Below is a pedigree of a rare human genetic disease. The filled in symbols indicate affected individuals. Assume that the disease is caused by a mutant allele of gene A. 1 2 3 4 If individuals 1 and 4 have a child together, what is the probability that the child will exhibit the disease? O 75% 30% 50% O 100%
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- The attached image is a pedigree of a family with a history of sickle cell anemia (the individuals with the filled-in symbols have the disease and no new mutations are occurring in any individual). Sickle cell anemia is inherited in an autosomal recessive manner. What is the probability that the individual with the question mark (?) will get the disease? a) 1/4 b) 1/2 c) 2/3 d) 1The pedigree below is of a rare single gene disease. Given the pedigree shown below, what is the most likely pattern of inheritance for this disease? Explain your answer. Note that closed symbols are for individuals affected by the disease and open symbols are unaffected individuals (limit 3-4 sentences).Duchenne muscular dystrophy is sex linked and usuallyaffects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brotherhas Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’sdisease, what is the probability that you have receivedthe allele?c. If your father’s brother had the disease, what is theprobability that you have received the allele?
- Below is a pedigree of a human genetic disease in which solid color indicates affected individuals. Assume that the disease is caused by a gene that can have the alleles A or a. a) Based on this pedigree, what is the most likely mode of inheritance? b) What is/are the possible genotype/s of person 1? c) What is/are the possible genotype (s) of person 4 ? Explain your answers.Galactosemia is a recessive human disease that istreatable by restricting lactose and glucose in the diet.Susan Smithers and her husband are both heterozygous for the galactosemia gene.a. Susan is pregnant with twins. If she hasfraternal (nonidentical) twins, what is theprobability both of the twins will be girls whohave galactosemia?b. If the twins are identical, what is the probabilitythat both will be girls and have galactosemia?For parts (c–g), assume that none of the children isa twin.c. If Susan and her husband have four children, whatis the probability that none of the four will havegalactosemia?d. If the couple has four children, what is the probability that at least one child will have galactosemia?e. If the couple has four children, what is the probability that the first two will have galactosemia andthe second two will not?f. If the couple has three children, what is the probability that two of the children will have galactosemia and one will not, regardless of order?g. If…In a particular human family, John and his motherboth have brachydactyly (a rare autosomal dominant allele causing short fingers). John’s father hasHuntington disease (another rare autosomal dominant allele). John’s wife is phenotypically normaland is pregnant. Two-thirds of people who inheritthe Huntington (HD) allele show symptoms by age50, and John is 50 and has no symptoms.Brachydactyly is 90% penetrant.a. What are the genotypes of John’s parents?b. What are the possible genotypes for John? Howlikely is John to have each of these genotypes?c. What is the probability the child will express bothbrachydactyly and Huntington disease by age 50 ifthe two genes are unlinked?d. How will your answer to part (c) change if insteadthese two loci are 20 m.u. apart?
- In letter B: If the map distance equals the number of recombinant/total of offspring, wouldn't it be 24/806 x 100? Wouldn't we add both recombinants? Can you explain letter C? I don't grasp that concept well. And since I'm using my question already, would you be able to answer D. Thank you!Tay-Sachs disease is a rare human disease in which toxic substances accumulate in nerve cells. The recessive allele responsible for the disease is inherited in a simple Mendelian manner. For unknown reasons, the allele is more common in populations of Ashkenazi Jews of eastern Europe. A woman is planning to marry her first cousin, but the couple discovers that their shared grandfather’s sister died in infancy of Tay-Sachs disease.a. Draw the relevant parts of the pedigree, and show all the genotypes as completely as possible. b. What is the probability that the cousins’ first child will have Tay-Sachs disease, assuming that all people who marry into the family are homozygous normal?. A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The second cross of pure lines is A/A • B/B • C/C • D/D •E/E × a/a • B/B • c/c • D/D • e/e.The geneticist crosses the F1 from this cross with arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?
- Cystic fibrosis is a genetic disease caused by a recessive allele. Two parents without the disease have a child with the disease, indicating they are both heterozygotes. Assuming that the child did not have a mutation giving it the disease, what is the probability that a second child of the same parents would have the disease? Group of answer choices a. 1 b. 0.25 c. 0 d. 0.5Sickle cell anemia is a recessive trait in humans. In a cross between a father who has sickle cell anemia and a mother who is heterozygous for the sickle cell allele, what is the probability that both of their first two children will be affected? A) none B) 1/16 C) 1/8 D) ¼ E) ½An individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriate