c) Assuming he does not run another endurance race, will the steady-state populations be affected one year later? If so, explain how.

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1 LS1a Fall 06 Problem Set #8 (100 points total) all questions including the (*extra*) one should be turned in 1. (18 points) Erythrocytes, mature red blood cells, are essential for transporting oxygen to the cells in your body. An adult human has around 2 X mature red blood cells. A single mature red blood cell has a lifespan of around 120 days. New red blood cells must be constantly produced from precursors to maintain the red blood cell populations. The developmental pathway for red blood cells is shown below. a) These precursor populations are also maintained within a narrow range. Give three reasons why this is an example of steady-state and not equilibrium. (3 points) 1. Continuous energy input is needed 2. The forward reactions that increase precursor cell populations are different from the reverse reactions that decrease the cell populations. 3. Catalyzing the development of a cell type will increase the rate of maturation and also increase the size of the more mature cell populations. 4. The development from one cell type to the next is energetically favorable ( G < 0) b) Extreme exercise can sometimes damage erythrocytes, shortening their lifespan. Consider an athlete that has completed a 100-mile endurance run. Assume he has damaged 10% of his erythrocytes, causing them to die early. How will this perturbation initially affect the populations of blood cells? (3 points) It will reduce the number of red blood cells to (1.8 x ) and have no immediate effect on the numbers of any of the other types of cells. c) Assuming he does not run another endurance race, will the steady-state populations be affected one year later? If so, explain how. (4 points) The population will look as if the damage had never occurred, since all the red cells that were present at the time of the race will have been replaced three times over. 1

2 d) Erythropoietin is a protein produced normally by the body that stimulates erythroblasts to develop into erythrocytes. In order to improve performance, athletes sometimes inject erythropoietin into their blood to increase the number of erythrocytes. How will this perturbation initially affect the number of red blood cells? (4 points) The initial response is to increase the rate of red cell production (or maturation from precursor cells), which will lead to an increase in the number of red cells since their birth now exceeds their death. [Given this information you would predict that initially the population of erythroblasts should fall since they are now differentiating faster than they are being born. In reality, EPO is likely to be increasing the generation of precursor cells from stem cells to prevent this fall.] e) If the athlete continues to inject erythropoietin weekly, will the steady-state populations of their red blood cells be affected one year later? If so, explain how. (4 points) Yes, weekly injections maintain the increased rate at which erythroblasts give rise to red blood cells and thus the population of red blood cells increases. The death rate will also increase and a new steady state will be reached when the death rate of the cells equals their birth rate. 2

3 2. (28 points) Development of anti-hiv drugs is challenging because of HIV's high rate of mutation. a) It is known that HIV reverse transcriptase (RT) has a much higher error rate than human DNA polymerase. How might this high error rate harm the virus? How might it benefit the virus? (4 points) A higher error rate can lead to mutations. Most mutations are deleterious, so a high error rate would normally be harmful to the virus. However, some mutations are beneficial, particularly ones that confer drug resistance. A high error rate would increase the chances of making one of these beneficial mutations, which can help the virus survive. b) Drugs that target RT are thought to decrease the rate at which drug-resistant mutant strains of HIV arise. Why does targeting RT prevent the generation of mutant strains while targeting HIV protease does not? (4 points) Mutations are introduced into the HIV genome when the HIV RNA is copied by RT into DNA. RT inhibitors inhibit the reverse transcription of HIV DNA from HIV RNA, and thus prevent the introduction of new mutations into the HIV genome and the subsequent rise of mutant strains. Targeting of HIV protease inhibits viral replication but does not prevent the generation of new mutations because protease activity occurs after RT activity in the viral life cycle. Therefore, if mutations are made that confer resistance to HIV protease inhibitors, the mutant strain will be able to propagate. c) In analyzing the genome of a particular strain of HIV, Mutant X, a single mutation was found in the RT gene. This mutation prevents RT from incorporating nucleotide analogues lacking a 3'-OH into the growing DNA strand. In the absence of any drug, Mutant X was observed to replicate as well as the wild-type virus. Is this mutation deleterious, neutral, or beneficial to the virus in the absence of drugs? Is it deleterious, neutral, or beneficial to an infected patient not taking any drugs? Briefly explain. (4 points) Neutral to virus, neutral to patient. In the absence of drug this mutant, Mutant X, replicates at the same rate as the wild-type virus. Therefore the mutation is neutral to the virus and to the infected patient. 3

