ANSWERS TO Problem set questions from Final Exam Human Genetics Nondisjunction and Cancer Mapping in humans using SSRs and LOD scores 1 You set out to genetically map the locus for color blindness with respect to SSR markers a A B b log c A a B c A cb cb B 8 2 0 25 10 2 1 2 0 4 0 1 1 2 0 4 0 1 A B d log c 9 1 1 1 2 0 4 0 1 1 2 0 4 0 1 0 25 a 8 9 10 2 You are conducting genetic linkage studies to search for a locus whose chromosomal location has not been firmly established associated with an autosomal recessive disease a log 4 0 0 1 2 0 49 0 01 1 2 0 49 0 01 0 25 4 4 0 87 b log 4 0 25 c log 4 0 0 4 0 87 1 1 3 0 82 4 3 1 2 0 49 0 01 1 2 0 49 0 01 2 log 4 4 0 0 1 2 0 49 0 01 1 2 0 49 0 01 0 25 f 0 87 4 4 0 25 e 0 1 2 0 49 0 01 1 2 0 49 0 01 0 25 d log 0 1 2 0 49 0 01 1 2 0 49 0 01 4 2 0 87 1 74 4 LOD 0 87 0 82 0 05 g no h yes i yes j no conclusions can be made that are publishable LOD score 3 3 Childhood deafness is often hereditary a autosomal recessive b the mating of these two people is like a human complementation test they had mutations in two different genes c no 4 Your colleague a human geneticist is conducting genetic linkage studies on the locus associated with an autosomal dominant disease a negative infinity there is a recombinant child and thus the two loci cannot be linked at theta 0 b negative infinity c log 4 1 0 25 d 0 1 2 0 5 0 00 1 2 0 5 0 00 0 903 4 4 2 log 4 1 0 1 2 0 5 0 00 1 2 0 5 0 00 0 25 4 2 0 903 1 806 4 5 As we have discussed in class SSR based genetic linkage studies in human families can be used to chromosomally localize the loci associated with heritable traits including diseases a log 1 0 05 0 45 0 25 4 log 5 1 0 05 0 45 0 25 4 0 644 5 b 0 2 LOD 2 log 1 0 1 0 4 4 0 837 5 0 25 c 20 cM 6 You are studying a mutation that causes an autosomal recessive phenotype of blindness in humans a you can t tell which of the genes encodes this mRNA because you don t know how much of the DNA length of each gene is introns and thus wouldn t be included in the final mRNA transcript b DNA sequencing find a gene by sequencing in this region that is wild type in all 10 individuals with normal vision but is mutated in all 10 who are blind Calculating Phenotypic concordance using twin studies 1 Congenital pyloric stenosis an obstruction to the stomach s outlet to the small intestine has a population incidence of 0 5 in newborn boys and of 0 1 in newborn girls a possibility 2 because if 1 were correct then girls should have been affected much much less often than boys b no c yes there is a genetic component it appears that two genes are involved and there is also an environmental component d because these offspring developed in the womb of a mother who was affected so the womb environment such as the mother s hormones influenced the development of the fetus 2 What phenotypic concordance rates approximate answers will suffice might you expect in MZ twins DZ twins and first cousins for each of the following diseases a Chicken pox MZ 100 DZ 100 1st cousins about the same as unrelated people but depending on the amount of time 1st cousins spend with each other b Tay Sachs disease MZ 100 DZ 25 1st cousins 0 c An autosomal dominant trait in which environment and a single gene are determinants MZ 80 DZ 40 1st cousins 10 Meiosis and chromosome loss gain by non disjunction 1 While working as a medical geneticist you encounter an unusual patient a 47 XXY girl a mother b meiosis I c i the cell in metaphase I 1A 2B 1A 2B 1C 2A 1C 2A 3B 4C 3B 4C 3A 4C 3A 4C ii the two cells in metaphase II 1A 2B 1A 2B 3B 4C 3B 4C 1C 2A 1C 2A 3A 4C 3A 4C iii the four final products of the meiosis 1A 2B 1C 2A 1A 2B 1C 2A 3B 4C 3A 4C 3B 4C 3A 4C d a deletion of the Sry gene e the father had a translocation between the X and Y chromosomes such that the X chromosome carried Sry and the Y chromosome carried SSR1 f in the father during meiosis II 2 Trisomy X that is XXX is one of the most common trisomies observed in human populations a father b meiosis II c i the cell in metaphase I 1A 1A 2B 3B 4A 2B 3B 4A ii the two cells in metaphase II 1A 1A 2B 3B 4A 2B 3B 4A iii the four final products of the meiosis 1A 1A 2B 3B 4A 2B 3B 4A d mother e meiosis II f i the cell in metaphase I 1A 1A 1C 1C 2A 2A 2B 2B 3C 4A 3C 4A 3A 4B 3A 4B ii the two cells in metaphase II 1A 1A 1C 1C 2A 2A 2B 2B 3C 4A 3A 4B 3C 4A 3A 4B iii the four final products of the meiosis 1A 1A 2A 2A 3A 4B 3C 4A 1C 1C 2B 2B 3C 4A 3A 4B 3 A married couple who already had a child with cystic fibrosis approach you because they wish to have another child but only if they can be assured that the child will not have cystic fibrosis a recombination the cell in metaphase I ii the two cells in metaphase II THE POLAR BODY b 2 or 3 because the remaining diploid cell that is about to undergo meiosis II only has wild type copies of the gene c 1 or 5 because then you can test the second polar body to see if it has the deletion and if it does then you know that the oocyte has the wild type allele 4 Trisomy 18 is one of the most common trisomies observed in human populations a mother b meiosis II c i the cell in metaphase I B B C C F F D D H J H J G K G K ii the two cells in metaphase II B B C C F F D D H J G K H …
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