Q6. Many different mutations (>1500 to date) in the CFTR gene can cause Cystic Fibrosis(CF). The disease shows allelic heterogeneity and many CF sufferers are compoundheterozygotes i.e. they have two different mutations, one in each of their copies of the CFTRgene. ~1 in 25 people in Australia are carriers for CF (an autosomal recessive disease). DNAtests are used to detect CF mutations in carriers but it is too expensive to test, or screen forall mutations so only the most common mutations are tested for. A typical CF carrier screenis 90% accurate, i.e. 10% of mutations are missed by the screen.a. Use Bayesian Analysis to calculate the probability that a woman who: a) has no familyhistory of CF; and b) tests negative (negative means no mutation is found) on a mutationscreen, is a carrier. Provide answers to 3 decimal places.Scenario A woman is aScenario B woman is not acarriercarrierPrior ProbabilityCond. ProbabilityJoint probabilityPosterior probabilityb. If this woman's partner is a known carrier for CF, the probability that the couple's first childwill have CF (to three decimal places): Q7. With multifactorial disorders, several genes (as well as environmental factors) arecontributing to the disease therefore Mendel's Laws cannot be used as anterior information tocalculate risk figures. In such cases, empiric recurrence risk is used to estimate the likelihoodof a disease occurring. These figures are calculated by comparing the incidence of a diseasein close relatives of an affected individual (familial incidence) to the frequency of the diseasein the general population (population incidence).Chance ofChance of affectedPopulationincidenceunaffected parentshaving a 2ndaffected childparents having anaffected child(%)Disorder(%)(%)Cleft lip/palate0.1-24Congenital heart0.81.42(father), 6(mother)defectSpina bifida0.254-54Schizophrenia1014a. Using the information provided in the table above (and from previous questions), rank thefollowing events in order of probability, from most likely to least likely: a) parents of a childwith cleft palate having a second child with same disease; b) someone with no familyhistory of the disease developing schizophrenia; c) parents of a child with thalassemiahaving a second child with thalassemia; d) a woman with a congenital heart defect havinga child with the same condition.b. An unaffected couple have one child with both cleft lip and congenital heart defects.Assuming that these two disorders are NOT related, what is the probability that this couplewill have a second child with BOTH these disorders (to five decimal places)? Show yourcalculation.

Name: Q6. Many different mutations (>1500 to date) in the CFTR gene can cau
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Description: Q6. Many different mutations (>1500 to date) in the CFTR gene can cause Cystic Fibrosis(CF). The disease shows allelic heterogeneity and many CF sufferers are compoundheterozygotes i.e. they have two different mutations, one in each of their copies

Q6. ANS- For two events, A and B, Bayes' theorem allows us tofigure out p(A. | B) (the probability…

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26. Many different mutations (>1500 to date) in the CFTR gene can cause Cystic Fibrosis (CF). The disease shows allelic heterogeneity and many CF sufferers are compound heterozygotes i.e. they have two different mutations, one in each of their copies of the CFTR gene. -1 in 25 people in Australia are carriers for CF (an autosomal recessive disease). DNA ests are used to detect CF mutations in carriers but it is too expensive to test, or screen for all mutations so only the most common mutations are tested for. A typical CF carrier screen s 90% accurate, i.e. 10% of mutations are missed by the screen. a. Use Bayesian Analysis to calculate the probability that a woman who: a) has no family history of CF; and b) tests negative (negative means no mutation is found) on a mutation screen, is a carrier. Provide answers to 3 decimal places. Scenario A woman is a carrier Scenario B woman is not a carrier Prior Probability Cond. Probability Joint probability Posterior probability b. If this woman's partner is a known carrier for CF, the probability that the couple's first child will have CF (to three decimal places):

26. Many different mutations (>1500 to date) in the CFTR gene can cause Cystic Fibrosis (CF). The disease shows allelic heterogeneity and many CF sufferers are compound heterozygotes i.e. they have two different mutations, one in each of their copies of the CFTR gene. -1 in 25 people in Australia are carriers for CF (an autosomal recessive disease). DNA ests are used to detect CF mutations in carriers but it is too expensive to test, or screen for all mutations so only the most common mutations are tested for. A typical CF carrier screen s 90% accurate, i.e. 10% of mutations are missed by the screen. a. Use Bayesian Analysis to calculate the probability that a woman who: a) has no family history of CF; and b) tests negative (negative means no mutation is found) on a mutation screen, is a carrier. Provide answers to 3 decimal places. Scenario A woman is a carrier Scenario B woman is not a carrier Prior Probability Cond. Probability Joint probability Posterior probability b. If this woman's partner is a known carrier for CF, the probability that the couple's first child will have CF (to three decimal places):

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 13RQ: The ABO blood groups in humans are expressed as the IAlB, and IAalleles. The allele encodes the A...

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Q. 3: In rats, the following genotypes of two independently assorting autosomal genes determine cont color: A-B- (gray) A-bb (yellow) aab (black) aabb (cream) A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Ce genotypes allow color according to the expression of the A and B alleles. However, the ce genotype results in albino rats regardless of the A and B alleles present. Determine the Fl phenotypic ratio of the following crosses: (a) AAbbCC x aaBBcc (b) AaBBCC x AABbce (c) AaBbCc x AaBbcc (d) AaBBCe x AaBBCe (e) AABbCe x AABbee
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