Practice set 3 lecs 1-7

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Binghamton University *

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Jan 9, 2024

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Practice set 3: Covers lectures 1-7 Complete each set of questions as we advance through the course material. Every question in this exam is connected! The questions relate to this scenario: Refer back to it as needed. Glucose Galactose Malabsorption (GGM) is a rare genetic disease that prevents the carbohydrates glucose and galactose from being absorbed from the lumen (inside) of the small intestine and into intestinal cells. The subsequent accumulation of glucose and galactose in the lumen of the small intestine pulls water out of the blood and into the lumen of the small intestine. This excess of water leads to severe diarrhea which ultimately causes the death of the infant. Questions 1-10 cover lectures 1-4: 1. Lucy has just given birth to a baby boy, but he quickly starts to show signs of GGM. A genetic test confirms this diagnosis. Lucy is sent home with an appointment with a specialist for the following week but she wants to learn more about GGM right away. With her first google search she finds a website promoting acupuncture for the treatment of digestive issues. The website says Heal all your digestive problems with a natural and ancient remedy. Just one session should bring relief by sealing disconnected energy meridians. Subsequent sessions will promote complete healing. As seen on Dr. Oz, acupuncture is known by doctors worldwide to aid in digestion and absorption and prevent the requirement of pharmaceuticals. Lucy is skeptical of this claim because (SELECT ALL) a) It lacks mechanism b) It lacks source c) It lacks evidence Lucy turns her attention to the website for the National Organization for Rare Disorders, an organization that accurately summarizes the peer reviewed literature. On the page for GGM, Lucy reads that 50 years old, physicians recommended eliminating all carbohydrates from the infant’s diet. But now physicians recommend eliminating all carbohydrates except fructose. What does this disparity in recommendations reveal about the scientific process? a) Mistakes are sometimes not caught during peer review b) Conclusions and recommendations change based on new evidence ¢) Scientists tends to have conflicting views d) The media does a poor job of describing scientific conclusions

When Lucy finally goes to the specialist appointment, the doctor takes an emotional approach to discussing the correct course of treatment (which includes the dietary change) for the baby. This means that the doctor a) Listens to Lucy’s experience. Validates her fears. Talks about the success she has been happy to witness with the course of treatment she is recommending. b) Uses highly scientific language to describe the cellular mechanism of disease and provides a diagram of the inside of a hypothetical baby’s small intestine. Lucy’s baby has his first ear infection. It is caused by the bacteria Streptococcus pneumoniae. Antibiotics are prescribed. The Streptococcus pneumoniae population in the baby’s ear develops resistance to the antibiotic. Resistance is conferred by a mutation in a specific gene. The original mutation that resulted in resistance a) Arose by chance. b) Arose because the bacteria were exposed to an antibiotic. The entire Streptococcus pneumoniae population acquiring the mutation and thus developing resistance a) Arose by chance. b) Arose because the bacteria were exposed to an antibiotic. 200 miles away in another city, Jasmin Washington PhD is doing new research on GGM. She hypothesizes that her new genetic engineering treatment will fix the genetic mutation which causes GGM and thus will permanently cure the disease. Which of the following are valid predictions that could be made from the hypothesis? SELECT ALL a) Infant GGM mice that are injected with the genetic engineering treatment within two hours of birth will have solid stool 7 days. b) Infant GGM mice that are injected with the genetic engineering treatment within 20 minutes of birth will not have excess fluid in the small intestine when samples are taken at day 2 and 4. ¢) Mice that are injected with the genetic engineering treatment will have mostly normal stools. Jasmin designs her experiment such that 4 mother mice (which all have GGM) are impregnated on the same day. Two are randomly assigned to the experimental group and two are assigned to the negative control group. In the experimental group, the babies are removed from each mother 30 minutes after the last mouse has been born (in each litter). They are placed in a holding cage and one-by-one are injected with the genetic engineering treatment. They are returned to the mother immediately and allowed to nurse. What is the most robust protocol for the negative control group? a) Baby mice are not removed from the mother and not injected. b) Baby mice are removed from the mother and injected with a medication currently used to treat GGM.

