Bio 206 HW5 key For 3-24 and 3-31 PDF

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Bio 206 HW5 Q1. Define the following terms: syntenic genes, complementation, recombinant class, centimorgan, synaptonemal complex Syntenic genes: Genes located on the same chromosome (also linked) Complementation: Process in which heterozygosity of chromosomes bearing recessive mutant alleles for two different genes produces a normal phenotype. Recombinant class: Reshuffling of gametes due to independent assortment or recombination that result in the production of classes of progeny unlike the parental classes Centimorgan: A unit of measure of recombination frequency. 1 cM is the same as 1 map unit. This refers to 1% chance that one gene will be separated from another linked gene during crossing over of meiosis in a single generation. Synaptonemal complex: structure that helps align homologous chromosomes become precisely paired and zipped together; occurs during zygotene of prophase 1. Q2. In Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sablecolored bodies. When F1 males and females were mated, the F2 generation was composed of 216 females with wild-type eyes and bodies, 223 females with wild-type eyes and sable bodies, 191 males with wild-type eyes and sable bodies, 188 males with raspberry eyes and wild-type bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses, and calculate any relevant map distances. The parents are from true breeding stocks: raspberry eye color male x sable body color female F1: wild type eye and body color female x sable body color male (if no mention is made of the eye color, then it is assumed to be wild type) F2: 216 wild type female 223 sable female 191 sable male: 188 raspberry male 23 wild type male 27 raspberry sable male 1. The phenotypes seem to be controlled by 2 genes, one for eye color and the other for body color. 2. The F1 female progeny show that the wild type allele is dominant for body color (s+ > s) and the wild type allele is also dominant for eye color (r+ > r). 3. Sable body color is seen in the F1 males but not in the F1 females, so the s gene is X-linked. 4. In the F2 generation the raspberry eye color is seen in the males but not in the females, so r is also an X-linked gene. We can now assign genotypes to the true breeding parents in this cross: r s+ / Y (raspberry male) x r+ s / r+ s female F1 r s+ / r+ s female (wild type) x r+ s / Y (sable male) Gametes: heterozygous F1 female can make the following gametes: (parentals) r s+, r+ s & (recombinants) r s, r+ s+; The F1 male can make Y and r+ s gametes]

F2 will be r s+ / r+ s (wild type females), r+ s / r+ s (sable females), r+ s+ / r+ s (wild type females), r+ s / Y (sable males), r s+ / Y (raspberry males), r s / Y (raspberry sable males), r+ s+ / Y (wild type males). The F1 female is heterozygous for both genes and will therefore make parental and recombinant gametes. The F1 male is not a true test cross parent, because he does not carry the recessive alleles for both of the X-linked genes. However, this sort of cross can be used for mapping, because F2 sons receive only the Y chromosome from the F1 male and are hemizygous for the X chromosome from the F1 female. Thus the phenotypes in the F2 males represent the array of parental and recombinant gametes generated by the F1 female as well as the frequencies of these gametes. Using the F2 males the recombination frequency between sable and raspberry = 23 (wild type males) + 27 (raspberry sable males) / 429 (total males) = 0.117 x 100 = 11.7 cM. Q3. CC DD and cc dd individuals were crossed to each other, and the F1 generation was backcrossed to the cc dd parent. 903 Cc Dd, 897 cc dd, 98 Cc dd, and 102 cc Dd offspring resulted. 1. How far apart are the c and d loci? 2. What progeny and in what frequencies would you expect to result from testcrossing the F1 generation from a CC dd × cc DD cross to cc dd? 3.1. CC DD x cc dd  F1 C D / c d x cc dd 903 Cc Dd, 897 cc dd, 98 Cc dd, 102 cc Dd. Because the gamete from the homozygous recessive parent is always c d, we can ignore one c and one d allele (the c d homolog) in each class of the F2 progeny. The remaining homolog in each class of F2 is the one contributed by the doubly heterozygous F1, the parent of interest when considering recombination. In the F2 the two classes of individuals with the greatest numbers represent parental gametes (C D or c d from the heterozygous F1 parent combining with the c d gamete from the homozygous recessive parent). The other two types of progeny result from a recombinant gametes (C d or c D combining with the c d gamete from the homozygous recessive parent). The number of recombinants divided by the total number of offspring x 100 gives the map distance: (98 + 102)/(903 + 897 + 98 + 102) = 200/200 = 0.01 x 100 = 10% rf or 10 map units (mu) or 10 cM. 3.2 CC dd x cc DD  F1 C d / c D x c d / c d  as determined in part a, c and d are 10 cM apart. Thus the gametes produced by the heterozygous F1 will be 45% C d, 45% c D, 5% C D, 5% c d. After fertilization with c d gametes, there would be 45% Cc dd, 45% cc Dd, 5% Cc Dd, 5% cc dd. Because this is a test cross, the gametes from the doubly heterozygous F1 parent determine the phenotypes of the progeny.

