In a family-based genetic study for a mental disorder called marijuana dependence, a total of 50 families, each with a pair of siblings, were recruited. The result showed that 11 families have the 1st siblings with marijuana dependence and the 2nd siblings without the disorder, while another 9 families are opposite (i.e. the 1st siblings without the disorder but the 2nd siblings with the disorder). In contrast, 13 families have both siblings exhibiting the disorder and another 17 families have both siblings without the disorder. What kind of heritability can you calculate using this family study and what is the heritability value for marijuana dependence? OA Broad-sense heritability, approximately 38% O B. Narrow-sense heritability, approximately 38% OC Narrow-sense heritability, approximately 76% O D. Broad-sense heritability, approximately 76%
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- Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?(7) In a family-based alcohol use disorder genetic study, a total of 50 families, each with a pair of siblings, were recruited. The result showed that 12 families have the 1" siblings with marijuana dependence and the 2nd siblings without the disorder, while another 11 families are opposite (i.e. the 1st siblings without the disorder but the 2nd siblings with the disorder). In contrast, 20 families have both siblings exhibiting the disorder and another 17 families have both siblings without the disorder. What kind of heritability can you calculate using this family study and what is the heritability value for alcohol use disorder? A) Narrow-sense heritability, approximately 88% B) Narrow-sense heritability, approximately 44% C) Broad-sense heritability, approximately 88% D) Broad-sense heritability, approximately 44%A woman diagnosed with early-onset Alzheimer's due to a mutation of the APP genehas children with a man that has no family history of familial Alzheimer's. Give the probability of each possible genotype with corresponding phenotype. (The woman has two possible allele combinations. You must show both possibilities)
- Concordance studies of twins for a neurodegenerative disorder show MZ= 46% and DZ= 15%. Further studies have shown a possible link to a gene on chromosome 9, however, there are some individuals in the study who have the allele but do not develop the disorder (group 1), and there are other individuals who do not have the allele yet develop the disorder (group 2). Amita's older sister and maternal uncle have this disorder. Currently, Amita & her 2 younger brothers do not show symptoms. Amita's paternal grandfather was rumored to have this disorder. 1. Draw the pedigree for Amita's family and determine the mode of inheritance if any. 2. Explain how the 2 groups in the study could be possible? 3. What would you tell Amita about the heritability of this disorder?Many genetic disorders exhibit locus heterogeneity. Define andgive two examples of locus heterogeneity. How does locus heterogeneityconfound a pedigree analysis?One particularly useful feature of the Hardy-Weinberg equation is that it allows us to estimate the frequency of heterozygotes for recessive genetic diseases, assuming that Hardy-Weinberg equilibrium exists. As an example, let’s consider cystic fibrosis, which is a human genetic disease involving a gene that encodes a chloride transporter. Persons with this disorder have an irregularity in salt and water balance. One of the symptoms is thick mucus in the lungs that can contribute to repeated lung infections. In populations of Northern European descent, the frequency of affected individuals is approximately 1 in 2500. Because this is a recessive disorder, affected individuals are homozygotes. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of individuals who are heterozygous carriers?
- In a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.A 20-year-old woman comes to your genetic counselling center because she knows that Huntington disease occurs in members of her family. Huntington is an autosomal dominant disease that often becomes apparent around 35–40 years of age. Her paternal grandfather was afflicted, but so far her 41 year old father shows no symptoms. Her two great-great grandmothers on her father's side were healthy well into their 90s, and one of her great-great grandfathers died of unknown causes at 45. Testing for Huntington disease is extremely expensive, but she is concerned that she may fall victim to this disease and wants to plan her life accordingly. After examining her pedigree you advise her to: not get tested because her father is only a carrier and it is very unlikely her mother is a carrier. not get tested because there is no possibility that she is homozygous. get tested because her father could be a carrier. not get tested because only males in her family get the disease. not get tested…The following genetic map describes three hypothetical human autosomal genes, each of which exhibits two alleles. Two-factor map distances are shown. A = Artistic (dominant) a = Inartistic (recessive) M = Moral (dominant) m = Immoral (recessive) G = Generous (dominant) g = Greedy (recessive) Assume that these traits exhibit simple Mendelian dominance/recessiveness. The coefficient of coincidence for this map is 0.4. An artistic, moral, generous heterozygous female of genotype AMG/amg marries an inartistic, immoral, greedy homozygous male of genotype amg/amg. What is the probability that their firstborn child will be inartistic, immoral and greedy? What is the probability that their firstborn child will be inartistic, moral and generous? What is the probability that their firstborn child will be artistic, immoral and generous?
- A woman has her personal genome analyzed for the BRCA1 mutation after learning that her father is heterozygous and carries one mutant allele. What is her chance of inheriting the mutant allele from her father? 0: men cannot transmit genes affecting breast cancer. 25% 50% 75% 100% Among the progeny of a heterozygous round (Aa) x homozygous wrinkled ( aa) cross, three seeds are chosen at random. What is the probability that all three seeds are round? (1/4)3 1/4 (1/2) 3 1/2 A single gene can produce different proteins. True FalseThe following pedigree illustrates the inheritance of a rare neurologicaldisease. What is the most likely mode of inheritance for this disorder?Explain your reasoning.The DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12