Let's break down the genetics problem step by step: ### 2. Tassel-less (ts), knotted (kn), and shrunken (sh) are linked genes in maize. A testcross is done with a heterozygote of cis conformation. The progeny is: | Phenotype | Number | |-----------|--------| | + + + | 709 | | kn sh ts | 680 | | kn + ts | 204 | | + sh + | 215 | | + + ts | 88 | | kn sh + | 95 | | + sh ts | 17 | | kn + + | 14 | | **Total** | 2,022 | #### a. **Determine the order of these genes on the chromosome.** **Step 1: Identify parental, single crossover (SCO), and double crossover (DCO) classes.** - **Parental classes:** Highest numbers (most frequent) - + + + (709) - kn sh ts (680) - **DCO classes:** Lowest numbers (least frequent) - + sh ts (17) - kn + + (14) - **SCO classes:** The rest - kn + ts (204) - + sh + (215) - + + ts (88) - kn sh + (95) **Step 2: Find the gene order** Compare parental and DCO classes to determine which gene is in the middle: Parental: + + +, kn sh ts DCO: + sh ts, kn + + Let's write the alleles for each gene: | Class | kn | sh | ts | |-------------|----|----|----| | + + + | + | + | + | | kn sh ts | kn | sh | ts | | + sh ts | + | sh | ts | | kn + + | kn | + | + | Compare parental (+ + +) to DCO (+ sh ts): - + → + (same) - + → sh (different) - + → ts (different) So, "sh" and "ts" are switched in the DCO compared to parental. Compare parental (kn sh ts) to DCO (kn + +): - kn → kn (same) - sh → + (different) - ts → + (different) So, again "sh" and "ts" are switched. **Conclusion:** The gene that differs between the parental and double crossover is **sh**. Therefore, **sh** is in the middle. **Gene order:** - kn --- sh --- ts --- #### b. **Calculate the map distances between genes.** **Map distance (in cM) = (Recombinants between the two genes / Total progeny) × 100** Let's group the recombinants: 1. **kn and sh:** - Single crossovers between kn and sh: - kn + ts (204) - + sh + (215) - Double crossovers (both types involve a crossover in both intervals, so count both DCO classes): - + sh ts (17) - kn + + (14) - **Total recombinants = 204 + 215 + 17 + 14 = 450** - **Map distance = (450 / 2022) × 100 = 22.26 cM** 2. **sh and ts:** - Single crossovers between sh and ts: - + + ts (88) - kn sh + (95) - Double crossovers (same as above): - + sh ts (17) - kn + + (14) - **Total recombinants = 88 + 95 + 17 + 14 = 214** - **Map distance = (214 / 2022) × 100 = 10.58 cM** --- #### c. **Determine the coefficient of coincidence (c.o.c.) and the interference.** **Expected DCO frequency = (recomb. freq. between kn-sh) × (recomb. freq. between sh-ts) × total progeny** - kn-sh recomb. freq. = 450 / 2022 = 0.2226 - sh-ts recomb. freq. = 214 / 2022 = 0.1058 **Expected DCO = 0.2226 × 0.1058 × 2022 = 47.59 ≈ 48** **Observed DCO = 17 (+ sh ts) + 14 (kn + +) = 31** **Coefficient of coincidence (c.o.c.):** \[ c.o.c. = \frac{\text{Observed DCO}}{\text{Expected DCO}} = \frac{31}{48} = 0.65 \] **Interference (I):** \[ I = 1 - \text{c.o.c.} = 1 - 0.65 = 0.35 \] **Explanation of interference:** Interference is the phenomenon where a crossover in one region of a chromosome reduces the probability of another crossover nearby. Here, an interference of 0.35 means that 35% of the expected double crossovers did not occur because a crossover in one region interfered with the occurrence of another nearby. --- ### **Summary Table** | Step | Result | |---------|----------------------------| | a) | kn — sh — ts | | b) | kn-sh: 22.3 cM <br> sh-ts: 10.6 cM | | c) | c.o.c. = 0.65 <br> interference = 0.35 | Let me know if you'd like a more detailed explanation or step-by-step working for any part!