cells? Why or why not? Neither of the organisms would be able to efficiently supply oxygen and nutrients to all of its internal cells, although the organism that is 1cm in diameter would do so better than the organism 100cm in diameter. As shown in question 5, nutrients and oxygen diffusing inwardfrom the surface of these organisms would take hours to reach cells more than a few mm from the
View Full Document
Unlocking...