Complicated Medical Genetics - Essay Example

Medical Genetics a) The inheritance pattern seen symbolized by gray shading above The affected individuals with certain genetic conditions b) The reasoning for the answer given in response to 1a The affected individual with certain genetic conditions is presented with solid or a filled-in squares or circles symbols depending on gender. However, the solid squares or circle present either an autosomal inheritance or an X-link inheritance as two copies of allele are required for an affected individual to express the susceptible acquired phenotype. In this pedigree, since the typical parents of a particular genetic condition are not affected, (open circles and squares) the affected individuals possess autosomal inherited condition (Nielsen & Slatkin, 2013). c) The percentage risk that the first-born child of individual II.5 will exhibit symptoms if she mates with a male (homozygous) who is wild type at the causative locus 50% d) The reasoning for the answer given in response to 1c Individual II.5 is a homozygous normal parent while the partner is a homozygous affected person for a particular DNA coding conferring to a wild type phenotype. However, in an autosomal dominant inheritance, only a single copy of the wild-type phenotype allele is required for an offspring to be susceptible to expressing the symptoms. Therefore, each offspring has an equal probability of 50% of acquiring the mutant gene allele symptoms, and the other half would not be affected by the autosomal inheritance according to Onkers, I. (2009). e) The risk that the first-born child of individual III.6 will exhibit symptoms if she mates with a male who is heterozygous for the causative mutation (express your answer as a fraction) 1/4 f) The reasoning for the answer given in response to 1e Since the type of inheritance in this particular genetic pedigree is an autosomal inheritance, a heterozygous male (Bb) for the DNA coding mating to an affected female (bb)individual would produce one out of four offsprings expressing altered symptoms. However, within the unaffected offspring there would be two carriers (Bb) who will not show undesired traits for a particular condition. This is because when a homozygous wild-type mate with a heterozygous individual, there is a 25% chance of acquiring mutant (affected) individuals (Relethford, 2012). g) Explanation of how Individual III.1 goes on to have 5 healthy children, even though their father suffers from the same disability as she does In a dominant, autosomal inheritance, individual III 1 (bb) can only acquire healthy offspring if she marries a homozygous normal man (BB). However, though the father (bb) of the individual III 1 (bb) suffers from the same disability as she does, the fathers’ disability does not affect her daughter’s subsequently genetic inheritance (Nielsen & Slatkin, 2013). This is because; the individual III 1 (bb) offspring can only acquire inherited characteristics only from the parents rather than the grandparents (wild-type conditions occur in one sibship per generation). Similarly, the genetic alterations need to be present in pairs of particular gene alleles to cause an observable or sufficient impairment to the offspring. However, during an autosomal genetic inheritance, all the offspring produced by the individual III 1 to a homozygous partner will provide healthy 5 children carriers for a particular autosomal recessive condition (Relethford, 2012). Question 2 a) Diagram explanation b) absolute carriers c) The type of inheritance pattern of Mia Autosomal recessive inheritance pattern d) The reasoning behind your answer to 2c Phenotypes associated with Mia inherited conditions in a particular version of genes are expressed only when individuals in a given generations have two copies of alleles with autosomal recessive traits. In this autosomal recessive inherited pattern, the phenotype traits are observed only when individuals at a particular generation are homozygous for undesirable allele’s pattern. However, the individuals’ in these generations with a single copy of undesirable alleles (carriers) would not show the phenotype expression but would instead pass the undesirable alleles on their subsequent generation (Relethford, 2012). e) The chance (expressed as a fraction) that Mia is a carrier of the cystic fibrosis mutation found in this family 1/50 f) The reasoning behind your answer to 2e Applying Hardy-Weinberg equation (p2 + 2pq + q2= 1) where