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 abc 460 ab+ 15 What is the map distance between the a and c genes? O 25.12 map units O 27.71 map units 29.39 map units O 32.53 map units None of the above
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 abc 460 ab+ 15 What is the map distance between the a and c genes? O 25.12 map units O 27.71 map units 29.39 map units O 32.53 map units None of the above
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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![### Progeny Analysis and Gene Mapping
A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results:
| Genotype | Progeny Count |
|---------|----------------|
| + + + | 450 |
| + + c | 10 |
| a + + | 70 |
| + b c | 210 |
| + b + | 65 |
| a + + | 200 |
| a b c | 460 |
| a b + | 15 |
**Question: What is the map distance between the a and c genes?**
**Options:**
- 25.12 map units
- 27.71 map units
- 29.39 map units
- 32.53 map units
- None of the above
**Explanation of the Calculation:**
Map distance between genes is calculated using the formula:
\[ \text{Map Distance} = \left( \frac{\text{Number of Recombinants}}{\text{Total Number of Offspring}} \right) \times 100 \]
For genes a and c, the recombinants are:
- + + c = 10
- a + + = 70
- + b + = 65
- a b + = 15
Thus, the number of recombinants is \(10 + 70 + 65 + 15 = 160\).
The total number of offspring is the sum of all progeny counts:
\[ 450 + 10 + 70 + 210 + 65 + 200 + 460 + 15 = 1480 \]
Now, applying the formula:
\[ \text{Map Distance} = \left( \frac{160}{1480} \right) \times 100 \approx 10.81 \text{ map units} \]
Since 10.81 map units is not an option provided, there's an oversight or error in given choices. Please review the progeny counts and sequences for a more accurate map distance.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e5efc23-a198-4327-bdc7-d2931bd5f43b%2F7d911931-e7e2-4b9d-9170-90d73fe684cd%2Fc0duwng_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Progeny Analysis and Gene Mapping
A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results:
| Genotype | Progeny Count |
|---------|----------------|
| + + + | 450 |
| + + c | 10 |
| a + + | 70 |
| + b c | 210 |
| + b + | 65 |
| a + + | 200 |
| a b c | 460 |
| a b + | 15 |
**Question: What is the map distance between the a and c genes?**
**Options:**
- 25.12 map units
- 27.71 map units
- 29.39 map units
- 32.53 map units
- None of the above
**Explanation of the Calculation:**
Map distance between genes is calculated using the formula:
\[ \text{Map Distance} = \left( \frac{\text{Number of Recombinants}}{\text{Total Number of Offspring}} \right) \times 100 \]
For genes a and c, the recombinants are:
- + + c = 10
- a + + = 70
- + b + = 65
- a b + = 15
Thus, the number of recombinants is \(10 + 70 + 65 + 15 = 160\).
The total number of offspring is the sum of all progeny counts:
\[ 450 + 10 + 70 + 210 + 65 + 200 + 460 + 15 = 1480 \]
Now, applying the formula:
\[ \text{Map Distance} = \left( \frac{160}{1480} \right) \times 100 \approx 10.81 \text{ map units} \]
Since 10.81 map units is not an option provided, there's an oversight or error in given choices. Please review the progeny counts and sequences for a more accurate map distance.
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