# (1-p) 2. p(1-p) From the table, frequency of DpyUnc = ¼ (p^2) = #DpyUnc = p^2 = ¼(1-p)^2 + ½(1-p)p + ¼(p^2) #Dpy + #DpyUnc

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Download "(1-p) 2. p(1-p) From the table, frequency of DpyUnc = ¼ (p^2) = #DpyUnc = p^2 = 0.0004 ¼(1-p)^2 + ½(1-p)p + ¼(p^2) #Dpy + #DpyUnc"

## Transcription

2 2B (5 points) In trans configuration #Dpy = p 2 #Dpy + #Dpy Therefore, we could simply count the Dpy animals and either the Dpy animals or animals to calculate recombination frequency. 2C (5 points) #Dpy = 20 = 20 = p 2 #Dpy + #Dpy p 2 = p = 0.02 = 2% = 2 map units 3 (5 points) When recombination frequency is p, phenotypes and frequencies of self-progenies of dpy-1 unc-1/+ + (cis configuration) will be as below. Parental (1-p) Recombinant (p) dpy-1 unc dpy unc-2 Dpy Dpy Parental dpy1 unc-1 (1-p) 2 (1-p)p (1-p) + + Dpy Dpy Recombinant dpy-1 + p(1-p) P 2 (p) + unc-1 According to the table, some of the Dpy non and non Dpy will be truly homozygous for recombinant alleles. Thus, it is not possible to distinguish true recombinant homozygotes from others based on their phenotype. However, in each combination of alleles, half consists of Dpy non and non Dpy except for parental homozygotes. Thus, 1-2*(fraction of Dpy non + non Dpy) = (1-p) (A) From the given data, fraction of Dpy non + non Dpy = (12 + 8) /( ) = Plugging this value into equation (A) above, 1-2X( ) = (1-p) 2 p = 0.01 = 1% = 1 m.u.

3 4A (5 points) rol-8 could be located either to the right of dpy-10 or to the left of dpy-10. dpy-10 rol-8 unc-9 unc-9 2 m.u. 1 m.u. rol-8 dpy-10 2 m.u. 3 m.u. unc-9 4B (5 points) For the 2-factor cis-cross, unc-9 rol-8 / unc9 rol-8 hermaphrodite is crossed to wild type male, and hermaphrodite progeny heterozygous for both mutations is self fertilized. If rol-8 is between dpy-10 and unc-9, the map distance between unc-9 and rol-8 would be 1m.u. On the other hand, if rol-8 is located outside of dpy-10 and unc-9, the map distance between unc-9 and rol-8 would be 5 m.u. unc-9 rol-8 / unc9 rol-8 X + +/ + + (hermaphrodite) (male) unc-9 rol-8 / + + (hermaphrodite) self fertilization investigate progenies Like in question 3, map unit can be determined by 1-2*(fraction of Dpy non Rol + Rol non Dpy) = (1-p)^2 4C (5 points) rol-8/rol-8 hermaphrodite is crossed to wild type male, and male progeny heterozygous for rol-8 is crossed to dpy-10 unc-9/dpy-10 unc-9. F2 hermaphrodite progeny heterozygous for all three mutations is self-fertilized. If rol-8 is between dpy-10 and unc-9, there will be some Dpy non that segregates rol-8 some non Dpy that that segregates rol-8. On the other hand, if rol-8 is located to the left of dpy-10, most of the non Dpy will segregate rol-8, while almost none of the Dpy non will segregate rol-8. 5A (5 points) Since only Dpy non Lon segregates Rol progeny, it is likely that rol-1 is located to the right of lon-2 or located very close to the left of lon-2. 5B (5 points) Since both non Lon and Lon non segregate Rol, rol-1 is located between unc-8 and lon-2. This suggests that rol-1 is located to the left of lon- 2, removing the possibility that rol-1 is to the right of lon-2 from question 5A. Since

4 there are more nonlon animals that segregate Rol than Lonnon animals that segregate Rol, rol-1 is closer to lon-2 than unc-8. 5C (5 points) Since almost all Rol segregates with Dpy non, rol-1 is located to the right of unc-8. nondpy animals that segregate Rol might be the outcome of double crossover one between dpy-3 and unc-8 and the other between unc-8 and rol-1. 5D (5 points) dpy-3 1 m.u. unc-8 lon-2 8 m.u. rol-1 6A (5 points) The first possibility is that am1 and lin-39(lf) mutations are actually occurring at two different genes, and lin-39 participates in the same pathway that control cell fate as am1. In this case two mutations in concert would result in intergenic complementation. Second possibility is that am1 and lin-39(lf) are separate mutations within the same gene. In this instance, am1 and lin-39(lf) in combination yield intragenic complementation. For example, am1 and lin-39(lf) can occur at distinct regions of lin-39 that encodes independent domains that contact each other. Each mutant domain alone would yield nonfunctional protein, but two mutant domains together might restore wildtype function. The second possibility is more likely because two mutations are mapped very closely to each other (only 0.5 m.u.). 6B (5 points) Wildtype lin-39 can be transfected into am1 mutant animal to see if exogenous lin-39 can rescue am1 phenotype. If am1 phenotype is rescued, this will suggest that am 1 and lin-39(lf) are the mutations in the same gene. 7A (5 points) In the S set, 26 mutant worms are found among 318 mutant worms studied in the F1 generation of EMS mutagenized worms. Thus the over all mutation rate is, 26 / (318 X 2 ) X 100 = 4.1% Next, in the M set, mutants are grouped into 77 different genes. Thus, the forward mutation rate induced by EMS is, / 77 = 5 X 10^-4 7B (5 points) To identify 4 mutations in genes responsible for detecting garlic, 5X10-4 mutations / 1 genome = 4 mutations / X genome X = 8000 genomes should be screened. Since worms are diploids, screening one animal equals to screening 2 genomes. Thus, 8000 genome X (1 animal/ 2 genome) X (1 plate/ 1 animal) = 4000 plates.

5 8A (5 points) Bristol N2 homozygous for dpy-5 is mated to Hawaiian male, and hermaphrodite progeny heterozygous for dpy-5 is self-fertilized to give a variety of progenies. From this F2 generation progenies homozygous for dpy-5 are picked based on their Dpy phenotype. 8B (5 points) Lanes from to 15.5 on chromosome LGI, specifically between 1 and 5, shows linkage of SNP markers located near dpy-5. In these lanes, restriction digestion with DNA from dpy-5 homozygote gives more N2 DNA pattern than that with DNA from the non-dpy5 animals. This indicates that dpy-5 is located on chromosome LGI. 8C (5 points) Lanes marked with an a indicate that the recombination occurred between W03D8 and D1007. Lanes marked with a b indicate that the recombination occurred between D1007 and B0205. Lanes marked with a c show that recombination occurred between dpy5 and B0205. The Lane marked with a * indicates that there was an animal that has one chromosome with single crossover and one chromosome with double crossover (homozygous Hawaiian at W03D8 and D1007 SNP loci and heterozygous at the B0205). Since recombination SNPs around dpy-5 are rare due to their tight linkage, this type of recombination is not likely to occur. Thus the paper re-examined the phenotype of this animal and confirmed that its Dpy phenotype is not as strong as that of real dpy-5 homozygote and concluded that the observed Dpy phenotype is not caused by dpy-5, a result of synthetic gene interaction.

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