Examples of independent inheritance

Transcription

1 GENETIC PROBLEMS

2 Examples of independent inheritance The amelogenesis imperfecta is is a disorder of tooth development that causes development of abnormally small, discolored teeth. The disease is caused by the recessive mutation of the enam gene linked to the 4 th chromosome. The phenylketonuria is caused by accumulation of toxic phenyl-piruvate due to the malfunction of PAH (phenylalanin hydroxilase) enzyme. The accumulating toxic metabolit results in severe and irreversible neurodegeneration, however the in-time recognised phenylketonuria can be treated by phenylalanin restricting diet. The PAH enzyme-coding pah gene is linked to the 12th chromosome, the malfunction causing allele is recessive. Jane and Joe want to have a child, but they worry about the abovementioned diseases, since both of their mothers suffers from amelogenesis imperfecta, and they also tell you that Jane s father like Joe himself is pyenylketonuric, although it s kept under control by strict diet. Their question is that what is the likelihood that they 5 month old fetus who is a girl will suffer from both diseases?

3 Examples of independent inheritance Jane ai + /ai; pvu/pvu + Joe ai + /ai; pvu/pvu ai + ;pvu ai + /ai + ; pvu/pvu ai;pvu ai/ai + ; pvu/pvu ai + ;pvu ai;pvu ai + ;pvu + ai;pvu + ai + /ai; pvu/pvu ai/ai; pvu/pvu 1/8 ai + /ai + ; pvu/pvu + ai/ai + ; pvu/pvu + ai + /ai; pvu/pvu + ai/ai; pvu/pvu +

4 Examples of independent inheritance The Hemophilia A develops in the absence of cloting factor VIII due to the recessive mutation in the coding gene that is linked to the X chromosome. The sickle cell anaemia is caused by the recessive mutation of the β- hemoglobin coding gene that is located on the 11 th chromosome. Jane and Joe wants to have a child again. They worry about the abovementioned diseases again. They tell you that Jane s mom as well as Joe s father suffers from sickle cell anaemia, and Jane s dad is hemophilic. Their question is that what is the likelihood that they 5 month old fetus who is a boy will suffer from both diseases? If they want an other child later, what is the likelihood that he/she will inherit both diseases.

5 Examples of independent inheritance Jane sca + /sca; X ha+ ;X ha Joe sca/sca + ; X ha+ ;Y sca + ;X ha+ sca + ;X ha sca;x ha+ sca;x ha sca + ;X ha+ sca + /sca + ; sca + /sca + ; sca + /sca + ; X ha+ ;X ha+ X ha+ ;X ha X ha+ ;X ha+ sca;x ha+ sca + /sca; X ha+ ;X ha+ sca + ;Y sca + /sca + ; X ha+ ;Y sca;y sca + /sca; X ha+ ;Y sca + /sca; X ha+ ;X ha sca + /sca + ; X ha ;Y sca + /sca; X ha ;Y sca/sca; X ha+ ;X ha+ sca + /sca; X ha+ ;Y sca/sca; X ha+ ;Y sca + /sca; X ha+ ;X ha sca/sca; X ha+ ;X ha sca + /sca; X ha ;Y sca/sca; X ha ;Y 1/8

6 Linkage Genetic mapping

7 Genetic mapping How can we rule out that the autosomal genes determining hair (BROWN-blonde) and eye (BROWN-blue) color are inherited independently or linked in man?

8 Genetic mapping We must look for couple, in which one of the parents has blue eyes and blonde hair (recessive homozygote), the other has BROWN eyes and hair, but one of his/her parents has (had) blue eyes and blonde hair (heterozygote) and have at least one child. In Csongrad county we have found 26 couples matching these criteria. Their children are with: BROWN eyes, blonde BROWN eyes and hair blue eyes, BROWN hair Blue eyes, blonde IIIII III IIIII IIIII IIIII II IIIII II IIIII IIIII IIIII III Are the two trait linked? If yes, what is the distance of the two genes?

9 Genetic mapping Yes, the two genes are linked, since the ratio of the different phenotypic categories amongst the offsprings is not 1:1:1:1 Genotype of one of the parents Br;B/br;b Genotype of the other parent br;b/br;b Recombinants BROWN eyes, blonde BROWN eyes and hair blue eyes, BROWN hair Blue eyes, blonde IIIII III IIIII IIIII IIIII II IIIII II IIIII IIIII IIIII III The distance of the two genes = (8+7)/( )=15/50=30cM

10 Linked inheritance and the marker mutations Colorblindness is caused by an X-linked recessive mutation of the coneopsin gene necessary for red-green color sensing. 20cM apart there is the g6pd gene that encodes for the glucose-6-phospho-dehydrogenase enzyme. The enzyme is necessary eliminating the free radicals. In the absence of the protein the cells of the individual are very sensitive to oxidative stress causing states, such as certain food, medicine and infection this is the so called favism syndrome. Of course, the appropriate diet including antioxidants the symptoms of favism are neglible. However, the function of the enzyme is easily detected: cells with proper G6PD are fluorescent under UV light. Interestingly, in certain cases only half of the cells of a woman show fluorescence, due to random X-chromosome inactivation (mosaicism). Jane and Joe wants to have a child again, and they afraid that their heir will be colorblind like Jane s dad. Jane s mom has favism. To make them calm, we take blood sample from the family members including Jane fetus and perform the G6PD test:

11 Linked inheritance and the marker mutations Fluorescent Non-fluorescent Jane X Joe Fetus What is the likelihood, that their child will be colorblind?

