Medical Genetics

Question: Discuss about theMedical Genetics. Answer: Pedigree Analysis Cystic fibrosis is an X- related recessive trait. Therefore, the presence of singe defective X chromosome will show diseased phenotypic characteristics in male, but in case of female, two defective X chromosome are needed for expressing the diseased phenotype. A female having a single defective X chromosome, would not show phenotypic characteristics. Ned is the son of Patty, who had a brother with CF. Therefore, it is likely that Pattys father or mother has CF. As no previous history has been provided, it can be considered that, none of Pattys father or mother was affected by CF; however, Pattys mother was a carrier of CF. It is because, if her mother carrier, one defective X chromosome, it can be replaced by the other X gene. However, if her father has a defective gene, he would express the disease characteristics. Now, from the carrier mother, Pattys brother got the defective X chromosome and became affected. From this information, Pattys chance of being a carrier of CF can be dete rmined in the following way. Male Female X Y XD XDX XDY X XX XY XD defective X chromosome In this case, Patty would have 50 % chance to be carrier for CF. b) Pattys son married Selmas daughter. Therefore, the mating was within cousins. With the information from the above answer, in the similar way, Selma have 50 % chance of being a carrier. Ned, Pattys son is not affected by CF and Selmas daughter is also not showing the disease characteristic; hence, she might be carrier or unaffected. Now, there are following possibilities of the new baby of Selmas daughter and pattys son to have CF: If Selmas daughter is a carrier, then: Male Female X Y XD XDX XDY X XX XY The new born baby will have 25% chance of having CF. On the other hand, if Selmas daughter is unaffected, then: Male Female X Y X XX XY X XX XY The new born baby would have no chance of having CF. c) Lisa is the daughter of Homer and Marge. It has been revealed from previous cross, that Homer has 50% chance of being a carrier of CF. Now, if Homer is considered as a carrier, then: Male Female X Y XD XDX XDY X XX XY From the above punnett square, it has been revealed that Lisa will also have 50% chance of being a carrier. For the next generation, if Lisa is a considered as a carrier, then, Male Female X Y XD XDX XDY X XX XY The Lisa and Millhouses baby would have 25% chance of having CF. In contrast, if Homer is considered as unaffected, Lisa will also be unaffected and there would be no chance of their baby, to have CF. d) Millhouse is unaffected, but if he has, a child with CF that means his ex-wife was a carrier of CF, as the son gets the defective X chromosome from mother (Rieger, Michaelis Green, 2012). It would not affect the chance of Lisa and Millhouses child to have CF. On the other hand, if Homer is unaffected, there would be no chance of Lisa to be a carrier and their baby will also be unaffected. Penny has a son and brother with Duchene Muscular Dystrophy. She has a daughter Amy having 4 sons. Pennys brother also has DMD. DMD is an X-linked recessive disorder. In case of daughters, the function of a defective gene in X chromosome can be supplemented by the other wild type gene in another X chromosome, but in case of son, the X chromosome is received only from the mother and there is no supplementary role of another X, as males have a Y chromosome. Pennys son has DMD, who got his X chromosome from his mother, thus it is clear that Penny is a carrier for DMD. Penny is showing normal characteristics, thus, her genotype would be either XDX (carrier) or XX (wild type). From the case scenario, it has been revealed that, he has Pennys son has DMD; as her son has only one X chromosome, he got this X chromosome from his mother Penny; DMD is an X-linked recessive trait and needs a defective X chromosome in males, to show the diseased phenotype (Snustad Simmons, 2012). Therefore, it is clear that, Pennys son got the defective X chromosome from his mother Penny. However, as it is a recessive trait, for females, two X genes have to be defective to show the diseased phenotype. In Pennys case, she has 100 % chance of being carrier for DMD, instead of having normal CK level in blood. It is because, Penny has one wild type X chromosome, which can supplement the function of the defective X chromosome, thereby showing normal phenotypic characteristics, instead of being a carrier for DMD. Amy is Pennys daughter, who does not show the characteristics of DMD, therefore, she can be either a carrier (XDX) or unaffected (XX) (Relethford, 2012). To be affected, Amy should have two defective x chromosome, within which one comes from penny and other one is from her father. It is lready known that Penny is a carrier for DMD, therefore, the chance of Amy to be a carrier can be determined from the following punnett square: Male Female X Y XD XDX XDY X XX XY As there is no history of Pennys husband to be affected y DMD, he can be considered as unaffected. From the above punnett square, it has been seen that Amy has 50 % chance of being a carrier for DMD. Amys husband is considered as normal unaffected. If Amy is a carrier, the chance of her next son to have DMD can be determined by the following Punnett sqauare: Male Female X Y XD XDX XDY X XX XY Therefore, Amys next son has 50 % chance of being a carrier, if Amy is carrier for DMD. However, if Amy is carrier, the chance would be altered: Male Female X Y X XX XY X XX XY Therefore, if Amy is unaffected, there is no chance of Amys next son, to have DMD. In the case scenario, the father is colorblind and the mother is an apparently normal female. The first child is daughter, who is colorblind and the second child is son, who has a normal vision. The defective gene, which determines the colorblindness, is located in X chromosome, thus, it is an X-related disorder (Klug, 2012). The gene for color blindness is denoted as Xc , whereas the wild type gene is denoted as X. Therefore, The father is colorblind and he has a single x chromosome, thus, his genotype must be- XcY. The mother is apparently normal. As the wild type version of the gene is dominant over the defective gene, the heterozygous condition shows the wild type characteristics. Therefore, the mother can be either XcX (carrier) or XX (unaffected). The daughter is colour blind, therefore, two X genes are defective; thus her genotype would be XcXc. The son has normal vision. As the disorder is X-linked and males have only single X chromosome, the son would have a genotype of XY. From the punnett square, mothers phenotype can be demonstrated: Male Female Xc Y Xc XcXc (daughter) - X - XY (son) From the punnett square, it has been revealed that the mother is a heterozygous carrier with the genotype Xc X. From a punnett square the frequency of the male childs genotype can be determined: Male Female Xc Y Xc XcXc (daughter) XcY X - XY (son) It has been revealed that the son will have 50 % chance of being affected. The first child of the couple is the color blind daughter with a geneotype of XcXc. If she marries a normal man (XY), the ratio of their offspring could be determined through the following punnett square. Male Female X Y Xc XcX (daughter) XcY (son) Xc XcX (daughter) XcY (son) Therefore, from the punntt square, it has been revealed that, all the daughters would be heterozygous carrier, with a genotype of XcX. Their phenotypic characteristics would include normal vision. In contrast, all the sons would be color blind with a genotype of XcY, as they would get the single X gene from their mother. Reference List Rieger, R., Michaelis, A., Green, M. M. (2012). Glossary of genetics: classical and molecular. Springer Science Business Media. Klug, W. (2012).Concepts of genetics. San Francisco: Pearson Education. Relethford, J. (2012).Human population genetics. Hoboken, N.J.: Wiley-Blackwell. Snustad, D. Simmons, M. (2012).Genetics. Singapore: Wiley.