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 (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter16: Human Genetics And The Human Genome

Section16.3: Genetic Diseases Caused By Single-gene Mutations

Problem 7LO: State whether each of the following genetic defects is inherited as an autosomal recessive,...

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Q11. One of the two genes known to be mutated in cases of Hypokalemic periodic paralysis (which is inherited in an autosomal dominant pattern but known to affect males more often than females) is the calcium voltage-gated channel subunit alpha1 S (CACNA1S). What is known about the gene is recorded here: https://www.ensembl.org/Homo_sapiens/Gene/Summary?db=core;g=ENSG00000081248;r=1:201039512-201112451 Please navigate to the link above and ensure that you click to reveal the transcript table. Then use the information in the table to answer the following question. What is the NCBI accession number (including the version) of the RefSeq Match for the first transcript (CACNA1S-201)? Answer: NCBI accession number (including the version) of the RefSeq Match for the first transcript (CACNA1S-201) is ...
. Mutations in an autosomal gene in humans cause aform of hemophilia called von Willebrand disease(vWD). This gene specifies a blood plasma proteincleverly called von Willebrand factor (vWF). vWFstabilizes factor VIII, a blood plasma protein specified by the wild-type hemophilia A gene. Factor VIIIis needed to form blood clots. Thus, factor VIII is rapidly destroyed in the absence of vWF.Which of the following might successfully be employed in the treatment of bleeding episodes in hemophiliac patients? Would the treatments workimmediately or only after some delay needed forprotein synthesis? Would the treatments have only ashort-term or a prolonged effect? Assume that allmutations are null (that is, the mutations result in thecomplete absence of the protein encoded by the gene)and that the plasma is cell-free.a. transfusion of plasma from normal blood into avWD patientb. transfusion of plasma from a vWD patient into adifferent vWD patientc. transfusion of plasma from a hemophilia A…
State whether each of the following genetic defects is inherited as an autosomal recessive, autosomal dominant, or X-linked recessive trait: phenylketonuria (PKU), sickle cell anemia, cystic fibrosis, Tay-Sachs disease, Huntingtons disease, and hemophilia A.
а. What type of inheritance is shown in the pedigree? II 2 II 1 2 4 IV Identify the genotypes of the following individuals: III- 2 b. |-1 Il- 2 III-4 IV-1 + v Paragraph В I
(19) WhatsApp 0 (542) H.E.R. - Changes X Why is it not possible X al%20School/OneDrive/Documents/Grade%2012%20variation%20and%20evolution%20 (D Page view A Read aloud V Draw Hig ii). iii). 13) 4. Sickle cell anemia is a favorable mutation. There is only one amino acid dillerent in Sickle hemoglobin (S) as compared with normal adult hemoglobin (A). Persons with sickle cell trait (AS) are able to survive in malarious areas. i) A couple each having sickle cell trait have five children. What is the probability thas they will have a child with sickle cell disease?
This is a blank question. Thank you in advance, Bloom Syndrome Bloom syndrome is a rare genetic disorder. It is characterized by short stature and a long narrow face with prominent nose and ears. There is also increased sensitivity to light. People who have the disorder often develop rashes on their face, forearms, and hands when they have been exposed to the sun. In addition, these people often suffer from chronic obstructive pulmonary disorder (COPD) and have a higher chance of developing cancer. The cause of this genetic disorder is a mutation in the BLM gene located on chromosome 15. The immediate effect of this mutation is that there is a defect in the functioning of the DNA helicase enzyme. What would be the effect of this mutation on DNA replication? What stage of the cell cycle would be most affected?
Talk about the challenges involved in determining the genetic components of polygenic illnesses. Explain complementation groups and how the biochemical underpinnings of disease are determined using them. Hereditary illnesses of genomic instability include Werner syndrome, Bloom syndrome, XP, ataxia-telangiectasia, and Fanconi anemia. Which of these ailments has molecular mechanisms behind it? Which kind of genetic instability is connected to which disorder?.
5. Sickle cell anemia is a genetically inherited autosomal recessive trait in which results in a condition that there are not enough healthy red blood cells to carry oxygen throughout the body due to the production of red blood cells that are sickle shaped. The frequency of the sickle cell condition is as high as 10% in Central Africa compared to 0.5% in the United States. (a) Calculate the frequency of the normal and carries of the sickle cell condition. (b) (b) Malaria is a disease caused by a parasite that infects healthy red blood cells and can lead to death. However, the parasite is not able to infect sickle cell red blood cells. Scientists claim that the higher frequency of sickle cell carriers is due to the presence of malaria. Provide reasoning to support the claim.
2. 235O 4) Q7. Haemophiliacs possess a non-functional form of the gene responsible for the production of blood clotting factors. Shown below is the occurrence of haemophilia in one family. = male = female = male haemophiliac 7. 8. 5. 9. 3. 11 12 Usingthe following symbols: H = dominant allele h = recessive allele 1) State the genotypes of the following individuals. Individual Genotype 1. 6. 2) On the basis of the information provided, is the inheritance of haemophilia: (i) autosomal or sex-linked? (ii) dominant or recessive? 3) State the probability of individual 8 being a carrier of haemophilia. 4) Explain why only females can be carriers of haemophilia.
Retinoblastoma: The Hits Just Keep Coming Part IV – Time to Reconvene Question Since Julie is indeed a carrier of the mutated RB1 allele and Chris is homozygous for the wild-type allele, what is the likelihood that their next child will inherit Julie’s RB1 mutation? Part V – A Different Kind of Hit Question Chris does not have an RB1 mutation, and is therefore homozygous wild-type. Julie is heterozygous for the mutation. However, Kay has inherited a different RB1 mutation than the one her mother carries. Therefore, Kay did not receive her mother’s mutant allele. Assuming that Chris really is the father, what other explanation might there be for how she got a germline mutation?
Mutation: Tan Body Pgeneration Phenotypes: Normal female X Tan male Fi generation Phenotvoe Females Males Total Ratio Normal Tan F, XF, Phenotypes: Normal female X Normal male F2 generation Phenotvoe Females Males Total Ratio Normal 8. 4. Tan 4 F Punnett square F. Punnett square Ths mutation is inhented aS 399F
Part 3C: Below is data from a pedigree trace of patients affected with mouse model of pancreatic cancer, and below are the observations regarding a new gene called 318 that is found on chromosome 5. Patient II-4 has two copies of loss of function variant alleles (or is homozygous for loss-of-function variant 318 allele). Patient I-2 has one copy of loss of function variant and one copy of unaffected allele (is heterozygous for loss-of-function variant 318 allele). Patient II-1 has two copies of unaffected allele (or is homozygous for unaffected 318 allele). |-1 1-2 Il-1 Il-2 Il-3 |l-4 II-5 II-1 III-2 III-3 Based upon data above, what is the mode of inheritance for mouse model for pancreatic cancer? Why?