7.03 Genetics: Answers to Problem Set 2 (1997)

7.03 Problem Set 2 ANSWERS 1. Suppose that there are two different mouse genes, 40 cM apart, that when mutated will produce mice with six toes. Further, suppose that the combination of mutations in both genes is lethal. You are given two different six-toed mice each from a different true-breeding line. A cross of the two mice produce F1 mice which all look normal. (a) What can you say about the two true-breeding lines that produced the mice that you were given? Because all of the F1 progeny are phenotypically normal, you know that the mutations in each of the parental strains complement each other. Therefore, true-breeding line is homozygous for a recessive mutation in different each genes. Calling the two linked genes A and B: _" .__... C A A b b C _ You now inter-cross X A b a B ,-. F1 mice to produce .- -c a a B B , '_ F2 progeny. (b) What fraction of the F2 mice that are born will look like wild-type? Remember you must take into account both recombination and the lethality of double mutants. This problem crossing should make it clear why mapping studies heterozygous individuals to a homozygous are, whenever possible, recessive test strain. done by Because the two genes are linked, they will not assort independently in the F2. Therefore, the relative frequencies of gametes produced by the heterozygous FI parents must be determined. Crossover classes, during and after meiosis I, are drawn on the next page (applies for both parents). Since the two genes are 40 cM apart, each crossover class should appear in 20% of total gametes. Possible Gametes: C) A b O A b 0.3 _B B _O © A a Bb 0.2 0.2 A ...... a /_ (7) _ a Frequency: b a B A Punnett square can then be used to represent all possible combinations gametes, and their relative frequencies, in the F2. AB 0.2 Ab 0.3 aB 0.3 ab 0.2 Wild-type AB 0.2 Ab 0.3 aB 0.3 ab 0.2 AABB 0.04 AAbB 0.06 aABB 0.06 aAbB 0.04 AABb 0.06 AAbb 0.09 aABb 0.09 aAbb 0.06 AaBb 0.06 AAbB 0.09 aaBB 0.09 aabB 0.06 AaBb 0.04 Aabb 0.06 aaBb 0.06 aabb 0.04 mice are in bold. figure into the percentage fraction wild-type= F2 progeny with genotype L of aabb are dead and do not of mice that are phenotypically (4+6+6+4+6+9+6+9+4) 96 _ 0.3 normal. = 54 96 = 56.25°'0 2. Consider the three hypothetical recessive traits in Drosophilia: bent wings (bt), yellow body (yl), and long bristles (Ig). You are given a fly that looks like wild-type, but is heterozygous for all three traits. You set up a test cross of your to a that is homozygous recessive for all three traits. From 500 progeny the following types are found (abbreviations are used for recessive traits and the corresponding dominant traits are indicated by a +). Phenotype bt yl Ig + + + bt+ + number 34 46 13 + yl Ig bt yl + + + Ig bt + Ig + yl + 7 202 193 3 2 ° (a) If the fly that you were given came from a cross of two true-breeding would be the phenotypes of the two true-breeding lines? lines, what The two true breeding lines would have the phenotypes of the 2 pare_ztaI classes. The two parental classes are the most common. Therefore the two true breedi'r_ li_zes are bt yl + and + + Ig bt yl + + + Ig (b) genes Produce a genetic map showing the order and the relative positions that control the three traits. of the To determine the gene order it is necessary to determine the double recombinant classes. The double recombinant classes are the lest frequent classes. They correspond to bt + Ig and + yl + The order that will give you the double recombinant recombination events occur is the correct order. .... bt--_--yl---V-+ ........... .... +--_--=.... +--_----'_--Ig........... ,.-_ classes when two bt .......... +........... + ....... +.......... yl........... Ig....... not doz_ble recombinant class .... bt---G-T-+----,q2---yl ........... bt.......... Ig.......... yl...... .... +__-_..__.__Ig_..z____.__+ ............ + ........... + ........... + ...... not do,_bIe recombinant class .... yl----KT---bt---_.i-2.