Problem Set 2 KEY

2006 7.012 Problem Set 2 KEY Due before 5 PM on FRIDAY, September 29, 2006. Turn answers in to te box outside of 68-120. PLEASE WRITE YOUR ASWERS O THIS PRITOUT. 1. You are doin a enetics experiment wit te fruitfly Drosopila melanoaster. In te P eneration, you cross two true-breedin flies. Te two flies are a lare-wined airless female and a puny-wined airy male. All of te flies in te F1 eneration are lare-wined and airy. Indicate te alleles associated wit dominant penotypes by a capital letter and alleles associated wit recessive penotypes by a lowercase letter. Assume te two traits you are followin are autosomal. Indicate te air alleles by te letters H and and indicate te win alleles as G and. (a) Te enotypes of te flies in te P eneration are: GG females and HH males. Te enotypes of te flies in te F1 eneration are: GH females and GH males. You now take an F1 eneration female and cross er to a true-breedin puny-wined airless male. Tis male s enotype is:. (b) You count 1600 offsprin in te F2 eneration. If te win and te air traits were unlinked, you would expect to count: 400 # of lare-wined airy flies (of te enotype GH ) 400 # of lare-wined airless flies (of te enotype G ) 400 # of puny-wined airless flies (of te enotype ) 400 # of puny-wined airy flies (of te enotype H ) Unlinked enes assort independently from eac oter, tis means tat te GH female would create te four possible ametes (GH,, G, and H) eac 25% of te time. Eac of tese 4 different kinds of es would be fertilized by a sperm containin from te fater (as e is doubly omozyous, tere is only one kind of amete e can produce). 1

(c) Instead, wen you count te F2 eneration, you really et: 85 lare-wined airy flies 712 lare-wined airless flies 75 puny-wined airless flies 728 puny-wined airy flies Wat is te enetic distance between te air and win enes? It is 10% or 10 cm. In te moter s cromosomes, te G and allele o toeter and te and H allele o toeter. (Tis is because of wo er parents were; see te diaram below.) Terefore any ametes tat se produces tat are G or H will be parental type and any ametes tat se produces tat are GH or will be recombinants. Te two recombinant types of ametes produce puny-wined airless flies and lare-wined airy flies. Tere are 85 lare wined-airy flies produced and 75 puny-wined airless flies. So te recombination frequency = # of recombinants x 100 Total proeny = 85 + 75 x 100 = 160 = 10% 85 + 75 + 712 + 728 1600 Terefore, te distance between te air and win enes is 10 cm. (d) A series of fruit fly matins sows tat te recombination frequency between te ene for win size and te ene for antenna lent is 5% (i.e. te enetic distance between tem is 5 centimorans). List all possible recombination frequencies between te ene for amount of air and te ene for antenna lent. It could be 5% or 15%. Tis is because we know tat: 10 5 l (l = antenna lent) Tese enes can be arraned in two ways: 5 l 5 10 OR 10 5 l 15 Distance between l and = 10-5 = 5 cm Terefore, te recombination frequency would be 5 %. Distance between l and = 10 + 5 = 15 cm Terefore, te recombination frequency would be 15 %. 2

(e) Draw out te arranement of te air and win enes in te nucleus of a randomly selected eye cell in te followin flies (wic are te same flies tat were described above). Draw te enotype inside te circles below, wic indicate a nucleus of an eye cell. Te lon lines represent te pair of omoloous cromosomes tat ave te enes for te air and win traits on tem. Te tick marks indicate te position of te two enes. Write in a letter next to eac tick mark in eac cell to indicate wic alleles exist at eac of te two enes on eac of te two omolos. Explanation of drawin: Te maternal omolo Te air ene Te win ene P eneration female: P eneration male: Te paternal omolo G H G H F1 eneration female: puny-wined airless male (wo was crossed to te F1 female): G H Te F1 female must look like tis because se ad to et one cromosome from eac of er parents. 3

