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Handout: tests
7.Bibliography: The main: 7.1. Biology. Yarygina V.N. Moscow.,2001, I volume,274-276pp. 7.2.Bochkov N.P. Clinical genetics.M.,2006,411-451pp. 7.3.Genetics. Ivanova V.I.M., 2006,569-591pp. 7.4. Inge-VechtomovS.G. Genetics on the basis of selection.M.,1989, 523-524pp. 7.5. Medical biology and genetics. Textbook. Kuandykov E.U. Almaty, 2004,169-178pp. 7.6.Kuandykov E.U. Fundamentals of general and medical genetics. Lecture course, Almaty,2010. Additional: 7.1. Kozlova S.I.,Selianova E., Demikova N.C., Blinnikova O.E. Hereditary syndromes and CIM.L.,1987,274-284. 7.2. Mutovin G.P. Fundamentals of clinical genetics.M.,2001,211-224. 7.3.Mutovin G.P. Clinical genetics. M.,2010,490-507pp. 8. Control: 7.1. Quiz on topics. 7.2.Tests. 3variants (each variant -10questions). 7.3. An algorithm for the consultation stages: diagnosis, prognosis, opinion, advice and to determine the physician’s tactics at each stage.
7.3. Compilation, genealogical analysis of family trees and calculation of genetic risk of the birth of the sick child: 1. To build a family tree, to determine the genotypes of all members, type of inheritance, and to calculate the genetic risk of having a sick child in a family in which the father and mother are healthy, two boys are sick, aunt and grandfather on the paternal side of children are sick. Penetrance of the gene, 80%. All patients were heterozygous with the mutant gene. 2. Alpha To build a family tree, to define genotypes of all members of a family tree, type of inheritance and to calculate genetic risk of the birth of the sick child in a family in which healthy parents consist in incestuous marriages (cousin siblings), the first pregnancy ended with the spontaneous abortion, the second – dead birth(the sex is unknown), the third birth of the sick girl. 3. To build a family tree, to determine the genotypes of all members, type of inheritance, and to calculate the genetic risk of having a sick child in a family in which parents are healthy, healthy daughter-bearer of the mutant gene, and two sons are sick. 4. To build a family tree, to determine the genotypes of all members, type of inheritance, and to calculate the genetic risk of having a sick child in a family in which the husband is healthy, the wife is-sick, 2 children are healthy: a son and a daughter, 2 children are sick: a son and a daughter. The mutant gene is localized in the X chromosome. 5. To build a family tree, to define genotypes of all members of a family tree, type of inheritance and to calculate genetic risk of the birth of the sick child in a family in which healthy parents consist in incestuous marriages(Uncle and niece), children: a son and a daughter are sick. 6. To build a family tree, to determine the genotypes of all members, type of inheritance, and to calculate the genetic risk of having a sick child in a family in which the father is healthy, mother is sick, the daughter is healthy, but the son is sick. Penetrance of the gene, 30%. 7.6. Fill in the table by prenatal methods of diagnosis:
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