Midterm IA Key - Biological Sciences 101-002 Dr Mark...

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Unformatted text preview: Biological Sciences 101 -002 Dr. Mark Sanders Winter Quarter 2010 First Midterm Exam Answer Key - FORM A 1 1A. Cross 1: TtAa x T tAa Both parent tall, axial /Efl Cross 11: TTAa x T -Aa Both parents tall, axial I I I/Efi 113. x2 = (46-45)2/45 + (15—15)2/15 + (14-15)2/15 + (5 -5)2/5 = 0. 0’6’6’ I / Df = 3; P value > .975, independent assortment between the gene is not rejected on the basis of the P value eing greater than 0.05. \ / I ' ‘ I 3 A98 1" 2A. Each prospective parent has a 2/3 chance of being a carrier (conditional probability). .3 “”1214 The chance the first child has galactosemia is (2/3)(2/3)(1/4) = 4/36 = 1/9 S 4, 2B. If the first child has galactosemia, then the parents are carriers. The second child is ,1 AP oz“, the product of a heterozygous mating and has a 1A chance of having the disease. ,1 w 2C. If the parents are carriers, there is a 1A chance of disease in each subsequent child and l 19703 1” a 3A chance of being disease-free. The probability that child 2 and child 3 are disease-free lufluc is (3/4)(3/4) = 9/16. 2 4 3A. The mutant allele is dominant. The heterozygous organism nor the mutant ‘ z homozygote fall short of producing sufficient enzymatic activity to produce the wild—type l phenotype and both are phenotypic mutants. 3B. Parents Pups ,l ffl Unaffected Unaffected 'Affected Affected Female Male Homoz ous Hetero ous Homo ous Hetero 4. The format of this answer is illustrated in Figure 3-8, page 104 of the text. many/4m 5 3/59 TEw/flflflSfiJ/ @713 0,1 [/39 5‘67’ #2: I < 5A. The F1 are wild-type because the plants are heterozyg'ous for the flower color mutation. The 3 :1 F2 ratio is consistent with inherited variation of a single gene causing .Z flower color mutation. 3 < 5B. Genetic complementation 3 < 5C. Both mutant lines carry mutations of the same gene. 1 4 5D. The F1 progeny are dihybrid and are expected to produce F2 progeny in the ratio 9/16 yellow to 7/16 white. The F2 progeny will be r+_d+_ (yellow) 9/16, r+_dd (white) 3/16, L < rrd+_ (white) 3/16, rrdd (white) 1/16. 6 . The trait is most likely X-linked recessive. The preponderance of males, the uniformity of affected males from an affected mother in generation 11, and the ease of X- linked recessive explanation of affected males in generations III and IV all argue in favor of this mode of inheritance. 6B. Woman A is most likely GG. Her mate is gY. 6C. The man C is GY. The woman B has a 1A chance of being a carrier (Gg). Her maternal grandmother was a carrier and there is a 1/2 chance of passing the X-linked recessive to Cs mother and another 1/2 chance her mother passes the allele to C. The son of B and C has a (1/2)(1/4) = 1/ 8 chance of being affected. 7A. 64 chromatids 7B. 16 chromosomes ,2 [6?! 7c. 64 chromatids 7D. 2 picograms 7E. l picogram 8A. Tall parent Short parent Tall F1 Lane 1 2 3 Kb ‘5 ( 8B. All three band patterns are possible. Homozygous tall will have a single band of 3.4 3 < kb, homozygous short will have a single band of 2.2 kb and heterozygous (tall) will have both bands. ...
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