P1: A female heterozygous for white ( w ), yellow ( y ) and miniature ( m ) – (a
ID: 187257 • Letter: P
Question
P1: A female heterozygous for white (w), yellow (y) and miniature (m) – (all 3 recessive mutations) is crossed to a male who is mutant for all three traits.
The F1 male progeny phenotypes are:
+ + m
2278
w y +
2157
w y m
1203
+ + +
1092
+ y m
49
w + +
41
+ y +
2
w + m
1
Total =
6823
Was there positive or negative interference in the number of double crossovers, if so, to what extent?
Negative, 0.9
Positive, 0.1
Zero interference
Positive, 0.9
Positive, 0.5
The F1 male progeny phenotypes are:
+ + m
2278
w y +
2157
w y m
1203
+ + +
1092
+ y m
49
w + +
41
+ y +
2
w + m
1
Total =
6823
Was there positive or negative interference in the number of double crossovers, if so, to what extent?
Negative, 0.9
Positive, 0.1
Zero interference
Positive, 0.9
Positive, 0.5
Explanation / Answer
Interference = 1- Coefficient of coincidence
Coefficient of coincidence:
number of observed double cross overs / no. of expected double cross overs
number of observed double cross overs: 2+1=3
no. of expected double cross overs = expected double crossover frequecy*total progeny
expected double crossover frequecy= frequency of single crossover in the first region*frequency of single crossover in the second region*
Frequency of single cross over:
a) single cross over between m and + = 1203+1092+2+1/6823=0.45
b) single cross over between y and + = 49+41+3/6823=0.0136
expected double cross over frequency =0.45*0.0136=0.00612
no. of expected double cross overs =0.00612*6823=41.7
which makes the coefficient of coincidence = 3/41.7=0.071
Interfernce = 1-coefficient of coincidence = 0.929
Thus, interfernce is negative
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