A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results + + + 450 + + C 10 a + C 70 + b c 210 + b 65 a + + 200 a bc 460 a b + 15 What is the map distance between the a and b genes? O 34.45 map units 36.23 map units • 38.51 map units O 40.20 map units 43.01 map units

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**Educational Content: Genetic Mapping Example**

**Note:** The next four questions are based on the same three-factor cross shown here. So please keep track of your calculations.

**Genetic Cross Analysis:**

A cross is made between a heterozygote, +++ / abc, and a recessive homozygote, abc / abc. Analysis of the progeny gave the following results:

| Genotype  | Number of Offspring |
|-----------|---------------------|
| + + +     | 450                 |
| + + c     | 10                  |
| a + c     | 70                  |
| + b c     | 210                 |
| + b +     | 65                  |
| a b +     | 200                 |
| a + +     | 460                 |
| a b c     | 15                  |

**Question:**
What is the map distance between the a and b genes?

- [ ] 34.45 map units
- [ ] 36.23 map units
- [ ] 38.51 map units
- [ ] 40.20 map units

---
**Graph/Diagram Explanation:**

- This dataset includes a table that lists different genotypes resulting from the genetic cross along with the corresponding number of offspring for each genotype.
  
- The frequencies of the recombinants can be used to calculate the genetic distances between the genes involved in the cross.

For illustration, the data demonstrates the number of offspring for eight different genotypes resulting from the cross between the heterozygote and the recessive homozygote. To determine the map distance between genes a and b, one would typically use the formula for recombination frequency and convert it to map units (1% recombination = 1 map unit). 

To calculate the map distance:

1. Determine the number of recombinant progeny for genes a and b.
2. Add the total number of progeny.
3. Use the formula:

\[ \text{Recombination Frequency} = \left(\frac{\text{Number of Recombinants}}{\text{Total Number of Progeny}}\right) \times 100 \]

4. Convert the recombination frequency to map units.

Please proceed with your calculations to determine the correct map distance between genes a and b.
Transcribed Image Text:**Educational Content: Genetic Mapping Example** **Note:** The next four questions are based on the same three-factor cross shown here. So please keep track of your calculations. **Genetic Cross Analysis:** A cross is made between a heterozygote, +++ / abc, and a recessive homozygote, abc / abc. Analysis of the progeny gave the following results: | Genotype | Number of Offspring | |-----------|---------------------| | + + + | 450 | | + + c | 10 | | a + c | 70 | | + b c | 210 | | + b + | 65 | | a b + | 200 | | a + + | 460 | | a b c | 15 | **Question:** What is the map distance between the a and b genes? - [ ] 34.45 map units - [ ] 36.23 map units - [ ] 38.51 map units - [ ] 40.20 map units --- **Graph/Diagram Explanation:** - This dataset includes a table that lists different genotypes resulting from the genetic cross along with the corresponding number of offspring for each genotype. - The frequencies of the recombinants can be used to calculate the genetic distances between the genes involved in the cross. For illustration, the data demonstrates the number of offspring for eight different genotypes resulting from the cross between the heterozygote and the recessive homozygote. To determine the map distance between genes a and b, one would typically use the formula for recombination frequency and convert it to map units (1% recombination = 1 map unit). To calculate the map distance: 1. Determine the number of recombinant progeny for genes a and b. 2. Add the total number of progeny. 3. Use the formula: \[ \text{Recombination Frequency} = \left(\frac{\text{Number of Recombinants}}{\text{Total Number of Progeny}}\right) \times 100 \] 4. Convert the recombination frequency to map units. Please proceed with your calculations to determine the correct map distance between genes a and b.
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