Q4. Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 males were then crossed to the F1 females to obtain the F2. Of the 400 F2 offspring obtained, 292 were wild type, 9 were cinnabar, 7 were reduced, and 92 were reduced, cinnabar. Explain these results and estimate the distance between the cn and rd loci. Wild type female x reduced cinnabar male  F1 female x F1 male  292 wild type, 9 cinnabar, 7 reduced, 92 reduced cinnabar. Two genes are involved in this cross, but the frequencies of the phenotypes in the second generation offspring do not look like frequencies expected for a cross between double heterozygous for two independently assorting genes (9: 3: 3: 1). The genes must be linked. Designate the alleles: cn+ = wild type, cn = cinnabar; rd+ = wild type, rd = reduced. The cross is: cn+ rd+ / cn+ rd+ female x cn rd / cn rd male  F1 cn+ rd+ / cn rd. Recombination occurs in Drosophila females but not in males. Thus males can only produce the parental cn+ rd+ or cn rd gametes. The females produce both the parental gametes and the recombinant gametes cn+ rd and cn rd+. Female Gametes cn+ rd+ (parental) cn rd (parental) cn+ rd (recombinant) cn rd+ (recombinants)

Male Gametes cn+ rd+ cn+ rd+ / cn+ rd+ (wild type) cn rd / cn+ rd+ (wild type) cn+ rd / cn+ rd+ (wild type) cn rd+ / cn+ rd+ (wild type)

Male Gametes cn rd cn+ rd+ / cn rd (wild type) cn rd / cn rd (cinnabar reduced) cn+ rd / cn rd (reduced) cn rd+ / cn rd (cinnabar)

The reduced flies and the cinnabar flies are recombinant classes. However, there should be an equal number of recombinant types that have a wild-type phenotype because they got a cn+ rd+ gamete from the male parent (genotypes shown in red in the Table above). If we assume that these recombinants are present in the same proportions, then rf = 2 x (7+9)/400 = 8% recombination. The genes are separated by 8 cM. Q5. The following results were obtained from a three point test cross (triple heterozygous female fly crossed with a male that is triple hemizygous for all mutations). Total flies counted = 20,785. Use the following symbols for the mutations: echinus = ec; scute = sc; crossveinless = cv Progeny phenotype scute echinus crossveinless wild type scute

Number 4 1 997

echinus crossveinless scute echinus crossveinless scute crossveinless echinus

1002 681 716 8808 8576

a. Do these numbers above follow the pattern predicted from Mendel’s rule of independent assortment? Justify your answer. b. What is the distance between the 3 genes? Show all your work (genotypes and phenotypes of all crosses and classes of progeny). a. Mendel’s law of independent assortment predicts that if 3 loci are not linked, the eight types of progeny from the cross will be in equal proportion. This is not the case in the given results and the data violate Mendel’s law of independent assortment. This observation predicts linkage. b. The most abundant cases are scute crossveinless (8808) and echinus (8576) Thus, the genotypes of parental cross is: sc sc cv cv ++ X + + + + ec ec (or sc cv +/sc cv + X + + ec /+ + ec). The heterozygous female in which the recombination is occurring is sc + cv ./+ ec + (triple heterozygotic but wild type in phenotype as all mutations are recessive. sc + cv / + ec + female X sc cv ec /Y hemizygous male Progeny phenotype scute echinus crossveinless wild type scute echinus crossveinless scute echinus crossveinless scute crossveinless echinus

Progeny # 4 1 997 1002 681 716 8808 8576

Progeny genotype Interpretation sc ec cv/sc ec cv DCO + + + /sc ec cv DCO sc + + /sc ec cv SCO between + ec cv/sc ec cv ec and cv sc ec +/sc ec cv SCO between + + cv/sc ec cv sc and ec sc + cv/sc ec cv Parental + ec +/sc ec cv Parental

1. Based on the DCO classes, the gene ec is in the middle. 2. The map distance between sc and ec is: [681 + 716 + 4 + 1 /20,785] X 100 = 6.74 cM 3. The map distance between ec and cv is: [1003 + 997 + 4 + 1 /20,785] X 100 = 9.65 cM 4. The map distance between sc and cv would be (997+1002+681+716+4+4+1+1)/20,785 =16.39...