Question 4 a) Below is the pedigree for a family with an autosomal recessive disease, disease X A ○= unaffected female affected female unaffected male affected male i)What is the genotype of individual a at the disease X locus? Use"+"to indicate the wildtype allele and "to indicate the mutant allele ii)What is the probability that individual B is a carrier of disease X? ii)Individuals C and d decide to have a child. What is the probability that the child will have disease x? iv) What is the probability that the child of individuals C and d will be a carrier of disease X? b) The most common mutant allele of the disease x gene is a deletion of three nucleotides which eliminates a phenylalanine at amino acid residue 508. Although the mutant X protein is made, it is not localized to the plasma membrane i)Assuming the altered X protein is stable, where might it be found? ii)Describe another mutation in this gene that could prevent the disease X protein from localizing to the plasma membrane Spring 2004 Final exam practice
Spring 2004 Final Exam Practice 6 Question 4 a) Below is the pedigree for a family with an autosomal recessive disease, disease X. ? = unaffected female = affected female = unaffected male = affected male A B C D i) What is the genotype of individual A at the disease X locus? Use “+” to indicate the wildtype allele and “-“ to indicate the mutant allele. ii) What is the probability that individual B is a carrier of disease X? iii) Individuals C and D decide to have a child. What is the probability that the child will have disease X? iv) What is the probability that the child of individuals C and D will be a carrier of disease X? b) The most common mutant allele of the disease X gene is a deletion of three nucleotides which eliminates a phenylalanine at amino acid residue 508. Although the mutant X protein is made, it is not localized to the plasma membrane. i) Assuming the altered X protein is stable, where might it be found? ii) Describe another mutation in this gene that could prevent the disease X protein from localizing to the plasma membrane
Question 4, continued c)Researchers are currently working on gene therapy for disease X patients. The most adenovirus is a double-stranded DNA virus that target wivip thelial cells, it can be used to promising therapy has involved incorporating the disease x gene into an adenovirus. Becau deliver the disease X gene to the lung cells of the affected individual display of MHC I molecules on the surface of cells. Why is this a desirable property of the virus used to deliver the disease x gene? ii)Using the plasmids and restriction enzymes provided, design a procedure to create a double-stranded dna to incorporate into the adenovirus particle. The final product should be linear, contain the majority of the virus genome and have the disease x gene under control of the El promoter(Pel). Nhel and Spel create the same sticky ends. All the other restriction enzymes create unique cuts P BamHI HindIll start Spel disease X cdnA BR-Ad2-7 Adenovirus pCMV-diseaseX HindII BamhI EcoRI Spring 2004 Final exam practice
Spring 2004 Final Exam Practice 7 Question 4, continued c) Researchers are currently working on gene therapy for disease X patients. The most promising therapy has involved incorporating the disease X gene into an adenovirus. Because adenovirus is a double-stranded DNA virus that targets lung epithelial cells, it can be used to deliver the disease X gene to the lung cells of the affected individual. i) The adenovirus used in these studies is able to produce gp19, a protein that inhibits the display of MHC I molecules on the surface of cells. Why is this a desirable property of the virus used to deliver the disease X gene? ii) Using the plasmids and restriction enzymes provided, design a procedure to create a, double-stranded DNA to incorporate into the adenovirus particle. The final product should be linear, contain the majority of the virus genome and have the disease X gene under control of the E1 promoter (PE1). NheI and SpeI create the same sticky ends. All the other restriction enzymes create unique cuts. pBR-Ad2-7 PE1 BamHI NheI HindIII SpeI BamHI Adenovirus genome pCMV-diseaseX EcoRI HindIII SpeI disease X cDNA start stop