12 colorblind Jane X Joe Fetus 20cM X cb+ X g6pd cb g6pd+ X cb+ Y g6pd+ cb+;g6pd cb;g6pd+ cb;g6pd cb+;g6pd+ cb+;g6pd+ sees color sees color sees color sees color 20% 20% 5% 5% Y sees color colorblind colorblind sees color 20% 20% 5% 40% 40% 10% 10% The fetus is colorblind = 20/( )=20/50=40% 5% 50% 50%

13 Linked inheritance and the marker mutations Fluorescent Non-fluorescent Jane X Joe Fetus What is the likelihood, that their child will be colorblind?

14 The Huntington syndrome is a dominantly inherited disease linked to the 4 th chromosome caused by the mutation of the Huntingtin protein. The symptoms of the disease progressive neurodegeneration and death within few years - develop late, usually in between 40 and 50. Although we don t have the necessary equipment in our office to test for the presence of the mutant allele, we know that the gene of Rh blood-group system is also linked to the 4 th chromosome and it is 10 cm apart from the huntingtin gene. Jane and Joe wants to have a child again, and they afraid their fetus inherits Huntington syndrome, because Joe s father died in it. Jane and Joe are in the middle of their twenties and thus show no symptoms of the disease. To answer their question, that what is the likelihood that their child gets the disease, we take blood from the family members and determine the Rh blood-group of the individuals: Jane: Rh+; Jane s father Rh+; Jane s mother Rh- Joe Rh+; Joe s mother Rh- The fetus Rh+ Linked inheritance and the marker mutations

15 Jane: Rh+; Jane s father Rh+; Jane s mother Rh- Joe Rh+; Joe s mother Rh- The fetus Rh+ Jane + D + d X Joe + D or + d H D D + d 50% 50% Their child is healthy for sure (100%)

16 Jane: Rh+; Jane s father Rh+; Jane s mother Rh- Joe Rh+; Joe s mother Rh- 50% 50% H D ;D +;d H D ;d +;D +;D Huntington Healthy Huntington Healthy +;d The fetus Rh+ Rh+ Huntington Rh+ Jane X Joe + D + D or + d + d Rh+ Healthy Rh- Rh+ Huntington Rh- Rh+ 22,5% 22,5% 2,5% 2,5% Healthy Rh+ 22,5% 22,5% 2,5% 2,5% 45% 45% 5% 5% Huntington=(22,5+22,5+2,5)/(22,5*3+2,5*3)=47,5/75=63,3% H D D + d 50% 50%

17 Lilla is heterozygous for f, an X-linked recessive mutation. The normal f + gene is activated by product of the Y-linked d+ gene. Product of the d+ gene - the D transcription factor - initiates development characteristic for the males. In absence of the f + gene (e.g. in the f /Y people) development follows the female pathway and one type of the pseudohermaphrodites develop. Kevin s Y chromosome carries the non-functional d mutant allele that does not encode synthesis of the D protein. However, one of Kevin s 17th chromosome carries a segment of the Y chromosome (as a consequence of a translocation) that includes the normal d+ gene. What types of children may Lilla and Kevin have? What proportion of their children will appear to be girl? Once grown up, what proportion of their daughters will be fertile? What percentage of their daughters will carry Barr body in their cells?

18 Lilla Kevin Normal X chromosome f- X chromosome d- Y chromosome Normal 17th chromosome d+ 17th chromosome

19 Lilla Kevin Normal female Sterile male Sterile female Normal male Normal female Sterile male Sterile female Sterile female

20 Sterile males and females are so called pseudohermaphrodites, meaning that their genotypic an phenotypic sex is different. What proportion of their children will appear to be girl? 5/8 Once grown up, what proportion of their daughters will be fertile? 2/5 What percentage of their daughters will carry Barr body in their cells? 2/5 Normal female Sterile male Sterile female Normal male Normal female Sterile male Sterile female Sterile female

21 Pedigree analysis Autosomal recessive, eg. amilogenesis imperfecta

22 Pedigree analysis Autosomal dominant, eg. neurofibromatosis

23 Pedigree analysis X-linked, eg. colorblindness

24 Pedigree analysis Y-linked, eg. Hairy ear

25 Pedigree analysis Maternal inheritance, eg. Leber s hereditary blindness