BioSc 231 General Genetics Exam 2 Name __________________________________Multiple Choice. (2 points each)1. ____ In a complementation test the number of complementation groups indicatesA. the number of genes required for a specific phenotypeB. the penetrance of a phenotypeC. the number of phenotypes for a geneD. the number of chromosomes in an organismE. the quantity of gene product required for a phenotype2. _____ The percentage of individuals with a given genotype who exhibit the phenotype associated with that genotype is calledA. penetranceB. incomplete dominanceC. co-dominanceD. lethality3. _____ 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 beA. 10%B. 25%C. 30%D. 40%E. 50%4. _____ A situation where each allele produces a phenotype (usually a protein) that can be detected in the heterozygote is calledA. penetranceB. expressivityC. incomplete dominanceD. co-dominanceE. lethality5. _____ 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. 1/4B. 3/4C. 1/16D. 7/16E. 9/166. _____An autosome is ___.A. a non-sex determining chromosomeB. an alternate form of a geneC. another term for epistasisD. present only in males and is responsible for sex determination7. _____ 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. 1/4B. 1/2C. 3/4D. 1/3E. 2/38. _____ The maximum recombination frequency between two genes isA. 100%B. 80%C. 50%D. 10%E. 1%9. _____ In a complementation test A. mutations that complement are allelicB. mutations that complement belong to the same complementation groupC. mutations that complement are in two different genes required for the wild-type phenotypeD. 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. 0.03B. 0.15C. 0.20D. 0.5011. _____ In sweet peas, the two allelic pairs C, c and P, p are known to affect pigment formation in the flowers. The dominants, C andP, 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. CcPp and CcppB. CCPP and CcppC. ccpp and CcppD. CcPp and ccppE. CcPp and ccPp12. _____ 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. CcPpRrB. CcppRRC. CcPPrrD. ccPPRRE. CcppRR13. _____ 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. DDee x ddEEB. DdEe x DdEeC. Ddee x DdEeD. DdEe x DDEe14. _____ 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. rrPP x RRppB. RrPp x RrPpC. RrPp x rrppD. RrPp x rrPPE. RrPp x RRpp15. _____ A person who has type O blood hasA. A antigens on the cell surfaceB. B antigens on the cell surfaceC. both A and B antigens on the cell surfaceD. no surface antigens16. _____ In crossing overA. Genetic exchange occurs before chromosome replicationB. The probability of its occurrence decreases with increasing distance between the genes exchangedC. Occurs more frequently between two loci very close togetherD. The reciprocal exchange between homologous chromosomes is random17. _____ A dihybrid cross results in a phenotypic ratio of 12:3:1. This type of ratio most likely results from A. incomplete dominanceB. co-dominanceC. epistasisD. co-penetranceE. variable penetranceShort Answer. (variable points)(4) Two-point testcrosses revealed the following map results:br__________ui 7 map unitsui___________ns 3 map unitsA. 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?(5) The table to the right shows the results of a series of experiments todetermine the sequence of intermediates in a biochemical pathway. 4independent auxotrophic mutants which all require compound E (an aminoacid) as a nutritional supplement were analyzed with 4 compounds that areprecursors in the synthesis of compound E. Each mutant was grown on aminimal medium supplemented with each of the indicated compounds. +indicates growth that is supported by the indicated precursor. Using thediagram below, show the order of the intermediates in the pathway andindicate which step in the pathway is catalyzed by each mutant by placingthe letter representing the appropriate compound in each box and thenumber of the appropriate mutant in each circle.Compound TestedMutantA B C D E1 + + -- + +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 - round, simple46 - jack-o-lantern, simple56 -