4 d) Do you think the mutation in Mutant X would be deleterious, neutral, or beneficial to the virus in the presence of AZT? Would it be deleterious, neutral, or beneficial to an infected patient taking AZT? Briefly explain. (4 points) Beneficial to virus, deleterious to patient. Since this mutation prevents RT from incorporating AZT into the growing DNA strand, Mutant X viruses would be able to reverse transcribe their genome in the presence of AZT. Thus, Mutant X viruses would be able replicate better than wild-type viruses making the mutation beneficial to the virus and deleterious to the infected patients who would have increased viral loads. e) Do you think the mutation in Mutant X would be deleterious, neutral, or beneficial to the virus in the presence of ritonavir? Would it be deleterious, neutral, or beneficial to an infected patient taking ritonavir? Briefly explain. (4 points) Neutral to virus, neutral to patient. Ritonavir is a protease inhibitor so the mutation in Mutant X does not affect its sensitivity to Ritonavir. Creation of infectious mature viral particles of Mutant X and wildtype virus would be similarly affected by ritonavir; so the mutation in the presence of ritonavir is neutral to the virus and to the patient. f) A second mutant strain of HIV, Mutant Y, was isolated that contains a single mutation in the HIV protease gene. This mutation makes HIV protease less efficient than wild-type protease and consequently slows viral replication. However, this mutation confers resistance to saquinavir. A third mutant strain of HIV, Mutant Z, contains both the RT mutation of Mutant X and the protease mutation of Mutant Y. If a population of cells is infected simultaneously with equal amounts of wild-type HIV and Mutants X, Y and Z, which strain(s) will eventually predominate under the following conditions, and which will eventually be lost? (8 points, 1 for each box) Treatment Predominant strain(s) Lost strain(s) None WT + Mutant X Mutant Y + Mutant Z AZT Mutant X WT+ Mutants Y and Z Saquinavir Mutant Y + Mutant Z WT + Mutant X AZT + Saquinavir Mutant Z WT + Mutants X and Y 4

5 3. (22 points) To generate mice with targeted genetic mutations, scientists first make chimeric mice. Embryonic stem cells of one genetic background with the desired mutation are injected into fertilized embryos of a different genetic background. These embryos are then implanted into a foster mother. When these chimeric embryos are born, all of their cells are either derived from the original embryo, or the injected cells. Any given cell has either the DNA of the original embryo or the injected DNA. (HINT: see Alberts page 358). a) The fertilized embryos parents both had black fur and the ES-cell injected embryos are implanted into a mouse with black fur. 5 of the born mice have all black fur, but 5 of them have black fur with patches of brown fur. What is the source of the brown fur? (4 points)the brown fur comes from the injected embryonic stem (ES) cells. b) You cross all ten of these potentially chimeric mice to pure bred black-furred mice and see the following results: Chimeric mouse Fur Color of Chimeric Mouse Fur Color of Progeny 1 Black Black 2 Black Black 3 Black Black 4 Black Black 5 Black Brown or black 6 Black/brown Black 7 Black/brown Black 8 Black/brown Black 9 Black/brown Brown 10 Black/brown Black or brown (10 points-2 point each) Why are the pups of mouse 1 all black? It is likely that mouse 1 is not chimeric. There are no ES cell derived germ cells in the chimera. Why are the pups of mouse 6 all black? It is likely that although the somatic cells of mouse 6 were chimeric the germline was not chimeric. Therefore, the DNA of injected ES cells is not being transmitted to the progeny. Why are the pups of mouse 5 brown or black? It is likely that although somatic cells were not visibly chimeric the germline was chimeric so now the DNA of the injected ES cells is being transmitted to the pups. [Note: Since chimeric mouse was crossed to a pure black mouse this indicates that brown fur color must be dominant over black fur color.] 5

6 Why are the pups of mouse 9 all brown? Mouse 9 s germline is chimeric (and possibly made up of mostly injected ES cells.) Why are the pups of mouse 10 brown or black? Mouse 9 s germline is chimeric. [Note: Chimeric mouse 10 with black/brown fur crossed to a pure black mouse produces progeny that are either black or brown. This also indicates that brown fur must be dominant over black fur.] c) If you cloned a mouse using the nucleus of a cell taken from a hair follicle of mouse 1, what color fur will the cloned mouse have? Explain. (4 points) Black, since you would be using the DNA, genetic material, that results in black fur color. d) If you cloned a mouse using the nucleus of a cell taken from a hair follicle of mouse 10, what color fur will the cloned mouse have? Explain. (4 points) Black or brown, since it would depend if you took the nucleus (and thus the DNA) of a black fur producing follicle cell or brown fur producing follicle cell. 6

7 4. (16 points) Two different cancer patients are seen by a physician with tumors in their stomach, pancreas, and kidney. A karyotype analysis of the three tumors is performed for each patient and the results are shown below. Healthy Patient 1 Patient 2 Stomach Stomach Stomach Pancreas Chr omosome 1 Pancreas Pancreas Kidney Chr omosome 1 Kidney Kidney Chr omosome 1 a) Do the results of patient 1 suggest that the tumors had a common origin or different origins? Explain. (2 points) Common origin. Analysis of chromosome 1 indicates that all of them have identical mutations. b) Do the results of patient 2 suggest that the tumors had a common origin or different origins? Explain. (2 points) Different origins. Analysis of chromosome 1 shows that the three different tumors have different mutations. Thus, the mutations have most likely been generated independently from one another. 7

