Handout 11

Handout 11 - 1. E.coli cells are simultaneously infected...

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1. E.coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype= h st c); another strain carries the wild-type alleles (genotype= h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The genotypes of the progeny phages are: Progeny phage genotype Number of plaques h+ c+ st+ h c st h+ c st h c+ st+ h+ c st+ h c+ st h+ c+ st h c st+ 321 338 26 30 106 110 5 6 a. Determine the order of the 3 genes on the phage chromosome Ans: h+ st+ c+; notice that phage is haploid. b. Determine the map distances between the genes Ans: h+ to st+ = [(26+ 30+ 5+60)/942] * 100% = 7.1% = 7.1 cM h+ to c+ = 31.2 cM st+ to c+ = 24.1 cM c. Determine the coefficient of coincidence and the interference Ans: COC= (6+ 5)/(.071 * .241 * 942) = 0.68 Interference = 1- 0.68 = 0.32 2. Three genes in Drosophila with the recessive alleles l, m, and n. You wish to determine the map distances between them and perform the cross shown below. ( Note: The genes are arbitrarily shown in alphabetical order - you need to determine the order of these genes on the chromosome) female male + m + x l m n l + n l m n Progeny with the following phenotypes are obtained: # progeny l m n l,m m,n l,n 99 l m n non non par 0 l m + non par non 901 l + n par par par 0 l + + par non non 0 + m n par non non 899 + m + par par par 0 + + n non par non 101 + + + non non par
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Total 2000 a) label the types of progeny that inherited nonparental type gametes, that is, the recombinants. Notice Drosophila is a diploid organism therefore pattern “l m n” indicates for phenotype, not genotype which require another copy of each gene. b) Determine the order of these markers on the chromosome and the distances between them in centimorgans. Ans: l to m: Parental gamete types are +m and l-, recombinant types (nonparental) are lm and ++ From table # non x100 = 99 + 101 + 0 + 0 x100 = 10% rf = 10 cM Total 2000 m to n: Parental gamete types are m+ and +n, recombinant types are mn and ++ From table # non x100 = 99 + 101 + 0 + 0 x100 = 10% rf = 10 cM Total 2000 l to n: Parental gamete types are ln and ++, recombinant types are l+ and +n From table # non x100 = 0 + 0 + 0 + 0 x100 = 0% rf for < 0.05 cM Total 2000 If you had gotten just one recombinant it would have yielded 1/2000 x 100 = .05cM so you write this distance as < 0.05 cM. You can show the two genes at the same position on the map: m [l,n] ________|_______________|___ 10cM d. Are these genes linked? Explain (l to m, m to n, l to n) Ans: they are all partially linked because two of their r.f. are less than 50%, the last one lies 2 different genes on the same location.
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3. The yeast cross a leu2 x A leu+ produces the following tetrads.
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This note was uploaded on 02/28/2009 for the course BICD BICD100 taught by Professor Smith during the Winter '09 term at UCSD.

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Handout 11 - 1. E.coli cells are simultaneously infected...

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