Answer:
BgD/BgD x bGd/bGd----Parents
BgD/bGd ------------------F1
BgD/bGd x bgd/bgd
A).
Expected double crossover progeny = 20% * 30% = 0.06
Class of gametes |
Phenotype |
Frequency of reciprocal pair |
Numbers |
Frequency of each class |
Number of progeny frequency * 1000 = % |
Parental |
BgD |
1- all recombinant progeny |
1-(0.14+0.24+0.06) = 0.56 |
0.28 |
280 (BgD/bgd) |
bGd |
0.28 |
280 (bGd/bgd) |
|||
SCO 1 |
BGd |
RF in region 1 = SCO1 – DCO |
SCO1 = 0.2 – 0.06 = 0.14 |
0.07 |
70 (BGd/bgd) |
bgD |
0.07 |
70 (bgD / bgd) |
|||
SCO 2 |
Bgd |
RF in region 2 = SCO2 – DCO |
SCO2 =0.3 – 0.06 = 0.24 |
0.12 |
120 (Bgd/bgd) |
bGD |
0.12 |
120 (bGD / bgd) |
|||
DCO |
BGD |
(RF in region 1) * (RF in region 2) |
0.06 |
0.03 |
30 (BGD/bgd) |
Bgd |
0.03 |
30 (bgd/bgd) |
B).
Interference = 1- COC
20% or 0.2 = 1 – COC
COC = 1 – 0.2 = 0.8
COC = Observed double crossover progeny / Expected double crossover progeny
Expected double crossover progeny = 20% * 30% = 0.06
0.8 = Observed double crossover progeny / 0.06
Observed double crossover progeny = 0.06 * 0.8 = 0.048
Class of gametes |
Phenotype |
Frequency of reciprocal pair |
Numbers |
Frequency of each class |
Number of progeny frequency * 1000 = % |
Parental |
BgD |
1- all recombinant progeny |
1-(0.152+0.252+0.048) = 0.548 |
0.274 |
274 (BgD/bgd) |
bGd |
0.274 |
274 (bGd/bgd) |
|||
SCO 1 |
BGd |
RF in region 1 = SCO1 – DCO |
SCO1 = 0.2 – 0.048 = 0.152 |
0.076 |
76 (BGd/bgd) |
bgD |
0.076 |
76 (bgD / bgd) |
|||
SCO 2 |
Bgd |
RF in region 2 = SCO2 – DCO |
SCO2 =0.3 – 0.048 = 0.252 |
0.126 |
126 (Bgd/bgd) |
bGD |
0.126 |
126 (bGD / bgd) |
|||
DCO |
BGD |
(RF in region 1) * (RF in region 2) |
0.048 |
0.024 |
24 (BGD/bgd) |
Bgd |
0.024 |
24 (bgd/bgd) |
please show all work clearly Disc 2, Bb# 3 In the morning glory plant, three genes...
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