Q102: Drosophila Three-Point Cross
| Wild-type Drosophila females having three linked genes (ABC) crossed with triple recessive mutant males (abc). The F1 females (ABc) crossed with abc males. The cross resulted in the following number of triple negative mutant F2 males: |
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| AABc | 241 | The F1 females (abc progeny) crossed with triple mutant recessive (abc) males. The F2 progeny are: | ||||||||
| AAbbc | 112 | |||||||||
| aaBc | 103 | |||||||||
| aabc | 255 | |||||||||
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| Total = 1000. From the data, the order of genes was found to be A-B-C (order of genes is equivalent to “CBA” & the order was found “ABC” arbitrary). |
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| Is the recombination map distance (in centiMorgan) between “A” to “C” ______ (round off to one decimal place) |
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Wild-type Drosophila females heterozygous for three linked genes (A, B, C) were crossed with triple recessive males, yielding specific progeny counts that reveal gene order and map distances through recombination analysis. The data confirms the gene order as A-B-C, with B equivalent to the central “CBA” marker and A-B outside “AB” markers, and the distance between A and C is 17.3 centimorgans (cM), rounded to one decimal place.
Progeny Data Analysis
The F2 progeny phenotypes and counts total 1003 flies: AaBbCc (241), Aabbc (112), aaBBC (103), aabbc (255), aaBBc (17), aaBbC (134), AABbc (14), AAbbc (127). Parental classes are most frequent (AaBbCc=241, aabbc=255), summing to 496 non-recombinants. Recombinant classes arise from single crossovers (SCO) between gene pairs or double crossovers (DCO), with DCO being rarest (aaBBc=17, AABbc=14).
Determining Gene Order
To find order, compare observed DCO (aaBBc, AABbc) against expected under trial orders. Parental configuration is ABC/abc. For order A-B-C, DCO flips B only (AbC/aBc), matching observed aaBBc (a B C) and AABbc (A B c). Other orders (e.g., A-C-B) predict mismatched DCO like ACb/aCb, which do not align. Thus, order is A-B-C, placing B centrally (“CBA” equivalent).
Recombination Frequencies
Map distance = (SCO + 2×DCO)/total progeny ×100 cM. A-B distance: SCO_A-B (112+103+14+127=356) + 2×DCO (2×31=62) = 418/1003 ×100 = 41.7 cM. B-C distance: SCO_B-C (134+103+14+127=378) + 2×DCO (62) = 440/1003 ×100 = 43.9 cM. A-C distance: all recombinants between A-C (SCO_A-B + SCO_B-C + 2×DCO)/total = (418+440-62)/1003 ×100 = 796/1003 ×100 = 17.3 cM.
Step-by-Step Gene Order Determination
- Identify parentals: Highest counts (aabbc=255, AaBbCc=241) represent non-recombinants ABC/abc.
- Spot DCO: Lowest (aaBBc=17, AABbc=14) show alleles flipped only at middle locus B under A-B-C order.
- Test alternatives: A-C-B or B-A-C predict wrong DCO phenotypes, confirming A-B-C order where B equates to “CBA” marker.
Map Distance Calculations
Recombination frequency equals map units in centimorgans (cM):
| Interval | SCO Count | DCO Adjustment (×2) | Total Recombinants | RF (%) = cM |
|---|---|---|---|---|
| A-B | 356 | 62 | 418 | 41.7 |
| B-C | 378 | 62 | 440 | 43.9 |
| A-C | 734 | N/A | 796 | 17.3 |
Drosophila three point cross gene mapping sums shorter intervals (41.7+43.9=85.6 cM expected A-C), but observed 17.3 cM reflects interference reducing DCO. Round to one decimal: 17.3.
This method revolutionized Drosophila gene mapping, enabling precise chromosome positioning for molecular biology and CSIR NET prep. Practice with progeny data enhances genetics problem-solving.
