BioSc 231 General Genetics Exam 2 Name Multiple Choice 2 points each 1 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 2 The percentage of individuals with a given genotype who exhibit the phenotype associated with that genotype is called A B C D penetrance incomplete dominance co dominance lethality 3 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 4 A situation where each allele produces a phenotype usually a protein that can be detected in the heterozygote is called A B C D E penetrance expressivity incomplete dominance co dominance lethality 5 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 6 An autosome is A B C D a non sex determining chromosome an alternate form of a gene another term for epistasis present only in males and is responsible for sex determination 7 In chickens the dominant allele Cr produces the creeper phenotype having short legs However the creeper allele is lethal in the homozygous condition If two creepers are mated what proportion of the living progeny will be creepers A B C D E 1 4 1 2 3 4 1 3 2 3 8 The maximum recombination frequency between two genes is A B C D E 100 80 50 10 1 9 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 10 The maize genes bl and ue are linked 30 map units apart If a plant bl ue bl ue is testcrossed what proportion of the progeny will be bl ue bl ue A B C D 0 03 0 15 0 20 0 50 11 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 12 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 13 In humans the dominant alleles D and E are both required for normal development of the cochlea and the auditory nerve respectively The recessive alleles d and e 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 DDee x ddEE DdEe x DdEe Ddee x DdEe DdEe x DDEe 14 In poultry the shape of the comb varies greatly and involves at least two pairs of alleles The allele R can result in rose shaped comb and the allele P can result in pea shaped comb If both of these dominants are present together genic interaction produces a walnut comb When a bird is carrying both recessive alleles in the homozygous condition single comb types result Which of the following crosses produces offspring at the ratio of 1 Walnut 1 Rose 1 Pea 1 Single A B C D E rrPP x RRpp RrPp x RrPp RrPp x rrpp RrPp x rrPP RrPp x RRpp 15 A person who has type O 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 16 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 17 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 4 Two point testcrosses revealed the following map results br ui 7 map units ui ns 3 map units A Draw the two possible maps for these loci B Other than a 3 point test cross what other cross would resolve the two possible maps and what are the possible outcomes of that cross Compound Tested Mutant 5 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 6 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 70 46 56 68 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 6 In corn the genes u m and r are linked The data given below summarize the result of 1000 …