p2 is a constant value of 1 representing the frequency of homozygous dominant (BB) and 1 of 10,000 babies born with cystic fibrosis while 2pq and q2 representing the frequency of heterozygous alleles and the frequency of homozygous recessive (Bb) respectively (Scott, 2012). The chance of Mia being a carrier would be equal to the frequency of heterozygous alleles (Bb). (p2 + 2pq + q2= 1) 1+√ 1/10,000+2(1) = 1/50 g) The chance (expressed as a percentage) that Mia’s unborn girl will suffer from cystic fibrosis, given that the background population career risk is 1/25 1% h) the reasoning behind your answer to 2g Provided that 1/25 of the general population at risks is carriers of the cystic fibrosis, Mias child with a different partner will have a 100% chances of being a carrier. However, since the husband changes at risk to be a carrier in the entire population is 1/25, the child will inherit the mutated alleles at 25%. Therefore, the formula for obtaining the risk of the child to be affected would be 1 x 1/25 x ¼ x 100=1% X chromosome inactivation process X chromosome inactivation process occurs where a single copy of the X-chromosomes present in a female individual is inactivated during the initial states of female embryonic development as illustrated by the diagram (mitosis) to maintain to equalize the X-linked encoded genes doses of the male and the female cells. For instance, upon differentiation stages A1 is inactivated while A2 remains active through life of the female individual after the X-chromosomes from one parent units to form a zygote with another parent X-chromosomes (Roy, 2012). However, during X-chromosomes inactivation the A1 sector of genes o the right side fails to express themselves in an individual throughout life allowing the A2 division of genes to express themselves eliciting a particular trait. Experiment 4 The LOD score considered evidence for linkage to the Neurofibromatosis 1 gene The LOD score is equal to logarithm probability of the birth sequence at a given value divided by the likelihood of the birth sequence without linkage in reference to Scott (2012). Therefore, since the combination frequency is equal to 1, the distance (d) would be d=r/N, which is equal to 1/10. LOD =log (1-0.1)/0.5=1.6. g)Knudson’s 2-Hit Hypothesis with respect to inherited (bilateral) retinoblastoma Knudson hypothesis states that an acquired abnormal results from an accumulated cells DNA mutations. However, inherited (bilateral) retinoblastoma occurs as an autosomal dominants inheritance with a high penetrative power. According to Knudson hypothesis, bilateral retinoblastoma occurs, with a 50% probability where one mutated allele is inherited through germinal cells while the subsequent mutations occur spontaneously in the somatic cells. Similarly, inherited traits in a bilateral retinoblastoma are sporadic possessing a high penetrative power according to Onkers (2009). i) Hardy-Weinberg equilibrium the genotype frequencies 0.734449 + 0.245102 + 0.020449 Allele frequency is equal to 0.240, 0.857 and 0.049 respectively. k) Explanation for a population having a higher than expected (according to the Hardy-Weinberg principle) frequency of heterozygotes Assuming a particular allele is A and the equation is (p2 + 2pq + q2= 1) In a population may have a population that has got a higher heterozygous genotype frequency as the mating during inheritance may be totally random, every individual in a particular generation may be producing same number of the offspring causing a greater deviation in of genetic frequencies from generation to generations according to ROY (2012). 1) Incidence of Phenylketonuria Provided that 25% of the general population at risks of Phenylketonuria, the individual would have a frequency 0.125. Therefore, the formula for obtaining the incidence risk of acquiring Phenylketonuria would be 1 x 0.125 x ¼=1/200 References NIELSEN, R., & SLATKIN, M. (2013). An introduction to population genetics: theory and applications. Sunderland, Mass, Sinauer Associates. ONKERS, I. (2009). X chromosome inactivation: activation of silencing. S.l, s.n.]. RELETHFORD, J. (2012). Human Population Genetics. Hoboken, John Wiley & Sons. http://public.eblib.com/choice/publicfullrecord.aspx?p=817499. ROY, D. (2012). Biometrical genetics: analysis of quanitative variation. Oxford, U.K., Alpha Science International. SCOTT, D. (2012). Solutions manual for Introduction to genetic analysis, tenth edition. New York, W H. Freeman & Co. Read More