3 is supposed to have cn in the middle can you help

Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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3 is supposed to have cn in the middle can you help

**Educational Content on Gene Linkage and Mapping**

1. **Gene Distance Measurements**

   - **Texture gene**: 14.6 map units (mu)
   - **Odor gene**: 15.4 map units (mu)
   - **Height gene**

2. **Gene Linkage Explanation**

   At first glance, it appears that only three genes are linked, as we have eight phenotypic classes. If all four were linked with recombination, we would have 16 phenotypic classes. For genes A and D, there are only two phenotypes: Ad and aD. This contrasts with the other three genes, which appear in all combinations (e.g., AB, Ab, aB, ab). In this case, genes A and D are so closely linked that no detectable recombination occurs between them, essentially acting as a single gene. Thus, although all four genes are linked, A and D behave as one gene.

   - **Linkage Map**:

     ```
     B —— 10 mu —— a/D —— 30 mu —— C
     ```

3. **Linkage Map Analysis**

   - **Experiment as analyzed on the Online_Linkage PowerPoint, slides 5 & 6**. The linkage map of the genes in the heterozygous parent is:

     ```
     b —— 10.3 mu —— cn —— 14.7 mu —— wx
     ```

4. **Further Linkage Analysis**

   - **This experiment analysis is similar to the one on slides 5 & 6**. The linkage map is:

     ```
     P —— 21 mu —— 0 —— 14 mu —— S
     ```

   - **Sequence and arrangement of genes in the heterozygous parent for 5a)**:

     ```
     +ts/po+
     ```

6. **Additional Experiment Analysis**

   - **This experiment, analyzed on slides 5 & 6**, shows the linkage map:

     ```
     v —— 13.1 mu —— ct —— 6.4 mu —— cv
     ```

7. **Final Experiment Analysis**

   - **This experiment, analyzed on slides 5 & 6**, yields the linkage map:

     ```
     an —— 16.7 mu —— f —— 4.8 mu —— br
     ```

This content helps in understanding the concepts of gene linkage, recombination, and genetic mapping of traits
Transcribed Image Text:**Educational Content on Gene Linkage and Mapping** 1. **Gene Distance Measurements** - **Texture gene**: 14.6 map units (mu) - **Odor gene**: 15.4 map units (mu) - **Height gene** 2. **Gene Linkage Explanation** At first glance, it appears that only three genes are linked, as we have eight phenotypic classes. If all four were linked with recombination, we would have 16 phenotypic classes. For genes A and D, there are only two phenotypes: Ad and aD. This contrasts with the other three genes, which appear in all combinations (e.g., AB, Ab, aB, ab). In this case, genes A and D are so closely linked that no detectable recombination occurs between them, essentially acting as a single gene. Thus, although all four genes are linked, A and D behave as one gene. - **Linkage Map**: ``` B —— 10 mu —— a/D —— 30 mu —— C ``` 3. **Linkage Map Analysis** - **Experiment as analyzed on the Online_Linkage PowerPoint, slides 5 & 6**. The linkage map of the genes in the heterozygous parent is: ``` b —— 10.3 mu —— cn —— 14.7 mu —— wx ``` 4. **Further Linkage Analysis** - **This experiment analysis is similar to the one on slides 5 & 6**. The linkage map is: ``` P —— 21 mu —— 0 —— 14 mu —— S ``` - **Sequence and arrangement of genes in the heterozygous parent for 5a)**: ``` +ts/po+ ``` 6. **Additional Experiment Analysis** - **This experiment, analyzed on slides 5 & 6**, shows the linkage map: ``` v —— 13.1 mu —— ct —— 6.4 mu —— cv ``` 7. **Final Experiment Analysis** - **This experiment, analyzed on slides 5 & 6**, yields the linkage map: ``` an —— 16.7 mu —— f —— 4.8 mu —— br ``` This content helps in understanding the concepts of gene linkage, recombination, and genetic mapping of traits
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