LINKAGE AND RECOMBINATION. Problem 4 An organism of genotype AaBbCc is test-crossed. The genotypes of the progeny were as follows:

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1 CHAPTER 6 LINKAGE AND RECOMBINATION Problem 1 If there is one crossover event per 50 meiotic divisions between the linked genes A and B, then the map distance between these genes is: a. 50 map units (m.u.) b. 1 m.u. c. 2 m.u. d. 5 m.u. e. 0.5 m.u. Problem 2 An plant of genotype AaBb is produced by crossing an AAbb and an aabb parent. These two genes are located on the same chromosome and are 10 map units apart. If this F1 is test-crossed, the fraction of the progeny that will be AaBb is: a. 0.1 b c. 0.9 d e. There is not enough information to answer this question Problem 3 Scarlet eyes and bent wings are recessive mutant traits that are located on the same autosome and are 20 map units apart A fly heterozygous for both traits is testcrossed. The fraction of F2 progeny that will be wildtype is: a. 0.4 b. 0.2 c. 0.8 d. 0.1 e. not enough informaiton to answer this questions. Problem 4 An organism of genotype AaBbCc is test-crossed. The genotypes of the progeny were as follows: Genotype # of progeny AaBbCc 90 AaBbcc 10 aabbcc 90 aabbcc 10 a. If these three genes were all assorting independently of each other, how many genotypic classes would you expect in the progeny of the testcross? b. If these three genes were so tightly linked that crossing-over never occurred between them, how many genotypic classes would you expect in the progeny of the testcross? c. What can you conclude from this data? 6.1

2 Problem 5 Mary is heterozygous for two autosomal dominant traits: myopia (nearsightness) and achondroplastic dwarfism. Both traits were inherited from her mother. Mary marries a man who is six-feet tall and has perfect vision. (i) Assuming that these traits are segregating independently, what fraction of her children will be normal with repect to both traits a. 1/4 b. 1/2 c. 0 d. 9/16 e. 1/16 (ii) Answer question in (i) assuming that the two traits are linked and 20 map units apart. a. 0.4 b. 0.1 c. 0.8 d. 0.2 e. 0 Problem 6 Drosophila melanogaster has one pair of sex chromosomes and three pairs of autosomes, referred to as chromosomes 2, 3, and 4. A male fly with short legs was discovered in a wild-type stock of flies by an observant genetics student. Using this male, the student was able to establish a pure breeding stock. Next the student constructed a homozygous strain carrying the short mutation (sh), the mutation black body (b gene located on chromosome 2) and the mutation cardinal eyes (cd gene located on chromosome 3). A female from the short, black, cardinal eye stock was mated to a wild-type male. All the progeny were wild type. The F1 females from this cross were then testcrossed. The F2 progeny were as follows: cardinal eye, wild-type cardinal eye short, black short, black females males Which chromosome is the short gene located on? 6.2

3 Problem 7 The ad- mutant of the haploid fungus Neurospora is an adenine auxotroph; the leumutant is a leucine auxotroph. An ad- mutant strain is crossed to a leu- mutant strain and the haploid progeny spores produced by the transient diploid are tested for growth on various media: Progeny Class # of progeny RM MM MM w/ad MM w/leu A B C D (+) = growth (- ) = no growth MM = minimal medium w/ad = adenine supplement w/leu = leucine supplement RM = rich medium (all the goodies for growth of any auxotroph) (i) With respect to progeny class D which statement is not true. a. It is a recombinant progeny. b. It is prototrophic. c. It will grow on medium supplemented with both adenine and leucine. (ii) From this set of data it can be concluded that: a. The leu and the ad genes show independent assortment. b. The leu and the ad genes are linked and 20 map units apart. c. The leu and the ad genes are linked and 10 map units apart. d. No conclusions about linkage can be drawn from this data. 6.3