Te four possible F2 eneration flies created by te cross of te F1 female to te puny airless male: -- te two possible non-recombinants: G H Tese F2s are non-recombinants because tey received a cromosome from mom tat did not require recombination to form. Te oter cromosome was received from teir fater. -- te two possible recombinants: H G Tese F2s are recombinants because te cromosome tey received from teir moter does not exist in tat arranement in te moter s cells. Tus it must ave required recombination to ave formed. Te oter cromosome was received from teir fater. 2. Eac of te families below exibits a different very rare enetic disorder were individuals expressin te disorder are sown by solid symbols. Give all possible modes of ineritance tat are consistent wit eac pediree (your coices are: autosomal recessive, X-linked recessive, X-linked dominant, or autosomal dominant). Also calculate te probability tat te next cild indicated by a (?) will be affected iven eac possible mode of ineritance. In te case of X-linked ineritance, calculate separate probabilities for sons and dauters. (a) (b)?? 4

(c) (d)?? (a) Tis pediree cannot represent a disorder followin an autosomal recessive or X-linked recessive ineritance because, if eiter of tose were te case, all te proeny will be affected, because bot parents would ave only possessed te mutant alleles so tat tey would bot sow a recessive mutant disorder. It cannot be an X-linked dominant disorder because all dauters would ave been affected since all of tem would ave received a mutant allele from dad. Tus, te only possible mode of ineritance is autosomal dominant. Bot parents are eterozyous (Aa) so te probability tat te next cild will be affected is ¾. (b) Tis disorder cannot be X-linked recessive. If it ad been, ten no F1 females would ave ever been affected because tey would always receive a normal allele from dad. However, te disorder can be X-linked dominant, in wic case te fater would be X a Y (if we use te capital letter to indicate te allele tat ives te dominant penotype) and te moter would be X A X a. Tis would lead to a alf of te dauters and a alf of te sons bein affected. Tus, te probability tat te next cild is affected is: ½ for females and ½ for males. Tis disorder can also be autosomal dominant. For tis to be true, mom ave to be eterozyous (Aa) and dad omozyous (aa), so ½ of te proeny will receive mutant allele from mom and will be affected. Te two possible modes of ineritance listed above are equally likely to be correct, since only 1 mutant allele total needs to be present in te parents in order to be consistent wit te pediree. In contrast, iven tat te disorder is very rare (as stated in te introduction), it is very unlikely (altou formally possible) tat te mode of ineritance is autosomal recessive. For tis to occur, te mom would ave to be omozyous for te mutant allele (aa) and te dad a carrier (Aa.) Terefore, 3 mutant alleles would ave to be present in te P eneration, wic is improbable iven tat te disorder is very rare. If it were to occur tat an affected female would appen to start a family wit a carrier male, ten te probability of avin anoter affected cild is ½. (c) Tis pediree most likely tracks an autosomal dominant disorder or an X-linked dominant disorder. For autosomal dominant to be true, mom as to be eterozyous (Aa) and dad omozyous (aa) so te probability of avin an affected cild is ½. Tis pediree can also represent an X-linked dominant disorder, in wic case mom will be X A X a and dad X a Y. A alf of te sons and a alf of te dauters will receive te mutant allele from mom and will ave te disorder. Tese 2 patterns of ineritance are te most likely (and are equally likely) since te presence of only 1 mutant allele in te P eneration is enou to produce te pediree as we see it. Altou less likely, X-linked recessive ineritance is also consistent wit te pediree, in wic case mom will be omozyous (X a X a ) and dad will ave te normal allele (X A Y). All 5

