BioSc 231 General Genetics Exam 2 Name Multiple Choice 4 points each In a complementation test the number of complementation groups indicates A B C D E the number of genes required for a specific phenotype the penetrance of a phenotype the number of phenotypes for a gene the number of chromosomes in an organism the quantity of gene product required for a phenotype A plant of genotype C D C D is crossed to c d c d and the resulting F1 testcrossed to c d c d If the genes are unlinked the percentage of c D recombinants will be A B C D E 10 25 30 40 50 In Drosophila the alleles for brown and for scarlet eyes resulting from two independent genes interact so that the double homozygous recessive is white A pure breeding brown BBss and pure breeding scarlet bbSS P generation are crossed What proportion of the F2 will be white A B C D E 1 4 3 4 1 16 7 16 9 16 The Arabidospis genes gr and een are linked 30 map units apart If a plant gr een gr een is testcrossed what proportion of the progeny will be gr een gr een A B C D 0 03 0 15 0 20 0 50 In humans the unlinked dominant alleles P and B are both required for normal development of the cochlea and the auditory nerve respectively Either of the recessive alleles p and b can result in deafness due to impairment of these essential parts of the ear Which of the following sets of parents would produce all hearing children A B C D PPbb x ppBB PpBb x PpBb Ppbb x PpBb PpBb x PPBb In sweet peas the two allelic pairs C c and P p are known to affect pigment formation in the flowers The dominants C and P are both necessary for colored flowers absence of either results in white A dihybrid plant with colored flowers is crossed to a white one which is heterozygous at the c locus What are the genotypes of these two plants A B C D E CcPp and Ccpp CCPP and Ccpp ccpp and Ccpp CcPp and ccpp CcPp and ccPp Assume that an additional allelic pair in sweet peas also affects pigment formation in addition to the genes mentioned in the previous question The presence of the dominant R allele is required for red flowers and the recessive r allele produces yellow flowers Which of the following genotypes would result in red flowers A B C D E CcPpRr CcppRR CcPPrr ccPPRR CcppRR In a complementation test A B C D mutations that complement are allelic mutations that complement belong to the same complementation group mutations that complement are in two different genes required for the wild type phenotype mutations that are allelic are required for complementation A person who has type AB blood has A B C D A antigens on the cell surface B antigens on the cell surface both A and B antigens on the cell surface no surface antigens In crossing over A B C D Genetic exchange occurs before chromosome replication The probability of its occurrence decreases with increasing distance between the genes exchanged Occurs more frequently between two loci very close together The reciprocal exchange between homologous chromosomes is random A dihybrid cross results in a phenotypic ratio of 12 3 1 This type of ratio most likely results from A B C D E incomplete dominance co dominance epistasis co penetrance variable penetrance Short Answer variable points Two point testcrosses revealed the following map results br ir 9 map units ir cs 5 map units A Draw the two possible maps for these loci 8 B What other cross would resolve the two possible maps and what are the possible outcomes of that cross 4 Compound Tested Mutant 8 The table to the right shows the results of a series of experiments to determine the sequence of intermediates in a biochemical pathway 4 independent auxotrophic mutants which all require compound E an amino acid as a nutritional supplement were analyzed with 4 compounds that are precursors in the synthesis of compound E Each mutant was grown on a minimal medium supplemented with each of the indicated compounds indicates growth that is supported by the indicated precursor Using the diagram below show the order of the intermediates in the pathway and indicate which step in the pathway is catalyzed by each mutant by placing the letter representing the appropriate compound in each box and the number of the appropriate mutant in each circle A B C D E 1 2 3 4 8 In pumpkins jack o lantern shaped fruit o is recessive to round shaped fruit o Branching vines s is recessive to simple vines s In the P generation plants from two different pure breeding lines are crossed One variety bears round fruit and has simple vines The other variety has jack o lantern fruits and branched vines The resulting F1 plants were testcrossed and the following 240 progeny were obtained 72 48 54 66 round simple jack o lantern simple round branched jack o lantern branched Calculate the chi square and P values based on the prediction that the genes are not linked chart is on the last page 12 In corn the genes m s and t are linked The data given below summarize the result of 1000 offspring from a three point testcross From the data construct a map showing the genes in the correct order and indicating the distances between each pair of genes m m m m m m m m s s s s s s s s t t t t t t t t 310 295 77 70 119 108 12 9 12 Based on the complementation data below A how many complementation groups exist and B which mutations belong to each group complementation no complementation Mutation Mutation 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 10 Bonus Question 5 pts An Arabidopsis thaliana flowering mutation has been generated in the Columbia Col line The mutant line was then crossed with a wild type Landsberg erectus Ler line to generate the F1 generation The F1 generation was allowed to self to produce the F2 generation F2 plants that displayed the mutant phenotype were assayed using the CAPS system to identify a molecular marker that is linked to the mutant flowering gene Two markers from each of the five Arabidopsis thaliana chromosomes were tested The results of those tests were as follows Marker Name Chromosome with Ler Markers with Col Markers m 235 m 305 1 top 1 bottom 43 11 51 59 PhylB hy3 m 429 2 top 2 bottom 38 39 36 29 g 4711 BGL1 3 top 3 bottom 29 20 21 30 GA1 AG 4 top 4 bottom 44 28 38 24 r 89998 DFR 5 top 5 bottom 41 50 45 34 Which marker is linked to the flowering mutation