Question 5 The figure below shows GDP in the binding pocket of a G protein. o|P|O HO OHOH y r HNN→C=NH O Arg Glu a) Circle the strongest interaction that exists between i)the side chain of Lys and the phosphate group of GDP van der waals covalent hydrogen bond lonIc ) the side chain of Glu and the ribose group of GDp van der waals covalent hydrogen bond Ionic ii)the side chain of Tyr and the guanine base of gDp van der waals covalent hydrogen bond ionIc b) You make mutations in the gDp-binding pocket of the g protein and examine their effects on the binding of GDP. Consider the size and the nature(e. g charge, polarity, hydrophilicity, hydrophobicity) of the amino acid side chains andand give the most likely reason why each mutation has the stated effect. Consider each mutation independently i)Arg is mutated to a Lys, resulting in a G protein that still binds GDP ii)Asp is mutated to a Tyr, resulting in a G protein that cannot bind GDP. Spring 2004 Final exam practice
Spring 2004 Final Exam Practice 8 Question 5 The figure below shows GDP in the binding pocket of a G protein. NH NH H N C NH C O C O HO Lys Arg Asp Glu Tyr 3 + 2 2 + P P O O O O O OH OH N N NH NH N O CH 2 O- OO- 2 O- O - a) Circle the strongest interaction that exists between: i) the side chain of Lys and the phosphate group of GDP van der Waals covalent hydrogen bond ionic ii) the side chain of Glu and the ribose group of GDP van der Waals covalent hydrogen bond ionic iii) the side chain of Tyr and the guanine base of GDP van der Waals covalent hydrogen bond ionic b) You make mutations in the GDP-binding pocket of the G protein and examine their effects on the binding of GDP. Consider the size and the nature (e.g. charge, polarity, hydrophilicity, hydrophobicity) of the amino acid side chains and and give the most likely reason why each mutation has the stated effect. Consider each mutation independently. i) Arg is mutated to a Lys, resulting in a G protein that still binds GDP. ii) Asp is mutated to a Tyr, resulting in a G protein that cannot bind GDP
Question 6 The bos/seven receptor is required for differentiation of a particular cell, called R7. It is a receptor tyrosine kinase with the structure below. As a monomer, the protein is inactive Binding of ligand causes the receptor to dimerize, causing phosphorylation of the intracellular domain, activating the protein. During processing of the protein, the extracellular domain is cleaved and a disulfide bridge forms between two cysteines, tethering the ligand-binding domain to the rest of the protein ligand-binding domain extracellular S-s S-s membraneITITIIIITTTIIiIli intracellular NACTIVE ACTIVE i) How would receptor activity be affected by changing one of the two cysteines shown above to an alanine? explain ii) What effect would this mutation have on the differentiation of r7? b) Name three amino acids that would be likely to be found in the transmembrane domain. What property do those amino acids have in common, and why do they cause the transmembrane domain to stay in the membrane? d) Draw a schematic of the receptor tyrosine kinase(discussed above) prior to any cleavage or modification using the template below. Include the domains of this protein that are required for targeting to and insertion in the plasma membrane. Also label the intracellular and extracellular domains C e)Activation of the above receptor causes Ras to exchange GDP for GTP, thereby activating it This activated Ras can activate a signal transduction cascade, which ultimately results in the Spring 2004 Final exam practice
Spring 2004 Final Exam Practice 9 Question 6 The bos/seven receptor is required for differentiation of a particular cell, called R7. It is a receptor tyrosine kinase with the structure below. As a monomer, the protein is inactive. Binding of ligand causes the receptor to dimerize, causing phosphorylation of the intracellular domain, activating the protein. During processing of the protein, the extracellular domain is cleaved and a disulfide bridge forms between two cysteines, tethering the ligand-binding domain to the rest of the protein. -S-Sligand-binding domain intracellular extracellular membrane -S-S- -S-Sligand INACTIVE ACTIVE P P a) i) How would receptor activity be affected by changing one of the two cysteines shown above to an alanine? Explain. ii) What effect would this mutation have on the differentiation of R7? b) Name three amino acids that would be likely to be found in the transmembrane domain. What property do those amino acids have in common, and why do they cause the transmembrane domain to stay in the membrane? d) Draw a schematic of the receptor tyrosine kinase (discussed above) prior to any cleavage or modification using the template below. Include the domains of this protein that are required for targeting to and insertion in the plasma membrane. Also label the intracellular and extracellular domains. N C e) Activation of the above receptor causes Ras to exchange GDP for GTP, thereby activating it. This activated Ras can activate a signal transduction cascade, which ultimately results in the