--+........... yl.......... +........... +....... j, .... +--J--_--+---_--_,-o-lg ........... +.......... bt........... lg....... Double recombinant class yl--bt--Ig The order is The distance between yl and bt: recombinant classes for yI and bt bt + + + yl Ig and bt + Ig + yl + distance = rec/totaI x I00 = (13+7+3+2)/500 x 100 = 5 cM The distance between bt and lg: recombinant classes for bt and Ig bt yl Ig + + + and bt + Ig + yl + distance = rec/total x 100 = (34+46+3+2)/500 .... yl ........ bt ................ I--5cM--I-17cM Ig ....... --I x 100 = 17 cM 3, You have isolated three new yeast mutants that are copper-resistant. Mutant 1 is dominant for resistance and is in mating type c_. Mutants 2 and 3 are recessive and are in mating type a. You mate mutant 1 to mutant 2, sporulate the resulting diploid and dissect tetrads. Out of 25 tetrads, 4 have two copper resistant spores and two copper sensitive (wild-type) spores, 5 have four copper-resistant spores, and 16 have three copper-resistant spores and one wild-type spore. (a) What can you conclude about the relative P.ositions of mutation 1 and mutation 2 ? mutlR,mut2S x mutlS,mut2R .... mulR ..... mut2S .... .... mulS ..... mut2R .... PD spore ratio # of tet. T NPD rnut l R,rnzlt2 s mutl R,mz_t2 s mzLtl S,mut2 R rnu t l R,rn_t2 s rnutl S,mut2 S mutl R,mz_t2 R rnut l R,m_t2 R rnutl R,mut2 R mut l S,mut2 s ;.nz_tl S,mut2 R mut ! S,mut2 R mz_tl S,mut 2 s 4 rest : 0 se_s 5 3 rest : 1 sens 16 2 rest : 2 sens 4 ratio of PD • T • NPD is I " 4 • 1 suggesting the two genes are _IiT_ked (b) If instead of dissecting tetrads you had examined 100 random spores, what fraction of the spores would you expect to be copper-resistant? What fraction of the copper-resistant spores would you expect to produce copper-resistant diploids when mated to wild-type? The genes are unlinked so they z_ndergo independent assortment i.e. yo_ have 4 eqztaIIy likely genotypes: mutlR,mut2 s mutlS,mut2 R mutlR,m_zt2 R mutlS,mz_t2 S The first three genotypes have a copper-resistant phenotype while the other is copper-sensitive, therefore, 3/4 will be copper resistant and 1/4 will be sensetive. Only the spores that are rout1R will give rise to copper-resistant diploids. Only 1st and 3rd genotypes are rnutl R, therefore, only 1/2 will give rise to resistant diploids. the Next, you mate mutant I to mutant 3, spcrulate the diploid and dissect tetrads. tetrads, 21 have four copper-resistant spores, and 4 have three copper-resistant spores and one wild-type spore. (c) What can you conclude mutlR,mut3 spore ratio # of _et. about the relative S x positions mutlS,m_t3 .... mul R..... rout3 s .... .... mulS ..... mut3R .... -_ PD T of mutation Of 25 1 and mutation 3 ? R NPD mut l R,mut3 S mut l R,mut3 s mut l R,mut3 S mut I S,ml_t3 s rnut l R,mut3 R mut l R,mut3 R mut l S,mut3R mut l S,mut3 R mut l R,mut3 R mutl S,mut3 R mut l S,mut3 S mut l S,mut3S 4 rest " O sens 21 3 rest ' l sens 4 2 rest " 2 sens 0 ratio of ?D " T ' NPD is not I • 4 • I suggesting the two ge:_es are Iinked distane = 100 x # tetratypes/(2 x total tetrads) = 4 / (2 x 25) = 8 cM (d) If instead of dissecting tetrads you had examined 100 random spores, what fraction of the spores would you expect to be copper-resistant? What fraction of the copper-resistant spores would you expect to produce copper-resistant diploids when mated to wild-type? The genes are i_ked so they do not undergo independent assortment i.e. you have 4 ge_otypes of varying _-enotypes: mut l R,ml_t3 s mut l S,m_t3 R mut l R,rnut3 R rnut l S, m_zt3 s The first two genotypes are parental classes and the other two are recombinant classes. When two genes are 8 cM apart then 92°'o of the offspring will be of the parental classes while 81/o will be recombinant. Therefore the fraction of each genotype is mutlR,mut3 s mutlS,mut3 R mutlR,mut3 R mutIS,mut35 46% 46% 4% 4% The first three genotypes have copper-sensitive, therefore, 96% Only the spores that are rout1R 1st and 3rd genotypes are rout1 diploids. a copper-resistant phenotype while the other is wiI1 be copper resistant a_d 4°'o wilI be sensetive. will give rise to copper-resistant diploids. Only the R, therefore, only 50%wili give rise to resistant