LECTURE 7: TETRAD ANALYSIS; GENE CONVERSION, continued
Reading: Ch. 5, p 142-151; Ch. 6, p 192-194
Problems: Ch. 5, solved problem III, also 28a, 29ab, 30, 32, 34, 35; Ch. 6, #30, 31
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First, let’s review how to calculate map distance in yeast:
Consider and example of two linked genes, ARG3 and URA2.
P: arg3 ura2 x ARG3 URA2
Diploid: arg3 ura2 / ARG3 URA2
Meiotic products:
PD (arg3 ura2; arg3 ura2; ARG3 URA2; ARG3 URA2) = 127
NPD (arg3 URA2; arg3 URA2; ARG3 ura2; ARG3 ura2) = 3
T: (arg3 ura2; arg3 URA2; ARG3 ura2; ARG3 URA2) = 70
Recombination frequency, RF = [(NPD + 1/2T) / total number of tetrads] x 100. In our example,
this is ([3 + 1/2(70)]/200) x 100 = 19 map units. We will modify the equation to obtain a better
estimate of map distance. If you draw out the possible crossover events between two linked
genes, you can see the different tetrads that result; see Fig. 5.17 (p. 147).
No crossovers --------> PD
Single crossover --------> T
Double crossover (2-strand) --------> PD
Double crossover (3-strand) --------> T
Double crossover (3-strand) --------> T
Double crossover (4-strand) --------> NPD
You can see how we can modify the equation to make it more accurate. Remember that half
(2/4) the strands recombine if there is a single crossover event and that 4 strands recombine if
there is a double crossover event (even if all of the strands don’t participate, some participate
more than once.)
Map distance = (total rec. events / total tetrads) x 100 = [(1/2[SCO] + DCO) / total tetrads] x 100
Map distance = ([1/2 (T – 2 NPD) + 4 NPD]/ total tetrads) x 100
Map distance = (1/2 T + 3 NPD) / total tetrads x 100
For our example above, map distance = ([1/2 (70) + 3 (3)] / 200) x 100 = 22 map units
This modified equation makes 2 assumptions: (1) there are no more than two crossovers in the
interval and (2) there is no chromosomal interference (all types of DCOs occur with equal
frequency.