¢) Baby mice are removed from the mother and injected with a benign solution that does not contain the genetic engineering ingredient. 8. Following the treatments, Jasmin’s lab assistants remove the baby mice from the mothers and place them in an empty holding cage for 20 minutes at a time, 3 times per day for 3 days. After 20 minutes the baby mice are returned to their mothers and the number of healthy stools that are left in the holding cage is counted. After this data collection, Jasmin calculates the average number of healthy stools per day per mouse in the two treatments. The results are shown below: Average number of healthy stool each day Hk 12 " Hk . =) Number of healthy stool o 2 o \é 1 2 3 —8—Experimental =@=Control What conclusion can Jasmin make based on this data? a) Fail to reject the null hypothesis. The genetic engineering treatment had no effect on the number of healthy stools each day and thus did not eliminate the main symptom of GGM. b) Fail to reject the null hypothesis. The genetic engineering treatment increased the number of healthy stools each day and thus eliminated the main symptom of GGM. ¢) Reject the null hypothesis. The genetic engineering treatment had no effect on the number of healthy stools each day and thus did not eliminate the main symptom of GGM. d) Reject the null hypothesis. The genetic engineering treatment increased the number of healthy stools each day and thus eliminated the main symptom of GGM. 9. Although much of this experiment was able to be carefully controlled, the number of mice in each litter varied slightly between the four mothers. In total, there were 10 babies in the experimental group and 12 in the control group. Which element of the results might this impact? a) Standard error

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b) Standard deviation c) Statistical significance 10. In which treatment was there likely a more uniform (consistent) response to the treatment? a) Experimental b) Control Questions 11-16 cover lectures 5-6: 11. The particular lab mice species used in the research evolved in a tropical climate. This species is related to a tundra mice species that evolved and lives in northern Canada and Alaska. Which species would you expect to contain a higher percentage of unsaturated phospholipids in the cell membranes of their feet skin? a) Lab mice b) Tundra mice 12. If JTasmin had used tundra mice in her research and kept the temperature of their cages ‘warm, what effect might that have had on her experiment? a) It would have increased the permeability of the cell membrane to glucose and galactose in GGM mice babies. b) It would have reduced the permeability of the cell membrane to glucose and galactose in GGM mice babies. ¢) It would have had no effect on the permeability of the cell membrane to glucose and galactose in GGM mice babies. 13. In healthy individuals, the ‘sodium/glucose cotransporter 1 transports glucose, galactose and sodium across the cell membrane and into intestinal cells. The genetic mutation of GGM causes the ‘sodium/glucose cotransporter 1° protein to be malformed, so glucose, galactose and sodium (Na+) remain in the lumen of the small intestine and do not travel into intestinal cells. Which of the following is true of the OH on the galactose? CH,OH ¢ o H0/| OH I/ W | : o T | OH H | N | H OH Galactose a) The O and H will form a bond other than a covalent bond because H has an almost-filled valence shell.