te sons will receive a mutant allele from mom; terefore te probability of a male cild to be affected is 1. Since all dauters will receive a normal allele from dad, te probability of avin a female cild wit te disorder is 0. Tis mode is less likely since it requires mom to ave two mutant alleles, wic is rare for an X-linked recessive trait. (Most people affected wit an X-linked trait are male.) It is also formally possible but very unlikely tat te mode of ineritance is autosomal recessive; tis mode is very unlikely since 3 mutant alleles would ave to ave met in te P eneration to ave tis pediree work wit tis mode of ineritance (mom would ave to be aa and dad would be Aa). If tis were te case, ten te probability of avin an affected cild would be ½. (d) Tis disorder can neiter be autosomal dominant nor X-linked dominant, oterwise one of te parents of an affected offsprin sould be affected. Te disorder is most likely X-linked recessive. Tis would mean tat mom is a carrier (X A X a ) and dad is not (X A Y). Tus, ½ of te sons will be affected. Tere will never be an affected dauter, since se will always receive an X cromosome from dad carryin te normal allele. It is formally possible tat tis disorder can also be autosomal recessive, wit bot parents bein eterozyous (Aa). Tis would make te probability of avin an affected cild of ¼. However tis mode of ineritance is less likely tan X-linked recessive because it requires a male carrier and a female carrier to appen to meet, wic is unlikely iven ow rare te allele is tat confers te disorder. 3. You are studyin a type of yeast tat as two different cromosomes in its enome. You ave isolated tree mutations, a, b and d, eac of wic causes te same penotype. Wen you mate a strain containin any one of tese tree mutations to wild-type, te resultin diploid exibits te wild-type penotype. You are in te process of doin complementation tests wit tese mutants. You discover tat a and d do complement eac oter, but a and b do not. Te correspondin wild-type alleles are A, B and D. Draw in te correct alleles tat exist at eac of tese loci (A, B, and D) in eac of te six yeast cells drawn below. Make sure to put te alleles in teir correct locations, as determined by tose already drawn in for you. Also make sure to draw in te cromosomes to any cell wose cromosome(s) is/are missin. First cross: You mate aploid yeast of enotype a to aploid yeast of enotype d. a D d A a A X = D d 6

Mutations a and d do complement, indicatin tat te two mutations lie in different enes. Eac mutation is in one of two enes necessary for te wild-type penotype. Two separate enes are involved; eac mutant will ave a lesion in one ene wile maintainin a wild type copy of te second ene. Wen te diploid cell is produced, it will express te mutant allele a and te wild-type allele D. Te diploid will also express te wild-type allele A and te mutant allele d. Because te diploid is expressin bot of te necessary wild-type alleles, te wild-type penotype is observed. Second cross: You mate aploid yeast of enotype a to aploid yeast of enotype b. a b X = a b Mutations a and b do not complement, indicatin tat tese two mutations occur in te same ene. Eac omolo will express a mutant version of te ene in te diploid cell. Witout any normal functionin ene product in te cell, a mutant penotype occurs. Te A/a alleles and te B/b alleles must be in te same location in te enome because tey are in te same ene. 4. Sown below is a ypotetical sceme for te formation of eye piment in Drosopila. Bl Pr Blue piment Purple piment Red piment Te enzyme encoded by te Pr ene converts a purple piment into te normal red piment in te eye. Te pr allele causes te recessive penotype of purple eyes. Te enzyme encoded by Bl ene converts a blue piment into te purple piment. Te bl allele causes te recessive penotype of blue eyes. Bot te Pr and Bl enes are on te X cromosome. A male from a true-breedin blue-eyed strain is crossed to a female from a true-breedin purple-eyed strain. (a) All of te F1 female proeny from tis cross ave normal eyes. Wat colored eyes sould te F1 male proeny ave? 7

Tey sould ave purple eyes. P bl pr+ x pr bl+ Y pr bl+ F1 pr bl+ and pr bl+ Y pr+ bl All males inerit te mutant pr- allele from te P female, so tey will ave purple eye color. (b) An F1 female fly (wit normal eyes) is crossed to a wild-type male, and a lare number of male proeny from tis cross are examined. Amon te male proeny, tere are flies wit normal red eyes, flies wit purple eyes, and flies wit blue eyes. You notice tat sinificantly more male proeny ave blue eyes tan ave purple eyes. In one sentence or less, state wy tis sould be te case. Tere are two acceptable ways to prase tis: 1) Tere are more blue eyed flies tan purple flies because purple eyed flies can only result from a sinle mutant class of offsprin, wereas blue eyed flies result from a sinle mutant class and from a double mutant class. 2) Tere are more blue eyed flies tan purple flies because purple eyed flies can only result from a parental class of offsprin, wereas blue eyed flies result from a parental class and from a recombinant class. See te cross drawn out below part (c) for an explanation of wy tis statement is true. (c) Given tat te Pr and Bl enes are 20 cm apart on te X cromosome (i.e. te recombination frequency between tem is 20%), determine te number out of 100 male proeny from te cross in part (b) tat sould ave purple eyes, blue eyes, or normal red eyes. umber Purple-eyed males: 40 Blue-eyed males: 50 Red-eyed males: 10 Total = 100 8

