Genetika molekuler (4)

1. Genetika molekuler (4) Sutarno

2. Lecture #3 Notes (Yeast Genetics) • LECTURE 3: TETRAD ANALYSIS • Tetrad analysis can give lots of information, but you have to know which questions to ask. They have limited numbers of answers, and the answers and their interpretation are known beforehand. • Question #1: Does the phenotype segregate 2:2? • Yes: the phenotype is due to segregation of a single gene • No: something else is happening • Phenotype is due to segregation of multiple loci • Non-Mendelian trait (plasmid, prion, mitochondria) • (especially important if the cells were mutagenized) • Once we know that a phenotype is due to a single mutation, we can test for linkage to other • genes. • Linkage • Used to determine whether two mutations are located near each other (or near a CEN) • Test of LOCATION. • Tetrad terminology: • When examining the segregation patterns of two mutations, there are only three possible types of tetrads: • Rb x rB  meiosis sporegenotypes • a Rb RB RB • b Rb RB Rb • r = drug resistant c rB rb rB • b = blue d rB rb rb • PD NPD T • 2 genotypes 2 genotypes 4 combinations • like both of unlike of genotypes • the parents either parent • PD, NPD, T only describes whether recombination has occurred between the two markers • A tetrad could be a PD for one pair of markers, and NPD or T for other pairs • Example: • ade1- his4-leu2- x ADE1+HIS4+LEU2+ • sporeAdeHisLeu • a - - - • b - - + • c + + - • d + + - • It is meaningless to simply ask if this tetrad is a PD, NPD, or T, • we need to know the parents • we need to know what phenotypes we are discussing • For Ade and Leu, this is a T • For Ade and His, this is a PD • For His and Leu, this is a T • Linkage is determined by counting the number of each type of tetrad. • ______________________________________________________ • Question #2: Do #PDs = #NPDs? (the most important question in determining linkage) • Yes: they are unlinked (should be 1:1:4 ratio of PD:NPD:T)(unless both are CEN linked) • No: they are linked (PDs should be much greater than NPDs) • (doesn’t have to be EXACT…a statistical tally of random events, so its like heads and tails.) • Why? The simple answer: If completely linked, there is no recombination, so all will be in the parental configuration (PD). Recombination events will generate NPD and T tetrads, depending on whether the genes are on different chromosomes, the number of recombination events, and the location of the genes and the recombination events relative to their centromeres. • In addition to a simple yes/no answer, we can also quantitate how tightly linked they are using the equation: • 1/2 (6NPD + T) (x 100) = cM (takes into account only single crossovers) • N + P + T • Usefulness of tetrad analysis • Is the phenotype due to mutation in a single gene? • Mapping genes relative to each other and to centromeres • Constructing strains (e.g. creating double mutant combinations) • Testing whether the correct gene has been cloned • Testing revertants • Are additional phenotypes due to the same mutation? (use drug resistance and Ts- as an • example) • Linked to both to • Examples: PDNPDTeach other?CEN? • 50 50 0 N Y • 45 45 10 N Y (5cM) • 90 0 10 Y X • 18 14 68 N N • 35 35 30 N Y (15 cM) • 80 4 16 Y X