Question

Wild-type Drosophila females having three linked genes (AABBCC) were crossed with triple recessive mutant (aabbcc) males. The F1 females (AaBbCc) were backcrossed with aabbcc males. The following F2 progeny numbers were obtained (total = 1000). The gene order is ABC. Find the recombination map distance (in cM) between A and C (round to one decimal place). \begin{tabular}{l r} AaBbCc & 241
Aabbcc & 112
aaBbCc & 103
aabbcc & 252
aaBbcc & 17
aabbCc & 134
AabbCc & 14
AaBbcc & 127
\end{tabular}

Hint:

For three-point mapping, {outer-gene distance} (A–C) = SCO\(_{AB}\) + SCO\(_{BC}\) \(+\) \(2\times\)DCO, because each double crossover contributes {two} crossovers between the outer loci.

Answer 1

Step 1: Identify nonrecombinants and double crossovers.
The two largest classes are the nonrecombinants (NR) (reflect parental gametes ABC and abc): NR = AaBbCc (241) and aabbcc (252) \(⇒\) \(241+252=493\).
The two smallest classes are the double crossovers (DCO) for the order A–B–C: DCO = AabbCc (AbC) = 14 and aaBbcc (aBc) = 17 \(⇒\) \(14+17=31\).
Step 2: Classify single crossovers (SCO).
For order A–B–C:
\(\bullet\) SCO in A–B interval: Abc (Aabbcc) = 112 and aBC (aaBbCc) = 103 \(⇒\) \(112+103=215\).
\(\bullet\) SCO in B–C interval: ABc (AaBbcc) = 127 and abC (aabbCc) = 134 \(⇒\) \(127+134=261\).
Step 3: Compute A–C map distance.
Map distance is the crossover frequency between the two loci (not just the recombinant frequency of outer markers). Each DCO contains two crossovers between A and C, so DCOs are counted twice: \[ \text{cM}_{A\text{–}C} = \frac{\text{SCO}_{AB} + \text{SCO}_{BC} + 2\times \text{DCO}}{\text{Total}}\times 100 = \frac{215 + 261 + 2\times 31}{1000}\times 100 = \frac{538}{1000}\times 100 = 53.8\ \text{cM}. \]