Since te recombination frequency is 20%, tere are 20% X 100 = 20 recombinants present out of 100 offsprin. Tere will be two recombinant classes, eac of wic will exist in equal frequencies, so tere will be 10 of one recombinant class and 10 of te oter recombinant class. Tat leaves us wit 80 parental offsprin, wic will fall into two equally probable classes, so tere will be 40 of one parental class and 40 of te oter parental class. F1 pr bl+ x pr+ bl+ pr+ bl- Y F2 pr bl+ 40 Parental purple eyes Y pr+ bl Y pr bl Y pr+ bl+ Y 40 Parental blue eyes 10 Recombinant blue eyes 10 Recombinant red eyes 50 blue eyes We are only lookin at sons, and te male always ives te Y to is sons. So te sons eye colors are determined by te moter. Te moter will ive one of er cromosomes to 40% of te males, te oter of er cromosomes to 40% of te males, and recombinant cromosomes tat se erself does not possess to 20% of er sons. 5. In many of te classic experiments we discussed in lecture, scientists use different radioisotopes to label different components of te cell. Be as specific as possible in your answers below reardin tese radioisotopes. (a) Wic specific part of DA is radioactively labeled by 15? Te nitroen-containin bases. (b) Wic specific part of protein is radioactively labeled by S 35? 9

Te two R roups of te 2 amino acids tat contain sulfur namely, ysteine and metionine. (c) Wic specific part of DA is radioactively labeled by P 32? Te posporous tat is part of te pospate roup(s) tat an off te 5 carbon of te suar tat makes up te nucleotide. 6. Imaine an oranism tat uses te four followin nitroenous bases, drawn below. Oter tan te identity of te bases, assume tat every oter aspect of te structure and function of tis oranism s DA is te same as te DA of oranisms tat use A, G,, and T. Just as wit te usual A, G,, and T, tere are two exclusive pairs of bases tat ydroen bond wit eac oter. O O Q: R: H H O H suar H2 suar H H H2 H2 S: T: H H H suar H H2 suar H O 10

(a) Wic base do you tink pairs wit eac base below? ircle one base for eac. Q base pairs wit? (circle one) Q R S T R base pairs wit? (circle one) Q R S T S base pairs wit? (circle one) Q R S T T base pairs wit? (circle one) Q R S T We know tat, in DA, one purine always basepairs wit one pyrimidine, so wen we look at T (a purine, because it as two rins), we know it will base-pair wit eiter R or S, te two pyrimidines (because tey ave one rin). If we flip S over and line it up wit T as it would in DA, tree sets of cemical roups line up wit eac oter: Two H2 roups An and a H A H2 wit a O one of tese tree pairs of cemical roups can ydroen bond wit eac oter. Terefore, T must bind wit R. If you flip R over and line it up wit T, you et te followin cemical roups lined up wit eac oter: O and H2 and H H2 and O Te top and te bottom sets can Hydroen bond wit eac oter, so R forms 2 H bonds wit T. ow you can flip S over and line it up wit Q. Te cemical roups tat alin are: O and H2 H and O and H2 Te top two sets can Hydroen bond wit eac oter, so S forms 2 H bonds wit Q. (b) Draw one pair of bases, alinin tem as tey would in a double stranded DA molecule, and sowin te ydroen bonds tat could occur between tem. (You do not ave to draw te entire nucleotides; you only ave to draw te nitroenous bases.) 11

Draw eac ydroen bond as a series of sort dases: Q S (c) Draw te oter pair of bases, alinin tem as tey would in a double stranded DA molecule, and sowin te ydroen bonds tat could occur between tem. Draw te ydroen bonds as a series of sort dases. T R 12