b) The O and H will form a bond other than a covalent bond because O has more valence electrons than H. ¢) The O and H will share electrons unevenly because the O has a higher electronegativity. d) The O and H, as a unit, is negatively charged and will be attracted to positive charges because the H is less electronegative than the O. 14. Why can’t galactose travel through the cell membrane on its own? a) Galactose is small and nonpolar so it will be repelled by the nonpolar tails of the phospholipids. b) Galactose is large and polar so it will be repelled by the nonpolar tails of the phospholipids. ¢) Galactose is small and nonpolar so it will get stuck interacting with the polar heads of the phospholipids. d) Galactose is large and polar so it will get stuck interacting with the polar heads of the phospholipids. 15. Remember that the accumulation of glucose and galactose in the small intestine (due to the defective ‘sodium/glucose cotransporter 1”) pulls water out of the blood and cells of the body and into the lumen of the small intestine. Why does this occur? a) Solutes move to an area of a lower solute concentration b) Solutes move to an area of a higher solute concentration ¢) Water moves to an area of a lower solute concentration d) Water moves to an area of a higher solute concentration Questions 16-20 cover lecture 6-7: 16. Which of the following would hypothetically be an effective immediate treatment for the symptoms of GGM? a) Provide oral rehydration fluid that contains a high concentration of sodium. b) Provide oral rehydration fluid that contains a low concentration of sodium. ¢) Provide intravenous fluid into the blood vessels that contains a high concentration of sodium. d) Provide intravenous fluid into the blood vessels that contains a low concentration of sodium. 17. As mentioned previously, mammalian milk contains lactose. When ingested, lactose binds to lactase and is broken down into glucose and galactose. Interestingly, lactase is also inhibited by galactose via allosteric inhibition. Which of the following is true? a) Lactase is the enzyme. Glucose and galactose are the products. Galactose binds to an area of lactase that is not the active site.

18. 19. 20. b) Lactase is the substrate. Glucose and galactose are the products. Galactose binds to the active site of lactose. c) Lactase is the substrate. Glucose is the product. Galactose binds to the active site of lactose. d) Lactase is the enzyme. Galactose is the product. Glucose binds to an area of lactase that is not the active site. In addition to her work with mice, Jasmin is also interested in studying the “‘sodium/glucose cotransporter 17 protein in an artificial cell environment. In healthy individuals, the ‘sodium/glucose cotransporter 1” allows glucose and galactose that has accumulated in the small intestine (from food) to move down their concentration gradients into the intestinal cells. Sodium also moves through this cotransporter. Jasmin constructs an artificial cell with a healthy ‘sodium/glucose cotransporter 1” protein embedded in the cell membrane. She places the cell in a beaker of water which contains 1 M glucose, 1 M galactose and 1 M sodium. She injects the artificial cell with 0 M glucose and galactose and 2 M sodium. Where will sodium move? a) Sodium will not move. b) Sodium will move from the artificial cell into the beaker water. ¢) Sodium will move from the beaker water into the artificial cell. Below is a graph of lactase activity by pH. Lactase activity by pH [ 7 14 If Jasmine wanted to incorporate lactase and lactose in some cell-in-a-flask experiments, would she have to adjust and maintain the pH of the beaker and cellular solution? a) Yes. Both should be maintained at about 7. b) Yes. Both should be maintained at about 14. ¢) No. Ph would have no effect on the experiment. The lactose metabolic pathway involves multiple enzymes and multiple steps. As seen in the diagram below, lactase converts lactose into galactose and glucose. Galactose is converted by galactokinase into glucose. Glucose is converted by hexokinase into ATP.

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Galactokinase Lactase Galafmse —_ \ Herokinase / Lactose *This metabolic pathway has been simplified for use in this exam. If the rate at which hexokinase operated slowed down, what would be the immediate effect on 1. the concentration of lactose; 2. the rate at which lactase operated? a) b) ©) a Key: The concentration of lactose would decrease. The rate at which lactase operated would speed up. The concentration of lactose would decrease. The rate at which lactase operated would slow down. The concentration of lactose would increase. The rate at which lactase operated would speed up. The concentration of lactose would increase. The rate at which lactase operated would slow down. Answer: A, B, C. There is no mechanism, no scientific explanation for how acupuncture helps. “sealing disconnected energy meridians” is nonsense. There is no source. Although a doctor, Dr. Oz is not a reliable source. Sources that are reliable include peer reviewed journal articles or organizations that accurately summarize peer reviewed journal article. There is also no evidence. Evidence would include quantification of the effects. Answer: B. As new information is revealed due to more scientific work, conclusions must change or be amended. This scenario does not describe a mistake, it does not emphasize differing views of scientists and it does not mention the media at all. Answer: A. An emotional approach means listening to a patient’s experience and using emotions and emotional language to convince them of the appropriate treatment. Answer: A. All genetic mutations arise by chance due to a mistake during DNA replication. Answer: B. The antibiotic is a change in environment. Only susceptible bacterial cells die, leaving the resistant cells alive and reproducing. While single mutations arise by chance, it’s the environment that will impact if the mutation sweeps through the

