Note for BIOLOGY 283 at UMass(3)
Note for BIOLOGY 283 at UMass(3)
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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 15 views.
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Date Created: 02/06/15
Name Student ID 4 You are a criminal lawyer defending Willy Winn a suspect in a murder case Early on in the case there were two suspects but unfortunately the other suspect Harry quotLuckyquot Bottoms was knifed while awaiting questioning and died A bureaucratic foulup allowed his body to be cremated before tissue and blood samples were taken Hence Lucky39s blood type is unknown Both Willy and Lucky have been placed by witnesses at the scene of the crime Tissue samples taken from under the victim39s ngernails indicated that his assailant was blood type 0 Willy s blood type is type 0 Somehow you have to prove that Lucky could be the murdererthat will cast reasonable doubt as to whether Willy was the real culprit You must nd out whether Lucky could have been type 0 if so he could have been the murderer Fortunately Lucky s parents and a few of siblings are aliveand will consent to ivin asam lefort in S g P may 3 The results arc a r 1quot k 399 39i Lucky s mother A tax A b 19 39 r you Jr Lucky s father B fiti l b A a 4 n 39 fl Lucky39s sisters A and A39B m w quot b C I I M I Iquot Lucky39s brother B 1 39 M 6023 s be I 39 X led 00M a 2 points Could Lucky be the murderer lei no 5 39 9 X 15 can f es s L 1 39 71 lquot My 33 by M roan poiggg Q iquot b 2 points The Jury aren39t buying it they all unked genetics hey insist that the emcesiof being type 0 considering his parents39 blood types is practically G 7 Explain how Lucky could have had type 0 blood i z quot W Lu Lquot quot39M 393 E I A v v Ti has at JAuJAiLU39 who r 39 a r 1 r 0 AC mus 6 c 2 points Late in the trial a woman comes forward who child The mother39s bl type is A and so is the child39s 7 Na rut 39 Luv x 4quot 097M a 9 ch EreM d2 pointsYouswintltemmwmygocsfree 39 i Lucky s other illegitimatechild comes forw 39 Should Willy make am n for the border Name Student ID 6 Albino CA black CB cream cc and sepiacs are all coat colors of guinea pigs Individual animals not necessarily from pure lines showing these colors were mated the results are tabulated below Cross Parental phenotypes Phenotypes of progeny albino black cream sepia l bla kXblacgk 39A 31 L c X c c 2 black X albino 392 2 Lacs en 3 cre39gm X cream 4 31 C C XC C 4 sepia X cream 11 V4 392 L SL q XL C 5 blacanlbinoH 392 12 a L6 C x ago 6 blackXcream 11L b5 12 12 C 5 K cccUw corc A I l 7 Bblagkxxsepi 4600 H he isVic L r c 12 a a L39 c C C occult bccsjc ltL 8 ablagszepisa 2 12 14 c V IL inc 4 Av a Sc ms C Is Xcm on r wc CM 39 Lcll Alohalg 9 se Xse ia 39A 3 S p5 p5 2 c or be WC 5 uf c C XL C 0quot C C hc o rf cm H Ac 10 creamXalbino 392 39 12 c A A C CH X6 6 a 8 points Write the genotypes of the parents for each of the crosses underneath their phenotypes If a genotype cannot be determined indicate any unknown alleles using cquot b 4 points Write the dominance ranking of the albino CA black CB cream c and sepiac alleles S C H c8gtc gtc gtc Name Student ID b 1 points Write the dominance ranking ofthe albino CA black CB cream cc and sepiacs alleles 3 6 points In sweet peas the intensity of purple pigment in the flowers varies widely depending on the genetic background of the plant Generally sweet peas could be categorized as one of these dark purple purple pale purple white In sweet peas the gene p encodes an enzyme called P that converts white precursor into a purple pigment This pigment is responsible for the typical purple color of sweet pea flowers In other words plants homozygous for the WT allele pp have purple flowers Notice that the flowers are not dark purple they are merely purple The activity of the enzyme can be measured in tissue extracts and in mixtures of tissue extracts In the table tissue extracts were prepared from flowers of plants that have particular genotypes see first column The activity of the enzyme P was measured in each extract and is given as percent activity In the third column 1 ml of each tissue extract was mixed with 1 ml of extract from a pp flower and the activity of P was measured Genotype Percent activity Percent activity when mixed 5050 with pp extract pp 100 100 pZpz 0 50 p3p3 300 200 a 2 points What flower color phenotype do you expect for a plant that has the genotype pZp Since the p2 allele does not produce a functional enzyme 0 activity in the table and since the mixture of P and p2 extracts has only 50 activity I expect about 50 of the normal pigment amount which would give me a pale purple color b 2 points What flower color phenotype do you expect for a plant that has the genotype p3p Since the p3 allele produces a functional enzyme with extra high activity 300 percent and since the mixture of p and p2 extracts has excess activity I expect about 200 of the normal pigment amount which would give me a dark purple color c 2 points What flower color phenotype do you expect for a plant that has the genotype pZpz Since the p2 allele does not produce a functional enzyme 0 activity in the table I expect that p2 homozygotes would have no pigment Thus they would be white Page 3 I39IUIIIHWUI K 3 1 m cpMV l mar that IES ln com colored aleurone R is dominant to colorless r and green plant color G is dominant to yellow 3 Two plants each heterozygous for both characteristics are testcrossed to homozy gous recessives and their progeny are combined to produce the following totals colored green 100 M l coloredyellow 97 AWN 6 nor M g i39 We u l L j I y C colorless green 103 I colorlessyellow 100 39 139 Galquot 54 L C l 5395 a Use Cid square analysis to best these data for independent assortment of the two characteristics 8 points Note The table for Chisquare P values is on the last page 1 1 0550 61PCCF A 0c ee 6 6 to o 0 0 0 l U 0 4 q Z I 0 0 3 q I00 I O 3 I 0 0 3 a 4 O I O 0 ag 0 O O L 1 00 0 Y z Loo BDF NOT sleightWHY DIFFEAEAT n U I l A I I b When the progeny of each of the two heterozygous plants are scoredose parately the followz39io gt LC 9 j ing results are obtained PHENOTYPES PLANT 1 PLANT 2 colored green 88 12 colored yellow 12 85 colorless green 8 95 colorless yellow 92 8 Use chisquare analysis to test each data set for independent assortment 8 points PIAill 0 E oEWE rut1H7 lt53 50 13quot 9 5A 39 am 9 2 So 1 J39W 7 H 8 50 35153 I a 4 7 511 50 351 M P240 E 6743 9L 5 5quot71 a39m 616 50 W5 N0 MWW i4 go 353 5ormf 2 7373737 I c Bxplah mresultsofthemmed uaxeanalyses 4 l The PM H Wat M a fe l iota a lmk a c quotany 39 P3 r 7
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