4 Problem 8 The following diagram is a map of the X chromosome of Drosophila:: cut wings singed bristles lozenge eyes vermilion eyes miniature wings sable-body garnet eyes Values are map units measured from the gene closest to the end of the chromosome. A true-breeding wild-type female is crossed with a vermilion-eyed, sable-bodied male. If the F1 males and F1 females are crossed to each other, what proportion of their male progeny will be: a. wild-type b. vermilion-eyed and sable-bodied c. vermilion-eyed d. sable-bodied 6.4

5 Problem 9 You isolate a new true-breeding mutant stock of Drosophila that has an abnormal eye color (e) due to a mutant gene on the X chromosome. A female fly heterozygous for abnormal eye (e), for cut wings (c) and forked bristles (f) is mated to a wild-type male and the phenotypes of the male progeny are examined (a "+" indicates the fly was normal with respect to that trait. The cut wings and forked bristle genes are shown on the map below. Phenotype Number of progeny e c + f e + c f 331 e + c + f 105 e c f e + c f + 49 e c + f 49 e + c + f + 15 e c f (i) The allele configuration (ignore gene order for the time being) of the heterozygous parent was: a. e+ c+ f+/ e c f b. e+ c f/ e c+ f+ c. e+ c+ f/ e c f+ d. e+ c f+/ e c+ f e. cannot be determined from this data 9. (ii) The linkage order of these three genes is: a. e c f b. e f c c. f e c d. can't tell because the e gene isn't linked to the others e. not enough data to answer this question 9. (iii)consult the genetic map shown on this page. The actual phenotype of abnormal eye strain is very likely to be: a. facet eyes b. ruby eyes c. lozenge eyes d. vermillion eyes e. garnet eyes 6.5

6 Problem 10 You isolate a new true-breeding mutant stock of Drosophila that has abnormal bristles. A female fly heterozygous for abnormal bristles (ab), for curled wings (cw) and hairy body (hb) is testcrossed and the phenotypes of the progeny are examined. Examine the data shown below. Phenotype Number of progeny wild type 74 ab, cw, hb 75 cw, hb 308 ab 308 hb 23 ab, cw 23 cw 95 ab, hb (i) The allele configuration (ignore gene order for the time being) of the heterozygous parent was: a. ab + cw + hb + / ab cw hb b. ab + cw hb/ ab cw + hb + c. ab + cw + hb/ ab cw hb + d. there is not enough information to answer this question 10. (ii) The linkage order of these three genes is: a. ab cw hb (or reverse: hb cw ab) b. ab hb cw (or reverse) c. hb ab cw (or reverse) d. can't tell because the ab gene isn't linked to the others e. cannot be determined from this data 10. (iii) The map distance between the ab gene and the cw gene is: a map units b map units c map units d. 50 map units e. none of these answers is correct 6.6

7 Problem 11 The black (b) and antennaless (al) genes are both located on the same autosome in Drosophila. The mutant phenotype for b is a black body (instead of "normal gray") and the mutant phenotype for al is that the antennae do not develop. A female from a true-breeding wild-type strain is mated to a black antennaless male. An F1 female is selected, and a testcross is performed. The F2 progeny are: 91 black antennaless, 89 gray w/ normal antennae, 11 black w/ normal antennae, 9 gray antennaless. a. Describe the testcross using appropriate allele symbols. b. What % of the progeny have "recombinant" phenotypes? c. What % of the gametes of the F1 female have recombinant genotypes? d. What % of the gametes of the F1 female have "parental" genotypes? (Parental genotype = same alleles on same chromosome as in the F1 parent.) e. What is the genetic map distance between the al and b genes (in map units?) f. What is the normal function of the al gene? Problem 12 The genes controlling colorblindness and hemophilia are located on the X chromosome 10 map units apart. Study the following pedigree and determine the probability that male III 1 will: a. be completely normal b. be colorblind only c. have hemophilia and normal vision d. have both traits I colorblind II hemophilia III?? 1 6.7