Medical Genetic Essay Example | Topics and Well Written Essays - 1000 words

Medical Genetic - Essay Example4. Giving the allele symbol M (dominant allele), m (recessive allele) the proband ge noneype is MM and the husbands 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 both Mm, Mm, mm or mm as shown in table 1.7. It is not practical 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 bodily 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 pedigre e. 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) take for granted 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, luck 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 ()=0viii) In independent linkage, the probability multiplies for each linkage probability is given by (0.25) because all four possible genotypes are equally probable.