Biosc 231 General Genetics Exam 2 Name ______

BioSc 231 General Genetics Exam 2 Name ______

Multiple Choice. (2 points each)

1. ____ In a complementation test the number of complementation groups indicates

A. the number of genes required for a specific phenotype B. the penetrance of a phenotype C. the number of phenotypes for a gene D. the number of chromosomes in an organism E. 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. penetrance B. incomplete dominance C. co-dominance D. 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. 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 called

A. penetrance B. expressivity C. incomplete dominance D. co-dominance E. 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. 1/4 B. 3/4 C. 1/16 D. 7/16 E. 9/16

6. _____An autosome is ___.

A. a non-sex determining chromosome B. an alternate form of a gene C. another term for epistasis D. 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. 1/4 B. 1/2 C. 3/4 D. 1/3 E. 2/3 8. _____ The maximum recombination frequency between two genes is

A. 100% B. 80% C. 50% D. 10% E. 1%

9. _____ In a complementation test

A. mutations that complement are allelic B. mutations that complement belong to the same complementation group C. mutations that complement are in two different genes required for the wild-type phenotype D. 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. 0.03 B. 0.15 C. 0.20 D. 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. CcPp and Ccpp B. CCPP and Ccpp C. ccpp and Ccpp D. CcPp and ccpp E. 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. CcPpRr B. CcppRR C. CcPPrr D. ccPPRR E. 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. DDee x ddEE B. DdEe x DdEe C. Ddee x DdEe D. 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. rrPP x RRpp B. RrPp x RrPp C. RrPp x rrpp D. RrPp x rrPP E. RrPp x RRpp 15. _____ A person who has type O blood has

A. A antigens on the cell surface B. B antigens on the cell surface C. both A and B antigens on the cell surface D. no surface antigens

16. _____ In crossing over

A. Genetic exchange occurs before chromosome replication B. The probability of its occurrence decreases with increasing distance between the genes exchanged C. Occurs more frequently between two loci very close together D. 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. incomplete dominance B. co-dominance C. epistasis D. co-penetrance E. 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? (5) The table to the right shows the results of a series of experiments to Compound Tested determine the sequence of intermediates in a biochemical pathway. 4 independent auxotrophic mutants which all require compound E (an amino A B C D E 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. + 1 + + -- + + indicates growth that is supported by the indicated precursor. Using the t n

diagram below, show the order of the intermediates in the pathway and a indicate which step in the pathway is catalyzed by each mutant by placing t 2 ------+ u the letter representing the appropriate compound in each box and the number of the appropriate mutant in each circle. M 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, simple 46 - jack-o-lantern, simple 56 - round, branched 68 - 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 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. u+ m+ r+ 304 u m r 295 u m+ r 70 u+ m r+ 64 u+ m r 119 u m+ r+ 108 u m r+ 18 u+ m+ r 22 (5) Based on the complementation data below, A) how many complementation groups exist, and B) which mutations belong to each group? (+ = complementation, -- = no complementation)

Mutation

1 2 3 4 5 6 7 8 9 10

2 + --

3 -- + --

4 + + + -- n o i t

a 5 + + + -- -- t u

M 6 + -- + + + --

7 -- + -- + + + --

8 + + + + + + + --

9 + + + + + + + -- --

10 + -- + + + -- + + + -- Bonus Question (4 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 # with Col Which marker is linked to the flowering Markers Markers mutation? m 235 1 top 43 51 m 305 1 bottom 27 27

PhylB/hy3 2 top 38 36 m 429 2 bottom 39 29

g 4711 3 top 29 21 BGL1 3 bottom 20 30

GA1 4 top 44 38 AG 4 bottom 11 59

r 89998 5 top 41 45 DFR 5 bottom 50 34