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A trihybrid cross is made between two pl...

A trihybrid cross is made between two plants with genotypes A/a B/b C/c how many offspring of such cross will have a genotype a/a b/b c/c-

A

`1//64`

B

`1//4`

C

`1//16`

D

`1//32`

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The correct Answer is:
To solve the problem of how many offspring will have the genotype a/a b/b c/c from a trihybrid cross between two plants with the genotypes A/a B/b C/c, we can follow these steps: ### Step 1: Identify the Genotypes The parent plants have the genotype A/a B/b C/c. This means each plant is heterozygous for three traits (A, B, and C). ### Step 2: Determine the Number of Gametes For each gene, the heterozygous genotype (A/a, B/b, C/c) can produce two types of gametes: - For A/a: Gametes are A and a - For B/b: Gametes are B and b - For C/c: Gametes are C and c Since there are three genes, the total number of different gametes that can be produced is calculated as: \[ \text{Number of gametes} = 2^n \] where \( n \) is the number of heterozygous traits. Here, \( n = 3 \). Thus, the number of gametes is: \[ 2^3 = 8 \] ### Step 3: Calculate the Total Offspring When two plants are crossed, the total number of offspring produced can be calculated as: \[ \text{Total offspring} = \text{Number of gametes from plant 1} \times \text{Number of gametes from plant 2} \] Since both plants can produce 8 types of gametes: \[ \text{Total offspring} = 8 \times 8 = 64 \] ### Step 4: Determine the Probability of the Recessive Genotype To find the probability of obtaining the genotype a/a b/b c/c, we need to consider the combinations that lead to this genotype: - For a/a: The probability of getting a from each parent is \( \frac{1}{2} \) (since each parent can contribute either A or a). - For b/b: Similarly, the probability of getting b from each parent is \( \frac{1}{2} \). - For c/c: The probability of getting c from each parent is \( \frac{1}{2} \). Thus, the combined probability of obtaining the genotype a/a b/b c/c is: \[ P(a/a) \times P(b/b) \times P(c/c) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \] ### Step 5: Calculate the Expected Number of Offspring with the Recessive Genotype Now, to find the expected number of offspring with the genotype a/a b/b c/c, we multiply the total number of offspring by the probability of the recessive genotype: \[ \text{Expected number of a/a b/b c/c} = \text{Total offspring} \times P(a/a b/b c/c) \] \[ = 64 \times \frac{1}{8} = 8 \] ### Step 6: Find the Fraction of Offspring with the Recessive Genotype Finally, to express this as a fraction of the total offspring: \[ \text{Fraction} = \frac{8}{64} = \frac{1}{8} \] ### Conclusion The probability of obtaining an offspring with the genotype a/a b/b c/c in this trihybrid cross is \( \frac{1}{64} \).
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ALLEN-GENETICS & MOLECULAR BASIS OF INHERITANCE AND MUTATION-EXERCISE-I (CONCEPTUAL QUESTIONS)
  1. A plant of F(1)-generation has genotype 'AABbCC'. On selfing of this p...

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  2. Which one of the following traits of garden pea studied by Mendel, was...

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  3. A trihybrid cross is made between two plants with genotypes A/a B/b C/...

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  4. How is the arrangement of Mendel's selected seven characters on four c...

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  5. When two different genotype produce the same phenotype due to environm...

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  6. When a red flower homozygous pea plant is crossed with a white flower ...

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  7. Heterozygous tall plant (Tt) is crossed with homozygous dwarf (tt) pla...

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  8. If a homozygous tall plant is crossed with a dwarf plant, what shall b...

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  9. How many different types of gametes can be formed by F(1) progent, res...

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  10. In order to find out the different types of gametes produced by a pea...

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  11. Law of independent assortment of Mendel was proved by :-

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  12. Mendel does not select which character in his experiment :-

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  13. Genes controlling seven traits in Pea studied by Mendel were actually ...

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  14. Two crosses between the same pair of genotypes or phenotypes in which ...

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  15. If selfing occurs in the plant having genotype RrYy, then ratio of giv...

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  16. The process of mating between closely related individuals is

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  17. Marriage between close relatives should be avoided because it induces...

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  18. A self-fertilizing trihybrid plant forms

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  19. Segregation of genes takes place during

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  20. An inherited character and its detectable variant is termed as :-

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