8 c) If the patient 2 is untreated and the tumors are re-analyzed a year from now, would you expect that the karyotypes will look the same as the one above? Explain. (4 points) The karyotypes are likely to change. Cancerous cells are genetically unstable, meaning the error rate during replication is high. Thus many more mutations would likely be generated in another year. d) Both patients are treated with a single cancer drug. In patient 1, all three tumors shrink, while in patient 2 only the tumor in the pancreas shrinks while the other two continue to grow. Based on your understanding of cancer as a disease and the results of the karyotype analysis, explain these results. (4 points) In patient 1, each of the tumors has the same mutation(s). Therefore, a single cancer drug that affects the modified genes in one tumor, should be able to affect all of the tumors with identical mutations. In patient 2, each of the tumors has a different set of mutations, so a single cancer drug may not affect them all equally. Since we know that cancer is actually many different diseases, it makes sense that 3 cancers of different origins may require 3 different treatments. e) After a year of treatment, the tumors in patient 1 begin to grow again and no amount of drug affects them. Explain. (4 points) Drug treatment is a selective pressure on the cancer cells. Over time cells that do not mutate and become resistant to treatment will die. If a random mutation occurs that gives a cancer cell a survival advantage in the presence of the drug it will now survive and proliferate. 8

9 5. (16 points) Coronary heart disease (CHD) is the leading cause of death of both men and women in the United States. It is caused through the hardening and narrowing of the arteries that supply blood to the heart. This occurs when plaque builds up on the inner walls of the arteries. A major component of these plaques is cholesterol. The disease can run in families, or it can occur sporadically. Take for example two families living in Massachusetts. Both families are originally of Irish ancestry and suffer from CHD. Symptoms appear in all of the male family members by the age of 40, and in women after age 50. a) Based on the information provided, can you determine if these families suffer from CHD due to a genetic or environmental component? Explain. (2 point) No. There are many factors involved in the genetic and environmental components that contribute to the development of CHD including diet. We have not been provided with enough information to determine which of these components is likely the cause. b) It was determined that all of the members of family 1 and 2 have elevated cholesterol levels. Does this piece of information allow you to determine if the cause of their CHD is environmental or genetic? Explain. (2 point) No. High cholesterol is a symptom that can lead to CHD. High cholesterol levels can be caused by poor diet or through genes. c) Relatives of family #1 moved to New Mexico 20 years ago and undertook a Vegan lifestyle. Each of those relatives has lived past the age of 70 with no signs of CHD (normal cholesterol levels, no narrowing arteries). Based on this information, can you determine if family #1 suffers from CHD due to an environmental or genetic component? Explain. (4 points) The data suggests that the environment is playing a large role in family #1 s CHD. The vegan lifestyle and moving to New Mexico has prevented the disease symptoms even though the genetic backgrounds of these two groups within the same family are highly similar. We cannot however determine if it is the vegan lifestyle or something else in New Mexico that is the cause of the CHD prevention. 9

10 d) The grandfather of family #2 had a brother that chose to raise his family in Hawaii and also partook in a vegan lifestyle. However, all of the relatives in this branch of the family still suffer from CHD. From this information can you absolutely determine if the cause of their disease is genetic or environment? Explain. (4 points) This data suggests that the cause of family #2 s CHD is genetic. Even though family members of similar genetic backgrounds undertook a lifestyle change and moved to a different state, they still suffered from the same disease. Thus genetics appear to play a large role in disease development, however you cannot absolutely say the cause of their disease is genetic. e) CHD often strikes later in life. Based on that piece of information, why hasn t coronary heart disease been selected against by evolution? (4 points) Evolution and natural selection only play a role while an organism or person is still reproducing. At the age of 40 and above, most humans have stopped having children, and thus any genetic diseases or defects that occur after our reproducing ages will not be selected against. This not only includes CHD, but other diseases of age such as Alzheimer s. f) (*extra*) CHD is often treated by cholesterol lowering drugs called statins. Statins are prescribed to people without knowing the basis for their disease. Why are statins effective for patients with familial and non-familial CHD? High serum cholesterol levels can lead to CHD. Serum cholesterol comes from cholesterol ingested with food and through biosynthesis by one s own cells. Regardless of the source of cholesterol, statins lower the overall cholesterol levels by inhibiting cholesterol biosynthesis. However, treatment with statins does not act to address the defect causing the disease such as ingestion of too much cholesterol or a mutation in a gene important for binding and clearing cholesterol from blood. 10

Name AP Biology Mrs. Laux Take home test #11 on Chapters 14, 15, and 17 DUE: MONDAY, DECEMBER 21, 2009

Name AP Biology Mrs. Laux Take home test #11 on Chapters 14, 15, and 17 DUE: MONDAY, DECEMBER 21, 2009 MULTIPLE CHOICE QUESTIONS 1. Inducible genes are usually actively transcribed when: A. the molecule degraded by the enzyme(s) is present in the cell. B. repressor molecules bind to the promoter. C. lactose