Biological meaning of “linkage b”
Genetic linkage occurs when two loci (for example A and the B locus) lie on the same chromosome and are inherited together more often than expected under independent assortment. Crossing over during meiosis produces recombinant gametes, and the fraction of recombinants estimates how far apart the loci are on the chromosome.
Given testcross data
Testcross: \(AB/ab \times ab/ab\). Each offspring phenotype reveals the gamete produced by the heterozygote.
| Gamete (offspring class) | Count | Type |
|---|---|---|
| \(AB\) | 410 | ? |
| \(ab\) | 390 | ? |
| \(Ab\) | 95 | ? |
| \(aB\) | 105 | ? |
Visualization: offspring class counts (parental vs recombinant)
Step 1: Identify parental and recombinant classes
- The largest counts are \(AB=410\) and \(ab=390\). These are taken as the parental classes, matching the heterozygote arrangement \(AB/ab\).
- The smaller counts \(Ab=95\) and \(aB=105\) are the recombinant classes, produced when a crossover occurs between loci A and B.
Step 2: Compute recombination frequency
- Total offspring: \[ N = 410 + 390 + 95 + 105 = 1000 \]
- Total recombinants: \[ R = 95 + 105 = 200 \]
- Recombination frequency: \[ r=\frac{R}{N}=\frac{200}{1000}=0.20 \] As a percentage: \[ r\% = 0.20\times 100 = 20\% \]
Recombination frequency: \(r=0.20\) (20%).
Under independent assortment, \(r\) is expected to be near \(0.50\) (50%). A value near 20% supports genetic linkage of the B locus to A (linkage b in the sense of the B locus being linked).
Step 3: Convert recombination frequency to map distance
For moderate distances, the genetic map distance in centiMorgans (cM) is approximated by recombination percentage: \[ \text{map distance (cM)} \approx 100r \] This is a practical approximation because multiple crossovers can cause the recombination frequency to underestimate true physical distance at larger separations.
\[ \text{distance} \approx 100(0.20) = 20\text{ cM} \]
Map distance between A and B: approximately \(20\text{ cM}\).
Optional statistical check: deviation from independent assortment
If A and B assort independently in a testcross, the four classes are expected in a \(1:1:1:1\) ratio. With \(N=1000\), the expected count per class is \(E=250\).
| Class | Observed \(O\) | Expected \(E\) | \((O-E)\) | \(\frac{(O-E)^2}{E}\) |
|---|---|---|---|---|
| \(AB\) | 410 | 250 | 160 | \(\frac{160^2}{250}=102.4\) |
| \(ab\) | 390 | 250 | 140 | \(\frac{140^2}{250}=78.4\) |
| \(Ab\) | 95 | 250 | -155 | \(\frac{155^2}{250}=96.1\) |
| \(aB\) | 105 | 250 | -145 | \(\frac{145^2}{250}=84.1\) |
\[ \chi^2 = 102.4 + 78.4 + 96.1 + 84.1 \approx 361.0 \]
With \(4\) classes, degrees of freedom \(df=4-1=3\). A \(\chi^2\) this large indicates a very strong departure from \(1:1:1:1\), consistent with linkage.
Conclusion
The B locus shows genetic linkage with A: parental classes \(AB\) and \(ab\) greatly exceed recombinant classes \(Ab\) and \(aB\). The recombination frequency is \(20\%\), giving an estimated map distance of about \(20\text{ cM}\) between loci A and B.