ANSWERS TO Problem set questions from Final Exam – Human Genetics,Nondisjunction, and CancerMapping in humans using SSRs and LOD scores1. You set out to genetically map the locus for color blindness with respect to SSRmarkers.(a) A c A a B a B c(b) log (1/2) (0.4)8 (0.1)2 + (1/2) (0.4)2 (0.1)8(0.25)10(c) A + A cb B cb B +(d) log (1/2) (0.4)9 (0.1)1 + (1/2) (0.4)1 (0.1)9(0.25)102. You are conducting genetic linkage studies to search for a locus (whosechromosomal location has not been firmly established) associated with an autosomalrecessive disease.(a) log (1/2) (0.49)4 (0.01)0 + (1/2) (0.49)0 (0.01)4 = 0.87(0.25)4(b) log (1/2) (0.49)4 (0.01)0 + (1/2) (0.49)0 (0.01)4 = 0.87(0.25)4(c) log (1/2) (0.49)4 (0.01)0 + (1/2) (0.49)0 (0.01)4 = 0.87(0.25)4(d) log (1/2) (0.49)3 (0.01)1 + (1/2) (0.49)1 (0.01)3 = - 0.82(0.25)4(e) 2 * log (1/2) (0.49)4 (0.01)0 + (1/2) (0.49)0 (0.01)4 = 2*(0.87) = 1.74(0.25)4(f) 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 hadmutations in two different genes(c) no4. Your colleague, a human geneticist, is conducting genetic linkage studies on thelocus associated with an autosomal dominant disease.(a) negative infinity (there is a recombinant child and thus the two loci cannot be linkedat theta = 0)(b) negative infinity(c) log (1/2) (0.5)4 (0.00)1 + (1/2) (0.5)0 (0.00)4 = 0.903(0.25)4(d) 2 * log (1/2) (0.5)4 (0.00)1 + (1/2) (0.5)0 (0.00)4 = 2*(0.903) = 1.806(0.25)45. As we have discussed in class, SSR-based genetic linkage studies in humanfamilies can be used to chromosomally localize the loci associated with heritable traits,including diseases.(a) log (0.05)1 (0.45)4 + log (0.05)1 (0.45)4 = 0.644(0.25)5 (0.25)5(b) θ = 0.2LOD = 2 * log (0.1)1 (0.4)4 = 0.837 (0.25)5(c) 20 cM6. You are studying a mutation that causes an autosomal recessive phenotype ofblindness in humans.(a) you can’t tell which of the genes encodes this mRNA because you don’t know howmuch of the DNA length of each gene is introns, and thus wouldn’t be included in thefinal mRNA transcript(b) DNA sequencing – find a gene by sequencing in this region that is wild-type in all 10individuals with normal vision, but is mutated in all 10 who are blindCalculating Phenotypic concordance using twin studies1. Congenital pyloric stenosis (an obstruction to the stomach’s outlet to the smallintestine) has a population incidence of 0.5% in newborn boys and of 0.1% in newborngirls.(a) possibility #2, because if #1 were correct then girls should have been affected muchmuch less often than boys(b) no(c) yes there is a genetic component, it appears that two genes are involved, and thereis also an environmental component(d) because these offspring developed in the womb of a mother who was affected, sothe womb environment (such as the mother’s hormones) influenced the development ofthe fetus2. What phenotypic concordance rates (approximate answers will suffice) might youexpect 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 time1st 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 aredeterminants:MZ – 80%DZ— 40%1st cousins – 10%Meiosis and chromosome loss/gain by non-disjunction1. While working as a medical geneticist, you encounter an unusual patient: a 47,XXYgirl.(a) mother(b) meiosis I(c)i) the cell in metaphase Iii) the two cells in metaphase II 1A 2B 3B 4C 1A 2B 3B 4C 1C 2A 3A 4C 1C 2A 3A 4C 1A 2B 3B 4C 1A 2B 3B 4C 1C 2A 3A 4C 1C 2A 3A 4Ciii) the four final products of the meiosis(d) a deletion of the Sry gene(e) the father had a translocation between the X and Y chromosomes such that the Xchromosome carried Sry and the Y chromosome carried SSR1(f) in the father, during meiosis II2. Trisomy X (that is, XXX) is one of the most common trisomies observed in humanpopulations.(a) father(b) meiosis II(c)i) the cell in metaphase I 1A 2B 3B 4C 1A 2B 3B 4C 1C 2A 3A 4C 1C 2A 3A 4C* 1A 2B 3B 4A 1A 2B 3B 4Aii) the two cells in metaphase IIiii) the four final products of the meiosis(d) mother(e) meiosis II* 1A 2B 3B 4A 1A 2B 3B 4A 1A 2B 3B 4A 1A 2B 3B 4A(f)i) the cell in metaphase Iii) the two cells in metaphase II 1A 2A 3C 4A 1A 2A 3C 4A 1C 2B 3A 4B 1C 2B 3A 4B 1A 2A 3C 4A 1A 2A 3A 4B 1C 2B 3C 4A 1C 2B 3A 4Biii) the four final products of the meiosis3. A married couple who already had a child with cystic fibrosis approach you becausethey wish to have another child, but only if they can be assured that the child will nothave cystic fibrosis.(a) recombinationthe cell in metaphase I: + +__* 1A 2A 3C 4A 1A 2A 3A 4B 1C 2B 3C 4A 1C 2B 3A 4B__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 onlyhas 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 allele4. 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 F H J B F H J C D G K C D G Kii) the two cells in metaphase IIiii) the four final products of the meiosis(d) father(e) meiosis I* B F H J B F G K C D H J C D G K B F H J B F G K C D H J C D G K(f)i) the cell in metaphase Iii) the two cells in metaphase II B E H J B E H J C F G L C F
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