Mapping of X-linked echinous (ec), scute (sc), and crossveinless (cv) loci
Gene order is not known
ec + + / + sc cv
X ec sc cv /Y|
Total Progeny |
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ec + + + sc cv |
8,576 8,808 |
Parentals |
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ec sc + + + cv |
681 716 |
recombinants |
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ec + cv + sc + |
1,002 997 |
recombinants |
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ec sc cv + + + |
4 1 |
double recombinants |
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20,785 |
total flies |
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Calculate map distances
:ec to sc interval
681 + 716 + 4 + 1 = 1402 / 20,785 = 6.74 %
ec to cv interval
1,002 + 997 + 4 + 1 = 2,004 / 20, 785 = 9.64 %
sc to cv interval
1,002 + 997+ 681 + 716 + 2 (4 + 1) = 3,406 / 20,785 = 16.4 %
Note: the map order is different than that drawn. Order is sc ec cv
Analyze double crossovers DCO:
Expect: (.067) (.096) X 20,785 total = 134
Coefficient of coincidence = # observed DCO / # expected DCO = 3.7 %
Interference = 1 - Coeff. of Coinc. = 0.96 96 % interference
Study questions:
1) Cite three ways to determine the relative map order.
2) Explain why double cross-overs are counted twice in determining the sc to cv interval.
3) How would the results and analysis change if wild-type males were used instead of sc ec cv males?