3. A population of 200 Mongolian gerbils living near Russia's Lake Baikal includes 80 brown gerbils with the AA genotype, 64 brown gerbils with the Aa genotype and 56 black gerbils with the aa genotype. Use this information to complete the table

Question

UpStudy AI · Accepted Answer

1. Count the number of individuals per genotype:

- $$ \text{AA (brown)} = 80 $$
- $$ \text{Aa (brown)} = 64 $$
- $$ \text{aa (black)} = 56 $$

2. Calculate the total number of individuals:

$$
\text{Total} = 80 + 64 + 56 = 200
$$

3. Compute the total number of alleles. Each individual has 2 alleles, so:

$$
\text{Total alleles} = 2 \times 200 = 400
$$

4. Determine the total number of $$A$$ and $$a$$ alleles contributed by each genotype:

- For genotype $$ \text{AA} $$: Each individual contributes 2 $$A$$ alleles.
  
  $$
  \text{Alleles from AA} = 80 \times 2 = 160 \quad (\text{all } A)
  $$

- For genotype $$ \text{Aa} $$: Each individual contributes 1 $$A$$ and 1 $$a$$ allele.
  
  $$
  \text{Alleles from Aa} = 64 \quad (\text{from } A) \quad \text{and} \quad 64 \quad (\text{from } a)
  $$

- For genotype $$ \text{aa} $$: Each individual contributes 2 $$a$$ alleles.
  
  $$
  \text{Alleles from aa} = 56 \times 2 = 112 \quad (\text{all } a)
  $$

5. Sum the alleles:

$$
\text{Total } A = 160 + 64 = 224
$$
$$
\text{Total } a = 64 + 112 = 176
$$

6. Calculate the allele frequencies:

$$
p = \text{frequency of } A = \frac{224}{400} = 0.56
$$
$$
q = \text{frequency of } a = \frac{176}{400} = 0.44
$$

7. Compile the results into the completed table:

| Genotype | Phenotype       | Number of Individuals | Genotype Frequency | Allele Contribution                                 |
|----------|-----------------|-----------------------|--------------------|-----------------------------------------------------|
| AA       | Brown           | 80                    | $$ \frac{80}{200} = 0.40 $$   | $$ A = 80 \times 2 = 160 $$                          |
| Aa       | Brown           | 64                    | $$ \frac{64}{200} = 0.32 $$   | $$ A = 64,\quad a = 64 $$                             |
| aa       | Black           | 56                    | $$ \frac{56}{200} = 0.28 $$   | $$ a = 56 \times 2 = 112 $$                          |
| **Total**| -               | 200                   | 1.00               | $$ A = 160 + 64 = 224,\quad a = 64 + 112 = 176 $$      |
|          |                 |                       |                    | **Allele Frequencies:** $$ p = 0.56,\quad q = 0.44 $$ |

This table summarizes the information for the population of Mongolian gerbils based on the given data.
| Genotype | Phenotype | Number | Genotype Frequency | Allele Contribution | |----------|-----------|---------|--------------------|---------------------| | AA | Brown | 80 | 0.40 | A: 160 | | Aa | Brown | 64 | 0.32 | A: 64, a: 64 | | aa | Black | 56 | 0.28 | a: 112 | | **Total**| - | 200 | 1.00 | A: 224, a: 176 | | | | | | Allele Frequencies: p=0.56, q=0.44 |

3. A population of 200 Mongolian gerbils living near Russia's Lake Baikal includes 80 brown gerbils with the AA genotype, 64 brown gerbils with the Aa genotype and 56 black gerbils with the aa genotype. Use this information to complete the table

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Biology Questions