KEY12005 7.03 Problem Set 7 KEYNO DUE DATE. This problem set is to provide practice on concepts from lectures 30-36.1. The following pedigree shows the inheritance of an autosomal recessive trait in aspecific family. This trait is caused by a specific allele “g” at the G/g locus. You have somereason to suspect that the G/g locus is linked to an SSR on chromosome 6 called SSR41.You obtain blood samples from each member of the family, and perform a PCR reaction onthe DNA of each individual that allows for the genotyping of SSR41. The results of the PCRreactions are shown below each family member in the pedigree, in a schematic of anagarose gel in which you have loaded the PCR reactions from each family member into aseparate well in the gel.paternally inheritedallele at SSR41CCDCCDmaternally inheritedallele at SSR41BCBBCCpaternally inheritedallele at G/g locusggggggmaternally inheritedallele at G/g locusggGGgg(a) Fill in the tables above to indicate which alleles have been passed on to each child fromtheir mother and father.A alleleB alleleC alleleD alleleKEY2(b) Whose alleles (the mother’s or the father’s or both) should you follow to calculate theLOD score for the linkage of the SSR to the G/g locus?You want to follow the alleles of the parent who is heterozygous at both the disease locusand the SSR locus (so you can tell which of the children are parental and which arerecombinant). In this case that parent is the mother.(c) Draw all possible phases for the parent(s) you listed in part (b).G B inherited from grandpag C inherited from grandmaSince you know the genotypes of the mother’s parents, there is only one possible phasefor the mother.(d) For each phase you drew you drew in part (c), state how many children arerecombinants and how many children are parentals given that phase.GB from mother = parentalgC from mother = parentalGC from mother = recombinantgB from mother = recombinantTherefore, child 1 is recombinant, the other 5 are all parental.(e) Calculate the LOD score for this family at theta = 0.04 for the linkage of the SSR to theG/g locus.For the numerator of the odds ratio, you assume that the SSR and G/g locus are linked.Since theta =0.04, this means you’d expect 4% recombinants and 96% parentals.There are two types of recombinants, GC and gB. Each type of recombinant will occur atequal frequencies, so there is a 2% chance of any child getting GC from the mother and a2% chance of any child getting gB from the mother. Each type of parental will occur atequal frequencies, so there is a 48% chance of any child getting GB from the mother anda 48% chance of any child getting gC from the mother.Chances of the mom creating each type of gamete:GB-48%gC-48%GC-2%gB-2%KEY3Child one got Bg. The chance of this is 2%.Child one got Cg. The chance of this is 48%.Child one got BG. The chance of this is 48%.Child one got BG. The chance of this is 48%.Child one got Cg. The chance of this is 48%.Child one got Cg. The chance of this is 48%.The chance that you saw these six kids would be 2% * 48% * 48% * 48% * 48% * 48% .For the denominator of the odds ratio, you assume that the SSR and G/g locus are NOTlinked. UN-linkage corresponds to theta = 0.5, because UN-linkage gives 50% parentalsand 50% recombinants.Since theta =0.5, this means you’d expect 50% recombinants and 50% parentals.There are two types of recombinants, GC and gB. Each type of recombinant will occur atequal frequencies, so there is a 25% chance of any child getting GC from the mother anda 25% chance of any child getting gB from the mother. Each type of parental will occur atequal frequencies, so there is a 25% chance of any child getting GB from the mother anda 25% chance of any child getting gC from the mother.Chances of the mom creating each type of gamete:GB-25%gC-25%GC-25%gB-25%Child one got Bg. The chance of this is 25%.Child one got Cg. The chance of this is 25%.Child one got BG. The chance of this is 25%.Child one got BG. The chance of this is 25%.Child one got Cg. The chance of this is 25%.Child one got Cg. The chance of this is 25%.The chance that you saw these six kids would be 25% * 25% * 25% * 25% * 25% * 25% .LOD = log of the odds ratio€ log10(0.48)5(0.02)1(0.25)6= 0.32KEY4(f) At what theta value would you achieve the maximal LOD score for this family, knowingeverything you know about them?You’d expect to achieve the maximal LOD score when theta = the fraction of recombinantchildren that you actually have in your family. In this case 1 in 6 children wererecombinant so you’d expect the maximum LOD score when theta = 1/6 = 0.17.(g) What is the LOD score value for the theta value you listed in part (f)?If theta = 0.17, you’d expect 83% total parentals and 17% total recombinants.€ log10(0.42)5(0.08)1(0.25)6= 0.63(h) If you had never seen the genotyping results for this family, and only had their pedigreeavailable, what would have been the theoretical maximum LOD score value that you couldhave ever calculated for this family? (Hint: Start by thinking about which theta value couldgive you the maximum possible LOD score.)The theoretical maximum LOD score for an SSR and a trait locus from a6-child family in which there is one relevant parent whose phase you do indeedknow occurs if all children are parentals. If all children are parentals, then the optimalLOD score value would occur at a theta = 0. Thus the theoreticalmaximum is:€ log10(0.5)6(0.25)6= 1.8(i) If you had never seen the genotyping results for this family, and only had their pedigreeavailable, what is the minimum number of kids that the family would have had to havecontained in order to reach a theoretical maximum LOD score that is > 3?€ log10(0.5)x(0.25)x≥ 3Solving for x, x = at least 10 children.KEY52. A tumor results when a cell in the body loses control over cell growth and division such that thecell divides many times, forming a ball of cells. Cancer can be extremely harmful to the organismwhen these balls of cells either physically interfere with function of an essential organ, or begin tosteal the nutrients away from cells of essential organs. Cells become capable of growing anddividing inappropriately when they have accumulated multiple mutations in genes (such asoncogenes and tumor suppressor genes) whose normal functions are to control cell growthand division (i.e. to control the cell cycle).(a) Why is the notion of there being “a cure for cancer” unreasonable?As we
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