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Handout: test questions

Прочитайте:
  1. Answer the questions.
  2. Handout: tests
  3. XIX Answer the following questions.
  4. XIX Answer the following questions.

 

7. References:

The main:

7.1. Biology. Edited by VN Yarygin. M: 2001, Part 1, pp. 264-285,569-591.

7.2. Bochkov NP Clinical genetics of M:, 2006, pp. 72-132,323-359,397- 447,365-395.

7.3. Genetics. Under Ivanov VI editor M:,2006,380-410,592-601.

7.4. Medical Biology and Genetics. Textbook edited by prof. Kuandykova EW, Almaty, 2004 - / pages 179-184.

7.5. Kuandykov EW Fundamentals of general and medical genetics. Lectures, Almaty/2010, pp. 192-200.

7.6. Schipkov VN, GN Krivosheina General and medical genetics. M, 2003, pp. 137-186,228.

 

Legal and ethical principles of conducting CIM (Article № 95. Code of the RK on people's health and the health care system. 2009).

6. Handout: tests

7.Bibliography:

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.

Diagnosis Prognosis Opinion Advice
       

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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