10. 11. 12. population. Here the mutation provides an advantage to the bacteria in this environment so individual cells with the mutation survive and reproduce more than those that don’t have the mutation. Answer: A and B. These are specific and measurable. C is not specific, nor is it measurable. It is just a restatement of the hypothesis. Answer: C. The negative control group must have the exact same treatment as the experimental group except for the one variable being tested, the genetic engineering ingredient. So the mice should be removed from the mother etc and also be injected with a benign solution (placebo) to account for the variables of being separated from the mother, being placed in a holding cage and being injected. If these variables are not accounted for there would be no way to say definitively that any improvement was due to the actual genetic engineering treatment, because it could have been due to the other variables. Answer: D. There are multiple asterisks over each day so there was a significant statistical difference between the experimental and control groups. Therefore, we reject the null hypothesis. The genetic engineering treatment increased the number of healthy stools. Answer: A. Standard error is impacted by sample size and standard deviation. In general, a larger sample size reduces standard error because there is a greater chance of sampling from everywhere under the curve...ie all the variation that exists in the population. Answer: B. Control. Standard error was very low for the control treatment. Although a small amount of this might be due to the slightly higher sample size, the large disparity between standard error in both groups must have also been influenced by standard deviation. Higher standard error bars in the experimental group indicates more variation...meaning less consistent measurements between babies compared to the control group. Answer: B. Tundra mice would have evolved more unsaturated phospholipids in the cell membrane of their feet because they were evolving in a cold climate. The unsaturated phospholipids would compensate for the lowered permeability due to a cold temperature and bring permeability back into homeostatic range. Answer: C. Increasing the temperature for a species that has a higher percentage of unsaturated phospholipids will increase permeability of the cell membrane to small molecules, especially small nonpolar molecules that can get across without a transport protein. But it will not have an effect on a large molecule like glucose or galactose. These

13. 14. 15. 16. 17. 18. 19. 20. cannot get through the cell membrane without a transport protein. They are too big and also polar. Thus, using this species would not change her results. Answer: C. O and H will form a covalent bond because neither of them have filled valence shells. The covalent bond will be polar because O has a higher electronegativity and will thus pull electrons closer to its nucleus. Answer: D. Glucose is large. It is composed of many atoms. It is also polar. It has many polar covalent bonds Answer: D. Because glucose and galactose stay in the lumen of the small intestine, it accumulates there, causing a high concentration of solutes. So water will move by osmosis into the lumen because water always moves into an area of a higher solute concentration. Glucose and galactose can’t move because of the defective cotransporter so solute movement is irrelevant here. Answer: C. Increasing the solute concentration of the blood should pull out water from the lumen of the small intestine because water will move to an area of a higher solute concentration. Answer: A. Lactose is a carbohydrate. It binds to lactase which breaks it down. So lactose is the substrate and lactase is the enzyme. Glucose and galactose are the products. Galactose must bind somewhere other than the active site because it is allosteric inhibition. Answer: C. The ‘sodium/glucose cotransporter 1” protein is working properly. So it will move glucose and galactose from outside the cell, a high concentration, to inside the cell, a low concentration. This provides the energy needed to power the movement of sodium against its concentration gradient. So sodium moves from the beaker water and into the artificial cell. Answer: A. lactase optimal activity occurs at a pH of 7. If Jasmine wanted to incorporate lactase and lactose in some experiments she would have to maintain the pH at about 7 so that lactase would function properly. Answer: D. If hexokinase slows down, then glucose and galactose will accumulate. . .thus lactase will slow down because there is more product than reactant. If lactase slows down, lactose will accumulate.

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