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Medical Genetic - Essay Example

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It is a non-Mendelian inheritance multifactorial inheritance pattern because a single parent can transmit the disorder to the child. In the pedigree, the fourth generation exhibits no signs of either IGT/IFG or DM. There are descriptions of diabetes in families resulting due…
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Medical Genetic
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Download file to see previous pages 4. Giving the allele symbol M (dominant allele), m (recessive allele) the proband genotype is MM and the husband’s genotype is mm. If II: 2 is heterozygous, her genotype will be Mm. III: 1 will be heterozygous or homozygous with equal possibilities. This because the genes cross and combines to produce a child that has a genotype of either Mm, Mm, mm or mm as shown in table 1.
7. It is not possible because II.12 has a possibility 75% possibility of having a healthy girl and 25% of having the one with the trait. The genotypes of the III.12 daughter either are Mm, mm, mm, or mm.
10. Yes, gene mutation leads to ether deletions or insertions caused by the uncertainties during the chromosomal crossover due to meiosis. This causes misalignment of homologous chromosomes leading to a different structure of the DNA.
3. The frequency of homozygous and heterozygous gametes expected from II.2 is computed using Punnett table 3 with the information on the pedigree. Consequently, the two genotypes have equal frequency of occurrence of 50%.
5. Typically, it is a Mendelian inheritance. This is because offspring with a dominant allele from either of the parents has the trait, and dominant allele dominates the recessive allele. The pedigree indicates phenotypic traits that show co-dominance of the dominant allele.
vii) Assuming No. of full linkage recombinant in the family, according to the pedigree=3. The no. of total progeny is 9 but one is exceptional hence the no. on non-recombinant offspring is 8. Consequently, probability of observing children of II.2 genotype is given by 0/2=infinity. Parental probability =hence linkage genotype probability (very small), which is the probability of sequence with linkage ()=0
viii) In independent linkage, the probability multiplies for each linkage probability is given by ¼ (0.25) because all four possible genotypes are equally probable. ...Download file to see next pagesRead More
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